Intermediate Algebra, 3rd Edition

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Algebra

Intermediate

Third Edition

Julie Miller Daytona State College

Molly O’Neill Daytona State College

Nancy Hyde Broward College—Professor Emeritus

TM

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INTERMEDIATE ALGEBRA, THIRD EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2011 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions © 2008 and 2004. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 QPD/QPD 1 0 9 8 7 6 5 4 3 2 1 0 ISBN 978–0–07–338422–1 MHID 0–07–338422–4 ISBN 978–0–07–729647–6 (Annotated Instructor’s Edition) MHID 0–07–729647–8

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Library of Congress Cataloging-in-Publication Data Miller, Julie, 1962Intermediate algebra / Julie Miller, Molly O’Neill, Nancy Hyde. — 3rd ed. p. cm. Includes index. ISBN 978–0–07–338422–1 — ISBN 0–07–338422–4 (hard copy : alk. paper) 1. Algebra—Textbooks. I. O’Neill, Molly, 1953- II. Hyde, Nancy. III. Title. QA154.3.M554 2011 512.9—dc22 2009020569

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Letter from the Authors

Dear Colleagues, We originally embarked on this textbook project because we were seeing a lack of student success in our developmental math sequence. In short, we were not getting the results we wanted from our students with the materials and textbooks that we were using at the time. The primary goal of our project was to create teaching and learning materials that would get better results. At Daytona State College, our students were instrumental in helping us develop the clarity of writing; the step-by-step examples; and the pedagogical elements, such as Avoiding Mistakes, Concept Connections, and Problem Recognition Exercises, found in our textbooks. They also helped us create the content for the McGraw-Hill video exercises that accompany this text. Using our text with a course redesign at Daytona State College, our student success rates in developmental courses have improved by 20% since 2006 (for further information, see The Daytona Beach News Journal, December 18, 2006). We think you will agree that these are the kinds of results we are all striving for in developmental mathematics courses. This project has been a true collaboration with our Board of Advisors and colleagues in developmental mathematics around the country. We are sincerely humbled by those of you who adopted the first edition and the over 400 colleagues around the country who partnered with us providing valuable feedback and suggestions through reviews, symposia, focus groups, and being on our Board of Advisors. You partnered with us to create materials that will help students get better results. For that we are immeasurably grateful. As an author team, we have an ongoing commitment to provide the best possible text materials for instructors and students. With your continued help and suggestions we will continue the quest to help all of our students get better results. Sincerely, Julie Miller [email protected]

Molly O’Neill [email protected]

Nancy Hyde [email protected]

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About the Authors

Julie Miller

Julie Miller has been on the faculty in the School of Mathematics at Daytona State College for 20 years, where she has taught developmental and upper-level courses. Prior to her work at DSC, she worked as a software engineer for General Electric in the area of flight and radar simulation. Julie earned a bachelor of science in applied mathematics from Union College in Schenectady, New York, and a master of science in mathematics from the University of Florida. In addition to this textbook, she has authored several course supplements for college algebra, trigonometry, and precalculus, as well as several short works of fiction and nonfiction for young readers. “My father is a medical researcher, and I got hooked on math and science when I was young and would visit his laboratory. I can remember using graph paper to plot data points for his experiments and doing simple calculations. He would then tell me what the peaks and features in the graph meant in the context of his experiment. I think that applications and hands-on experience made math come alive for me and I’d like to see math come alive for my students.” —Julie Miller

Molly O’Neill

Molly O’Neill is also from Daytona State College, where she has taught for 22 years in the School of Mathematics. She has taught a variety of courses from developmental mathematics to calculus. Before she came to Florida, Molly taught as an adjunct instructor at the University of Michigan– Dearborn, Eastern Michigan University, Wayne State University, and Oakland Community College. Molly earned a bachelor of science in mathematics and a master of arts and teaching from Western Michigan University in Kalamazoo, Michigan. Besides this textbook, she has authored several course supplements for college algebra, trigonometry, and precalculus and has reviewed texts for developmental mathematics. “I differ from many of my colleagues in that math was not always easy for me. But in seventh grade I had a teacher who taught me that if I follow the rules of mathematics, even I could solve math problems. Once I understood this, I enjoyed math to the point of choosing it for my career. I now have the greatest job because I get to do math every day and I have the opportunity to influence my students just as I was influenced. Authoring these texts has given me another avenue to reach even more students.” —Molly O’Neill

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Nancy Hyde served as a full-time faculty member of the Mathematics Department at Broward College for 24 years. During this time she taught the full spectrum of courses from developmental math through differential equations. She received a bachelor of science degree in math education from Florida State University and a master’s degree in math education from Florida Atlantic University. She has conducted workshops and seminars for both students and teachers on the use of technology in the classroom. In addition to this textbook, she has authored a graphing calculator supplement for College Algebra. “I grew up in Brevard County, Florida, with my father working at Cape Canaveral. I was always excited by mathematics and physics in relation to the space program. As I studied higher levels of mathematics I became more intrigued by its it ab b stract t t nature t and infinite possibilities. It is enjoyable and rewarding to convey this perspective to students while helping them to understand mathematics.”

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Nancy Hyde

—Nancy Hyde

Dedication To Warren C. Tucker —Julie Miller To Stephen and Samantha —Molly O’Neill In memory of my aunt and uncle, Nancy and John Garvey —Nancy Hyde

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Get Better Results with Miller/O’Neill/Hyde About the Cover A mosaic is made up of pieces placed together to create a unified whole. Similarly, an intermediate algebra course provides an array of topics that together create a solid mathematical foundation for the developmental mathematics student. The Miller/O’Neill/Hyde developmental mathematics series helps students see the whole picture through better pedagogy and supplemental materials. In this Intermediate Algebra textbook, Julie Miller, Molly O’Neill, and Nancy Hyde focused their efforts on guiding students successfully through core topics, building mathematical proficiency, and getting better results!

“We originally embarked on this textbook project because we were seeing a lack of student success in courses beyond our developmental sequence. We wanted to build a better bridge between developmental algebra and higher level math courses. Our goal has been to develop pedagogical features to help students achieve better results in mathematics.” —Julie Miller, Molly O’Neill, Nancy Hyde vi

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Get Better Results

How Will Miller/O’Neill/Hyde Help Your Students Get Better Results? Better Clarity, Quality, and Accuracy

“I think the level of rigor is perfect for my students.

Julie Miller, Molly O’Neill, and Nancy Hyde know what I have examined other textbooks that would have students need to be successful in mathematics. Better been placed at the extremes of a continuum. This results come from clarity in their exposition, quality of book is, in the words of Goldilocks, ‘just right.’” step-by-step worked examples, and accuracy of their exercises ercises —Angie McCombs, Illinois State University sets; but it takes more than just great authors to build a textbook series to help students achieve success in mathematics. Our authors worked with a strong mathematical team of instructors from around the country to ensure that the clarity, quality, and accuracy you expect from the Miller/O’Neill/Hyde series was included in this edition.

Better Exercise Sets! Comprehensive sets of exercises are available for every student level. Julie Miller, Molly O’Neill, and Nancy Hyde worked with a board of advisors from across the country to offer the appropriate depth and breadth of exercises for your students. Problem Recognition Exercises were created to improve prove “Plenty of exercises covering all concepts. The mixed student performance while testing. Our practice exercise sets help students progress from skill development to conceptual understanding. Student tested and instructor approved, the Miller/O’Neill/ ill/ Hyde exercise sets will help your student get better results. esults. ▶

Problem Recognition Exercises

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Skill Practice Exercises

▶

Study Skills Exercises

▶

Mixed Exercises

▶

Expanding Your Skills Exercises

Better Step-By-Step Pedagogy! Intermediate Algebra provides enhanced step-by-step learning tools to help students get better results.

exercises help students to realize they have to be aware of the difference between types of problems. The quality of exercises range from basic to more difficult concepts with a good transition between the two, and they are relevant to the concepts taught in the section.” —Natalie Weaver, Daytona State College

“MOH does a much better job in a clear step-by-step approach to the examples. Where MOH has the edge is in identifying common mistakes, and discussing how to avoid them.” —Don York, Danville Area Community College

▶

Worked Examples provide an “easy-to-understand” nd” approach, clearly guiding each student through a step-by-step approach to master each practice exercise for better comprehension.

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TIPs offer students extra cautious direction to help improve understanding through hints and further insight.

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Avoiding Mistakes boxes alert students to common errors and provide practical ways to avoid them. Both of these learning aids will help students dents get better results by showing how to work through hrough a problem using a clearly defined step-by-step tep methodology that has been class tested and student approved.

“This text provides a great step-by-step approach to how to work problems in Intermediate Algebra and gives students a solid foundation on which to build. Students using this textbook should come out not just knowing how to work problems but why the methods are used.” —Brianna Killian, Daytona State College

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Formula for Student Success Step-by-Step Worked Examples ▶ ▶ ▶

Do you get the feeling that there is a disconnection between your students’ class work and homework? Do your students have trouble finding worked examples that match the practice exercises? Do you prefer that your students see examples in the textbook that match the ones you use in class?

Miller/O’Neill/Hyde’s Worked Examples offer a clear, concise methodology that replicates the mathematical processes used in the authors’ classroom lectures!

“MOH has very high quality examples; easy, medium and hard problems are worked out. Very comprehensive coverage of concepts.” —Shawna Mahan, Pikes Peak Community College

Example 4

Solving a Dependentt System y

Solve by using the substitution method.

Classroom Example: p. 248, Exercise 30

4x  2y  6 y  3  2x

Solution: 4x  2y  6 y  2x  3 v

y  3  2x

4x  212x  32  6 4x  4x  6  6

Step 1:

Step 2:

Solve for one of the variables. ables. Substitute the quantity 2x  3 for y in the other equation.

“Examples in the text have clarity, readability and should be useful to the students and there are plenty to follow.” —Judy McBride, Indiana University-Purdue University at Indianapolis

Step 3: Solve for x. Apply the distributive stributive property to clear the parentheses.

6  6 The system reduces to the identity 6  6. Therefore, the original origin two equations are equivalent, and the system em m is i dependent. The solution con consists of all points on the common line, giving ngg us u an a infinite number of solutions. B Because the equations 4x  2y  6 and nd d y  3  2x represent the same line, the ssolution set is or 51x, illustrate y2 0 4x  2y 2 the  66 5 1x, y2 0 y I 3often  2x6 “The worked examples mechanics very well. find myself saying, “If you need more help, the textbook has a Skill Practice Solve the system by using substitution. very nice example on page [x].” It would have been difficult to 4. 3x  6y  12 improve upon them. Again, the text and tips included with these 2y 2  x  4 examples often include things I have said to my own students, so I feel like the authors “gel” with my approach to teaching.” —Angie McCombs, Illinois State University

To ensure that the classroom experience also matches the examples in the text and the practice exercises, we have included references to even-numbered exercises to be used as Classroom Examples. These exercises are highlighted in the Practice Exercises at the end of each section. viii

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Get Better Results

Better Learning Tools Chapter Openers

Chapter 8

Tired of students not being prepared? The Miller/ O’Neill/Hyde Chapter Openers help students get better results through engaging Puzzles and Games that introduce the chapter concepts and ask “Are You Prepared?”

“I liked how the MOH puzzle asked questions that made use of many of the concepts students will encounter in the chapter.” —Michelle Jackson, Bowling Green Community College at WKU

This chapter is devoted to the study of exponential and logarithmic functions. These functions are used to study many naturally occurring phenomena such as population growth, exponential decay of radioactive matter, and growth of investments. Are You Prepared? To prepare yourself for this chapter, practice evaluating the expressions with exponents. Use the clues to fill in the boxes labeled A–H. Then fill in the remaining part of the grid so that every row, every column, and every 2 3 box contains the digits 1 through 6. A. B. C. D. E. F. G. H.

Value of x in the equation 5x  25. Value of x in the equation 3x  81. Value of x in the equation 81/x  2. Real value of x in the equation x 4  16 (assume that x 7 0). 1 Absolute value of x in the equation 2 x  . 16 x Value of x in the equation 2  64. Value of x in the equation 50  x. Value of x in the equation e0  x.

B

F

C

4

2 D

G

4

A

4

3 E

2 H

3 2

“Students would probably be inclined to do this activity (the first step!!). They are so familiar with soduko, that I think they could jump right in!” —Janice Rech, University of Nebraska at Omaha

TIP and Avoiding Mistakes Boxes TIP and Avoiding Mistakes boxes have been created based on the authors’ classroom experiences—they have also been integrated into the Worked Examples. These pedagogical tools will help students get better results by learning how to work through a problem using a clearly defined step-by-step methodology. Example 5

Factoring a Four-Term Polynomial by Grouping Three Terms

Factor completely.

Avoiding Mistakes Boxes:

x2  y2  6y  9

Solution: Grouping “2 by 2” will not work to factor this polynomial. However, if we factor out 1 from the last three terms, the resulting trinomial will be a perfect square trinomial. x2

Avoiding Mistakes When factoring the expression x 2  1y  32 2 as a difference of squares, be sure to use parentheses around the quantity 1y  32. This will help you remember to "distribute the negative” in the expression ession 3 x  1y  322 4 .

 y2  6y  9

Avoiding Mistakes boxes are integrated throughout the textbook to alert students to common errors and how to avoid them.

Group the last three terms.

 x2  11y2  6y  92

Factor out 1 from the last three terms.

 x2  1y  32 2

Factor thee perfect square trinomial all y2  6y  9 as 1y  32 2.

“I love the tips and avoiding mistakes boxes. These help the students identify common mistakes that are made.”

The quantity tity x2  11y  322 2 is a difference di diff d ffe ferrence of squares, s, a2  b2, where here ere a  x and an b  1y  32.

冤 x  1y  32 冥  1x  y  32

 冤冤x  1y  32冥 2冥冤冤x  1y  322 冥

Factor as a2  b2  11a  b21a  b2 b b2. 2.

 1x  y  321x 32 1x  y  32

Apply thee distributive property p to o clear clear the inner parentheses. p

—Andrea Reese, Daytona State College

Skilll Practice Skil Practice Practi ctice Factor F tor ccompletely. Fact ompl pletel etely t ly.

“Without question, Avoiding Mistakes is the h he most helpful for me in the classroom.” —Joseph Howe, St. Charles Community College

TIP Boxes Teaching tips are usually revealed only in the classroom. Not anymore! TIP boxes offer students helpful hints and extra direction to help improve understanding and further insight.

TIP: The inequalities in Example 1 are strict inequalities. Therefore, x  1 and

x  5 (where f1x2  0) are not included in the solution set. However, the corresponding inequalities using the symbols  and  do include the values where f1x2  0. The solution to x2  6x  5  0 is 5x 0 1  x  56 or equivalently, 3 1, 5 4. The solution to x2  6x  5  0 is 5x 0 x  1 or x  56 or 1, 14 ´ 3 5,  2.

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Better Exercise Sets! Better Practice! Better Results! ▶ ▶ ▶

Do your students have trouble with problem solving? Do you want to help students overcome math anxiety? Do you want to help your students improve performance on math assessments?

Problem Recognition Exercises Problem Recognition Exercises present a collection of problems that look similar to a student upon first glance, but are actually quite different in the manner of their individual solutions. Students sharpen critical thinking skills and better develop their “solution recall” to help them distinguish the method needed to solve an exercise—an essential skill in developmental mathematics.

Problem Recognition Exercises, tested in a developmental mathematics classroom, were created to improve student performance while testing.

“I think that the PRE format is both innovative as well as productive in strengthening the students’ expertise in problem solving. Good job!” —Mary Dennison, University of Nebraska at Omaha

Problem Recognition n Exercises Simplifying Radical Expressions ssions For Exercises 1–20, simplify the expressions. 1. a. 224

2. a. 254

3. a. 2200y6

4. a. 232z15

3 b. 224

3 b. 254

3 b. 2200y6

3 b. 232z15

5. a. 280

6. a. 248

7. a. 2x5y6

b. 280

3

b. 248

3

5 6

b. 2x y

3 b. 2a10b12

4

4

4

5 6

4 c. 2a10b12

c. 280 3 9. a. 232s5t6 4

c. 248 3 10. a. 296v7w20

b. 232s5t6

4 b. 296v7w20

5 c. 232s5t6

5 c. 296v7w20

13. a. 226  526

14. a. 327  1027

2 ⴢ 526 26 2 b. 2226 5218 2  428 2 17. a. 52 b 55218 218 ⴢ 428 2 2 b.

c. 2x y

11. a. 25  25

b. 210 ⴢ 210

15. a. 28  22

327 2 ⴢ 102 1027 2 b. 32 250  2 2 272 18. a. 25

12. a. 210  210

b. 25 ⴢ 25

16. a. 212  23

b. 28 ⴢ 22

b. 212 ⴢ 23

3

3

23 19. a. 4224  62

b. 2 250 ⴢ 2 272 b.

“These are so important to test whether a student can recognize different types of problems and the method of solving each. They seem very unique—I have not noticed this feature in many other texts or at least your presentation of the problems is very organized and unique.” —Linda Kuroski, Erie Community College

x

8. a. 2a10b12

3

3

3

b. 4224 ⴢ 6 23

3 3 20 a. 222 2 54  552 2 2 20. 3 3 b. 222 52 2 b. 254 ⴢ 52

“REALLY like this feature! So many algebra students do not recognize the differences that these sections highlight.” —Gayle Krzemien, Pikes Peak Community College

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Get Better Results Student Centered Applications! The Miller/O’Neill/Hyde Board of Advisors partnered with our authors to bring the best applications from every region in the country! These applications include real data and topics that are more relevant and interesting to today’s student. 16. The figure represents the winning time for the men’s 100-m freestyle swimming event for selected Olympic games. Winning Times for Men's 100-m Freestyle Swimming for Selected Olympics

y

Time (sec)

60 50

(0, 57.3)

40

(48, 48.7)

30 20 10 0

x

0

10 20 30 40 50 Year (x  0 corresponds to 1948)

60

Group Activities! Each chapter concludes with a Group Activity to promote classroom discussion and collaboration—helping students not only to solve problems but to explain their solutions for better mathematical mastery. Group Activities are great for instructors and adjuncts—bringing a more interactive approach to teaching mathematics! All required materials, activity time, and suggested group sizes are provided in the end-of-chapter material. Deciphering a Coded Message, Computing the Future Value of an Investment, Creating a Population Model and more!

Group Activity Deciphering a Coded Message Materials: A calculator Estimated time: 20–25 minutes Group Size: 4 (two pairs) Cryptography is the study of coding and decoding messages. One type of coding process assigns a number to each letter of the alphabet and to the space character. For example: A 1

B 2

C 3

D 4

E 5

F 6

G 7

H 8

I 9

J 10

K 11

L 12

M 13

O 15

P 16

Q 17

R 18

S 19

T 20

U 21

V 22

W 23

X 24

Y 25

Z 26

space 27

N 14

According to the numbers assigned to each letter, the message “Decimals have a point” would be coded as follows: D E C I M A L S

__ H A V E __ A __

4 5 3 9 13 1 12 19 27 8 1 22 5 27 1

P O I

N T

27 16 15 9 14 20

Now suppose each letter is encoded by applying a function such as f(x)  2x  5, where x is the numerical value of each letter. For example: The letter “a” a would be coded as:

f(1) f  2(1) 2 57

The letter “b” would be coded as:

f(2)  2(2) 59 f 2

Usingg this encodingg function, function, we have

“A great idea, and beneficial have, Message: when you D E C as I M our school, multiple part-timers teaching Original: 4 5 3 9 13 sections. Brings an aspect of useful uniformity to Coded Form: 13 15 11 23 31 the different sections.”

A L

S

__ H A V E __ A ___

1 12 1 19 27

8

1 22

5

7 29 2 43 59 21 7 49 15

27 1 277

“I like this, because often hands-on application are left out in P O I N activities T Intermediate Algebra. I would 16 15 9 14 20 definitely use this in my classroom.”

59 7 599 37 35 23 33 45 Daytona State College —Brianna Killian,

—Don York, Danville Area Community College To decode decode this this message, messag mes sag ge, e, the the receiver receiv rec eiver eiv er would would need neeed to ne t reverse the operations assigned byy f(x) f x)  2x + 5. f( 5 Since the function f i f multiplies l i li x by b 2 and d then h adds dd 5, 5 we can ca reverse this process by subtracting 5 and dividing by 2. This is represented by g1x2  x 2 5.

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Dynamic Math Animations The Miller/O’Neill/Hyde author team has developed a series of Flash animations to illustrate difficult concepts where static images and text fall short. The animations leverage the use of on-screen movement and morphing shapes to enhance conceptual learning. For example, one animation “cuts” a triangle into three pieces and rotates the pieces to show that the sum of the angular measures equals 180º (below).

Triangles Tria Tr angle es a and nd tthe he eP Pythagorean ythag Theorem

Section 8.6

1. Triangles

Objectives

A triangle is a three-sided polygon. Furt Furthermore, the sum of the measures of the angles within a triangle is 180°. Teachers often demonstrate this fact by tearing a triangular sheet of paper as shown in F Figure 8-24. Then they align the vertices straight angle. (points) of the triangle to form a straigh

1. Triangles 2. Square Roots 3. Pythagorean Theorem

98 60

60 98

22

22

Figure 8-24

PROPERTY P ROPERTY Angles of a Tria Triangle anglee The sum of the measures of the angles angle of a triangle equals 180º.

Example 1

Finding the Measure of Angles Within a Triangle

a.

Skill Practice Find the measures of angles a and b. 1.

Find the measure of angles a and b. b.

a

b a 38

43

a

130

42

2.

Solution: a. Recall that the ⵧ symbol represents a 90° angle. 38°  90°  m1⬔a2  180° 128°  m1⬔a2  180° 128°128°  m1⬔a2  180°  128°

The sum of the angles within a triangle is 180°.

b

39 a 100

Add the measures of the two known angles. Solve for m1⬔a2 .

m1⬔a2  52°

Through their classroom experience, the authors recognize that such media assets are great teaching tools for the classroom and excellent for online learning. The Miller/O’Neill/Hyde animations are interactive and quite diverse in their use. Some provide a virtual laboratory for which an application is simulated and where students can collect data points for analysis and modeling. Others provide interactive question-and-answer sessions to test conceptual learning. For word problem applications, the animations ask students to estimate answers and practice “number sense.”

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Get Better Results

The animations were created by the authors based on over 75 years of combined teaching experience! To facilitate the use of the animations, the authors have placed icons in the text to indicate where animations are available. Students and instructors can access these assets online in MathZone or ALEKS.

2. Graphing Linear Equations tion ns in Two Variables In the introduction to this section, we found several solutions to the equation x  y  4. If we graph these solutions, ons, notice that the points all line up. (See Figure 9-9.) Equation:

y

xy4

(1, ( 1, 5)

12, 22

Several solutions:

11, 32

xy4

14, 02

5 4 3 2

55 4 3 3 2 1 0 1 11 22

11, 52

((1, 1, 3) ((2, 2, 2) ((4, 4, 0)

1 2

3 4

5

x

3 44

The equation actually has infinitely many any solutions.  5 5 This is because there are infinitely many any combinaFigure 9-9 tions of x and y whose sum is 4. Thee graph of all solutions to this equation makes up the he line shown in Figure 9-9. The arrows at each end indicate that the line extends ds infinitely. This is called the graph of the equation. The graph of a linear equation is a line. Therefore, we need to plot at least two points and then draw the line between them. them This is demonstrated in Example 4.

Skill Practice

x  y  2

Graph the equation.

Solution:

y 5 4 3 2

We will find three ordered pairs that are solutions to x  y  2. To find the ordered pairs, choose arbitrary values for x or y, such as those shown in the table. Then complete the table.

1 5 4 3 2 1 0 1 1 2

Graphing a Linear Equation

Example 4

Graph the equation. 7. x  y  4

2

3 4

5

x

x

3

y 13,

3

4 5

1

2

11,

1

Complete: 13,

2

x  y  2 132  y  2 3  3  y  2  3 y5

Answer y

7.

2 , 22 2

Complete: 1

, 22

x  y  2

x  122  2

112  y  2

x  2  2  2  2 x  4 x  4

5

Complete: 11,

x  y  2

2

1y2 11y21 y1

x  y  4 4 3 2 1 5 4 3 2 1 0 1 1 2

2

3 4

5

x

3 4 5

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Get Better Results Experience Student Success! ALEKS is a unique online math tool that uses adaptive questioning and artificial intelligence to correctly place, prepare, and remediate students . . . all in one product! Institutional case studies have shown that ALEKS has improved pass rates by over 20% versus traditional online homework, and by over 30% compared to using a text alone. By offering each student an individualized learning path, ALEKS directs students to work on the math topics that they are ready to learn. Also, to help students keep pace in their course, instructors can correlate ALEKS to their textbook or syllabus in seconds. To learn more about how ALEKS can be used to boost student performance, please visit www.aleks.com/highered/math or contact your McGraw-Hill representative.

Easy Graphing Utility! ALEKS Pie

Students can answer graphing problems with ease!

Each student is given her or his own individualized learning path.

Course Calendar Instructors can schedule assignments and reminders for students.

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With New ALEKS Instructor Module Enhanced Functionality and Streamlined Interface Help to Save Instructor Time The new ALEKS Instructor Module features enhanced functionality and a streamlined interface based on research with ALEKS instructors and homework management instructors. Paired with powerful assignment-driven features, textbook integration, and extensive content flexibility, the new ALEKS Instructor Module simplifies administrative tasks and makes ALEKS more powerful than ever.

New Gradebook! Instructors can seamlessly track student scores on automatically graded assignments. They can also easily adjust the weighting and grading scale of each assignment.

Gradebook view for all students

Gradebook view for an individual student

Track Student Progress Through Detailed Reporting Instructors can track student progress through automated reports and robust reporting features.

Automatically Graded Assignments Instructors can easily assign homework, quizzes, tests, and assessments to all or select students. Deadline extensions can also be created for select students.

Learn more about ALEKS by visiting

www.aleks.com/highered/math or contact your McGraw-Hill representative. Select topics for each assignment

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360° Development Process McGraw-Hill’s 360° Development Process is an ongoing, never-ending, market-oriented approach to building accurate and innovative print and digital products. It is dedicated to continual large-scale and incremental improvement that is driven by multiple customer-feedback loops and checkpoints. This is initiated during the early planning stages of our new products, and intensifies during the development and production stages—then begins again upon publication, in anticipation of the next edition. A key principle in the development of any mathematics text is its ability to adapt to teaching specifications in a universal way. The only way to do so is by contacting those universal voices—and learning from their suggestions. We are confident that our book has the most current content the industry has to offer, thus pushing our desire for accuracy to the highest standard possible. In order to accomplish this, we have moved through an arduous road to production. Extensive and open-minded advice is critical in the production of a superior text. Here is a brief overview of the initiatives included in the Intermediate Algebra, 360° Development Process:

Board of Advisors A hand-picked group of trusted teachers active in the Intermediate Algebra course served as chief advisors and consultants to the author and editorial team with regards to manuscript development. The Board of Advisors reviewed parts of the manuscript; served as a sounding board for pedagogical, media, and design concerns; consulted on organizational changes; and attended a focus group to confirm the manuscript’s readiness for publication.

Would you like to inquire about becoming a BOA member? If so, email the editor, David Millage at [email protected].

Prealgebra

Beginning Algebra

Intermediate Algebra

Beginning and Intermediate Algebra

Vanetta Grier-Felix, Seminole

Anabel Darini, Suffolk County

Connie Buller, Metropolitan

Annette Burden, Youngstown

State College of Florida Teresa Hasenauer, Indian River State College Shelbra Jones, Wake Technical Community College Nicole Lloyd, Lansing Community College Kausha Miller, Bluegrass Community and Technical College Linda Schott, Ozarks Technical Community College Renee Sundrud, Harrisburg Area Community College

Community College Sabine Eggleston, Edison State College Brandie Faulkner, Tallahassee Community College Kelli Hammer, Broward College–South Joseph Howe, St. Charles Community College Laura Iossi, Broward College– Central DiDi Quesada, Miami Dade College

Community College Nancy Carpenter, Johnson County Community College Pauline Chow, Harrisburg Area Community College Donna Gerken, Miami Dade College Gayle Krzemien, Pikes Peak Community College Judy McBride, Indiana University–Purdue University at Indianapolis Patty Parkison, Ball State University

xvi

State University

Lenore Desilets, DeAnza College

Gloria Guerra, St. Philip’s College

Julie Turnbow, Collin County Community College

Suzanne Williams, Central Piedmont Community College Janet Wyatt, Metropolitan Community College– Longview

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Get Better Results Better Development! Question:

How do you build a better developmental mathematics textbook series?

Answer:

Employ a developmental mathematics instructor from the classroom to become a McGraw-Hill editor!

Emilie Berglund joined the developmental mathematics team at McGraw-Hill, bringing her extensive classroom experience to the Miller/O’Neill/Hyde textbook series. A former developmental mathematics instructor at Utah Valley State College, Ms. Berglund has won numerous teaching awards and has served as the beginning algebra course coordinator for the department. Ms. Berglund’s experience teaching developmental mathematics students from the Miller/O’Neill/Hyde translates into more well-developed pedagogy throughout the textbook series and can be seen in everything from the updated Worked Examples to the Exercise Sets.

Listening to You . . . This textbook has been reviewed by over 300 teachers across the country. Our textbook is a commitment to your students, providing a clear explanation, a concise writing style, step-by-step learning tools, and the best exercises and applications in developmental mathematics. How do we know? You told us so!

Teachers Just Like You are saying great things about the Miller/O’Neill/Hyde developmental mathematics series:

“As we matched MOH against many other Intermediate Algebra books, the reading level and writing style, combined with the appropriate use of color and good “looks” of the pages, made this book rise to the top.” —Connie Buller, Metropolitan Community College

“I think the level of rigor is perfect for my students. I have examined other textbooks that would have been placed at the extremes of a continuum. This book is, in the words of Goldilocks, “just right.” ” —Angie McCombs, Illinois State University

“I believe that MOH has separated and grouped the content into sections that will make it easier for students to digest.” —Gayle Krzemien, Pikes Peak Community College

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Acknowledgments and Reviewers The development of this textbook series would never have been possible without the creative ideas and feedback offered by many reviewers. We are especially thankful to the following instructors for their careful review of the manuscript.

Symposia Every year McGraw-Hill conducts general mathematics symposia that are attended by instructors from across the country. These events provide opportunities for editors from McGraw-Hill to gather information about the needs and challenges of instructors teaching these courses. This information helped to create the book plan for Intermediate Algebra. A forum is also offered for the attendees to exchange ideas and experiences with colleagues they otherwise might not have met.

Advisors Symposium—Barton Creek, Texas Connie Buller, Metropolitan Community College Pauline Chow, Harrisburg Area Community College Anabel Darini, Suffolk County Community College Maria DeLucia, Middlesex County College Sabine Eggleston, Edison State College Brandie Faulkner, Tallahassee Community College Vanetta Grier-Felix, Seminole State College of Florida Gloria Guerra, St. Philip’s College Joseph Howe, St. Charles Community College Laura Iossi, Broward College–Central

Gayle Krzemien, Pikes Peak Community College Nicole Lloyd, Lansing Community College Judy McBride, Indiana University–Purdue University at Indianapolis Kausha Miller, Bluegrass Community and Technical College Patty Parkison, Ball State University Linda Schott, Ozarks Technical and Community College Renee Sundrud, Harrisburg Area Community College Janet Wyatt, Metropolitan Community College–Longview

Napa Valley Symposium Antonio Alfonso, Miami Dade College Lynn Beckett-Lemus, El Camino College Kristin Chatas, Washtenaw Community College Maria DeLucia, Middlesex County College Nancy Forrest, Grand Rapids Community College Michael Gibson, John Tyler Community College Linda Horner, Columbia State College Matthew Hudock, St. Philip’s College

Judith Langer, Westchester Community College Kathryn Lavelle, Westchester Community College Scott McDaniel, Middle Tennessee State University Adelaida Quesada, Miami Dade College Susan Shulman, Middlesex County College Stephen Toner, Victor Valley College Chariklia Vassiliadis, Middlesex County College Melanie Walker, Bergen Community College

Myrtle Beach Symposium Patty Bonesteel, Wayne State University Zhixiong Chen, New Jersey City University Latonya Ellis, Bishop State Community College Bonnie Filer, Tubaugh University of Akron Catherine Gong, Citrus College

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Marcia Lambert, Pitt Community College Katrina Nichols, Delta College Karen Stein, The University of Akron Walter Wang, Baruch College

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Get Better Results La Jolla Symposium Darryl Allen, Solano Community College Yvonne Aucoin, Tidewater Community College–Norfolk Sylvia Carr, Missouri State University Elizabeth Chu, Suffolk County Community College Susanna Crawford, Solano Community College Carolyn Facer, Fullerton College

Terran Felter, California State University–Bakersfield Elaine Fitt, Bucks County Community College John Jerome, Suffolk County Community College Sandra Jovicic, The University of Akron Carolyn Robinson, Mt. San Antonio College Carolyn Shand-Hawkins, Missouri State University

Class Tests Multiple class tests provided the editorial team with an understanding of how content and the design of a textbook impact a student’s homework and study habits in the general mathematics course area.

Special “thank you” to our Manuscript Class-Testers

Manuscript Review Panels Over 200 teachers and academics from across the country reviewed the various drafts of the manuscript to give feedback on content, design, pedagogy, and organization. This feedback was summarized by the book team and used to guide the direction of the text.

Reviewers of Miller/O’Neill/Hyde Developmental Mathematics Series Max Aeschbacher, Utah Valley University Ali Ahmad, Dona Ana Community College James Alsobrook, Southern Union State Community College Lisa Angelo, Bucks County Community College Peter Arvanites, Rockland Community College Holly Ashton, Pikes Peak Community College Tony Ayers, Collin County Community College–Plano Tom Baker, South Plains College Lynn Beckett-Lemus, El Camino College Chris Bendixen, Lake Michigan College Mary Benson, Pensacola Junior College Vickie Berry, Northeastern Oklahoma A&M College Abraham Biggs, Broward College–South Erika Blanken, Daytona State College Andrea Blum, Suffolk County Community College Steven Boettcher, Estrella Mountain Community College Gabriele Booth, Daytona State College Charles Bower, Saint Philip’s College Cherie Bowers, Santa Ana College Lee Brendel, Southwestern Illinois College Ellen Brook, Cuyahoga Community College Debra Bryant, Tennessee Tech University Robert Buchanan, Pensacola Junior College Gail Butler, Erie Community College–North

Susan Caldiero, Consumnes River College Kimberly Caldwell, Volunteer State Community College Jose Castillo, Broward College–South Chris Chappa, Tyler Junior College Timothy Chappell, Penn Valley Community College Dianna Cichocki, Erie Community College–South William Clarke, Pikes Peak Community College David Clutts, Southeast Kentucky Community & Technical College De Cook, Okaloosa-Walton College Susan Costa, Broward College–Central Mark Crawford, Waubonsee Community College Patrick Cross, University of Oklahoma Imad Dakka, Oakland Community College–Royal Oak Shirley Davis, South Plains College Nelson De La Rosa, Miami Dade College Mary Dennison, University of Nebraska at Omaha Donna Densmore , Bossier Parish Community College David DeSario, Georgetown College Michael Divinia, San Jose City College Dennis Donohue, College of Southern Nevada Jay Driver, South Plains College Laura Dyer, Southwestern Illinois College Sabine Eggleston, Edison State College Mike Everett, Santa Ana College

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Reviewers of the Miller/O’Neill/Hyde Developmental Mathematics Series Elizabeth Farber, Bucks County Community College Nerissa Felder, Polk Community College Rhoderick Fleming, Wake Technical Community College Carol Ford, Copiah-Lincoln Community College–Wesson Marion Foster, Houston Community College–Southeast Kevin Fox, Shasta College Matt Gardner, North Hennepin Community College Sunshine Gibbons, Southeast Missouri State University Antonnette Gibbs, Broward College–North Barry Gibson, Daytona State College Jeremiah Gilbert, San Bernardino Valley College Sharon Giles, Grossmont College Elizabeth Gore, Georgia Highlands College Brent Griffin, Georgia Highlands College Albert Guerra, Saint Philip’s College Lucy Gurrola, Dona Ana Community College Elizabeth Hamman, Cypress College Mark Harbison, Sacramento City College Pamela Harden, Tennessee Tech University Sherri Hardin, East Tennessee State University Cynthia Harris, Triton College Christie Heinrich, Broward College–North Linda Henderson, Ocean County College Rodger Hergert, Rock Valley College Max Hibbs, Blinn College Terry Hobbs, Maple Woods Community College Richard Hobbs, Mission College Michelle Hollis, Bowling Green Community College at WKU Kathy Holster, South Plains College Mark Hopkins, Oakland Community College–Auburn Hills Robert Houston, Rose State College Steven Howard, Rose State College Joe Howe, St. Charles Community College Glenn Jablonski, Triton College Michelle Jackson, Bowling Green Community College at WKU Pamela Jackson, Oakland Community College– Orchard Ridge Thomas Jay, Houston Community College–Northwest Michael Jones, Suffolk County Community College Diane Joyner, Wayne Community College Maryann Justinger, Erie Community College–South Cheryl Kane, University of Nebraska–Lincoln Susan Kautz, Lone Star College–CyFair Joseph Kazimir, East Los Angeles College Eliane Keane, Miami Dade College Mandy Keiner, Iowa Western Community College Maria Kelly, Reedley College

xx

Brianna Killian, Daytona State College Harriet Kiser, Georgia Highlands College Daniel Kleinfelter, College of the Desert Linda Kuroski, Erie Community College Catherine LaBerta, Erie Community College–North Debra Landre, San Joaquin Delta College Cynthia Landrigan, Erie Community College–South Melanie Largin, Georgia Highlands College Betty Larson, South Dakota State University Kathryn Lavelle, Westchester Community College Karen Lee, Oakland Community College–Southfield Paul Lee, St. Philip’s College Richard Leedy, Polk Community College Julie Letellier, University of Wisconsin–Whitewater Nancy Leveille, University of Houston–Downtown Janna Liberant, Rockland Community College Joyce Lindstrom, St. Charles Community College John Linnen, Ferris State University Mark Littrell, Rio Hondo College Linda Lohman, Jefferson Community & Technical College Tristan Londre, Metropolitan Community College– Blue River Wanda Long, St. Charles Community College Yixia Lu, South Suburban College Shawna Mahan, Pikes Peak Community College Vincent Manatsa, Georgia Highlands College Dorothy Marshall, Edison State College Melvin Mays, Metropolitan Community College William Mays, Salem Community College Angela McCombs, Illinois State University Paul Mccombs, Rock Valley College Robert McCullough, Ferris State University Raymond McDaniel, Southern Illinois University Edwardsville Jamie McGill, East Tennessee State University Vicki McMillian, Ocean County College Lynette Meslinsky, Erie Community College–City Gabrielle Michaelis, Cumberland County College John Mitchell, Clark College Chris Mizell, Okaloosa-Walton College Daniel Munton, Santa Rosa Junior College Revathi Narasimhan, Kean Community College Michael Nasab, Long Beach City College Elsie Newman, Owens Community College Charles Odion, Houston Community College Jean Olsen, Pikes Peak Community College Jason Pallett, Metropolitan Community College–Longview Alan Papen, Ozarks Technical Community College

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Reviewers of the Miller/O’Neill/Hyde Developmental Mathematics Series Victor Pareja, Daytona State College Linda Parrish, Brevard College Mari Peddycoart, Lone Star College–Kingwood Joanne Peeples, El Paso Community College Matthew Pitassi, Rio Hondo College Froozen Pourboghrat-Afiat, College of Southern Nevada Jay Priester, Horry-Georgetown Technical College Gail Queen, Shelton State Community College Adelaida Quesada, Miami Dade College Jill Rafael, Sierra College Janice Rech, University of Nebraska at Omaha George Reed, Angelina College Pamelyn Reed, Lone Star College–CyFair Andrea Reese, Daytona State College Donna Riedel, Jefferson Community & Technical College Donald Robertson, Olympic College Cosmin Roman, The Ohio State University Tracy Romesser, Erie Community College Suzanne Rosenberger, Harrisburg Area Community College Connie Rost, South Louisiana Community College Richard Rupp, Del Mar College Angela Russell, Wenatchee Valley College Kristina Sampson, Lone Star College–CyFair Jenell Sargent, Tennessee Tech University Vicki Schell, Pensacola Junior College Linda Schott, Ozarks Technical Community College Rebecca Schuering, Metropolitan Community College– Blue River Christyn Senese, Triton College

Alicia Serfaty De Markus, Miami Dade College Angie Shreckhise, Ozarks Technical Community College Abdallah Shuaibi, Harry S Truman College Julia Simms, Southern Illinois University Edwardsville Azar Sioshansi, San Jose City College Leonora Smook, Suffolk County Community College Carol St. Denis, Okaloosa-Walton College Andrew Stephan, St. Charles Community College Sean Stewart, Owens Community College Arcola Sullivan, Copiah-Lincoln Community College Nader Taha, Kent State University Michael Tiano, Suffolk County Community College Roy Tucker, Palo Alto College Clairie Vassiliadis, Middlesex County College Rieken Venema, University of Alaska Anchorage Sherry Wallin, Sierra College Kathleen Wanstreet, Southern Illinois University Edwardsville Natalie Weaver, Daytona State College Greg Wheaton, Kishwaukee College Deborah Wolfson, Suffolk County Community College Rick Woodmansee, Sacramento City College Kevin Yokoyama, College of the Redwoods Donald York, Danville Area Community College Vivian Zabrocki, Montana State University–Billings Ruth Zasada, Owens Community College Loris Zucca, Kingwood College Diane Zych, Erie Community College–North

Special thanks go to Jon Weertz for preparing the Instructor’s Solutions Manual and the Student’s Solution Manual and to Carrie Green, Rebecca Hubiak, and Hal Whipple for their work ensuring accuracy. Many thanks to Cindy Reed for her work in the video series, and to Kelly Jackson for advising us on the Instructor Notes. Finally, we are forever grateful to the many people behind the scenes at McGraw-Hill without whom we would still be on page 1. To our developmental editor (and math instructor extraordinaire), Emilie Berglund, thanks for your day-to-day support and understanding of the world of developmental mathematics. To David Millage, our executive editor and overall team captain, thanks for keeping the train on the track. Where did you find enough hours in the day? To Torie Anderson

and Sabina Navsariwala, we greatly appreciate your countless hours of support and creative ideas promoting all of our efforts. To our director of development and champion, Kris Tibbetts, thanks for being there in our time of need. To Pat Steele, where would we be without your watchful eye over our manuscript? To our publisher, Stewart Mattson, we’re grateful for your experience and energizing new ideas. Thanks for believing in us. To Jeff Huettman and Amber Bettcher, we give our greatest appreciation for the exciting technology so critical to student success, and to Peggy Selle, thanks for keeping watch over the whole team as the project came together. Most importantly, we give special thanks to all the students and instructors who use Intermediate Algebra in their classes.

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Get Better Results A COMMITMENT TO ACCURACY You have a right to expect an accurate textbook, and McGraw-Hill invests considerable time and effort to make sure that we deliver one. Listed below are the many steps we take to make sure this happens.

Our Accuracy Verification Process 1st Round: Author’s Manuscript

✓

Multiple Rounds of Review by College Math Instructors

2nd Round: Typeset Pages

Accuracy Checks by: ✓ Authors ✓ Professional Mathematician ✓ 1st Proofreader

First Round Step 1: Numerous college math instructors review the manuscript and report on any errors that they may find. Then the authors make these corrections in their final manuscript.

Second Round Step 2: Once the manuscript has been typeset, the authors check their manuscript against the first page proofs to ensure that all illustrations, graphs, examples, exercises, solutions, and answers have been correctly laid out on the pages, and that all notation is correctly used. Step 3: An outside, professional mathematician works through every example and exercise in the page proofs to verify the accuracy of the answers. Step 4: A proofreader adds a triple layer of accuracy assurance in the first pages by hunting for errors, then a second, corrected round of page proofs is produced.

Third Round 3rd Round: Typeset Pages

Accuracy Checks by: ✓ Authors ✓ 2nd Proofreader

4th Round: Typeset Pages

Accuracy Checks by: ✓ 3rd Proofreader ✓ Test Bank Author ✓ Solutions Manual Author ✓ Consulting Mathematicians for MathZone site ✓ Math Instructors for text’s video series

Final Round: Printing

✓

xxii

Accuracy Check by 4th Proofreader

Step 5: The author team reviews the second round of page proofs for two reasons: (1) to make certain that any previous corrections were properly made, and (2) to look for any errors they might have missed on the first round. Step 6: A second proofreader is added to the project to examine the new round of page proofs to double check the author team’s work and to lend a fresh, critical eye to the book before the third round of paging.

Fourth Round Step 7: A third proofreader inspects the third round of page proofs to verify that all previous corrections have been properly made and that there are no new or remaining errors. Step 8: Meanwhile, in partnership with independent mathematicians, the text accuracy is verified from a variety of fresh perspectives: • The test bank authors check for consistency and accuracy as they prepare the computerized test item file. • The solutions manual author works every exercise and verifies his/her answers, reporting any errors to the publisher. • A consulting group of mathematicians, who write material for the text’s MathZone site, notifies the publisher of any errors they encounter in the page proofs. • A video production company employing expert math instructors for the text’s videos will alert the publisher of any errors it might find in the page proofs.

Final Round Step 9: The project manager, who has overseen the book from the beginning, performs a fourth proofread of the textbook during the printing process, providing a final accuracy review. What results is a mathematics textbook that is as accurate and error-free as is humanly possible, and our authors and publishing staff are confident that our many layers of quality assurance have produced textbooks that are the leaders in the industry for their integrity and correctness.

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Brief Contents

Chapter R

Review of Basic Algebraic Concepts

1

Chapter 1

Linear Equations and Inequalities in One Variable

43

Chapter 2

Linear Equations in Two Variables and Functions

127

Chapter 3

Systems of Linear Equations and Inequalities

Chapter 4

Polynomials

Chapter 5

Rational Expressions and Rational Equations

Chapter 6

Radicals and Complex Numbers

Chapter 7

Quadratic Equations and Functions

Chapter 8

Exponential and Logarithmic Functions and Applications 655

Chapter 9

Conic Sections

233

313 415

491 573

745

Chapter 10 Binomial Expansions, Sequences, and Series

805

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Contents Chapter R

Review of Basic Algebraic Concepts R.1

Study Skills

2

Group Activity: Becoming a Successful Student R.2 Sets of Numbers and Interval Notation R.3 Operations on Real Numbers Chapter R Summary

5

30

39

Chapter R Review Exercises Chapter R Test

3

16

R.4 Simplifying Algebraic Expressions

Chapter 1

1

41

42

Linear Equations and Inequalities in One Variable 1.1

Linear Equations in One Variable

44

Problem Recognition Exercises: Equations Versus Expressions 1.2 Applications of Linear Equations in One Variable 1.3 Applications to Geometry and Literal Equations 1.4 Linear Inequalities in One Variable 1.5 Compound Inequalities

43 56

57 68

76

85

1.6 Absolute Value Equations

97

1.7 Absolute Value Inequalities

103

Problem Recognition Exercises: Identifying Equations and Inequalities Group Activity: Understanding the Symbolism of Mathematics Chapter 1 Summary

Chapter 2

122

125

Linear Equations in Two Variables and Functions 2.1

114

115

Chapter 1 Review Exercises Chapter 1 Test

Linear Equations in Two Variables

2.2 Slope of a Line and Rate of Change 2.3 Equations of a Line

127

128 145

156

Problem Recognition Exercises: Characteristics of Linear Equations 2.4 Applications of Linear Equations and Modeling 2.5 Introduction to Relations

182

2.6 Introduction to Functions

191

2.7 Graphs of Functions

170

202

Problem Recognition Exercises: Characteristics of Relations

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113

214

169

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Group Activity: Deciphering a Coded Message Chapter 2 Summary

216

Chapter 2 Review Exercises Chapter 2 Test

222

227

Chapters 1–2 Cumulative Review Exercises

Chapter 3

215

230

Systems of Linear Equations and Inequalities 3.1

233

Solving Systems of Linear Equations by the Graphing Method

234

3.2 Solving Systems of Linear Equations by the Substitution Method 3.3 Solving Systems of Linear Equations by the Addition Method

243

249

Problem Recognition Exercises: Solving Systems of Linear Equations 3.4 Applications of Systems of Linear Equations in Two Variables

256

3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables 3.6 Systems of Linear Equations in Three Variables and Applications 3.7 Solving Systems of Linear Equations by Using Matrices

256 265

278

288

Group Activity: Creating a Quadratic Model of the Form y ⴝ at2 ⴙ bt ⴙ c Chapter 3 Summary

298

Chapter 3 Review Exercises Chapter 3 Test

305

309

Chapters 1–3 Cumulative Review Exercises

Chapter 4

Polynomials 4.1

297

311

313

Properties of Integer Exponents and Scientific Notation

314

4.2 Addition and Subtraction of Polynomials and Polynomial Functions 4.3 Multiplication of Polynomials 4.4 Division of Polynomials

334

343

Problem Recognition Exercises: Operations on Polynomials 4.5 Greatest Common Factor and Factoring by Grouping 4.6 Factoring Trinomials

362

4.7 Factoring Binomials

376

Problem Recognition Exercises: Factoring Summary 4.8 Solving Equations by Using the Zero Product Rule Group Activity: Investigating Pascal’s Triangle Chapter 4 Summary

353

354

385

388

402

403

Chapter 4 Review Exercises Chapter 4 Test

323

408

412

Chapters 1–4 Cumulative Review Exercises

413

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Chapter 5

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Rational Expressions and Rational Equations 5.1

Rational Expressions and Rational Functions

416

5.2 Multiplication and Division of Rational Expressions

426

5.3 Addition and Subtraction of Rational Expressions 5.4 Complex Fractions

415

431

441

Problem Recognition Exercises: Operations on Rational Expressions 5.5 Solving Rational Equations

449

Problem Recognition Exercises: Rational Equations vs. Expressions 5.6 Applications of Rational Equations and Proportions 5.7 Variation

469

Chapter 5 Summary Chapter 5 Test

478

479

Chapter 5 Review Exercises

484

487

Chapters 1–5 Cumulative Review Exercises

Radicals and Complex Numbers 6.1

Definition of an nth Root

6.2 Rational Exponents

488

491

492

503

6.3 Simplifying Radical Expressions

510

6.4 Addition and Subtraction of Radicals 6.5 Multiplication of Radicals

517

522

Problem Recognition Exercises: Simplifying Radical Expressions 6.6 Division of Radicals and Rationalization 6.7 Solving Radical Equations 6.8 Complex Numbers

530

540

550

Group Activity: Margin of Error of Survey Results Chapter 6 Summary

561

Chapter 6 Review Exercises Chapter 6 Test

567

570

Chapters 1–6 Cumulative Review Exercises

xxvi

457

458

Group Activity: Computing the Future Value of an Investment

Chapter 6

449

571

559

530

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Chapter 7

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Quadratic Equations and Functions 7.1

573

Square Root Property and Completing the Square

7.2 Quadratic Formula

574

583

7.3 Equations in Quadratic Form

598

Problem Recognition Exercises: Quadratic and Quadratic Type Equations 7.4 Graphs of Quadratic Functions

604

7.5 Vertex of a Parabola: Applications and Modeling 7.6 Nonlinear Inequalities

618

628

Problem Recognition Exercises: Recognizing Equations and Inequalities

640

Group Activity: Creating a Quadratic Model of the Form y ⴝ a(x ⴚ h)2 ⴙ k

641

Chapter 7 Summary

642

Chapter 7 Review Exercises Chapter 7 Test

647

650

Chapters 1–7 Cumulative Review Exercises

Chapter 8

603

652

Exponential and Logarithmic Functions and Applications 655 8.1

Algebra of Functions and Composition

8.2 Inverse Functions

656

663

8.3 Exponential Functions

672

8.4 Logarithmic Functions

682

Problem Recognition Exercises: Identifying Graphs of Functions 8.5 Properties of Logarithms

695

696

8.6 The Irrational Number e and Change of Base

704

Problem Recognition Exercises: Logarithmic and Exponential Forms 8.7 Logarithmic and Exponential Equations and Applications Group Activity: Creating a Population Model Chapter 8 Summary

718

730

731

Chapter 8 Review Exercises Chapter 8 Test

717

736

740

Chapters 1–8 Cumulative Review Exercises

742

xxvii

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Conic Sections 9.1

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745

Distance Formula, Midpoint Formula, and Circles

9.2 More on the Parabola

746

757

9.3 The Ellipse and Hyperbola

766

Problem Recognition Exercises: Formulas and Conic Sections 9.4 Nonlinear Systems of Equations in Two Variables

775

776

9.5 Nonlinear Inequalities and Systems of Inequalities

783

Group Activity: Investigating the Graphs of Conic Sections on a Calculator 792 Chapter 9 Summary

793

Chapter 9 Review Exercises Chapter 9 Test

798

801

Chapters 1–9 Cumulative Review Exercises

Chapter 10

803

Binomial Expansions, Sequences, and Series 10.1

Binomial Expansions

10.2 Sequences and Series

805

806 812

10.3 Arithmetic Sequences and Series

820

10.4 Geometric Sequences and Series

826

Problem Recognition Exercises: Identifying Arithmetic and Geometric Sequences 834 Group Activity: Investigating Mean and Standard Deviation Chapter 10 Summary

836

Chapter 10 Review Exercises Chapter 10 Test

839

840

Chapters 1–10 Cumulative Review Exercises

Additional Topics Appendix A.1

Determinants and Cramer’s Rule

Student Answer Appendix Index

xxviii

I-1

A-1 A-1

SA-1

842

835

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Section R.1

Study Skills

1

R

Review of Basic Algebraic Concepts

CHAPTER OUTLINE R.1 Study Skills 2 Group Activity: Becoming a Successful Student

3

R.2 Sets of Numbers and Interval Notation 5 R.3 Operations on Real Numbers 16 R.4 Simplifying Algebraic Expressions 30

Chapter R In this chapter, we begin our study of algebra by reviewing the sets of numbers used in day-to-day life. We also review how to simplify numerical expressions and algebraic expressions. To prepare for this chapter, practice the following operations on whole numbers, decimals, fractions, and mixed numbers. Are You Prepared? For Exercises 1–12, simplify each expression. Then find the answer on the right and record the corresponding letter at the bottom of the page. When you are finished, you will have a key definition for this chapter. Exercises 1. 36,636 ⫼ 43 3. 24.0842 ⫹ 365.7 7 ⫼ 14 3 2 7 7. ⫺ 3 12 1 1 9. 3 ⫻ 2 5 2 11. 582 ⫼ 0.01 5.

2. 0.25 ⫻ 6340 4 21 4. ⫻ 7 10 5 1 6. ⫺ 3 2 5 1 8. 4 ⫺ 1 9 3 1 10. 3.75 ⫹ 8 5 12. 582 ⫻ 0.01

Answers B. 1585 I. 389.7842 1 6 D. R. 6 5 T. 852 E. 3

2 9

7 6 V. 58,200 Y.

S. 11.95 P.

1 12

O. 8 U. 5.82

In this chapter we will show that a(b ⫹ c) ⫽ ab ⫹ ac. This important property is called the: ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ 4 3 10 1 5 3 2 12 1 3 11 8

___ ___ ___ ___ ___ ___ ___ ___. 7 5 9 7 8 5 1 6

1

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Chapter R Review of Basic Algebraic Concepts

Section R.1 Concepts 1. 2. 3. 4.

Before the Course During the Course Preparation for Exams Where to Go for Help

Study Skills In taking a course in algebra, you are making a commitment to yourself, your instructor, and your classmates. Following some or all of the study tips below can help you be successful in this endeavor. The features of this text that will assist you are printed in blue.

1. Before the Course • Purchase the necessary materials for the course before the course begins or on the first day. • Obtain a three-ring binder to keep and organize your notes, homework, tests, and any other materials acquired in the class. We call this type of notebook a portfolio. • Arrange your schedule so that you have enough time to attend class and to do homework. A common rule is to set aside at least 2 hours for homework for every hour spent in class. That is, if you are taking a 4-credit-hour course, plan on at least 8 hours a week for homework. If you experience difficulty in mathematics, plan for more time. A 4-credit-hour course will then take at least 12 hours each week—about the same as a part-time job. • Communicate with your employer and family members the importance of your success in this course so that they can support you. • Be sure to find out the type of calculator (if any) that your instructor requires. Also determine if there will be online homework or other computer requirements.

2. During the Course • To prepare yourself for the next day’s class, read the section in the text before coming to class. This will help you familiarize yourself with the material and terminology. • Attend every class, and be on time. • Take notes in class. Write down all of the examples that the instructor presents. Read the notes after class, and add any comments to make your notes clearer to you. Use a tape recorder to record the lecture if the instructor permits the recording of lectures. • Ask questions in class. • Read the section in the text after the lecture, and pay special attention to the Tip boxes and Avoiding Mistakes boxes. • After you read an example, try the accompanying Skill Practice problem. The skill practice problem mirrors the example and tests your understanding of what you have read. • Do homework every night. Even if your class does not meet every day, you should still do some work every night to keep the material fresh in your mind. • Check your homework with the answers that are supplied in the back of this text. Analyze what you did wrong and correct the exercises that do not match. Circle or star those that you cannot correct yourself. This way you can easily find them and ask your instructor the next day. • Write the definition and give an example of each Key Term found at the beginning of the Practice Exercises. • The Problem Recognition Exercises are located in most chapters. These provide additional practice distinguishing among a variety of problem types. Sometimes the most difficult part of learning mathematics is retaining all that you learn. These exercises are excellent tools for retention of material.

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Group Activity

• Form a study group with fellow students in your class, and exchange phone numbers. You will be surprised by how much you can learn by talking about mathematics with other students. • If you use a calculator in your class, read the Calculator Connections boxes to learn how and when to use your calculator. • Ask your instructor where you might obtain extra help if necessary.

3. Preparation for Exams •

Look over your homework and rework exercises that gave you trouble. Pay special attention to the exercises you have circled or starred to be sure that you have learned that concept. • Read through the Summary at the end of the chapter. Be sure that you understand each concept and example. If not, go to the section in the text and reread that section. • Give yourself enough time to take the Chapter Test uninterrupted. Then check the answers. For each problem you answered incorrectly, go to the Review Exercises and do all of the problems that are similar. • To prepare for the final exam, complete the Cumulative Review Exercises at the end of each chapter, starting with Chapter 2. If you complete the cumulative reviews after finishing each chapter, then you will be preparing for the final exam throughout the course. The Cumulative Review Exercises are another excellent tool for helping you retain material.

4. Where to Go for Help • At the first sign of trouble, see your instructor. Most instructors have specific office hours set aside to help students. Don’t wait until after you have failed an exam to seek assistance. • Get a tutor. Most colleges and universities have free tutoring available. • When your instructor and tutor are unavailable, use the Student Solutions Manual for step-by-step solutions to the odd-numbered problems in the exercise sets. • Work with another student from your class. • Work on the computer. Many mathematics tutorial programs and websites are available on the Internet, including the website that accompanies this text.

Group Activity Becoming a Successful Student Materials: Computer with Internet access Estimated time: 15 minutes Group Size: 4 Good time management, good study skills, and good organization will help you be successful in this course. Answer the following questions and compare your answers with your group members. 1. To motivate yourself to complete a course, it is helpful to have clear reasons for taking the course. List your goals for taking this course and discuss them with your group.

3

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Chapter R Review of Basic Algebraic Concepts

2. Taking 12 credit-hours is the equivalent of a full-time job. Often students try to work too many hours while taking classes at school. a. Write down the number of hours you work per week and the number of credit hours you are taking this term. number of hours worked per week number of credit-hours this term b. The table gives a recommended limit to the number of hours you should work for the number of credit hours you are taking at school. (Keep in mind that other responsibilities in your life such as your family might also make it necessary to limit your hours at work even more.) How do your numbers from part (a) compare to those in the table? Are you working too many hours?

Number of Credit-Hours

Maximum Number of Hours of Work per Week

3

40

6

30

9

20

12

10

15

0

c. It is often suggested that you devote two hours of study and homework time outside of class for each credit-hour you take at school. For example: 12 credit-hours ⫹ 24 study hours 36 total hours

full-time job!

Based on the number of credit-hours you are taking, how many study hours should you plan for? What is the total number of hours (class time plus study time) that you should devote to school? 3. For the following week, write down the times each day that you plan to study math. Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

4. Write down the date of your next math test.

5. Look through the book in Chapter 2 and find the page number corresponding to each feature in the book. Discuss with your group members how you might use each feature. Problem Recognition Exercises: page Chapter Summary: page Chapter Review Exercises: page Chapter Test: page Cumulative Review Exercises: page

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Section R.2

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5

6. Look at the Skill Practice exercises after each example (for instance, find Skill Practice exercises 1–4 in Section R.2).Where are the answers to these exercises located? Discuss with your group members how you might use the Skill Practice exercises.

7. Discuss with your group members places where you can go for extra help in math. Then write down three of the suggestions.

8. Do you keep an organized notebook for this class? Can you think of any suggestions that you can share with your group members to help them keep their materials organized? 9. Do you think that you have math anxiety? Read the following list for some possible solutions. Check the activities that you can realistically try to help you overcome this problem. Read a book on math anxiety. Search the Web for help tips on handling math anxiety. See a counselor to discuss your anxiety. See your instructor to inform him or her about your situation. Evaluate your time management to see if you are trying to do too much. Then adjust your schedule accordingly. 10. Some students favor different methods of learning over others. For example, you might prefer: • Learning through listening and hearing. • Learning through seeing images, watching demonstrations, and visualizing diagrams and charts. • Learning by experience through a hands-on approach. • Learning through reading and writing. Most experts believe that the most effective learning comes when a student engages in all of these activities. However, each individual is different and may benefit from one activity more than another. You can visit a number of different websites to determine your “learning style.” Try doing a search on the Internet with the key words “learning styles assessment.” Once you have found a suitable website, answer the questionnaire and the site will give you feedback on what method of learning works best for you.

Sets of Numbers and Interval Notation

Section R.2

1. The Set of Real Numbers

Concepts

Algebra is a powerful mathematical tool that is used to solve real-world problems in science, business, and many other fields. We begin our study of algebra with a review of basic definitions and notations used to express algebraic relationships.

1. 2. 3. 4.

The Set of Real Numbers Inequalities Interval Notation Translations Involving Inequalities

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Chapter R Review of Basic Algebraic Concepts

In mathematics, a collection of items (called elements) is called a set, and the set braces { } are used to enclose the elements of the set. For example, the set {a, e, i, o, u} represents the vowels in the English alphabet. The set {1, 3, 5, 7} represents the first four positive odd numbers. Another method to express a set is to describe the elements of the set by using set-builder notation. Consider the set {a, e, i, o, u} in set-builder notation. description of set

Set-builder notation: {x | x is a vowel in the English alphabet}

“the set of ”

“all x”

“such that”

“x is a vowel in the English alphabet”

Consider the set {1, 3, 5, 7} in set-builder notation. description of set

Set-builder notation: {x | x is an odd number between 0 and 8}

“the set of ”

“all x”

“such that”

“x is an odd number between 0 and 8”

Several sets of numbers are used extensively in algebra. The numbers you are familiar with in day-to-day calculations are elements of the set of real numbers. These numbers can be represented graphically on a horizontal number line with a point labeled as 0. Positive real numbers are graphed to the right of 0, and negative real numbers are graphed to the left. Each point on the number line corresponds to exactly one real number, and for this reason, the line is called the real number line (Figure R-1). ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 0 Negative numbers

1 2 3 4 Positive numbers

5

Figure R-1

Several sets of numbers are subsets (or part) of the set of real numbers. These are The set of natural numbers The set of whole numbers The set of integers The set of rational numbers The set of irrational numbers

DEFINITION Natural Numbers, Whole Numbers, and Integers The set of natural numbers is {1, 2, 3, . . . }. The set of whole numbers is {0, 1, 2, 3, . . . }. The set of integers is { . . . , ⫺3, ⫺2, ⫺1, 0, 1, 2, 3, . . . }.

The set of rational numbers consists of all the numbers that can be defined as a ratio of two integers.

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Section R.2

Sets of Numbers and Interval Notation

DEFINITION Rational Numbers

The set of rational numbers is 5 q 0 p and q are integers and q does not equal zero}. p

Example 1

Identifying Rational Numbers

Show that each number is a rational number by finding two integers whose ratio equals the given number. a.

4 7

b. 8

c. 0.6

d. 0.87

Solution: a.

4 7

is a rational number because it can be expressed as the ratio of the integers 4 and 7.

b. 8 is a rational number because it can be expressed as the ratio of the integers 8 and 1 18  81 2. In this example we see that an integer is also a rational number. c. 0.6 represents the repeating decimal 0.6666666    and can be expressed as the ratio of 2 and 3 10.6  23 2. In this example we see that a repeating decimal is a rational number.

87 d. 0.87 is the ratio of 87 and 100 10.87  100 2. In this example we see that a terminating decimal is a rational number.

Skill Practice Show that the numbers are rational by writing them as a ratio of integers. 1.

9 8

2. 0

3. 0.3

4. 0.45

TIP: Any rational number can be represented by a terminating decimal or by a repeating decimal.

Some real numbers such as the number p (pi) cannot be represented by the ratio of two integers. In decimal form, an irrational number is a nonterminating, nonrepeating decimal. The value of p, for example, can be approximated as p ⬇ 3.1415926535897932. However, the decimal digits continue indefinitely with no pattern. Other examples of irrational numbers are the square roots of nonperfect squares, such as 13 and 111.

DEFINITION Irrational Numbers The set of irrational numbers is a subset of the real numbers whose elements cannot be written as a ratio of two integers. Note: An irrational number cannot be written as a terminating decimal or as a repeating decimal.

The set of real numbers consists of both the rational numbers and the irrational numbers. The relationships among the sets of numbers discussed thus far are illustrated in Figure R-2.

Answers 9 8 1 3. 3 1.

0 1 9 45 or 4. 100 20

2.

7

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Chapter R Review of Basic Algebraic Concepts

Real numbers Rational numbers 

0.25

2 7

0.3

Irrational numbers √2

Integers … 3, 2, 1

√17 ␲

Whole numbers 0 Natural numbers 1, 2, 3, …

Figure R-2

Classifying Numbers by Set

Example 2

Check the set(s) to which each number belongs. The numbers may belong to more than one set. Natural Numbers

Whole Numbers

Rational Numbers

Integers

Irrational Numbers

Real Numbers

6 123 27 3 2.3

Solution: Natural Numbers

Whole Numbers

6

Integers

Rational Numbers

✓

✓

123 27 3

✓

✓

✓

2.3

Natural

✓

Whole

✓

Integer

✓

Rational

✓

Integer ✓

✓

✓

✓

Rational Irrational

✓ ✓

Natural Whole

Irrational Real

1

✓

Real

✓

✓

✓

✓

5. Check the set(s) to which each number belongs. 0.47 15 12

✓

✓ ✓

Answer 1

Real Numbers

✓ ✓

Skill Practice

5.

Irrational Numbers

0.47 15 ⴚ12

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Section R.2

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2. Inequalities The relative value of two numbers can be compared by using the real number line. We say that a is less than b (written mathematically as a  b) if a lies to the left of b on the number line. a

b ab

We say that a is greater than b (written mathematically as a  b) if a lies to the right of b on the number line. b

a ab

From looking at the number line, note that a  b is the same as b  a. Table R-1 summarizes the relational operators that compare two real numbers a and b. Table R-1 Mathematical Expression

Translation

Other Meanings

a 6 b

a is less than b

b exceeds a

a 7 b

a is greater than b

a b

a is less than or equal to b

ab

a is greater than or equal to b

ab

a is equal to b

ab

a is not equal to b

a⬇b

a is approximately equal to b

b is greater than a a exceeds b b is less than a a is at most b a is no more than b a is no less than b a is at least b

The symbols , , , , and  are called inequality signs, and the expressions a 6 b, a 7 b, a b, a  b, and a  b are called inequalities.

Example 3

Ordering Real Numbers

Fill in the blank with the appropriate inequality symbol:  or  a. 2 ______ 5

b.

4 3 ______ 7 5

c. 1.3 ______ 1.3

Solution: a. 2

5

7

6 5 4 3 2 1

0

1

2

3

4

5

6

b. To compare 74 and 35, write the fractions as equivalent fractions with a common denominator. 4 5 20 ⴢ  7 5 35

and

3 7 21 ⴢ  5 7 35 4 7

Because

20 21 4 6 , then 35 35 7

6

3 5

6 5 4 3 2 1

3 5

0

1

2

3

4

5

6

9

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Chapter R Review of Basic Algebraic Concepts

c. 1.3



1.33333 . . .

1.3

1.3

2

1

0

1

2

Skill Practice Fill in the blanks with the appropriate symbol,  or . 6. 2 ______ 12

7.

1 2 ______ 4 9

8. 7.2 ______ 7.2

3. Interval Notation

The set {x 0 x  3} represents all real numbers greater than or equal to 3. This set can be illustrated graphically on the number line. 5 4 3 2 1

0

1

2

3

4

5

5 4 3 2 1

0

1

2

3

4

5

By convention, a closed circle ● or a square bracket [ is used to indicate that an “endpoint” (x  3) is included in the set.

The set {x 0 x  3} represents all real numbers strictly greater than 3. This set can be illustrated graphically on the number line. 5 4 3 2 1

0

1

2

3

5 4 3 2 1

0

1

2

3

4

5

4

5

By convention, an open circle 䊊 or a parenthesis ( is used to indicate that an “endpoint” (x  3) is not included in the set.

(

Notice that the sets {x 0 x  3} and {x 0 x  3} consist of an infinite number of elements that cannot all be listed. Another method to represent the elements of such sets is by using interval notation. To understand interval notation, first consider the real number line, which extends infinitely far to the left and right. The symbol  is used to represent infinity. The symbol  is used to represent negative infinity. 

 0

To express a set of real numbers in interval notation, sketch the graph first, using the symbols ( ) or [ ]. Then use these symbols at the endpoints to define the interval.

Example 4

Expressing Sets by Using Interval Notation

Graph each set on the number line, and express the set in interval notation. a. {x 0 x  3}

b. { p 0 p  3}

Solution: a. Set-builder notation: Graph:

7. 

5 4 3 2 1

Interval notation:

Answers 6. 



8. 

{x 0 x  3}

33, 2

 0

1

2

3 [3

4

5 )

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Section R.2

Sets of Numbers and Interval Notation

The graph of the set {x 0 x  3} “begins” at 3 and extends infinitely far to the right. The corresponding interval notation “begins” at 3 and extends to . Notice that a square bracket [ is used at 3 for both the graph and the interval notation to include x  3. A parenthesis is always used at and at 

because there is no endpoint. b. Set-builder notation: Graph:



{ p 0 p  3}

5 4 3 2 1

Interval notation:

0

1

2

(

3

4

5

(3

13, 2



)

Skill Practice Graph each set, and express the set in interval notation. 9. 5w 0 w  76

10. 5x 0 x 6 06

In general, we use the following guidelines when applying interval notation.

PROCEDURE Using Interval Notation • The endpoints used in interval notation are always written from left to right. That is, the smaller number is written first, followed by a comma, followed by the larger number. • Parentheses ) or ( indicate that an endpoint is excluded from the set. • Square brackets ] or [ indicate that an endpoint is included in the set. • Parentheses are always used with and  .

Example 5

Expressing Sets by Using Interval Notation

Graph each set on the number line, and express the set in interval notation. a. 5z 0 z 32 6

b. 5x 0 4 6 x 26

Solution: a. Set-builder notation:

5z 0 z 32 6  32

Graph: 

5 4 3 2 1

(

Interval notation:

1 ,

32 4

0

1

2

3

4

5



 32 ]

The graph of the set 5z 0 z 32 6 extends infinitely far to the left. Interval notation is always written from left to right. Therefore,  is written first, followed by a comma, and then followed by the right-hand endpoint 32. b. The inequality 4 6 x 2 means that x is greater than 4 and also less than or equal to 2. More concisely, we can say that x represents the real numbers between 4 and 2, including the endpoint, 2. Set-builder notation: Graph:

9. 7 37, 2

{x 0 4  x 2}

10.

(

5 4 3 2 1

Interval notation:

Answers

(4, 2]

0

1

2

3

4

5

(

0 1 , 02

11

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Chapter R Review of Basic Algebraic Concepts

Skill Practice Graph the set on the number line, and express the set in interval notation. 11. {w 0 w ⱖ ⫺53 }

12. {y 0 ⫺7 ⱕ y ⬍ 4}

Table R-2 summarizes interval notation. Table R-2 Interval Notation

Interval Notation

Graph

(a, ⬁)

Graph

[a, ⬁)

(

a

a

(

(⫺⬁, a)

(⫺⬁, a]

a

(a, b) (a, b]

a

(

(

a

b

( a

[a, b] a

b

a

b

(

[a, b) b

4. Translations Involving Inequalities In Table R-1, we learned that phrases such as at least, at most, no more than, no less than, and between can be translated into mathematical terms by using inequality signs. Example 6

Translating Inequalities

The intensity of a hurricane is often defined according to its maximum sustained winds, for which wind speed is measured to the nearest mile per hour. Translate the italicized phrases into mathematical inequalities. a. A tropical storm is updated to hurricane status if the sustained wind speed, w, is at least 74 mph. b. Hurricanes are categorized according to intensity by the Saffir-Simpson scale. On a scale of 1 to 5, a category 5 hurricane is the most destructive. A category 5 hurricane has sustained winds, w, exceeding 155 mph. c. A category 4 hurricane has sustained winds, w, of at least 131 mph but no more than 155 mph.

Solution: a. w ⱖ 74 mph Answers

c. 131 mph ⱕ w ⱕ 155 mph

Skill Practice Translate the italicized phrase to a mathematical inequality.

11. ⫺53 3 ⫺53 , ⬁2

(

12. ⫺7

b. w ⬎ 155 mph

3 ⫺7, 42

13. m ⱖ 30 14. m 7 45 15. 10 ⱕ m ⱕ 20

4

13. The gas mileage, m, for a Honda Civic is at least 30 mpg. 14. The gas mileage, m, for a Harley Davidson motorcycle is more than 45 mpg. 15. The gas mileage, m, for a Ford Explorer is at least 10 mpg, but no more than 20 mpg.

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Section R.2

13

Sets of Numbers and Interval Notation

Section R.2 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercises 1. In this text, we will provide skills for you to enhance your learning experience. Many of the Practice Exercises will begin with an activity that focuses on one of the following areas: learning about your course, using your text, taking notes, doing homework, and taking an exam. In subsequent chapters we will insert skills pertaining to the specific material in the chapter. Each activity requires only a few minutes and will help you pass this class and become a better math student. To begin, write down the following information. a. Instructor’s name

b. Days of the week that the class meets

c. The room number in which the class meets

d. Is there a lab requirement for this course? If so, what is the requirement and what is the location of the lab?

e. Instructor’s office number

f. Instructor’s telephone number

g. Instructor’s email address

h. Instructor’s office hours

2. Define the key terms. a. Set

b. Set-builder notation

c. Real numbers

d. Real number line

e. Subset

f. Natural numbers

g. Whole numbers

h. Integers

i. Rational numbers

j. Irrational numbers

k. Inequality

l. Interval notation

Concept 1: The Set of Real Numbers 3. Plot the numbers on the number line.

4. Plot the numbers on the number line.

51.7, p, 5, 4.26

6 5 4 3 2 1

0

1

2

1 1 3 e 1 , 0, 3,  , f 2 2 4 3

4

5

6 5 4 3 2 1

0

1

2

3

4

5

6

7

For Exercises 5–10, show that each number is a rational number by finding a ratio of two integers equal to the given number. (See Example 1.) 5. 10

6. 7

8. 0.1

9. 0

3 4

7. 

3 5

10. 0.35

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Chapter R Review of Basic Algebraic Concepts

11. Check the sets to which each number belongs. (See Example 2.) Real Numbers

Irrational Numbers

Rational Numbers

Integers

Whole Numbers

Natural Numbers

Integers

Whole Numbers

Natural Numbers

5 19 1.7 1 2 17 0 4 0.2

12. Check the sets to which each number belongs. Real Numbers

Irrational Numbers

Rational Numbers

6 8 112 p 0 0.8 8 2 4.2

Concept 2: Inequalities For Exercises 13–20, fill in the blanks with the appropriate symbol: 6 or 7. 13. 9 ___ 1 17.

14. 0 ___ 6

5 10 ___ 3 7

18. 

21 17 ___  5 4

(See Example 3.)

15. 0.15 ___ 0.15

16. 2.5 ___ 2.5

5 1 19.  ___  8 8

20. 

13 17 ___  15 12

Concept 3: Interval Notation For Exercises 21–28, express the set in interval notation. 21.

22.

(

25. 9

26.

)

5

0

28. 4.7

0

5 6

24.

27.

23.

(

2

12.8

1

(

15

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Section R.2

Sets of Numbers and Interval Notation

15

For Exercises 29–46, graph the sets and express each set in interval notation. (See Examples 4–5.) 29. 5x 0 x 7 16

30. 5x 0 x 6 36

31. 5y 0 y 26

32. 5z 0 z  46

33. 5w 0 w 6 92 6

34. 5p 0 p  73 6

35. 5x 0 2.5 6 x 4.56

36. 5x 0 6 x 6 06

37. All real numbers less than 3.

38. All real numbers greater than 2.34.

39. All real numbers greater than 52. 40. All real numbers less than 47.

41. All real numbers not less than 2.

42. All real numbers no more than 5.

43. All real numbers between 4 and 4.

44. All real numbers between 7 and 1.

45. All real numbers between 3 and 0, inclusive.

46. All real numbers between 1 and 6, inclusive.

For Exercises 47–54, write an expression in words that describes the set of numbers given by each interval. (Answers may vary.) 47. ( , 4)

48. [2, )

49. (2, 7]

50. (3.9, 0)

51. [180, 90]

52. (3.2, )

53. ( , )

54. ( , 1]

Concept 4: Translations Involving Inequalities For Exercises 55–64, write the expressions as an inequality. (See Example 6.) 55. The age, a, to get in to see a certain movie is at least 18 years old.

56. Winston is a cat that was picked up at the Humane Society. His age, a, at the time was no more than 2 years.

57. The cost, c, to have dinner at Jack’s Café is at most $25.

58. The number of hours, h, that Katlyn spent studying was no less than 40.

59. The wind speed, s, for an F-5 tornado is no less than 261 mph.

60. The high temperature, t, for a certain December day in Albany is at most 26F.

61. After a summer drought, the total rainfall, r, for June, July, and August was no more than 4.5 in.

62. Jessica works for a networking firm. Her salary, s, is at least $85,000 per year.

63. To play in a certain division of a tennis tournament, a player’s age, a, must be at least 18 years but not more than 25 years.

64. The average age, a, of students at Central Community College is estimated to be between 25 years and 29 years.

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Chapter R Review of Basic Algebraic Concepts

The following chart defines the ranges for normal blood pressure, high normal blood pressure, and high blood pressure (hypertension). All values are measured in millimeters of mercury (mm Hg). (Source: American Heart Association.) Normal

Systolic less than 130

High normal

Systolic 130–139, inclusive

Diastolic 85–89, inclusive

Hypertension

Systolic 140 or greater

Diastolic 90 or greater

Diastolic less than 85

For Exercises 65–68, write an inequality using the variable p that represents each condition. 65. Normal systolic blood pressure

66. Diastolic pressure in hypertension

67. High normal range for systolic pressure

68. Systolic pressure in hypertension

A pH scale determines whether a solution is acidic or alkaline. The pH scale runs from 0 to 14, with 0 being the most acidic and 14 being the most alkaline. A pH of 7 is neutral (distilled water has a pH of 7). For Exercises 69–72, write the pH ranges as inequalities and label the substances as acidic or alkaline. 69. Lemon juice: 2.2 through 2.4, inclusive

70. Eggs: 7.6 through 8.0, inclusive

71. Carbonated soft drinks: 3.0 through 3.5, inclusive

72. Milk: 6.6 through 6.9, inclusive

Section R.3

Operations on Real Numbers

Concepts

1. Opposite and Absolute Value

1. Opposite and Absolute Value 2. Addition and Subtraction of Real Numbers 3. Multiplication and Division of Real Numbers 4. Exponential Expressions 5. Square Roots 6. Order of Operations 7. Evaluating Expressions

Several key definitions are associated with the set of real numbers and constitute the foundation of algebra. Two important definitions are the opposite of a real number and the absolute value of a real number.

DEFINITION Opposite of a Real Number Two numbers that are the same distance from 0 but on opposite sides of 0 on the number line are called opposites of each other. Symbolically, we denote the opposite of a real number a as ⫺a.

3 2

The numbers ⫺4 and 4 are opposites of each other. Similarly, the numbers and ⫺32 are opposites. ⫺4

Opposites

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺ 32

0

4

1

Opposites

2

3

4

5

6

3 2

The absolute value of a real number a, denoted 0a 0 , is the distance between a and 0 on the number line. Note: The absolute value of any real number is nonnegative. For example:

05 0 ⫽ 5 and 0⫺5 0 ⫽ 5

|⫺5| ⫽ 5 5 units ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

|5| ⫽ 5 5 units 0

1

2

3

4

5

6

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Section R.3

Example 1

17

Operations on Real Numbers

Evaluating Absolute Value Expressions

Simplify the expressions. a. 02.5 0

5 b. ` ` 4

c.  0 4 0

Calculator Connections Topic: Using the Absolute Value Function

Solution:

a. 02.5 0  2.5

5 4

4 units

5 5 b. ` `  4 4

6 5 4 3 2 1

c.  04 0  142  4

0

units 1

2

3

4

5

6

Some calculators have an absolute value function. For example,

2.5 units

Skill Practice Simplify. 1. 0 9.2 0

7 2. ` ` 6

3.  0 2 0

The absolute value of a number a is its distance from zero on the number line. The definition of 0a 0 may also be given algebraically depending on whether a is negative or nonnegative.

DEFINITION Absolute Value of a Real Number Let a be a real number. Then

1. If a is nonnegative (that is, if a  02, then 0a 0  a. 2. If a is negative (that is, if a 6 02, then 0 a 0  a.

This definition states that if a is positive or zero, then 0a 0 equals a itself. If a is a negative number, then 0a 0 equals the opposite of a. For example, 09 0  9

07 0  7

Because 9 is positive, 09 0 equals the number 9 itself.

Because 7 is negative, 07 0 equals the opposite of 7, which is 7.

2. Addition and Subtraction of Real Numbers PROCEDURE Adding Real Numbers • To add two numbers with the same sign, add their absolute values and apply the common sign to the sum. • To add two numbers with different signs, subtract the smaller absolute value from the larger absolute value. Then apply the sign of the number having the larger absolute value.

Answers 1. 9.2

2.

7 6

3. 2

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Chapter R Review of Basic Algebraic Concepts

Example 2

Adding Real Numbers

Perform the indicated operations. a. ⫺2 ⫹ (⫺6)

b. ⫺10.3 ⫹ 13.8

c.

1 5 ⫹ a⫺1 b 6 4

Solution:

a. ⫺2 ⫹ 1⫺62

First find the absolute value of the addends. 0 ⫺2 0 ⫽ 2 and 0 ⫺6 0 ⫽ 6

⫽ ⫺12 ⫹ 62

Add their absolute values and apply the common sign. In this case, the common sign is negative.

Common sign is negative.

⫽ ⫺8

The sum is ⫺8.

b. ⫺10.3 ⫹ 13.8

First find the absolute value of the addends. 0 ⫺10.3 0 ⫽ 10.3 and 0 13.8 0 ⫽ 13.8 The absolute value of 13.8 is greater than the absolute value of ⫺10.3. Therefore, the sum is positive.

⫽ ⫹113.8 ⫺ 10.32

Subtract the smaller absolute value from the larger absolute value.

Apply the sign of the number with the larger absolute value.

⫽ 3.5

c.

5 1 ⫹ a⫺1 b 6 4 ⫽

5 5 ⫹ a⫺ b 6 4

1 Write ⫺1 as a fraction. 4

⫽

5ⴢ2 5ⴢ3 ⫹ a⫺ b 6ⴢ2 4ⴢ3

The LCD is 12. Write each fraction with the LCD.

⫽

10 15 ⫹ a⫺ b 12 12

Find the absolute value of the addends. 10 10 15 15 ` ` ⫽ and `⫺ ` ⫽ 12 12 12 12 The absolute value of ⫺15 12 is greater than the absolute value of 10 . Therefore, the sum is negative. 12

⫽⫺a

15 10 ⫺ b 12 12

Subtract the smaller absolute value from the larger absolute value.

Apply the sign of the number with the larger absolute value.

⫽⫺

5 12

Skill Practice Perform the indicated operations. 4. ⫺4 ⫹ 1⫺12

Answers 4. ⫺5

5. ⫺0.8

6. ⫺

10 7

5. ⫺2.6 ⫹ 1.8

3 6. ⫺1 ⫹ a⫺ b 7

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Section R.3

19

Operations on Real Numbers

Subtraction of real numbers is defined in terms of the addition process. To subtract two real numbers, add the opposite of the second number to the first number.

DEFINITION Subtraction of Real Numbers If a and b are real numbers, then

a ⫺ b ⫽ a ⫹ (⫺b)

Subtracting Real Numbers

Example 3

Perform the indicated operations. a. ⫺13 ⫺ 5

b. 2.7 ⫺ (⫺3.8)

c.

5 2 ⫺4 2 3

Solution: a. ⫺13 ⫺ 5

⫽ ⫺13 ⫹ 1⫺52

⫽ ⫺18

Add the opposite of the second number to the first number. Add.

b. 2.7 ⫺ 1⫺3.82

⫽ 2.7 ⫹ 13.82

Add the opposite of the second number to the first number.

⫽ 6.5

c.

Add.

5 2 ⫺4 2 3 ⫽

5 2 ⫹ a⫺4 b 2 3

Add the opposite of the second number to the first number.

⫽

5 14 ⫹ a⫺ b 2 3

Write the mixed number as a fraction.

⫽

5ⴢ3 14 ⴢ 2 ⫹ a⫺ b 2ⴢ3 3ⴢ2

The common denominator is 6.

⫽

15 28 ⫹ a⫺ b 6 6

Get a common denominator and add.

⫽⫺

13 1 or ⫺2 6 6

Skill Practice Subtract. 7. ⫺9 ⫺ 8

8. 1.1 ⫺ 1⫺4.22

9.

1 1 ⫺2 6 4

Answers 7. ⫺17

8. 5.3

9. ⫺2

25 1 or ⫺ 12 12

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Chapter R Review of Basic Algebraic Concepts

3. Multiplication and Division of Real Numbers The sign of the product of two real numbers is determined by the signs of the factors.

PROPERTY Multiplication of Real Numbers 1. The product of two real numbers with the same sign is positive. 2. The product of two real numbers with different signs is negative. 3. The product of any real number and zero is zero.

Example 4

Multiplying Real Numbers

Multiply the real numbers. a. 122 15.12

2 9 b.  ⴢ 3 8

1 3 c. a3 b a b 3 10

Solution: a. 122 15.12

 10.2

Different signs. The product is negative.

2 9 b.  ⴢ 3 8 

18 24

Different signs. The product is negative.



3 4

Simplify to lowest terms.

1 3 c. a3 b a b 3 10  a 

10 3 b a b 3 10

30 30

Write the mixed number as a fraction. Same signs. The product is positive.

1

Simplify to lowest terms.

Skill Practice Multiply.

TIP: A number and its reciprocal have the same sign. For example: a

10 3 b a b  1 3 10 1 and 3ⴢ 1 3

Answers 10. 11

11. 

2 3

12. 14

10. 15212.22

11.

5 14 ⴢ a b 7 15

1 8 12. a5 ba b 4 3

Notice from Example 4(c) that 1103 21103 2  1. If the product of two numbers is 1, then the numbers are reciprocals. That is, the reciprocal of a real number a is 1a. Furthermore, a ⴢ 1a  1.

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Section R.3

Operations on Real Numbers

Recall that subtraction of real numbers was defined in terms of addition. In a similar way, division of real numbers can be defined in terms of multiplication.

PROCEDURE Dividing Real Numbers To divide two real numbers, multiply the first number by the reciprocal of the second number. For example: Multiply

10 ⫼ 5 ⫽ 2

or equivalently

10 ⴢ

1 ⫽2 5

Reciprocal

Because division of real numbers can be expressed in terms of multiplication, the sign rules that apply to multiplication also apply to division.

⎫⎪ ⎪ ⎬ 1 ⎪ ⫺10 ⫼ 1⫺22 ⫽ ⫺10 ⴢ a⫺ b ⫽ 5 ⎪ 2 ⎭ 10 ⫼ 2 ⫽ 10 ⴢ

1 ⫽5 2

1 10 ⫼ 1⫺22 ⫽ 10 ⴢ a⫺ b ⫽ ⫺5 2 ⫺10 ⫼ 2 ⫽ ⫺10 ⴢ

1 ⫽ ⫺5 2

⎫⎪ ⎪ ⎬ ⎪ ⎪⎭

Dividing two numbers of the same sign produces a positive quotient.

Dividing two numbers of opposite signs produces a negative quotient.

PROPERTY Division of Real Numbers Assume that a and b are real numbers such that b ⫽ 0. 1. If a and b have the same sign, then the quotient ba is positive. 2. If a and b have different signs, then the quotient ba is negative. 3.

0 ⫽ 0. b

4.

b is undefined. 0

The relationship between multiplication and division can be used to investigate properties 3 and 4 from the preceding box. For example, 0 ⫽0 6

Because 6 ⴢ 0 ⫽ 0 ✓

6 is undefined 0

Because there is no number that when multiplied by 0 will equal 6

Note: The quotient of 0 and 0 cannot be determined. Evaluating an expression of 0 the form 0 ⫽ ? is equivalent to asking, “What number times zero will equal 0?” That is, (0)(?) ⫽ 0. Any real number will satisfy this requirement; however, expressions involving 00 are usually discussed in advanced mathematics courses.

21

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Chapter R Review of Basic Algebraic Concepts

Dividing Real Numbers

Example 5

Divide the real numbers. Write the answer as a fraction or whole number.

TIP: Multiplication may

⫺42 7

be used to check a division problem.

Solution:

a.

⫺42 ⫽ ⫺6 7

b.

1 2 ⫼ a⫺ b 10 5

d.

⫺8 ⫺7

⫺42 ⫽ ⫺6 7

Different signs. The quotient is negative.

b.

⫺96 2 ⫽ ⫺144 3

Same signs. The quotient is positive. Simplify.

c. 3

1 2 ⫼ a⫺ b 10 5 ⫽

31 5 a⫺ b 10 2

TIP: If the numerator and

⫺

c. 3

a.

Check: 1721⫺62 ⫽ ⫺42 ✓

denominator of a fraction have opposite signs, then the quotient will be negative. Therefore, a fraction has the same value whether the negative sign is written in the numerator, in the denominator, or in front of the fraction.

⫺96 ⫺144

Write the mixed number as an improper fraction, and multiply by the reciprocal of the second number.

1

31 5 ⫽ a⫺ b 10 2 2

⫽⫺

d.

31 ⫺31 31 ⫽ ⫽ 4 4 ⫺4

31 4

Different signs. The quotient is negative.

⫺8 8 ⫽ ⫺7 7

Same signs. The quotient is positive. Because 7 does not divide into 8 evenly, the answer can be left as a fraction.

Skill Practice Divide. 13.

42 ⫺2

14.

⫺28 ⫺4

2 15. ⫺ ⫼ 4 3

16.

⫺1 ⫺2

4. Exponential Expressions To simplify the process of repeated multiplication, exponential notation is often used. For example, the quantity 3 ⴢ 3 ⴢ 3 ⴢ 3 ⴢ 3 can be written as 35 (3 to the fifth power).

DEFINITION b n Let b represent any real number and n represent a positive integer. Then

⎫ ⎪⎪ ⎪⎪ ⎬ ⎪⎪ ⎪ ⎭⎪

bn ⫽ b ⴢ b ⴢ b ⴢ b ⴢ . . . b n factors of b

bn is read as “b to the nth power.” b is called the base and n is called the exponent, or power. b2 is read as “b squared,” and b3 is read as “b cubed.” Answers 13. ⫺21 1 15. ⫺ 6

14. 7 1 16. 2

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Section R.3

Example 6

Operations on Real Numbers

23

Evaluating Exponential Expressions

Simplify the expression. a. 53

b. (2)4

c. 24

1 2 d. a b 3

Solution: a. 53  5 ⴢ 5 ⴢ 5

The base is 5, and the exponent is 3.

 125 b. 122 4  122122122122  16 c. 24   32 ⴢ 2 ⴢ 2 ⴢ 24  16

The base is 2, and the exponent is 4. The exponent 4 applies to the entire contents of the parentheses. The base is 2, and the exponent is 4. Because no parentheses enclose the negative sign, the exponent applies to only 2.

TIP: The quantity 24 can also be interpreted as 1 ⴢ 24. 24  1 ⴢ 24  1 ⴢ 12 ⴢ 2 ⴢ 2 ⴢ 22  16

1 2 1 1 d. a b  a b a b 3 3 3 

The base is 13, and the exponent is 2.

1 9

Calculator Connections Topic: Using the Exponent Keys On many calculators, the key is used to square a number.The key is used to raise a base to any power.

Skill Practice Simplify. 17. 23

18. 1102 2

19. 102

3 3 20. a b 4

5. Square Roots The inverse operation to squaring a number is to find its square roots. For example, finding a square root of 9 is equivalent to asking, “What number when squared equals 9?” One obvious answer is 3, because (3)2  9. However, 3 is also a square root of 9 because 132 2  9. For now, we will focus on the principal square root, which is always taken to be nonnegative. The symbol 1 , called a radical sign, is used to denote the principal square root of a number. Therefore, the principal square root of 9 can be written as 19. The expression 164 represents the principal square root of 64.

Answers 17. 8 19. 100

18. 100 27 20.  64

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Chapter R Review of Basic Algebraic Concepts

Evaluating Square Roots

Example 7

Evaluate the expressions, if possible.

Calculator Connections Topic: Using the Square Root Key The key is used to find the square root of a nonnegative real number.

a. 181

b.

25 A 64

c. 116

Solution: a. 181  9

because

(9)2  81

25 5  A 64 8

because

5 2 25 a b  8 64

b.

c. 116

is not a real number because no real number when squared will be negative.

Skill Practice Evaluate the expressions, if possible. 21. 125

22.

49 B 100

23. 14

Example 7(c) illustrates that the square root of a negative number is not a real number because no real number when squared will be negative.

PROPERTY The Square Root of a Negative Number Let a be a negative real number. Then 1a is not a real number.

6. Order of Operations When algebraic expressions contain numerous operations, it is important to use the proper order of operations. Parentheses ( ), brackets [ ], and braces { } are used for grouping numbers and algebraic expressions. It is important to recognize that operations must be done first within parentheses and other grouping symbols.

PROCEDURE Order of Operations Step 1 First, simplify expressions within parentheses and other grouping symbols. These include absolute value bars, fraction bars, and radicals. If embedded parentheses are present, start with the innermost parentheses. Step 2 Evaluate expressions involving exponents, radicals, and absolute values. Step 3 Perform multiplication or division in the order in which they occur from left to right. Step 4 Perform addition or subtraction in the order in which they occur from left to right.

Answers 7 10 23. Not a real number 21. 5

22.

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Section R.3

Example 8

Operations on Real Numbers

25

Applying the Order of Operations

Simplify the expression.

10  32  416  82 4 2  216  7

Solution: 10  3 2  416  82 4 2  216  7  10  32  4 122 4 2  29  10  32  84 2  29

Avoiding Mistakes Simplify inside the innermost parentheses and inside the radical. Simplify within square brackets. Perform multiplication before addition or subtraction.

 10  3104 2  29  10  100  3

Simplify the exponential expression and the radical.

 90  3

Perform addition or subtraction in the order in which they appear from left to right.

Don’t try to perform too many steps at once. Taking a shortcut may result in a careless error. For each step rewrite the entire expression, changing only the operation being evaluated.

 87 Skill Practice Simplify the expression. 24. 36  22 ⴢ 3  3 118  52 ⴢ 2  64

Example 9

Applying the Order of Operations

Simplify the expression.

0 132 3  152  32 0 15  132122

Solution: 0 132 3  152  32 0 15  132122 

 

0 132 3  125  32 0 5122

0 132 3  1222 0 10

027  22 0 10

Calculator Connections Topic: Order of Operations

Simplify numerator and denominator separately. Numerator: Simplify within the inner parentheses. Denominator: Perform division and multiplication (left to right). Numerator: Simplify inner parentheses. Denominator: Multiply. Simplify exponent.



05 0 10

Add within the absolute value.



5 1 or 10 2

Evaluate the absolute value and simplify.

To evaluate the expression 0 132 3  152  32 0 15  132122

on a graphing calculator, use parentheses to enclose the absolute value expression. Likewise, it is necessary to use parentheses to enclose the entire denominator.

Skill Practice Simplify the expression. 25.

 0 5  7 0  11 11  22 2

Answers 24. 5

25. 1

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Chapter R Review of Basic Algebraic Concepts

7. Evaluating Expressions

b2 h

Subscripts

b1

Figure R-3

An algebraic expression or formula involves operations on numbers and variables. A variable is a letter that may represent any numerical value.To evaluate an expression or formula, we substitute known values of the variables into the expression. Then we follow the order of operations. A list of common geometry formulas is given in the inside back cover of the text. It is important to note that some geometric formulas use Greek letters (such as p) and some use variables with subscripts. A subscript is a number or letter written to the right of and below a variable. For example, the area of a trapezoid is given by A ⫽ 12 1b1 ⫹ b2 2h. The values b1 and b2 (read as “b sub 1” and “b sub 2”) represent two different bases of the trapezoid (Figure R-3).

Example 10

Evaluating an Algebraic Expression

A homeowner in North Carolina wants to buy protective film for a trapezoidshaped window. The film will adhere to shattered glass in the event that the glass breaks during a bad storm. Find the area of the window whose dimensions are given in Figure R-4.

b1 ⫽ 4.0 ft b2 ⫽ 2.5 ft h ⫽ 5.0 ft

Figure R-4

Solution: A⫽

TIP: Subscripts should not be confused with superscripts, which are written above a variable. Superscripts are used to denote powers. b2 ⫽ b

2

1 1b ⫹ b2 2h 2 1

⫽

1 14.0 ft ⫹ 2.5 ft215.0 ft2 2

⫽

1 16.5 ft215.0 ft2 2

Substitute b1 ⫽ 4.0 ft, b2 ⫽ 2.5 ft, and h ⫽ 5.0 ft. Simplify inside parentheses.

⫽ 16.25 ft2

Multiply from left to right.

The area of the window is 16.25 ft2. Skill Practice 26. Use the formula given in Example 10 to find the area of the trapezoid. b2 ⫽ 5 in.

h ⫽ 10 in.

b1 ⫽ 12 in.

Answer 26. The area is 85 in.2

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Section R.3

27

Operations on Real Numbers

Section R.3 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercises 1. Sometimes you may run into a problem with homework, or you may find that you are having trouble keeping up with the pace of the class. A tutor can be a good resource. Answer the following questions. a. Does your college offer tutoring?

b. Is it free?

c. Where should you go to sign up for a tutor?

d. Is there tutoring available online?

2. Define the key terms: a. Opposite

b. Absolute value

c. Reciprocal

d. Base

e. Exponent

f. Power

g. Principal square root

h. Radical sign

i. Order of operations

j. Subscript

k. Variable

Review Exercises For Exercises 3–6, express each set in interval notation. 3. 5z 0 z 6 46

4. 5w 0 w ⱖ ⫺26

5. 5x 0 ⫺3 ⱕ x 6 ⫺16

6. 5 p 0 0 6 p ⱕ 86

Concept 1: Opposite and Absolute Value 7. If the absolute value of a number can be thought of as its distance from zero, explain why an absolute value can never be negative. 8. If a number is negative, then its opposite will be a. Positive b. Negative. 9. If a number is negative, then its reciprocal will be a. Positive b. Negative. 10. If a number is negative, then its absolute value will be a. Positive b. Negative. 12. Complete the table.

11. Complete the table. (See Example 1.) Number

Opposite

Reciprocal

Absolute Value

Number

Opposite

Reciprocal

⫺9

6 ⫺111

2 3

⫺18

14 ⫺1

13 10

0

0 ⫺0.3

219

For Exercises 13–20, fill in the blank with the appropriate symbol 1 6, 7, ⫽2. (See Example 1.) 13. ⫺ 06 0 ______ 0⫺6 0 14. ⫺1⫺52 ______ ⫺ 0 ⫺5 0 15. 0 ⫺4 0 ______ 04 0 16. ⫺ 02 0 ______ 1⫺22

17. ⫺ 0⫺1 0 ______ 1

19. 02 ⫹ 1⫺52 0 ______ 02 0 ⫹ 0⫺5 0

20. 0 4 ⫹ 3 0 ______ 04 0 ⫹ 03 0

18. ⫺3 _______ ⫺ 0 ⫺7 0

Absolute Value

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Concept 2: Addition and Subtraction of Real Numbers For Exercises 21–36, add or subtract as indicated. (See Examples 2–3.) 21. 8  4

22. 3  172

23. 12  172

24. 5  1112

25. 17  1102

26. 14  122

27. 5  192

28. 8  142

29. 6  15

30. 21  4

31. 1.5  9.6

32. 4.8  10

4 4 34.   a1 b 7 7

5 14 35.   9 15

36. 6 

33.

2 1  a2 b 3 3

2 9

Concept 3: Multiplication and Division of Real Numbers For Exercises 37–52, perform the indicated operation. (See Examples 4–5.) 37. 4182

39.

2 12 ⴢ 9 7

5 7 40. a b ⴢ a1 b 9 11

42.

15 24

1 5 43. 2  4 8

5 2 44.   a1 b 3 7

45. 7  0

46.

1 0 16

47. 0  132

48. 0  11

49. 11.2213.12

50. 14.6212.252

41.

6 10

38. 21132

51.

5 11

3 13

52.

Concept 4: Exponential Expressions For Exercises 53–60, evaluate the expression. (See Example 6.) 53. 43

54. 23

55. 72

56. 34

57. 172 2

58. 152 2

5 3 59. a b 3

60. a

10 2 b 9

Concept 5: Square Roots For Exercises 61–68, evaluate the expression, if possible. (See Example 7.) 61. 19 65.

1 B4

62. 11 66.

9 B4

63. 14

64. 136

67. 149

68. 1100

Concept 6: Order of Operations For Exercises 69–96, simplify by using the order of operations. (See Examples 8–9.) 69. 5  33

70. 10  24

71. 5 ⴢ 23

72. 12  22

73. 12  32 2

74. 14  12 3

75. 22  32

76. 43  13

77. 6  10  2 ⴢ 3  4

78. 12  3 ⴢ 4  18

79. 42  15  22 2 ⴢ 3

80. 5  318  42 2

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Section R.3

81. 2  519  42252 2

82. 52  129  4  22

3 2 3 5 7 83. a b  ⴢ  5 5 9 10

85. 1.75  0.25  11.252 2

86. 5.4  10.32 2  0.09

87.

2102  82 32

84.

1 2 5 5 a  b 2 3 9 6

88.

116  7  32 116  14

89.  011  5 0  0 7  2 0

90.  0 8  3 0  18  32

91. 25  2 3 17  32 2  44  218  2

92. 229  22  38  316  22 4  4 ⴢ 5

93.

0 110  72  23 0

94.

16  8 ⴢ 3

1 2 64 2 52 2 b a b 95. a b  a 2 5 10

29

Operations on Real Numbers

0 12  17  32 2 2 0

96. a

62  8  2

2 8  122 2 23 b  a b 23  1 32

For Exercises 97–98, find the average of the set of data values by adding the values and dividing by the number of values. 97. Find the average low temperature for a week in January in St. John’s, Newfoundland. Round to the nearest tenth of a degree.

98. Find the average high temperature for a week in January in St. John’s, Newfoundland. Round to the nearest tenth of a degree.

Day

Mon. Tues. Wed. Thur. Fri.

Sat. Sun.

Low temperature 18C 16C 20C 11C 4C 3C 1C

Day

Mon. Tues. Wed. Thur. Fri. Sat. Sun.

High temperature 2C 6C 7C

0C

1C 8C 10C

Concept 7: Evaluating Expressions 99. The formula C  59 1F  322 converts temperatures in the Fahrenheit scale to the Celsius scale. Find the equivalent Celsius temperature for each Fahrenheit temperature. a. 77°F

b. 212°F

c. 32°F

d. 40°F

100. The formula F  95 C  32 converts Celsius temperatures to Fahrenheit temperatures. Find the equivalent Fahrenheit temperature for each Celsius temperature. a. 5°C

b. 0°C

c. 37°C

d. 40°C

Use the geometry formulas found in the inside back cover of the book to answer Exercises 101–110. For Exercises 101–104, find the area. (See Example 10.) 101. Trapezoid 5 in. 2 in.

102. Parallelogram

103. Triangle

104. Rectangle

8.5 m 6m

3 4

3.1 cm 1

4 in.

7 6 yd 5.2 cm

yd

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For Exercises 105–110, find the volume. (Use the p key on your calculator, and round the final answer to one decimal place.) 105. Sphere

106. Sphere

r  1.5 ft

107. Right circular cone

r  12 yd h  4.1 ft r  2.5 ft

108. Right circular cone

109. Right circular cylinder

h  12 cm

110. Right circular cylinder

h  9.5 m

h  5 in.

r4m

r  5 cm r  3 in.

Graphing Calculator Exercises 111. Which expression when entered into a graphing calculator will yield the correct value of 12 6  2

or

12 16  22

112. Which expression when entered into a graphing calculator will yield the correct value of 124  62 3

or

12 ? 62 24  6 ? 3

24  6 3

113. Verify your solution to Exercise 87 by entering the expression into a graphing calculator: 1 1 1102  82 22 32 114. Verify your solution to Exercise 88 by entering the expression into a graphing calculator: 1 1 116  72  32 2 1 1 1162  1 1422

Section R.4

Simplifying Algebraic Expressions

Concepts

1. Recognizing Terms, Factors, and Coefficients

1. Recognizing Terms, Factors, and Coefficients 2. Properties of Real Numbers 3. Simplifying Expressions

A term is a constant or the product of a constant and one or more variables. An algebraic expression is a single term or a sum of two or more terms. For example, the expression 6x2  5xyz  11

or

consists of the terms 6x2, 5xyz, and 11.

6x2  5xyz  1112

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Simplifying Algebraic Expressions

The terms 6x2 and 5xyz are variable terms, and the term 11 is called a constant term. It is important to distinguish between a term and the factors within a term. For example, the quantity 5xyz is one term, but the values 5, x, y, and z are factors within the term. The constant factor in a term is called the numerical coefficient or simply coefficient of the term. In the terms 6x2, 5xyz, and 11, the coefficients are 6, 5, and 11, respectively. A term containing only variables such as xy has a coefficient of 1. Terms are called like terms if they each have the same variables and the corresponding variables are raised to the same powers. For example: Like Terms 6t 1.8ab 1 2 3 2c d 4

and and and and

Example 1

Unlike Terms 4t 3ab c2d3 6

6t 1.8xy 1 2 3 2c d 4p

and and and and

4s 3x c2d 4

(different (different (different (different

variables) variables) powers) variables)

Identifying Terms, Factors, Coefficients, and Like Terms

a. List the terms of the expression.

4x2  7x  23

b. Identify the coefficient of the term.

yz3

c. Identify the pair of like terms.

16b, 4b2

or

1 2 c,

16 c

Solution: a. The terms of the expression 4x2  7x  23 are 4x2, 7x, and 23. b. The term yz3 can be written as 1yz3; therefore, the coefficient is 1. c.

16 c are like terms because they have the same variable raised to the same power. 1 2 c,

Skill Practice Given: 2x 2  5x 

1 y2 2

1. List the terms of the expression. 2. Which term is the constant term? 3. Identify the coefficient of the term y 2.

2. Properties of Real Numbers Simplifying algebraic expressions requires several important properties of real numbers that are stated in Table R-3. Assume that a, b, and c represent real numbers or real-valued algebraic expressions.

Answers 1. 2x 2, 5x, 2.

1 2

1 , y 2 2

3. 1

31

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Chapter R Review of Basic Algebraic Concepts

Table R-3 Property Name

Algebraic Representation

Commutative property a  b  b  a of addition Commutative property a ⴢ b  b ⴢ a of multiplication

Example

Description/Notes

5335

The order in which two real numbers are added or multiplied does not affect the result.

152 132  132 152

1a  b2  c  a  1b  c2

12  32  7  2  13  72

1a ⴢ b2 c  a 1b ⴢ c2

12 ⴢ 327  213 ⴢ 72

Distributive property of multiplication over addition

a1b  c2  ab  ac

315  22 3ⴢ53ⴢ2

Identity property of addition

5  0  0  5  5 Any number added 0 is the identity element for to the identity addition because element 0 will a00aa remain unchanged.

Identity property of multiplication

5ⴢ11ⴢ55 1 is the identity element for multiplication because aⴢ11ⴢaa

Associative property of addition Associative property of multiplication

Inverse property of addition

Inverse property of multiplication

3  132  0 a and 1a2 are additive inverses because a  1a2  0 and 1a2  a  0 5 ⴢ 15  1 a and a1 are multiplicative inverses because 1 a ⴢ  1 and a 1 ⴢa1 a

The manner in which real numbers are grouped under addition or multiplication does not affect the result. A factor outside the parentheses is multiplied by each term inside the parentheses.

Any number multiplied by the identity element 1 will remain unchanged. The sum of a number and its additive inverse (opposite) is the identity element 0. The product of a number and its multiplicative inverse (reciprocal) is the identity element 1.

(provided a  0)

The properties of real numbers are used to multiply algebraic expressions. To multiply a term by an algebraic expression containing more than one term, we apply the distributive property of multiplication over addition.

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Section R.4

Example 2

Simplifying Algebraic Expressions

33

Applying the Distributive Property

Apply the distributive property. a. 412x  52

b. 13.4q  5.7r2

c. 31a  2b  5c2

2 3 d.  a9x  y  5b 3 8

Solution: a. 412x  52  412x2  4152

Apply the distributive property.

 8x  20

Simplify, using the associative property of multiplication.

b. 13.4q  5.7r2

The negative sign preceding the parentheses can be interpreted as a factor of 1.

 113.4q  5.7r2

 113.4q2  11215.7r2

Apply the distributive property.

 3.4q  5.7r

TIP: When applying the

c. 31a  2b  5c2

 31a2  13212b2  13215c2

Apply the distributive property.

 3a  6b  15c

Simplify.

31a  2b  5c2

2 3 d.  a9x  y  5b 3 8 2 2 3 2   19x2  a b a yb  a b 152 3 3 8 3 

18 6 10 x y 3 24 3

1 10  6x  y  4 3

distributive property, a negative factor preceding the parentheses will change the signs of the terms within the parentheses.

Apply the distributive property.

3a  6b  15c

Simplify. Simplify to lowest terms.

Skill Practice Apply the distributive property. 4. 10130y  402 6. 214x  3y  62

5. 17t  1.6s  9.22 1 7.  14a  72 2 Answers

Notice that the parentheses are removed after the distributive property is applied. Sometimes this is referred to as clearing parentheses. Two terms can be added or subtracted only if they are like terms. To add or subtract like terms, we use the distributive property, as shown in Example 3.

4. 300y  400 5. 7t  1.6s  9.2 6. 8x  6y  12 7 7. 2a  2

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Chapter R Review of Basic Algebraic Concepts

Using the Distributive Property to Add and Subtract Like Terms

Example 3

Add and subtract as indicated. a. 8x  3x

b. 4.75y2  9.25y2  y2

Solution: a. 8x  3x

 18  32x

Apply the distributive property.

 152x

Simplify.

 5x b. 4.75y2  9.25y2  y2  4.75y2  9.25y2  1y2

Notice that y2 is interpreted as 1y2.

 14.75  9.25  12y2

Apply the distributive property.

 13.52y

2

Simplify.

 3.5y2 Skill Practice Combine like terms. 8. 4y  7y

9. a 2  6.2a 2  2.8a 2

Although the distributive property is used to add and subtract like terms, it is tedious to write each step. Observe that adding or subtracting like terms is a matter of combining the coefficients and leaving the variable factors unchanged. This can be shown in one step. This shortcut will be used throughout the text. For example: 4w  7w  11w

8ab2  10ab2  5ab2  13ab2

3. Simplifying Expressions Clearing parentheses and combining like terms are important tools to simplifying algebraic expressions. This is demonstrated in Example 4.

Example 4

Clearing Parentheses and Combining Like Terms

Simplify by clearing parentheses and combining like terms. a. 4  312x  82  1

b. 13s  11t2  512t  8s2  10s

Solution: a. 4  312x  82  1

Answers 8. 3y

9. 2.4a 2

 4  6x  24  1

Apply the distributive property.

 6x  4  24  1

Group like terms.

 6x  27

Combine like terms.

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35

Simplifying Algebraic Expressions

b. 13s  11t2  512t  8s2  10s  3s  11t  10t  40s  10s

Apply the distributive property.

 3s  40s  10s  11t  10t

Group like terms.

 53s  t

Combine like terms.

Skill Practice Simplify by clearing parentheses and combining like terms. 10. 7  213x  42  5

Example 5

11. 16z  10y2  413z  y2  8y

Clearing Parentheses and Combining Like Terms

Simplify by clearing parentheses and combining like terms. a. 2 31.5x  4.71x2  5.2x2  3x4

1 1 b.  13w  62  a w  4b 3 4

TIP: By using the

Solution: a. 231.5x  4.71x2  5.2x2  3x4  23 1.5x  4.7x2  24.44x  3x4  231.5x  24.44x  3x  4.7x2 4

Apply the distributive property to the inner parentheses. Group like terms.

 2325.94x  4.7x 4

Combine like terms.

 51.88x  9.4x2

Apply the distributive property.

2

 9.4x  51.88x 2

commutative property of addition, the expression 51.88x  9.4x 2 can also be written as 9.4x 2  151.88x2 or simply 9.4x 2  51.88x. Although the expressions are all equal, it is customary to write the terms in descending order of the powers of the variable.

1 1 b.  13w  62  a w  4b 3 4 3 6 1  w  w4 3 3 4

Apply the distributive property.

1  w  2  w  4 4

Simplify fractions.

4 1  w w24 4 4

Group like terms and find a common denominator.

5  w2 4

Combine like terms.

Skill Practice Simplify by clearing parentheses and combining like terms. 12. 431.4a  2.2 1a 2  6a2 4  5.1a 2

5 1 13.  14p  12  1p  22 2 2

Answers 10. 6x  10 12. 3.7a  47.2a 2

11. 2y  18z 11 9 13.  p  2 2

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Chapter R Review of Basic Algebraic Concepts

Section R.4 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. It is very important to attend class every day. Math is cumulative in nature and you must master the material learned in the previous class to understand a new day’s lesson. Because this is so important, many instructors tie attendance to the final grade. Write down the attendance policy for your class. 2. Define the key terms. a. Term

b. Variable term

c. Constant term

d. Factor

e. Coefficient

f. Like terms

Review Exercises For Exercises 3–4, simplify. 3.

32  05  172 0

4. 36  122 ⴢ 3  416  82

26  22

For Exercises 5–8, write the set in interval notation. 5. 5x 0 x 7 03 0 6

4 6. e x 0 x  `  ` f 3

7. e w 0 

8. e z 0 2  z 6

5 6 w  29 f 2

11 f 3

Concept 1: Recognizing Terms, Factors, and Coefficients For Exercises 9–12: a. Determine the number of terms in the expression. b. Identify the constant term. c. List the coefficients of each term, separated by commas. (See Example 1.) 9. 2x3  5xy  6

10. a2  4ab  b2  8

11. pq  7  q2  4q  p

12. 7x  1  3xy

Concept 2: Properties of Real Numbers For Exercises 13–30, match each expression with the appropriate property. 13. 3 

1 1  3 2 2

14. 7.214  12  7.2142  7.2112

a. Commutative property of addition

15. 10  0  10

16. 7 ⴢ 1  7

17. 16  82  2  6  18  22

18. 14  192  7  119  42  7

b. Associative property of multiplication c. Distributive property of multiplication over addition

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Section R.4

19. 6 ⴢ

1 1 6

Simplifying Algebraic Expressions

20. 2  122  0

21. 914 ⴢ 122  19 ⴢ 4212

1 22. a  2b 20  5  40 4

23. 42 ⴢ 1  42

1 24. 4 ⴢ  1 4

d. Commutative property of multiplication e. Associative property of addition f. Identity property of addition g. Identity property of multiplication

25. 113 ⴢ 4126  141 ⴢ 1326

26. 61x  32  6x  18

27. 8  182  0

28. 21  0  21

h. Inverse property of addition

29. 31y  102  3110  y2

30. 513 ⴢ 72  15 ⴢ 327

i. Inverse property of multiplication

For Exercises 31–42, clear parentheses by applying the distributive property. (See Example 2.) 31. 21x  3y  82

32. 512a  4b  9c2

33. 1014s  9t  32

34. 418x  6y  3z2

35. 17w  5z2

36. 122a  17b2

1 5 37.  a a  10b  8b 5 2

3 4 38.  a6x  4y  b 4 9

39. 312.6x  4.12

40. 517.2y  2.32

41. 217c  82  516d  f 2

42. 213q  r2  715s  2t2

Concept 3: Simplifying Expressions For Exercises 43–80, clear parentheses and combine like terms. (See Examples 3–5.) 43. 8y  2x  y  5y

44. 9a  a  b  5a

45. 4p2  2p  3p  6  2p2

46. 6q  9  3q2  q2  10

47. 2p  7p2  5p  6p2

48. 5a2  2a  7a2  6a  4

49. m  4n3  3  5n3  9

50. x  2y3  2x  8y3

51. 5ab  2ab  8a

52. 6m2n  3mn2  2m2n

53. 14xy2  5y2  2xy2

54. 9uv  3u2  5uv  4u2

55. 81x  32  1

56. 41b  22  3

57. 21c  32  2c

58. 41z  42  3z

59. 110w  12  9  w

60. 12y  72  4  3y

61. 9  412  z2  1

62. 3  314  w2  11

63. 412s  72  1s  22

64. 21t  32  1t  72

65. 315  2w2  8w  21w  12

66. 5  14t  72  t  9

67. 8x  41x  22  212x  12  6

68. 61y  22  312y  52  3

69.

71. 3.112x  22  411.2x  12

72. 4.515  y2  311.9y  12

70.

2 13d  62  4d 3

1 14  2c2  5c 2

37

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Chapter R Review of Basic Algebraic Concepts

1 73. 2 c 5a a  3b  1a2  a2  4 d 2

2 74. 3 c 3ab  b  21b  42  6b2 d 3

75. 12y  52  21y  y2 2  3y

76. 1x  62  31x2  12  2x 78. 3.2  56.1y  43 9  12y  2.52 4  7y6

77. 2.254  836x  1.51x  42  64  7.5x6 79.

1 2 2 124n  16m2  13m  18n  22  8 3 3

80.

1 4 1 125a  20b2  121a  14b  22  5 7 7

Expanding Your Skills 81. What is the identity element for addition? Use it in an example. 82. What is the identity element for multiplication? Use it in an example. 83. What is another name for a multiplicative inverse? 84. What is another name for an additive inverse? 85. Is the operation of subtraction commutative? If not, give an example. 86. Is the operation of division commutative? If not, give an example. 87. Given the rectangular regions:

x

A

yz

x

B

C

y

z

a. Write an expression for the area of region A. (Do not simplify.) b. Write an expression for the area of region B. c. Write an expression for the area of region C. d. Add the expressions for the area of regions B and C. e. Show that the area of region A is equal to the sum of the areas of regions B and C. What property of real numbers does this illustrate?

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Summary

Summary

Chapter R

Section R.2

Sets of Numbers and Interval Notation

Key Concepts

Natural numbers: 51, 2, 3, . . .6 Whole numbers: 50, 1, 2, 3, . . .6 Integers: 5. . . , ⫺3, ⫺2, ⫺1, 0, 1, 2, 3, . . .6 Rational numbers: e

p q

p and q are integers and q ⫽ 0 f

Examples

Example 1 Some rational numbers are: 1 7,

0.5, 0.3

Rational numbers include all terminating and repeating decimals. Irrational numbers: A subset of the real numbers whose elements cannot be written as a ratio of two integers. Irrational numbers cannot be written as repeating or terminating decimals. Real numbers: 5x 0 x is rational or x is irrational6 a a a a a

⬍ ⬎ ⱕ ⱖ ⬍

b b b b x⬍b

“a “a “a “a “x

is is is is is

less than b” greater than b” less than or equal to b” greater than or equal to b” between a and b”

Section R.3

Example 2 Set-Builder Notation 5x 0 x 7 36

Interval Notation 13, ⬁2

5x 0 x ⱖ 36

33, ⬁2

5x 0 x 6 36

1⫺⬁, 32

5x 0 x ⱕ 36

1⫺⬁, 3 4

Examples

The reciprocal of a number a ⫽ 0 is The opposite of a number a is ⫺a. The absolute value of a, denoted 0a 0 , is its distance from zero on the number line. 1 a.

0a 0 ⫽ ⫺a

17, 12, p

Operations on Real Numbers

Key Concepts

0a 0 ⫽ a

Some irrational numbers are:

if a ⱖ 0

Example 1 Given: ⫺5 The reciprocal is ⫺15. The opposite is 5. The absolute value is 5.

if a 6 0

Addition of Real Numbers Same Signs: Add the absolute values of the numbers, and apply the common sign to the sum. Unlike Signs: Subtract the smaller absolute value from the larger absolute value. Then apply the sign of the number having the larger absolute value.

Example 2 ⫺3 ⫹ 1⫺42 ⫽ ⫺7 ⫺5 ⫹ 7 ⫽ 2

Graph ( 3 3

( 3 3

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Chapter R Review of Basic Algebraic Concepts

Subtraction of Real Numbers

Example 3

Add the opposite of the second number to the first number. a ⫺ b ⫽ a ⫹ 1⫺b2

7 ⫺ 1⫺52 ⫽ 7 ⫹ 152 ⫽ 12

Multiplication and Division of Real Numbers

Example 4

Same Signs: Product or quotient is positive. Opposite Signs: Product or quotient is negative.

1⫺321⫺42 ⫽ 12

⫺15 ⫽5 ⫺3

1⫺22152 ⫽ ⫺10

6 1 ⫽⫺ ⫺12 2

The product of any real number and 0 is 0. The quotient of 0 and a nonzero number is 0. The quotient of a nonzero number and 0 is undefined.

1⫺72102 ⫽ 0

0⫼9⫽0

⫺3 ⫼ 0 is undefined Exponents and Radicals

Example 5

b ⫽ b ⴢ b ⴢ b ⴢ b (b is the base, 4 is the exponent) 1b is the principal square root of b (1 is the radical sign).

63 ⫽ 6 ⴢ 6 ⴢ 6 ⫽ 216

Order of Operations

Example 6

4

1. Simplify expressions within parentheses and other grouping symbols first. 2. Evaluate expressions involving exponents, radicals and absolute values. 3. Perform multiplication or division in order from left to right. 4. Perform addition or subtraction in order from left to right.

Section R.4

1100 ⫽ 10

10 ⫺ 513 ⫺ 12 2 ⫹ 116 ⫽ 10 ⫺ 5122 2 ⫹ 116 ⫽ 10 ⫺ 5142 ⫹ 4 ⫽ 10 ⫺ 20 ⫹ 4 ⫽ ⫺10 ⫹ 4 ⫽ ⫺6

Simplifying Algebraic Expressions

Key Concepts

Examples

A term is a constant or the product or quotient of a constant and one or more variables.

Example 1

• A variable term contains at least one variable. • A constant term has no variable. The coefficient of a term is the numerical factor of the term.

⫺2x 2

Variable term has coefficient ⫺2.

xy

Variable term has coefficient 1.

6

Constant term has coefficient 6.

Example 2

Like terms have the same variables, and the corresponding variables are raised to the same powers.

4ab3 and 2ab3 are like terms.

Distributive Property of Multiplication over Addition

Example 3

a1b ⫹ c2 ⫽ ab ⫹ ac

21x ⫹ 4y2 ⫽ 2x ⫹ 8y ⫺1a ⫹ 6b ⫺ 5c2 ⫽ ⫺a ⫺ 6b ⫹ 5c

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Review Exercises

Two terms can be added or subtracted if they are like terms. Sometimes it is necessary to clear parentheses before adding or subtracting like terms.

41

Example 4 ⫺4d ⫹ 12d ⫹ d ⫽ 9d Example 5 ⫺23w ⫺ 41w ⫺ 22 4 ⫹ 3 ⫽ ⫺23w ⫺ 4w ⫹ 84 ⫹ 3 ⫽ ⫺23⫺3w ⫹ 84 ⫹ 3 ⫽ 6w ⫺ 16 ⫹ 3 ⫽ 6w ⫺ 13

Chapter R

Review Exercises

Section R.2

Section R.3

1. Find a number that is a whole number but not a natural number. For Exercises 2–3, answers may vary.

For Exercises 14–15, find the opposite, reciprocal, and absolute value. 14. ⫺8

15.

4 9

2. List three rational numbers that are not integers. 3. List five integers, two of which are not whole numbers. For Exercises 4–9, write an expression in words that describes the set of numbers given by each interval. (Answers may vary.) 4. (7, 16)

5. 10, 2.64

6. 3 ⫺6, ⫺34

7. 18, ⬁2

8. 1⫺⬁, 13 4

9. 1⫺⬁, ⬁2

For Exercises 10–12, graph each set and write the set in interval notation. 10. 5x 0 x 6 26

16. 42, 14

17. 252, 125

For Exercises 18–31, perform the indicated operations. 18. 6 ⫹ 1⫺82

19. 1⫺22 ⫺ 1⫺52

20. 81⫺2.72

21. 1⫺1.1217.412

22.

5 13 ⫼ a⫺ b 8 40

1 11 23. a⫺ b ⫼ a⫺ b 4 16

24.

2 ⫺ 413 ⫺ 72 ⫺4 ⫺ 511 ⫺ 32

25.

26. 24 ⫼ 8 ⴢ 2

12122 ⫺ 8 41⫺32 ⫹ 2152

27. 40 ⫼ 5 ⴢ 6

28. 32 ⫹ 21 0 ⫺10 ⫹ 5 0 ⫼ 52

11. 5x 0 x ⱖ 06

29. ⫺91 ⫹ 141 125 ⫺ 132 2

12. 5x 0 ⫺1 6 x 6 56 13. True or false?

For Exercises 16–17, simplify the exponents and the radicals.

30. x 6 3 is equivalent to 3 7 x

313 ⫺ 82 2 0 8 ⫺ 32 0

31.

415 ⫺ 22 2

03 ⫺ 7 ⫺ 50

32. Given h ⫽ 12gt 2 ⫹ v0 t ⫹ h 0, find h if g ⫽ ⫺32, v0 ⫽ 64, h 0 ⫽ 256, and t ⫽ 4.

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42

Chapter R Review of Basic Algebraic Concepts

33. Find the area of a parallelogram with base 42 in. and height 18 in.

For Exercises 38–41, clear parentheses if necessary, and combine like terms. 38. 5 ⫺ 6q ⫹ 13q ⫺ 19 40. 7 ⫺ 31y ⫹ 42 ⫺ 3y

18 in.

41.

42 in.

Section R.4

1 3 18x ⫺ 42 ⫹ 16x ⫹ 42 4 2

For Exercises 42–43, answers may vary.

For Exercises 34–37, apply the distributive property and simplify. 1 1x ⫹ 8y ⫺ 52 2

34. 31x ⫹ 5y2

35.

36. ⫺1⫺4x ⫹ 10y ⫺ z2

37. ⫺113a ⫺ b ⫺ 5c2

42. Write an example of the commutative property of addition. 43. Write an example of the associative property of multiplication.

Test

Chapter R

1. a. List the integers between ⫺5 and 2, inclusive.

11. Given z ⫽

2. Explain the difference between the interval 34, ⬁2 and 14, ⬁ 2. 3. Answer true or false: The set 5x 0 x ⱖ 56 is the same as 5x 0 5 ⱕ x6. For Exercises 4–5, graph the inequality and express the set in interval notation. 4. 5y 0 y 6 ⫺43 6

5. 5p 0 12 ⱕ p6

6. Write the opposite, reciprocal, and absolute value for each number. 1 2

b. 4

c. 0

x⫺m

, find z when n ⫽ 16, x ⫽ 18, s/ 2n s ⫽ 1.8, and m ⫽ 17.5. (Round the answer to 1 decimal place.)

b. List three rational numbers between 1 and 2. (Answers may vary.)

a. ⫺

39. 18p ⫹ 3 ⫺ 17p ⫹ 8p

For Exercises 12–15, answer true or false.

12. 1x ⫹ y2 ⫹ 2 ⫽ 2 ⫹ 1x ⫹ y2 is an example of the associative property of addition. 13. 12 ⴢ 32 ⴢ 5 ⫽ 13 ⴢ 22 ⴢ 5 is an example of the commutative property of multiplication. 14. 1x ⫹ 324 ⫽ 4x ⫹ 12 is an example of the distributive property of multiplication over addition. 15. 110 ⫹ y2 ⫹ z ⫽ 10 ⫹ 1y ⫹ z2 is an example of the associative property of addition.

For Exercises 16–18, simplify the expressions. 16. 5b ⫹ 2 ⫺ 7b ⫹ 6 ⫺ 14

For Exercises 7–10, simplify the expression. ⫺62 ⫺ 10 2 7. 0⫺8 0 ⫺ 412 ⫺ 32 ⫼ 24 8. ⫺1 ⫹ 32

17. ⫺314 ⫺ x2 ⫹ 91x ⫺ 12 ⫺ 512x ⫺ 42

2

1 4 9. a⫺ ⫹ b 6 B9

2

10. ⫺8 ⫼ 3 ⴢ 2

18.

1 3 12x ⫺ 12 ⫺ a3x ⫺ b 2 2

For Exercises 19–20, write each English phrase as an algebraic statement. 19. x is no more than 5.

20. p is at least 7.

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1

Linear Equations and Inequalities in One Variable CHAPTER OUTLINE 1.1 Linear Equations in One Variable 44 Problem Recognition Exercises: Equations Versus Expressions

56

1.2 Applications of Linear Equations in One Variable 57 1.3 Applications to Geometry and Literal Equations 68 1.4 Linear Inequalities in One Variable 76 1.5 Compound Inequalities 85 1.6 Absolute Value Equations 97 1.7 Absolute Value Inequalities 103 Problem Recognition Exercises: Identifying Equations and Inequalities Group Activity: Understanding the Symbolism of Mathematics

113

114

Chapter 1 In this chapter, we study linear equations and inequalities and their applications. Are You Prepared? To prepare yourself, try the crossword puzzle. The clues in the puzzle review formulas from geometry and other important mathematical facts that you will encounter in this chapter as you work through the application problems. Across 2. What is the next consecutive integer after 1306? 4. Given that distance  rate  time, what is the distance between Atlanta and Los Angeles if it takes 33 hr, traveling 60 mph? 6. What is the next consecutive odd integer after 7803? 7. What is 10% of 64,780? 9. If an angle measures 41, what is the complement? 10. If an angle measures 70, what is the supplement?

1

2

4

3

5

6

7

8

Down 1. 2. 3. 4. 5. 8.

What is 40% of 32,640? 9 What is the sum of the measures of the angles in a triangle? Evaluate 07729  262 0 . What number is 50 more than twice 8707? If the area of a rectangle is 6370 ft2, and the width is 65 ft, what is the length? What is the amount of simple interest earned on $4000 at 5% interest for 4 yr?

10

43

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44

Chapter 1 Linear Equations and Inequalities in One Variable

Section 1.1

Linear Equations in One Variable

Concepts

1. Definition of a Linear Equation in One Variable

1. Definition of a Linear Equation in One Variable 2. Solving Linear Equations 3. Clearing Fractions and Decimals 4. Conditional Equations, Contradictions, and Identities

An equation is a statement that indicates that two quantities are equal.The following are equations. x  4

p  3  11

2z  20

All equations have an equal sign. Furthermore, notice that the equal sign separates the equation into two parts, the left-hand side and the right-hand side. A solution to an equation is a value of the variable that makes the equation a true statement. Substituting a solution to an equation for the variable makes the right-hand side equal to the left-hand side. Equation

Solution

p  3  11

8

Check p  3  11 Substitute 8 for p. Right-hand side equals left-hand side.

8  3  11 ✓ 2z  20

10

2z  20 21102  20 ✓

Substitute 10 for z. Right-hand side equals left-hand side.

The solution set to an equation is the set of all solutions to an equation. We write the solution set using set brackets. For example: Equation

Solution set

p  3  11

This equation has one solution.

w2  16

This equation has two solutions.

586

54, 46

Throughout this text we will learn to recognize and solve several different types of equations, but in this chapter, we will focus on the specific type of equation called a linear equation in one variable.

DEFINITION Linear Equation in One Variable Let a and b be real numbers such that a  0. A linear equation in one variable is an equation that can be written in the form ax  b  0

Notice that a linear equation in one variable will contain only one variable. Furthermore, because the variable has an implied exponent of 1, a linear equation is sometimes called a first-degree equation. Linear equation in one variable

Not a linear equation in one variable

4x  3  0

4x2  8  0

4 5p



3 10

0

5x  2  0

4 5p



3 10 q

0

51x  2  0

(exponent for x is not 1) (more than one variable) (Equation is not in the form ax  b  0. The variable is in the denominator.)

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Section 1.1

Linear Equations in One Variable

2. Solving Linear Equations To solve a linear equation, the goal is to simplify the equation to isolate the variable. Each step used in simplifying an equation results in an equivalent equation. Equivalent equations have the same solution set. For example, the equations 2x  3  7 and 2x  4 are equivalent because 526 is the solution set for both equations. To solve an equation, we may use the addition, subtraction, multiplication, and division properties of equality. These properties state that adding, subtracting, multiplying, or dividing the same quantity on each side of an equation results in an equivalent equation.

PROPERTY Addition and Subtraction Properties of Equality Let a, b, and c represent real numbers. Addition property of equality:

If a  b, then a  c  b  c.

*Subtraction property of equality:

If a  b, then a  c  b  c.

*The subtraction property of equality follows directly from the addition property, because subtraction is defined in terms of addition.

If a  1c2  b  1c2 then, acbc

PROPERTY Multiplication and Division Properties of Equality Let a, b, and c represent real numbers. Multiplication property of equality:

If a  b, then a ⴢ c  b ⴢ c.

*Division property of equality:

If a  b, then

b a  1provided c  02. c c

*The division property of equality follows directly from the multiplication property, because division is defined as multiplication by the reciprocal.

Example 1

If a ⴢ

1 1 bⴢ c c

then,

a b  c c

1c  02

Solving a Linear Equation

Solve the equation.

12  x  40

Solution: 12  x  40 12  12  x  40  12 x  28

To isolate x, subtract 12 from both sides. Simplify.

45

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Chapter 1 Linear Equations and Inequalities in One Variable

Check:

12 ⫹ x ⫽ 40

12 ⫹ 1282 ⱨ 40

40 ⱨ 40 

The solution set is 5286 .

True.

Skill Practice Solve the equation. 1. x ⫺ 5 ⫽ ⫺11

Example 2

Solving Linear Equations

Solve each equation. 3 4 a. ⫺ p ⫽ 5 15

b. 4 ⫽

w 2.2

c. ⫺x ⫽ 6

Solution: a.

3 4 ⫺ p⫽ 5 15 5 3 5 4 a⫺ ba⫺ pb ⫽ a⫺ ba b 3 5 3 15

To isolate p, multiply both sides by the 3 reciprocal of ⫺ . 5

1

5 4 p ⫽ a⫺ ba b 3 15

Multiply fractions.

3

p⫽⫺

4 9

4 The value ⫺ checks in the original 9 equation.

4 The solution set is e ⫺ f . 9 b.

4⫽

w 2.2

2.2142 ⫽ a

w b ⴢ 2.2 2.2

To isolate w, multiply both sides by 2.2.

8.8 ⫽ w

The value 8.8 checks in the original equation.

The solution set is 58.86 . c.

⫺x ⫽ 6 ⫺11⫺x2 ⫽ ⫺1162

To isolate x, multiply both sides by ⫺1.

x ⫽ ⫺6

The value ⫺6 checks in the original equation.

The solution set is 5⫺66 .

Skill Practice Solve the equations. Answers 1. 5⫺66 3. 5806

1 2. e f 2 4. 526

6 3 2. ⫺ y ⫽ ⫺ 5 5

3. 5 ⫽

t 16

4. ⫺a ⫽ ⫺2

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Section 1.1

Linear Equations in One Variable

For more complicated linear equations, several steps are required to isolate the variable. These steps are listed below.

PROCEDURE Solving a Linear Equation in One Variable Step 1 Simplify both sides of the equation. • Clear parentheses. • Consider clearing fractions or decimals (if any are present) by multiplying both sides of the equation by a common denominator of all terms. • Combine like terms. Step 2 Use the addition or subtraction property of equality to collect the variable terms on one side of the equation. Step 3 Use the addition or subtraction property of equality to collect the constant terms on the other side of the equation. Step 4 Use the multiplication or division property of equality to make the coefficient of the variable term equal to 1. Step 5 Check your answer and write the solution set.

Example 3

Solving a Linear Equation

Solve the linear equation and check the answer. 3x  1  7

Solution: 3x  1  7 3x  1  1  7  1

Subtract 1 from both sides.

3x  8

Combine like terms.

3x 8  3 3

To isolate x, divide both sides of the equation by 3.

x

8 3

Simplify. Check:

3x  1 ⱨ 7 8 3a b  1 ⱨ 7 3 8 3a b  1 ⱨ 7 3 8  1 ⱨ 7 7 ⱨ 7 ✓

True.

8 The solution set is e  f . 3 Skill Practice Solve the linear equation and check the answer. 5. 5x  19  23

Answer 4 5. e  f 5

47

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Chapter 1 Linear Equations and Inequalities in One Variable

Example 4

Solving a Linear Equation

Solve the linear equation and check the answer. 11z  2  51z  22

Solution: 11z  2  51z  22 11z  2  5z  10 11z  5z  2  5z  5z  10 6z  2  10 6z  2  2  10  2

Apply the distributive property to clear parentheses. Subtract 5z from both sides. Combine like terms. Subtract 2 from both sides.

6z  12

Combine like terms.

6z 12  6 6

To isolate z, divide both sides of the equation by 6.

z  2

Simplify. 11z  2  51z  22

Check:

11122  2 ⱨ 512  22 22  2 ⱨ 5142 20 ⱨ 20 ✓

The solution set is 526 .

True.

Skill Practice Solve the equations. 6. 7  21y  32  6y  3

Example 5

Solving a Linear Equation

Solve the equation.

31x  42  2  7  1x  12

Solution: 31x  42  2  7  1x  12 3x  12  2  7  x  1 3x  14  x  6 3x  x  14  x  x  6 2x  14  6 2x  14  14  6  14

Answer 1 6. e  f 2

Clear parentheses. Combine like terms. Add x to both sides of the equation. Combine like terms. Subtract 14 from both sides.

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Section 1.1

2x  8

Combine like terms.

8 2x  2 2

To isolate x, divide both sides by 2.

x4 The solution set is 546 .

Linear Equations in One Variable

Simplify. The solution checks in the original equation.

Skill Practice Solve the equation. 7. 412t  22  61t  12  6  t

Example 6

Solving a Linear Equation 43y  31y  52 4  216  5y2

Solve the equation.

Solution:

43y  31y  52 4  216  5y2 43y  3y  154  12  10y 432y  154  12  10y 8y  60  12  10y 8y  10y  60  12  10y  10y 18y  60  12 18y  60  60  12  60

Clear parentheses. Combine like terms. Clear parentheses. Add 10y to both sides of the equation. Combine like terms. Add 60 to both sides of the equation.

18y  72 72 18y  18 18 y4

The solution set is 546 .

To isolate y, divide both sides by 18. The solution checks.

Skill Practice Solve the equation. 8. 33p  21p  22 4  41p  32

3. Clearing Fractions and Decimals When an equation contains fractions or decimals, it is sometimes helpful to clear the fractions and decimals. This is accomplished by multiplying both sides of the equation by the least common denominator (LCD) of all terms within the equation. This is demonstrated in Example 7.

Answers 8 7. e  f 3

8. 506

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Chapter 1 Linear Equations and Inequalities in One Variable

Solving a Linear Equation by Clearing Fractions

Example 7

1 1 1 w ⫹ w ⫺ 1 ⫽ 1w ⫺ 42 4 3 2

Solve the equation.

Solution: 1 1 1 w ⫹ w ⫺ 1 ⫽ 1w ⫺ 42 4 3 2 1 1 1 w⫹ w⫺1⫽ w⫺2 4 3 2

Clear parentheses.

1 1 1 12 ⴢ a w ⫹ w ⫺ 1b ⫽ 12 ⴢ a w ⫺ 2b 4 3 2 1 1 1 12 ⴢ w ⫹ 12 ⴢ w ⫹ 12 ⴢ 1⫺12 ⫽ 12 ⴢ w ⫹ 12 ⴢ 1⫺22 4 3 2

Multiply both sides of the equation by the LCD of all terms. In this case, the LCD is 12. Apply the distributive property.

3w ⫹ 4w ⫺ 12 ⫽ 6w ⫺ 24 7w ⫺ 12 ⫽ 6w ⫺ 24 w ⫺ 12 ⫽ ⫺24 The solution set is 5⫺126 .

Subtract 6w.

w ⫽ ⫺12

The solution checks.

Skill Practice Solve the equation by first clearing the fractions. 9.

3 1 2 1 a⫹ ⫽ a⫹ 4 2 3 3

Example 8 Solve.

Solving a Linear Equation by Clearing Fractions

x⫺2 x⫺4 x⫹4 ⫺ ⫽2⫹ 5 2 10

Solution: x⫺4 2 x⫹4 x⫺2 ⫺ ⫽ ⫹ 5 2 1 10 10 a 2

TIP: Clearing fractions is an application of the multiplication property of equality. We are multiplying both sides of the equation by the same number.

Answer 9. 5⫺26

The LCD of all terms in the equation is 10.

x⫺2 x⫺4 2 x⫹4 ⫺ b ⫽ 10 a ⫹ b 5 2 1 10 5

Multiply both sides by 10.

1

10 x⫺2 10 x⫺4 10 2 10 x⫹4 ⴢa b⫺ ⴢa b⫽ ⴢa b⫹ ⴢa b 1 5 1 2 1 1 1 10 1

1

1

Apply the distributive property.

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Section 1.1

21x  22  51x  42  20  11x  42 2x  4  5x  20  20  x  4

Linear Equations in One Variable

Clear fractions. Apply the distributive property.

3x  16  x  24

Simplify both sides of the equation.

4x  16  24

Subtract x from both sides.

4x  8 x  2 The solution set is 526.

Subtract 16 from both sides. The value 2 checks in the original equation.

Skill Practice Solve. 10.

1 x3 3x  2   8 4 2

The same procedure used to clear fractions in an equation can be used to clear decimals.

Example 9

Solving a Linear Equation by Clearing Decimals

Solve the equation.

0.55x  0.6  2.05x

Solution: Recall that any terminating decimal can be written as a fraction. Therefore, the equation 0.55x  0.6  2.05x is equivalent to 55 6 205 x  x 100 10 100 A convenient common denominator for all terms in this equation is 100. Multiplying both sides of the equation by 100 will have the effect of “moving” the decimal point 2 places to the right. 10010.55x  0.62  10012.05x2

Multiply both sides by 100 to clear decimals.

55x  60  205x 60  150x

Subtract 55x from both sides.

60 x 150

To isolate x, divide both sides by 150.



2 x    0.4 5

The solution checks.

The solution set is 50.46 . Skill Practice Solve the equation by first clearing the decimals. 11. 2.2x  0.5  1.6x  0.2 Answers 10. e

3 f 14

11. 50.56

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Chapter 1 Linear Equations and Inequalities in One Variable

4. Conditional Equations, Contradictions, and Identities The solution to a linear equation is the value of x that makes the equation a true statement. While linear equations have one unique solution, some equations have no solution, and others have infinitely many solutions. I. Conditional Equations An equation that is true for some values of the variable but false for other values is called a conditional equation. The equation x  4  6 is a conditional equation because it is true on the condition that x  2. For other values of x, the statement x  4  6 is false. II. Contradictions Some equations have no solution, such as x  1  x  2. There is no value of x that when increased by 1 will equal the same value increased by 2. If we tried to solve the equation by subtracting x from both sides, we get the contradiction 1  2. x1x2 xx1xx2 12

1contradiction2

This indicates that the equation has no solution. An equation that has no solution is called a contradiction. The solution set for a contradiction is the empty set and is denoted by the symbol 5 6 or . III. Identities An equation that is true for all real numbers is called an identity. For example, consider the equation x  4  x  4. Because the left- and right-hand sides are identical, any real number substituted for x will result in equal quantities on both sides. If we solve the equation, we get the identity 4  4. In such a case, the solution set is the set of real numbers. x  4x4 xx4xx4 44

1identity2

The solution set is the set of real numbers.

In set-builder notation, we have, 5x 0 x is a real number6. Example 10

Identifying Conditional Equations, Contradictions, and Identities

Identify each equation as a conditional equation, a contradiction, or an identity. Then give the solution set. a. 33x  1x  12 4  2

b. 513  c2  2  2c  3c  17 c. 4x  3  17

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Section 1.1

Linear Equations in One Variable

53

Solution: a. 33x  1x  12 4  2 33x  x  14  2

Clear parentheses.

3314  2

Combine like terms.

3  2

Contradiction

This equation is a contradiction. The solution set is 5 6. b. 513  c2  2  2c  3c  17 15  5c  2  5c  17

Clear parentheses and combine like terms.

5c  17  5c  17

Identity

TIP: Interval notation can

00

This equation is an identity. The solution set is 5x 0 x is a real number6 . c.

also be used to express the set of real numbers, 1 , 2 .

4x  3  17 4x  3  3  17  3

Add 3 to both sides.

4x  20 4x 20  4 4

To isolate x, divide both sides by 4.

x5

This equation is a conditional equation. The solution set is 556. Skill Practice Identify each equation as a conditional equation, an identity, or a contradiction. Then give the solution set. 12. 215x  12  2x  12x  6 13. 213x  12  61x  12  8 14. 4x  1  x  6x  2

Section 1.1

Answers

12. Contradiction; 5 6 13. Identity; 5x 0 x is a real number6 14. Conditional equation; 516

Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. Some instructors allow the use of calculators. Does your instructor allow the use of a calculator? If so, what kind? Will you be allowed to use a calculator on tests or just for occasional calculator problems in the text? Helpful Hint: If you are not permitted to use a calculator on tests, you should do your homework in the same way, without the calculator. 2. Define the key terms. a. Equation

b. Solution to an equation

c. Linear equation in one variable

d. Solution set

e. Conditional equation

f. Contradiction

g. Empty set

h. Identity

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Chapter 1 Linear Equations and Inequalities in One Variable

Review Exercises For Exercises 3–6, clear parentheses and combine like terms. 3. 8x  3y  2xy  5x  12xy

4. 5ab  5a  13  2a  17

5. 213z  42  1z  122

6. 16w  52  314w  52

Concept 1: Definition of a Linear Equation in One Variable For Exercises 7–12, label the equation as linear or nonlinear. 7. 2x  1  5

8. 10  x  6

10. 3  x3  x  4

9. x2  7  9

11. 3  x

12. 5.2  7x  0

13. Use substitution to determine which value is the solution to 2x  1  5. a. 2

b. 3

c. 0

d. 1

14. Use substitution to determine which value is the solution to 2y  3  2. a. 1

b.

1 2

c. 0

d. 

1 2

Concept 2: Solving Linear Equations For Exercises 15–44, solve the equation and check the solution. (See Examples 1–6.) 16. 3  y  28

7 5 19.    z 8 6

20. 

23. 2.53  2.3t

24. 4.8  6.1  y

25. p  2.9  3.8

26. 4.2a  4.494

27. 6q  4  62

28. 2w  15  15

29. 4y  17  35

30. 6z  25  83

31. b  5  2

32. 6  y  1

33. 31x  62  2x  5

34. 13y  4  51y  42

12 4  b 13 3

17. x  2

21.

a  8 5

18. t 

3 4

15. x  7  19

22.

x 1  8 2

35. 6  1t  22  513t  42

36. 1  51p  22  21p  132

37. 61a  32  10  21a  42

38. 81b  22  3b  91b  12

39. 235  12z  12 4  4  213  z2

40. 33w  110  w2 4  71w  12

41. 61y  42  312y  32  y  5  5y

42. 13  4w  51w  62  21w  12

43. 14  2x  5x  412x  52  6

44. 8  1p  22  6p  7  p  13

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Concept 3: Clearing Fractions and Decimals For Exercises 45–56, solve the equations. (See Examples 7–9.) 47.

1 3 1 1p  52  p  p  1 5 5 10

3x  7 3  5x 3  6x   2 3 5

50.

2y  4 5y  13 y   5 4 2

3a  9 2a  5 a  2   0 15 5 10

53. 6.3w  1.5  4.8

45.

2 1 5 3 1 x  x  x 3 6 12 2 6

1 9 2 46.  y  4   y  2 10 5

48.

5 7 1 1q  22   q   2 6 9 3

49.

51.

5q  6 q 4 12q  62   0 3 6 3

52.

54. 0.2x  53.6  x

55. 0.751m  22  0.25m  0.5

56. 0.41n  102  0.6n  2

Concept 4: Conditional Equations, Contradictions, and Identities 57. What is a conditional equation? 58. Explain the difference between a contradiction and an identity. For Exercises 59–64, identify the equation as a conditional equation, a contradiction, or an identity. Then give the solution set. (See Example 10.) 59. 4x  1  212x  12  1

60. 3x  6  3x

61. 11x  41x  32  2x  12

62. 51x  22  7  3

63. 2x  4  8x  7x  8  3x

64. 7x  8  4x  31x  32  1

Mixed Exercises For Exercises 65–96, solve the equations. 65. 5b  9  71

66. 3x  18  66

67. 16  10  13x

68. 15  12  9x

69. 10c  3  3  12c

70. 2w  21  6w  7

71. 12b  15b  8  6  4b  6  1

72. 4z  2  3z  5  3  z  4

73. 51x  22  2x  3x  7

74. 2x  31x  52  15

75.

2 77. 0.7518x  42  16x  92 3

1 78.  14z  32  z 2

79. 71p  22  4p  3p  14

80. 61z  22  3z  8  3z

81. 4 33  513  b2  2b4  6  2b

82.

1 1 1 1x  32   12x  52 3 6 6

3 83. 3  x  9 4

84.

9 5  4w  10 2

85.

y3 2y  1 5   4 8 2

87.

2y  9 3  y 10 2

88.

2 5 1 x x3 x5 3 6 2

86.

2 x2 5x  2   3 6 2

89. 0.48x  0.08x  0.121260  x2

c c 3c   1 2 4 8

90. 0.07w  0.061140  w2  90

76.

d d 5d 7    5 10 20 10

1 5 91. 0.5x  0.25  x  3 4

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92. 0.2b ⫹

1 7 ⫽ 3 15

93. 0.3b ⫺ 1.5 ⫽ 0.251b ⫹ 22

1 1 3 7 95. ⫺ y ⫹ ⫽ a5 ⫺ yb 8 4 2 4

94. 0.71a ⫺ 12 ⫽ 0.25 ⫹ 0.7a

96. 5x ⫺ 18 ⫺ x2 ⫽ 23⫺4 ⫺ 13 ⫹ 5x2 ⫺ 134

Expanding Your Skills ⫺21y ⫺ 12 ⫹ 31y ⫹ 22

97. a. Simplify the expression. b. Solve the equation.

98. a. Simplify the expression.

⫺21y ⫺ 12 ⫹ 31y ⫹ 22 ⫽ 0

c. Explain the difference between simplifying an expression and solving an equation.

b. Solve the equation.

4w ⫺ 812 ⫹ w2

4w ⫺ 812 ⫹ w2 ⫽ 0

c. Explain the difference between simplifying an expression and solving an equation.

Problem Recognition Exercises Equations Versus Expressions For Exercises 1–20, identify each exercise as an expression or an equation. Then simplify the expressions and solve the equations. 1. 4x ⫺ 2 ⫹ 6 ⫺ 8x

2. ⫺3y ⫺ 3 ⫺ 4y ⫹ 8

3. 7b ⫺ 1 ⫽ 2b ⫹ 4

4. 10t ⫹ 2 ⫽ 2 ⫺ 7t

5. 41b ⫺ 82 ⫺ 712b ⫹ 12

6. 1012x ⫹ 32 ⫺ 815 ⫺ x2

7. 712 ⫺ s2 ⫽ 5s ⫹ 8

8. 1513 ⫺ 2y2 ⫽ 21 ⫹ 2y

9. 213x ⫺ 42 ⫺ 415x ⫹ 12 ⫽ ⫺8x ⫹ 7 1 3 2 7 v⫹ ⫺ v⫺ 2 5 3 10

10. 612 ⫺ 3a2 ⫺ 218a ⫹ 32 ⫽ ⫺12a ⫺ 19

11.

7 4 5 11 12. ⫺ t ⫺ u ⫺ t ⫹ u 8 3 4 6

14. 7 ⫹ 8b ⫺ 12 ⫽ 3b ⫺ 8 ⫹ 5b

15.

5 7 1 3 y⫺ ⫽ y⫹ 6 8 2 4

18. 0.45k ⫺ 1.67 ⫹ 0.89 ⫺ 1.456k

13. 20x ⫺ 8 ⫹ 7x ⫹ 28 ⫽ 27x ⫺ 9

16.

4 1 ⫹ 3z ⫽ z ⫹ 1 5 2

19. 0.12512p ⫺ 82 ⫽ 0.251p ⫺ 42

20. 0.5u ⫹ 1.2 ⫺ 0.74u ⫽ 0.8 ⫺ 0.24u ⫹ 0.4

17. 0.29c ⫹ 4.495 ⫺ 0.12c

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Applications of Linear Equations in One Variable

Section 1.2

1. Introduction to Problem Solving

Concepts

One of the important uses of algebra is to develop mathematical models for understanding real-world phenomena. To solve an application problem, relevant information must be extracted from the wording of a problem and then translated into mathematical symbols. This is a skill that requires practice. The key is to stick with it and not to get discouraged.

1. Introduction to Problem Solving 2. Applications Involving Consecutive Integers 3. Applications Involving Percents and Rates 4. Applications Involving Principal and Interest 5. Applications Involving Mixtures 6. Applications Involving Distance, Rate, and Time

Problem-Solving Flowchart for Word Problems Step 1

Read the problem carefully.

Step 2

Assign labels to unknown quantities.

Step 3

Develop a verbal model.

Step 4

Write a mathematical equation.

Step 5

Solve the equation.

Step 6

Interpret the results and write the final answer in words.

• Familiarize yourself with the problem. Estimate the answer, if possible. • Identify the unknown quantity or quantities. Let x represent one of the unknowns. Draw a picture and write down relevant formulas.

57

• Write an equation in words.

• Replace the verbal model with a mathematical equation using x or another variable.

• Solve for the variable, using the steps for solving linear equations. • Once you’ve obtained a numerical value for the variable, recall what it represents in the context of the problem. Can this value be used to determine other unknowns in the problem? Write an answer to the word problem in words.

To write an English statement as an algebraic expression, review the list of key terms given in Table 1-1. Table 1-1 Addition: a ⫹ b • the sum of a and b • a plus b • b added to a • b more than a • a increased by b • the total of a and b

Subtraction: a ⫺ b • the difference of a and b • a minus b • b subtracted from a • a decreased by b • b less than a

Multiplication: a ⴢ b • the product of a and b • a times b • a multiplied by b

Division: a ⫼ b, ba • the quotient of a and b • a divided by b • b divided into a • the ratio of a and b • a over b • a per b

Avoiding Mistakes Once you have reached a solution to a word problem, verify that it is reasonable in the context of the problem.

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Example 1

Translating and Solving a Linear Equation

The sum of two numbers is 39. One number is 3 less than twice the other. What are the numbers?

Solution: Step 1:

Read the problem carefully.

Step 2:

Let x represent one number. Let 2x ⫺ 3 represent the other number.

Step 3: (One number) ⫹ (other number) ⫽ 39 Step 4:

Replace the verbal model with a mathematical equation. (One number) ⫹ (other number) ⫽ 39 ⫹

x Step 5:

(2x ⫺ 3)

⫽ 39

Solve for x.

x ⫹ 12x ⫺ 32 ⫽ 39 3x ⫺ 3 ⫽ 39 3x ⫽ 42 3x 42 ⫽ 3 3 x ⫽ 14

Step 6:

Interpret your results. Refer back to step 2. One number is x:

14

The other number is 2x ⫺ 3: 2(14) ⫺ 3

25

Answer: The numbers are 14 and 25. Skill Practice 1. One number is 5 more than 3 times another number. The sum of the numbers is 45. Find the numbers.

2. Applications Involving Consecutive Integers The word consecutive means “following one after the other in order.” • The numbers ⫺2, ⫺1, 0, 1, 2, and so on are examples of consecutive integers. Notice that two consecutive integers differ by 1. Therefore, if x represents an integer, then x ⫹ 1 represents the next consecutive integer. • The numbers 2, 4, 6, 8, and so on are consecutive even integers. Consecutive even integers differ by 2. Therefore, if x represents an even integer, then x ⫹ 2 represents the next consecutive even integer. • The numbers 15, 17, 19, and so on are consecutive odd integers. Consecutive odd integers also differ by 2. Therefore, if x represents an odd integer, then x ⫹ 2 represents the next consecutive odd integer. Answer 1. The numbers are 10 and 35.

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Solving a Linear Equation Involving Consecutive Integers

Example 2

Three times the sum of two consecutive odd integers is 516. Find the integers.

Solution: Step 1:

Read the problem carefully.

Step 2:

Label the unknown: Let x represent the first odd integer. Then x ⫹ 2 represents the next odd integer.

Step 3:

Write an equation in words. 3[(first odd integer) ⫹ (second odd integer)] ⫽ 516 33x ⫹ 1x ⫹ 22 4 ⫽ 516

Step 4:

Write a mathematical equation.

312x ⫹ 22 ⫽ 516

Step 5:

Solve for x.

6x ⫹ 6 ⫽ 516 6x ⫽ 510 x ⫽ 85 Step 6:

Interpret your results. The first odd integer is x:

85

The second odd integer is x ⫹ 2: 85 ⫹ 2

87

Answer: The integers are 85 and 87. Skill Practice

Avoiding Mistakes After completing a word problem, it is always a good idea to check that the answer is reasonable. Notice that 85 and 87 are consecutive odd integers, and three times their sum is 3(85 ⫹ 87), which equals 516.

2. Four times the sum of three consecutive integers is 264. Find the integers.

3. Applications Involving Percents and Rates In many real-world applications, percents are used to represent rates. • • • •

The sales tax rate for a certain county is 6%. An ice cream machine is discounted 20%. A real estate sales broker receives a 4 12% commission on sales. A savings account earns 7% simple interest.

The following models are used to compute sales tax, commission, and simple interest. In each case the value is found by multiplying the base by the percentage. Sales tax ⫽ (cost of merchandise)(tax rate) Commission ⫽ (dollars in sales)(commission rate) Simple interest ⫽ (principal)(annual interest rate)(time in years) I ⫽ Prt

Answer 2. The integers are 21, 22, and 23.

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Chapter 1 Linear Equations and Inequalities in One Variable

Solving a Percent Application

Example 3

A woman invests $5000 in an account that earns 5 14% simple interest. If the money is invested for 3 years (yr), how much money is in the account at the end of the 3-yr period?

Solution: Let x represent the total money in the account. P  $5000

(principal amount invested)

r  0.0525

(interest rate)

t3

(time in years)

Label the variables.

The total amount of money includes principal plus interest.

TIP: Remember to use

(Total money)  (principal)  (interest)

the decimal form of a percent when the number is used in a calculation.

Verbal model

x  P  Prt

Mathematical model

x  $5000  ($5000)(0.0525)(3) x  $5000  $787.50

Substitute for P, r, and t.

x  $5787.50

Solve for x.

The total amount of money in the account is $5787.50.

Interpret the results.

Skill Practice 3. Markos earned $340 in 1 yr on an investment that paid a 4% dividend. Find the amount of money invested.

As consumers, we often encounter situations in which merchandise has been marked up or marked down from its original cost. It is important to note that percent increase and percent decrease are based on the original cost. For example, suppose a microwave oven originally priced at $305 is marked down 20%. The discount is determined by 20% of the original price: (0.20)($305)  $61.00. The new price is $305.00  $61.00  $244.00.

Example 4

Solving a Percent Increase Application

A college bookstore uses a standard markup of 40% on all books purchased wholesale from the publisher. If the bookstore sells a calculus book for $179.20, what was the original wholesale cost?

Answer 3. $8500

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Solution: Let x ⫽ original wholesale cost.

Label the variables.

The selling price of the book is based on the original cost of the book plus the bookstore’s markup. 1Selling price2 ⫽ 1original price2 ⫹ 1markup2

Verbal model

1Selling price2 ⫽ 1original price2 ⫹ 1original price ⴢ markup rate2 179.20 ⫽

x

⫹ 1x210.402

Mathematical model

179.20 ⫽ x ⫹ 0.40x 179.20 ⫽ 1.40x

Combine like terms.

179.20 ⫽x 1.40 x ⫽ 128

Simplify.

The original wholesale cost of the textbook was $128.00.

Interpret the results.

Skill Practice 4. An online bookstore gives a 20% discount on paperback books. Find the original price of a book that has a selling price of $5.28 after the discount.

4. Applications Involving Principal and Interest Example 5

Solving an Investment Growth Application

Miguel had $10,000 to invest in two different mutual funds. One was a relatively safe bond fund that averaged 4% return on his investment at the end of 1 yr. The other fund was a riskier stock fund that averaged 7% return in 1 yr. If at the end of the year Miguel’s portfolio grew to $10,625 ($625 above his $10,000 investment), how much money did Miguel invest in each fund?

Solution: This type of word problem is sometimes categorized as a mixture problem. Miguel is “mixing” his money between two different investments. We have to determine how the money was divided to earn $625. The information in this problem can be organized in a chart. (Note: There are two sources of money: the amount invested and the amount earned.) 4% Bond Fund Amount invested ($) Amount earned ($)

7% Stock Fund

x

(10,000 ⫺ x)

0.04x

0.07(10,000 ⫺ x)

Total 10,000 625

Because the amount of principal is unknown for both accounts, we can let x represent the amount invested in the bond fund. If Miguel spends x dollars in the bond fund, then he has (10,000 ⫺ x) left over to spend in the stock fund. The return for each fund is found by multiplying the principal and the percent growth rate.

Answer 4. $6.60

61

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To establish a mathematical model, we know that the total return ($625) must equal the earnings from the bond fund plus the earnings from the stock fund: 1Earnings from bond fund2 ⫹ 1earnings from stock fund2 ⫽ 1total earnings2 ⫹

0.04x

0.07110,000 ⫺ x2

0.04x ⫹ 0.07(10,000 ⫺ x) ⫽ 625

⫽

625

Mathematical model

4x ⫹ 7(10,000 ⫺ x) ⫽ 62,500

Multiply by 100 to clear decimals.

4x ⫹ 70,000 ⫺ 7x ⫽ 62,500 ⫺3x ⫹ 70,000 ⫽ 62,500 ⫺3x ⫽ ⫺7500

Combine like terms. Subtract 70,000 from both sides.

⫺3x ⫺7500 ⫽ ⫺3 ⫺3 x ⫽ 2500

Solve for x and interpret the results.

The amount invested in the bond fund is $2500. The amount invested in the stock fund is $10,000 ⫺ x, or $7500. Skill Practice 5. Jonathan borrowed $4000 in two loans. One loan charged 7% interest, and the other charged 1.5% interest. After 1 yr, Jonathan paid $225 in interest. Find the amount borrowed in each loan.

5. Applications Involving Mixtures Example 6

TIP: To understand the role of the concentration rate within a mixture problem, consider this example. Suppose you had 30 gal of a 10% antifreeze mixture. How much pure antifreeze is in the mixture?

Solving a Mixture Application

How many liters (L) of a 40% antifreeze solution must be added to 4 L of a 10% antifreeze solution to produce a 35% antifreeze solution?

Solution: The given information is illustrated in Figure 1-1.

40% Antifreeze solution

pure antifreeze ⫽ 0.10(30 gal)

35% Antifreeze solution

10% Antifreeze solution

⫽ 3 gal Multiply the concentration rate by the amount of mixture.

⫹

x L of solution

0.40x L of pure antifreeze

⫽

4 L of solution

0.10(4) L of pure antifreeze

Figure 1-1

Answer 5. $3000 was borrowed at 7% interest, and $1000 was borrowed at 1.5% interest.

(4 ⫹ x) L of solution

0.35(4 ⫹ x) L of pure antifreeze

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The information can also be organized in a table.

40% Antifreeze

10% Antifreeze

Final Solution: 35% Antifreeze

x

4

(4  x)

0.40x

0.10(4)

0.35(4  x)

Number of liters of solution Number of liters of pure antifreeze

Notice that an algebraic equation is obtained from the second row of the table relating the number of liters of pure antifreeze in each container. a

Pure antifreeze pure antifreeze pure antifreeze ba ba b from solution 1 from solution 2 in the final solution 

0.40x

0.10142



0.40x  0.10142  0.351x  42

0.3514  x2 Mathematical equation

0.4x  0.4  0.35x  1.4

Apply the distributive property.

0.4x  0.35x  0.4  0.35x  0.35x  1.4

Subtract 0.35x from both sides.

0.05x  0.4  1.4 0.05x  0.4  0.4  1.4  0.4

Subtract 0.4 from both sides.

0.05x  1.0 1.0 0.05x  0.05 0.05

Divide both sides by 0.05.

x  20 Therefore, 20 L of a 40% antifreeze solution is needed. Skill Practice 6. Find the number of ounces (oz) of 30% alcohol solution that must be mixed with 10 oz of a 70% solution to obtain a solution that is 40% alcohol.

6. Applications Involving Distance, Rate, and Time The fundamental relationship among the variables distance, rate, and time is given by Distance  1rate21time2

or

d  rt

For example, a motorist traveling 65 mph (miles per hour) for 3 hr (hours) will travel a distance of da

65 mi b 13 hr2  195 mi hr

Answer 6. 30 oz of the 30% solution is needed.

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Example 7

Solving a Distance, Rate, Time Application

A hiker can hike 1 mph faster downhill to Moose Lake than she can hike uphill back to the campsite. If it takes her 3 hr to hike to the lake and 4.5 hr to hike back, what is her speed hiking back to the campsite?

Solution: The information given in the problem can be organized in a table. Distance (mi) Trip to the lake Return trip

Rate (mph)

Time (hr)

x1

3

x

4.5

Column 2: Let the rate of the return trip be represented by x. Then the trip to the lake is 1 mph faster and can be represented by x  1. Column 3: The times hiking to and from the lake are given in the problem. Column 1: To express the distance, we use the relationship d  rt. That is, multiply the quantities in the second and third columns.

Campsite

d  (rate)(time) d  (x  1)(3) d  (rate)(time) d  x(4.5)

Distance (mi) Trip to the lake

Time (hr)

31x  12

x1

3

4.5x

x

4.5

Return trip

Moose Lake

Rate (mph)

To create a mathematical model, note that the distances to and from the lake are equal. Therefore, (Distance to lake)  (return distance) 31x  12  4.5x 3x  3  4.5x 3x  3x  3  4.5x  3x

Verbal model Mathematical model Apply the distributive property. Subtract 3x from both sides.

3  1.5x 1.5x 3  1.5 1.5 2x

Divide by 1.5 to isolate the variable. Solve for x.

The hiker’s speed on the return trip to the campsite is 2 mph. Skill Practice

Answer 7. Jody normally drives 60 mph.

7. During a bad rainstorm, Jody drove 15 mph slower on a trip to her mother’s house than she normally would when the weather is clear. If a trip to her mother’s house takes 3.75 hr in clear weather and 5 hr in a bad storm, what is her normal driving speed during clear weather?

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Section 1.2

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65

Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. After doing a section of homework, check the odd-numbered answers in the back of the text. Choose a method to identify the exercises that gave you trouble (i.e., circle the number or put a star by the number). List some reasons why it is important to label these problems. 2. Define the key terms. a. Sum

b. Difference

c. Product

d. Quotient

e. Sales tax

f. Commission

g. Simple interest

Review Exercises For Exercises 3–8, solve the equations. 3. 7a ⫺ 2 ⫽ 11

4. 2z ⫹ 6 ⫽ ⫺15

6. ⫺31y ⫺ 52 ⫹ 4 ⫽ 1

7.

3 3 3 p⫹ ⫽p⫺ 8 4 2

5. 41x ⫺ 32 ⫹ 7 ⫽ 19 8.

1 ⫺ 2x ⫽ 5 4

For the remaining exercises, follow the steps outlined in the Problem-Solving Flowchart found on page 57.

Concept 1: Introduction to Problem Solving 9. If x represents a number, write an expression for 5 more than the number.

10. If n represents a number, write an expression for 10 less than the number.

11. If t represents a number, write an expression for 7 less than twice the number.

12. If y represents a number, write an expression for 4 more than 3 times the number.

13. The larger of two numbers is 3 more than twice the smaller. The difference of the larger number and the smaller number is 8. Find the numbers. (See Example 1.)

14. One number is 3 less than another. Their sum is 15. Find the numbers.

15. The sum of 3 times a number and 2 is the same as the difference of the number and 4. Find the number.

16. Twice the sum of a number and 3 is the same as 1 subtracted from the number. Find the number.

17. The sum of two integers is 30. Ten times one integer is 5 times the other integer. Find the integers. (Hint: If one number is x, then the other number is 30 ⫺ x.)

18. The sum of two integers is 10. Three times one integer is 3 less than 8 times the other integer. Find the integers. (Hint: If one number is x, then the other number is 10 ⫺ x.)

Concept 2: Applications Involving Consecutive Integers 19. The sum of two consecutive page numbers in a book is 223. Find the page numbers. (See Example 2.)

20. The sum of the numbers on two consecutive raffle tickets is 808,455. Find the numbers on the tickets.

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21. The sum of two consecutive odd integers is 148. Find the two integers.

22. The sum of three consecutive integers is 57. Find the integers.

23. Three times the smaller of two consecutive even integers is the same as 146 minus 4 times the larger integer. Find the integers.

24. Four times the smaller of two consecutive odd integers is the same as 73 less than 5 times the larger. Find the integers.

25. Two times the sum of three consecutive odd integers is the same as 23 more than 5 times the largest integer. Find the integers.

26. Five times the smallest of three consecutive even integers is 10 more than twice the largest. Find the integers.

Concept 3: Applications Involving Percents and Rates 27. Belle had the choice of taking out a 4-yr car loan at 8.5% simple interest or a 5-yr car loan at 7.75% simple interest. If she borrows $15,000, how much interest would she pay for each loan? Which option will require less interest?

28. Robert can take out a 3-yr loan at 8% simple interest or a 2-yr loan at 812% simple interest. If he borrows $7000, how much interest will he pay for each loan? Which option will require less interest?

(See Example 3.)

29. An account executive earns $600 per month plus a 3% commission on sales. The executive’s goal is to earn $2400 this month. How much must she sell to achieve this goal? 30. A salesperson earns $50 a day plus 12% commission on sales over $200. If her daily earnings are $76.88, how much money in merchandise did she sell? 31. J. W. is an artist and sells his pottery each year at a local Renaissance Festival. He keeps track of his sales and the 8.05% sales tax he collects by making notations in a ledger. Every evening he checks his records by counting the total money in his cash drawer. After a day of selling pottery, the cash totaled $1293.38. How much is from the sale of merchandise and how much is sales tax? 32. Wayne County has a sales tax rate of 7%. How much does Mike’s used Honda Civic cost before tax if the total cost of the car plus tax is $13,888.60? 33. The price of a swimsuit after a 20% markup is $43.08. What was the price before the markup? (See Example 4.) 34. The price of a used textbook after a 35% markdown is $29.25. What was the original price? 35. For a recent year, 1800 medical degrees were awarded to women. This represents a 5.5% increase over the number awarded the previous year. How many women were awarded a medical degree the previous year? 36. For a recent year, Americans spent approximately $69 billion on weddings. This represents a 50% increase from the amount spent in 2001. What amount did Americans spend on weddings in 2001?

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Concept 4: Applications Involving Principal and Interest 37. Tony has a total of $12,500 in two accounts. One account pays 2% simple interest per year and the other pays 5% simple interest. If he earned $370 in interest in the first year, how much did he invest in each account? (See Example 5.)

38. Lillian had $15,000 invested in two accounts, one paying 9% simple interest and one paying 10% simple interest. How much was invested in each account if the interest after 1 yr is $1432?

39. Jason borrowed $18,000 in two loans. One loan charged 11% simple interest and the other charged 6% simple interest. After 1 yr, Jason paid a total of $1380. Find the amount borrowed in each loan.

40. Amanda borrowed $6000 from two sources: her parents and a credit union. Her parents charged 3% simple interest and the credit union charged 8% simple interest. If after 1 yr, Amanda paid $255 in interest, how much did she borrow from her parents, and how much did she borrow from the credit union?

41. Donna invested money in two accounts: one paying 4% simple interest and the other paying 3% simple interest. She invested $4000 more in the 4% account than in the 3% account. If she received $720 in interest at the end of 1 yr how much did she invest in each account?

42. Mr. Hall had some money in his bank earning 4.5% simple interest. He had $5000 more deposited in a credit union earning 6% simple interest. If his total interest for 1 yr was $1140, how much did he deposit in each account?

43. Ms. Riley deposited some money in an account paying 5% simple interest and twice that amount in an account paying 6% simple interest. If the total interest from the two accounts is $765 for 1 yr, how much was deposited into each account?

44. Sienna put some money in a certificate of deposit earning 4.2% simple interest. She deposited twice that amount in a money market account paying 4% simple interest. After 1 yr her total interest was $488. How much did Sienna deposit in her money market account?

Concept 5: Applications Involving Mixtures 45. Ahmed mixes two plant fertilizers. How much fertilizer with 15% nitrogen should be mixed with 2 oz of fertilizer with 10% nitrogen to produce a fertilizer that is 14% nitrogen? (See Example 6.)

46. How much 8% saline solution should Kent mix with 80 cc (cubic centimeters) of an 18% saline solution to produce a 12% saline solution?

47. Jacque has 3 L of a 50% antifreeze mixture. How much 75% mixture should be added to get a mixture that is 60% antifreeze?

48. One fruit punch has 40% fruit juice and another is 70% fruit juice. How much of the 40% punch should be mixed with 10 gal of the 70% punch to create a fruit punch that is 45% fruit juice?

49. How many liters of an 18% alcohol solution must be added to a 10% alcohol solution to get 20 L of a 15% alcohol solution?

50. How many milliliters of a 2.5% bleach solution must be mixed with a 10% bleach solution to produce 600 mL of a 5% bleach solution?

51. Ronald has a 12% solution of the fertilizer Super Grow. How much pure Super Grow should he add to the mixture to get 32 oz of a 17.5% concentration?

52. How many ounces of water must be added to 20 oz of an 8% salt solution to make a 2% salt solution?

53. Two different teas are mixed to make a blend that will be sold at a fair. Black tea sells for $2.20 per pound and green tea sells for $3.00 per pound. How much of each should be used to obtain 4 lb of a blend selling for $2.50?

54. A nut mixture consists of almonds and cashews. Almonds are $4.98 per pound, and cashews are $6.98 per pound. How many pounds of each type of nut should be mixed to produce 16 lb selling for $5.73 per pound?

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Concept 6: Applications Involving Distance, Rate, and Time 55. A Piper Cub airplane has an average speed that is 30 mph faster than a Cessna 150 airplane. It takes the Cessna 5 hr to fly from Fort Lauderdale to Atlanta, while it takes the Piper Cub only 4 hr to make the same trip. What is the average speed of each plane? (See Example 7.) 56. A woman can hike 1 mph faster down a trail to Archuletta Lake than she can on the return trip uphill. It takes her 3 hr to get to the lake and 6 hr to return. What is her speed hiking down to the lake? 57. Two cars are 192 mi apart and travel toward each other on the same road. They meet in 2 hr. One car travels 4 mph faster than the other. What is the average speed of each car? 58. Two cars are 190 mi apart and travel toward each other along the same road. They meet in 2 hr. One car travels 5 mph slower than the other car. What is the average speed of each car? 59. Two boats traveling the same direction leave a harbor at noon. After 3 hr they are 60 mi apart. If one boat travels twice as fast as the other, find the average rate of each boat. 60. Two canoes travel down a river, starting at 9:00. One canoe travels twice as fast as the other. After 3.5 hr, the canoes are 5.25 mi apart. Find the average rate of each canoe.

Section 1.3

Applications to Geometry and Literal Equations

Concepts

1. Applications Involving Geometry

1. Applications Involving Geometry 2. Literal Equations

Some word problems involve the use of geometric formulas such as those listed in the inside back cover of this text. Example 1

Solving an Application Involving Perimeter

The length of a rectangular corral is 2 ft more than 3 times the width. The corral is situated such that one of its shorter sides is adjacent to a barn and does not require fencing. If the total amount of fencing is 774 ft, then find the dimensions of the corral.

Solution: Read the problem and draw a sketch (Figure 1-2). 3x ⫹ 2 x

3x ⫹ 2

Figure 1-2

TIP: In Example 1, the length of the field is given in terms of the width. Therefore, we let x represent the width.

x

Let x represent the width. Let 3x ⫹ 2 represent the length.

Label variables.

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To create a verbal model, we might consider using the formula for the perimeter of a rectangle. However, the formula P ⫽ 2l ⫹ 2w incorporates all four sides of the rectangle.The formula must be modified to include only one factor of the width. a

2 times 1 times Distance around b⫽a b⫹a b three sides the length the width ⫽ 2(3x ⫹ 2) ⫹

774

x

Verbal model Mathematical model

774 ⫽ 213x ⫹ 22 ⫹ x

Solve for x.

774 ⫽ 6x ⫹ 4 ⫹ x

Apply the distributive property.

774 ⫽ 7x ⫹ 4

Combine like terms.

770 ⫽ 7x

Subtract 4 from both sides.

110 ⫽ x

Divide by 7 on both sides.

x ⫽ 110 Because x represents the width, the width of the corral is 110 ft. The length is given by 3x ⫹ 2

or

311102 ⫹ 2 ⫽ 332

Interpret the results.

Avoiding Mistakes

The width of the corral is 110 ft, and the length is 332 ft.

To check the answer to Example 1, verify that the three sides add to 774 ft.

Skill Practice 1. The length of Karen’s living room is 2 ft longer than the width. The perimeter is 80 ft. Find the length and width.

110 ft ⫹ 332 ft ⫹ 332 ft ⫽ 774 ft 

Recall some important facts involving angles. • • •

Two angles are complementary if the sum of their measures is 90°. Two angles are supplementary if the sum of their measures is 180°. The sum of the measures of the angles within a triangle is 180°. Example 2

Solving an Application Involving Angles

Two angles are complementary. One angle measures 10° less than 4 times the other angle. Find the measure of each angle (Figure 1-3). (4x ⫺ 10)°

Solution:

x°

Let x represent the measure of one angle. Let 4x ⫺ 10 represent the measure of the other angle.

Figure 1-3

Recall that two angles are complementary if the sum of their measures is 90°. Therefore, a verbal model is 1One angle2 ⫹ 1the complement of the angle2 ⫽ 90° x ⫹ 14x ⫺ 102 ⫽ 90

5x ⫺ 10 ⫽ 90

Verbal model Mathematical equation Solve for x. Answer 1. The length is 21 ft, and the width is 19 ft.

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5x  100 x  20 If x  20, then 4x  10  41202  10  70. The two angles are 20 and 70. Skill Practice 2. Two angles are supplementary, and the measure of one is 16° less than 3 times the other. Find their measures.

2. Literal Equations Literal equations are equations that contain several variables. A formula is a literal equation with a specific application. For example, the perimeter of a rectangle can be found by the formula P  2l  2w. In this equation, P is expressed in terms of l and w. However, in science and other branches of applied mathematics, formulas may be more useful in alternative forms. For example, the formula P  2l  2w can be manipulated to solve for either l or w: Solve for l

Solve for w

P  2l  2w

P  2l  2w

P  2w  2l

Subtract 2w.

P  2l  2w

Subtract 2l.

P  2w l 2

Divide by 2.

P  2l w 2

Divide by 2.

l

P  2w 2

w

P  2l 2

To solve a literal equation for a specified variable, use the addition, subtraction, multiplication, and division properties of equality. Example 3

Applying a Literal Equation

Buckingham Fountain is one of Chicago’s most familiar landmarks. With 133 jets spraying a total of 14,000 gal of water per minute, Buckingham Fountain is one of the world’s largest fountains. The circumference of the fountain is approximately 880 ft. a. The circumference of a circle is given by C  2pr. Solve the equation for r. b. Use the equation from part (a) to find the radius and diameter of the fountain. Use 3.14 for p and round to the nearest foot.

Solution: a.

C  2pr C 2pr  2p 2p C r 2p r

Answer 2. 49° and 131°

C 2p

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Section 1.3

b. r ⬇

880 ft 213.142

71

Applications to Geometry and Literal Equations

Substitute 880 ft for C and 3.14 for p.

Calculator Connections

⬇ 140 ft

The radius is approximately 140 ft. The diameter is twice the radius 1d  2r2 . Therefore, the diameter is 280 ft.

The p key on the calculator can also be used in the calculation for Example 3(b).

Skill Practice The formula to compute the surface area S of a sphere is given by S  4pr 2. 3. Solve the equation for p. 4. A sphere has a surface area of 113 in.2 and a radius of 3 in. Use the formula found in part (a) to approximate p. Round to two decimal places.

Example 4

Solving a Literal Equation b2

The formula to find the area of a trapezoid is given by A  12 1b1  b2 2h, where b1 and b2 are the lengths of the parallel sides and h is the height. (See Figure 1-4.) Solve this formula for b1.

Solution: A  12 1b1  b2 2h

2A  2 ⴢ 12 1b1  b2 2h 2A  1b1  b2 2h

h b1

Figure 1-4

The goal is to isolate b1. Multiply by 2 to clear fractions. Apply the distributive property.

2A  b1h  b2h 2A  b2h  b1h

Subtract b2h from both sides.

2A  b2h b1h  h h

Divide by h.

2A  b2h  b1 h r

Skill Practice 5. The formula for the volume of a right circular cylinder is V  pr 2h. Solve for h.

h

TIP: When solving a literal equation for a specified variable, there is sometimes more than one way to express your final answer. This flexibility often presents difficulty for students. Students may leave their answer in one form, but the answer given in the text looks different. Yet both forms may be correct. To know if your answer is equivalent to the form given in the text you must try to manipulate it to look like the answer in the book, a process called form fitting. The literal equation from Example 4 can be written in several different forms. The quantity 12A  b2h2 h can be split into two fractions. 2A  b2h b2h 2A 2A b1      b2 h h h h

Answers 3. p 

S 4r 2

4. 3.14

5. h 

V pr 2

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Solving a Literal Equation

Example 5

Given 2x  3y  5, solve for y.

Solution: 2x  3y  5 3y  2x  5

Add 2x to both sides.

3y 2x  5  3 3

Divide by 3 on both sides.

y

2x  5 3

or

2 5 y x 3 3

Skill Practice Solve for y. 6. 5x  2y  11

Sometimes the variable we want to isolate may appear in more than one term in a literal equation. In such a case, isolate all terms with that variable on one side of the equation. Then apply the distributive property as demonstrated in Example 6. Example 6

Solving a Literal Equation

Solve the equation for x.

ax  3  cx  7

Solution: ax  3  cx  7 ax  cx  10

Collect the terms containing x on one side of the equation. Collect the remaining terms on the other side. The variable x appears twice in the equation. To isolate x, we want x to appear in only one term. To accomplish this, we apply the distributive property in reverse.

TIP: Applying the distributive property in reverse is called factoring. Factoring will be studied in detail in Chapter 4.

x1a  c2  10

Apply the distributive property. The variable x now appears one time in the equation.

x1a  c2

Divide both sides by 1a  c2 .

1a  c2



x

10 1a  c2 10 ac

Skill Practice Solve for t. 7. mt  4  nt  9

Answers 11  5x 5 11 or y   x  2 2 2 5 7. t  mn 6. y 

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73

Practice Exercises

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Study Skills Exercise 1. In your next math class, take notes by drawing a vertical line about three-fourths of the way across the paper, as shown. On the left side, write down what your instructor puts on the board or overhead. On the right side, make your own comments about important words, procedures, or questions that you have.

Review Exercises For Exercises 2– 6, solve the equations.

2. 7 ⫹ 5x ⫺ 12x ⫺ 62 ⫽ 61x ⫹ 12 ⫹ 21

3.

4. 33z ⫺ 12 ⫺ 3z2 ⫺ 44 ⫽ z ⫺ 7

5. 2a ⫺ 4 ⫹ 8a ⫽ 7a ⫺ 8 ⫹ 3a

3 y ⫺ 3 ⫹ 2y ⫽ 5 5

6. 31t ⫹ 62 ⫹ t ⫹ 2 ⫽ 51t ⫹ 42 ⫺ t

Concept 1: Applications Involving Geometry For Exercises 7–18, use the geometry formulas listed in the inside back cover of the text. 7. A volleyball court is twice as long as it is wide. If the perimeter is 177 ft, find the dimensions of the court. (See Example 1.)

8. The length of a rectangular picture frame is 4 in. less than twice the width. The perimeter is 112 in. Find the length and the width.

9. The lengths of the sides of a triangle are given by three consecutive even integers. The perimeter is 24 m. What is the length of each side?

10. A triangular garden has sides that can be represented by three consecutive integers. If the perimeter of the garden is 15 ft, what are the lengths of the sides?

11. Raoul would like to build a rectangular dog run in the rear of his backyard, away from the house. 1 The width of the yard is 122 yd, and Raoul wants 2 an area of 100 yd for his dog.

12. Joanne wants to plant a flower garden in her backyard in the shape of a trapezoid, adjacent to her house (see the figure). She also wants a front yard garden in the same shape, but with sides one-half as long. What should the dimensions be for each garden if Joanne has only a total of 60 ft of fencing?

a. Find the dimensions of the dog run. b. How much fencing would Raoul need to enclose the dog run? x

2x x

x Back

House

12 12 yd

Dog run

House

Front

1 2x

1 2x

x

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13. George built a rectangular pen for his rabbit such that the length is 7 ft less than twice the width. If the perimeter is 40 ft, what are the dimensions of the pen?

14. Antoine wants to put edging in the form of a square around a tree in his front yard. He has enough money to buy 18 ft of edging. Find the dimensions of the square that will use all the edging.

15. The measures of two angles in a triangle are equal. The third angle measures 2 times the sum of the equal angles. Find the measures of the three angles.

16. The smallest angle in a triangle is one-half the size of the largest. The middle angle measures 25 less than the largest. Find the measures of the three angles.

17. Two angles are complementary. One angle is 5 times as large as the other angle. Find the measure of each angle. (See Example 2.)

18. Two angles are supplementary. One angle measures 12 less than 3 times the other. Find the measure of each angle.

In Exercises 19–26, solve for x, and then find the measure of each angle. 19.

20. (7x  1)°

(2x  1)°

(10x  36)° [2(x  15)]°

21.

22. (3x  3)°

(2x  5)°

[3(5x  1)]°

(x  2.5)°

23.

24. (10x)°

(2 x)°

(x  2)° (5x  1)°

(20x  4)°

(x  35)°

25.

26. (2x  4)°

(x  2)°

[3(x  7)]°

[4(x  8)]°

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75

Concept 2: Literal Equations 27. In 2008, Scott Dixon won the Indianapolis 500 car race in 3 hr 28 min 58 sec (⬇ 3.483 hr). (See Example 3.) a. The relationship among the variables distance, rate, and time is given by d  rt. Solve this formula for the rate, r. b. Determine Scott Dixon’s average rate of speed if the total distance for the race is 500 mi. Round to one decimal place. 28. In 2008, Ryan Newman won the 50th running of the Daytona 500 car race with an average speed of 156.672 mph. a. Solve the formula d  rt for the time, t. b. Determine the total time of the race if the race is 500 mi long. Round to one decimal place. 29. The amount of simple interest earned or borrowed is the product of the principal, the annual interest rate and the time invested (in years). This is given by I  Prt. a. Solve I  Prt for t. b. Determine the amount of time necessary for the interest on $5000 invested at 4% to reach $1400. 30. The force of an object is equal to its mass times the acceleration, or F  ma. a. Solve F  ma for m. b. The force on an object is 24.5 N (newtons), and the acceleration is 9.8 m/sec2. Find the mass of the object (the answer will be in kilograms). For Exercises 31–48, solve for the indicated variable. (See Example 4.) 31. A  lw

34. a  b  c  P 37. F  95 C  32 40. I  Prt

32. C1  52R

for l for b for C

for P

35. W  K2  K1

for K1

36. y  mx  b

38. C  59 1F  322

for F

39. K  12 mv2

41. v  v0  at

for r

33. I  Prt

for R

43. w  p1v2  v1 2

for v2

44. A  lw

46. P  2L  2W

for L

1 47. V  Bh 3

for a for w for B

for x for v2

42. a2  b2  c2

for b2

45. ax  by  c

for y

1 2 pr h 3

for h

48. V 

In Chapter 2 it will be necessary to change equations from the form Ax  By  C to y  mx  b. For Exercises 49–60, express each equation in the form y  mx  b by solving for y. (See Example 5.) 49. 3x  y  6

50. x  y  4

51. 5x  4y  20

52. 4x  5y  25

53. 6x  2y  13

54. 5x  7y  15

55. 3x  3y  6

56. 2x  2y  8

4 57. 9x  y  5 3

1 58. 4x  y  5 3

2 59. x  y  0 3

1 60. x  y  0 4

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In statistics, the z-score formula z  61. a. Solve z 

xm is used in studying probability. Use this formula for Exercises 61–62. s

xm for x. s

62. a. Solve z 

b. Find x when z  2.5, m  100, and s  12.

a. 

5 x3

5 3x

b.

c.

65. Which expressions are equivalent to a. 

x7 y

x7 y

b.

b. Find s when x  150, z  2.5, and m  110.

5 ? x3

63. Which expressions are equivalent to

64. Which expressions are equivalent to

5 x  3

a.

x  7 ? y

c.

xm for s. s

1z 2

b. 

z1 2

a. 

3w x  y

b.

z  1 2

c.

66. Which expressions are equivalent to

x  7 y

z1 ? 2

3w ? x  y

3w xy

c. 

3w xy

For Exercises 67–75, solve for the indicated variable. (See Example 6.) 67. 6t  rt  12 70. cx  4  dx  9 73. T  mg  mf

68. 5  4a  ca

for t for x for m

Section 1.4

69. ax  5  6x  3

for a

71. A  P  Prt

for P

74. T  mg  mf

for f

72. A  P  Prt

for x for r

75. ax  by  cx  z

for x

Linear Inequalities in One Variable

Concepts

1. Solving Linear Inequalities

1. Solving Linear Inequalities 2. Applications of Inequalities

In Sections 1.1–1.3, we learned how to solve linear equations and their applications. In this section, we will learn the process of solving linear inequalities. A linear inequality in one variable, x, is defined as any relationship of the form: ax  b 6 0, ax  b 0, ax  b 7 0, or ax  b  0, where a  0. The solution to the equation x  3 can be graphed as a single point on the number line. 5 4 3 2 1

0

1

2

3

4

5

Now consider the inequality x 3. The solution set to an inequality is the set of real numbers that makes the inequality a true statement. In this case, the solution set is all real numbers less than or equal to 3. Because the solution set has an infinite number of values, the values cannot be listed. Instead, we can graph the solution set or represent the set in interval notation or in set-builder notation. A complete discussion of set-builder notation and interval notation is given in Section R.2. Graph 5 4 3 2 1

0

1

2

3

4

5

Interval Notation 1 , 3 4

Set-Builder Notation 5x 0 x 36

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Linear Inequalities in One Variable

The addition and subtraction properties of equality indicate that a value added to or subtracted from both sides of an equation results in an equivalent equation. The same is true for inequalities.

PROPERTY Addition and Subtraction Properties of Inequality Let a, b, and c represent real numbers. *Addition property of inequality:

*Subtraction property of inequality:

If then

a 6 b ac 6 bc

If then

a 6 b ac 6 bc

*These properties may also be stated for a  b, a 7 b, and a  b.

Example 1

Solving a Linear Inequality

Solve the inequality. Graph the solution and write the solution set in interval notation. 3x  7 7 21x  42  1

Solution: 3x  7 7 21x  42  1 3x  7 7 2x  8  1

Apply the distributive property.

3x  7 7 2x  9 3x  2x  7 7 2x  2x  9

Subtract 2x from both sides.

x  7 7 9 x  7  7 7 9  7

Add 7 to both sides.

x 7 2 Graph (

4 3 2 1

0

1

2

3

4

Interval Notation 12, 2

Skill Practice Solve the inequality. Graph the solution and write the solution set in interval notation. 1. 412x  12 7 7x  1

Multiplying both sides of an equation by the same quantity results in an equivalent equation. However, the same is not always true for an inequality. If you multiply or divide an inequality by a negative quantity, the direction of the inequality symbol must be reversed. For example, consider multiplying or dividing the inequality 4 6 5 by 1. Multiply/divide by 1:

45 4  5

6 5 4 3 2 1 4  5

0

1

2

3

4 5 6 45

The number 4 lies to the left of 5 on the number line. However, 4 lies to the right of 5. Changing the signs of two numbers changes their relative position on the number line.This is stated formally in the multiplication and division properties of inequality.

Answer 1.

( 15, 2

5

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PROPERTY Multiplication and Division Properties of Inequality Let a, b, and c represent real numbers. *If c is positive and a 6 b, then

ac 6 bc

and

a b 6 c c

*If c is negative and a 6 b, then

ac 7 bc

and

a b 7 c c

The second statement indicates that if both sides of an inequality are multiplied or divided by a negative quantity, the inequality sign must be reversed. *These properties may also be stated for a ⱕ b, a 7 b, and a ⱖ b.

Example 2

Solving a Linear Inequality

Solve the inequality. Graph the solution and write the solution set in interval notation. ⫺2x ⫺ 5 6 2

Solution: ⫺2x ⫺ 5 6 2 ⫺2x ⫺ 5 ⫹ 5 6 2 ⫹ 5

Add 5 to both sides.

⫺2x 6 7 ⫺2x 7 7 ⫺2 ⫺2 x 7 ⫺

Divide by ⫺2 (reverse the inequality sign).

7 2

x 7 ⫺3.5

or

Graph

Interval Notation

⫺27

)

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

7 a⫺ , ⬁b 2

The inequality ⫺2x ⫺ 5 6 2 could have been solved by isolating x on the right-hand side of the inequality. This creates a positive coefficient on the x term and eliminates the need to divide by a negative number.

TIP:

⫺2x ⫺ 5 6 2 ⫺5 6 2x ⫹ 2 ⫺7 6 2x

Subtract 2 from both sides.

2x ⫺7 6 2 2

Divide by 2 (because 2 is positive, do not reverse the inequality sign).

⫺

7 6 x 2

(Note that the inequality ⫺72 6 x is equivalent to x 7 ⫺72.2

Skill Practice Solve the inequality. Graph the solution set and write the solution in interval notation.

Answer

2. ⫺4x ⫺ 12 ⱖ 20

2. 1⫺⬁, ⫺84

Add 2x to both sides.

⫺8

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Example 3

79

Linear Inequalities in One Variable

Solving a Linear Inequality

Solve the inequality. Graph the solution and write the solution set in interval notation. 61x  32  2  21x  82

Solution: 61x  32  2  21x  82 6x  18  2  2x  16

Apply the distributive property.

6x  18  18  2x

Combine like terms.

6x  2x  18  18  2x  2x

Add 2x to both sides.

4x  18  18 4x  18  18  18  18

Subtract 18 from both sides.

4x  0 4x 0  4 4

Divide by 4 (reverse the inequality sign).

x0 Graph

Interval Notation

5 4 3 2 1

0

1

2

3

4

5

1, 0 4

Skill Practice Solve the inequality. Graph the solution and write the solution set in interval notation. 3. 513x  12 6 415x  52

Example 4

Solving a Linear Inequality

5x  2 7 x  2. Graph the solution and write the solution 3 set in interval notation. Solve the inequality

Solution: 5x  2 7 x2 3 3a

5x  2 b 6 31x  22 3

Multiply by 3 to clear fractions (reverse the inequality sign).

5x  2 6 3x  6 2x  2 6 6

Add 3x to both sides.

2x 6 8

Subtract 2 from both sides.

8 2x 7 2 2

Divide by 2 (the inequality sign is reversed again).

x 7 4

Simplify.

Answer 3.

( 15, 2

5

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Graph

Interval Notation

3 2 1

0

1

2

3

(

4

5

14, 2

6

Skill Practice Solve the inequality. Graph the solution set and write the solution in interval notation. 4.

x1  x  1 3

In Example 4, the inequality sign was reversed twice: once for multiplying the inequality by 3 and once for dividing by 2. If you are in doubt about whether you have the inequality sign in the correct direction, you can check your final answer by using the test point method. That is, pick a point in the proposed solution set, and verify that it makes the original inequality true. Furthermore, any test point picked outside the solution set should make the original inequality false. 3 2 1

0

1

Pick x ⴝ 0 as a test point

2

3

3

6

5x  2 7 x2 3

7 102  2 ?

2 ? 7 2 3

5

Pick x ⴝ 5 as a test point

5x  2 7 x2 3 5102  2

(

4

5152  2 3

7 152  2 ?

23 ? 7 7 3

False

?

723 7 7

True

Because a test point to the right of x 4 makes the inequality true, we have shaded the correct part of the number line.

2. Applications of Inequalities Example 5

Solving a Linear Inequality Application

Beth received grades of 97%, 82%, 89%, and 99% on her first four algebra tests. To earn an A in the course, she needs an average of 90% or more. What scores can she receive on the fifth test to earn an A?

Solution: Let x represent the score on the fifth test. The average of the five tests is given by

Answer 4. 2 3 2, 2

97  82  89  99  x 5

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To earn an A, we have:

1Average of test scores2  90

Verbal model

97  82  89  99  x  90 5

Mathematical model

367  x  90 5 5a

Simplify the numerator.

367  x b  51902 5 367  x  450

Clear fractions. Simplify.

x  83 To earn an A, Beth would have to score at least 83% on her fifth test. Skill Practice 5. Jamie is a salesman who works on commission, so his salary varies from month to month. To qualify for an automobile loan, his monthly salary must average at least $2100 for 6 months. His salaries for the past 5 months have been $1800, $2300, $1500, $2200, and $2800. What amount does he need to earn in the last month to qualify for the loan?

Example 6

Solving a Linear Inequality Application

The number of registered passenger cars, N (in millions), in the United States has risen since 1960 according to the equation N 2.5t  64.4, where t represents the number of years after 1960 (t 0 corresponds to 1960, t 1 corresponds to 1961, and so on) (Figure 1-5).

N

Number of cars (millions)

200 150

Number of Registered Passenger Cars, United States N 2.5t  64.4

100 50 0 0

10 20 30 40 Year (t 0 corresponds to 1960)

50

t

Figure 1-5 Source: U.S. Department of Transportation

For what years was the number of registered passenger cars less than 89.4 million?

Answer 5. Jamie’s salary must be at least $2000.

81

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Solution: We require N 6 89.4 million. N

6 89.4

d

2.5t ⫹ 64.4 6 89.4

Substitute the expression 2.5t ⫹ 64.4 for N.

2.5t ⫹ 64.4 ⫺ 64.4 6 89.4 ⫺ 64.4

Subtract 64.4 from both sides.

2.5t 6 25 2.5t 25 6 2.5 2.5

Divide both sides by 2.5. t ⫽ 10 corresponds to the year 1970.

t 6 10

Before 1970, the number of registered passenger cars was less than 89.4 million. Skill Practice

Answer 6. The population was less than 417 thousand for t 6 30. This corresponds to the years before 1980.

Section 1.4

6. The population of Alaska has steadily increased since 1950 according to the equation P ⫽ 10t ⫹ 117, where t represents the number of years after 1950 and P represents the population in thousands. For what years since 1950 was the population less than 417 thousand people?

Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. Look over the notes that you took today. Do you understand what you wrote? If there were any rules, definitions, or formulas, highlight them so that they can be easily found when studying for the test. You may want to begin by highlighting the rule indicating when the direction of an inequality sign must be reversed. 2. Define the key terms. a. Linear inequality

b. Test point method

Review Exercises For Exercises 3–4, solve the equation. 3. 4 ⫹ 514 ⫺ 2x2 ⫽ ⫺21x ⫺ 12 ⫺ 4 5. Solve for v.

d ⫽ vt ⫺ 16t2

7. a. The area of a triangle is given by A ⫽ 12 bh. Solve for h. b. If the area of a triangle is 10 cm2 and the base is 3 cm, find the height.

4.

1 1 2 3 1 1 t⫺ ⫺ t⫹ ⫽ t⫹ 5 2 10 5 10 2

6. Solve for y.

5x ⫹ 3y ⫹ 6 ⫽ 0

8. Five more than 3 times a number is 6 less than twice the number. Find the number.

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Linear Inequalities in One Variable

83

Concept 1: Solving Linear Inequalities For Exercises 9–46, solve the inequalities. Graph the solution and write the solution set in interval notation. Check each answer by using the test point method. (See Examples 1–4.) 9. 2y  6  4

10. 3y  11 7 5

11. 2x  5  25

13. 6z  3 7 16

14. 8w  2  13

1 16. 4  p 5

17.

19. 0.8a  0.5  0.3a  11

20. 0.2w  0.7 6 0.4  0.9w

21. 5x  7 6 22

22. 3w  6 7 9

3 5 23.  x   6 4

3 21 24.  y 7  2 16

25.

27. 0.2t  1 7 2.4t  10

1 28. 20  8  x 3

29. 3  41y  22  6  412y  12

30. 1  41b  22 6 21b  52  4

31. 7.2k  5.1  5.7

32. 6h  2.92  16.58

33. 6p  1 7 17

34. 4y  1  11

35.

2 36.  a  3 7 5 5

37. 1.2b  0.4  0.4b

38. 0.4t  1.2 6 2

3 5 39.  c   2c 4 4

2 1 1 40.  q  7 q 3 3 2

41. 4  41y  22 6 5y  6

42. 6  61k  32  4k  12

43. 612x  12 6 5  1x  42  6x

44. 214p  32  p  5  31p  32

45. 6a  19a  12  31a  12  2

46. 81q  12  12q  12  5 7 12

12. 4z  2 7 22

15. 8 7

18.

2 t 3

2 12x  12 7 10 5

3p  1 7 5 2

26.

3 18y  92 6 3 4

3k  2 4 5

3 x81 4

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Concept 2: Applications of Inequalities 47. Nadia received quiz grades of 80%, 86%, 73%, and 91%. (See Example 5.) a. What grade would she need to make on the fifth quiz to get a B average, that is, at least 80% but less than 90%? b. Is it possible for Nadia to get an A average for her quizzes (at least 90%)? 48. Ty received test grades of 78%, 75%, 71%, 83%, and 73%. a. What grade would he need to make on the sixth test to get a C if a C is at least 75% but less than 80%? b. Is it possible for Ty to get a B or better for his test average (at least 80%)? For Exercises 49–52, use the graph that shows the average height for boys based on age. Let a represent a boy’s age (in years) and let h represent his height( in inches). (See Example 6.)

50. Determine the age range for which the average height of boys is greater than or equal to 41 in.

Height (in.)

49. Determine the age range for which the average height of boys is at least 51 in.

Average Height for Boys Based on Age

60

h 2.5a  31

50 40 30 20 10 0

0

1

2

3

51. Determine the age range for which the average height of boys is no more than 46 in.

4 5 6 Age (yr)

7

8

9

10

52. Determine the age range for which the average height of boys is at most 53.5 in. 53. Nolvia sells copy machines, and her salary is $25,000 plus a 4% commission on sales. The equation S 25,000  0.04x represents her salary S in dollars in terms of her total sales x in dollars. a. How much money in sales does Nolvia need to earn a salary that exceeds $40,000? b. How much money in sales does Nolvia need to earn a salary that exceeds $80,000? c. Why is the money in sales required to earn a salary of $80,000 more than twice the money in sales required to earn a salary of $40,000? 54. The amount of money A in a savings account depends on the principal P, the interest rate r, and the time in years t that the money is invested. The equation A P  Prt shows the relationship among the variables for an account earning simple interest. If an investor deposits $5000 at 6 12% simple interest, the account will grow according to the formula A 5000  500010.0652t. a. How many years will it take for the investment to exceed $10,000? (Round to the nearest tenth of a year.) b. How many years will it take for the investment to exceed $15,000? (Round to the nearest tenth of a year.) 55. The revenue R for selling x fleece jackets is given by the equation R 49.95x. The cost to produce x jackets is C 2300  18.50x. Find the number of jackets that the company needs to sell to produce a profit. (Hint: A profit occurs when revenue exceeds cost.) 56. The revenue R for selling x mountain bikes is R 249.95x. The cost to produce x bikes is C 56,000  140x. Find the number of bikes that the company needs to sell to produce a profit.

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Compound Inequalities

85

Expanding Your Skills For Exercises 57–60, assume a ⬎ b. Determine which inequality sign (⬎ or ⬍) should be inserted to make a true statement. Assume a ⫽ 0 and b ⫽ 0. 57. a ⫹ c ______ b ⫹ c, 59. ac ______ bc,

58. a ⫹ c ______ b ⫹ c,

for c 7 0

60. ac ______ bc,

for c 6 0

for c 6 0

for c 7 0

Compound Inequalities

Section 1.5

1. Union and Intersection of Sets

Concepts

Two or more sets can be combined by the operations of union and intersection.

1. Union and Intersection of Sets 2. Solving Compound Inequalities: And 3. Solving Inequalities of the Form a ⬍ x ⬍ b 4. Solving Compound Inequalities: Or 5. Applications of Compound Inequalities

DEFINITION A Union B and A Intersection B The union of sets A and B, denoted A  B, is the set of elements that belong to set A or to set B or to both sets A and B. The intersection of two sets A and B, denoted A  B, is the set of elements common to both A and B.

The concepts of the union and intersection of two sets are illustrated in Figures 1-6 and 1-7. A

B

B

AB A union B The elements in A or B or both

AB A intersection B The elements in A and B

Figure 1-6

Figure 1-7

Example 1

Finding the Union and Intersection of Sets

Given the sets: A ⫽ {a, b, c, d, e, f} Find:

A

a. A ´ B

b. A ¨ B

B ⫽ {a, c, e, g, i, k}

C ⫽ {g, h, i, j, k}

c. A ¨ C

Solution: a. A ´ B ⫽ {a, b, c, d, e, f, g, i, k}

The union of A and B includes all the elements of A along with all the elements of B. Notice that the elements a, c, and e are not listed twice.

b. A ¨ B ⫽ {a, c, e}

The intersection of A and B includes only those elements that are common to both sets.

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TIP: The empty set may

c. A ¨ C  { } (the empty set)

Because A and C share no common elements, the intersection of A and C is the empty set (also called the null set).

be denoted by the symbol { } or by the symbol .

Skill Practice Given: A  {r, s, t, u, v, w} Find:

1. B ´ C

Example 2 Given the sets:

2. A ¨ B

B  {s, v, w, y, z}

C  {x, y, z}

3. A ¨ C

Finding the Union and Intersection of Sets A  {x 0 x  3}

B  {x 0 x 2}

C  {x 0 x 5}

Graph the following sets. Then express each set in interval notation. a. A ¨ B

b. A ´ C

Solution: It is helpful to visualize the graphs of individual sets on the number line before taking the union or intersection. a. Graph of A  {x 0 x  3} Graph of B  {x 0 x 2}

6 5 4 3 2 1

0

1

2

6 5 4 3 2 1

0

1

2

6 5 4 3 2 1

0

1

2

Graph of A ¨ B (the “overlap”)

(

3

4

5

6

3

4

5

6

4

5

6

(

3

Interval notation: [2, 3) Note that the set A ¨ B represents the real numbers greater than or equal to 2 and less than 3. This relationship can be written more concisely as a compound inequality: 2 x  3. We can interpret this inequality as “x is between 2 and 3, including x  2.” b. Graph of A  {x 0 x  3} Graph of C  {x 0 x 5} Graph of A ´ C Interval notation: (, 3) ´ [5,  )

6 5 4 3 2 1

0

1

2

6 5 4 3 2 1

0

1

2

6 5 4 3 2 1

0

1

2

(

3

4

5

6

3

4

5

6

4

5

6

(

3

A  C includes all elements from set A along with the elements from set C.

Skill Practice Given the sets: A  5x 0 x 16, B  5x 0 x 6 46, and C  5x 0 x 96, determine the union or intersection and express the answer in interval notation. 4. A ¨ B

5. B ´ C

In Example 3, we find the union and intersection of sets expressed in interval notation. Answers 1. {s, v, w, x, y, z } 2. {s, v, w } 3. { } 4. [1, 4) 5. (, 4) ´ [9, )

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Section 1.5

Example 3

Compound Inequalities

Finding the Union and Intersection of Two Intervals

Find the union or intersection as indicated. Write the answer in interval notation. a. 1⫺⬁, ⫺22 ´ 3 ⫺4, 32

b. 1⫺⬁, ⫺22 ¨ 3⫺4, 32

Solution:

a. 1⫺⬁, ⫺22 ´ 3⫺4, 32

To find the union, graph each interval separately. The union is the collection of real numbers that lie in the first interval, the second interval, or both intervals. 4

5

1⫺⬁, ⫺22

4

5

3⫺4, 32

3

4

5

3

4

5

1⫺⬁, ⫺22 3⫺4, 32

) ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

2

)

)

The union is 1⫺⬁, 32. b. 1⫺⬁, ⫺22 傽 3 ⫺4, 32 )

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

) ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

0

1

2

3

4

5

) ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

The union consists of all real numbers in the red interval along with the real numbers in the blue interval: 1⫺⬁, 32

The intersection is the “overlap” of the two intervals: [⫺4, ⫺2).

The intersection is [⫺4, ⫺2). Skill Practice Find the union or intersection. Write the answer in interval notation. 6. 1⫺⬁, ⫺5 4 ´ 1⫺7, 02

7. 1⫺⬁, ⫺5 4 傽 1⫺7, 02

2. Solving Compound Inequalities: And The solution to two inequalities joined by the word and is the intersection of their solution sets. The solution to two inequalities joined by the word or is the union of their solution sets.

PROCEDURE Solving a Compound Inequality Step 1 Solve and graph each inequality separately. Step 2 • If the inequalities are joined by the word and, find the intersection of the two solution sets. • If the inequalities are joined by the word or, find the union of the two solution sets. Step 3 Express the solution set in interval notation or in set-builder notation. Answers 6. (⫺⬁, 0)

7. (⫺7, ⫺5]

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As you work through the examples in this section, remember that multiplying or dividing an inequality by a negative factor reverses the direction of the inequality sign. Example 4

Solving a Compound Inequality: And

Solve the compound inequality. ⫺2x 6 6 and

x⫹5ⱕ7

Solution: ⫺2x 6 6

and

x⫹5ⱕ7

⫺2x 6 7 ⫺2 ⫺2

and

xⱕ2

x 7 ⫺3

and

xⱕ2

Solve each inequality separately. Reverse the first inequality sign.

( ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

5x 0 x 7 ⫺36

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

5x 0 x ⱕ 26

( ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

Take the intersection of the solution sets: 5x 0 ⫺3 6 x ⱕ 26

The solution is 5x 0 ⫺3 6 x ⱕ 26, or equivalently in interval notation, 1⫺3, 2 4. Skill Practice Solve the compound inequality. 8. 5x ⫹ 2 ⱖ ⫺8 and ⫺4x 7 ⫺24

Example 5

Solving a Compound Inequality: And

Solve the compound inequality. 4.4a ⫹ 3.1 6 ⫺12.3 and

⫺2.8a ⫹ 9.1 6 ⫺6.3

Solution: 4.4a ⫹ 3.1 6 ⫺12.3

and

4.4a 6 ⫺15.4

and

⫺2.8a 6 ⫺15.4

Solve each inequality separately.

4.4a ⫺15.4 6 4.4 4.4

and

⫺2.8a ⫺15.4 7 ⫺2.8 ⫺2.8

Reverse the second inequality sign.

a 6 ⫺3.5

Answer

8. 5x 0 ⫺2 ⱕ x 6 66; 3 ⫺2, 62

⫺2.8a ⫹ 9.1 6 ⫺6.3

a 7 5.5

and

( ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6 (

5a 0 a 6 ⫺3.56

6

5a 0 a 7 5.56

6

The intersection of the solution sets is the empty set: { }

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Section 1.5

Compound Inequalities

There are no real numbers that are simultaneously less than 3.5 and greater than 5.5. There is no solution. The solution set is 5 6.

Skill Practice Solve the compound inequality. 9. 3.2y  2.4 7 16.8 and 4.1y 8.2

Example 6

Solving a Compound Inequality: And

Solve the compound inequality. 2  x 6 and 3

Solution: 2  x 6 3 3 2 3  a xb  162 2 3 2 x 9

1  x 6 1 2

and

1  x 6 1 2

and

1 2a xb 7 2112 2 x 7 2

and

Solve each inequality separately.

10 9 8 7 6 5 4 3 2 1

0

1

2

3

5x 0 x 96

) 10 9 8 7 6 5 4 3 2 1

0

1

2

3

5x 0 x 7 26

) 10 9 8 7 6 5 4 3 2 1

0

1

2

3

Take the intersection of the solution sets: 5x 0 x 7 26

The solution set is 5x 0 x 7 26 , or in interval notation, 12, 2. Skill Practice Solve the compound inequality. 5 1 1 10.  z 6 and z  1 3 4 8 2

3. Solving Inequalities of the Form a < x < b An inequality of the form a 6 x 6 b is a type of compound inequality, one that defines two simultaneous conditions on x. a

a 6 x

x

and

b

x 6 b

The solution set to the compound inequality a 6 x 6 b is the intersection of the solution sets to the inequalities a 6 x and x 6 b.

Answers

9. 5 6 10. 5z 0 z 46; 34,  2

89

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Chapter 1 Linear Equations and Inequalities in One Variable

Solving an Inequality of the Form a < x < b

Example 7

⫺4 6 3x ⫹ 5 ⱕ 10

Solve the inequality.

Solution: ⫺4 6 3x ⫹ 5 ⱕ 10 ⫺4 6 3x ⫹ 5

3x ⫹ 5 ⱕ 10

and

⫺9 6 3x

and

3x ⱕ 5

⫺9 3x 6 3 3

and

3x 5 ⱕ 3 3

⫺3 6 x

and

xⱕ

⫺3 6 x ⱕ ) ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

Set up the intersection of two inequalities. Solve each inequality.

5 3

5 3

Take the intersection of the solution sets.

2

3

4

5

5 3

The solution is 5x 0 ⫺3 6 x ⱕ 53 6, or equivalently in interval notation, 1⫺3, 53 4. Skill Practice Solve the inequality. 11. ⫺6 ⱕ 2x ⫺ 5 6 1

To solve an inequality of the form a < x < b, we can also work with the inequality as a “three-part” inequality and isolate x. This is demonstrated in Example 8.

Solving an Inequality of the Form a < x < b

Example 8

2ⱖ

Solve the inequality.

p⫺2 ⱖ ⫺1 ⫺3

Solution: 2ⱖ

p⫺2 ⱖ ⫺1 ⫺3

⫺3122 ⱕ ⫺3a

Isolate the variable in the middle part.

p⫺2 b ⱕ ⫺31⫺12 ⫺3

Multiply all three parts by ⫺3. Remember to reverse the inequality signs.

⫺6 ⱕ p ⫺ 2 ⱕ 3

Simplify.

⫺6 ⫹ 2 ⱕ p ⫺ 2 ⫹ 2 ⱕ 3 ⫹ 2

Add 2 to all three parts to isolate p.

⫺4 ⱕ p ⱕ 5 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

The solution is 5p 0 ⫺4 ⱕ p ⱕ 56, or equivalently in interval notation [⫺4, 5]. Skill Practice Solve the inequality. Answers

11. 5x 0 ⫺12 ⱕ x 6 36; 3 ⫺12 , 32 12. 5t 0 ⫺20 6 t 6 66; 1⫺20, 62

12. 8 7

t⫹4 7 ⫺5 ⫺2

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Section 1.5

Compound Inequalities

4. Solving Compound Inequalities: Or In Examples 9 and 10, we solve compound inequalities that involve inequalities joined by the word “or.” In such a case, the solution to the compound inequality is the union of the solution sets of the individual inequalities. Example 9

Solving a Compound Inequality: Or

Solve the compound inequality. ⫺3y ⫺ 5 7 4 or

4⫺yⱕ6

Solution: ⫺3y ⫺ 5 7 4

or

4⫺yⱕ6

⫺3y 7 9

or

⫺y ⱕ 2

⫺3y 9 6 ⫺3 ⫺3

or

⫺y 2 ⱖ ⫺1 ⫺1

y 6 ⫺3

or

y ⱖ ⫺2

Solve each inequality separately. Reverse the inequality signs.

( ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

5y 0 y 6 ⫺36

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

5y 0 y ⱖ ⫺26

( ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

Take the union of the solution sets: 5y 0 y 6 ⫺3 or y ⱖ ⫺26

The solution is 5y 0 y 6 ⫺3 or y ⱖ ⫺26 or, equivalently in interval notation, 1⫺⬁, ⫺32 ´ 3⫺2, ⬁2. Skill Practice Solve the compound inequality. 13. ⫺10t ⫺ 8 ⱖ 12 or 3t ⫺ 6 7 3

Example 10

Solving a Compound Inequality: Or

Solve the compound inequality. 4x ⫹ 3 6 16 or

⫺2x 6 3

Solution: 4x ⫹ 3 6 16

or

4x 6 13

or

x 7 ⫺

3 2

13 4

or

x 7 ⫺

3 2

x 6

⫺2x 6 3 Solve each inequality separately.

Answer

13. 5t 0 t ⱕ ⫺2 or t 7 36; 1⫺⬁, ⫺2 4 ´ 13, ⬁2

91

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⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

(

3

4

5

6

5x 0 x 6

13 46

5x 0 x 7 ⫺32 6

13 4

( ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

0

1

2

3

4

5

6

⫺32 ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

Take the union of the solution sets.

The union of the solution sets is {x 0 x is a real number}, or equivalently, 1⫺⬁, ⬁2. Skill Practice Solve the compound inequality. 14. x ⫺ 7 7 ⫺2 or ⫺6x 7 ⫺48

5. Applications of Compound Inequalities Compound inequalities are used in many applications, as shown in Examples 11 and 12. Example 11

Translating Compound Inequalities

The normal level of thyroid-stimulating hormone (TSH) for adults ranges from 0.4 to 4.8 microunits per milliliter 1mU/mL2 , inclusive. Let x represent the amount of TSH measured in microunits per milliliter. a. Write an inequality representing the normal range of TSH, inclusive. b. Write a compound inequality representing abnormal TSH levels.

Solution: a. 0.4 ⱕ x ⱕ 4.8

b. x 6 0.4

or

x 7 4.8

Skill Practice The length of a normal human pregnancy, w, is from 37 to 41 weeks, inclusive. 15. Write an inequality representing the normal length of a pregnancy. 16. Write a compound inequality representing an abnormal length for a pregnancy.

TIP: • In mathematics, the word “between” means strictly between two values. That is, the endpoints are excluded. Example: x is between 4 and 10 1 (4, 10). • If the word “inclusive” is added to the statement, then we include the endpoints. Example: x is between 4 and 10, inclusive 1 [4, 10].

Answers

14. 5x 0 x is a real number 6; 1⫺⬁, ⬁2 15. 37 ⱕ w ⱕ 41 16. w 6 37 or w 7 41

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Example 12

Compound Inequalities

93

Translating and Solving a Compound Inequality

The sum of a number and 4 is between ⫺5 and 12. Find all such numbers.

Solution: Let x represent a number. ⫺5 6 x ⫹ 4 6 12

Translate the inequality.

⫺5 ⫺ 4 6 x ⫹ 4 ⫺ 4 6 12 ⫺ 4

Subtract 4 from all three parts of the inequality.

⫺9 6 x 6 8

The number may be any real number between ⫺9 and 8: 5x 0 ⫺9 6 x 6 86. Skill Practice 17. The sum of twice a number and 11 is between 21 and 31. Find all such numbers.

Section 1.5

Answer 17. Any real number between 5 and 10: 5n 0 5 6 n 6 106

Practice Exercises

Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercises 1. Which activities might you try when working in a study group to help you learn and understand the material? Quiz one another by asking one another questions. Practice teaching one another. Share and compare class notes. Support and encourage one another. Work together on exercises and sample problems. 2. Define the key terms. a. Compound inequality

b. Intersection

c. Unionc

Review Exercises For Exercises 3–6, solve the linear inequality. Write the solution in interval notation. 3. ⫺6u ⫹ 8 ⬎ 2

4. 2 ⫺ 3z ⱖ ⫺4

Concept 1: Union and Intersection of Sets 7. Given: M ⫽ {⫺3, ⫺1, 1, 3, 5} and N ⫽ {⫺4, ⫺3, ⫺2, ⫺1, 0}. (See Example 1.)

3 5. ⫺12 ⱕ p 4

6. 5 7

1 w 3

8. Given: P ⫽ 5a, b, c, d, e, f, g, h, i6 and Q ⫽ 5a, e, i, o, u6 .

List the elements of the following sets:

List the elements of the following sets.

a. M ¨ N

a. P ¨ Q

b. M ´ N

b. P ´ Q

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For Exercises 9–20, refer to the sets A, B, C, and D. Determine the union or intersection as indicated. Express the answer in interval notation, if possible. (See Example 2.) A  5x 0 x 6 46, B  5x 0 x 7 26, C  5x 0 x 76, D  5x 0 0 x 6 56

9. A  C

10. B  C

11. A ´ B

12. A ´ D

13. A  B

14. A  D

15. B ´ C

16. B ´ D

17. C  D

18. B  D

19. C ´ D

20. A ´ C

For Exercises 21–26, find the intersection and union of sets as indicated. Write the answers in interval notation. (See Example 3.)

21. a. 12, 52 ¨ 31,  2

22. a. 1, 42 ¨ 31, 52

5 9 23. a. a , 3b ¨ a1, b 2 2

b. 12, 52 ´ 31,  2

b. 1, 42 ´ 31, 52

5 9 b. a , 3b ´ a1, b 2 2

24. a. 13.4, 1.62 ¨ 12.2, 4.12

b. 13.4, 1.62 ´ 12.2, 4.12

25. a. 14, 5 4 ¨ 10, 24

b. 14, 5 4 ´ 10, 24

26. a. 31, 52 ¨ 10, 32

b. 31, 52 ´ 10, 32

Concept 2: Solving Compound Inequalities: And For Exercises 27–36, solve the inequality and graph the solution. Write the answer in interval notation. (See Examples 4–6.) 27. y  7 9 and

y2 5

28. a  6 7 2 and

5a 6 30

29. 2t  7 6 19 and 5t  13 7 28

30. 5p  2p 21 and 9p  3p 24

31. 2.1k  1.1 0.6k  1.9 and

32. 0.6w  0.1 7 0.3w  1.1 and

0.3k  1.1 6 0.1k  0.9

33.

2 12p  12 10 and 3

35. 2 6 x  12 and

2.3w  1.5 0.3w  6.5

4 13p  42 20 5

34.

5 1a  22 6 6 and 2

3 1a  22 6 1 4

14 6 51x  32  6x

36. 8 3y  2 and 31y  72  16 7 4y

Concept 3: Solving Inequalities of the Form a < x < b 37. Write 4 t 6

3 4

as two separate inequalities.

38. Write 2.8 6 y 15 as two separate inequalities.

39. Explain why 6 6 x 6 2 has no solution.

40. Explain why 4 6 t 6 1 has no solution.

41. Explain why 5 7 y 7 2 has no solution.

42. Explain why 3 7 w 7 1 has no solution.

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For Exercises 43–54, solve the inequality and graph the solution set. Write the answer in interval notation. (See Examples 7–8.) a

1 6

43. 0 2b  5 6 9

44. 6 6 3k  9 0

1 46. 3 x 6 0 2

47. 

49. 5 3x  2 8

50. 1 6 2x  4 5

51. 12 7 6x  3 0

52. 4 2x  5 7 7

53. 0.2 6 2.6  7t 6 4

54. 1.5 6 0.1x 8.1

45. 1 6

y4 2 1 6 6 3 6 3

48.

1 t4 7 7 2 3 3

Concept 4: Solving Compound Inequalities: Or For Exercises 55–64, solve the inequality and graph the solution set. Write the answer in interval notation. (See Examples 9–10.)

55. 2y  1 3 or y 6 2

56. x 6 0 or

57. 1 7 6z  8 or

58. 22 7 4t  10 or

7 7 2t  5

60. p  7 10 or

31p  12 12

8z  6 10

59. 51x  12 5 or

61.

5 v 5 or 3

5  x 11

v  6 6 1

63. 0.5w  5 6 2.5w  4 or

62.

0.3w 0.1w  1.6

3 u1 7 0 8

3x  1 7

or

2u 4

64. 1.25a  3 0.5a  6 or

2.5a  1 9  1.5a

Mixed Exercises For Exercises 65–74, solve the inequality. Write the answer in interval notation. 65. a. 3x  5 6 19 and b. 3x  5 6 19 or

2x  3 6 23 2x  3 6 23

67. a. 8x  4 6.4 or 0.31x  62 0.6 b. 8x  4 6.4 and 69. 4

2  4x 6 8 3

0.31x  62 0.6

66. a. 0.516x  82 7 0.8x  7 and 41x  12 6 7.2 b. 0.516x  82 7 0.8x  7 68. a. 2r  4 8 or b. 2r  4 8 and 70. 1 6

3x

0 2

or

41x  12 6 7.2

3r  5 8 3r  5 8

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71. 5 ⱖ ⫺41t ⫺ 32 ⫹ 3t 6 6 12t ⫹ 814 ⫺ t2 73.

⫺x ⫹ 3 4⫹x 7 2 5

72. 3 7 ⫺1w ⫺ 32 ⫹ 4w or ⫺5 ⱖ ⫺31w ⫺ 52 ⫹ 6w

or

or

1⫺x 2⫺x 7 4 3

74.

y⫺7 1 6 ⫺3 4

or

y⫹1 1 7 ⫺ ⫺2 3

Concept 5: Applications of Compound Inequalities 75. The normal number of white blood cells for human blood is between 4800 and 10,800 cells per cubic millimeter, inclusive. Let x represent the number of white blood cells per cubic millimeter. (See Example 11.) a. Write an inequality representing the normal range of white blood cells per cubic millimeter. b. Write a compound inequality representing abnormal levels of white blood cells per cubic millimeter. 76. Normal hemoglobin levels in human blood for adult males are between 13 and 16 grams per deciliter (g/dL), inclusive. Let x represent the level of hemoglobin measured in grams per deciliter. a. Write an inequality representing normal hemoglobin levels for adult males. b. Write a compound inequality representing abnormal levels of hemoglobin for adult males. 77. The normal number of platelets in human blood is between 200,000 and 350,000 platelets per cubic millimeter, inclusive. Let x represent the number of platelets per cubic millimeter. a. Write an inequality representing a normal platelet count per cubic millimeter. b. Write a compound inequality representing abnormal platelet counts per cubic millimeter. 78. Normal hemoglobin levels in human blood for adult females are between 12 and 15 g/dL, inclusive. Let x represent the level of hemoglobin measured in grams per deciliter. a. Write an inequality representing normal hemoglobin levels for adult females. b. Write a compound inequality representing abnormal levels of hemoglobin for adult females. 79. Twice a number is between ⫺3 and 12. Find all such numbers. (See Example 12.) 80. The difference of a number and 6 is between 0 and 8. Find all such numbers. 81. One plus twice a number is either greater than 5 or less than ⫺1. Find all such numbers. 82. One-third of a number is either less than ⫺2 or greater than 5. Find all such numbers. 83. Amy knows from reading her syllabus in intermediate algebra that the average of her chapter tests accounts for 80% (0.8) of her overall course grade. She also knows that the final exam counts as 20% (0.2) of her grade. a. Suppose that the average of Amy’s chapter tests is 92%. Determine the range of grades that she would need on her final exam to get an “A” in the class. (Assume that a grade of “A” is obtained if Amy’s overall average is 90% or better.) b. Determine the range of grades that Amy would need on her final exam to get a “B” in the class. (Assume that a grade of “B” is obtained if Amy’s overall average is at least 80% but less than 90%.)

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97

84. Robert knows from reading his syllabus in intermediate algebra that the average of his chapter tests accounts for 60% (0.6) of his overall course grade. He also knows that the final exam counts as 40% (0.4) of his grade. a. Suppose that the average of Robert’s chapter tests is 89%. Determine the range of grades that he would need on his final exam to get an “A” in the class. (Assume that a grade of “A” is obtained if Robert’s overall average is 90% or better.) b. Determine the range of grades that Robert would need on his final exam to get a “B” in the class. (Assume that a grade of “B” is obtained if Robert’s overall average is at least 80% but less than 90%.) 85. The average high and low temperatures for Vancouver, British Columbia, in January are 5.6⬚C and 0⬚C, respectively. The formula relating Celsius temperatures to Fahrenheit temperatures is given by C ⫽ 59 1F ⫺ 322. Convert the inequality 0.0° ⱕ C ⱕ 5.6° to an equivalent inequality using Fahrenheit temperatures. 86. For a day in July, the temperatures in Austin, Texas, ranged from 20⬚C to 29⬚C. The formula relating Celsius temperatures to Fahrenheit temperatures is given by C ⫽ 59 1F ⫺ 322. Convert the inequality 20° ⱕ C ⱕ 29° to an equivalent inequality using Fahrenheit temperatures.

Absolute Value Equations

Section 1.6

1. Solving Absolute Value Equations

Concepts

An equation of the form 0x 0 ⫽ a is called an absolute value equation. For example, consider the equation 0x 0 ⫽ 4. From the definition of absolute value, the solutions are found by solving the equations x ⫽ 4 and ⫺x ⫽ 4. This gives the equivalent equations x ⫽ 4 and x ⫽ ⫺4. Also recall from Section R.3 that the absolute value of a number is its distance from zero on the number line. Therefore, geometrically, the solutions to the equation 0x 0 ⫽ 4 are the values of x that are 4 units from zero on the number line (Figure 1-8). 0x 0 ⫽ 4

x ⫽ ⫺4

or

4 units

x⫽4

⫺4 ⫺3 ⫺2 ⫺1

4 units

0

1

2

3

4

Figure 1-8

PROCEDURE Solving Absolute Value Equations of the Form 0 x 0 ⴝ a If a is a real number, then

• If a ⱖ 0, the solutions to the equation 0 x 0 ⫽ a are given by x ⫽ a and x ⫽ ⫺a. • If a 6 0, there is no solution to the equation 0x 0 ⫽ a.

1. Solving Absolute Value Equations 2. Solving Equations Containing Two Absolute Values

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To solve an absolute value equation of the form 0 x 0 ⫽ a 1a ⱖ 02, rewrite the equation as x ⫽ a or x ⫽ ⫺a.

Solving Absolute Value Equations

Example 1

Solve the absolute value equations. a. 0x 0 ⫽ 5

b. 0w 0 ⫺ 2 ⫽ 12

Solution: a.

x⫽5

0x 0 ⫽ 5 x ⫽ ⫺5

The solution set is 55, ⫺56. b. 0w 0 ⫺ 2 ⫽ 12

Rewrite the equation as x ⫽ a or x ⫽ ⫺a.

Isolate the absolute value to write the equation in the form 0 w 0 ⫽ a.

0w 0 ⫽ 14

or

d. 0 x 0 ⫽ ⫺6

The equation is in the form 0 x 0 ⫽ a, where a ⫽ 5.

or

w ⫽ 14

c. 0 p 0 ⫽ 0

w ⫽ ⫺14

The solution set is 514, ⫺146.

Rewrite the equation as w ⫽ a or w ⫽ ⫺a.

0p 0 ⫽ 0

c.

p⫽0

or

p ⫽ ⫺0

The solution set is 506 . d. 0x 0 ⫽ ⫺6

Rewrite as two equations. Notice that the second equation p ⫽ ⫺0 is the same as the first equation. Intuitively, p ⫽ 0 is the only number whose absolute value equals 0. This equation is of the form 0 x 0 ⫽ a, but a is negative. There is no number whose absolute value is negative.

No solution, 5 6

Skill Practice Solve the absolute value equations. 1. 0y 0 ⫽ 7

2. 0 v 0 ⫹ 6 ⫽ 10

3. 0w 0 ⫽ 0

4. 0z 0 ⫽ ⫺12

We have solved absolute value equations of the form 0 x 0 ⫽ a. Notice that x can represent any algebraic quantity. For example, to solve the equation 02w ⫺ 3 0 ⫽ 5, we still rewrite the absolute value equation as two equations. In this case, we set the quantity 2w ⫺ 3 equal to 5 and to ⫺5, respectively. 02w ⫺ 3 0 ⫽ 5

2w ⫺ 3 ⫽ 5 Answers 1. 57, ⫺76 3. 506

2. 54, ⫺46 4. 5 6

or

2w ⫺ 3 ⫽ ⫺5

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PROCEDURE Solving an Absolute Value Equation Step 1 Isolate the absolute value. That is, write the equation in the form 0x 0 ⫽ a, where a is a real number. Step 2 If a 6 0, there is no solution. Step 3 Otherwise, if a ⱖ 0, rewrite the absolute value equation as x ⫽ a or x ⫽ ⫺a. Step 4 Solve the individual equations from step 3. Step 5 Check the answers in the original absolute value equation.

Example 2

Solving an Absolute Value Equation 02w ⫺ 3 0 ⫽ 5

Solve the equation.

Solution:

02w ⫺ 3 0 ⫽ 5

The equation is already in the form 0x 0 ⫽ a, where x ⫽ 2w ⫺ 3.

2w ⫺ 3 ⫽ 5

or

2w ⫺ 3 ⫽ ⫺5

2w ⫽ 8

or

2w ⫽ ⫺2

w⫽4

or

w ⫽ ⫺1

Check: w ⫽ 4 02w ⫺ 3 0 ⫽ 5

Check: w ⫽ ⫺1 02w ⫺ 3 0 ⫽ 5

08 ⫺ 3 0 ⱨ 5

0⫺2 ⫺ 3 0 ⱨ 5

02 142 ⫺ 3 0 ⱨ 5

02 1⫺12 ⫺ 3 0 ⱨ 5

05 0 ⱨ 5 ✔

Rewrite as two equations. Solve each equation.

Check the solutions in the original equation.

0⫺5 0 ⱨ 5 ✔

The solution set is 54, ⫺16. Skill Practice Solve the equation. 5. 0 4x ⫹ 1 0 ⫽ 9

Example 3

Solving an Absolute Value Equation

Solve the equation.

02c ⫺ 5 0 ⫹ 6 ⫽ 2

Solution:

02c ⫺ 5 0 ⫹ 6 ⫽ 2

02c ⫺ 5 0 ⫽ ⫺4

No solution, 5 6

Avoiding Mistakes Isolate the absolute value. The equation is in the form 0 x 0 ⫽ a, where x ⫽ 2c ⫺ 5 and a ⫽ ⫺4. Because a 6 0, there is no solution.

Always isolate the absolute value first. Otherwise you will get answers that do not check.

There are no numbers c that will make an absolute value equal to a negative number.

Skill Practice Solve the equation. 6. 03z ⫹ 10 0 ⫹ 3 ⫽ 1

Answers 5 5. e 2, ⫺ f 2

6. No solution, 5 6

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Example 4

Solving an Absolute Value Equation 2 ⫺2 ` p ⫹ 3 ` ⫺ 7 ⫽ ⫺19 5

Solve the equation.

Solution: 2 ⫺2 ` p ⫹ 3 ` ⫺ 7 ⫽ ⫺19 5 2 ⫺2 ` p ⫹ 3 ` ⫽ ⫺12 5

Isolate the absolute value.

2 ⫺2 ` p ⫹ 3 ` 5 ⫺12 ⫽ ⫺2 ⫺2

Divide both sides by ⫺2.

2 ` p⫹3` ⫽6 5 2 p⫹3⫽6 5

or

2 p ⫹ 3 ⫽ ⫺6 5

2p ⫹ 15 ⫽ 30

or

2p ⫹ 15 ⫽ ⫺30

2p ⫽ 15

or

2p ⫽ ⫺45

15 2

or

p⫽⫺

p⫽

The solution set is e

45 2

Rewrite as two equations. Multiply by 5 to clear fractions.

Both solutions check in the original equation.

15 45 , ⫺ f. 2 2

Skill Practice Solve the equation. 3 7. 3 ` a ⫹ 1 ` ⫹ 2 ⫽ 14 2

Example 5

Solving an Absolute Value Equation 6.9 ⫽ 04.1 ⫺ p 0 ⫹ 6.9

Solve the equation.

Solution:

6.9 ⫽ 0 4.1 ⫺ p 0 ⫹ 6.9

0 4.1 ⫺ p 0 ⫹ 6.9 ⫽ 6.9

First write the absolute value on the left. Then subtract 6.9 from both sides to write the equation in the form 0 x 0 ⫽ a.

04.1 ⫺ p 0 ⫽ 0 4.1 ⫺ p ⫽ 0

Answer 7. e 2, ⫺

10 f 3

or

Isolate the absolute value. 4.1 ⫺ p ⫽ ⫺0

Rewrite as two equations. Notice that the equations are the same.

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Section 1.6

⫺p ⫽ ⫺4.1

Absolute Value Equations

101

Subtract 4.1 from both sides.

p ⫽ 4.1

Check: p ⫽ 4.1

0 4.1 ⫺ p 0 ⫹ 6.9 ⫽ 6.9

04.1 ⫺ 4.1 0 ⫹ 6.9 ⱨ 6.9 0 0 0 ⫹ 6.9 ⱨ 6.9

The solution set is 54.16.

6.9 ⱨ 6.9 ✔

Skill Practice Solve the equation. 8. ⫺3.5 ⫽ 0 1.2 ⫹ x 0 ⫺ 3.5

2. Solving Equations Containing Two Absolute Values

Some equations have two absolute values such as 0x 0 ⫽ 0y 0 . If two quantities have the same absolute value, then the quantities are equal or the quantities are opposites.

PROPERTY

Equality of Absolute Values

Example 6

0x 0 ⫽ 0 y 0 implies that x ⫽ y or x ⫽ ⫺y.

Solving an Equation Having Two Absolute Values 02w ⫺ 3 0 ⫽ 0 5w ⫹ 1 0

Solve the equation.

Solution:

02w ⫺ 3 0 ⫽ 05w ⫹ 1 0

2w ⫺ 3 ⫽ 5w ⫹ 1

2w ⫺ 3 ⫽ 5w ⫹ 1

or

2w ⫺ 3 ⫽ ⫺15w ⫹ 12

or

2w ⫺ 3 ⫽ ⫺5w ⫺ 1

⫺3w ⫺ 3 ⫽ 1

or

7w ⫺ 3 ⫽ ⫺1

⫺3w ⫽ 4

or

7w ⫽ 2

or

w⫽

w⫽⫺

4 3

2 7

Rewrite as two equations, x ⫽ y or x ⫽ ⫺y. Solve for w.

Avoiding Mistakes To take the opposite of the quantity 5w ⫹ 1, use parentheses and apply the distributive property.

Both values check in the original equation.

4 2 The solution set is e ⫺ , f . 3 7 Skill Practice Solve the equation. 9. 0 3 ⫺ 2x 0 ⫽ 03x ⫺ 1 0

Answers 8. 5⫺1.26

4 9. e , ⫺2 f 5

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Chapter 1 Linear Equations and Inequalities in One Variable

Example 7

Solving an Equation Having Two Absolute Values

Solve the equation.

0x ⫺ 4 0 ⫽ 0 x ⫹ 8 0

Solution:

0 x ⫺ 4 0 ⫽ 0x ⫹ 8 0 x⫺4⫽x⫹8 ⫺4 ⫽ 8

or

x ⫺ 4 ⫽ ⫺1x ⫹ 82

Rewrite as two equations, x ⫽ y or x ⫽ ⫺y.

or

x ⫺ 4 ⫽ ⫺x ⫺ 8

Solve for x.

2x ⫺ 4 ⫽ ⫺8 contradiction

2x ⫽ ⫺4 x ⫽ ⫺2

x ⫽ ⫺2 checks in the original equation.

The solution set is 5⫺26. Skill Practice Solve the equation.

Answer

10. 04t ⫹ 3 0 ⫽ 04t ⫺ 5 0

1 10. e f 4

Section 1.6

Practice Exercises • Practice Problems • Self-Tests • NetTutor

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• e-Professors • Videos

Study Skills Exercises 1. One way to know that you really understand a concept is to explain it to someone else. In your own words, explain how to solve an absolute value equation. 2. Define the key term absolute value equation.

Review Exercises For Exercises 3–6, solve the inequalities. Write the answers in interval notation. 3. 31a ⫹ 22 ⫺ 6 7 2 and 5. 5 ⱖ

x⫺4 7 ⫺3 ⫺2

⫺21a ⫺ 32 ⫹ 14 7 ⫺3

4. 3x ⫺ 5 ⱖ 7x ⫹ 3 or

2x ⫺ 1 ⱕ 4x ⫺ 5

1 6. 4 ⱕ x ⫺ 2 6 7 3

Concept 1: Solving Absolute Value Equations For Exercises 7–38, solve the absolute value equations. (See Examples 1–5.) 7. 0 p 0 ⫽ 7

8. 0q 0 ⫽ 10

9. 0x 0 ⫹ 5 ⫽ 11

10. 0x 0 ⫺ 3 ⫽ 20

11. 0y 0 ⫽ 12

12. 0y 0 ⫽ 15

13. 0 w 0 ⫺ 3 ⫽ ⫺5

14. 0w 0 ⫹ 4 ⫽ ⫺8

15. 03q 0 ⫽ 0

16. 04p 0 ⫽ 0

17. 0 3x ⫺ 4 0 ⫽ 8

18. 0 4x ⫹ 1 0 ⫽ 6

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Section 1.7

Absolute Value Inequalities

7z 1 ⫺ ` ⫹3⫽6 3 3

22. `

103

19. 5 ⫽ 0 2x ⫺ 4 0

20. 10 ⫽ 0 3x ⫹ 7 0

21. `

w 3 ⫹ ` ⫺2⫽7 2 2

23. 0 0.2x ⫺ 3.5 0 ⫽ ⫺5.6

24. 0 1.81 ⫹ 2x 0 ⫽ ⫺2.2

1 25. 1 ⫽ ⫺4 ⫹ ` 2 ⫺ w ` 4

26. ⫺12 ⫽ ⫺6 ⫺ 06 ⫺ 2x 0

27. 10 ⫽ 4 ⫹ 02y ⫹ 1 0

28. ⫺1 ⫽ ⫺ 0 5x ⫹ 7 0

29. ⫺2 0 3b ⫺ 7 0 ⫺ 9 ⫽ ⫺9

30. ⫺3 05x ⫹ 1 0 ⫹ 4 ⫽ 4

31. ⫺2 0x ⫹ 3 0 ⫽ 5

32. ⫺3 0 x ⫺ 5 0 ⫽ 7

33. 0 ⫽ 0 6x ⫺ 9 0

34. 7 ⫽ 04k ⫺ 6 0 ⫹ 7

1 1 9 35. ` ⫺ ⫺ k ` ⫽ 5 2 5

1 2 1 36. ` ⫺ ⫺ h ` ⫽ 6 9 2

37. ⫺3 0 2 ⫺ 6x 0 ⫹ 5 ⫽ ⫺10 38. 5 01 ⫺ 2x 0 ⫺ 7 ⫽ 3

Concept 2: Solving Equations Containing Two Absolute Values For Exercises 39–52, solve the absolute value equations. (See Examples 6–7.) 39. 04x ⫺ 2 0 ⫽ 0⫺8 0

40. 03x ⫹ 5 0 ⫽ 0 ⫺5 0

41. 0 4w ⫹ 3 0 ⫽ 0 2w ⫺ 5 0

42. 03y ⫹ 1 0 ⫽ 02y ⫺ 7 0

43. 02y ⫹ 5 0 ⫽ 0 7 ⫺ 2y 0

44. 0 9a ⫹ 5 0 ⫽ 0 9a ⫺ 1 0

45. `

4w ⫺ 1 2w 1 ` ⫽ ` ⫹ ` 6 3 4

46. `

3p ⫹ 2 1 ` ⫽ ` p⫺2` 4 2

47. 0x ⫹ 2 0 ⫽ 0 ⫺x ⫺ 2 0

48. 02y ⫺ 3 0 ⫽ 0 ⫺2y ⫹ 3 0

49. 03.5m ⫺ 1.2 0 ⫽ 08.5m ⫹ 6 0

51. 04x ⫺ 3 0 ⫽ ⫺ 0 2x ⫺ 1 0

52. ⫺ 0 3 ⫺ 6y 0 ⫽ 08 ⫺ 2y 0

50. 0 11.2n ⫹ 9 0 ⫽ 07.2n ⫺ 2.1 0

Expanding Your Skills 53. Write an absolute value equation whose solution is the set of real numbers 6 units from zero on the number line.

54. Write an absolute value equation whose solution is the set of real numbers 72 units from zero on the number line.

55. Write an absolute value equation whose solution is the set of real numbers 43 units from zero on the number line.

56. Write an absolute value equation whose solution is the set of real numbers 9 units from zero on the number line.

Absolute Value Inequalities

Section 1.7

1. Solving Absolute Value Inequalities by Definition

Concepts

In Section 1.6, we studied absolute value equations in the form 0x 0 ⫽ a. In this section, we will solve absolute value inequalities. An inequality in any of the forms 0x 0 6 a, 0x 0 ⱕ a, 0x 0 7 a, or 0x 0 ⱖ a is called an absolute value inequality. Recall that an absolute value represents distance from zero on the real number line. Consider the following absolute value equation and inequalities. 1.

0x 0 ⫽ 3 x⫽3

or

x ⫽ ⫺3

Solution: The set of all points 3 units from zero on the number line 3 units ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

3 units 0

1

2

3

4

5

6

1. Solving Absolute Value Inequalities by Definition 2. Solving Absolute Value Inequalities by the Test Point Method 3. Translating to an Absolute Value Expression

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Chapter 1 Linear Equations and Inequalities in One Variable

2.

0x 0 7 3 x 6 ⫺3

Solution: x 7 3

or

The set of all points more than 3 units from zero 3 units

(

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

3.

3 units 0

1

2

(

3

4

5

6

0x 0 6 3

Solution:

⫺3 6 x 6 3

The set of all points less than 3 units from zero 3 units

(

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

3 units 0

1

2

(

3

4

5

6

PROCEDURE Solving Absolute Value Equations and Inequalities Let a be a real number such that a 7 0. Then Equation/ Inequality 0x 0 ⫽ a

Solution (Equivalent Form)

Graph

x ⫽ ⫺a or x ⫽ a ⫺a

0x 0 7 a

x 6 ⫺a or x 7 a

0x 0 6 a

⫺a 6 x 6 a

a

(

(

⫺a

a

(

(

⫺a

a

To solve an absolute value inequality, first isolate the absolute value and then rewrite the absolute value inequality in its equivalent form. Example 1

Solving an Absolute Value Inequality 0 3w ⫹ 1 0 ⫺ 4 6 7

Solve the inequality.

Solution:

03w ⫹ 1 0 ⫺ 4 6 7

0 3w ⫹ 1 0 6 11

⫺11 6 3w ⫹ 1 6 11 ⫺12 6 3w 6 10

TIP: Recall that a strict inequality (using the symbols > and 0 translates to “a is greater than zero.” This also means that a is positive. Read and interpret the following conditions imposed on the variables a, b, c, d, and x. Then determine whether the statements in Exercises 1–12 are true or false. a 7 0

b 6 0

⫺1 6 c 6 1

d 7 1

x⫽0

1. ab 7 0

2. bd 6 0

3. a ⫹ d 7 0

4. b2 6 0

5. a2 7 0

6. c ⫹ d ⱖ 0

7. c2 6 1

8. d2 7 1

9. 0c 0 6 1

10. bx 6 0

11. b ⫹ x 7 0

12.

x ⫽0 d

For Exercises 13–24, suppose that m represents an odd integer and n represents an even integer. Determine whether the statements are true or false. 13. m ⫹ 1 is an odd integer.

14. m ⫹ 2 is an odd integer.

15. m ⫺ 1 is an odd integer.

16. m ⫺ 2 is an odd integer.

17. m ⫹ n is an even integer.

18. m ⫺ n is an even integer.

19. m2 is an even integer.

20. n2 is an even integer.

21. 1m ⫹ n2 2 is an odd integer.

22. 1n ⫹ n2 2 is an even integer.

23. 1m ⫹ n ⫹ 12 3 is an odd integer.

24. 1m ⫹ m ⫺ 22 3 is an odd integer.

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Summary

Chapter 1

Summary

Section 1.1

Linear Equations in One Variable

Key Concepts

Examples

A linear equation in one variable can be written in the form ax ⫹ b ⫽ 0 1a ⫽ 02. Steps to Solve a Linear Equation in One Variable 1. Simplify both sides of the equation. • Clear parentheses. • Consider clearing fractions or decimals (if any are present) by multiplying both sides of the equation by a common denominator of all terms. • Combine like terms. 2. Use the addition or subtraction property of equality to collect the variable terms on one side of the equation. 3. Use the addition or subtraction property of equality to collect the constant terms on the other side. 4. Use the multiplication or division property of equality to make the coefficient on the variable term equal to 1. 5. Check your answer and write the solution set.

Example 1 1 3 1 1x ⫺ 42 ⫺ 1x ⫹ 22 ⫽ 2 4 4 3 3 1 1 x⫺2⫺ x⫺ ⫽ 2 4 2 4 1 3 3 1 4a x ⫺ 2 ⫺ x ⫺ b ⫽ 4a b 2 4 2 4 2x ⫺ 8 ⫺ 3x ⫺ 6 ⫽ 1 ⫺x ⫺ 14 ⫽ 1 ⫺x ⫽ 15 x ⫽ ⫺15 The solution ⫺15 checks in the original equation. The solution set is 5⫺156.

An equation that has no solution is called a contradiction.

Example 2 3x ⫹ 6 ⫽ 31x ⫺ 52 3x ⫹ 6 ⫽ 3x ⫺ 15 6 ⫽ ⫺15

Contradiction

There is no solution, { }. An equation that has all real numbers as its solutions is called an identity.

Example 3 ⫺15x ⫹ 122 ⫺ 3 ⫽ 51⫺x ⫺ 32 ⫺5x ⫺ 12 ⫺ 3 ⫽ ⫺5x ⫺ 15 ⫺5x ⫺ 15 ⫽ ⫺5x ⫺ 15 ⫺15 ⫽ ⫺15

Identity

All real numbers are solutions. The solution set is 5x 0 x is a real number6.

115

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Chapter 1 Linear Equations and Inequalities in One Variable

Section 1.2

Applications of Linear Equations in One Variable

Key Concepts

Examples

Problem-Solving Steps for Word Problems

Example 1

1. 2. 3. 4. 5. 6.

Read the problem carefully. Assign labels to unknown quantities. Develop a verbal model. Write a mathematical equation. Solve the equation. Interpret the results and write the final answer in words.

Sales tax: (cost of merchandise)(tax rate) Commission: (dollars in sales)(rate) Simple interest: I  Prt Distance  (rate)(time) d  rt

1. Estella needs to borrow $8500. She borrows part of the money from a friend and agrees to pay the friend 6% simple interest. She borrows the rest of the money from a bank that charges 10% simple interest. If she pays back the money at the end of 1 yr and also pays $750 in interest, find the amount that Estella borrowed from each source. 2. Let x represent the amount borrowed at 6%. Then 8500  x is the amount borrowed at 10%.

3. a

6% Account

10% Account

Total

Principal

x

8500  x

8500

Interest

0.06x

0.10(8500  x)

750

interest b  a total b Interest ba owed at 10% interest owed at 6%

4. 0.06x  0.1018500  x2  750 5. 6x  1018500  x2  75,000 6x  85,000  10x  75,000 4x  10,000 x  2500 6. x  2500 8500  2500  6000 $2500 was borrowed at 6% and $6000 was borrowed at 10%.

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Summary

Section 1.3

117

Applications to Geometry and Literal Equations

Key Concepts

Examples

Some useful formulas for word problems:

Example 1

Perimeter

A border of marigolds is to enclose a rectangular flower garden. If the length is twice the width and the perimeter is 25.5 ft, what are the dimensions of the garden?

Rectangle: P  2l  2w Area Rectangle: A  lw Square:

A  s2

Triangle:

1 A  bh 2

Trapezoid: A 

x 2x

P  2l  2w 25.5  212x2  21x2

1 1b  b2 2h 2 1

25.5  4x  2x 25.5  6x 4.25  x The width is 4.25 ft, and the length is 2(4.25) ft or 8.5 ft.

Angles Two angles whose measures total 90 are complementary angles. Two angles whose measures total 180 are supplementary angles. The sum of the measures of the angles of a triangle is 180.

x  y  z  180

x

y

z

Literal equations (or formulas) are equations with several variables. To solve for a specific variable, follow the steps to solve a linear equation.

Example 2 Solve for y. 4x  5y  20 5y  4x  20 5y 4x  20  5 5 y

4x  20 5

or

4 y x4 5

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Chapter 1 Linear Equations and Inequalities in One Variable

Section 1.4

Linear Inequalities in One Variable

Key Concepts

Examples

A linear inequality in one variable can be written in the form

Example 1

ax  b 6 0, ax  b  0

ax  b 7 0,

ax  b  0,

Properties of Inequalities 1. If a 6 b, then a  c 6 b  c. 2. If a 6 b, then a  c 6 b  c. 3. If c is positive and a 6 b, then ac 6 bc and a b 6 . c c 4. If c is negative and a 6 b, then ac 7 bc and b a 7 . c c Properties 3 and 4 indicate that if we multiply or divide an inequality by a negative value, the direction of the inequality sign must be reversed.

Section 1.5

Solve.

or

14  x 6 3x 2 2 a

14  x b 7 213x2 2

(Reverse the inequality sign.)

14  x 7 6x 7x 7 14 7x 14 6 7 7

(Reverse the inequality sign.)

x 6 2 ( Interval notation: 1 , 22

2

Compound Inequalities

Key Concepts

Examples

A ´ B is the union of A and B. This is the set of elements that belong to set A or set B or both sets A and B.

Example 1 Union A

A ¨ B is the intersection of A and B. This is the set of elements common to both A and B.

Intersection B

AB

A

B

AB

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Summary

•

Solve two or more inequalities joined by and by finding the intersection of their solution sets.

•

Solve two or more inequalities joined by or by finding the union of their solution sets.

Example 2

119

Example 3

⫺7x ⫹ 3 ⱖ ⫺11

and

1 ⫺ x 6 4.5

5y ⫹ 1 ⱖ 6

or

⫺7x ⱖ ⫺14

and

⫺x 6 3.5

5y ⱖ 5

or

2y ⱕ ⫺6

xⱕ2

and

yⱖ1

or

y ⱕ ⫺3

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

(

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

(

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

1

2

x 7 ⫺3.5 3

4

5

0

1

2

3

4

5

0

1

2

3

4

5

xⱕ2

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

x 7 ⫺3.5

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

The solution is 5x ƒ ⫺3.5 6 x ⱕ 26 or equivalently 1⫺3.5, 24.

Example 4

The inequality a 6 x 6 b is represented by

Solve.

(

a

b

or, in interval notation, (a, b).

yⱖ1 y ⱕ ⫺3

The solution is 5y ƒ y ⱖ 1 or y ⱕ ⫺36 or equivalently 1⫺⬁, ⫺3 4 ´ 31, ⬁2.

Inequalities of the form a < x < b :

(

2y ⫺ 5 ⱕ ⫺11

⫺13 ⱕ 3x ⫺ 1 6 5 ⫺13 ⫹ 1 ⱕ 3x ⫺ 1 ⫹ 1 6 5 ⫹ 1 ⫺12 ⱕ 3x 6 6 3x 6 ⫺12 ⱕ 6 3 3 3 ⫺4 ⱕ x 6 2 ⫺4

(

2

Interval notation: 3⫺4, 22

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Chapter 1 Linear Equations and Inequalities in One Variable

Section 1.6

Absolute Value Equations

Key Concepts

The equation 0x 0 ⫽ a is an absolute value equation. For a ⱖ 0, the solution to the equation 0x 0 ⫽ a is x ⫽ a or x ⫽ ⫺a. Steps to Solve an Absolute Value Equation 1. Isolate the absolute value to write the equation in the form 0x 0 ⫽ a. 2. If a 6 0, there is no solution. 3. Otherwise, if a ⱖ 0, rewrite the equation 0 x 0 ⫽ a as x ⫽ a or x ⫽ ⫺a. 4. Solve the equations from step 3. 5. Check answers in the original equation.

Examples Example 1 0 2x ⫺ 3 0 ⫹ 5 ⫽ 10 02x ⫺ 3 0 ⫽ 5

Isolate the absolute value.

2x ⫺ 3 ⫽ 5 or

2x ⫺ 3 ⫽ ⫺5

2x ⫽ 8 or

2x ⫽ ⫺2

x ⫽ 4 or

x ⫽ ⫺1

The solution set is 54, ⫺16. Example 2 0x ⫹ 2 0 ⫹ 5 ⫽ 1

0x ⫹ 2 0 ⫽ ⫺4

The equation 0x 0 ⫽ 0y 0 implies x ⫽ y or x ⫽ ⫺y.

No solution, 5 6

Example 3 02x ⫺ 1 0 ⫽ 0x ⫹ 4 0 2x ⫺ 1 ⫽ x ⫹ 4 or x⫽5

2x ⫺ 1 ⫽ ⫺1x ⫹ 42

or 2x ⫺ 1 ⫽ ⫺x ⫺ 4 or

3x ⫽ ⫺3

or

x ⫽ ⫺1

The solution set is 55, ⫺16.

Section 1.7

Absolute Value Inequalities

Key Concepts

Examples

Solutions to Absolute Value Inequalities

Example 1

For a 7 0, we have: 0x 0 7 a 1 x 6 ⫺a or x 7 a 0x 0 6 a 1 ⫺a 6 x 6 a

0 5x ⫺ 2 0 6 12 ⫺12 6 5x ⫺ 2 6 12 ⫺10 6 5x 6 14 ⫺2 6 x 6

14 5

The solution is a⫺2,

14 b. 5

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Summary

Test Point Method to Solve Inequalities 1. Find the boundary points of the inequality. (Boundary points are the real solutions to the related equation and points where the inequality is undefined.) 2. Plot the boundary points on the number line. This divides the number line into intervals. 3. Select a test point from each interval and substitute it into the original inequality. • If a test point makes the original inequality true, then that interval is part of the solution set. 4. Test the boundary points in the original inequality. • If the original inequality is strict (< or >), do not include the boundary in the solution set. • If the original inequality is defined using  or , then include the boundary points that are well defined within the inequality. Note: Any boundary point that makes an expression within the inequality undefined must always be excluded from the solution set.

Example 2 0x  3 0  2  7 0x  3 0  5

Isolate the absolute value.

0x  3 0  5 x  3  5 or x8

Solve the related equation. x  3  5 x  2

or I

Boundary points

II

III

2

8

Interval I: Test x  3:

0 132  3 0  2  7

True

0 102  3 0  2  7

False

0 192  3 0  2  7

True

?

Interval II: ?

Test x  0: Interval III:

?

Test x  9: True

False 2

True 8

The solution is 1 , 2 4 ´ 38, 2. If a is negative (a < 0), then

1. 0x 0 6 a has no solution. 2. 0x 0 7 a is true for all real numbers.

Example 3 0x  5 0 7 2 The solution is all real numbers because an absolute value will always be greater than a negative number. 1 , 2 Example 4 0x  5 0 6 2 There is no solution because an absolute value cannot be less than a negative number. The solution set is 5 6.

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Chapter 1 Linear Equations and Inequalities in One Variable

Chapter 1

Review Exercises

Section 1.1 1. Describe the solution set for a contradiction. 2. Describe the solution set for an identity. For Exercises 3–12, solve the equations and identify each as a conditional equation, a contradiction, or an identity. 7 ⫽1 8

3. x ⫺ 27 ⫽ ⫺32

4.

y⫹

5. 7.23 ⫹ 0.6x ⫽ 0.2x

6.

0.1y ⫹ 1.122 ⫽ 5.2y

7. ⫺14 ⫹ 3m2 ⫽ 913 ⫺ m2 8. ⫺215n ⫺ 62 ⫽ 31⫺n ⫺ 32 9.

x⫺3 2x ⫹ 1 ⫺ ⫽1 5 2

10. 31x ⫹ 32 ⫺ 2 ⫽ 3x ⫹ 2 11.

10 7 3 m ⫹ 18 ⫺ m ⫽ m ⫹ 25 8 8 8

2 1 1 1 12. m ⫹ 1m ⫺ 12 ⫽ ⫺ m ⫹ 14m ⫺ 12 3 3 3 3

Section 1.2 13. Explain how you would label three consecutive integers. 14. Explain how you would label two consecutive odd integers. 15. Explain what the formula d ⫽ rt means. 16. Explain what the formula I ⫽ Prt means. 17. a. Cory made $30,403 in taxable income. If he pays 28% in federal income tax, determine the amount of tax he must pay.

19. For a recent year, there were 17,430 deaths due to alcohol-related accidents in the United States. This was a 5% increase over the number of alcohol-related deaths in 1999. How many such deaths were there in 1999? 20. Of three consecutive even integers, the sum of the smallest two integers is equal to 6 less than the largest. Find the integers. 21. To do a rope trick, a magician needs to cut a piece of rope so that one piece is one-third the length of the other piece. If she begins with a 223-ft rope, what will be the lengths of the two pieces of rope? 22. Sharyn invests $2000 more in an account that earns 9% simple interest than she invests in an account that earns 6% simple interest. How much did she invest in each account if her total interest is $405 after 1 yr? 23. How much 10% acid solution should be mixed with 1 L of 25% acid solution to produce a solution that is 15% acid? 24. Two friends plan to meet at a restaurant for lunch. They both leave their homes at 11:30 A.M. and between the two of them, they drive a total of 37.5 mi. Lynn drives in from a neighboring town and averages 15 mph faster than her friend Linda. If they meet at noon, find the average driving speed for each.

Section 1.3 25. The length of a rectangle is 2 ft more than the width. Find the dimensions if the perimeter is 40 ft. For Exercises 26–27, solve for x, and then find the measure of each angle. 26. x a ⫹ 1b ⬚ 2

b. What is his net income (after taxes)? 18. For a recent year, approximately 7.2 million men were in college in the United States. This represents an 8% increase over the number of men in college in 2000. Approximately how many men were in college in 2000? (Round to the nearest tenth of a million.)

(x ⫺ 25)⬚

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Review Exercises

123

Section 1.5

27.

39. Explain the difference between the union and intersection of two sets. You may use the sets C and D in the following diagram to provide an example.

(x ⫺ 1)⬚

(2x ⫹ 1)⬚

For Exercises 28–31, solve for the indicated variable. 28. 3x ⫺ 2y ⫽ 4 for y C

29. ⫺6x ⫹ y ⫽ 12 for y 30. S ⫽ 2pr ⫹ pr2h

for h

1 31. A ⫽ bh for b 2 32. a. The circumference of a circle is given by C ⫽ 2pr. Solve this equation for p. b. Tom measures the radius of a circle to be 6 cm and the circumference to be 37.7 cm. Use these values to approximate p. (Round to 2 decimal places.)

Section 1.4 For Exercises 33–38, solve the inequality. Graph the solution and write the solution set in interval notation. 33. ⫺6x ⫺ 2 7 6 34. ⫺10x ⱕ 15 35. 5 ⫺ 71x ⫹ 32 7 19x 36. 4 ⫺ 3x ⱖ 101⫺x ⫹ 52 5 ⫺ 4x 37. ⱖ9 8 38.

3 ⫹ 2x ⱕ8 4

D

Let X ⫽ 5x 0 x ⱖ ⫺106, Y ⫽ 5x 0 x 6 16, Z ⫽ 5x 0 x 7 ⫺16, and W ⫽ 5x 0 x ⱕ ⫺36. For Exercises 40–45, find the intersection or union of the sets X, Y, Z, and W. Write the answers in interval notation. 40. X ¨ Y

41. X ´ Y

42. Y ´ Z

43. Y ¨ Z

44. Z ´ W

45. Z ¨ W

For Exercises 46–55, solve the compound inequalities. Write the solutions in interval notation. 46. 4m 7 ⫺11 and

4m ⫺ 3 ⱕ 13

47. 4n ⫺ 7 6 1 and 7 ⫹ 3n ⱖ ⫺8 48. ⫺3y ⫹ 1 ⱖ 10 and

⫺2y ⫺ 5 ⱕ ⫺15

49.

1 h 7 ⫺ ⱕ⫺ 2 12 12

1 h 1 ⫺ 7 ⫺ 2 10 5

50.

2 t ⫺ 3 ⱕ 1 or 3

and

3 t⫺2 7 7 4

51. 213x ⫹ 12 6 ⫺10 or 52. ⫺7 6 ⫺712w ⫹ 32

312x ⫺ 42 ⱖ 0

or

53. 51p ⫹ 32 ⫹ 4 7 p ⫺ 1

⫺2 6 ⫺4 13w ⫺ 12

or

54. 2 ⱖ ⫺1b ⫺ 22 ⫺ 5b ⱖ ⫺6 55. ⫺4 ⱕ

3 1 1x ⫺ 12 6 ⫺ 2 2

4 1p ⫺ 12 ⫹ 2 7 p ⫹ 8

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Chapter 1 Linear Equations and Inequalities in One Variable

56. The product of 13 and the sum of a number and 3 is between 1 and 5. Find all such numbers. 57. Normal levels of total cholesterol vary according to age. For adults between 25 and 40 yr old, the normal range is generally accepted to be between 140 and 225 mg/dL (milligrams per deciliter), inclusive. a. Write an inequality representing the normal range for total cholesterol for adults between 25 and 40 yr old. b. Write a compound inequality representing abnormal ranges for total cholesterol for adults between 25 and 40 yr old. 58. Normal levels of total cholesterol vary according to age. For adults younger than 25 yr old, the normal range is generally accepted to be between 125 and 200 mg/dL, inclusive. a. Write an inequality representing the normal range for total cholesterol for adults younger than 25 yr old. b. Write a compound inequality representing abnormal ranges for total cholesterol for adults younger than 25 yr old. 59. One method to approximate your maximum heart rate is to subtract your age from 220. To maintain an aerobic workout, it is recommended that you sustain a heart rate of between 60% and 75% of your maximum heart rate. a. If the maximum heart rate h is given by the formula h  220  A, where A is a person’s age, find your own maximum heart rate. (Answers will vary.) b. Find the interval for your own heart rate that will sustain an aerobic workout. (Answers will vary.) 60. Dave earned the following test scores in his biology class: 82%, 88%, 92%, and 93%. How high does he have to score on the fifth test to have an average of 90% or more?

Section 1.6

65. 16  0x  2 0  9

66. 5  0x  2 0  4

67. 04x  1 0  6  4

68. 0 3x  1 0  7  3

69. `

7x  3 ` 44 5

70. `

4x  5 `  3  3 2

71. 0 3x  5 0  0 2x  1 0 72. 08x  9 0  0 8x  1 0 73. Which absolute value expression represents the distance between 3 and 2 on the number line? 03  122 0

02  3 0

Section 1.7 74. Write the compound inequality x 6 5 or x 7 5 as an absolute value inequality. 75. Write the compound inequality 4 6 x 6 4 as an absolute value inequality. For Exercises 76–77, write an absolute value inequality that represents the solution sets graphed here. 76. 77.

(

(

6

6

( 2  3

( 2 3

For Exercises 78–91, solve the absolute value inequalities. Graph the solution set and write the solution in interval notation. 78. 0x  6 0  8

79. 0x  8 0  3

80. 2 0 7x  1 0  4 7 4 81. 4 05x  1 0  3 7 3 82. 03x  4 0  6  4

83. 05x  3 0  3  6

84. `

85. `

For Exercises 61–72, solve the absolute value equations. 61. 0x 0  10

62. 0x 0  17

63. 08.7  2x 0  6.1

64. 05.25  5x 0  7.45

x 6` 6 5 2

86. 04  2x 0  8  8

x 2` 6 2 3

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87. 09 ⫹ 3x 0 ⫹ 1 ⱖ 1 88. ⫺2 05.2x ⫺ 7.8 0 6 13 89. ⫺ 02.5x ⫹ 15 0 6 7 90. 03x ⫺ 8 0 6 ⫺1

91. 0 x ⫹ 5 0 6 ⫺4

92. State one possible situation in which an absolute value inequality will have no solution. 93. State one possible situation in which an absolute value inequality will have a solution of all real numbers.

Chapter 1 x ⫹ 1 ⫽ 20 7

2. 8 ⫺ 514 ⫺ 3z2 ⫽ 214 ⫺ z2 ⫺ 8z 3. 0.121x2 ⫹ 0.08160,000 ⫺ x2 ⫽ 10,500 4.

94. The Neilsen ratings estimated that the percent, p, of the television viewing audience watching American Idol was 20% with a 3% margin of error. Solve the inequality 0 p ⫺ 0.20 0 ⱕ 0.03 and interpret the answer in the context of this problem. 95. The length, L, of a screw is supposed to be 338 in. Due to variation in the production equipment, there is a 41-in. margin of error. Solve the inequality 0L ⫺ 338 0 ⱕ 14 and interpret the answer in the context of this problem.

Test

For Exercises 1–4, solve the equations. 1.

125

5⫺x 2x ⫺ 3 x ⫺ ⫽ 6 2 3

5. Label each equation as a conditional equation, an identity, or a contradiction. a. 15x ⫺ 92 ⫹ 19 ⫽ 51x ⫹ 22 b. 2a ⫺ 211 ⫹ a2 ⫽ 5

c. 14w ⫺ 32 ⫹ 4 ⫽ 315 ⫺ w2 6. The difference between two numbers is 72. If the larger is 5 times the smaller, find the two numbers.

8. Shawnna banks at a credit union. Her money is distributed between two accounts: a certificate of deposit (CD) that earns 5% simple interest and a savings account that earns 3.5% simple interest. Shawnna has $100 less in her savings account than in the CD. If after 1 year her total interest is $81.50, how much did she invest in the CD? 9. A yield sign is in the shape of an equilateral triangle (all sides have equal length). Its perimeter is 81 in. Find the length of the sides. 10. The sum of three consecutive odd integers is 41 less than four times the largest. Find the numbers. 11. How many gallons of a 20% acid solution must be mixed with 6 gal of a 30% acid solution to make a 22% solution? For Exercises 12–13, solve the equations for the indicated variable. 12. 4x ⫹ 2y ⫽ 6 for y

7. Joëlle is determined to get some exercise and walks to the store at a brisk rate of 4.5 mph. She meets her friend Yun Ling at the store, and together they walk back at a slower rate of 3 mph. Joëlle’s total walking time was 1 hr. a. How long did it take her to walk to the store? b. What is the distance to the store?

13. x ⫽ m ⫹ zs for z

For Exercises 14–16, solve the inequalities. Graph the solution and write the solution set in interval notation. 14. x ⫹ 8 7 42 16. ⫺2 6 3x ⫺ 1 ⱕ 5

3 15. ⫺ x ⫹ 6 ⱖ x ⫺ 3 2

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Chapter 1 Linear Equations and Inequalities in One Variable

17. An elevator can accommodate a maximum weight of 2000 lb. If four passengers on the elevator have an average weight of 180 lb each, how many additional passengers of the same average weight can the elevator carry before the maximum weight capacity is exceeded?

23. The normal range in humans of the enzyme adenosine deaminase (ADA), is between 9 and 33 IU (international units), inclusive. Let x represent the ADA level in international units. a. Write an inequality representing the normal range for ADA. b. Write a compound inequality representing abnormal ranges for ADA. For Exercises 24–26, solve the absolute value equations. 1 24. ` x  3 `  4  4 2 25. 03x  4 0  0x  12 0 26. 5  8  0 2y  3 0

For Exercises 18–22, solve the compound inequalities. Write the answers in interval notation. 18. 2  3x  1  5 2 or  x  16 3

3 19.  x  1  8 5 20. 2x  3 7 3

and x  3  0

21. 5x  1  6 or 2x  4 7 6 22. 2x  3 7 1

and x  4 6 1

For Exercises 27–31, solve the absolute value inequalities. Write the answers in interval notation. 27. 0 3  2x 0  6 6 2

28. 0 3x  8 0  9

29. 00.4x  0.3 0  0.2 6 7 30. 07  3x 0  1 7 3

31. 6  0 2x  5 0  5

32. The mass of a small piece of metal is measured to be 15.41 g. If the measurement error is at most

0.01 g, write an absolute value inequality that represents the possible mass, x, of the piece of metal.

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Linear Equations in Two Variables and Functions CHAPTER OUTLINE 2.1 Linear Equations in Two Variables 128 2.2 Slope of a Line and Rate of Change 145 2.3 Equations of a Line 156 Problem Recognition Exercises: Characteristics of Linear Equations

169

2.4 Applications of Linear Equations and Modeling 170 2.5 Introduction to Relations 182 2.6 Introduction to Functions 191 2.7 Graphs of Functions 202 Problem Recognition Exercises: Characteristics of Relations Group Activity: Deciphering a Coded Message

214

215

Chapter 2 In this chapter, we cover graphing and the applications of graphing. Graphs appear in magazines and newspapers and in other aspects of day-to-day life. In many fields of study such as the sciences and business, graphs are used to display data (information). Are You Prepared? Try answering questions 1–5 below. These problems will refresh your skills at reading and interpreting graphs. Match each answer with the correct letter and record the letter in the space below.

Average Height for Girls, Ages 2–9

Height (in.)

1. Use the graph to approximate the average height of a 5-yr-old girl. 2. Use the graph to estimate a girl’s age if her height is 39 in. tall. 3. The equation h ⫽ 2.5a ⫹ 31 can be used to 60 approximate a girl’s height, h, in inches according to 50 her age, a, in years. Use the formula h ⫽ 2.5a ⫹ 31 40 to estimate a girl’s height at 6 yr old. 30 4. Use the formula h ⫽ 2.5a ⫹ 31 to estimate the age 20 of a girl if her height is 50 in. 10 5. Use the formula h ⫽ 2.5a ⫹ 31 to predict the height 0 0 1 of an 11-yr-old girl. A graph intersects the x-axis at an

T. 7.6 yr X. 44 in.

2

3

4

5 6 Age (yr)

E. 46 in. C. 58.5 in.

7

8

9

10

I. 3 yr

N ___ ___ ___ R ___ ___ ___ P ___. ___ - ___ ___ 1 2 4 3 5 3 4 127

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Section 2.1

Linear Equations in Two Variables

Concepts

1. The Rectangular Coordinate System

1. The Rectangular Coordinate System 2. Linear Equations in Two Variables 3. Graphing Linear Equations in Two Variables 4. x- and y-Intercepts 5. Horizontal and Vertical Lines

One application of algebra is the graphical representation of numerical information (or data). For example, Table 2-1 shows the percentage of individuals who participate in leisure sports activities according to the age of the individual. Table 2-1 Age (years)

Percentage of Individuals Participating in Leisure Sports Activities

20

59%

30

52%

40

44%

50

34%

60

21%

70

18%

Source: U.S. National Endowment for the Arts

Information in table form is difficult to picture and interpret. However, when the data are presented in a graph, there appears to be a downward trend in the participation in leisure sports activities as age increases (Figure 2-1).

Percent

Percentage of Individuals Who Participate in Leisure Sports Activities Versus Age 70 60 50 40 30 20 10 0 0

10

20 30 40 50 60 Age of Participant (years)

70

80

Figure 2-1

In this example, two variables are related: age and the percentage of individuals who participate in leisure sports activities. To picture two variables simultaneously, we y-axis use a graph with two number lines drawn at 6 right angles to each other (Figure 2-2). This 5 forms a rectangular coordinate system.The hor4 Quadrant II 3 Quadrant I izontal line is called the x-axis, and the vertical 2 line is called the y-axis.The point where the lines x-axis 1 Origin intersect is called the origin. On the x-axis, the ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 0 1 2 3 4 5 6 ⫺1 numbers to the right of the origin are positive, ⫺2 and the numbers to the left are negative. On the Quadrant III ⫺3 Quadrant IV y-axis, the numbers above the origin are positive, ⫺4 ⫺5 and the numbers below are negative. The x- and ⫺6 y-axes divide the graphing area into four Figure 2-2 regions called quadrants.

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Linear Equations in Two Variables

Points graphed in a rectangular coordinate system are defined by two numbers as an ordered pair (x, y). The first number (called the x-coordinate or abscissa) is the horizontal position from the origin. The second number (called the y-coordinate or ordinate) is the vertical position from the origin. Example 1 shows how points are plotted in a rectangular coordinate system. Example 1

Plotting Points

Plot each point and state the quadrant or axis where it is located. a. (4, 1) d.

(⫺52,

⫺2)

b. (⫺3, 4)

c. (4, ⫺3)

e. (0, 3)

f. (⫺4, 0)

Solution: y

a. The point (4, 1) is in quadrant I.

(⫺3, 4)

b. The point (⫺3, 4) is in quadrant ⌱⌱. c. The point (4, ⫺3) is in quadrant IV.

1

(⫺4, 0)

d. The point (⫺52, ⫺2) can also be written as (⫺2.5, ⫺2). This point is in quadrant III.

5 4 3 (0, 3) 2

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 0 1 ⫺1 ⫺2 5 ⫺2 , ⫺2 ⫺3

e. The point (0, 3) is on the y-axis. f. The point (⫺4, 0) is located on the x-axis.





2

⫺4 ⫺5

(4, 1) 3 4

5

x

(4, ⫺3)

Figure 2-3

TIP: Notice that the points (⫺3, 4) and (4, ⫺3) are in different quadrants. Changing the order of the coordinates changes the location of the point. That is why points are represented by ordered pairs (Figure 2-3).

Skill Practice Plot the point and state the quadrant or axis where it is located. 1. a. (3, 5) d. (0, 4)

b. (⫺2, 0) e. (⫺2, ⫺2)

c. (2, ⫺1) f. (⫺5, 2)

2. Linear Equations in Two Variables Recall from Section 1.1 that an equation in the form ax ⫹ b ⫽ 0 is called a linear equation in one variable. In this section, we will study linear equations in two variables.

DEFINITION Linear Equation in Two Variables Let A, B, and C be real numbers such that A and B are not both zero. A linear equation in two variables is an equation that can be written in the form Ax ⫹ By ⫽ C

This form is called standard form.

Answers 1. a. b. c. d. e. f.

(3, 5); quadrant I (⫺2, 0); x-axis (2, ⫺1); quadrant IV (0, 4); y-axis (⫺2, ⫺2); quadrant III (⫺5, 2); quadrant II y 5 (0, 4) 4 3 2 (⫺5, 2)

(3, 5)

(⫺2, 0) 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 0 1 2 3 4 ⫺1 ⫺2 (2, ⫺1)

(⫺2, ⫺2) ⫺3

⫺4 ⫺5

5

x

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Chapter 2 Linear Equations in Two Variables and Functions

A solution to a linear equation in two variables is an ordered pair 1x, y2 that makes the equation a true statement. Example 2

Determining Solutions to a Linear Equation

For the linear equation ⫺2x ⫹ 3y ⫽ 8, determine whether the ordered pair is a solution. a. 1⫺4, 02

b. 12, ⫺42

Solution: a.

⫺2x ⫹ 3y ⫽ 8 ⫺21⫺42 ⫹ 3102 ⱨ 8 8 ⫹ 0 ⱨ 8  (true)

b.

⫺2x ⫹ 3y ⫽ 8 ⫺2122 ⫹ 31⫺42 ⱨ 8 ⫺4 ⫹ 1⫺122 ⱨ 8

⫺16 ⱨ 8 (false)

c.

⫺2x ⫹ 3y ⫽ 8 ⫺2112 ⫹ 3a

10 ⱨ b 8 3

⫺2 ⫹ 10 ⱨ 8  (true)

c. a1,

10 b 3

The ordered pair 1⫺4, 02 indicates that x ⫽ ⫺4 and y ⫽ 0. Substitute x ⫽ ⫺4 and y ⫽ 0 into the equation.

The ordered pair 1⫺4, 02 makes the equation a true statement. The ordered pair is a solution to the equation. Test the ordered pair 12, ⫺42 . Substitute x ⫽ 2 and y ⫽ ⫺4 into the equation. The ordered pair 12, ⫺42 does not make the equation a true statement. The ordered pair is not a solution to the equation. Test the ordered pair 11,

10 3 2.

Substitute x ⫽ 1 and y ⫽ 103. The ordered pair 11, 103 2 is a solution to the equation.

Skill Practice Determine whether each ordered pair is a solution for the equation x ⫹ 4y ⫽ ⫺8. 2. 1⫺2, ⫺12

3. 14, ⫺32

3 4. a⫺14, b 2

3. Graphing Linear Equations in Two Variables Consider the linear equation x ⫺ y ⫽ 3. The solutions to the equation are ordered pairs such that the difference of x and y is 3. Several solutions are given in the following list: Answers 2. Not a solution 3. Solution 4. Solution

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Linear Equations in Two Variables

y

Solution (x, y) (3, 0) (4, 1) (0, 3) (1, 4) (2, 1)

5 4 3 2

Check xy3 (3)  (0)  3 ✔ (4)  (1)  3 ✔ (0)  (3)  3 ✔ (1)  (4)  3 ✔ (2)  (1)  3 ✔

(4, 1) (3, 0)

1 5 4 3 2 1 1 2

(1, 4)

3 4 5

1

2

3 4

5

x

(2, 1) (0, 3)

Figure 2-4

By graphing these ordered pairs, we see that the solution points line up (see Figure 2-4). There are actually an infinite number of solutions to the equation x  y  3. The graph of all solutions to a linear equation forms a line in the xy-plane. Conversely, each ordered pair on the line is a solution to the equation.

DEFINITION The Graph of an Equation in Two Variables To graph a linear equation in two variables means that we will graph all ordered pair solutions to the equation.

To graph a linear equation, it is sufficient to find two solution points and draw the line through them. We will find three solution points and use the third point as a check point. This is demonstrated in Example 3. Example 3

Graphing a Linear Equation in Two Variables

Graph the equation 3x  5y  15.

Solution: We will find three ordered pairs that are solutions to the equation. In the table, we have selected arbitrary values for x or y and must complete the ordered pairs. x

y

(0,

0

( , 2)

2

(5,

5

From the first row, substitute x  0.

)

From the second row, substitute y  2.

)

From the third row, substitute x  5.

3x  5y  15

3x  5y  15

3x  5y  15

3102  5y  15

3x  5122  15

3152  5y  15

5y  15

3x  10  15

15  5y  15

y3

3x  5

5y  0

5 3

y0

x

131

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Chapter 2 Linear Equations in Two Variables and Functions

The completed list of ordered pairs is shown.To graph the equation, plot the three solutions and draw the line through the points (Figure 2-5). Arrows on the ends of the line indicate that points on the line extend infinitely in both directions. y 5 4 (0, 3) 3 3x  5y  15 ( 53 , 2) 2

x

y

0

3

5 3

2

Q3 , 2 R

5

0

(5, 0)

(5, 0)

1

(0, 3)

5 4 3 2 1 1 2

5

1

2

3 4

5

x

3 4 5

Figure 2-5

Skill Practice Given 2x  y  1, complete the table and graph the line through the points. 5.

x

y

0 5 1

Example 4

Graphing a Linear Equation in Two Variables

Graph the equation y 

1 x  2. 2

Solution: Because the y-variable is isolated in the equation, it is easy to substitute a value for x and simplify the right-hand side to find y. Since any number can be used for x, choose numbers that are multiples of 2 that will simplify easily when multiplied by 12. Substitute x  0. x 0 2 4

Answer y

5.

(3, 5)

5 4 2x  y  1 3 2 1

(1, 1)

5 4 3 2 1 1 2 3 4 1 (0, 1) 2 3 4 5

5

x

y

y

1 102  2 2

Substitute x  2. y

1 122  2 2

Substitute x  4. y

1 142  2 2

y02

y12

y22

y  2

y  1

y0

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Linear Equations in Two Variables

The completed list of ordered pairs is as follows. To graph the equation, plot the three solutions and draw the line through the points (Figure 2-6). y

x

y

0

2

2

1

4

0

5 4 3 1 y  2x  2 2

10, 22 12, 12 (4, 0)

1 5 4 3 2 1 1 (0, 2)2

(4, 0) 1

2

3 4

5

x

(2, 1)

3 4 5

Figure 2-6

Skill Practice 1 6. Graph the equation y   x  1. 3 Hint: Select values of x that are multiples of 3.

4. x- and y-Intercepts For many applications of graphing, it is advantageous to know the points where a graph intersects the x- or y-axis. These points are called the x- and y-intercepts. In Figure 2-5, the x-intercept is (5, 0). In Figure 2-6, the x-intercept is (4, 0). In general, a point on the x-axis must have a y-coordinate of zero. In Figure 2-5, the y-intercept is (0, 3). In Figure 2-6, the y-intercept is (0, 2). In general, a point on the y-axis must have an x-coordinate of zero. y

DEFINITION x- and y -Intercepts

An x-intercept* is a point 1a, 02 where a graph intersects the x-axis. (See Figure 2-7.)

(0, b)

(a, 0)

A y-intercept is a point 10, b2 where a graph intersects the y-axis. (See Figure 2-7.)

*In some applications, an x-intercept is defined as the x-coordinate of a point of intersection that a graph makes with the x-axis. For example, if an x-intercept is at the point (3, 0), it is sometimes stated simply as 3 (the y-coordinate is understood to be zero). Similarly, a y-intercept is sometimes defined as the y-coordinate of a point of intersection that a graph makes with the y-axis. For example, if a y-intercept is at the point (0, 7), it may be stated simply as 7 (the x-coordinate is understood to be zero).

x

Figure 2-7

To find the x- and y-intercepts from an equation in x and y, follow these steps: Answer

PROCEDURE Determining the x- and y-Intercepts from an Equation Given an equation in x and y, • Find the x-intercept(s) by substituting y  0 into the equation and solving for x. • Find the y-intercept(s) by substituting x  0 into the equation and solving for y.

6.

y 5 4 3 2 y  1 x  1 3 1 5 4 3 2 1 1 2 3 4 5

1

2

3 4

5

x

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Example 5

Finding the x- and y-Intercepts of a Line

Given 2x  4y  8, find the x- and y-intercepts. Then graph the equation.

Solution: To find the x-intercept, substitute y  0. 2x  4y  8

2x  4y  8

2x  4102  8

2102  4y  8

2x  8

4y  8

x4

y2

The x-intercept is (4, 0).

y

2x  4y  8

1

(4, 0) 1

The y-intercept is (0, 2).

In this case, the intercepts are two distinct points and may be used to graph the line. A third point can be found to verify that the points all fall on the same line (points that lie on the same line are said to be collinear). Choose a different value for either x or y, such as y  4.

5 (4, 4) 4 3 (0, 2) 2 5 4 3 2 1 1 2

To find the y-intercept, substitute x  0.

2

3 4

5

x

2x  4142  8 2x  16  8

2x  4y  8

Substitute y  4. Solve for x.

2x  8

3 4 5

x  4

Figure 2-8

The point (4, 4) lines up with the other two points (Figure 2-8).

Skill Practice 7. Given 2x  y  4, find the x- and y-intercepts. Then graph the equation.

Example 6

Finding the x- and y-Intercepts of a Line

Given y  14x, find the x- and y-intercepts. Then graph the equation.

Solution: To find the x-intercept, substitute y  0. 1 y x 4 1 102  x 4

Answer

0x

y

7. 2x + y  4

5 4 3 2

The x-intercept is (0, 0).

1 5 4 3 2 1 1 2

(2, 0) 1

2

3 4 (0, 4) 5

3 4

5

x

To find the y-intercept, substitute x  0. 1 y x 4 y

1 102 4

y0 The y-intercept is (0, 0).

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Linear Equations in Two Variables

Notice that the x- and y-intercepts are both located at the origin (0, 0). In this case, the intercepts do not yield two distinct points. Therefore, another point is necessary to draw the line. We may pick any value for either x or y. However, for this equation, it would be particularly convenient to pick a value for x that is a y multiple of 4 such as x  4. 5 4 3 2

1 y x 4 y

1 142 4

1

y  4x (4, 1)

1

Substitute x  4.

5 4 3 2 1 1 2 3 4 1 (0, 0) 2

y1

5

x

3 4 5

The point (4, 1) is a solution to the equation (Figure 2-9).

Figure 2-9

Skill Practice 8. Given y  5x, find the x- and y-intercepts. Then graph the equation.

Example 7

Avoiding Mistakes You can always find a third point on a line to check the accuracy of your graph. For example, the point 14, 12 satisfies the equation y  14 x and lines up with the other points from Example 6.

Interpreting the x- and y-Intercepts of a Line

Companies and corporations are permitted to depreciate assets that have a known useful life span. This accounting practice is called straight-line depreciation. In this procedure the useful life span of the asset is determined, and then the asset is depreciated by an equal amount each year until the taxable value of the asset is equal to zero. The J. M. Gus trucking company purchases a new truck for $65,000. The truck will be depreciated at $13,000 per year. The equation that describes the depreciation line is y  65,000  13,000x where y represents the value of the truck in dollars and x is the age of the truck in years. a. Find the x- and y-intercepts. Plot the intercepts on a rectangular coordinate system, and draw the line that represents the straight-line depreciation. b. What does the x-intercept represent in the context of this problem? c. What does the y-intercept represent in the context of this problem?

Solution: a. To find the x-intercept, substitute y  0. 0  65,000  13,000x 13,000x  65,000

To find the y-intercept, substitute x  0. y  65,000  13,000102 y  65,000

y

8.

x5 The x-intercept is (5, 0).

Answer

The y-intercept is (0, 65,000).

5 4 3 2

y  5x

(0, 0) 1 5 4 3 2 1 1 2 3 4 5

1

2

3 4

5

x

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Taxable Value ($)

y

TIP: In Example 7 we graphed the line only in the first quadrant where both the x- and y-coordinates are positive. (A negative x-coordinate would imply a negative age, and a negative y-coordinate would imply a negative value of a car, neither of which makes sense.)

70,000 60,000 50,000 40,000 30,000 20,000 10,000 0 0

Taxable Value of a Truck Versus the Age of the Vehicle

1

2

3 4 Age (years)

5

6

x

b. The x-intercept (5, 0) indicates that when the truck is 5 yr old, the taxable value of the truck will be $0. c. The y-intercept (0, 65,000) indicates that when the truck was new (0 yr old), its taxable value was $65,000. Skill Practice 9. Acme motor company tests the engines of its trucks by running the engines in a laboratory. The engines burn 4 gal of fuel per hour. The engines begin the test with 30 gal of fuel. The equation y  30  4x represents the amount of fuel y left in the engine after x hours. a. Find the x- and y-intercepts. b. Interpret the x-intercept in the context of this problem. c. Interpret the y-intercept in the context of this problem.

5. Horizontal and Vertical Lines Recall that a linear equation can be written in the form Ax  By  C, where A and B are not both zero. If either A or B is 0, then the resulting line is horizontal or vertical, respectively.

DEFINITION Vertical and Horizontal Lines 1. A vertical line is a line whose equation can be written in the form x  k, where k is a constant. 2. A horizontal line is a line whose equation can be written in the form y  k, where k is a constant.

Example 8

Graphing a Vertical Line

Graph the equation x  6. Answer

Solution:

9. a. x-intercept: (7.5, 0); y-intercept:

Because this equation is in the form x  k, the line is vertical and must cross the x-axis at x  6. We can also construct a table of solutions to the equation x  6. The choice for the x-coordinate must be 6, but y can be any real number (Figure 2-10).

(0, 30) b. The x-intercept (7.5, 0) represents the amount of fuel in the truck after 7.5 hr. After 7.5 hr the tank contains 0 gal. It is empty. c. The y-intercept (0, 30) represents the amount of fuel in the truck initially (after 0 hr). After 0 hr, the tank contains 30 gal of fuel.

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Linear Equations in Two Variables

y 5 4 3 2

x

1

y

6

2 1 1 2

–4

6

1

6

5

x6

1

2 3 4

5

6 7

8

x

3 4 5

Figure 2-10

Skill Practice 10. Graph the equation x  4.

Example 9

Graphing a Horizontal Line

Graph the equation 4y  7.

Solution: The equation 4y  7 is equivalent to y  74. Because the equation is in the form y  k, the line must be horizontal and must pass through the y-axis at y  74 (Figure 2-11). We can also construct a table of solutions to the equation 4y  7. The choice for the y-coordinate must be 74 , but x can be any real number. y 5 4 3 2

x

y

0

 74

3

 74

2

 74

1 5 4 3 2 1 1 2

y   74 1

2

3 4

5

x

3 4 5

Figure 2-11

Answers y

10. 5 4 x  4 3 2

Skill Practice 11. Graph the equation 2y  9.

1 5 4 3 2 1 1 2

1

2

3 4

5

1

2

3 4

5

x

3 4 5

TIP: Notice that horizontal and vertical lines that do not pass through the origin have only one intercept. For instance, • The vertical line in Example 8 has an x-intercept but no y-intercept. • The horizontal line in Example 9 has a y-intercept but no x-intercept.

11.

y 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

2y  9

x

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Chapter 2 Linear Equations in Two Variables and Functions

Calculator Connections Topic: Using the Table and Graph Features A viewing window of a graphing calculator shows a portion of a rectangular coordinate system. The standard viewing window for most calculators shows both the x- and y-axes between ⫺10 and 10. Furthermore, the scale defined by the tick marks on both axes is usually set to 1.

The standard viewing window is shown here.

Linear equations can be analyzed with a graphing calculator. •

It is important to isolate the y-variable in the equation. Then enter the equation in the calculator. For example, to enter the equation from Example 5, we have: 2x ⫹ 4y ⫽ 8

4y ⫽ ⫺2x ⫹ 8 4y ⫺2x 8 ⫽ ⫹ 4 4 4 1 y⫽⫺ x⫹2 2

•

A Table feature can be used to find many solutions to an equation. Several solutions to 1 y ⫽ ⫺ x ⫹ 2 are shown here. 2

•

A Graph feature can be used to graph a line.

Sometimes the standard viewing window does not provide an adequate display for the graph of an equation. For example, in the standard viewing window, the graph of y ⫽ ⫺x ⫹ 15 is visible only in a small portion of the upper right corner.

To see the x- and y-intercepts of this line, we can change the viewing window to accommodate larger values of x and y. Most calculators have a Range or Window feature that enables the user to change the minimum and maximum x- and y-values. In this case, we changed the values of x to range between ⫺5 and 20, and the values of y to range between ⫺10 and 20. We also changed the scaling for the x- and y-axes so that the tick marks appear in increments of 5 (Xscl ⫽ 5 and Yscl ⫽ 5).

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Section 2.1

Section 2.1

139

Linear Equations in Two Variables

Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. After getting a test back, it is a good idea to correct the test so that you do not make the same errors again. One recommended approach is to use a clean sheet of paper, and divide the paper down the middle vertically, as shown. For each problem that you missed on the test, rework the problem correctly on the left-hand side of the paper. Then give a written explanation on the righthand side of the paper. To reinforce the correct procedure, return to the section of text from which the problem was taken and do several more problems.

Perform the correct math here.

Explain the process here.

2 ⫹ 4(5) ⫽ 2 ⫹ 20 ⫽ 22

Do multiplication before addition.

Take the time this week to make corrections from your last test. 2. Define the key terms. a. Rectangular coordinate system

b. x-Axis

c. y-Axis

d. Origin

e. Quadrant

f. Ordered pair

g. x-Coordinate

h. y-Coordinate

i. Linear equation in two variables

j. x-Intercept

k. y-Intercept

l. Vertical line

m. Horizontal line

Concept 1: The Rectangular Coordinate System 3. Given the coordinates of a point, explain how to determine in which quadrant the point lies. 4. What is meant by the word ordered in the term ordered pair? 5. Plot the points on a rectangular coordinate system. (See Example 1.) a. 1⫺2, 12

b. 10, 42

3 7 e. a , ⫺ b 2 3

f. 1⫺4.1, ⫺2.72

c. 10, 02

d. 1⫺3, 02

6. Plot the points on a rectangular coordinate system. a. 1⫺2, 52 c. 14, ⫺32 e. 12, 22

5 b. a , 0b 2

d. 10, ⫺22

f. 1⫺3, ⫺32

7. A point on the x-axis will have what y-coordinate? 8. A point on the y-axis will have what x-coordinate?

y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1

2

3 4

5

1

2

3 4

5

x

⫺3 ⫺4 ⫺5 y 6 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4

x

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Chapter 2 Linear Equations in Two Variables and Functions

For Exercises 9–10, give the coordinates of the labeled points, and state the quadrant or axis where the point is located. 9.

10.

y 5 4 3 2

A

A C

1

B

5 4 3 2 1 1 2

1

B

1 2

3 4

5

x

5 4 3 2 1 1 2 E

D

3

E

y 5 4 3 2 1

3 C

4

4

5

5

2

3 4

5

x

D

Concept 2: Linear Equations in Two Variables For Exercises 11–14, determine if the ordered pair is a solution to the linear equation. (See Example 2.) 11. 2x  3y  9

1 13. x  y  1 3

12. 5x  2y  6

3 14. y   x  4 2

a. 10, 32

a. 10, 32

a. 11, 02

a. 10, 42

b. 16, 12

6 b. a , 0b 5

b. 12, 32

b. 12, 72

7 c. a1,  b 3

c. 12, 22

c. 16, 12

c. 14, 22

Concept 3: Graphing Linear Equations in Two Variables For Exercises 15–18, complete the table. Then graph the line defined by the points. (See Examples 3–4.) y

15. 3x  2y  4 x

y

0

1

1

5 4 3 2 1 1 2

4

y

16. 4x  3y  6

5 4 3 2

x 1

2

3 4

5

x

5 4 3 2

y 2

1

1

5 4 3 2 1 1 2

3

3

x

y

y

1

5

5 4 3 2 1 1 2

5

3 4 5

3 4

5

x

4 5

y

1 18. y  x 3

5 4 3 2

0

2

3

4 5

1 17. y   x 5

1

x 1

2

3 4

5

x

0 3 6

y

5 4 3 2 1 43 2 1 1 2 3 4 5

1 2 3 4 5 6

x

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141

Linear Equations in Two Variables

In Exercises 19–30, graph the linear equation. (See Examples 3–4.) 19. x  y  5

20. x  y  8

21. 3x  4y  12

y

y

7 6

1

5 4 3 2

8 7 6 5 4 3 2 1 1

1

4 5

3 2 1 1

y 5 4 3 2

2

1

2

x

2 3

1

2

3 4

5

6 7

x

1 5 4 3 2 1 1 2

23. y  3x  5

y

5 4 3 2 1 1 2

1

2

3 4

5

x

3

5 4 3 2 1 1 2

4

3

2 25. y  x  1 5

5 26. y  x  1 3

2

3 4

5

x

1

2

3 4

5

5 4 3 2 1 1 2

4 5

y

1

2

3 4

5

x

5 4 3 2 1 1 2

3 4

5

x

1

2

3 4

5

30. x  3y y 5 4 3 2

1 2

x

4 5

y

1

5

3

5 4 3 2

1 5 4 3 2 1 1 2

3 4

1

29. x  2y

5 4 3 2

2

y 5 4 3 2

3

28. x  4y  2

1

x

27. x  5y  5

1

3

5

4 5

5 4 3 2

x

3 4

3

7 6

1 5 4 3 2 1 1 2

1

5 4 3 2 1 1 2

y

y

2

1

1

5 4 3 2

1

y

5 4 3 2

1

x

5 4 3 2

7 6

5 4 3 2

5

24. y  2x  2

y

6

3 4

4 5

8

22. 5x  3y  15

2

3

6 7

2 3

1

5 4 3 2 1 1 2

1 1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

x

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Chapter 2 Linear Equations in Two Variables and Functions

Concept 4: x - and y -Intercepts 31. Given a linear equation, how do you find an x-intercept? How do you find a y-intercept? 32. Can the point (4, 1) be an x- or y-intercept? Why or why not? For Exercises 33–44, a. find the x-intercept, b. find the y-intercept, and c. graph the equation. (See Examples 5–6.) 33. 2x  3y  18

34. 2x  5y  10 y

y 8

5

7

4

6

3 2 1

5 4 3 2

3 2 1 1

1

2

3 4

5

6 7 8 9

y 5 4 3 2 1 1

2

x

5

6 7

x

5 4 3 2 1 1 2

37. 5x  3y y

6

5 4 3 2

5 4 3 2

5 4 3 2 1 1 2

7

1

2 3

4

5

6 7 8 9

39. y  2x  4

3 4

5

x

5 4 3 2 1 1 2 4 5

40. y  3x  1

4 41. y   x  2 3

y

y

5 4 3 2 1 1 2

2

3

1

2

3 4

5

5 4 3 2 1 1 2

1

2

3 4

5

5 4 3 2 1 1 2

3

3

4 5

4 5

4 5

1 43. x  y 4

2

3 4

5

x

5

1

2

3 4

5

x

5 4 3 2

1 1

3 4

y

5 4 3 2

1 5 4 3 2 1 1 2

2

2 44. x  y 3 y

y 5 4 3 2

1

x

1 x

3

2 42. y   x  1 5

5

5 4 3 2

1 x

3 4

y

5 4 3 2

1

2

1 1

4 5

5 4 3 2

1

x

y

3 x

5

5 4 3 2

1

1

3 4

38. 3y  5x

y 8

2

4 5

5

9

1

3

4

36. x  y  8

1 1

3 4

2 3

1 1 1 2

35. x  2y  4

5 4 3 2 1 1 2

1 1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

x

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Linear Equations in Two Variables

45. A salesperson makes a base salary of $15,000 a year plus an 8% commission on total sales for the year. The yearly salary can be expressed as a linear equation. (See Example 7.) y  15,000  0.08x a. What is the salesperson’s salary for a year in which his sales total $500,000? b. What is the salary for a year in which sales total $300,000? c. What does the y-intercept mean in the context of this problem?

Yearly Salary ($)

100,000

Yearly Salary Versus Sales

y

80,000

y  15,000  0.08x

60,000 40,000 20,000

d. Why is it unreasonable to use negative values for x in this equation?

0

0

200,000 400,000 600,000 Total Yearly Sales ($)

x

800,000

46. A taxi company in Portland charges $3.50 for any distance up to the first mile and $2.50 for every mile thereafter. The cost of a cab ride can be modeled graphically. a. Explain why the first part of the model is represented by a horizontal line.

35

b. What does the y-intercept mean in the context of this problem?

25

d. How much would it cost to take a cab 3.5 mi?

Cost of Cab Ride Versus Number of Miles

30 Cost ($)

c. Explain why the line representing the cost of traveling more than 1 mi is not horizontal.

y

20 15

3.50

10 5 0

0

1

2

3

4 5 6 7 8 Number of Miles

9

Concept 5: Horizontal and Vertical Lines For Exercises 47–54, determine if the equation represents a horizontal line or a vertical line. Then graph the line and identify the x- and y-intercepts. (See Examples 8–9.) 47. y  1

48. y  3

y

y

5 4 3 2

5 4 3 2

1

1 5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

4 5

4 5

1

2

3 4

5

1

2

3 4

5

x

50. x  5

49. x  2 y

y

5 4 3 2

5 4 3 2

1 5 4 3 2 1 1 2 3 4 5

1 1

2

3 4

5

x

5 4 3 2 1 1 2 3 4 5

x

x

10 11 12

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Chapter 2 Linear Equations in Two Variables and Functions

51. 2x  6  5

52. 3x  12 y

y

5 4 3 2

5 4 3 2

1 5 4 3 2 1 1 2

1 1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

4 5

4 5

53. 2y  1  9

2

3 4

5

1

2

3 4

5

x

54. 5y  10 y

y

5 4 3 2

5 4 3 2

1

1 5 4 3 2 1 1 2

1

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

4 5

4 5

x

55. Explain why not every line has both an x-intercept and a y-intercept. 56. Which of the lines defined here has only one unique intercept? a. 2x  3y  8

b. x  7

c. 3y  9

d. x  y  0

57. Which of the lines defined here has only one unique intercept? a. y  5

b. x  2y  0

c. 3x  4  2

d. x  3y  6

Expanding Your Skills For Exercises 58–61, find the x- and y-intercepts. 58.

y x  1 2 3

59.

y x  1 7 4

60.

y x  1 a b

61. Ax  By  C

Graphing Calculator Exercises For Exercises 62–65, solve the equation for y. Use a graphing calculator to graph the equation on the standard viewing window. 62. 2x  3y  7

63. 4x  2y  2

64. 3y  9

65. 2y  10  0

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Slope of a Line and Rate of Change

145

For Exercises 66–69, use a graphing calculator to graph the lines on the suggested viewing window. 1 66. y   x  10 2

1 67. y   x  12 3

30  x  10 15  y  5

68. 2x  4y  1

10  x  40 10  y  20

1  x  1 1  y  1

69. 5y  4x  1 0.5  x  0.5 0.5  y  0.5

For Exercises 70–71, graph the lines in parts (a)–(c) on the same viewing window. Compare the graphs. Are the lines exactly the same? 70. a. y  x  3

71. a. y  2x  1

b. y  x  3.1

b. y  1.9x  1

c. y  x  2.9

c. y  2.1x  1

Slope of a Line and Rate of Change

Section 2.2

1. Introduction to the Slope of a Line

Concepts

In Section 2.1, we learned how to graph a linear equation and to identify its x- and y-intercepts. In this section, we learn about another important feature of a line called the slope of a line. Geometrically, slope measures the “steepness” of a line. Figure 2-12 shows a set of stairs with a wheelchair ramp to the side. Notice that the stairs are steeper than the ramp.

1. Introduction to the Slope of a Line 2. The Slope Formula 3. Parallel and Perpendicular Lines 4. Applications and Interpretation of Slope

3 ft

3 ft

18 ft

4 ft

Figure 2-12

To measure the slope of a line quantitatively, consider two points on the line. The slope is the ratio of the vertical change between the two points to the horizontal change. That is, the slope is the ratio of the change in y to the change in x. As a memory device, we might think of the slope of a line as “rise over run.” Slope 

change in y rise  run change in x

Change in x (run) Change in y (rise)

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Chapter 2 Linear Equations in Two Variables and Functions

To move from point A to point B on the stairs, rise 3 ft and move to the right 4 ft (Figure 2-13). B 3-ft change in y

Slope ⫽

A

change in y 3 ft 3 ⫽ ⫽ change in x 4 ft 4

4-ft change in x

Figure 2-13

To move from point A to point B on the wheelchair ramp, rise 3 ft and move to the right 18 ft (Figure 2-14). B 3-ft change in y A

18-ft change in x

Figure 2-14

Slope ⫽

change in y 3 ft 1 ⫽ ⫽ change in x 18 ft 6

The slope of the stairs is 34 , which is greater than the slope of the ramp, which is 16.

Example 1

Finding the Slope in an Application

Find the slope of the ladder against the wall.

Solution: Slope ⫽

change in y change in x

⫽

15 ft 5 ft

⫽

3 or 3 1

15 ft

5 ft

The slope is 31 , which indicates that a person climbs 3 ft vertically for every 1 ft traveled horizontally. Skill Practice 1. Find the slope of the roof.

Answer 1.

2 5

8 ft 20 ft

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Slope of a Line and Rate of Change

2. The Slope Formula

y

The slope of a line may be found by using any two points on the line—call these points (x1, y1) and (x2, y2). The change in y between the points can be found by taking the difference of the y-values: y2  y1. The change in x can be found by taking the difference of the x-values in the same order: x2  x1. The slope of a line is often symbolized by the letter m and is given by the following formula.

(x2, y2) Change in y y2  y1 x

(x1, y1) x2  x1 Change in x

FORMULA

Slope of a Line

The slope of a line passing through the distinct points (x1, y1) and (x2, y2) is m

Example 2

y2  y1 x2  x1

provided

x2  x1  0

Finding the Slope of a Line Through Two Points

Find the slope of the line passing through the points (1, 1) and (7, 2).

Solution: To use the slope formula, first label the coordinates of each point, and then substitute their values into the slope formula. 11, 12 1x1, y1 2

and

17, 22 1x2, y2 2

2  112 y2  y1 m  x2  x1 71  

3 6

Label the points.

y 5 4 3 2

Apply the slope formula.

1 3

Simplify.

2 1 1 2

1 2

1

2

3 4

5 6 7 8

(1, 1)

3 4 5

The slope of the line can be verified from the graph (Figure 2-15).

Figure 2-15

TIP: The slope formula does not depend on which point is labeled (x1, y1) and which point is labeled (x2, y2). For example, reversing the order in which the points are labeled in Example 2 results in the same slope: 3 1 1  2   then m and 17, 22 11, 12 17 6 2 1x1, y1 2 1x2, y2 2

Skill Practice 2. Find the slope of the line that passes through the points (–4, 5) and (6, 8).

When you apply the slope formula, you will see that the slope of a line may be positive, negative, zero, or undefined. • Lines that “increase,” or “rise,” from left to right have a positive slope. • Lines that “decrease,” or “fall,” from left to right have a negative slope.

(7, 2) 6

Answer 2.

3 10

x

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Chapter 2 Linear Equations in Two Variables and Functions

• Horizontal lines have a zero slope. • Vertical lines have an undefined slope. Positive slope

Example 3

Negative slope

Zero slope

Undefined slope

Finding the Slope of a Line Through Two Points

Find the slope of the line passing through the points (3, 4) and (5, 1). y

Solution: 13, 42 1x1, y1 2

5 4 3 2 1 5 4 3 2

(5, 1)

8

1 2

1

2 3

3 4 5

4

5

x

3

(3, 4) m   38 The line slopes downward from left to right.

Figure 2-16

m 

and

15, 12 1x2, y2 2

1  142 y2  y1  x2  x1 5  3

3 3  8 8

Label points. Apply the slope formula. Simplify.

The two points can be graphed to verify that 38 is the correct slope (Figure 2-16). Skill Practice Find the slope of the line passing through the given points. 3. (1, –8) and (–5, –4)

Example 4

Finding the Slope of a Line Through Two Points

Find the slope of the line passing through the points (3, 4) and (3, 2).

Solution: 13, 42 1x1, y1 2

m

and

13, 22 1x2, y2 2

y2  y1 2  4  x2  x1 3  132



6 3  3



6 0

Label points. Apply slope formula.

y 5

(3, 4)

3 2 1

Undefined

The slope is undefined. The points form a vertical line (Figure 2-17).

5 4 3 2 1 1 (3, 2) 2 3

1

2 3

4 5

4 5

Figure 2-17

Skill Practice Find the slope of the line passing through the given points. Answers 3. 

2 3

4. Undefined

4. (5, –2) and (5, 5)

x

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Section 2.2

Slope of a Line and Rate of Change

Finding the Slope of a Line Through Two Points

Example 5

Find the slope of the line passing through the points (0, 2) and (4, 2).

Solution: 10, 22 1x1, y1 2

m 

and

14, 22 1x2, y2 2

y2  y1 2 2  x2  x1 4  0

y 5 4 3

Label the points. (4, 2)

Apply the slope formula.

2 (0, 2) 1

5 4 3 2 1 1 2 3

0 4

1

2 3

4 5

x

4

0

Simplify.

5

Figure 2-18

The slope is zero. The line through the two points is a horizontal line (Figure 2-18).

Skill Practice Find the slope of the line passing through the points. 5. (1, 6) and (–7, 6)

3. Parallel and Perpendicular Lines Lines in the same plane that do not intersect are parallel. Nonvertical parallel lines have the same slope and different y-intercepts (Figure 2-19). Lines that intersect at a right angle are perpendicular. If two lines are perpendicular, then the slope of one line is the opposite of the reciprocal of the slope of the other (provided neither line is vertical) (Figure 2-20). These lines are perpendicular. m1  72

These two lines are parallel. m1  53

2 cm

m 2  53

7 cm

m 2  27 2 cm

5 ft

7 cm

5 ft 3 ft 3 ft

Figure 2-19

Figure 2-20

PROPERTY Slopes of Parallel Lines If m1 and m2 represent the slopes of two parallel (nonvertical) lines, then m1  m2 See Figure 2-19.

Answer 5. 0

149

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PROPERTY Slopes of Perpendicular Lines If m1 ⫽ 0 and m2 ⫽ 0 represent the slopes of two perpendicular lines, then m1 ⫽ ⫺

Example 6

1 or, equivalently, m1 ⴢ m2 ⫽ ⫺1 m2

See Figure 2-20.

Determining the Slope of Parallel and Perpendicular Lines

Suppose a given line has a slope of ⫺5. a. Find the slope of a line parallel to the given line. b. Find the slope of a line perpendicular to the given line.

Solution: a. The slope of a line parallel to the given line is m ⫽ ⫺5 (same slope). b. The slope of a line perpendicular to the given line is m ⫽ 15 (the opposite of the reciprocal of ⫺5). 4 Skill Practice The slope of line L1 is ⫺ . 3 6. Find the slope of a line parallel to L1. 7. Find the slope of a line perpendicular to L1.

Example 7

Determining Whether Two Lines Are Parallel, Perpendicular, or Neither

Two points are given from each of two lines: L1 and L2. Without graphing the points, determine if the lines are parallel, perpendicular, or neither. L1: 12, ⫺32 and 14, 12

L2: 15, ⫺62 and 1⫺3, ⫺22

Solution: First determine the slope of each line. Then compare the values of the slopes to determine if the lines are parallel or perpendicular.

TIP: You can also verify that the lines in Example 7 are perpendicular by noting that the product of their slopes is ⫺1. 1 2 ⴢ a⫺ b ⫽ ⫺1 2

For line 1:

For line 2:

L1:

L2:

12, ⫺32 and 14, 12 1x1, y1 2 1x2, y2 2

m⫽ ⫽

1 ⫺ 1⫺32 4⫺2

4 2

⫽2 Answers 6. ⫺

4 3

7.

3 4

15, ⫺62 and 1⫺3, ⫺22 1x1, y1 2 1x2, y2 2

m⫽ ⫽

⫺2 ⫺ 1⫺62 ⫺3 ⫺ 5

Label the points. Apply the slope formula.

4 ⫺8

⫽⫺

1 2

The slope of L1 is 2. The slope of L2 is ⫺12. The slope of L1 is the opposite of the reciprocal of L2. By comparing the slopes, the lines must be perpendicular.

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151

Skill Practice Two points are given for lines L1 and L 2 . Determine if the lines are parallel, perpendicular, or neither. 8. L 1: (4, 1) and (3, 6) L 2: (1, 3) and (2, 0)

4. Applications and Interpretation of Slope In applications, the slope of a line represents a rate of change between the y variable and the x variable. For example, a hiker walking up a hill with a slope of 16 means that 1 ft of elevation is gained for every 6 ft traveled horizontally. Using this rate of change, we can also say that a hiker gains 10 ft of elevation for 60 ft traveled horizontally. See Figure 2-21.

1 ft

10 ft

6 ft 60 ft

Figure 2-21

Interpreting the Slope of a Line in an Application

Example 8

The number of males 20 years old or older who were employed full-time in the United States has grown linearly since 1970. Approximately 43.0 million males 20 years old or older were employed full-time in 1970. By 2005, this number had grown to 65.4 million (Figure 2-22).

Number (millions)

Number of Males 20 Years or Older Employed Full-Time in the United States (2005, 65.4)

70 60 50 40 (1970, 43.0) 30 20 10 0 1965 1970 1975 1980 1985 1990 1995 2000 2005 2010

Figure 2-22 Source: U.S. Census Bureau

a. Find the slope of the line, using the points (1970, 43.0) and (2005, 65.4). b. Interpret the meaning of the slope in the context of this problem.

Solution:

a. 11970, 43.02 1x1, y1 2

and

12005, 65.42 1x2, y2 2

y2  y1 65.4  43.0 m  x2  x1 2005  1970 m

22.4 35

or

Label the points. Apply the slope formula.

m  0.64

b. The slope is approximately 0.64, meaning that the full-time workforce has increased at a rate of 0.64 million men (or 640,000 men) per year between 1970 and 2005. Skill Practice The number of people per square mile (called population density) in Alaska has increased linearly since 1990. In 1990, there were 0.96 people per square mile. By 2010, this increased to 1.32 people per square mile. 9. Find the slope of the line passing through the points (1990, 0.96) and (2010, 1.32). 10. Interpret the meaning of the slope in the context of this problem.

22.4 , 35 means that the workforce increased by 22.4 million men over 35 yr. This is the same

TIP: The slope, m 

0.64 , meaning that 1 the workforce increased by 0.64 million men per year. rate as

Answers 8. Parallel 9. 0.018 10. The population density increased at a rate of 0.018 people per square mile per year.

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Section 2.2 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. Instructors vary in what they emphasize on tests. For example, test material may come from the textbook, notes, handouts, and homework. What did your instructor emphasize on the last test? 2. Define the key term slope.

Review Exercises 3. Find the missing coordinate so that the ordered pairs are solutions to the equation 12x  y  4. a. 1 0, 2

b. 1 , 0 2

c. 14, 2

For Exercises 4–6, find the x- and y-intercepts (if possible) for each equation, and sketch the graph. 4. 2x  8  0

5. 4  2y  0 y

y

y

5 4 3 2

5 4 3 2

5 4 3 2

1 6 5 4 3 2 1 1 2

6. 2x  2y  6  0

1

1 1

2

3 4

x

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

1

2

3 4

5

Concept 1: Introduction to the Slope of a Line 7. A 25-ft ladder is leaning against a house, as shown in the diagram. Find the slope of the ladder. (See Example 1.)

8. Find the slope shown in the figure.

4 ft 10 ft

24 ft 7 ft

x

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9. Find the slope of the treadmill.

153

Slope of a Line and Rate of Change

10. Find the average slope of the hill. 150 yd 500 yd

8 in. 72 in.

11. The road sign shown in the figure indicates the percent grade of a hill. This gives the slope of the road as the change in elevation per 100 horizontal ft. Given a 4% grade, write this as a slope in fractional form.

4% Grade

12. If a plane gains 1000 ft in altitude over a distance of 12,000 horizontal ft, what is the slope? Explain what this value means in the context of the problem.

Concept 2: The Slope Formula For Exercises 13–30, use the slope formula to determine the slope of the line containing the two points. (See Examples 2–5.)

13. 16, 02 and 10, ⫺32

14. 1⫺5, 02 and 10, 42

15. 1⫺2, 32 and 14, ⫺72

16. 1⫺5, ⫺42 and 11, ⫺72

17. 1⫺2, 52 and 12, ⫺32

18. 14, ⫺22 and 16, ⫺82

19. 10.3, ⫺1.12 and 1⫺0.1, ⫺0.82

20. 10.4, ⫺0.22 and 10.3, ⫺0.12

21. 12, 32 and 12, 72

22. 1⫺1, 52 and 1⫺1, 02

23. 15, ⫺12 and 1⫺3, ⫺12

24. 1⫺8, 42 and 11, 42

25. 1⫺4.6, 4.12 and 10, 6.42

26. 11.1, 42 and 1⫺3.2, ⫺0.32

3 4 7 27. a , b and a , 1b 2 3 2

2 1 1 3 28. a , ⫺ b and a⫺ , ⫺ b 3 2 6 2

3 7 1 1 29. a , b and a , 2 b 4 3 2 3

9 2 1 1 30. a , b and a2 , b 4 5 4 10

31. Explain how to use the graph of a line to determine whether the slope of a line is positive, negative, zero, or undefined. 32. If the slope of a line is 43, how many units of change in y will be produced by 6 units of change in x? For Exercises 33–38, estimate the slope of the line from its graph. 33.

34.

y

35.

y

y 5

5 4 3

5 4 3

4 3

2 1

2 1

2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1

2 3

4 5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1

2

3 4 5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1

2

3 4 5

x

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y

36.

y

37.

5

y

38.

5

5

4 3

4 3

4 3

2 1

2 1

2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3 4 5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3 4 5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

2

3 4 5

x

⫺2 ⫺3 ⫺4 ⫺5

⫺2 ⫺3 ⫺4 ⫺5

⫺2 ⫺3 ⫺4 ⫺5

1

Concept 3: Parallel and Perpendicular Lines For Exercises 39–44, the slope of a line is given. a. Find the slope of a line parallel to the given line. b. Find the slope of a line perpendicular to the given line. (See Example 6.) 39. m ⫽ 5 42. m ⫽ ⫺

2 11

4 7

40. m ⫽ 3

41. m ⫽ ⫺

43. m ⫽ 0

44. m is undefined.

45. Can the slopes of two perpendicular lines both be positive? Explain your answer. 46. Suppose a line is defined by the equation x ⫽ 2. What is the slope of a line perpendicular to this line? 47. Suppose a line is defined by the equation y ⫽ ⫺5. What is the slope of a line perpendicular to this line? 48. Suppose a line is defined by the equation x ⫽ ⫺3. What is the slope of a line parallel to this line? 49. What is the slope of a line parallel to the x-axis? 50. What is the slope of a line perpendicular to the y-axis? 51. What is the slope of a line perpendicular to the x-axis? 52. What is the slope of a line parallel to the y-axis? In Exercises 53–60, two points are given from each of two lines L1 and L2. Without graphing the points, determine if the lines are parallel, perpendicular, or neither. (See Example 7.) 53. L1: 12, 52 and 14, 92 L2: 1⫺1, 42 and 13, 22

54. L1: 1⫺3, ⫺52 and 1⫺1, 22 L2: 10, 42 and 17, 22

55. L1: 14, ⫺22 and 13, ⫺12 L2: 1⫺5, ⫺12 and 1⫺10, ⫺162

56. L1: 10, 02 and 12, 32 L2: 1⫺2, 52 and 10, ⫺22

57. L1: 15, 32 and 15, 92 L2: 14, 22 and 10, 22

58. L1: 13, 52 and 12, 52 L2: 12, 42 and 10, 42

59. L1: 1⫺3, ⫺22 and 12, 32 L2: 1⫺4, 12 and 10, 52

60. L1: 17, 12 and 10, 02 L2: 1⫺10, ⫺82 and 14, ⫺62

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Concept 4: Applications and Interpretation of Slope 61. The graph shows the number of cellular phone subscriptions (in millions) purchased in the United States for selected years. (See Example 8.) a. Use the coordinates of the given points to find the slope of the line, and express the answer in decimal form.

62. The U.S. population (in millions) has grown approximately linearly since 1980. a. Find the slope of the line defined by the two given points. b. Interpret the meaning of the slope in the context of this problem.

b. Interpret the meaning of the slope in the context of this problem. U.S. Population by Year Number of Cellular Phone Subscriptions 400 350 Millions

300 250

(2006, 232)

200 150

Population (millions)

350 300 250

(20, 281)

200

(0, 227)

150 100 50 0

100

0

5 10 15 20 25 Year (x ⫽ 0 represents 1980)

(1998, 70)

50 0 1996

2000

2004 Year

2008

30

2012

63. The data in the graph show the average weight for boys based on age. a. Use the coordinates of the given points to find the slope of the line. b. Interpret the meaning of the slope in the context of this problem.

64. The data in the graph show the average weight for girls based on age. a. Use the coordinates of the given points to find the slope of the line, and write the answer in decimal form. b. Interpret the meaning of the slope in the context of this problem.

Average Weight for Boys by Age Average Weight for Girls by Age

80 (10, 74.5)

60 (5, 44.5)

40 20 0

0

1

2

3

4

5 6 7 Age (yr)

8

9

10 11 12

100 90 80 70 60 50 40 30 20 10 0

(11, 87.5)

Weight (lb)

Weight (lb)

100

(5, 42.5)

0

1

2

3

4

5 6 7 Age (yr)

8

9

10 11 12

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Expanding Your Skills For Exercises 65–70, given a point P on a line and the slope m of the line, find a second point on the line (answers may vary). Hint: Graph the line to help you find the second point. 65. P10, 02 and m  2

66. P12, 12 and m  

y

y

5 4 3 2 1 1

2

3 4

5

x

5 4 3 2 1 1 2 3

4 5

5 4 3 2 1 1

2

3 4

5

x

5 4 3 2 1 1 2 3

4 5

68. P12, 42 and m  0

2 3

1

2

3 4

5

x

3 4

5

x

4 5

y

5 4 3 2 1 5 4 3 2 1 1 2 3

2

70. P11, 42 and m 

y

y

1

4 5

69. P11, 22 and m  

7 6 5 4 3 2 1 5 4 3 2 1 1 2 3

67. P12, 32 and m is undefined

y

5 4 3 2 1 5 4 3 2 1 1 2 3

1 3

5 4 3 2 1 1

2

3 4

5

x

5 4 3 2 1 1 2 3

4 5

1

2

3 4

5

x

4 5

71. Given that (2, y) and (4, 6) are points on a line whose slope is 32 , find y. 72. Given that (x, 4) and (3, 2) are points on a line whose slope is 67, find x. 73. The pitch of a roof is defined as

rise . rafter

E

a. Determine the pitch of the roof shown. b. Determine the slope of the line segment from point D to point E.

Section 2.3

4 ft D

F Rafter 24 ft

Equations of a Line

Concepts

1. Slope-Intercept Form

1. Slope-Intercept Form 2. The Point-Slope Formula 3. Equations of a Line: A Summary

In Section 2.1, we learned that an equation of the form Ax  By  C (where A and B are not both zero) represents a line in a rectangular coordinate system. An equation of a line written in this way is in standard form. In this section, we will learn a new form, called the slope-intercept form, which is useful in determining the slope and y-intercept of a line.

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Let (0, b) represent the y-intercept of a line. Let (x, y) represent any other point on the line where x  0. Then the slope of the line through the two points is m

y2  y1 x2  x1

m

yb x

mⴢxa

S

m

yb x0

Simplify.

yb bⴢx x

Clear fractions.

mx  y  b mx  b  y

Apply the slope formula.

Simplify. or

y  mx  b

Solve for y: slope-intercept form

DEFINITION Slope-Intercept Form of a Line y  mx  b is the slope-intercept form of a line. m is the slope and the point (0, b) is the y-intercept. The equation y  4x  7 is written in slope-intercept form. By inspection, we can see that the slope of the line is 4 and the y-intercept is (0, 7). Example 1

Finding the Slope and y-Intercept of a Line

Given 3x  4y  4, write the equation of the line in slope-intercept form. Then find the slope and y-intercept.

Solution: Write the equation in slope-intercept form, y  mx  b, by solving for y. 3x  4y  4 4y  3x  4

Subtract 3x from both sides.

4y 3x 4   4 4 4

To isolate y, divide both sides by 4.

3 y x1 4

3 The slope is  and the y-intercept is (0, 1). 4

Skill Practice Write the equation in slope-intercept form. Determine the slope and the y-intercept. 1. 2x  4y  3

The slope-intercept form is a useful tool to graph a line. The y-intercept is a known point on the line, and the slope indicates the “direction” of the line and can be used to find a second point. Using slope-intercept form to graph a line is demonstrated in Example 2. Answer 1 2

1. y  x 

3 4

3 1 Slope: ; y-intercept: a0,  b 2 4

157

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Example 2

Graphing a Line Using the Slope and y-Intercept

Graph the equation y  34x  1 using the slope and y-intercept.

Solution: First plot the y-intercept (0, 1). The slope m  34 can be written as

m

y

left 4

The change in y is 3.

3 4

y

The change in x is 4.

To find a second point on the line, start at the y-intercept and move down 3 units and to the right 4 units. Then draw the line through the two points (Figure 2-23). Similarly, the slope can be written as

3 4x

1

5 4 3 2 1

Start at up 3 y-intercept (0, 1)

5 4 32 1 1 2 3 4 5 1 down 3 2 right 4 3 4 5

x

Figure 2-23

The change in y is 3.

3 m 4

The change in x is 4.

To find a second point on the line, start at the y-intercept and move up 3 units and to the left 4 units. Then draw the line through the two points (see Figure 2-23). Skill Practice 2. Graph the equation y  15x  2 using the slope and y-intercept.

As we have seen earlier, two lines are parallel if they have the same slope and different y-intercepts. Two lines are perpendicular if the slope of one line is the opposite of the reciprocal of the slope of the other line. Otherwise, the lines are neither parallel nor perpendicular. Example 3

Determining if Two Lines Are Parallel, Perpendicular, or Neither

Given the equations for two lines, L1 and L2, determine if the lines are parallel, perpendicular, or neither. a. L1: y  2x  7 L2: y  2x  1

L2: 4x  6y  12

c. L1: x  y  6 L2: y  6

Solution:

Answer 2.

b. L1: 2y  3x  2

a. The equations are written in slope-intercept form.

y 5 4 3 2

L1: y  2x  7 y

1 5 4 3 2 1 1 2 3 4 5

1

2

1 5x

L2: y  2x  1

2

3 4

5

x

The slope is 2 and the y-intercept is 10, 7 2.

The slope is 2 and the y-intercept is 10, 12.

Because the slopes are the same and the y-intercepts are different, the lines are parallel.

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Equations of a Line

b. Write each equation in slope-intercept form by solving for y. L1: 2y  3x  2 2y 3x 2   2 2 2

L2: 4x  6y  12 Divide by 2.

3 y x1 2

6y  4x  12

Add 4x to both sides.

6y 4 12  x 6 6 6

Divide by 6.

2 y x2 3 3 The slope of L1 is  . 2

2 The slope of L2 is . 3

The value 32 is the opposite of the reciprocal of 23. Therefore, the lines are perpendicular. c. L1: x  y  6 is equivalent to y  x  6. The slope is 1. L2: y  6 is a horizontal line, and the slope is 0. The slopes are not the same. Therefore, the lines are not parallel. The slope of one line is not the opposite of the reciprocal of the other slope. Therefore, the lines are not perpendicular. The lines are neither parallel nor perpendicular. Skill Practice Given the pair of equations, determine if the lines are parallel, perpendicular, or neither. 3 3. y   x  1 4. 3x  y  4 5. x  y  7 4 6x  6  2y x1 4 y x3 3

Example 4

Using Slope-Intercept Form to Find an Equation of a Line

Use slope-intercept form to find an equation of the line with slope 3 and passing through the point (1, 4).

Solution: To find an equation of a line in slope-intercept form, y  mx  b, it is necessary to find the slope, m, and the y-intercept, b. The slope is given in the problem as m  3. Therefore, the slope-intercept form becomes y  mx  b y  3x  b

Answers 3. Perpendicular 5. Neither

4. Parallel

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Furthermore, because the point (1, 4) is on the line, it is a solution to the equation. Therefore, if we substitute (1, 4) for x and y in the equation, we can solve for b. 4  3112  b 4  3  b 1  b

Thus, the slope-intercept form is y  3x  1. Skill Practice 6. Use slope-intercept form to find an equation of the line with slope 2 and passing through (3, 5).

Calculator Connections Topic: Using the Value Feature The Value feature of a graphing calculator prompts the user for a value of x, and then returns the corresponding y-value of an equation. We can check the answer to Example 4 by graphing the equation, y  3x  1. Using the Value feature, we see that the line passes through the point (1, 4) as expected.

2. The Point-Slope Formula In Example 4, we used the slope-intercept form of a line to construct an equation of a line given its slope and a known point on the line. Here we provide another tool called the point-slope formula that (as its name suggests) can accomplish the same result. Suppose a nonvertical line passes through a given point (x1, y1) and has slope m. If (x, y) is any other point on the line, then m m1x  x1 2 

y  y1 x  x1

Slope formula

y  y1 1x  x1 2 x  x1

Clear fractions.

m1x  x1 2  y  y1 or

Answer 6. y  2x  1

y  y1  m1x  x1 2

Point-slope formula

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FORMULA The Point-Slope Formula The point-slope formula is given by

y ⫺ y1 ⫽ m1x ⫺ x1 2

where m is the slope of the line and 1x1, y1 2 is a known point on the line. The point-slope formula is used specifically to find an equation of a line when a point on the line is known and the slope is known. To illustrate the point-slope formula, we will repeat the problem from Example 4.

Using the Point-Slope Formula to Find an Equation of a Line

Example 5

Use the point-slope formula to find an equation of the line having a slope of ⫺3 and passing through the point (1, ⫺4). Write the answer in slope-intercept form.

Solution: m ⫽ ⫺3

and

1x1, y12 ⫽ 11, ⫺42

y ⫺ y1 ⫽ m1x ⫺ x1 2

y ⫺ 1⫺42 ⫽ ⫺31x ⫺ 12 y ⫹ 4 ⫽ ⫺31x ⫺ 12

Apply the point-slope formula. Simplify.

To write the answer in slope-intercept form, clear parentheses and solve for y. y ⫹ 4 ⫽ ⫺3x ⫹ 3 y ⫽ ⫺3x ⫺ 1

Clear parentheses. Solve for y. The answer is written in slopeintercept form. Notice that this is the same equation as in Example 4.

Skill Practice 7. Use the point-slope formula to write an equation for the line passing through the point (⫺2, ⫺6 ) and having a slope of ⫺5. Write the answer in slope-intercept form.

TIP: The solution to Example 5 can also be written in standard form, Ax ⫹ By ⫽ C. y ⫽ ⫺3x ⫺ 1

Slope-intercept form

3x ⫹ y ⫽ ⫺3x ⫹ 3x ⫺ 1

Add 3x to both sides.

3x ⫹ y ⫽ ⫺1

Standard form

In general, we will write the solution in slope-intercept form, because the slope and y-intercept can be easily identified.

Answer 7. y ⫽ ⫺5x ⫺ 16

161

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Example 6

Finding an Equation of a Line Given Two Points

Find an equation of the line passing through the points (5, 1) and (3, 1). Write the answer in slope-intercept form.

Solution: The slope formula can be used to compute the slope of the line between two points. Once the slope is known, the point-slope formula can be used to find an equation of the line. First find the slope. m

1  112 y2  y1 2    1 x2  x1 35 2

Next, apply the point-slope formula. y  y1  m1x  x1 2

y  1  11x  32

Substitute m  1 and use either point for (x1, y1). We will use (3, 1) for (x1, y1).

y  1  x  3

Clear parentheses.

y  x  4

Solve for y. The final answer is in slopeintercept form.

Skill Practice 8. Use the point-slope formula to write an equation of the line that passes through the points 15, 22 and 11, 12. Write the answer in slope-intercept form.

TIP: In Example 6, the point (3, 1) was used for (x1, y1) in the point-slope formula. However, either point could have been used. Using the point (5, 1) for (x1, y1) produces the same final equation: y  112  11x  52 y  1  x  5

y  x  4

Example 7

Finding an Equation of a Line Parallel to Another Line

Find an equation of the line passing through the point 12, 32 and parallel to the line 4x  y  8. Write the answer in slope-intercept form.

Solution: To find an equation of a line, we must know a point on the line and the slope. The known point is 12, 32. Because the line is parallel to 4x  y  8, the two lines must have the same slope. Writing the equation 4x  y  8 in slopeintercept form, we have y  4x  8. Therefore, the slope of both lines must be 4. Answer 3 7 8. y   x  4 4

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Now find an equation of the line passing through (2, 3) having a slope of 4. y  y1  m1x  x1 2

Apply the point-slope formula.

y  132  4 3x  122 4

Substitute m  4 and 12, 32 for 1x1, y1 2.

y  3  41x  22 y  3  4x  8 y  4x  11

Clear parentheses.

y 12 10 8 6 4

Write the answer in slope-intercept form.

Skill Practice

y  4x  11

9. Find an equation of a line containing (4, 1) and parallel to 2x  y  7. Write the answer in slope-intercept form.

2 12108 6 4 2 2 (2, 3) 4 6

Solution: The slope of the given line can be found from its slope-intercept form. 2x  3y  3 Solve for y.

3y 2x 3   3 3 3 2 The slope is  . 3

The slope of a line perpendicular to this line must be the opposite of the reciprocal of 23; hence, m  32. Using m  32 and the known point (4, 3), we can apply the point-slope formula to find an equation of the line. y  y1  m1x  x1 2

Apply the point-slope formula.

3 y  3  1x  42 2

Substitute m  32 and (4, 3) for (x1, y1).

3 y3 x6 2

Clear parentheses.

3 y x3 2

Solve for y.

Skill Practice 10. Find an equation of the line passing through the point (1,6) and perpendicular to the line x  2y  8. Write the answer in slope-intercept form.

y  4x  8

12

Find an equation of the line passing through the point (4, 3) and perpendicular to the line 2x  3y  3. Write the answer in slope-intercept form.

2 y x1 3

6 8 10 12

Figure 2-24

Finding an Equation of a Line Perpendicular to Another Line

3y  2x  3

4

8 10

We can verify the answer to Example 7 by graphing both equations. We see that the line defined by y  4x  11 passes through the point (2, 3) and is parallel to the line y  4x  8. See Figure 2-24. Example 8

2

Answers 9. y  2x  9 10. y  2x  8

x

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Calculator Connections From Example 8, the line defined by y  32x  3 should be perpendicular to the line y  23x  1 and should pass through the point (4, 3). Note: In this example, we are using a square window option, which sets the scale to display distances on the x- and y-axes as equal units of measure.

y  23 x  1

y  32 x  3

3. Equations of a Line: A Summary A linear equation can be written in several different forms, as summarized in Table 2-2. Table 2-2 Form

Example

Standard Form Ax  By  C

2x  3y  6

Comments A and B must not both be zero.

Horizontal Line yk (k is constant)

y3

Vertical Line xk (k is constant)

x  2

The slope is undefined and the x-intercept is (k, 0).

y  2x  5

Solving a linear equation for y results in slope-intercept form. The coefficient of the x-term is the slope, and the constant defines the location of the y-intercept.

Slope-Intercept Form y  mx  b Slope is m. y-Intercept is (0, b). Point-Slope Formula y  y1  m1x  x1 2 Slope is m and 1x1, y1 2 is a point on the line.

Slope  2 y-Intercept is (0, 5). m  2 1x1, y1 2  13, 12 y  1  21x  32

The slope is zero, and the y-intercept is (0, k).

This formula is typically used to build an equation of a line when a point on the line is known and the slope is known.

Although it is important to understand and apply slope-intercept form and the point-slope formula, they are not necessarily applicable to all problems. Example 9 illustrates how a little ingenuity may lead to a simple solution. Example 9

y

Finding an Equation of a Line

5

Find an equation of the line passing through the point (4, 1) and perpendicular to the x-axis.

4 3 2

(4, 1) 1 5 4 3 2 1 1

x  4

1

2

2 3 4 5

Figure 2-25

Answer 11. y  50

3 4 5

x

Solution: Any line perpendicular to the x-axis must be vertical. Recall that all vertical lines can be written in the form x  k, where k is constant. A quick sketch can help find the value of the constant (Figure 2-25). Because the line must pass through a point whose x-coordinate is 4, the equation of the line is x  4. Skill Practice 11. Write an equation of the line through the point (20, 50) and having a slope of 0.

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Section 2.3 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

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Study Skills Exercises 1. Test yourself. Yes _____ No _____ Did you have sufficient time to study for the test on the last chapter? If not, what could you have done to create more time for studying? Yes _____ No _____ Did you work all the assigned homework problems in this chapter? Yes _____ No _____ If you encountered difficulty in this chapter, did you see your instructor or tutor for help? Yes _____ No _____ Have you taken advantage of the textbook supplements such as the Student Solution Manual? 2. Define the key terms. a. Standard form

b. Slope-intercept form

c. Point-slope formula

Review Exercises 3. Given

y x  1 2 3

y 5 4 3 2

a. Find the x-intercept.

1

b. Find the y-intercept.

5 4 3 2 1 1 2

c. Sketch the graph.

1

2

3 4

5

x

3 4 5

4. Using slopes, how do you determine whether two lines are parallel? 5. Using the slopes of two lines, how do you determine whether the lines are perpendicular? 6. Write the formula to find the slope of a line given two points 1x1, y1 2 and 1x2, y2 2 .

Concept 1: Slope-Intercept Form For Exercises 7–18, determine the slope and the y-intercept of the line. (See Example 1.) 2 7. y   x  4 3

3 8. y  x  1 7

9. y  2  3x

10. y  5  7x

11. 17x  y  0

12. x  y  0

13. 18  2y

1 14. 7  y 2

15. 8x  12y  9

16. 9x  10y  4

17. y  0.625x  1.2

18. y  2.5x  1.8

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In Exercises 19–24, match the equation with the correct graph. 3 19. y  x  2 2 22. y  x 

1 2

a.

21. y 

23. x  2

1 24. y   x  2 2

b.

y

c.

y

5 4 3 2

5 4 3 2

1 1

2

3 4

5

x

y 5 4 3 2

1

5 4 3 2 1 1 2

13 4

20. y  x  3

5 4 3 2 1 1 2

1 1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

d.

e.

y

5 4 3 2

1 5 4 3 2 1 1 2

f.

y

5 4 3 2 2

3 4

5

x

5 4 3 2 1 1 2

2

3 4

5

1

2

3 4

5

x

y 5 4 3 2

1 1

1

1 1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

x

For Exercises 25–30, write the equations in slope-intercept form (if possible). Then graph each line, using the slope and y-intercept. (See Example 2.) 25. y  2  4x

26. 3x  5  y y

1

2

3 4

5

x

28. x  2y  8

6

5 4 3 2 1 1 2 3

1

2

3 4

5

x

1

2

3 4

5

x

4 5

1

2

3 4

5

1

2

3 4

5

x

30. 3x  y  0 y 5 4 3 2 1

5 4 3 2 1 3 2 1 1 2 3

5 4 3 2 1 1 2 3 4 5

y

4 3 2 1

4 5

5 4 3 2 1

29. 2x  5y  0

y

5 4 3 2 1 1 2 3

y

y 7 6 5 4 3 2 1

8 7 6 5 4 3 2 1 5 4 3 2 1 1 2

27. 3x  2y  6

1

2

3 4

5 6 7

x

5 4 3 2 1 1 2 3 4 5

x

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31. Given the standard form of a linear equation Ax  By  C, B  0, solve for y and write the equation in slope-intercept form. What is the slope of the line? What is the y-intercept? 32. Use the result of Exercise 31 to determine the slope and y-intercept of the line 3x  7y  9. For Exercises 33–38, the equations of two lines are given. Determine if the lines are parallel, perpendicular, or neither. (See Example 3.) 33. 3y  5x  1 6x  10y  12

36. 4.8x  1.2y  3.6 y  1  4x

34. x  6y  3

35. 3x  4y  12

1 3x  y  0 2

1 2 x y1 2 3

37. 3y  5x  6

38. y  3x  2

5x  3y  9

6x  2y  6

For Exercises 39–44, use the slope-intercept form of a line to find an equation of the line having the given slope and passing through the given point. (See Example 4.) 39. m  3, 10, 52

40. m  4, 10, 32

41. m  2, 14, 32

42. m  3, 11, 52

4 43. m   , 110, 02 5

2 44. m   , 13, 12 7

Concept 2: The Point-Slope Formula For Exercises 45–74, write an equation of the line satisfying the given conditions. Write the answer in slopeintercept form or standard form. 45. The line passes through the point (0, 2) and has a slope of 3.

46. The line passes through the point (0, 5) and has a slope of 12.

47. The line passes through the point (2, 7) and has a slope of 2. (See Example 5.)

48. The line passes through the point (3, 10) and has a slope of 2.

49. The line passes through the point (2, 5) and has a slope of 3.

50. The line passes through the point (1, 6) and has a slope of 4.

51. The line passes through the point (6, 3) and has a slope of 45.

52. The line passes through the point (7, 2) and has a slope of 72.

53. The line passes through (0, 4) and (3, 0).

54. The line passes through (1, 1) and (3, 7).

(See Example 6.)

55. The line passes through (6, 12) and (4, 10).

56. The line passes through (2, 1) and (3, 4).

57. The line passes through (5, 2) and (1, 2).

58. The line passes through (4, 1) and (2, 1).

59. The line contains the point (3, 2) and is parallel to a line with a slope of 34. (See Example 7.)

60. The line contains the point (1, 4) and is parallel to a line with a slope of 12.

61. The line contains the point (3, 2) and is perpendicular to a line with a slope of 34.

62. The line contains the point (2, 5) and is perpendicular to a line with a slope of 12.

(See Example 8.)

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63. The line contains the point (2, 5) and is parallel to 3x  4y  7.

64. The line contains the point 16, 12 and is parallel to 2x  3y  12.

65. The line contains the point (8, 1) and is perpendicular to 15x  3y  9.

66. The line contains the point 14, 22 and is perpendicular to 4x  3y  6.

67. The line contains the point (4, 0) and is parallel to the line defined by 3x  2y.

68. The line contains the point 13, 02 and is parallel to the line defined by 5x  6y.

69. The line is perpendicular to the line defined by 3y  2x  21 and passes through the point (2, 4).

70. The line is perpendicular to 7y  x  21 and passes through the point 114, 82.

71. The line is perpendicular to 12y  x and passes through (3, 5).

72. The line is perpendicular to 14y  x and passes through 11, 52.

73. The line is parallel to the line 3x  y  7 and passes through the origin.

74. The line is parallel to the line 2x  y  5 and passes through the origin.

Concept 3: Equations of a Line: A Summary For Exercises 75–82, write an equation of the line satisfying the given conditions. 75. The line passes through 12, 32 and has a zero slope.

76. The line contains the point ( 52, 0) and has an undefined slope.

77. The line contains the point 12, 32 and has an undefined slope. (See Example 9.)

78. The line contains the point ( 52, 0) and has a zero slope.

79. The line is parallel to the x-axis and passes through (4, 5).

80. The line is perpendicular to the x-axis and passes through (4, 5).

81. The line is parallel to the line x  4 and passes through (5, 1).

82. The line is parallel to the line y  2 and passes through (3, 4).

Expanding Your Skills 83. Is the equation x  2 in slope-intercept form? Identify the slope and y-intercept.

84. Is the equation x  1 in slope-intercept form? Identify the slope and y-intercept.

85. Is the equation y  3 in slope-intercept form? Identify the slope and y-intercept.

86. Is the equation y  5 in slope-intercept form? Identify the slope and y-intercept.

Graphing Calculator Exercises For Exercises 87–90, use a graphing calculator to graph the lines on the same viewing window. Then explain how the lines are related. 1 87. y1  x  4 2

1 88. y1   x  5 3

1 y2  x  2 2

1 y2   x  3 3

89. y1  x  2

90. y1  2x  1

y2  2x  2

y2  3x  1

y3  3x  2

y3  4x  1

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For Exercises 91–92, use a graphing calculator to graph the lines on a square viewing window. Then explain how the lines are related. 1 92. y1 ⫽ x ⫺ 3 91. y1 ⫽ 4x ⫺ 1 2 1 y2 ⫽ ⫺2x ⫺ 3 y2 ⫽ ⫺ x ⫺ 1 4 93. Use a graphing calculator to graph the equation from Exercise 53. Use the Value feature to verify that the line passes through the points (0, 4) and (3, 0).

94. Use a graphing calculator to graph the equation from Exercise 54. Use the Value feature to verify that the line passes through the points (1, 1) and (3, 7).

Problem Recognition Exercises Characteristics of Linear Equations For Exercises 1–20, choose the equation(s) from column B whose graph satisfies the condition described in column A. Give all possible answers. Column A

Column B

1. Line whose slope is positive.

2. Line whose slope is negative.

a. y ⫽ ⫺4x

3. Line that passes through the origin.

4. Line that contains the point (2, 0).

b. 2x ⫺ 4y ⫽ 4

5. Line whose y-intercept is (0, ⫺3).

6. Line whose y-intercept is (0, 0).

1 7. Line whose slope is ⫺ . 3

1 8. Line whose slope is . 2

9. Line whose slope is 0.

10. Line whose slope is undefined.

11. Line that is parallel to the line with equation x ⫹ 3y ⫽ 6.

12. Line perpendicular to the line with equation x ⫺ 4y ⫽ ⫺4.

13. Line that is vertical.

14. Line that is horizontal.

15. Line whose x-intercept is (12, 0).

1 16. Line whose x-intercept is a , 0b. 5

17. Line that has no x-intercept.

18. Line that is perpendicular to the x-axis.

19. Line with a negative slope and positive y-intercept.

20. Line with a positive slope and negative y-intercept.

1 c. y ⫽ ⫺ x ⫺ 3 3 d. 3x ⫹ 5y ⫽ 10 e. 3y ⫽ ⫺9 f. y ⫽ 5x ⫺ 1 g. 4x ⫹ 1 ⫽ 9 h. x ⫹ 3y ⫽ 12

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Section 2.4

Applications of Linear Equations and Modeling

Concepts

1. Writing a Linear Model

1. Writing a Linear Model 2. Interpreting a Linear Model 3. Finding a Linear Model from Observed Data Points

A mathematical model is a formula or equation that represents a relationship between two or more variables in a real-world application. Algebra (or some other field of mathematics) can then be used to solve the problem. The use of mathematical models is found throughout the physical and biological sciences, sports, medicine, economics, business, and many other fields. For an equation written in slope-intercept form, y  mx  b, the term mx is called the variable term. The value of this term changes with different values of x. The term b is called the constant term and it remains unchanged regardless of the value of x. The slope of the line, m, is called the rate of change. A linear equation can be created if the rate of change and the constant are known.

Example 1

Writing a Linear Model

Buffalo, New York, had 2 ft (24 in.) of snow on the ground before a snowstorm. During the storm, snow fell at an average rate of 58 in./hr. a. Write a linear equation to compute the total snow depth y after x hr of the storm. b. Graph the equation. c. Use the equation to compute the depth of snow after 8 hr. d. If the snow depth was 31.5 in. at the end of the storm, determine how long the storm lasted.

Solution: a. The constant or base amount of snow before the storm began is 24 in. The rate of change is given by 85 in. of snow per hour. If m is replaced by 58 and b is replaced by 24, we have the linear equation y  mx  b 5 y  x  24 8 b. The equation is in slopeintercept form, and the corresponding graph is shown in the figure. 5 c. y  x  24 8 y

5 182  24 8

y  5  24

Snow Depth (in.)

Snow Depth Versus Time 40 35 30 25 20 15 10 5 0 0

y  58 x  24 5 in. 8 hr

2

4

6

8 10 12 14 16 18 20 Time (hr)

Substitute x  8. Solve for y.

y  29 in. The snow depth was 29 in. after 8 hr. The corresponding ordered pair is (8, 29) and can be confirmed from the graph.

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d.

Applications of Linear Equations and Modeling

5 y  x  24 8 5 31.5  x  24 8

Substitute y  31.5.

5 8131.52  8 a x  24b 8 252  5x  192 60  5x

Multiply by 8 to clear fractions. Clear parentheses. Solve for x.

12  x The storm lasted for 12 hr. The corresponding ordered pair is (12, 31.5) and can be confirmed from the graph. Skill Practice When Joe graduated from college, he had $1000 in his savings account. When he began working, he decided he would add $120 per month to his savings account. 1. Write a linear equation to compute the amount of money y in Joe’s account after x months of saving. 2. Use the equation to compute the amount of money in Joe’s account after 6 months. 3. Joe needs $3160 for a down payment for a car. How long will it take for Joe’s account to reach this amount?

2. Interpreting a Linear Model Example 2

Interpreting a Linear Model

In 1938, President Franklin D. Roosevelt signed a bill enacting the Fair Labor Standards Act of 1938 (FLSA). In its final form, the act banned oppressive child labor and set the minimum hourly wage at 25 cents and the maximum workweek at 44 hr. Over the years, the minimum hourly wage has been increased by the government to meet the rising cost of living. The minimum hourly wage y (in dollars per hour) in the United States since 1970 can be approximated by the equation y  0.14x  1.60

x0

Minimum Wage ($/hr)

where x represents the number of years since 1970 (x  0 corresponds to 1970, x  1 corresponds to 1971, and so on) (Figure 2-26). 8 7 6 5 4 3 2 1 0 0

Federal Minimum Hourly Wage by Year y  0.14x  1.60

5 10 15 20 25 30 35 Year (x  0 represents the year 1970)

Figure 2-26

40

Answers 1. y  120x  1000 2. $1720 3. 18 months

171

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a. Determine the slope of the line and interpret the meaning of the slope as a rate of change. b. Find the y-intercept of the line and interpret the meaning of the y-intercept in the context of this problem. c. Use the linear equation to approximate the minimum wage in 1985.

Solution: a. The equation y ⫽ 0.14x ⫹ 1.60 is written in slope-intercept form. The slope is 0.14 and indicates that minimum hourly wage rose an average of $0.14 per year since 1970. b. The y-intercept is (0, 1.60). The y-intercept indicates that the minimum wage in the year 1970 1x ⫽ 02 was approximately $1.60 per hour. c. The year 1985 is 15 years after the year 1970. Substitute x ⫽ 15 into the linear equation. y ⫽ 0.14x ⫹ 1.60 y ⫽ 0.141152 ⫹ 1.60

Substitute x ⫽ 15.

y ⫽ 2.1 ⫹ 1.60 y ⫽ 3.70 According to the linear model, the minimum wage in 1985 was approximately $3.70 per hour. (The actual minimum wage in 1985 was $3.35 per hour.) Skill Practice One cell-phone plan charges a monthly fee plus a charge per minute for the number of minutes used beyond 400 min. If a customer exceeds the 400-min cap, then the monthly fee, y (in dollars), is given by y ⫽ 0.40x ⫹ 49.99, where x is the number of minutes used beyond 400 min. 4. What is the slope? Interpret its meaning in the context of the problem. 5. What is the y-intercept? 6. Use the equation to determine the cost of using 445 min in this plan.

3. Finding a Linear Model from Observed Data Points Graphing a set of data points offers a visual method to determine whether the points follow a linear pattern. When two variables are related, it is often desirable to find a mathematical equation (or model) to describe the relationship.

Example 3

Answers 4. The slope is 0.40. The customer is charged $0.40 for each minute used beyond 400 min. 5. The y-intercept is (0, 49.99). This means that if 0 min is used beyond 400 min, the customer is charged $49.99. 6. $67.99

Writing a Linear Model from Observed Data

Figure 2-27 represents the number of women, y (in thousands), who graduated from law school in the United States by year. Let x represent the number of years since 1980.

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Number of Women Graduating from Law School by Year 24 Thousands

20 16

(15, 16.8)

12

(25, 19.7)

8 4 0

0

5

10 15 20 25 30 Year (x ⫽ 0 represents 1980) Source: U.S. National Center for Education Statistics

35

Figure 2-27

a. Use the given ordered pairs to find a linear equation to model the number of women graduating from law school by year. b. Determine the slope of the line and interpret the meaning of the slope as a rate of change. c. Use the linear equation to predict the number of women who will graduate from law school in the year 2015. d. Would it be practical to use the linear model to predict the number of women who would graduate from law school for the year 2050?

Solution: a. The slope formula can be used to compute the slope of the line between the points. 115, 16.82 (x1, y1)

m⫽

and

125, 19.72 (x2, y2)

y2 ⫺ y1 19.7 ⫺ 16.8 2.9 ⫽ ⫽ ⫽ 0.29 x2 ⫺ x1 25 ⫺ 15 10

y ⫺ y1 ⫽ m1x ⫺ x1 2

Label the points. Apply the slope formula. Apply the point-slope formula.

y ⫺ 16.8 ⫽ 0.291x ⫺ 152

Use the point (15, 16.8) and m ⫽ 0.29.

y ⫺ 16.8 ⫽ 0.29x ⫺ 4.35

Clear parentheses.

y ⫽ 0.29x ⫹ 12.45

Solve for y. The equation is in slope-intercept form.

b. The slope is 0.29. This means that the number of women who graduate from law school increased at a rate of 0.29 thousand (290) per year during this time period. c. The year 2015 is 35 years after the year 1980. Therefore, substitute x ⫽ 35. y ⫽ 0.291352 ⫹ 12.45 ⫽ 22.6 In the year 2015, we predict that approximately 22,600 women will graduate from law school.

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d. It would not be practical to use this equation to predict the number of women who will graduate from law school in the year 2050. It is unreasonable to assume that the linear trend will continue this far beyond the observed data. Skill Practice The figure shows data relating the cost of college textbooks (in dollars) to the number of pages in the book. Let y represent the cost of the book, and let x represent the number of pages.

Answers 7. y  0.25x  7 8. m  0.25; The cost of the textbook goes up by $0.25 per page. 9. $97

Cost of Textbook Versus Number of Pages y

Cost ($)

7. Use the ordered pairs indicated in the figure to write a linear equation to model the cost of textbooks (in dollars) versus the number of pages. 8. What is the slope of this line and what does it mean in the context of this problem? 9. Use the equation to predict the cost of a textbook that has 360 pages.

120 100 80 60 40 20 0

(400, 107) (200, 57)

0

100

200

300

400

x

500

Number of Pages

Section 2.4 Practice Exercises • Practice Problems • Self-Tests • NetTutor

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Study Skills Exercises 1. On test day, take a look at any formulas or important points that you had to memorize before you enter the classroom. Then when you sit down to take your test, write these formulas on the test or on scrap paper. This is called a memory dump. Write down the formulas from Sections 2.1–2.4. 2. Define the key term mathematical model.

Review Exercises For Exercises 3–5, a. Find the slope (if possible) of the line passing through the two points. b. Find an equation of the line passing through the two points. Write the answer in slope-intercept form (if possible) and in standard form. c. Graph the line by using the slope and y-intercept. Verify that the line passes through the two given points. 3. 13, 02 and 13, 22

4. 11, 12 and 13, 52

y

y

5 4 3 2

y

5 4 3 2

1 5 4 3 2 1 1 2

5. 14, 32 and 12, 32

5 4 3 2

1 1

2

3 4

5

x

5 4 3 2 1 1 2

1 1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

1

2

3 4

5

x

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6. What is the slope of the line defined by x ⫽ ⫺2?

Concept 1: Writing a Linear Model 7. A car rental company charges a flat fee of $59.95 plus $0.45 per mile. (See Example 1.) a. Write an equation that expresses the cost y (in dollars) of renting a car driven x miles. b. Graph the equation. Cost of Rental Car Versus Miles Driven 160 140 Cost ($)

120 100 80 60 40 20 0

0

40

80 120 Number of Miles

160

200

c. What is the y-intercept and what does it mean in the context of this problem? d. Using the equation from part (a), find the cost of driving the rental car 50 mi, 100 mi, and 200 mi. e. What is the slope and what does it mean in the context of this problem? f. Find the total cost of driving the rental car 100 mi if the sales tax is 6%. g. Is it reasonable to use negative values for x in the equation? Why or why not? 8. Alex is a sales representative and earns a base salary of $1000 per month plus a 4% commission on his sales for the month. a. Write a linear equation that expresses Alex’s monthly salary y in terms of his sales x. b. Graph the equation. y 3000

Monthly Salary Versus Sales

Salary ($)

2500 2000 1500 1000 500 10,000

20,000

30,000

40,000

x 50,000

Sales ($)

c. What is the y-intercept and what does it represent in the context of this problem? d. What is the slope of the line and what does it represent in the context of this problem? e. How much will Alex make if his sales for a given month are $30,000?

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9. Ava recently purchased a home in Crescent Beach, Florida. Her property taxes for the first year are $2742. Ava estimates that her taxes will increase at a rate of $52 per year. a. Write an equation to compute Ava’s yearly property taxes. Let y be the amount she pays in taxes, and let x be the time in years.

Taxes ($)

b. Graph the line. Yearly Taxes Versus Year

y 4000 3500 3000 2500 2000 1500 1000 500 2

4

6

8

10 12 14 16 18 20

x

Time (Years)

c. What is the slope of this line? What does the slope of the line represent in the context of this problem? d. What is the y-intercept? What does the y-intercept represent in the context of this problem? e. What will Ava’s yearly property tax be in 10 years? In 15 years? 10. Luigi Luna has started a chain of Italian restaurants called Luna Italiano. He has 19 restaurants in various locations in the northeast United States and Canada. He plans to open five new restaurants per year. a. Write a linear equation to express the number of restaurants, y, Luigi plans to open in terms of the time in years, x. b. How many restaurants will he have in 4 yr? c. How many years will it take him to have 100 restaurants?

Concept 2: Interpreting a Linear Model 11. Sound travels at approximately one-fifth of a mile per second. Therefore, for every 5-sec difference between seeing lightning and hearing thunder, we can estimate that a storm is approximately 1 mi away. Let y represent the distance (in miles) that a storm is from an observer. Let x represent the difference in time between seeing lightning and hearing thunder. Then the distance of the storm can be approximated by the equation y ⫽ 0.2x, where x ⱖ 0. (See Example 2.) a. Use the linear model provided to determine how far away a storm is for the following differences in time between seeing lightning and hearing thunder: 4 sec, 12 sec, and 16 sec. b. If a storm is 4.2 mi away, how many seconds will pass between seeing lightning and hearing thunder? 12. The force y (in pounds) required to stretch a particular spring x in. beyond its rest (or “equilibrium”) position is given by the equation y ⫽ 2.5x, where 0 ⱕ x ⱕ 20. a. Use the equation to determine the amount of force necessary to stretch the spring 6 in. from its rest position. How much force is necessary to stretch the spring twice as far? b. If 45 lb of force is exerted on the spring, how far will the spring be stretched?

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13. The figure represents the median cost of new privately owned, one-family houses sold in the midwest since 1980.

Price ($1000)

250

y

Median Cost of New One-Family Houses Sold in the Midwest y  5.3x  63.4

200 150 100 50 0

0

5 10 15 20 25 Year (x  0 corresponds to 1980)

30

x

Source: U.S. Bureau of the Census and U.S. Department of Housing and Urban Development

Let y represent the median cost of a new privately owned, one-family house sold in the midwest. Let x represent the year, where x  0 corresponds to the year 1980, x  1 represents 1981, and so on. Then the median cost of new privately owned, one-family houses sold in the midwest can be approximated by the equation y  5.3x  63.4, where x  0. a. Use the linear equation to approximate the median cost of new privately owned, one-family houses in the midwest for the year 2005. b. Use the linear equation to approximate the median cost for the year 1988, and compare it with the actual median cost of $101,600. c. What is the slope of the line and what does it mean in the context of this problem? d. What is the y-intercept and what does it mean in the context of this problem?

Miles Driven

14. Let y represent the average number of miles driven per year for passenger cars in the United States since 1980. Let x represent the year where x  0 corresponds to 1980, x  1 corresponds to 1981, and so on. The average yearly mileage for passenger cars can be approximated by the equation y  142x  9060, where x  0.

14,000 12,000 10,000 8,000 6,000 4,000 2,000 0

y

Average Yearly Mileage for Passenger Cars, United States, 1980–2005

y  142x  9060

0

5

10 15 20 25 Year (x  0 corresponds to 1980)

30

x

a. Use the linear equation to approximate the average yearly mileage for passenger cars in the United States in the year 2005. b. Use the linear equation to approximate the average mileage for the year 1985, and compare it with the actual value of 9700 mi. c. What is the slope of the line and what does it mean in the context of this problem? d. What is the y-intercept and what does it mean in the context of this problem?

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Concept 3: Finding a Linear Model from Observed Data Points 15. Windchill is the apparent temperature felt on exposed skin. This is a relationship between air temperature and wind speed. At a fixed air temperature of 5⬚F, windchill temperature is approximately linear for speeds between 10 mph and 110 mph. At 20 mph the windchill is ⫺15⬚F and at 50 mph it is ⫺24⬚F. (See Example 3.) a. Make a graph with wind speed on the x-axis and windchill on the y-axis. Plot the points (20, ⫺15) and (50, ⫺24), and draw the line through the points. Windchill Versus Wind Speed for Fixed Temperature of 5 Fahrenheit Windchill (°F)

30 20 10 0 ⫺10

10

20

30

40

50

60

⫺20 ⫺30

Wind Speed (mph)

b. Write an equation of the line through the given points. Write the equation in slope-intercept form. c. Use the equation from part (b) to estimate the windchill for a wind speed of 40 mph. d. Use the equation from part (b) to estimate the windchill for a wind speed of 46 mph. e. What is the slope of the line and what does it mean in the context of this problem? 16. The figure represents the winning time for the men’s 100-m freestyle swimming event for selected Olympic games.

Time (sec)

60 50

y

(0, 57.3)

40 30 20 10 0

Winning Times for Men's 100-m Freestyle Swimming for Selected Olympics

0

(48, 48.7)

10 20 30 40 50 Year (x ⫽ 0 corresponds to 1948)

60

x

a. Let y represent the winning time (in seconds). Let x represent the number of years since 1948 (where x ⫽ 0 corresponds to the year 1948, x ⫽ 4 represents 1952, and so on). Use the ordered pairs given in the graph (0, 57.3) and (48, 48.7) to find a linear equation to estimate the winning time for the men’s 100-m freestyle versus the year. (Round the slope to two decimal places.) b. Use the linear equation from part (a) to approximate the winning 100-m time for the year 1972, and compare it with the actual winning time of 51.2 sec. c. Use the linear equation to approximate the winning time for the year 1988. d. What is the slope of the line and what does it mean in the context of this problem? e. Interpret the meaning of the x-intercept of this line in the context of this problem. Explain why the men’s swimming times will never “reach” the x-intercept. Do you think this linear trend will continue for the next 50 yr, or will the men’s swimming times begin to “level off” at some time in the future? Explain your answer.

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17. The graph displays the number of associate degrees conferred in the United States at the end of selected academic years. The variable x represents the number of years since 1970, and the variable y represents the number of associate degrees in thousands. Number of Associate Degrees Awarded in the United States (in thousands) for Selected Years Degrees (thousands)

700

y

600 500

(34, 665)

(20, 455)

400 300 200 100 0

0

10 20 30 Year (x ⫽ 0 represents 1970)

40

x

a. Use the points (20, 455) and (34, 665) to create a linear model of the data. b. What does the slope mean in the context of this problem? c. If this linear trend continues, predict the number of associate degrees that will be conferred in the United States in the year 2010. 18. The number of prisoners in federal or state correctional facilities is shown in the figure by year. (Source: U.S. Department of Justice) Number of Prisoners Under Jurisdiction of Federal or State Correctional Authorities

a. Use the given points to create a linear model of the data. 2500

c. Use the equation in part (a) to predict the number of prisoners in federal or state correctional facilities for the year 2012.

Number (1000’s)

b. What does the slope mean in the context of this problem?

y

2000 1500 1000

(4, 1003)

(8, 1220)

500 0

4 8 12 16 Year (x ⫽ 0 represents 1990)

0

20

19. At a concession stand at a high school football game, the owner notices that the relationship between the price of a hot dog and the number of hot dogs sold is linear. If the price is $2.50 per hot dog, then approximately 650 are sold each night. If the price is raised to $3.50, then the number sold drops to 475 per night.

b. Find an equation of the line through the points. Write the equation in slope-intercept form. c. Use the equation from part (b) to predict the number of hot dogs that would sell if the price were raised to $4.00. Round to the nearest whole unit.

1200 Number Sold

a. Make a graph with the price of hot dogs on the x-axis and the number of hot dogs sold on the y-axis. Use the points (2.50, 650) and (3.50, 475). Then graph the line through the points with x ⱖ 0.

y

Number of Hot Dogs Sold Versus Price

1000 800 600 400 200 0

0

x

0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 Price of Hot Dogs ($)

x

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20. Sales at a concession stand indicate that the relationship between the price of a drink and the number of drinks sold is linear. If drinks are sold at $1.00 each, then approximately 1020 are sold each night. If the price is raised to $1.50, then the number sold drops to 820 per night.

Number Sold

y

1600 1400 1200 1000 800 600 400 200 0 0.00

Number of Drinks Sold Versus Price

x

0.50

1.00 1.50 Price per Drink ($)

2.00

a. Make a graph with the price of drinks on the x-axis and the number of drinks sold on the y-axis. Graph the points (1.00, 1020) and (1.50, 820). Then graph the line through the points with x ⱖ 0. b. Find an equation of the line through the points. Write the equation in slope-intercept form. c. Use the equation from part (b) to predict the number of drinks that would sell if the price were $2.00 per drink. 21. In order to advise students properly, a college advisor is interested in the relationship between the number of hours a student studies in an average week and the student’s GPA. The data are shown in the table.

GPA

a. Let x represent study time and let y represent GPA. Graph the points.

Grade Point Average vs. Weekly Study Time y 4 3.5 3 2.5 2 1.5 1 0.5 x 0 0 5 10 15 20 25 30 35 40 45 50 Weekly Study Time (hr)

b. Does there appear to be a linear trend?

Student

Study Time (in hours)

GPA

1

15

2.5

2

38

3.9

3

10

2.1

4

24

2.8

5

35

3.3

6

15

2.7

7

45

4.0

8

28

3.1

9

35

3.4

10

10

2.2

11

6

1.8

c. Use the data points (28, 3.1) and (10, 2.2) to find an equation of the line through these points. d. Use the equation in part (c) to estimate the GPA of a student who studies for 30 hours a week. e. Why would the linear model found in part (c) not be realistic for a student who studies more than 46 hours per week?

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22. Loraine is enrolled in an algebra class that meets 5 days per week. Her instructor gives a test every Friday. Loraine has a study plan and keeps a portfolio with notes, homework, test corrections, and vocabulary. She also records the amount of time per day that she studies and does homework. The following data represent the amount of time she studied per day and her weekly test grades.

Test Score (%)

a. Graph the points on a rectangular coordinate system. Do the data points appear to follow a linear trend? y 90 80 70 60 50 40 30 20 10 10

20 30 40 50 60 70

x 80 90 100

Minutes

Time Studied per Day (min) x

Weekly Test Grade (percent) y

60

69

70

74

80

79

90

84

100

89

b. Find a linear equation that relates Loraine’s weekly test score y to the amount of time she studied per day x. (Hint: Pick two ordered pairs from the observed data, and find an equation of the line through the points.) c. How many minutes should Loraine study per day in order to score at least 90% on her weekly examination? Would the equation used to determine the time Loraine needs to study to get 90% work for other students? Why or why not? d. If Loraine is only able to spend 12 hr/day studying her math, predict her test score for that week.

Expanding Your Skills Points are collinear if they lie on the same line. For Exercises 23–26, use the slope formula to determine if the points are collinear. (Hint: Three points are collinear if the slope calculated using one pair of points is equal to the slope calculated using a different pair of points.) 23. 13, 42 10, 52 19, 22

24. 14, 32 14, 12 12, 22

25. 10, 22 12, 122 11, 62

26. 12, 22 10, 32 14, 12

Graphing Calculator Exercises 27. Use a Table feature to confirm your answers to Exercise 11(a).

28. Use a Table feature to confirm your answers to Exercise 12.

29. Graph the equation y  175x  1087.5 on the viewing window 0  x  5 and 0  y  1200. Use the Eval feature to support your answer to Exercise 19 by showing that the line passes through the points (2.5, 650) and (3.5, 475).

30. Graph the equation y  400x  1420 on the viewing window 0  x  3 and 0  y  1600. Use the Eval feature to support your answer to Exercise 20 by showing that the line passes through the points (1, 1020) and (1.5, 820).

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Section 2.5

Introduction to Relations

Concepts

1. Definition of a Relation

1. Definition of a Relation 2. Domain and Range of a Relation 3. Applications Involving Relations

In many naturally occurring phenomena, two variables may be linked by some other type of relationship. Table 2-3 shows a correspondence between the length of a woman’s femur and her height. (The femur is the large bone in the thigh attached to the knee and hip.) Table 2-3 Length of Femur (cm) x

Height (in.) y

Ordered Pair

45.5

65.5

(45.5, 65.5)

48.2

68.0

(48.2, 68.0)

41.8

62.2

(41.8, 62.2)

46.0

66.0

(46.0, 66.0)

50.4

70.0

(50.4, 70.0)

Each data point from Table 2-3 may be represented as an ordered pair. In this case, the first value represents the length of a woman’s femur and the second, the woman’s height. The set of ordered pairs {(45.5, 65.5), (48.2, 68.0), (41.8, 62.2), (46.0, 66.0), (50.4, 70.0)} defines a relation between femur length and height.

2. Domain and Range of a Relation DEFINITION Relation in x and y A set of ordered pairs (x,y) is called a relation in x and y. Furthermore, • The set of first components in the ordered pairs is called the domain of the relation. • The set of second components in the ordered pairs is called the range of the relation.

Example 1

Finding the Domain and Range of a Relation

Find the domain and range of the relation linking the length of a woman’s femur to her height {(45.5, 65.5), (48.2, 68.0), (41.8, 62.2), (46.0, 66.0), (50.4, 70.0)}.

Solution: Domain:

{45.5, 48.2, 41.8, 46.0, 50.4}

Set of first components

Range:

{65.5, 68.0, 62.2, 66.0, 70.0}

Set of second components

Skill Practice Find the domain and range of the relation. 1 1. e 10, 02, 18, 42, a , 1b, 13, 42, 18, 02 f 2 Answer 1. Domain: e 0, 8, range: 50, 4, 16

1 , 3 f ; 2

The x and y components that constitute the ordered pairs in a relation do not need to be numerical. This is demonstrated in Example 2.

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Introduction to Relations

183

Writing a Relation and Finding Its Domain and Range

Table 2-4 gives five states in the United States and the corresponding number of representatives in the House of Representatives for a recent year.

Table 2-4 Number of Representatives y

State x

a. The data in the table define a relation. Write a list of ordered pairs for this relation.

Alabama

7

California

53

Colorado

7

b. Write the domain and range.

Florida

25

Kansas

4

Solution:

a. {(Alabama, 7), (California, 53), (Colorado, 7), (Florida, 25), (Kansas, 4)} b. Domain: Range:

{Alabama, California, Colorado, Florida, Kansas} {7, 53, 25, 4}

(Note: The element 7 is not listed twice.)

Skill Practice The table depicts six types of animals and their average longevity. 2. Write the ordered pairs indicated by the relation in the table. 3. Find the domain and range of the relation.

Longevity (Years) y

Animal x Bear

22.5

Cat

11

Cow

20.5

Deer

12.5

Dog

11

Elephant

35

A relation may consist of a finite number of ordered pairs or an infinite number of ordered pairs. Furthermore, a relation may be defined by several different methods. • A relation may be defined as a set of ordered pairs. {(1, 2), (3, 4), (1, 4), (3, 4)} • A relation may be defined by a correspondence (Figure 2-28). The corresponding ordered pairs are {(1, 2), (1, 4), (3, 4), (3, 4)}.

x

y

1

2

3

4

3

4

Domain

Range

Figure 2-28 y

• A relation may be defined by a graph (Figure 2-29). The corresponding ordered pairs are {(1, 2), (3, 4), (1, 4), (3, 4)}.

(3, 4)

5 4 3 2 1

5 4 3 2 1 1 2 3 4 5

(3, 4) (1, 2) 1

2

3

4

(1, 4)

Figure 2-29

5

x

Answers 2. {(Bear, 22.5), (Cat, 11), (Cow, 20.5), (Deer, 12.5), (Dog, 11), (Elephant, 35)} 3. Domain: {Bear, Cat, Cow, Deer, Dog, Elephant}; range: {22.5, 11, 20.5, 12.5, 35}

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y

• A relation may be expressed by an equation such as x  y 2. The solutions to this equation define an infinite set of ordered pairs of the form 51x, y2 ƒ x  y 2 6. The solutions can also be represented by a graph in a rectangular coordinate system (Figure 2-30).

4 3

x  y2

2 1 4 3 2 1 1

1

2

3

2 3 4

Figure 2-30

Example 3

Finding the Domain and Range of a Relation

Write the relation as a set of ordered pairs. Then find the domain and range. a.

x

b.

y

y 4 3

3

2

9

2

1 4 3 2 1 1

7

1

2

3

4

x

2 3 4

Solution: a. From the figure, the relation defines the set of ordered pairs: {(3, 9), (2, 9), (7, 9)} Domain: {3, 2, 7} Range:

{9}

b. The points in the graph make up the set of ordered pairs: {(2, 3), (1, 0), (0, 1), (1, 0), (2, 3)} Domain: {2, 1, 0, 1, 2} Range:

{3, 0, 1}

Skill Practice Find the domain and range of the relations. 4.

5. 5

0 8

2 15 4

16

y 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

Answers 4. Domain: {5, 2, 4};

range: {0, 8, 15, 16} 5. Domain: {4, 0, 1, 4}; range: {5, 3, 1, 2, 4}

1

2

3 4

5

x

4

x

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Finding the Domain and Range of a Relation

Example 4

Use interval notation to express the domain and range of the relation. a.

b.

y

y

10 8 6 4 2

5 4 3 2 1

108 6 4 2 2 4 6

2

4

6 8 10

x

5 4 3 2 1 1 2 3

8 10

1

2

3 4

5

x

4 5

Solution: a.

y 10 8 6 4 2 108 6 4 2 2 4 6

2

4

x

6 8 10

Domain: 38, 8 4

8 10

b.

Range:

y

35, 54

The arrow on the curve indicates that the graph extends infinitely far up and to the right. The open circle means that the graph will end at the point (4, 2), but not include that point.

5 4 3 2 1 5 4 3 2 1 1 2 3

The domain consists of an infinite number of x-values extending from 8 to 8 (shown in red). The range consists of all y-values from 5 to 5 (shown in blue). Thus, the domain and range must be expressed in set-builder notation or in interval notation.

1

2

3 4

5

x

Domain: 30, 2 Range:

12, 2

4 5

Skill Practice Use interval notation to express the domain and range of the relations. 6.

7.

y 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

y 5 4 3 2 1

1

2

3 4

5

x

5 4 3 2 1 1 2 3

1

2

3 4

5

x

4 5

Answers 6. Domain: [4, 0] Range: [2, 2] 7. Domain: (, 0] Range: (, )

185

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3. Applications Involving Relations Example 5

Analyzing a Relation

The data in Table 2-5 depict the length of a woman’s femur and her corresponding height. Based on these data, a forensics specialist can find a linear relationship between height y (in inches) and femur length x (in centimeters):

Table 2-5 Length of Femur (cm) x

Height (in.) y

45.5

65.5

48.2

68.0

a. Find the height of a woman whose femur is 46.0 cm.

41.8

62.2

b. Find the height of a woman whose femur is 51.0 cm.

46.0

66.0

50.4

70.0

y ⫽ 0.91x ⫹ 24

40 ⱕ x ⱕ 55

From this type of relationship, the height of a woman can be inferred based on skeletal remains.

c. Why is the domain restricted to 40 ⱕ x ⱕ 55?

Solution: a. y ⫽ 0.91x ⫹ 24 ⫽ 0.91146.02 ⫹ 24

Substitute x ⫽ 46.0 cm.

⫽ 65.86

The woman is approximately 65.9 in. tall.

b. y ⫽ 0.91x ⫹ 24 ⫽ 0.91151.02 ⫹ 24

Substitute x ⫽ 51.0 cm.

⫽ 70.41

The woman is approximately 70.4 in. tall.

c. The domain restricts femur length to values between 40 cm and 55 cm inclusive. These values are within the normal lengths for an adult female and are in the proximity of the observed data (Figure 2-31).

Height (in.)

y

80 70 60 50 40 30 20 10 0 0

Height of an Adult Female Based on the Length of the Femur y ⫽ 0.91x ⫹ 24

5 10 15 20 25 30 35 40 45 50 55 60 Length of Femur (cm)

x

Figure 2-31

Skill Practice The linear equation, y ⫽ ⫺0.014x ⫹ 64.5, for 1500 ⱕ x ⱕ 4000, relates the weight of a car, x (in pounds), to its gas mileage, y (in mpg). 8. Find the gas mileage in miles per gallon for a car weighing 2550 lb. 9. Find the gas mileage for a car weighing 2850 lb. 10. Why is the domain restricted to 1500 ⱕ x ⱕ 4000?

Answers 8. 28.8 mpg 9. 24.6 mpg 10. The relation is valid only for cars weighing between 1500 lb and 4000 lb, inclusive.

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Section 2.5 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. A good way to determine what will be on a test is to look at both your notes and the exercises assigned by your instructor. List five kinds of problems that you think will be on the test for this chapter. 2. Define the key terms. a. Relation in x and y

b. Domain of a relation

c. Range of a relation

Review Exercises For Exercises 3–6, a. Determine if the equation represents a horizontal line, a vertical line, or a slanted line. b. Determine the slope of the line (if it exists). c. Determine the x-intercept (if it exists). d. Determine the y-intercept (if it exists). 3. 2x  3  4

4. 2x  3y  4

5. 3x  2y  4

6. 2  3y  4

Concept 2: Domain and Range of a Relation For Exercises 7–14, a. Write the relation as a set of ordered pairs. b. Determine the domain and range. (See Examples 1–3.) 7. State

% of Populution Uninsured

8.

x

y

0

3

Alabama

15.3

2

1 2

Florida

20.2

7

1

Massachusetts

13.4

2

8

California

18.8

5

1

Minnesota

7.9

9.

Year of First Man or Woman in Space

10.

Year of Statehood, y

USSR

1961

State, x

USA

1962

Maine

1820

Poland

1978

Nebraska

1823

Vietnam

1980

Utah

1847

Cuba

1980

Hawaii

1959

Alaska

1959

Country

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Chapter 2 Linear Equations in Two Variables and Functions

x

y

A

1

12.

x

y

5 2 10

B

2 15

C

3

D

4

E

5

3 20

y

13.

y

14.

5 4 3 2

5 4 3 2

1

1

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

4

4

5

5

1

2

3 4

5

x

For Exercises 15–30, find the domain and range of the relations. Use interval notation where appropriate. (See Example 4.)

15.

16.

y 5 4 3 2 1

5 4 3 2 1 1 2 3

1

2

3 4

5

x

5 4 3 2 1 1 2 3

4 5

19.

17.

y 5 4 3 2 1

(3, 2.8)

1

2

3 4

5

x

5 4 3 2 1 1

20.

4

(0, 3) 2

1

2

3 4

5

5 4 3 2 1 1 2

(3.1, 1.8)

5

5

4 3 2

4 3 2

4 3 2

4 3 2

1

1 1

2

3 4 5

x

5 4 3 2 1 1 2

3 4

3 4

5

5

1

2

3 4 5

x

5 4 3 2 1 1 2

3 4 5

x

y

22.

5

1

2

3 (0, 3) 4

5

5 4 3 2 1 1 2

1

5

y

21.

(3.2, 2)

1 x

4 5

y

y 5

2 3 (1.3, 2.1)

4 5

y

18.

y 5 4 3 2 1

1 1

2

3 4 5

x

5 4 3 2 1 1 2

3 4

3 4

5

5

1

2

3 4 5

x

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23.

24.

y

25.

y

26.

y

y

5

5

5

5

4 3 2

4 3 2

4 3 2

4 3 2

1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1

2

3 4 5

⫺3 ⫺4

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

⫺5

27.

1

1 x

2

3 4 5

1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

(0, ⫺1.3)

2

3 4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

⫺2

⫺2

⫺3 ⫺4

⫺3 ⫺4

⫺3 ⫺4

⫺5

⫺5

⫺5

28.

y

1

x

y

29.

y 5

5

5

4 3 2

4 3 2

4 3 2

4 3 2

1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1

2

3 4 5

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1

2

3 4 5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

2

3 4

5

1

2

3 4 5

1

2

3 4 5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

⫺3 ⫺4

⫺3 ⫺4

⫺3 ⫺4

⫺3 ⫺4

⫺5

⫺5

⫺5

⫺5

Concept 3: Applications Involving Relations 31. The table gives a relation between the month of the year and the average precipitation for that month for Miami, Florida. (See Example 5.)

Month x

Precipitation (in.) y

Month x

Precipitation (in.) y

Jan.

2.01

July

5.70

a. What is the range element corresponding to April?

Feb.

2.08

Aug.

7.58

b. What is the range element corresponding to June?

Mar.

2.39

Sept.

7.63

Apr.

2.85

Oct.

5.64

May

6.21

Nov.

2.66

June

9.33

Dec.

1.83

c. Which element in the domain corresponds to the least value in the range? d. Complete the ordered pair: ( , 2.66) e. Complete the ordered pair: (Sept., )

Source: U.S. National Oceanic and Atmospheric Administration

f. What is the domain of this relation? 32. The table gives a relation between a person’s age and the person’s maximum recommended heart rate.

Age (years) x

Maximum Recommended Heart Rate (Beats per Minute) y

20

200

30

190

c. The range element 200 corresponds to what element in the domain?

40

180

50

170

d. Complete the ordered pair: (50,

60

160

a. What is the domain? b. What is the range?

e. Complete the ordered pair: (

)

x

1

1

1

1

y

30.

5

x

189

Introduction to Relations

, 190)

33. The population of Canada, y (in millions), can be approximated by the relation y ⫽ 0.146x ⫹ 31, where x represents the number of years since 2000. a. Approximate the population of Canada in the year 2006. b. In what year will the population of Canada reach approximately 32,752,000?

x

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34. As of April 2006, the world record times for selected women’s track and field events are shown in the table. The women’s world record time y (in seconds) required to run x meters can be approximated by the relation y  0.159x  10.79. Distance (m)

Time (sec)

Winner’s Name and Country

100

10.49

Florence Griffith Joyner (United States)

200

21.34

Florence Griffith Joyner (United States)

400

47.60

Marita Koch (East Germany)

800

113.28

Jarmila Kratochvilova (Czechoslovakia)

1000

148.98

Svetlana Masterkova (Russia)

1500

230.46

Qu Yunxia (China)

a. Predict the time required for a 500-m race. b. Use this model to predict the time for a 1000-m race. Is this value exactly the same as the data value given in the table? Explain.

Expanding Your Skills 35. a. Define a relation with four ordered pairs such that the first element of the ordered pair is the name of a friend and the second element is your friend’s place of birth. b. State the domain and range of this relation. 36. a. Define a relation with four ordered pairs such that the first element is a state and the second element is its capital. b. State the domain and range of this relation. 37. Use a mathematical equation to define a relation whose second component y is 1 less than 2 times the first component x. 38. Use a mathematical equation to define a relation whose second component y is 3 more than the first component x. 39. Use a mathematical equation to define a relation whose second component is the square of the first component. 40. Use a mathematical equation to define a relation whose second component is one-fourth the first component.

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Introduction to Functions

Introduction to Functions

Section 2.6

1. Definition of a Function

Concepts

In this section, we introduce a special type of relation called a function.

1. 2. 3. 4.

DEFINITION Function Given a relation in x and y, we say “y is a function of x” if, for each element x in the domain, there is exactly one value of y in the range. Note: This means that no two ordered pairs may have the same first coordinate and different second coordinates.

To understand the difference between a relation that is a function and a relation that is not a function, consider Example 1. Example 1

Determining Whether a Relation Is a Function

Determine which of the relations define y as a function of x. a.

b. x

y

2

3 4 ⫺1

3

⫺2

1

c. x

y

x

1

2

1

2

⫺1

2

3

4

3

Solution: a. This relation is defined by the set of ordered pairs same x

511, 32, 11, 42, 12, ⫺12, 13, ⫺226 different y-values

When x ⫽ 1, there are two possible range elements: y ⫽ 3 and y ⫽ 4. Therefore, this relation is not a function.

b. This relation is defined by the set of ordered pairs 511, 42, 12, ⫺12, 13, 226.

Notice that no two ordered pairs have the same value of x but different values of y. Therefore, this relation is a function.

c. This relation is defined by the set of ordered pairs 511, 42, 12, 42, 13, 426.

Notice that no two ordered pairs have the same value of x but different values of y. Therefore, this relation is a function.

y

4

Definition of a Function Vertical Line Test Function Notation Finding Function Values from a Graph 5. Domain of a Function

191

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Skill Practice Determine if the relations define y as a function of x. 1.

x

y

2

12

6

13

7

10

2. {(4, 2), (⫺5, 4), (0, 0), (8, 4)}

3. {(⫺1, 6), (8, 9), (⫺1, 4), (⫺3, 10)}

2. Vertical Line Test y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

Points align vertically 1 2

(4, 2)

3 4 5

x

(4, ⫺2)

⫺3 ⫺4 ⫺5

Figure 2-32

A relation that is not a function has at least one domain element x paired with more than one range value y. For example, the set {(4, 2), (4, ⫺2)} does not define a function because two different y-values correspond to the same x. These two points are aligned vertically in the xy-plane, and a vertical line drawn through one point also intersects the other point (see Figure 2-32). If a vertical line drawn through a graph of a relation intersects the graph in more than one point, the relation cannot be a function. This idea is stated formally as the vertical line test.

PROCEDURE The Vertical Line Test Consider a relation defined by a set of points (x, y) in a rectangular coordinate system. The graph defines y as a function of x if no vertical line intersects the graph in more than one point. The vertical line test can be demonstrated by graphing the ordered pairs from the relations in Example 1. a. 511, 32, 11, 42, 12, ⫺12, 13, ⫺226

b. 511, 42, 12, ⫺12, 13, 226

y

y

5

5

Intersects more than once

4 3 2

4 3 2

1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1 1

3 4 5

⫺3 ⫺4

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1 2

3 4 5

x

⫺3 ⫺4

⫺5

⫺5

Not a Function A vertical line intersects in more than one point.

Example 2

x

Function No vertical line intersects more than once.

Using the Vertical Line Test

Use the vertical line test to determine whether the relations define y as a function of x. a. b. y y

x

Answers 1. Yes

2. Yes

3. No

x

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193

Solution: a.

b.

y

y

x

x

Not a Function A vertical line intersects in more than one point.

Function No vertical line intersects in more than one point.

Skill Practice Use the vertical line test to determine whether the relations define y as a function of x. 4.

5.

y

y

x

x

3. Function Notation A function is defined as a relation with the added restriction that each value in the domain must have only one corresponding y value in the range. In mathematics, functions are often given by rules or equations to define the relationship between two or more variables. For example, the equation y  2x defines the set of ordered pairs such that the y value is twice the x value. When a function is defined by an equation, we often use function notation. For example, the equation y  2x may be written in function notation as f 1x2  2x

• • •

f is the name of the function. x is an input value from the domain of the function. f (x) is the function value (y value) corresponding to x.

The notation f (x) is read as “f of x” or “the value of the function f at x.” A function may be evaluated at different values of x by substituting x-values from the domain into the function. For example, to evaluate the function defined by f1x2  2x at x  5, substitute x  5 into the function.

Avoiding Mistakes Be sure to note that f (x ) is not f ⴢ x.

f 1x2  2x

f 152  2152

f 152  10

TIP: f 152  10 can be

written as the ordered pair (5, 10).

Answers 4. Yes

5. No

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Chapter 2 Linear Equations in Two Variables and Functions

Thus, when x  5, the corresponding function value is 10. We say: • • •

f of 5 is 10. f at 5 is 10. f evaluated at 5 is 10.

The names of functions are often given by either lowercase or uppercase letters, such as f, g, h, p, K, and M. The input variable may also be a letter other than x. For example, the function y  P(t) might represent population as a function of time.

Evaluating a Function

Example 3

Given the function defined by g1x2  12x  1, find the function values. a. g102

b. g122

c. g142

d. g122

Solution: 1 b. g1x2  x  1 2

1 a. g1x2  x  1 2 g102 

1 102  1 2

g122 

01

11

 1

0

We say that “g of 0 is 1.” This is equivalent to the ordered pair 10, 12.

We say that “g of 2 is 0.” This is equivalent to the ordered pair 12, 02.

1 c. g1x2  x  1 2 g142 

d.

1 142  1 2

21

 1  1

1

 2 We say that “g of 2 is 2.” This is equivalent to the ordered pair 12, 22.

Skill Practice Given the function defined by f (x)  2x  3, find the function values.

y 5 4 g(x)  12 x  1 3 2 1

6. f (1)

7. f (0)

8. f (3)

1 9. f a b 2

(4, 1)

5 4 3 2 1 1 (2, 0) 1 (0, 1) 2

5

(2, 2) 4 5

Figure 2-33

Answers 7. 3

1 g1x2  x  1 2 1 g122  122  1 2

We say that “g of 4 is 1.” This is equivalent to the ordered pair 14, 12.

6. 5

1 122  1 2

8. 3

9. 4

x

Notice that g102, g122, g142, and g122 correspond to the ordered pairs 10, 12, 12, 02, 14, 12, and 12, 22. In the graph, these points “line up.” The graph of all ordered pairs defined by this function is a line with a slope of 12 and y-intercept of 10, 12 (Figure 2-33). This should not be surprising because the function defined by g1x2  12x  1 is equivalent to y  12x  1.

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Calculator Connections The values of g1x2 in Example 3 can be found using a Table feature. Y1 ⫽ 12x ⫺ 1

Function values can also be evaluated by using a Value (or Eval) feature. The value of g142 is shown here.

A function may be evaluated at numerical values or at algebraic expressions, as shown in Example 4.

Evaluating Functions

Example 4

Given the functions defined by f1x2 ⫽ x2 ⫺ 2x and g1x2 ⫽ 3x ⫹ 5, find the function values. b. g1w ⫹ 42

a. f1t2

c. f 1⫺t2

Solution: a. f1x2 ⫽ x2 ⫺ 2x

f1t2 ⫽ 1t2 2 ⫺ 21t2 ⫽ t2 ⫺ 2t

Substitute x ⫽ t for all values of x in the function. Simplify.

g1x2 ⫽ 3x ⫹ 5

b.

g1w ⫹ 42 ⫽ 31w ⫹ 42 ⫹ 5

Substitute x ⫽ w ⫹ 4 for all values of x in the function.

⫽ 3w ⫹ 12 ⫹ 5 ⫽ 3w ⫹ 17 c.

f 1x2 ⫽ x2 ⫺ 2x

f 1⫺t2 ⫽ 1⫺t2 ⫺ 21⫺t2

Simplify. Substitute ⫺t for x.

2

⫽ t 2 ⫹ 2t

Simplify.

Skill Practice Given the function defined by g(x ) ⫽ 4x ⫺ 3, find the function values. 10. g(a)

11. g(x ⫹ h)

12. g (⫺x ) Answers 10. 4a ⫺ 3 12. ⫺4x ⫺ 3

11. 4x ⫹ 4h ⫺ 3

195

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Chapter 2 Linear Equations in Two Variables and Functions

4. Finding Function Values from a Graph We can find function values by looking at a graph of the function. The value of f(a) refers to the y-coordinate of a point with x-coordinate a.

Example 5

Finding Function Values from a Graph

Consider the function pictured in Figure 2-34.

h(x) 5

a. Find h112 .

y  h(x) 4 3 2 1

b. Find h122 . c. Find h152 .

5 4 3 2 1 1 2

d. For what value of x is h1x2  3?

1

2

3 4

5

x

3

e. For what values of x is h1x2  0?

4 5

Solution:

Figure 2-34

a. h112  2

This corresponds to the ordered pair (1, 2).

b. h122  1

This corresponds to the ordered pair (2, 1).

c. h152 is not defined.

The value 5 is not in the domain.

d. h1x2  3

for x  4

This corresponds to the ordered pair (4, 3).

e. h1x2  0

for x  3 and x  4

These are the ordered pairs (3, 0) and (4, 0).

Skill Practice Refer to the function graphed here. 13. 14. 15. 16. 17.

Find f (0). Find f (2). Find f (5). For what value(s) of x is f (x )  0? For what value(s) of x is f (x )  4?

y 5 4 3 2 1 5 4 3 2 1 1 2

y  f(x)

1

2

3 4

5

x

3 4 5

5. Domain of a Function A function is a relation, and it is often necessary to determine its domain and range. Consider a function defined by the equation y  f 1x2 . The domain of f is the set of all x-values that when substituted into the function produce a real number. The range of f is the set of all y-values corresponding to the values of x in the domain. To find the domain of a function defined by y  f 1x2, keep these guidelines in mind. • • Answers 13. 15. 16. 17.

3 14. 1 not defined x  4 and x  4 x  5

Exclude values of x that make the denominator of a fraction zero. Exclude values of x that make the expression within a square root negative.

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197

Finding the Domain of a Function

Example 6

Write the domain in interval notation. a. f1x2 ⫽

x⫹7 2x ⫺ 1

c. k1t2 ⫽ 1t ⫹ 4

b. h1x2 ⫽

x⫺4 x2 ⫹ 9

d. g1t2 ⫽ t2 ⫺ 3t

Solution:

a. f 1x2 ⫽ 2xx ⫹⫺ 71 will not be a real number when the denominator is zero, that is, when 2x ⫺ 1 ⫽ 0 2x ⫽ 1 x⫽

1 2

The value x ⫽ 12 must be excluded from the domain. 1 2

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

((( 0

1

2

3

4

5

6

1 1 Interval notation: a⫺⬁, b ´ a , ⬁b 2 2 b. For h1x2 ⫽ xx2 ⫺⫹ 49 the quantity x2 is greater than or equal to 0 for all real numbers x, and the number 9 is positive. The sum x2 ⫹ 9 must be positive for all real numbers x. The denominator will never be zero; therefore, the domain is the set of all real numbers. Interval notation: 1⫺⬁, ⬁ 2

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

c. The function defined by k1t2 ⫽ 1t ⫹ 4 will not be a real number when the radicand is negative. The domain is the set of all t values that make the radicand greater than or equal to zero: t⫹4ⱖ0 t ⱖ ⫺4

Interval notation: 3⫺4, ⬁ 2

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

d. The function defined by g1t2 ⫽ t2 ⫺ 3t has no restrictions on its domain because any real number substituted for t will produce a real number. The domain is the set of all real numbers. Interval notation: 1⫺⬁, ⬁2

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

Skill Practice Write the domain in interval notation. 18. f 1x2 ⫽

2x ⫹ 1 x⫺9

20. g 1x2 ⫽ 1x ⫺ 2

19. k 1x 2 ⫽

⫺5 4x 2 ⫹ 1

21. h 1x2 ⫽ x ⫹ 6

Answers

18. 1⫺⬁, 92 ´ 19, ⬁2 20. 3 2, ⬁2

19. 1⫺⬁, ⬁2 21. 1⫺⬁, ⬁2

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Chapter 2 Linear Equations in Two Variables and Functions

Section 2.6 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercises 1. Look back over your notes for this chapter. Have you highlighted the important topics? Have you underlined the key terms? Have you indicated the places where you are having trouble? If you find that you have problems with a particular topic, write a question that you can ask your instructor either in class or in the instructor’s office. 2. Define the key terms. a. Function

b. Function notation

c. Domain

d. Range

e. Vertical line test

Review Exercises For Exercises 3–4, a. write the relation as a set of ordered pairs, b. identify the domain, and c. identify the range. 3.

Parent, x

Child, y

Kevin

Kayla

Kevin

Kira

Kathleen

Katie

Kathleen

y

4. 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

Kira

1

2

3 4 5

x

⫺3 ⫺4 ⫺5

Concept 1: Definition of a Function For Exercises 5–10, determine if the relation defines y as a function of x. (See Example 1.) 5.

x

6.

y

x

y

⫺2

1

⫺2

1

2

2

2

2

0

8

0

8

9. 511, 22, 13, 42, 15, 42, 1⫺9, 326

8. 3

10 4

5 6

12

7. u

21

w

23

x

24

y

25

z

26

1 1 10. e10, ⫺1.12, a , 8b, 11.1, 82, a4, bf 2 2

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Introduction to Functions

Concept 2: Vertical Line Test For Exercises 11–16, use the vertical line test to determine whether the relation defines y as a function of x. (See Example 2.)

11.

12.

y

13.

y

x

14.

x

15.

y

y

x

16.

y

x

y

x

x

Concept 3: Function Notation

Consider the functions defined by f1x2  6x  2, g1x2  x2  4x  1, h1x2  7, and k1x2  0 x  2 0 . For Exercises 17–48, find the following. (See Examples 3–4.) 17. g122

18. k122

19. g102

20. h102

21. k102

22. f 102

23. f 1t2

24. g1a2

25. h1u2

26. k1v2

27. g132

28. h152

29. k122

30. f 162

31. f 1x  12

32. h1x  12

33. g12x2

34. k1x  32

35. g1p2

36. g1a2 2

37. h1a  b2

38. f 1x  h2

39. f 1a2

40. g1b2

41. k1c2

42. h1x2

1 43. f a b 2

1 44. g a b 4

1 45. h a b 7

3 46. k a b 2

47. f 12.82

48. k15.42

Consider the functions p  51 12, 62, 12, 72, 11, 02, 13, 2p26 and q  516, 42, 12, 52, 1 34, 15 2, 10, 926. For Exercises 49–56, find the function values. 49. p122

50. p112

51. p132

1 52. pa b 2

53. q122

3 54. qa b 4

55. q162

56. q102

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Chapter 2 Linear Equations in Two Variables and Functions

Concept 4: Finding Function Values from a Graph 57. The graph of y  f 1x2 is given. (See Example 5.)

y 5

a. Find f102 . b. Find f132. c. Find f122. d. For what value(s) of x is f1x2  3?

y  f(x)

4 3 2 1 5 4 3 2 1 1 2

1

2

3 4

5

x

3

e. For what value(s) of x is f1x2  3?

4 5

f. Write the domain of f. g. Write the range of f. 58. The graph of y  g1x2 is given.

y 5 4 3 2 1

a. Find g112. b. Find g112. c. Find g142. d. For what value(s) of x is g1x2  3?

y  g(x)

5 4 3 2 1 1

1

2 3

1

2

4 5

x

2 3

e. For what value(s) of x is g1x2  0?

4 5

f. Write the domain of g. g. Write the range of g. 59. The graph of y  H1x2 is given. a. Find H132. b. Find H142. c. Find H132. d. For what value(s) of x is H1x2  3? e. For what value(s) of x is H1x2  2?

y

y  H(x)

5 4 3 2 1

5 4 3 2 1 1 2 3

3 4

5

x

4 5

f. Write the domain of H. g. Write the range of H. 60. The graph of y  K1x2 is given. a. Find K102. b. Find K152. c. Find K112. d. For what value(s) of x is K1x2  0? e. For what value(s) of x is K1x2  3? f. Write the domain of K. g. Write the range of K.

y 5 4 3 y  K(x) 2 1 5 4 3 2 1 1 2 3 4 5

1

2

3 4

5

x

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Introduction to Functions

201

For Exercises 61–70, refer to the functions y ⫽ f 1x2 and y ⫽ g1x2 , defined as follows: f ⫽ 51⫺3, 52, 1⫺7, ⫺32, 1⫺32, 42, 11.2, 526

g ⫽ 510, 62, 12, 62, 16, 02, 11, 026 61. Identify the domain of f.

62. Identify the range of f.

63. Identify the range of g.

64. Identify the domain of g.

65. For what value(s) of x is f(x) ⫽ 5?

66. For what value(s) of x is f(x) ⫽ ⫺3?

67. For what value(s) of x is g(x) ⫽ 0?

68. For what value(s) of x is g(x) ⫽ 6?

69. Find f (⫺7).

70. Find g(0).

Concept 5: Domain of a Function 71. Explain how to determine the domain of the function defined by f 1x2 ⫽

x⫹6 . x⫺2

72. Explain how to determine the domain of the function defined by g1x2 ⫽ 1x ⫺ 3. For Exercises 73–88, find the domain. Write the answers in interval notation. (See Example 6.) 73. k1x2 ⫽

77. h1p2 ⫽

x⫺3 x⫹6 p⫺4 p2 ⫹ 1

74. m1x2 ⫽

78. n1p2 ⫽

x⫺1 x⫺4 p⫹8 p2 ⫹ 2

75. f1t2 ⫽

5 t

76. g1t2 ⫽

t⫺7 t

79. h1t2 ⫽ 1t ⫹ 7

80. k1t2 ⫽ 1t ⫺ 5

81. f1a2 ⫽ 1a ⫺ 3

82. g1a2 ⫽ 1a ⫹ 2

83. m1x2 ⫽ 11 ⫺ 2x

84. n1x2 ⫽ 112 ⫺ 6x

85. p1t2 ⫽ 2t2 ⫹ t ⫺ 1

86. q1t2 ⫽ t3 ⫹ t ⫺ 1

87. f1x2 ⫽ x ⫹ 6

88. g1x2 ⫽ 8x ⫺ p

Mixed Exercises 89. The height (in feet) of a ball that is dropped from an 80-ft building is given by h1t2 ⫽ ⫺16t2 ⫹ 80, where t is the time in seconds after the ball is dropped. a. Find h(1) and h(1.5) b. Interpret the meaning of the function values found in part (a). 90. A ball is dropped from a 50-m building. The height (in meters) after t sec is given by h1t2 ⫽ ⫺4.9t2 ⫹ 50. a. Find h(1) and h(1.5). b. Interpret the meaning of the function values found in part (a). 91. If Alicia rides a bike at an average speed of 11.5 mph, the distance that she rides can be represented by d1t2 ⫽ 11.5t, where t is the time in hours. a. Find d(1) and d(1.5). b. Interpret the meaning of the function values found in part (a).

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Chapter 2 Linear Equations in Two Variables and Functions

92. Brian’s score on an exam is a function of the number of hours he spends studying. The function defined by P1x2 

100x2 (x  0) indicates that he will achieve a score of P% if he studies for x hours. 50  x2 P(x) 100 90 80 70 60 50 40 30 20 10 0 A 0

Student Score (Percent) as a Function of Study Time D

Percent

Evaluate P(0), P(5), P(10), P(15), P(20), and P(25) and confirm the values on the graph. (Round to one decimal place.) Interpret P(25) in the context of this problem.

F

E

C B

5

10 15 20 Study Time (hr)

25

30

x

For Exercises 93–96, write a function defined by y  f(x) subject to the conditions given. 93. The value of f(x) is three more than two times x.

94. The value of f(x) is four less than the square of x.

95. The value of f(x) is ten less than the absolute value of x.

96. The value of f(x) is sixteen times the square root of x.

Expanding Your Skills For Exercises 97–98, write the domain in interval notation. 97. q1x2 

2

98. p1x2 

2x  2

8 2x  4

Graphing Calculator Exercises 99. Graph k1t2  1t  5. Use the graph to support your answer to Exercise 80.

Section 2.7

100. Graph h1t2  1t  7. Use the graph to support your answer to Exercise 79.

Graphs of Functions

Concepts

1. Linear and Constant Functions

1. Linear and Constant Functions 2. Graphs of Basic Functions 3. Definition of a Quadratic Function 4. Finding the x- and y-Intercepts of a Graph Defined by y ⫽ f ( x)

A function may be expressed as a mathematical equation that relates two or more variables. In this section, we will look at several elementary functions. We know from Section 2.1 that an equation in the form y  k, where k is a constant, is a horizontal line. In function notation, this can be written as f1x2  k. For example, the graph of the function defined by f1x2  3 is a horizontal line, as shown in Figure 2-35. We say that a function defined by f1x2  k is a constant function because for any value of x, the function value is constant.

f(x) 5

f(x)  3

4 3 2 1 5 4 3 2 1 1 2

1

2

3 4 5

Figure 2-35

3 4

5

x

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Section 2.7

An equation of the form y  mx  b is represented graphically by a line with slope m and y-intercept (0, b). In function notation, this may be written as f(x)  mx  b. A function in this form is called a linear function. For example, the function defined by f(x)  2x  3 is a linear function with slope m  2 and y-intercept (0, 3) (Figure 2-36).

DEFINITION Linear Function and Constant Function Let m and b represent real numbers such that m  0. Then A function that can be written in the form f1x2  mx  b is a linear function. A function that can be written in the form f1x2  b is a constant function.

203

Graphs of Functions

f(x) 5 4 3 2 1 5 4 3 2 1 1 2

1

2

3 4

5

x

3 f(x)  2x  3 4 5

Figure 2-36

Note: The graphs of linear and constant functions are lines.

2. Graphs of Basic Functions At this point, we are able to recognize the equations and graphs of linear and constant functions. In addition to linear and constant functions, the following equations define six basic functions that will be encountered in the study of algebra: Equation

Function Notation f 1x2  x

yx

⎫ yx ⎪ yx ⎪ ⎪ y  0x 0 ⎬ ⎪ y  1x ⎪ 1 ⎪ y x ⎭

f 1x2  x

2

2

3

equivalent function notation

f 1x2  x3

f 1x2  0 x 0

Type of Function Identity function Quadratic function Cubic function Absolute value function

f 1x2  1x

Square root function

f 1x2 

Reciprocal function

1 x

The graph of the function defined by f1x2  x is linear, with slope m  1 and y-intercept (0, 0) (Figure 2-37). To determine the shapes of the other basic functions, we can plot several points to establish the pattern of the graph. Analyzing the equation itself may also provide insight to the domain, range, and shape of the graph. To demonstrate this, we will graph f1x2  x2 and g1x2  x1 . Example 1

Graphing Basic Functions

Graph the function defined by f1x2  x2.

Solution:

The domain of the function given by f1x2  x2 1 or equivalently y  x2 2 is all real numbers. To graph the function, choose arbitrary values of x within the domain of the function. Be sure to choose values of x that are positive and values that are negative to determine the behavior of the function to the right and left of the origin (Table 2-6). The graph of f1x2  x2 is shown in Figure 2-38. The function values are equated to the square of x, so f1x2 will always be greater than or equal to zero. Hence, the y-coordinates on the graph will never

y 5 4 3 2 1 5 4 3 2 1 1 2

f(x)  x 1

2

3 4 5

Figure 2-37

3 4

5

x

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Chapter 2 Linear Equations in Two Variables and Functions

be negative. The range of the function is 30, 2. The arrows on each branch of the graph imply that the pattern continues indefinitely. f (x)

Table 2-6 11

x

f(x) ⴝ x2

0

0

1

1

7 6

2

4

5

3

9

1

1

4 3 2

2

4

3

9

10 9 8

f(x)  x2

1 7 6 5 4 3 2 1 1

1

2

3 4 5

6 7

x

2

Figure 2-38

Skill Practice

1. Graph f 1x 2  x 2 by first making a table of points.

Graphing Basic Functions

Example 2

1 Graph the function defined by g1x2  . x

Solution: g1x2 

1 x

Notice that x  0 is not in the domain of the function. From the equation y  x1 , the y values will be the reciprocal of the x-values. The graph defined by g1x2  x1 is shown in Figure 2-39. g(x)

x

Answers 1.

f (x) 5 4 3 2 1 5 4 3 2 1 1 2

1

2

3 4

5

h(x) 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

h(x)  |x|  1

1

2

3 4

5

1

1

2

1 2

1 3

3

1 3

1

1

2

12

3

13

1 g(x) ⴝ x 2 3

1 4

4

12

2

13

3

14

4

5 4 3 2

1

g(x)  x

1 5

1 1

1

2

3 4 5

x

5

Figure 2-39

Notice that as x approaches  and , the y values approach zero, and the graph approaches the x-axis. In this case, the x-axis is called a horizontal asymptote. Similarly, the graph of the function approaches the y-axis as x gets close to zero. In this case, the y-axis is called a vertical asymptote. In general terms, an asymptote is a line that a curve approaches more and more closely but never reaches.

4 5

2.

x 1 2

x

f (x)  x2

3

1 g(x) ⴝ x

x

Skill Practice

2. Graph h 1x2  ƒ x ƒ 1 by first making a table of points.

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Section 2.7

Calculator Connections Topic: Graphing Nonlinear Functions

To enter a function, let f 1x2  y. That is, to graph f 1x2  x2, enter Y1  x2. 1 To graph g1x2  , enter Y1  1/x. x

Summary of Six Basic Functions and Their Graphs Function

1. f1x2  x

Graph

Domain and Range

Domain 1, 2

y

Identity function x

2. f1x2  x2

Domain 1, 2

y

Quadratic function x

3. f1x2  x3

x

5. f1x2  1x

x

1 x Reciprocal function

6. f1x2 

Range 30, 2

Domain 30, 2

y

Square root function

Range 1, 2

Domain 1, 2

y

Absolute value function

Range 30, 2

Domain 1, 2

y

Cubic function

4. f1x2  0x 0

Range 1, 2

x

Range 3 0, 2

Domain 1, 02 ´ 10, 2

y

x

Range 1, 02 ´ 10, 2

Graphs of Functions

205

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The shapes of these six graphs will be developed in the homework exercises. These functions are used often in the study of algebra. Therefore, we recommend that you associate an equation with its graph and commit each to memory.

3. Definition of a Quadratic Function

In Example 1 we graphed the function defined by f 1x2  x2 by plotting points. This function belongs to a special category called quadratic functions.

DEFINITION Quadratic Function A quadratic function is a function defined by f 1x2  ax2  bx  c

where a, b, and c are real numbers and a  0.

The graph of a quadratic function is in the shape of a parabola. The leading coefficient, a, determines the direction of the parabola. If a 7 0, then the parabola opens upward, and the vertex is the minimum point on the parabola. For example, the graph of f 1x2  x2 is shown in Figure 2-40. If a 6 0, then the parabola opens downward, and the vertex is the maximum point on the parabola. For example, the graph of f 1x2  x2 is shown in Figure 2-41.

• •

y

y

5

5

4 3 2 1 5 4 3 2 1 1

f (x)  x2 Vertex

Vertex 1

2

3 4

5

x

5 4 3 2 1 1

2 3 4

f(x)  x2 1

2

3 4

5

x

2 3 4

5

5

Figure 2-40

Example 3

4 3 2 1

Figure 2-41

Identifying Functions

Identify each function as linear, constant, quadratic, or none of these. a. f 1x2  4 d. f 1x2 

4x  8 8

b. f 1x2  x2  3x  2 e. f 1x2 

c. f 1x2  7  2x

6 2 x

Solution:

a. f 1x2  4 is a constant function. It is in the form f 1x2  b, where b  4. b. f 1x2  x2  3x  2 is a quadratic function. It is in the form f 1x2  ax2  bx  c, where a  0.

c. f 1x2  7  2x is linear. Writing it in the form f 1x2  mx  b, we get f 1x2  2x  7, where m  2 and b  7.

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Section 2.7

d. f 1x2 ⫽ f 1x2 ⫽

Graphs of Functions

4x ⫹ 8 is linear. Writing it in the form f 1x2 ⫽ mx ⫹ b, we get 8 4x 8 ⫹ 8 8

1 1 ⫽ x ⫹ 1, where m ⫽ and b ⫽ 1. 2 2 e. f 1x2 ⫽

6 ⫹ 2 fits none of these categories. x

Skill Practice Identify each function as constant, linear, quadratic, or none of these. 3. m1x2 ⫽ ⫺2x 2 ⫺ 3x ⫹ 7 4 1 5. W 1x2 ⫽ x ⫺ 3 2

4. n1x2 ⫽ ⫺6 4 1 6. R 1x2 ⫽ ⫺ 3x 2

4. Finding the x- and y-Intercepts of a Graph Defined by y ⴝ f( x) In Section 2.1, we learned that to find the x-intercept, we substitute y ⫽ 0 and solve the equation for x. Using function notation, this is equivalent to finding the real solutions of the equation f(x) ⫽ 0. To find the y-intercept, substitute x ⫽ 0 and solve the equation for y. In function notation, this is equivalent to finding f (0).

PROCEDURE Finding Intercepts Using Function Notation Given a function defined by y ⫽ f(x), Step 1 The x-intercepts are the real solutions to the equation f (x) ⫽ 0. Step 2 The y-intercept is given by f (0).

Example 4

Finding x- and y-Intercepts

Given the function defined by f (x) ⫽ 2x ⫺ 4: a. Find the x-intercept(s). b. Find the y-intercept. c. Graph the function.

Solution: a. To find the x-intercept(s), find the real solutions to the equation f (x) ⫽ 0. f1x2 ⫽ 2x ⫺ 4 0 ⫽ 2x ⫺ 4 4 ⫽ 2x 2⫽x

Substitute f 1x2 ⫽ 0. The x-intercept is 12, 02.

Answers 3. 4. 5. 6.

Quadratic Constant Linear None of these

207

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Chapter 2 Linear Equations in Two Variables and Functions

b. To find the y-intercept, evaluate f(0).

y 5

f102  2102  4

4 3 2 1

f102  4

5 4 3 2 1 1

1

2

3 4

5

x

2 3 4 f(x)  2x  4 5

Substitute x  0.

The y-intercept is 10, 42.

c. This function is linear, with a y-intercept of 10, 42, an x-intercept of (2, 0), and a slope of 2 (Figure 2-42). Skill Practice Consider f (x)  5x  1. 7. Find the x-intercept. 8. Find the y-intercept. 9. Graph the function.

Figure 2-42

Calculator Connections Topic: Finding x- and y-Intercepts Refer to Example 4 with f(x)  2x  4. To find the y-intercept, let x  0. We can do this using the Value key.

To find the x-intercept, use the Zero feature. To use this feature you must give bounds for the x-intercept. To find a left bound, place the curser to the left of the x-intercept and press Enter. For the right bound, place the curser to the right of the x-intercept and press Enter.

Then make a guess by placing the curser near the x-intercept. Press Enter and the x-intercept will be displayed. Answers 7. 1 15 , 02 9.

8. 10, 12 y

5 4

f (x)  5x  1 3 2 1 5 4 3 2 1 1 2 3 4 5

1

2

3 4

5

x

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Section 2.7

Example 5

209

Graphs of Functions

Finding x- and y-Intercepts

For the function pictured in Figure 2-43, estimate y

a. The real values of x for which f (x)  0.

7 6 5 4 3 2 1

b. The value of f(0).

Solution: a. The real values of x for which f (x)  0 are the x-intercepts of the function. For this graph, the x-intercepts are located at x  2, x  2, and x  3.

5 4 3 2 1 1

b. The value of f(0) is the value of y at x  0. That is, f(0) is the y-intercept, f(0)  6.

2

3 4

5

x

2 3

Figure 2-43

y

Skill Practice Use the function pictured. 5

10. a. Estimate the real value(s) of x for which f (x)  0 b. Estimate the value of f (0).

y  f (x)

4 3 2 1

5 4 3 2 1 1 2

1

2

3 4

5

x

3

Answer

4 5

Section 2.7

1

10. a. x  2, x  1, and x  4 b. f 102  4

Practice Exercises

Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercises 1. Make up a practice test for yourself for this chapter. Use examples or exercises from the text. Be sure to cover each concept that was presented. 2. Define the key terms. a. Linear function

b. Constant function

Review Exercises

3. Given: g  516, 12, 15, 22, 14, 32, 13, 426

c. Quadratic function

d. Parabola

4. Given: f  517, 32, 12, 32, 15, 326

a. Is this relation a function?

a. Is this relation a function?

b. List the elements in the domain.

b. List the elements in the domain.

c. List the elements in the range.

c. List the elements in the range.

5. Given: f 1x2  2x  4

a. Evaluate f 102, f 132, f 142, and f 152, if possible. b. Write the domain of this function in interval notation.

6. The force (measured in pounds) to stretch a certain spring x inches is given by f 1x2  3x. Evaluate f(3) and f(10), and interpret the results in the context of this problem.

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Chapter 2 Linear Equations in Two Variables and Functions

Concept 1: Linear and Constant Functions 7. Fill in the blank with the word vertical or horizontal. The graph of a constant function is a ______ line. 8. For the linear function defined by f1x2  mx  b, identify the slope and y-intercept. 9. Graph the constant function f 1x2  2. Then use the graph to identify the domain and range of f.

10. Graph the linear function g1x2  2x  1. Then use the graph to identify the domain and range of g.

y

g

5

5

4 3 2 1

4 3 2 1

5 4 3 2 1 1 2

1

2

3 4

x

5

5 4 3 2 1 1 2

3

3

4 5

4 5

1

2

3 4

5

x

Concept 2: Graphs of Basic Functions For Exercises 11–16, sketch a graph by completing the table and plotting the points. (See Examples 1–2.) 11. f1x2  x

1 x f(x)

2

x

12

1

14

2

13. h1x2  x3 x

h(x)

2 1 1 2

1 0 1 2

x

q(x)

y 5 4 3 2 1

g(x)

2

5 4 3 2 1 1 2 3

1

2

3 4

5

x

1

5 4 3 2 1 1 2 3

0 1 2

4 5

10 8 6 4 2

x

4

6 8 10

x

1

5 4 3 2 1 1 2 3

0 2

5 4 3 2 1 1 2 3 4 5

x 0 1

2

3 4

5

x

1 4 9 16

5

x

1

2

3 4

5

x

4 5

16. p1x2  1x

5 4 3 2 1

3 4

y

k(x)

1

y

2

5 4 3 2 1

2 2

1

4 5

14. k1x2  x

y

8 10

15. q1x2  x2

2

f(x)

108 6 4 2 2 4 6

0

x

5 4 3 2 1

1 4 1 2

1

12. g1x2  0 x 0

y

p(x)

y 10 8 6 4 2 2 2 4 6 8 10

2 4 6 8 10 12 14 16 18

x

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Section 2.7

211

Graphs of Functions

Concept 3: Definition of a Quadratic Function For Exercises 17–28, determine if the function is constant, linear, quadratic, or none of these. (See Example 3.) 17. f 1x2  2x2  3x  1 21. m1x2 

4 3

2 1 25. t1x2  x  3 4

18. g1x2  x2  4x  12

19. k1x2  3x  7

22. n1x2  0.8

23. p1x2 

1 26. r1x2  x  3 5

27. w1x2  24  x

20. h1x2  x  3

2 1  3x 4

24. Q1x2 

1 3 5x

28. T1x2   0 x  10 0

Concept 4: Finding the x- and y-Intercepts of a Graph Defined by y ⴝ f (x ) For Exercises 29–36, find the x- and y-intercepts, and graph the function. (See Example 4.) 29. f 1x2  5x  10

30. f 1x2  3x  12

y 10 8 6 4 2 108 6 4 2 2 4

2

4

6 8 10

x

5 4 3 2

6 4 2

1

108 6 4 2 2

8 10

4 6

2

4

6 8 10

x

y

3 4

5

x

6 8 10

2 35. g1x2  x  2 3 y

5 4 3 2 1 4 8

3 4

5

x

6

16 20

8 10

1 3 4

5

x

5 4 3 2 1 2 4

12

y

1 2

1 2

3 4

5

x

2 1 2

5 4 3 2

1

5

y

3 36. h1x2   x  3 5

5 4 3 2

3 4

10 8 6 4

4 1 2

2

34. g1x2  7 20 16 12 8

2

1

4 5

y

10 8 6 4

5 4 3 2 1 1 2 3

33. f 1x2  18

32. h1x2  2x  9

5 4 3 2 1 1 2

y

14 12 10 8

6

5 4 3 2 1 2 4

31. g1x2  6x  5

y

5 4 3 2 1 1 2

3

3

4 5

4 5

1 2

3 4

5

x

x

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212

Chapter 2 Linear Equations in Two Variables and Functions

For Exercises 37–42, use the graph to estimate

a. The real values of x for which f 1x2  0. y

37.

b. The value of f 102. (See Example 5.) y

38.

39.

y

5

5

5

4 3 2

4 3 2

4 3 2

y  f(x)

1 5 4 3 2 1 1 2

1

2

y  f(x)

1 x

3 4 5

5 4 3 2 1 1 2

1

2

3 4 5

5 4 3 2 1 1 2

3 4

3 4

3 4

5

5

5

40.

41.

y

42.

y

5

4 3 2

y  f(x)

1 1

2

x

3 4 5

5 4 3 2 1 1 2

2

3 4 5

1

2

3 4 5

x

y 4 3 2

y  f(x)

1

1

5 4 3 2 1 1 2

1

5

5

4 3 2

y  f(x)

1 x

1

2

3 4 5

x

5 4 3 2 1 1 2

3 4

3 4

3 4

5

5

5

x

y  f(x)

Mixed Exercises For Exercises 43–52, a. Identify the domain of the function. b. Identify the y-intercept. c. Match the function with its graph by recognizing the basic shape of the graph and using the results from parts (a) and (b). Plot additional points if necessary. 43. q1x2  2x2

44. p1x2  2x2  1

45. h1x2  x3  1

46. k1x2  x3  2

47. r1x2  2x  1

48. s1x2  2x  4

49. f1x2 

1 x3

50. g1x2 

51. k1x2  0x  2 0

1 x1

52. h1x2  0x  1 0  2 i.

ii.

y

iii.

y

iv.

y

y

5

5

5

5

4 3 2

4 3 2

4 3 2

4 3 2

1

1 5 4 3 2 1 1 2

1

2

3 4 5

x

5 4 3 2 1 1 2

1 1

2

3 4 5

x

5 4 3 2 1 1 2

1 1

2

3 4 5

x

5 4 3 2 1 1 2

3 4

3 4

3 4

3 4

5

5

5

5

1

2

3 4 5

x

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Section 2.7

y

v.

vi.

vii.

y

viii.

y

y

5

5

5

5

4 3 2

4 3 2

4 3 2

4 3 2

1

1

5 4 3 2 1 1 2

1

2

3 4 5

x

3 4 5 y

ix.

2

3 4 5

x

5 4 3 2 1 1 2

1 1

2

3 4 5

x

5 4 3 2 1 1 2

3 4

3 4

5

5

5

1

2

3 4 5

y

x. 5

4 3 2

4 3 2

5 4 3 2 1 1 2

1

3 4

5

1

1

5 4 3 2 1 1 2

213

Graphs of Functions

1 1

2

3 4 5

x

5 4 3 2 1 1 2

3 4

3 4

5

5

1

2

3 4 5

x

53. Suppose that a student has an 80% average on all of her chapter tests in her Intermediate Algebra class. This counts as 34 of the student’s overall grade. The final exam grade counts as the remaining 14 of the student’s overall grade. The student’s overall course grade, G(x), can be computed by 3 1 G(x)  1802  x, where x is the student’s grade on the final exam. 4 4 a. Is this function linear, quadratic, or neither? b. Evaluate G(90) and interpret its meaning in the context of the problem. c. Evaluate G(50) and interpret its meaning in the context of the problem. 54. The median weekly earnings, E(x) in dollars, for women 16 yr and older working full time can be approximated by E(x)  0.14x2  7.8x  540. For this function, x represents the number of years since 2000. (Source: U.S. Department of Labor) a. Is this function linear, quadratic, or neither? b. Evaluate E(5) and interpret its meaning in the context of this problem. c. Evaluate E(10) and interpret its meaning in the context of this problem.

Graphing Calculator Exercises For Exercises 55–60, use a graphing calculator to graph the basic functions. Verify your answers from the table on page 205. 55. f 1x2  x

56. f 1x2  x2

57. f 1x2  x3

58. f 1x2  冟x冟

59. f 1x2  1x

60. f 1x2 

1 x

For Exercises 61–64, find the x- and y-intercepts using a graphing calculator and the Value and Zero features. 1 61. y   x  1 8

1 62. y   x  3 2

4 63. y  x  4 5

7 64. y  x  7 2

x

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Chapter 2 Linear Equations in Two Variables and Functions

Problem Recognition Exercises Characteristics of Relations Exercises 1–15 refer to the sets of points (x, y) described in a–h. a. {(0, 8), (1, 4), (12 , 4), (–3, 5), (2, 1)}

b. {(6, 4), (2, 3), (9, 6), (2, 1), (0, 10)}

c. c1x2  3x2  2x  1

d. d1x2  5x  9

e.

f.

y

y 5 4 3 2 1

5 4 3 2 1 5 4 3 2 1 1 2 3

1

2

3 4

5

x

5 4 3 2 1 1 2 3

h.

y 5 4 3 2 1 5 4 3 2 1 1 2 3

1

2

3 4

5

x

4 5

4 5

g.

y  f (x)

y  g(x)

1

2

3 4

5

x

x

y

0

1

1

3

2

6

4

8

4 5

1. Which relations define y as a function of x?

2. Which relations contain the point (2, 1)?

3. Use relation (c) to find c(–1).

4. Use relation (f) to find f(–4).

5. Find the domain of relation (a).

6. Find the range of relation (b).

7. Find the domain of relation (g).

8. Find the range of relation (f).

9. Use relation (h) to complete the ordered pair (__, 3).

10. Find the x-intercept(s) of the graph of relation (d).

11. Find the y-intercept(s) of the graph shown in relation (e).

12. Use relation (g) to determine the value of x such that g(x)  2.

13. Which relation describes a quadratic function?

14. Which relation describes a linear function?

15. Use relation (d) to find the value of x such that d(x)  6.

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Group Activity

Group Activity Deciphering a Coded Message Materials: A calculator Estimated time: 20–25 minutes Group Size: 4 (two pairs) Cryptography is the study of coding and decoding messages. One type of coding process assigns a number to each letter of the alphabet and to the space character. For example: A 1

B 2

C 3

D 4

E 5

F 6

G 7

H 8

I 9

J 10

K 11

L 12

M 13

N 14

O 15

P 16

Q 17

R 18

S 19

T 20

U 21

V 22

W 23

X 24

Y 25

Z 26

space 27

According to the numbers assigned to each letter, the message “Decimals have a point” would be coded as follows: D E C I M A L S

__ H A V E __ A __

4 5 3 9 13 1 12 19 27 8 1 22 5 27 1

P O I

N T

27 16 15 9 14 20

Now suppose each letter is encoded by applying a function such as f(x)  2x  5, where x is the numerical value of each letter. For example: The letter “a” would be coded as:

f(1)  2(1)  5  7

The letter “b” would be coded as:

f(2)  2(2)  5  9

Using this encoding function, we have Message:

D E C

I

Original:

4

9 13 1 12 19 27

Coded Form:

13 15 11 23 31 7 29 43 59 21 7 49 15

5

3

M A L

S

__ H A V E __ A __ 8

1 22

5

P

O

27 1 27 16 15

I

N

T

9 14 20

59 7 59 37 35 23 33 45

To decode this message, the receiver would need to reverse the operations assigned by f(x)  2x + 5. Since the function f multiplies x by 2 and then adds 5, we can reverse this process by subtracting 5 and dividing by 2. This is represented by g1x2  x 2 5. 1. a. One pair of students will encode the follow message according to f1x2  4x  2. MATH IS THE KEY TO THE SCIENCES b. The second pair of students will encode the follow message according to f1x2  3x  1. MATH IS NOT A SPECTATOR SPORT 2. With each message encoded, the pairs will exchange papers. Each pair will then decode the message.

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Chapter 2 Linear Equations in Two Variables and Functions

Chapter 2

Summary

Section 2.1

Linear Equations in Two Variables

Key Concepts

Examples

A linear equation in two variables can be written in the form Ax  By  C, where A, B, and C are real numbers and A and B are not both zero. The graph of a linear equation in two variables is a line and can be represented in a rectangular coordinate system.

Example 1 To graph the equation 3x  4y  12, we can construct a table of points. y

x 0 4 1

5

y 3 0 9  4

4 3 2 1 5 4 3 2 1 1

1

2

3 4 5

x

2 3 4 5

Example 2 An x-intercept is a point (a, 0) where a graph intersects the x-axis. Given an equation of a graph, to find an x-intercept, substitute 0 for y and solve for x.

Given 2x  3y  8, find the x- and y-intercepts. 2x  3102  8

x-intercept:

2x  8 x4 A y-intercept is a point (0, b) where a graph intersects the y-axis. Given an equation of a graph, to find a y-intercept, substitute 0 for x and solve for y.

2102  3y  8

y-intercept:

3y  8 y

An equation of a vertical line can be written in the form x  k. An equation of a horizontal line can be written in the form y  k.

14, 02

8 3

8 a0, b 3

Example 3

Example 4

Graph x  2.

Graph y  3.

y

y

5

5

4 3

4 3

2 1

2 1

5 4 3 2 1 1

1

2

3 4 5

x

5 4 3 2 1 1

2

2

3 4

3 4

5

5

x  2

y3

1

2 3 4 5

x

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Summary

Section 2.2

217

Slope of a Line and Rate of Change

Key Concepts

Examples

The slope m of a line passing through two distinct points (x1, y1) and (x2, y2) is given by

Example 1 The slope of the line passing through (1,⫺3) and (⫺3,7) is

m⫽

y2 ⫺ y1 , x2 ⫺ x1

x2 ⫺ x1 ⫽ 0

The slope of a line may be positive, negative, zero, or undefined.

m⫽

7 ⫺ 1⫺32 10 5 ⫽ ⫽⫺ ⫺3 ⫺ 1 ⫺4 2

Example 2

Positive slope

Negative slope

Zero slope

Undefined slope

Two parallel (nonvertical) lines have the same slope: m1 ⫽ m2. Two lines are perpendicular if the slope of one line is the opposite of the reciprocal of the slope of the other line: 1 or equivalently, m1m2 ⫽ ⫺1 m1 ⫽ ⫺ m2

Example 3 The slopes of two lines are given. Determine whether the lines are parallel, perpendicular, or neither. a. m1 ⫽ ⫺7

and

m2 ⫽ ⫺7

Parallel

1 5

and

m2 ⫽

5

Perpendicular

3 2

and

m2 ⫽ ⫺

2 3

Neither

b. m1 ⫽ ⫺

c. m1 ⫽ ⫺

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Chapter 2 Linear Equations in Two Variables and Functions

Section 2.3

Equations of a Line

Key Concepts

Examples

Standard Form: Ax  By  C (A and B are not both zero)

Example 1

Horizontal line: y  k

Find the slope and y-intercept. Then graph the equation.

Vertical line: x  k

7x  2y  4

Slope-intercept form: y  mx  b

Point-slope formula: y  y1  m1x  x1 2 Slope-intercept form is used to identify the slope and y-intercept of a line when the equation is given. Slope-intercept form can also be used to graph a line.

Solve for y.

2y  7x  4 7 y x2 2

Slope-intercept form

The slope is 72; the y-intercept is (0, 2). Right 2

y 5 4 3

Up 7

2

5 4 3 2 1

1

2

3 4 5

x

(0, 2) 2 Start here 3 4 5

The point-slope formula can be used to construct an equation of a line, given a point and a slope.

Example 2 Find an equation of the line passing through the point (2, 3) and having slope m  4. Using the point-slope formula gives y  y1  m1x  x1 2

y  132  41x  22 y  3  4x  8 y  4x  5

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Summary

Section 2.4

Applications of Linear Equations and Modeling

Key Concepts

Examples

A linear model can be constructed to describe data for a given situation.

Example 1 The per capita income in the United States has been rising linearly since 1980. In the graph, x represents the number of years since 1980, and y represents average income in dollars. Average per Capita Yearly Income in United States 30,000

(25, 26,100)

Dollars ($)

25,000 20,000 15,000 10,000 5,000 0

• Given two points from the data, use the point-slope formula to find an equation of the line.

(5, 11,000) 0

5 10 15 20 25 Year (x  0 represents 1980)

30

Write an equation of the line, using the points (5, 11,000) and (25, 26,100). Slope:

15,100 26,100  11,000   755 25  5 20

y  11,000  7551x  52 y  11,000  755x  3775 y  755x  7225 • Interpret the meaning of the slope and y-intercept in the context of the problem.

• The slope 755 indicates that the average income has increased at a rate of $755 per year. • The y-intercept (0, 7225) means that the average income in 1980 (year x = 0) was $7225.

• Use the equation to predict values.

By substituting different values of x, the equation can be used to approximate the average income for that year. For the year 2010 (x  30), we have: y  7551302  7225 y  29,875 The average per capita income in 2010 would be approximately $29,875.

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Chapter 2 Linear Equations in Two Variables and Functions

Section 2.5

Introduction to Relations

Key Concepts

Examples

A set of ordered pairs (x, y) is called a relation in x and y.

Example 1 Let A ⫽ 510, 02, 11, 12, 12, 42, 13, 92, 1⫺1, 12, 1⫺2, 426. Domain of A: 50, 1, 2, 3, ⫺1, ⫺26 Range of A: 50, 1, 4, 96

The domain of a relation is the set of first components in the ordered pairs in the relation. The range of a relation is the set of second components in the ordered pairs.

Example 2 Domain: 3⫺5, 54 Range: 30, 44

y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

x

3 4 5

⫺2 ⫺3 ⫺4 ⫺5

Section 2.6

Introduction to Functions

Key Concepts

Examples

Given a relation in x and y, we say “y is a function of x” if, for each element x in the domain, there is exactly one value of y in the range.

Example 1

Note: This means that no two ordered pairs may have the same first coordinate and different second coordinates.

The Vertical Line Test for Functions Consider a relation defined by a set of points (x, y) in a rectangular coordinate system. The graph defines y as a function of x if no vertical line intersects the graph in more than one point.

Function 511, 32, 12, 52, 16, 326 Not a function 511, 32, 12, 52, 11, 426

Example 2 y

y

x

Function

x

Not a function

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Summary

Function Notation

Example 3

f(x) is the value of the function f at x.

Given f (x)  3x 2  5x, find f (2). f 122  3122 2  5122  12  10  22

The domain of a function defined by y  f (x) is the set of x-values that when substituted into the function produces a real number. In particular, • Exclude values of x that make the denominator of a fraction zero. • Exclude values of x that make the expression within a square root negative.

Section 2.7

A function of the form f 1x2  mx  b 1m  02 is a linear function. Its graph is a line with slope m and y-intercept (0, b). A function of the form f 1x2  k is a constant function. Its graph is a horizontal line. A function of the form f 1x2  ax2  bx  c 1a  02 is a quadratic function. Its graph is a parabola.

f(x)

f(x)

f(x)  x2

x

1. f 1x2 

x4 ; Domain: 1, 52 ´ 15,  2 x5

2. f 1x2  1x  3; Domain: 33,  2

3. f 1x2  3x2  5; Domain: 1,  2

f(x)  x3

x

Examples Example 1 f 1x2  3

f1x2  2x  3 y 5

y

f(x)  2x  3

5

4 3 2 1 5 4 3 2 1 1

f(x)  3 1

2

3 4

x

5

4 3 2 1

5 4 3 2 1 1

2 3 4

2 3 4

5

5

f (x)

x

5 4 3 2 (0, 1) 1

(1, 0) (5, 0)

f(x)

f(x)

f(x)  |x|

x

3 2 1 1 2

f(x)

3

f(x) ⫽ 1x

f(x) ⫽ 1x x

x

The x-intercepts of a function are determined by finding the real solutions to the equation f 1x2  0. The y-intercept of a function is at f (0).

1

2

3 4

5

x

Example 2 Find the x- and y-intercepts for the function pictured.

Graphs of basic functions:

f(x)  x

Find the domain.

Graphs of Functions

Key Concepts

f(x)

Example 4

4 5

1 2 3 4

5 6 7

y  f(x)

f1x2  0, when x  1 and x  5. The x-intercepts are (1, 0) and (5, 0). f102  1. The y-intercept is (0, 1).

x

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Chapter 2 Linear Equations in Two Variables and Functions

Review Exercises

Chapter 2

6. 2y  3  10

Section 2.1

y

1. Label the following on the diagram:

x

a. Origin

7 6 5 4 3 2

y

0

b. x-Axis

5

c. y-Axis

1 5 4 3 2 1 1

4

d. Quadrant I

7. 6  x  2 x

0 1

5 4 3 2 1 1 2 3

2

4. Determine the coordinates of the points labeled in the graph.

5 4 3 2 1

5 4 3 2 1 1 2

D

1

2

3 4

x

5

4 5

For Exercises 8–11, graph the lines. In each case find at least three points and identify the x- and y-intercepts (if possible).

y

E

x

5

5 4 3 2 1

y

2. Determine if (2, 5) is a solution to 2x  4y  16.

F

3 4

y

g. Quadrant IV

3. Determine if (3, 4) is a solution to 5x  15.

2

2 3

e. Quadrant II f. Quadrant III

1

8. 2x  3y  6

B A 1

2

3 4

3

5

x

y

5 4 3 2 1

G

4 5

9. 5x  2y  0

y

C

For Exercises 5–7, complete the table and graph the line defined by the points. 5. 3x  2y  6

5 4 3 2 1 1 2 3

5 4 3 2 1 1

2

3 4

5

x

4 5

5 4 3 2 1 1 2 3

1

2

3 4

5

1

2

3 4

5

x

4 5

y

x

y

0 0 1

y

1 5 4 3 2 1 1 2 3 4 5

11. 3x  6

10. 2y  6

5 4 3 2

1

2

3 4

5

x

y

6 5 4 3 2 1 5 4 3 2 1 1 2 3 4

5 4 3 2 1

1

2

3 4

5

x

5 4 3 2 1 1 2 3 4 5

x

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Review Exercises

Section 2.2

19. Two points for each of two lines are given. Determine if the lines are parallel, perpendicular, or neither.

12. Find the slope of the line. a.

L1: 14, 62 and 13, 22

y 5 4 3

L2: 13, 12 and (7, 0)

2 1 5 4 3 2 1 1 2

1

2

3 4

5

For Exercises 20–22, the slopes of two lines are given. Based on the slopes, are the lines parallel, perpendicular, or neither?

x

3 4 5

b.

1 20. m1   , m2  3 3

5 4 21. m1  , m2  4 5

y 5 4 3

22. m1  7, m2  7

2 1

23. The graph indicates that the enrollment for a small college has been increasing linearly since 1990.

5 4 3 2 1 1 2

1

2

3 4

5

x

a. Use the two data points to find the slope of the line.

3 4 5

c.

b. Interpret the meaning of the slope in the context of this problem.

y 5 4 3 2 1 1

2

3 4

5

x

Number of Students

5 4 3 2 1 1 2 3 4 5

14. Draw a line with slope 34 (answers may vary).

13. Draw a line with slope 2 (answers may vary).

5 4 3 2 1

5 4 3 2 1 1

2

3 4

4 5

5

x

5 4 3 2 1 1 2 3

3000 2500 2000 1500

(2010, 3080) (1990, 2020)

1000 500 0 1985

1990

1995

2000

2005

2010

24. Approximate the slope of the stairway pictured here.

y

y

5 4 3 2 1 1 2 3

College Enrollment by Year

3500

1

2

3 4

5

x

36 in. 48 in.

4 5

Section 2.3 For Exercises 15–18, find the slope of the line that passes through each pair of points. 15. 12, 62, 11, 02

16. 17, 22, 13, 52

17. 18, 22, 13, 22

1 18. a4, b, 14, 12 2

25. Write a formula. a. Horizontal line b. Point-slope formula c. Standard form d. Vertical line e. Slope-intercept form

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Chapter 2 Linear Equations in Two Variables and Functions

For Exercises 26–30, write your answer in slope-intercept form or in standard form. 26. Write an equation of the line that has slope 91 and y-intercept (0, 6). 27. Write an equation of the line that has slope ⫺23 and x-intercept (3, 0).

a. Write a relationship for the daily cost y in terms of the number of ice cream products sold per day x. b. Graph the equation from part (a) by letting the horizontal axis represent the number of ice cream products sold per day and letting the vertical axis represent the daily cost.

28. Write an equation of the line that passes through the points (⫺8, ⫺1) and (⫺5, 9).

400 300 200 100

30. Write an equation of the line that passes through the point (0, ⫺3) and is parallel to the line 4x ⫹ 3y ⫽ ⫺1. 31. For each of the given conditions, find an equation of the line

y

500 Cost ($)

29. Write an equation of the line that passes through the point (6, ⫺2) and is perpendicular to the line y ⫽ ⫺13x ⫹ 2.

600

0

0

x

75 150 225 300 375 450 Number of Ice Cream Products

525

c. What does the y-intercept represent in the context of this problem?

a. Passing through the point (⫺3, ⫺2) and parallel to the x-axis.

d. What is her cost if she sells 450 ice cream products?

b. Passing through the point (⫺3, ⫺2) and parallel to the y-axis.

e. What is the slope of the line?

c. Passing through the point (⫺3, ⫺2) and having an undefined slope. d. Passing through the point (⫺3, ⫺2) and having a zero slope. 32. Are any of the lines in Exercise 31 the same?

f. What does the slope of the line represent in the context of this problem? 34. The margin of victory for a certain college football team seems to be linearly related to the number of rushing yards gained by the star running back. The table shows the statistics.

Section 2.4

Yards Rushed

Margin of Victory

100

20

33. Ally loves the beach and decides to spend the summer selling various ice cream products on the beach. From her accounting course, she knows that her total cost is calculated as

She estimates that her fixed cost for the summer season is $50 per day. She also knows that each ice cream product costs her $0.75 from her distributor.

10 24

50

7

a. Graph the data to determine if a linear trend exists. Let x represent the number of yards rushed by the star running back and y represent the points in the margin of victory. 45 Margin of Victory

Total cost ⫽ fixed cost ⫹ variable cost

60 120

y

35 25 15 5 20

40

60 80 Yards Rushed

100

x

120

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Review Exercises

40. Sketch a relation that is a function. (Answers may vary.)

b. Find an equation for the line through the points (50, 7) and (100, 20). c. Based on the equation, what would be the result of the football game if the star running back did not play?

y 5 4 3 2 1

Section 2.5 For Exercises 35–38, find the domain and range.

5 4 3

3 4

x

5

c. Find the range. 41.

42.

38. y

y 5 4

(9, 60)

1

20

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

10 4

6 8 10

x

1 2 3 4

5

Section 2.6

3

2 1

2 1 (1.25, 0.35) 1 2 3 4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

44. 510, 22, 10, 32, 14, 42, 10, 526 45.

y

x

y

4

2

9

⫺2

5 4 3 2 1

⫺4 ⫺5

3

43. 511, 32, 12, 32, 13, 32, 14, 326

39. Sketch a relation that is not a function. (Answers may vary.)

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

5 4

⫺4 ⫺5

x

⫺4 ⫺5

⫺20 ⫺30

5 4

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

3 2

2

y

y

37.

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺10

x

b. Find the domain.

(2, 1)

⫺4 ⫺5

(⫺3, 60)

5

a. Determine whether the relation defines y as a function of x.

(3, 52 )

2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 (0, 0) ⫺1 ⫺2 (⫺3, ⫺2) ⫺3

70 60 50 40 30

3 4

For Exercises 41–46:

y

(⫺1, 1)

2

⫺4 ⫺5

1 1 1 2 35. e a , 10b, a6, ⫺ b, a , 4b, a7, b f 3 2 4 5 36.

1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

3 ⫺3 1

2

3 4

5

x

46.

x

y

6

9

7

10

8

11

9

12

1 2 3 4

5

x

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Chapter 2 Linear Equations in Two Variables and Functions

For Exercises 47–54, find the function values given f1x2  6x2  4. 47. f102

48. f112

49. f112

50. f1t2

51. f1b2

52. f1p2

53. f 1a2

54. f 122

y

5 4 3 2 1 1 2 3

56. h1x2 

57. k1x2  1x  8

x  10 x  11

y

3 4

5

x

4 5

5

1

2

3 4

5

x

5 4 3 2 1 1 2 3 4 5

y

70. Given: h1x2 

b. What is the domain of h? 71. Given: k1x2   0 x  3 0

5 4 3 2 1 3 4

5

x

3 x3

a. Find h132, h102, h122, and h152.

y

5 4 3 2 1

a. Find s(4), s(3), s(2), s(1), and s(0).

x

63. w 1x2  0 x 0

62. g1x2  x

4 5

3 4

4 5

3

5 4 3 2 1 1 2 3 4 5

a. Find k152, k142, k132, and k122. b. What is the domain of k? 1

2

5

x

y

b. What is the domain of r? 2

3 4

5 4 3 2 1

69. Given: r 1x2  2 1x  4

1

2

67. k1x2  2x  1

a. Find r(2), r(4), r(5), and r(8).

5 4 3 2 1 1 2 3

1

4 5

b. What is the domain of s?

y

2

5 4 3 2 1 1 2 3

4 5

c. 20 tables

5 4 3 2 1

1

x

4 5

5 4 3 2 1 1 2 3

61. f1x2  x2

5 4 3 2 1

5 4 3 2 1 1 2 3

5

68. Given: s1x2  1x  22 2

60. h1x2  x

2

3 4

y

For Exercises 60–65, sketch the functions from memory.

1

2

5 4 3 2 1

Section 2.7

5 4 3 2 1 1 2 3

1

66. q1x2  3

58. w1x2  1x  2

b. 15 tables

y

For Exercises 66–67, sketch the functions.

59. Anita is a waitress and makes $6 per hour plus tips. Her tips average $5 per table. In one 8-hr shift, Anita’s pay can be described by p1x2  48  5x, where x represents the number of tables she waits on. Determine how much Anita will earn if she waits on a. 10 tables

1 x 5 4 3 2 1

5 4 3 2 1

For Exercises 55–58, write the domain of each function in interval notation. 55. g1x2  7x3  1

65. r1x2 

64. s1x2  1x

3 4

5

x

1

2

3 4

5

x

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Test

For Exercises 72–73, find the x- and y-intercepts.

For Exercises 75–80, refer to the graph.

72. p1x2  4x  7

y 5 4

73. q1x2  2x  9

3 2 1

74. The function defined by b1t2  0.7t  4.5 represents the per capita consumption of bottled water in the United States since 1985. The values of b(t) are measured in gallons, and t  0 corresponds to the year 1985. (Source: U.S. Department of Agriculture)

5 4 3 2 1 1 2 3

y  g(x) 1 2 3 4

5

x

4 5

a. Evaluate b(0) and b(7) and interpret the results in the context of this problem.

75. Find g122 .

76. Find g142 .

b. Determine the slope and interpret its meaning in the context of this problem.

78. For what value(s) of x is g1x2  4?

77. For what value(s) of x is g1x2  0?

79. Write the domain of g. 80. Write the range of g.

Test

Chapter 2

1. Given the equation x  23y  6, complete the ordered pairs and graph the corresponding points. 10, 2 1 , 02 1 , 32 y 41 2 1 1

1

2

3 4

5 6 7

8

x

2 3 4 5 6 7 8 9

2. Determine whether the following statements are true or false and explain your answer. a. The product of the x- and y-coordinates is positive only for points in quadrant I. b. The quotient of the x- and y-coordinates is negative only for points in quadrant IV. c. The point (2, 3) is in quadrant III. d. The point (0, 0) lies on the x-axis.

3. Determine if 14, 12 is a solution to 1 y   x  3. 2 4. Explain the process for finding the x- and y-intercepts. For Exercises 5–8, identify the x- and y-intercepts (if possible) and graph the line. 5. 6x  8y  24

6. x  4

y

y

5 4 3 2 1 5 4 3 2 1 1

5 4 3 2 1 1

2

3 4

5

x

2 3

7 6 5 4 3 2 1 1 2 3

4 5

4 5

1

2

3

x

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Chapter 2 Linear Equations in Two Variables and Functions

7. 3x  5y

8. 2y  6

y

4 3 2 1

5 4 3 2 1 5 4 3 2 1 1 2 3

14. Determine if the lines are parallel, perpendicular, or neither.

y

1

2

3 4

5

5 4 3 2 1 1 2 3

x

1

2

3 4

5

x

b. 9x  3y  1 15x  5y  10

c. 3y  6 x  0.5

d. 5x  3y  9 3x  5y  10

15. Write an equation that represents a line subject to the following conditions. (Answers may vary.)

4 5

4 5

a. y  x  4 yx3

6

9. Find the slope of the line, given the following information:

a. A line that does not pass through the origin and has a positive slope b. A line with an undefined slope

a. The line passes through the points (7, 3) and (1, 8).

c. A line perpendicular to the y-axis. What is the slope of such a line?

b. The line is given by 6x  5y  1.

d. A slanted line that passes through the origin and has a negative slope

10. Describe the relationship of the slopes of 16. Write an equation of the line that passes through the point (8, 12) with slope 2. Write the answer in slope-intercept form.

a. Two parallel lines b. Two perpendicular lines 11. The slope of a line is 7. a. Find the slope of a line parallel to the given line. b. Find the slope of a line perpendicular to the given line. 12. Two points are given for each of two lines. Determine if the lines are parallel, perpendicular, or neither. L1: 14, 42 and 11, 62

17. Write an equation of the line containing the points (2, 3) and (4, 0). 18. Write an equation of a line containing (4, 3) and parallel to 6x  3y  1. 19. Write an equation of the line that passes through the point (10, 3) and is perpendicular to 3x  y  7. Write the answer in slope-intercept form. 20. Jack sells used cars. He is paid $800 per month plus $300 commission for each automobile he sells.

L2: 12, 02 and 10, 32

13. Given the equation 3x  4y  4, a. Write the line in slope-intercept form. b. Determine the slope and y-intercept. c. Graph the line, using the slope and y-intercept.

a. Write an equation that represents Jack’s monthly earnings y in terms of the number of automobiles he sells x. b. Graph the linear equation you found in part (a).

y

5 4 3 2 1 1 2 3 4 5

1

2

3 4

5

x

Monthly Earnings ($)

y

5 4 3 2 1

6400 5600 4800 4000 3200 2400 1600 800 2

4

6

8 10 12 14 16 18 Number of Cars

x

20

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Test

c. What does the y-intercept mean in the context of this problem? d. How much will Jack earn in a month if he sells 17 automobiles?

24. Explain how to find the x- and y-intercepts of the graph defined by y ⫽ f 1x2. For Exercises 25–28, graph the functions. 25. f 1x2 ⫽ ⫺3x ⫺ 1

21. The graph represents the life expectancy for females in the United States according to birth year. The value x ⫽ 0 represents a birth year of 1940.

y

Life Expectancy for Females in the United States According to Birth Year Life Expectancy (years)

90 (0, 66)

70

70

10 20 30 40 50 60 Year of Birth (x ⫽ 0 corresponds to 1940)

Source: National Center for Health Statistics

a. Determine the y-intercept from the graph. What does the y-intercept represent in the context of this problem? b. Using the two points (0, 66) and (30, 75), determine the slope of the line. What does the slope of the line represent in the context of this problem? c. Use the y-intercept and the slope found in parts (a) and (b) to write an equation of the line by letting x represent the year of birth and y represent the corresponding life expectancy.

22.

1 2 3 4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

3

(3, 1)

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 1 2 3 4 ⫺1 ⫺2 (1, ⫺1) (⫺1, ⫺2) ⫺3

5

x

(⫺1, 0)

⫺4 ⫺5

3

3

2 1

2 1 5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1 2 3 4

5

⫺4 ⫺5

⫺4 ⫺5

For Exercises 29–31, write the domain in interval notation. x⫺5 x⫹7

b. What is the domain of r? 33. The function defined by s(t) ⫽ 1.6t ⫹ 36 approximates the per capita consumption of soft drinks in the United States since 1985. The values of s(t) are measured in gallons, and t ⫽ 0 corresponds to the year 1985. (Source: U.S. Department of Agriculture) a. Evaluate s(0) and s(7) and interpret the results in the context of this problem.

2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

y

5 4

a. Find r (⫺2), r(0), and r(3).

5 4

(3, 3)

28. w 1x2 ⫽ 冟 x冟

5 4

1 2 3 4

x

5

⫺4 ⫺5

y

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1 2 3 4

32. Given: r(x) ⫽ x2 ⫺ 2x ⫹ 1

y

y 5 4

⫺4 ⫺5

2 1

31. h1x2 ⫽ 1x ⫹ 721x ⫺ 52

23.

2 1

3

2 1

30. f 1x2 ⫽ 1x ⫹ 7

For Exercises 22–23, a. determine if the relation defines y as a function of x, b. identify the domain, and c. identify the range.

3

3

27. p 1x2 ⫽ x2

29. f 1x2 ⫽

d. Using the linear equation from part (c), approximate the life expectancy for women born in the United States in 1994. How does your answer compare with the reported life expectancy of 79 yr?

(⫺3, 3)

5 4

⫺4 ⫺5

60 50 0

y

5 4

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

(30, 75) 80

26. k1x2 ⫽ ⫺2

1 2 3 4

5

x

b. Determine the slope and interpret its meaning in the context of this problem.

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Chapter 2 Linear Equations in Two Variables and Functions

For Exercises 34–41, refer to the graph. y

40. For what value(s) of x is f 1x2  0?

y  f (x)

5 4

41. For what value(s) of x is f 1x2  1?

3 2 1 2 1 1 2 3

39. Find the x-intercept.

1 2 3 4

5

6

7 8

x

4 5

34. Find f 112.

35. Find f 142.

36. Write the domain of f.

For Exercises 42–45, determine if the function is constant, linear, quadratic, or none of these. 42. f 1x2  3x2

43. g1x2  3x

44. h1x2  3

45. k1x2  

3 x

3 46. Find the x- and y-intercepts for f 1x2  x  9. 4

37. Write the range of f. 38. Answer true or false. The value y  5 is in the range of f.

Chapters 1–2

Cumulative Review Exercises

For Exercises 1–3, simplify the expressions. 5  23 4  7 1. 1  314  12

10. 02x  1 0  3 6 12 11. `

x3 ` 2 5

2. 3  225  81 292 6 3. 433x  51y  2x2  34  716y  x2 For Exercises 4–6, solve the equation. 4.

2x  3 x1   2 6 4

5. z  13  2z2  5  z  5 6. 0x  5 0  7  10 For Exercises 7–11, solve the inequality. Write the answer in interval notation. 7. 4 

x1 6 3 2

8. 3x  2  11 and 4  2x 9. 3x  2  11 or 4  2x

12. Given: f1x2  12x  1 and g1x2  3x2  2x a. Find f(4).

b. Find g(3).

13. Find the slope of the line that passes through the points (4, 5) and (6, 3). 14. Determine if (13, 7) is a solution to y  6x  5. For Exercises 15–16, a. find the x- and y-intercepts, b. find the slope, and c. graph the line. 15. 3x  5y  10 y 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

1

2

3 4

5

x

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Cumulative Review Exercises

16. 2y  4  10

21. Find an equation of the line passing through (1, 4) and perpendicular to y  14x  2. Write the answer in slope-intercept form.

y 5 4 3 2 1 5 4 3 2 1 1 2 3

231

1

2

3 4

5

x

4 5

17. Find an equation for the vertical line that passes through the point (7, 11). 18. Find the slope of the line 5x  2y  10. 19. State the domain and range of the relation. Is the relation a function? {(3, 1), (4, 5), (3, 8)} 20. Find an equation of the line passing through (1, 4) and parallel to 2x  y  6. Write the answer in slope-intercept form.

22. At the movies, Laquita paid for drinks and popcorn for herself and her two children. She spent twice as much on popcorn as on drinks. If her total bill came to $17.94, how much did she spend on drinks and how much did she spend on popcorn? 23. Write the domain of the function in interval notation. f 1x2 

1 x  15

24. A chemist mixes a 20% salt solution with 15 L of a 50% salt solution to get a 38% salt solution. How much of the 20% solution does she use? 25. The yearly rainfall for Seattle, Washington, is 0.7 in. less than twice the yearly rainfall for Los Angeles, California. If the total yearly rainfall for the two cities is 50 in., how much rain does each city get per year?

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3

Systems of Linear Equations and Inequalities CHAPTER OUTLINE 3.1 Solving Systems of Linear Equations by the Graphing Method 234 3.2 Solving Systems of Linear Equations by the Substitution Method 243 3.3 Solving Systems of Linear Equations by the Addition Method 249 Problem Recognition Exercises: Solving Systems of Linear Equations

256

3.4 Applications of Systems of Linear Equations in Two Variables 256 3.5 Linear Inequalities and Systems of Linear Inequalities in Two Variables 265 3.6 Systems of Linear Equations in Three Variables and Applications 278 3.7 Solving Systems of Linear Equations by Using Matrices 288 Group Activity: Creating a Quadratic Model of the Form y ⴝ at2 ⴙ bt ⴙ c

297

Chapter 3 y

In this chapter, we solve systems of linear equations in two and three variables. One of the skills we need is to find the point of intersection between two lines. Are You Prepared? To practice, graph the five lines on the grid provided. Then answer the questions and place the letter corresponding to each answer in the space below. This will define an important vocabulary word for this chapter. Line Line Line Line Line 1. 2. 3. 4. 5. 6. 7.

The The The The The The The

1. 2. 3. 4. 5.

y y x y x

    

5 4 3 2 1 5 4 3 2 1 1

2

3

4

5

x

2 3 4

x2 x  4 y1 2 41

point of intersection between lines point of intersection between lines point of intersection between lines line that is parallel to line 3. line that is vertical. line that has (3, 1) as a solution. point of intersection between lines

1

5

1 and 2. 3 and 4. 5 and 1.

1 and 3.

T (1, 2) I (3, 1) U Line 2 O (1, 3) S Line 1 N no point of intersection L Line 5

A ___ ___ ___ ___ ___ ___ ___ ___ to a system of linear equations is a point of 4 1 5 6 2 3 1 7 intersection between two lines. 233

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234

Chapter 3 Systems of Linear Equations and Inequalities

Section 3.1

Solving Systems of Linear Equations by the Graphing Method

Concepts

1. Solutions to Systems of Linear Equations

1. Solutions to Systems of Linear Equations 2. Dependent and Inconsistent Systems of Linear Equations 3. Solving Systems of Linear Equations by Graphing

A linear equation in two variables has an infinite number of solutions that form a line in a rectangular coordinate system. Two or more linear equations form a system of linear equations. For example: x ⫺ 3y ⫽ ⫺5 2x ⫹ 4y ⫽ 10 A solution to a system of linear equations is an ordered pair that is a solution to each individual linear equation. Example 1

Determining Solutions to a System of Linear Equations

Determine whether the ordered pairs are solutions to the system. x ⫹ y ⫽ ⫺6 a. 1⫺2, ⫺42

b. 10, ⫺62

3x ⫺ y ⫽ ⫺2

Solution:

a. Substitute the ordered pair 1⫺2, ⫺42 into both equations: x ⫹ y ⫽ ⫺6 3x ⫺ y ⫽ ⫺2

1⫺22 ⫹ 1⫺42 ⱨ ⫺6  True

31⫺22 ⫺ 1⫺42 ⱨ ⫺2  True

Because the ordered pair 1⫺2, ⫺42 is a solution to each equation, it is a solution to the system of equations. b. Substitute the ordered pair 10, ⫺62 into both equations: x ⫹ y ⫽ ⫺6 3x ⫺ y ⫽ ⫺2

102 ⫹ 1⫺62 ⱨ ⫺6  True

3102 ⫺ 1⫺62 ⱨ ⫺2

False

Because the ordered pair 10, ⫺62 is not a solution to the second equation, it is not a solution to the system of equations. Skill Practice Determine whether the ordered pairs are solutions to the system. 3x ⫹ 2y ⫽ ⫺8 1. 1⫺2, ⫺12

2. 14, ⫺102

y ⫽ 2x ⫺ 18

A solution to a system of two linear equations can be interpreted graphically as a point of intersection between the two lines. Answers 1. No

2. Yes

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Section 3.1

235

Solving Systems of Linear Equations by the Graphing Method

Graphing the lines from Example 1 we see that the point of intersection is 12, 42. Therefore, we say that the solution set is 512, 426. See Figure 3-1.

2. Dependent and Inconsistent Systems of Linear Equations When two lines are drawn in a rectangular coordinate system, three geometric relationships are possible:

y 3 2 1 5 4 3 2 1 1 2

3x  y  2 1

2

3

(2,4)

4 5 6 7

x + y = 6

1. Two lines may intersect at exactly one point.

Figure 3-1

2. Two lines may intersect at no point. This occurs if the lines are parallel.

3. Two lines may intersect at infinitely many points along the line. This occurs if the equations represent the same line (the lines are coinciding). If a system of linear equations has one or more solutions, the system is said to be a consistent system. If a linear system has no solution, it is said to be an inconsistent system. If two equations represent the same line, then all points along the line are solutions to the system of equations. In such a case, the system is characterized as a dependent system. An independent system is one in which the two equations represent different lines. The different possibilities for solutions to systems of linear equations are given in Table 3–1. Table 3-1

Solutions to Systems of Linear Equations in Two Variables

One Unique Solution

No Solution

Infinitely Many Solutions

One point of intersection

Parallel lines

Coinciding lines

System is consistent. System is independent.

System is inconsistent. System is independent.

System is consistent. System is dependent.

3. Solving Systems of Linear Equations by Graphing Example 2

Solving a System of Linear Equations by Graphing

Solve the system by graphing both linear equations and finding the point(s) of intersection. 1 y x2 2 4x  2y  6

3 4

5

x

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Chapter 3 Systems of Linear Equations and Inequalities

Solution: To graph each equation, write the equation in slope-intercept form y  mx  b. First equation 1 y x2 2

Second equation 4x  2y  6

Slope: 12

2y  4x  6 2y 4x 6   2 2 2 y  2x  3

Slope: 2

From their slope-intercept forms, we see that the lines have different slopes, indicating that the lines must intersect at exactly one point.We can graph the lines using the slope and y-intercept to find the point of intersection (Figure 3-2). y 5 4 4x  2y  6 3 2 1 5 4 3 2 1 1 2 1

Avoiding Mistakes

y 2x  2

Using graph paper may help you be more accurate when graphing lines. There are many websites from which you can print graph paper.

1

2

x 3 4 5 Point of intersection

(2, 1)

3 4 5

Figure 3-2

The point (2, 1) appears to be the point of intersection. This can be confirmed by substituting x  2 and y  1 into both equations. 1 y x2 2 1 1 ⱨ 122  2 2 1 ⱨ 1  2

4x  2y  6 4122  2112 ⱨ 6 82ⱨ6 6 ⱨ 6 ✔ True

1 ⱨ 1 ✔ True

The solution set is 512, 126. Skill Practice Solve by using the graphing method. 3. y  3x  5 x  2y  4

TIP: In Example 2, the lines could also have been graphed by using the x- and y-intercepts or by using a table of points. However, the advantage of writing the equations in slope-intercept form is that we can compare the slope and y-intercept of each line. 1. 2.

Answer

3. 512, 12 6

3.

If the slopes differ, the lines are different and nonparallel and must cross in exactly one point. If the slopes are the same and the y-intercepts are different, the lines are parallel and do not intersect. If the slopes are the same and the y-intercepts are the same, the two equations represent the same line.

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Section 3.1

Example 3

Solving Systems of Linear Equations by the Graphing Method

237

Solving a System of Linear Functions by Graphing

Solve the system.

f 1x2  3 g1x2  2x  1

Solution: This first function can be written as y  3. This is an equation of a horizontal line. Writing the second equation as y  2x  1, we have a slope of 2 and a y-intercept of (0, 1). The graphs of the functions are shown in Figure 3-3. The point of intersection is (1, 3). Therefore, the solution set is 511, 326.

y 5 4 3

f(x)  3

5 4 3 2 1 1 2

g(x)  2x  1

1

2

3 4

5

x

3 4 5

Skill Practice Solve the system by graphing. 4. f 1x2  1 g 1x2  3x  4

Example 4

(1, 3)

2 1

Figure 3-3

Solving a System of Linear Equations by Graphing

Solve the system by graphing. x  3y  6 6y  2x  6

Solution: To graph the lines, write each equation in slope-intercept form. x  3y  6

6y  2x  6

3y  x  6 3y x 6   3 3 3 1 y x2 3

6y 2x 6   6 6 6 1 y x1 3

Because the lines have the same slope but different y-intercepts, they are parallel (Figure 3-4). Two parallel lines do not intersect, which implies that the system has no solution. The system is inconsistent. The solution set is the empty set, 5 6.

y 5 4 3

y  13 x  1

2 1 5 4 3 2 1 1 2

1

2

3 4

x 5 y 1x  2 3

3 4 5

Figure 3-4

Skill Practice Solve the system by graphing. 5. 2y  2x x  y  3 Answers 4. {(1, 1)} 5. The solution set is { }. The system is inconsistent.

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Chapter 3 Systems of Linear Equations and Inequalities

Example 5

Solving a System of Linear Equations by Graphing

Solve the system by graphing. x  4y  8 1 y x2 4

Solution: Write the first equation in slope-intercept form. The second equation is already in slope-intercept form. First equation

Second equation

x  4y  8

1 y x2 4

4y  x  8

y

4y x 8   4 4 4

5 4 3 2 1

1 y x2 4

5 4 3 2 1 1 2

y   14 x  2 1

2

3 4

5

x

Notice that the slope-intercept forms of 3 the two lines are identical. Therefore, the 4 equations represent the same line (Figure 3-5). 5 The system is dependent, and the solution Figure 3-5 to the system of equations is the set of all points on the line. Because the ordered pairs in the solution set cannot all be listed, we can write the solution in set-builder notation. Furthermore, the equations x  4y  8 and y  14x  2 represent the same line. Therefore, the solution set may be written as 51x, y2 0 y  14x  26 or 51x, y2 0 x  4y  86. Skill Practice Solve the system by graphing. 1 6. y  x  1 2 x  2y  2

Calculator Connections Topic: Using the Intersect Feature The solution to a system of equations can be found by using either a Trace feature or an Intersect feature on a graphing calculator to find the point of intersection between two curves. For example, consider the system 2x  y  6 Answer 6. Infinitely many solutions; 1 {(x, y ) 0 y  x  1}; 2 dependent system

5x  y  1

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Section 3.1

Solving Systems of Linear Equations by the Graphing Method

239

First graph the equations together on the same viewing window. Recall that to enter the equations into the calculator, the equations must be written with the y variable isolated. Isolate y.

⫺2x ⫹ y ⫽ 6

y⫽

5x ⫹ y ⫽ ⫺1

2x ⫹ 6

y ⫽ ⫺5x ⫺ 1

By inspection of the graph, it appears that the solution is (⫺1, 4 ). The Trace option on the calculator may come close to (⫺1, 4) but may not show the exact solution (Figure 3-6). However, an Intersect feature on a graphing calculator may provide the exact solution (Figure 3-7). See your user’s manual for further details. Using Trace

Using Intersect

Figure 3-6

Figure 3-7

Section 3.1

Practice Exercises

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Study Skills Exercises 1. Before you proceed further in Chapter 3, make your test corrections for the Chapter 2 test. See Exercise 1 of Section 2.1 for instructions. 2. Define the key terms. a. System of linear equations

b. Solution to a system of linear equations

c. Consistent system

d. Inconsistent system

e. Dependent system

f. Independent system

Concept 1: Solutions to Systems of Linear Equations For Exercises 3–8, determine which points are solutions to the given system. (See Example 1.) 1 3. y ⫽ 8x ⫺ 5 4. y ⫽ ⫺ x ⫺ 5 5. 2x ⫺ 7y ⫽ ⫺30 2 y ⫽ 4x ⫹ 3 y ⫽ 3x ⫹ 7 3 y ⫽ x ⫺ 10 1⫺1, 132, 1⫺1, 12, (2, 11) 3 4 10, ⫺302 , a , 5b, 1⫺1, 42 2 9 14, ⫺72, 10, ⫺102, a3, ⫺ b 2

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Chapter 3 Systems of Linear Equations and Inequalities

6. x  2y  4

7.

x y6

8. x  3y  3

4x  3y  4

1 y x2 2

2x  9y  1

14, 22, 16, 02, 12, 42

1 12, 32, 14, 02, a3, b 2

10, 12, 14, 12, 19, 22

Concept 2: Dependent and Inconsistent Systems of Linear Equations For Exercises 9–14, the graph of a system of linear equations is given. a. Identify whether the system is consistent or inconsistent. b. Identify whether the system is dependent or independent. c. Identify the number of solutions to the system. 9. y  x  3

10. 5x  3y  6

3x  y  1

11. 2x  y  4

3y  2x  3

y

4x  2y  2

y

y

5 4 3

5 4 3

5 4 3

2 1

2 1

2 1

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

4x  2y  0

x  3y  6

y

5 4 3

2 1

2 1

2 1

2

3 4

5

x

5

2

3 4

5

3 4

5

x

y

5 4 3

1

3 4

4x  6y  6

y

5 4 3

5 4 3 2 1 1 2

2

2 14. y   x  1 3

1 13. y  x  2 3

12. y  2x  3

1

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

1

x

Concept 3: Solving Systems of Linear Equations by Graphing For Exercises 15–32, solve the systems of equations by graphing. (See Examples 2–5.) 15. 2x  y  3

16. 4x  3y  12

x  y  3

17. f1 x2  2x  3 g1x 2 

3x  4y  16

y

y

5x  4 y

5 4 3

5 4 3

5 4 3

2 1

2 1

2 1

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

1

2

x

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Section 3.1

18. h1 x2  2x  5

1 x5 3

1 20. f1x2  x  2 2

2 f 1 x2   x  2 3

5 g1x2  x  2 2

19. k1x2 

g1x2  x  2

y

y

y

5 4 3

5 4 3

5 4 3

2 1

2 1

2 1

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

21. x  4

22. 3x  2y  6

y  2x  3

5 4 3

2 1

2 1

2 1

2

3 4

5

x

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

1 24. y  x  2 3

25.

x  3y  9

2 y x1 3

y

5 4 3

2 1

2 1

2 1

2

3 4

5

x

2

3 4

5

5 4 3 2 1 1 2

2

3 4

5

x

y

5 4 3

1

1

x

1 y x2 2

5 4 3

5 4 3 2 1 1 2

5

26. 4x  16  8y

2x  3y  3

y

3 4

y

5 4 3

1

2

2x  y  1 y

5 4 3

5 4 3 2 1 1 2

1

23. y  2x  3

y  3

y

241

Solving Systems of Linear Equations by the Graphing Method

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

1

x

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Chapter 3 Systems of Linear Equations and Inequalities

27. 2x  4

28. y  7  6

29. x  3y  6 6y  2x  12

5  2x

1 y  1 2 y

y

y

5 4 3

5 4 3

5 4 3

2 1

2 1

2 1

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

30. 3x  2y  4

31. 2x  y  4

4y  6x  8

y

5 4 3

2 1

2 1

2 1

2

3 4

5

x

5

3 4

5

x

y

5 4 3

1

3 4

2x  8y  16

5 4 3

5 4 3 2 1 1 2

2

32. x  4y  4

4x  2  2y

y

1

5 4 3 2 1 1 2

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

3

4 5

4 5

4 5

1

2

x

For Exercises 33–36, identify each statement as true or false. 33. A consistent system is a system that always has a unique solution.

34. A dependent system is a system that has no solution.

35. If two lines coincide, the system is dependent.

36. If two lines are parallel, the system is independent.

Graphing Calculator Exercises For Exercises 37–42, use a graphing calculator to graph each linear equation on the same viewing window. Use a Trace or Intersect feature to find the point(s) of intersection. 37. y  2x  3 y  4x  9

40.

x  2y  2 3x  2y  6

1 38. y   x  2 2 1 y x3 3 41. x  3y  6 6y  2x  6

39.

xy4 2x  y  5

42. x  4y  8 1 y x2 4

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Section 3.2

Solving Systems of Linear Equations by the Substitution Method

Solving Systems of Linear Equations by the Substitution Method

Section 3.2

1. The Substitution Method

Concepts

Graphing a system of equations is one method to find the solution of the system. However, sometimes it is difficult to determine the solution using this method because of limitations in the accuracy of the graph. This is particularly true when the coordinates of a solution are not integer values or when the solution is a point not sufficiently close to the origin. Identifying the coordinates of the point 1 173 , ⫺239 2 or 1⫺251, 83492 , for example, might be difficult from a graph.

1. The Substitution Method 2. Solving Inconsistent Systems and Dependent Systems

In this section and Section 3.3, we will present two algebraic methods to solve a system of equations. The first is called the substitution method. This technique is particularly important because it can be used to solve more advanced problems including nonlinear systems of equations. The first step in the substitution process is to isolate one of the variables from one of the equations. Consider the system x ⫹ y ⫽ 16 x⫺y⫽4 Solving the first equation for x yields x ⫽ 16 ⫺ y. Then, because x is equal to 16 ⫺ y, the expression 16 ⫺ y may replace x in the second equation. This leaves the second equation in terms of y only. Solve for x.

x ⫹ y ⫽ 16

Second equation:

116 ⫺ y2 ⫺ y ⫽ 4

x ⫽ 16 ⫺ y v

First equation:

16 ⫺ 2y ⫽ 4

Substitute x ⫽ 16 ⫺ y. Solve for y.

⫺2y ⫽ ⫺12 y⫽6 x ⫽ 16 ⫺ y x ⫽ 16 ⫺ 162 x ⫽ 10

To find x, substitute y ⫽ 6 back into the expression x ⫽ 16 ⫺ y.

Check the ordered pair (10, 6) in both original equations. x ⫹ y ⫽ 16

1102 ⫹ 162 ⱨ 16 ✔ True The solution set is 5110, 626.

x⫺y⫽4

1102 ⫺ 162 ⱨ 4 ✔ True

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Chapter 3 Systems of Linear Equations and Inequalities

PROCEDURE Solving a System of Equations by the Substitution Method Step 1 Step 2 Step 3 Step 4

Isolate one of the variables from one equation. Substitute the quantity found in step 1 into the other equation. Solve the resulting equation. Substitute the value found in step 3 back into the equation in step 1 to find the value of the remaining variable. Step 5 Check the solution in both equations, and write the answer as an ordered pair within set notation.

Example 1

Using the Substitution Method to Solve a System of Linear Equations

Solve the system by using the substitution method. 3x  2y  7 6x  y  6

Solution: The y variable in the second equation is the easiest variable to isolate because its coefficient is 1. 3x  2y  7

Avoiding Mistakes

6x  y  6

6x  16x  62  6

y  6x  6

Step 1:

Solve the second equation for y.

Step 2:

Substitute the quantity 6x  6 for y in the other equation.

Step 3:

Solve for x.

Step 4:

Substitute x  13 into the expression y  6x  6.

Step 5:

Check the ordered pair 1 13, 42 in each original equation.

¶

6x  y  6

Do not substitute y  6x  6 into the same equation from which it came. This mistake will result in an identity:

3x  216x  62  7 3x  12x  12  7

6x  6x  6  6

15x  12  7

66

15x  5 x

1 3

y  6x  6 1 y  6 a b  6 3 y  2  6 y4 3x  2y  7

6x  y  6

1 3 a b  2142 ⱨ 7 3

1 6a b  4 ⱨ 6 3

1  8 ⱨ 7 ✔

The solution set is 51 13, 426.

24ⱨ6✔

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Section 3.2

Solving Systems of Linear Equations by the Substitution Method

Skill Practice Solve by using the substitution method. 1. 3x  y  8 x  2y  12

Example 2

Using the Substitution Method to Solve a System of Linear Equations

Solve the system by using the substitution method. 3x  4y  9 1 x y2 3

Solution: 3x  4y  9 Step 1:

¶

1 x y2 3

In the second equation, x is already isolated.

1 3a y  2b  4y  9 3

Step 2:

1 Substitute the quantity  y  2 for 3 x in the other equation.

y  6  4y  9

Step 3:

Solve for y.

5y  15 y3 Now use the known value of y to solve for the remaining variable x. 1 x y2 3 1 x   132  2 3 x  1  2

Step 4:

Substitute y  3 into the equation 1 x   y  2. 3

x1 Step 5:

Check the ordered pair (1, 3) in each original equation. 3x  4y  9

3112  4132 ⱨ 9 3  12 ⱨ 9 ✔ True The solution set is 511, 326.

1 x y2 3 1 1 ⱨ  132  2 3 1 ⱨ 1  2 ✔ True

Skill Practice Solve by using the substitution method. 2. x  2y  3 4x  2y  0 Answers 1. {(4, 4)}

2. {(1, 2)}

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Chapter 3 Systems of Linear Equations and Inequalities

2. Solving Inconsistent Systems and Dependent Systems Example 3

Solving an Inconsistent System

Solve the system by using the substitution method. x  2y  4 2x  4y  6

Solution: x  2y  4

Step 1:

The x variable is already isolated.

212y  42  4y  6

Step 2:

Substitute the quantity x  2y  4 into the other equation.

4y  8  4y  6

Step 3:

Solve for y.

2x  4y  6

86

The equation reduces to a contradiction, indicating that the system has no solution. The lines never intersect and must be parallel. The system is inconsistent.

There is no solution. The system is inconsistent. The solution set is 5 6.

TIP: The answer to Example 3 can be verified by writing each equation in slope-intercept form and graphing the equations.

Equation 1 x  2y  4

2x  4y  6

2y  x  4

4y  2x  6

2y x 4   2 2 2

4y 2x 6   4 4 4

y

1 x2 2

y

Equation 2

5 x  2y  4 4 3 2x  4y  6 2 1 x 1 2 3 4 5 5 4 3 2 1 1

1 3 y x 2 2

2 3 4

Notice that the equations have the same slope, but different y-intercepts; therefore, the lines must be parallel. There is no solution to this system of equations.

Skill Practice Solve by using the substitution method. 3. 8x  16y  3 1 y x1 2

Answer 3. No solution; { }; inconsistent system

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Section 3.2

Solving Systems of Linear Equations by the Substitution Method

Solving a Dependent System

Example 4

Solve by using the substitution method. 4x  2y  6 y  3  2x

Solution: 4x  2y  6 y  2x  3 v

y  3  2x

4x  212x  32  6 4x  4x  6  6

Step 1:

Solve for one of the variables.

Step 2:

Substitute the quantity 2x  3 for y in the other equation.

Step 3: Solve for x. Apply the distributive property to clear the parentheses.

6  6 The system reduces to the identity 6  6. Therefore, the original two equations are equivalent, and the system is dependent. The solution consists of all points on the common line, giving us an infinite number of solutions. Because the equations 4x  2y  6 and y  3  2x represent the same line, the solution set is 51x, y2 0 4x  2y  66

or

51x, y2 0 y  3  2x6

Skill Practice Solve the system by using substitution. 4. 3x  6y  12 2y  x  4

TIP: We can confirm the results of Example 4 by writing each equation in slopeintercept form. The slope-intercept forms are identical, indicating that the lines are the same. slope-intercept form

4x  2y  6

2y  4x  6

y  3  2x

y  2x  3

Answer 4. Infinitely many solutions; 5 1x, y2 0 3x  6y  126 ; dependent system

y  2x  3

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Review Exercises For Exercises 1–4, use the slope-intercept form of the lines to determine the number of solutions to the system. 1. y  8x  1 2x  16y  3

2.

4x  6y  1

3. 2x  4y  0

5 2

x  2y  9

10x  15y 

4. 6x  3y  8 8x  4y  1

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Chapter 3 Systems of Linear Equations and Inequalities

5. Determine if the ordered pair (4, 3) is a solution to the system. x  2y  10 2x  y  11 For Exercises 6–7, solve the system by graphing. 6.

x y4

7. y  2x  3

y 5 4 3 2

3x  4y  12

y 5 4 3 2

6x  3y  9

1 5 4 3 2 1 1 2

1 1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

4 5

4 5

1

2

3 4

5

x

Concept 1: The Substitution Method For Exercises 8–21, solve by using the substitution method. (See Examples 1–2.) 8. 4x  12y  4

9. y  3x  1

y  5x  11 11. 3x  8y  1

2x  3y  8 12. 12x  2y  0

4x  11  y 14.

x  3y  3 2x  3y  6

17. 2x  y  1 y  2x 20. 2x  3y  7 5x  2y  12

7x  y  1 15.

10. 10y  34  x 7x  y  31 13. 3x  12y  36 x  5y  12

x y8

16. 5x  2y  10

3x  2y  9

yx1

18. 1  3y  10 5x  2y  6

19. 2x  3  7 3x  4y  6

21. 4x  5y  14 3y  x  7

22. Describe the process of solving a system of linear equations by using substitution.

Concept 2: Solving Inconsistent Systems and Dependent Systems For Exercises 23–30, solve the systems. (See Examples 3–4.) 23. 2x  6y  2

24. 2x  4y  22

x  3y  1

x  2y  11

3 1 26. x   y  2 2 4x  6y  7 29. 3x  y  7 14  6x  2y

27. 5x  y  10 2y  10x  5

25. y 

1 7

x3

x  7y  4 28. x  4y  8 3x  3  12y

30. x  4y  1 12y  3x  3

31. When using the substitution method, explain how to determine whether a system of linear equations is dependent. 32. When using the substitution method, explain how to determine whether a system of linear equations is inconsistent.

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Section 3.3

Solving Systems of Linear Equations by the Addition Method

Mixed Exercises For Exercises 33–58, solve the system by using the substitution method. 33. x ⫽ 1.3y ⫹ 1.5 y ⫽ 1.2x ⫺ 4.6

1 5 36. x ⫽ y ⫺ 6 3 1 21 y⫽ x⫹ 5 5 39. 3x ⫹ 2y ⫽ 6 y⫽x⫹3 42. 200y ⫽ 150x y⫺4⫽1

45. y ⫽

200x ⫺ 320

y ⫽ ⫺150x ⫹ 1080 48. y ⫽

6.8x ⫹ 2.3

y ⫽ ⫺4.1x ⫹ 56.8 51. 21x ⫹ 2y2 ⫽ 12 ⫺6x ⫽ 5y ⫺ 8

34. y ⫽ 0.8x ⫺ 1.8 1.1x ⫽ ⫺y ⫹ 9.6

37. ⫺2x ⫹ y ⫽ 4

2 1 35. y ⫽ x ⫺ 3 3 1 17 x⫽ y⫹ 4 4 38. 8x ⫺

1 1 1 ⫺ x⫹ y⫽ 4 8 4 40. ⫺x ⫹ 4y ⫽ ⫺4

1 1 1 x⫺ y⫽ 3 24 2 41. ⫺300x ⫺ 125y ⫽ 1350

y⫽x⫺1 43. 2x ⫺

y⫹2⫽8

y⫽6

44.

1 1 1 x⫺ y⫽ 6 12 2 46. y ⫽ ⫺54x ⫹ 300 y⫽

49. 4x ⫹ 4y ⫽ 5

x ⫺ 4y ⫽ 8 1 1 1 x⫺ y⫽ 16 4 2

47. y ⫽ ⫺2.7x ⫺ 5.1

20x ⫺ 70

x ⫺ 4y ⫽ ⫺

y⫽8

y⫽

3.1x ⫺ 63.1

50. ⫺2x ⫹

6x ⫺ 13y ⫽ ⫺12

5 2

52. 5x ⫺ 2y ⫽ ⫺25

y ⫽ ⫺6

53. 513y ⫺ 22 ⫽ x ⫹ 4

10x ⫽ 31y ⫺ 102

4y ⫽ 7x ⫺ 3

54. 2x ⫽ ⫺31y ⫹ 32

55. 2x ⫺ 5 ⫽ 7

56. ⫺2 ⫽ 4 ⫺ 2y

3x ⫺ 4y ⫽ ⫺22

4 ⫽ 3y ⫹ 1

7x ⫺ 5 ⫽ ⫺5

57. 0.01y ⫽ 0.02x ⫺ 0.11 0.3x ⫺ 0.5y ⫽ 2

58. 0.3x ⫺ 0.4y ⫽ 1.3 0.01x ⫽ 0.03y ⫹ 0.01

Solving Systems of Linear Equations by the Addition Method

Section 3.3

1. The Addition Method

Concepts

The next method we present to solve systems of linear equations is the addition method (sometimes called the elimination method). With the addition method, begin by writing both equations in standard form Ax ⫹ By ⫽ C. Then multiply one or both equations by appropriate constants to create opposite coefficients on either the x or the y variable. Next, add the equations to eliminate the variable having opposite coefficients. This process is demonstrated in Example 1.

1. The Addition Method 2. Solving Inconsistent Systems and Dependent Systems

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Chapter 3 Systems of Linear Equations and Inequalities

Example 1

Solving a System by the Addition Method

Solve the system by using the addition method. 3x  4y  2 4x  y  9

Solution: 3x  4y  2

Avoiding Mistakes

4x  y  9

3x  4y  2 Multiply by 4.

16x  4y  36

Multiply the second equation by 4. This makes the coefficients of the y variables opposite.

3x  4y  2

Now if the equations are added, the y variable will be eliminated.

Be sure to multiply both sides of the equation by 4: 414x  92  4192

16x  4y  36  38

19x

x2

Solve for x.

3x  4y  2 3122  4y  2 6  4y  2

Substitute x  2 back into one of the original equations and solve for y.

4y  4 y1 Check the ordered pair (2, 1) in each original equation:

TIP: Substituting x  2 into the other equation, 4x  y  9, produces the same value for y. 4x  y  9 4122  y  9 8y9 y1

3x  4y  2 3122  4112 ⱨ 2 ✔ True

The solution set is 512, 126.

4x  y  9

4122  112 ⱨ 9 ✔ True

Skill Practice Solve by using the addition method. 1. 2x  3y  13 x  2y  3

The steps to solve a system of linear equations in two variables by the addition method is outlined in the following box.

PROCEDURE Solving a System of Linear Equations by the Addition Method

Answer 1. {(5, 1)}

Step 1 Write both equations in standard form: Ax  By  C. Step 2 Clear fractions or decimals (optional). Step 3 Multiply one or both equations by nonzero constants to create opposite coefficients for one of the variables. Step 4 Add the equations from step 3 to eliminate one variable. Step 5 Solve for the remaining variable. Step 6 Substitute the known value found in step 5 into one of the original equations to solve for the other variable. Step 7 Check the ordered pair in both equations and write the solution set.

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Section 3.3

Example 2

Solving Systems of Linear Equations by the Addition Method

Solving a System by the Addition Method

Solve the system by using the addition method. 4x ⫹ 5y ⫽ 2 3x ⫽ 1 ⫺ 4y

Solution: 4x ⫹ 5y ⫽ 2

4x ⫹ 5y ⫽ 2

3x ⫽ 1 ⫺ 4y

3x ⫹ 4y ⫽ 1

Step 1: Write both equations in standard form. There are no fractions or decimals.

We may choose to eliminate either variable. To eliminate x, change the coefficients to 12 and ⫺12. 4x ⫹ 5y ⫽ 2 3x ⫹ 4y ⫽ 1

Multiply by 3.

12x ⫹ 15y ⫽ 6

Multiply by ⫺4.

Step 3: Multiply the first equation by 3.

⫺12x ⫺ 16y ⫽ ⫺4 12x ⫹ 15y ⫽ 6

Multiply the second equation by ⫺4. Step 4: Add the equations.

⫺12x ⫺ 16y ⫽ ⫺4 ⫺y ⫽ 2

Step 5: Solve for y.

y ⫽ ⫺2 Step 6: Substitute y ⫽ ⫺2 back into one of the original equations and solve for x.

4x ⫹ 5y ⫽ 2 4x ⫹ 51⫺22 ⫽ 2 4x ⫺ 10 ⫽ 2 4x ⫽ 12 x⫽3

The solution set is 513, ⫺226.

Step 7: Check the ordered pair (3, ⫺2) in both original equations.

TIP: To eliminate the x variable in Example 2, both equations were multiplied by appropriate constants to create 12x and ⫺12x. We chose 12 because it is the least common multiple of 4 and 3. We could have solved the system by eliminating the y variable. To eliminate y, we would multiply the top equation by 4 and the bottom equation by ⫺5. This would make the coefficients of the y variable 20 and ⫺20, respectively. 4x ⫹ 5y ⫽ 2

Multiply by 4.

3x ⫹ 4y ⫽ 1

Multiply by ⫺5.

16x ⫹ 20y ⫽ 8 ⫺15x ⫺ 20y ⫽ ⫺5

Skill Practice Solve by using the addition method. 2. 2y ⫽ 5x ⫺ 4 3x ⫺ 4y ⫽ 1 Answer 1 2. e a1, b f 2

251

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Example 3

Solving a System by the Addition Method

Solve the system by using the addition method. x ⫺ 2y ⫽ 6 ⫹ y 0.05y ⫽ 0.02x ⫺ 0.10

Solution: x ⫺ 2y ⫽ 6 ⫹ y

x ⫺ 3y ⫽ 6

0.05y ⫽ 0.02x ⫺ 0.10 x

⫺0.02x ⫹ 0.05y ⫽ ⫺0.10

Step 1: Write both equations in standard form.

⫺ 3y ⫽ 6

⫺0.02x ⫹ 0.05y ⫽ ⫺0.10

Multiply by 100.

Multiply by 2.

x ⫺ 3y ⫽ 6 ⫺2x ⫹ 5y ⫽ ⫺10

⫺2x ⫹ 5y ⫽ ⫺10

2x ⫺ 6y ⫽ 12 ⫺2x ⫹ 5y ⫽ ⫺10 ⫺y ⫽ 2 y ⫽ ⫺2

x ⫺ 2y ⫽ 6 ⫹ y

x ⫺ 21⫺22 ⫽ 6 ⫹ 1⫺22

Step 2: Clear decimals. Step 3: Create opposite coefficients. Step 4: Add the equations. Step 5: for y.

Solve

Step 6: To solve for x, substitute y ⫽ ⫺2 into one of the original equations.

x⫹4⫽4 x⫽0 The solution set is 510, ⫺226.

Step 7: The ordered pair (0, ⫺2) checks in each original equation.

Skill Practice Solve by using the addition method. 3. 0.2x ⫹ 0.3y ⫽ 1.5 5x ⫹ 3y ⫽ 20 ⫺ y

2. Solving Inconsistent Systems and Dependent Systems Example 4

Solving a Dependent System

Solve the system by using the addition method. 1 1 x⫺ y⫽1 5 2 ⫺4x ⫹ 10y ⫽ ⫺20

Solution: Answer 3. {(0, 5)}

1 1 x ⫺ y⫽1 5 2 ⫺4x ⫹ 10y ⫽ ⫺20

Step 1:

Equations are in standard form.

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1 1 10 a x  yb  10 ⴢ 1 5 2

Multiply by 10.

253

Solving Systems of Linear Equations by the Addition Method

Step 2: Clear fractions.

2x  5y  10

4x  10y  20 2x  5y  10

Multiply by 2.

4x  10y  20

4x  10y  20

Step 3: Multiply the first equation by 2.

4x  10y  20 00

Step 4: Add the equations.

Notice that both variables were eliminated. The system of equations is reduced to the identity 0  0. Therefore, the two original equations are equivalent and the system is dependent. The solution set consists of an infinite number of ordered pairs (x, y) that fall on the common line of intersection 4x  10y  20, or equivalently 15x  12y  1. The solution set is 51x, y2 0 4x  10y  206

or

e 1x, y2 `

1 1 x  y  1f 5 2

Skill Practice Solve by the addition method. 4. 3x 

y4 1 4 x y 3 3

Example 5

Solving an Inconsistent System

Solve the system by using the addition method. 2y  3x  4 2016x  5y2  40  20y

Solution: Step 1:

Write the equations in standard form.

2y  3x  4

3x  2y  4

2016x  5y2  40  20y

120x  100y  40  20y

120x  80y  40

With both equations now in standard form,we can proceed with the addition method. Step 2:

y

There are no decimals or fractions. 3x  2y  4

Multiply by 40.

120x  80y  40

120x  80y  160 120x  80y 

Step 3:

40

0  120

Step 4:

Multiply the top equation by 40. Add the equations.

5 4 3 2

5 4 3 2 1 1 2

2

3 4

5

x

4 5

Figure 3-8

There is no solution, 5 6. 5. 18  10x  6y 5x  3y  9

1

3

The equations reduce to a contradiction, indicating that the system has no solution. The system is inconsistent. The two equations represent parallel lines, as shown in Figure 3-8.

Skill Practice Solve by using the addition method.

3x  2y  4

120x  80y  40 1

Answers 4. Infinitely many solutions; {(x, y ) 0 3x  y  4}; dependent system 5. No solution; { }; inconsistent system

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Section 3.3 Practice Exercises • Practice Problems • Self-Tests • NetTutor

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Review Exercises For Exercises 1–4, use the slope-intercept form of the lines to determine the number of solutions for the system of equations. 1 1. y ⫽ x ⫺ 4 2

2. y ⫽ 2.32x ⫺ 8.1

3. 4x ⫽ y ⫹ 7

y ⫽ 1.46x ⫺ 8.1

1 y⫽ x⫹1 2

⫺2y ⫽ ⫺8x ⫹ 14

4. 3x ⫺ 2y ⫽ 9 3 y⫽ x 2

Concept 1: The Addition Method For Exercises 5–16, solve the system by using the addition method. (See Examples 1–3.) 5.

3x ⫺ y ⫽ ⫺1 ⫺3x ⫹ 4y ⫽ ⫺14

9.

6. 5x ⫺ 2y ⫽ 15 3x ⫹ 2y ⫽ ⫺7

3x ⫹ 7y ⫽ ⫺20

10. 6x ⫺ 9y ⫽ ⫺15

⫺5x ⫹ 3y ⫽ ⫺84

5x ⫺ 2y ⫽ ⫺40

2x ⫹ 3y ⫽ 3

7.

⫺10x ⫹ 2y ⫽ ⫺32 11. 3x ⫽ 10y ⫹ 13 7y ⫽ 4x ⫺ 11

13. 1.2x ⫺ 0.6y ⫽ 3

14. 1.8x ⫹ 0.8y ⫽ 1.4

0.8x ⫺ 1.4y ⫽ 3

1.2x ⫹ 0.6y ⫽ 1.2

15. 3x ⫹ 2 ⫽ 4y ⫹ 2 7x ⫽ 3y

8. 2x ⫺ 5y ⫽ 7 3x ⫺ 10y ⫽ 13 12. ⫺5x ⫽ 6y ⫺ 4 5y ⫽ 1 ⫺ 3x 16. ⫺4y ⫺ 3 ⫽ 2x ⫺ 3 5y ⫽ 3x

Concept 2: Solving Inconsistent Systems and Dependent Systems For Exercises 17–24, solve the systems. (See Examples 4–5.) 17.

3x ⫺ 2y ⫽ 1 ⫺6x ⫹ 4y ⫽ ⫺2

21. 12x ⫺ 4y ⫽ 2 6x ⫽ 1 ⫹ 2y

18. 3x ⫺ y ⫽ 4

19. 6y ⫽ 14 ⫺ 4x

6x ⫺ 2y ⫽ 8 22. 10x ⫺ 15y ⫽ 5 3y ⫽ 2x ⫺ 1

23.

20. 2x ⫽ 4 ⫺ y

2x ⫽ ⫺3y ⫺ 7

⫺y ⫽ 2x ⫺ 2

1 7 x⫹ y⫽ 2 6

24. 0.2x ⫺ 0.1y ⫽ ⫺1.2

x ⫹ 2y ⫽ 4.5

x⫺

1 y⫽ 3 2

Mixed Exercises 25. Describe a situation in which you would prefer to use the substitution method over the addition method. 26. If you used the addition method to solve the given system, would it be easier to eliminate the x or y variable? Explain. 3x ⫺ 5y ⫽ 4 7x ⫹ 10y ⫽ 31

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For Exercises 27–52, solve by using either the addition method or the substitution method. 27. 2x ⫺ 4y ⫽ 8

28. 8x ⫹ 6y ⫽ ⫺8

y ⫽ 2x ⫹ 1

x ⫽ 6y ⫺ 10

30. 0.1x ⫹ 0.5y ⫽ 0.7

31. 0.2x ⫺ 0.1y ⫽ 0.8

0.2x ⫹ 0.7y ⫽ 0.8

0.1x ⫺ 0.1y ⫽ 0.4

33. 4x ⫺ 6y ⫽ 5

34. 3x ⫹ 6y ⫽ 7

2x ⫺ 3y ⫽ 7

2x ⫹ 4y ⫽ 5

36.

1 1 x⫹ y⫽7 3 5

37.

1 2 x ⫺ y ⫽ ⫺4 6 5 39.

⫺71x ⫺ y2 ⫽ 16 ⫹ 3y

41x ⫹ 2y2 ⫽ 50 ⫹ 3y

1 45. 3x ⫺ 2 ⫽ 111 ⫹ 5y2 3 2 x ⫹ 12y ⫺ 32 ⫽ ⫺2 3 1 1 8 x⫺ y⫽⫺ 10 2 5 1 11 x⫹ y⫽⫺ 4 2 51. 4x ⫽ 3y 4 y⫽ x⫹2 3

4x ⫺ 7y ⫽ ⫺16 1 32. y ⫽ x ⫺ 3 2 4x ⫹ y ⫽ ⫺3 35.

43.

0.04x ⫽ ⫺0.05y ⫹ 1.7

38.

1 x ⫹ y ⫽ 17 ⫹ y2 5

49. 4x ⫹ y ⫽ ⫺2 5x ⫺ y ⫽ ⫺7

2 2 x⫺ y⫽0 5 3 3 y⫽ x 5

41. ⫺4y ⫽ 10 4x ⫹ 3 ⫽ 1 44. ⫺0.01x ⫽ ⫺0.06y ⫹ 3.2

⫺0.03y ⫽ ⫺2.4 ⫹ 0.07x 46. 212y ⫹ 32 ⫺ 2x ⫽ 1 ⫺ x

1 1 x ⫺ y ⫽ ⫺2 4 6 1 1 ⫺ x⫹ y⫽4 6 5

3 x⫽ y 2 40. ⫺31x ⫹ y2 ⫽ 10 ⫺ 4y

3y ⫹ 2 ⫽ 1

48.

1 1 x⫺ y⫽0 3 2

21x ⫹ 2y2 ⫽ 20 ⫺ y

42. ⫺9x ⫽ 15

29. 2x ⫹ 5y ⫽ 9

0.08y ⫽ 0.03x ⫹ 4.6 47.

1 1 11 x⫹ y⫽ 4 2 4 2 1 7 x⫹ y⫽ 3 3 3

50. 4y ⫽ 8x ⫹ 20 8x ⫽ 24

52. 4x ⫺ 2y ⫽ 6 1 3 x⫽ y⫹ 2 2

Expanding Your Skills Sometimes the solution to a system of equations is an ordered pair containing fractions. In such a case, it is often easier to solve for each coordinate separately using the addition method. That is, use the addition method to solve for x. Then, rather than substituting x back into one of the original equations, repeat the addition method and solve for y. For Exercises 53–55, solve each system. 53. 9x ⫹ 11y ⫽ 47 ⫺5x ⫹ 3y ⫽ 23

54. ⫺6x ⫹ 7y ⫽ ⫺4 4x ⫺ 9y ⫽ 31

55. 4x ⫺ 10y ⫽ 19 5x ⫹ 12y ⫽ ⫺41

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Problem Recognition Exercises Solving Systems of Linear Equations For Exercises 1–4, solve each system by using three different methods. a. Use the graphing method. b. Use the substitution method. c. Use the addition method. 1. ⫺3x ⫹ y ⫽ ⫺2 4x ⫺ y ⫽ 4

2. 3x ⫺ 2y ⫽ 4 2 x⫽ y⫺2 3

3. 5x ⫽ 2y

4. 2y ⫽ 3x ⫹ 1 4x ⫽ 4

5 y⫽ x⫹1 2

For Exercises 5–8, use the systems of equations shown. Use each choice only once. a. y ⫽ ⫺4x ⫺ 9 8x ⫹ 3y ⫽ ⫺29

b. 5x ⫺ 2y ⫽ ⫺17 x ⫹ 5y ⫽ 2

c. 5x ⫺ 3y ⫽ 2 7x ⫹ 4y ⫽ ⫺30

d.

1 2 3 x⫺ y⫽⫺ 10 5 5 1 13 3 x⫹ y⫽ 4 3 6

5. For which system would you clear fractions? Solve the system. 6. Which system would be solved most efficiently by using the substitution method? Solve the system. 7. Which system would be solved most efficiently by using the addition method? Solve the system. 8. Which system would be solved efficiently using either the addition or the substitution method? Solve the system.

Section 3.4

Applications of Systems of Linear Equations in Two Variables

Concepts

1. Applications Involving Cost

1. Applications Involving Cost 2. Applications Involving Mixtures 3. Applications Involving Principal and Interest 4. Applications Involving Uniform Motion 5. Applications Involving Geometry

In Chapter 1 we solved numerous application problems using equations that contained one variable. However, when an application has more than one unknown, sometimes it is more convenient to use multiple variables. In this section, we will solve applications containing two unknowns. When two variables are present, the goal is to set up a system of two independent equations.

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Solving a Cost Application

Example 1

At an amusement park, five hot dogs and one drink cost $16. Two hot dogs and three drinks cost $9. Find the cost per hot dog and the cost per drink.

Solution: Let h represent the cost per hot dog.

Label the variables.

Let d represent the cost per drink. a

cost of 1 Cost of 5 b ⫹ a drink b ⫽ $16 hot dogs

5h ⫹ d ⫽ 16

a

cost of 3 Cost of 2 b ⫹ a drinks b ⫽ $9 hot dogs

2h ⫹ 3d ⫽ 9

Write two equations.

This system can be solved by either the substitution method or the addition method. We will solve by using the substitution method. The d variable in the first equation is the easiest variable to isolate. 5h ⫹ d ⫽ 16

d ⫽ ⫺5h ⫹ 16

Solve for d in the first equation.

2h ⫹ 3d ⫽ 9 2h ⫹ 31⫺5h ⫹ 162 ⫽ 9 2h ⫺ 15h ⫹ 48 ⫽ 9 ⫺13h ⫹ 48 ⫽ 9

Substitute the quantity ⫺5h ⫹ 16 for d in the second equation. Clear parentheses. Solve for h.

⫺13h ⫽ ⫺39 h⫽3 d ⫽ ⫺5132 ⫹ 16

d⫽1

Substitute h ⫽ 3 in the equation d ⫽ ⫺5h ⫹ 16.

Because h ⫽ 3, the cost per hot dog is $3.00. Because d ⫽ 1, the cost per drink is $1.00. Skill Practice 1. At the movie theater, Tom spent $15.50 on 3 soft drinks and 2 boxes of popcorn. Carly bought 5 soft drinks and 1 box of popcorn for a total of $16.50. Use a system of equations to find the cost of a soft drink and the cost of a box of popcorn.

TIP: A word problem can be checked by verifying that the solution meets the conditions specified in the problem. 5 hot dogs ⫹ 1 drink ⫽ 5($3.00) ⫹ 1($1.00) ⫽ $16.00 ✔ 2 hot dogs ⫹ 3 drinks ⫽ 2($3.00) ⫹ 3($1.00) ⫽ $9.00 ✔

Answer 1. Soft drinks cost $2.50 and popcorn costs $4.00.

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2. Applications Involving Mixtures Solving an Application Involving Chemistry

Example 2

One brand of cleaner used to etch concrete is 25% acid. A stronger industrialstrength cleaner is 50% acid. How many gallons of each cleaner should be mixed to produce 20 gal of a 40% acid solution?

Solution: Let x represent the amount of 25% acid cleaner. Let y represent the amount of 50% acid cleaner.

Avoiding Mistakes Do not forget to write the percent as a decimal.

25% Acid

50% Acid

40% Acid

Number of gallons of solution

x

y

20

Number of gallons of pure acid

0.25x

0.50y

0.40(20), or 8

From the first row of the table, we have a

total amount amount of Amount of b ⫹ a 50% solution b ⫽ a of solution b 25% solution

x ⫹ y ⫽ 20

From the second row of the table we have Amount of amount of amount of ° pure acid in ¢ ⫹ ° pure acid in ¢ ⫽ ° pure acid in ¢ 25% solution 50% solution resulting solution x⫹

y ⫽ 20

x⫹

0.25x ⫹ 0.50y ⫽ 8

x⫹ 25x ⫹

y ⫽ 20 50y ⫽ 800

y ⫽ 20

25x ⫹ 50y ⫽ 800 Multiply by ⫺25.

⫺25x ⫺ 25y ⫽ ⫺500 25x ⫹ 50y ⫽ 800 25y ⫽ 300 y ⫽ 12

x ⫹ y ⫽ 20

x ⫹ 1122 ⫽ 20 x⫽8

0.25x ⫹ 0.50y ⫽ 8

Multiply by 100 to clear decimals. Create opposite coefficients of x. Add the equations to eliminate x. Substitute y ⫽ 12 back into one of the original equations.

Therefore, 8 gal of 25% acid solution must be added to 12 gal of 50% acid solution to create 20 gal of a 40% acid solution. Skill Practice 2. A pharmacist needs 8 ounces (oz) of a solution that is 50% saline. How many ounces of 60% saline solution and 20% saline solution must be mixed to obtain the mixture needed?

Answer 2. The pharmacist should mix 6 oz of 60% solution and 2 oz of 20% solution.

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3. Applications Involving Principal and Interest Example 3

Solving an Application Involving Finance

Serena invested money in two accounts: a savings account that yields 4.5% simple interest and a certificate of deposit (CD) that yields 7% simple interest. The amount invested at 7% was twice the amount invested at 4.5%. How much did Serena invest in each account if the total interest at the end of 1 yr was $1017.50?

Solution: Let x represent the amount invested in the savings account (the 4.5% account). Let y represent the amount invested in the certificate of deposit (the 7% account). 4.5% Account

7% Account

Principal

x

y

Interest

0.045x

0.07y

Total 1017.50

Because the amount invested at 7% was twice the amount invested at 4.5%, we have Amount amount ° invested ¢ ⫽ 2 ° invested ¢ at 7% at 4.5%

y ⫽ 2x

From the second row of the table, we have Interest interest total ° earned from ¢ ⫹ ° earned from ¢ ⫽ a interest b 4.5% account 7% account

0.045x ⫹ 0.07y ⫽ 1017.50

y ⫽ 2x 45x ⫹ 70y ⫽ 1,017,500

Multiply by 1000 to clear decimals. Because the y variable in the first equation is isolated, we will use the substitution method.

45x ⫹ 7012x2 ⫽ 1,017,500 45x ⫹ 140x ⫽ 1,017,500

Substitute the quantity 2x into the second equation. Solve for x.

185x ⫽ 1,017,500 x⫽

1,017,500 185

x ⫽ 5500 y ⫽ 2x y ⫽ 2155002

Substitute x ⫽ 5500 into the equation y ⫽ 2x to solve for y.

y ⫽ 11,000 Because x ⫽ 5500, the amount invested in the savings account is $5500. Because y ⫽ 11,000, the amount invested in the CD is $11,000. Skill Practice 3. Seth invested money in two accounts, one paying 5% interest and the other paying 6% interest. The amount invested at 6% was $1000 more than the amount invested at 5%. He earned a total of $830 interest in 1 yr. Use a system of equations to find the amount invested in each account.

Answer 3. Seth invested $7000 at 5% and $8000 at 6%.

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TIP: To check Example 3, note that $11,000 is twice $5500. Furthermore, Interest interest ° earned from ¢ ⫹ ° earned from ¢ ⫽ $550010.0452 ⫹ $11,00010.072 ⫽ 1017.50 ✔ 4.5% account 7% account

4. Applications Involving Uniform Motion Example 4

Solving a Distance, Rate, and Time Application

A plane flies 660 mi from Atlanta to Miami in 1.2 hr when traveling with a tailwind. The return flight against the same wind takes 1.5 hr. Find the speed of the plane in still air and the speed of the wind.

Solution: Let p represent the speed of the plane in still air. Let w represent the speed of the wind. The speed of the plane with the wind: (Plane’s still airspeed) ⫹ (wind speed)

p⫹w

The speed of the plane against the wind: (Plane’s still airspeed) ⫺ (wind speed)

p⫺w

Set up a chart to organize the given information: Distance

Rate

Time

With a tailwind

660

p⫹w

1.2

Against the wind

660

p⫺w

1.5

Two equations can be found by using the relationship d ⫽ rt.

TIP: In Section 3.3 we used the multiplication property of equality to create opposite coefficients. Example 4 demonstrates that we can also use the division property of equality to create opposite coefficients.

°

Distance speed time with ¢ ⫽ ° with ¢ ° with ¢ wind wind wind

Distance speed time ° against ¢ ⫽ ° against ¢ ° against ¢ wind wind wind 660 ⫽ 1p ⫹ w211.22

660 ⫽ 1p ⫺ w211.52 660 ⫽ 1p ⫹ w211.22 660 ⫽ 1p ⫺ w211.52

660 ⫽ 1p ⫹ w211.22 660 ⫽ 1p ⫺ w211.52

Notice that the first equation may be divided by 1.2 and still leave integer coefficients. Similarly, the second equation may be simplified by dividing by 1.5. Divide by 1. 2.

Divide by 1.5.

1p ⫹ w21.2 660 ⫽ 1.2 1.2

1p ⫺ w21.5 660 ⫽ 1.5 1.5

550 ⫽ p ⫹ w 440 ⫽ p ⫺ w

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261

550 ⫽ p ⫹ w 440 ⫽ p ⫺ w 990 ⫽ 2p

Add the equations.

p ⫽ 495

550 ⫽ 14952 ⫹ w

Substitute p ⫽ 495 into the equation 550 ⫽ p ⫹ w.

55 ⫽ w

Solve for w.

The speed of the plane in still air is 495 mph, and the speed of the wind is 55 mph. Skill Practice 4. A plane flies 1200 mi from Orlando to New York in 2 hr with a tailwind. The return flight against the same wind takes 2.5 hr. Find the speed of the plane in still air and the speed of the wind.

5. Applications Involving Geometry Example 5

Solving a Geometry Application

The sum of the two acute angles in a right triangle is 90⬚. The measure of one angle is 6⬚ less than 2 times the measure of the other angle. Find the measure of each angle.

Solution:

x

Let x represent one acute angle. Let y represent the other acute angle.

y

The sum of the two acute angles is 90⬚

x ⫹ y ⫽ 90

One angle is 6⬚ less than 2 times the other angle

x ⫽ 2y ⫺ 6

¶

x ⫹ y ⫽ 90 x ⫽ 2y ⫺ 6 12y ⫺ 62 ⫹ y ⫽ 90

Because one variable is already isolated, we will use the substitution method. Substitute x ⫽ 2y ⫺ 6 into the first equation.

3y ⫺ 6 ⫽ 90 3y ⫽ 96 y ⫽ 32 x ⫽ 2y ⫺ 6 x ⫽ 21322 ⫺ 6

To find x, substitute y ⫽ 32 into the equation x ⫽ 2y ⫺ 6.

x ⫽ 64 ⫺ 6 x ⫽ 58 The two acute angles in the triangle measure 32⬚ and 58⬚. Skill Practice 5. Two angles are supplementary. The measure of one angle is 16° less than 3 times the measure of the other. Use a system of equations to find the measures of the angles.

Answers 4. The speed of the plane is 540 mph, and the speed of the wind is 60 mph. 5. The angles are 49° and 131°.

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Section 3.4 Practice Exercises • Practice Problems • Self-Tests • NetTutor

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Review Exercises 1. State three methods that can be used to solve a system of linear equations in two variables. For Exercises 2– 4, state which method you would prefer to use to solve the system. Then solve the system. 2. y ⫽ 9 ⫺ 2x

3. 7x ⫺ y ⫽ ⫺25

3x ⫺ y ⫽ 16

2x ⫹ 5y ⫽ 14

4.

5x ⫹ 2y ⫽ 6 ⫺2x ⫺ y ⫽ 3

Concept 1: Applications Involving Cost 5. A 1200-seat theater sells two types of tickets for a concert. Premium seats sell for $30 each and regular seats sell for $20 each. At one event $30,180 was collected in ticket sales with 10 seats left unsold. How many of each type of ticket was sold? (See Example 1.)

6. John and Ariana bought school supplies. John spent $10.65 on 4 notebooks and 5 pens. Ariana spent $7.50 on 3 notebooks and 3 pens. What is the cost of 1 notebook and what is the cost of 1 pen?

7. Mickey bought lunch for his fellow office workers on Monday. He spent $7.35 on 3 hamburgers and 2 fish sandwiches. Chloe bought lunch on Tuesday and spent $7.15 for 4 hamburgers and 1 fish sandwich. What is the price of 1 hamburger, and what is the price of 1 fish sandwich?

8. A group of four golfers paid $150 to play a round of golf. Of the golfers, one is a member of the club and three are nonmembers. Another group of golfers consists of two members and one nonmember. They paid a total of $75. What is the cost for a member to play a round of golf, and what is the cost for a nonmember?

9. Meesha has a pocket full of change consisting of dimes and quarters. The total value is $3.15. There are 7 more quarters than dimes. How many of each coin are there?

10. Jessica has several dimes and quarters in her purse, totaling $2.70. The number of dimes is one less than the number of quarters. How many of each coin are there?

Concept 2: Applications Involving Mixtures 11. A jar of one face cream contains 18% moisturizer, and another type contains 24% moisturizer. How many ounces of each should be combined to get 12 oz of a cream that is 22% moisturizer? (See Example 2.)

12. Logan wants to mix an 18% acid solution with a 45% acid solution to get 16 L of a 36% acid solution. How many liters of the 18% solution and how many liters of the 45% solution should be mixed?

13. How much fertilizer containing 8% nitrogen should be mixed with a fertilizer containing 12% nitrogen to get 8 L of a fertilizer containing 11% nitrogen?

14. How much 30% acid solution should be added to 10% acid solution to make 100 mL of a 12% acid solution?

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15. How much pure bleach should Tim combine with a solution that is 4% bleach to make 12 oz of a 12% bleach solution?

263

16. A fruit punch that contains 25% fruit juice is combined with 100% fruit juice. How many ounces of each should be used to make 48 oz of a mixture that is 75% fruit juice?

Concept 3: Applications Involving Principal and Interest 17. Mr. Coté invested 3 times as much money in a stock fund that returned 8% interest after 1 yr as he did in a bond fund that earned 5% interest. If his total earnings came to $435 after 1 yr, how much did he invest in each fund? (See Example 3.)

18. Aliya deposited half as much money in a savings account earning 2.5% simple interest as she invested in a money market account that earns 3.5% simple interest. If the total interest after 1 yr is $247, how much did she invest in each account?

19. A credit union offers 5.5% simple interest on a certificate of deposit (CD) and 3.5% simple interest on a savings account. If Mr. Levy invested $200 more in the CD than in the savings account and the total interest after the first year was $245, how much was invested in each account?

20. Jody invested $5000 less in an account paying 3% simple interest than she did in an account paying 4% simple interest. At the end of the first year, the total interest from both accounts was $725. Find the amount invested in each account.

21. Alina borrowed a total of $15,000 from two banks to buy a new boat. Because of her excellent credit, one bank charged only 6% simple interest and the other charged 7% simple interest. At the end of 5 yr, the total amount of money she paid in interest was $4750. How much did she borrow from each bank?

22. Didi plans to take a trip to the Galapagos Islands in 4 yr and knows that she needs approximately $3500 for the trip. She invests a total of $15,500 in two funds. One fund has a 6% return, and the other has a 5% return. How much should she invest in each fund so that she earns $3500 after 4 yr?

Concept 4: Applications Involving Uniform Motion 23. It takes a boat 2 hr to travel 16 mi downstream with the current and 4 hr to return against the current. Find the speed of the boat in still water and the speed of the current. (See Example 4.)

24. A plane flew 720 mi in 3 hr with the wind. It would take 4 hr to travel the same distance against the wind. What is the speed of the plane in still air and the speed of the wind?

25. A plane flies from Atlanta to Los Angeles against the wind in 5 hr. The return trip back to Atlanta with the wind takes only 4 hr. If the distance between Atlanta and Los Angeles is 3200 km, find the speed of the plane in still air and the speed of the wind.

26. The Gulf Stream is a warm ocean current that extends from the eastern side of the Gulf of Mexico up through the Florida Straits and along the southeastern coast of the United States to Cape Hatteras, North Carolina. A boat travels with the current 100 mi from Miami, Florida, to Freeport, Bahamas, in 2.5 hr. The return trip against the same current takes 313 hr. Find the speed of the boat in still water and the speed of the current.

27. A moving sidewalk in the Atlanta airport moves people between gates. It takes Molly’s 8-year-old son Stephen 20 sec to travel 100 ft walking with the sidewalk. It takes him 30 sec to travel 60 ft walking against the moving sidewalk (in the opposite direction). Find the speed of the moving sidewalk and Stephen’s walking speed on nonmoving ground.

28. Kim rides a total of 48 km in the bicycle portion of a triathlon. The course is an “out and back” route. It takes her 3 hr on the way out against the wind. The ride back takes her 2 hr with the wind. Find the speed of the wind and Kim’s speed riding her bike in still air.

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Concept 5: Applications Involving Geometry For Exercises 29–34, solve the applications involving geometry. If necessary, refer to the geometry formulas listed in the inside back cover of the text. 29. In a right triangle, one acute angle measures 6⬚ more than 3 times the other. If the sum of the measures of the two acute angles must equal 90⬚, find the measures of the acute angles. (See

30. An isosceles triangle has two angles of the same measure. If the angle represented by y measures 3⬚ less than the angle x, find the measures of all angles of the triangle.

Example 5.) y⬚

y⬚ x⬚

x⬚

x⬚

31. Two angles are supplementary. One angle measures 2⬚ less than 3 times the other. What are the measures of the two angles?

32. The measure of one angle is 5 times the measure of another. If the two angles are supplementary, find the measures of the angles.

33. One angle measures 6⬚ more than twice another. If the two angles are complementary, find the measures of the angles.

34. Two angles are complementary. One angle measures 15° more than 2 times the measure of the other. What are the measures of the two angles?

Mixed Exercises 35. How much pure gold (24K) must be mixed with 60% gold to get 20 grams(g) of 75% gold? 36. Connie is the head of maintenance at a large hospital. She received news of a new state mandate indicating that the minimum strength for disinfectant was to be 17%, up from the old of requirement of 15%. Connie had plenty of barrels of 15% disinfectant left over, and also lots of the strong 55% disinfectant used in rooms for patients with highly contagious diseases. How many gallons of each disinfectant should be mixed to get 50 gal of 17% disinfectant? 37. A rowing team trains on the Halifax River. It can row upstream 10 mi against the current in 2.5 hr and 16 mi downstream with the current in the same amount of time. Find the speed of the boat in still water and the speed of the current.

38. In her kayak, Bonnie can travel 31.5 mi downstream with the current in 7 hr. The return trip against the current takes 9 hr. Find the speed of the kayak in still water and the speed of the current.

39. There are two types of tickets sold at the Canadian Formula One Grand Prix race. The price of 6 grandstand tickets and 2 general admissions tickets costs $2330. The price of 4 grandstand tickets and 4 general admission tickets cost $2020. What is the price of each type of ticket?

40. A basketball player scored 19 points by shooting two-point and three-point baskets. If she made a total of eight baskets, how many of each type did she make?

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265

41. A bank offers two accounts, a money market account at 2% simple interest and a regular savings account at 1.3% interest. If Svetlana deposits $3000 between the two accounts and receives $51.25 in total interest in the first year, how much did she invest in each account? 42. Angelo invested $8000 in two accounts: one that pays 3% and one that pays 1.8%. At the end of the first year, his total interest earned was $222. How much did he deposit in the account that pays 3%? 43. The perimeter of a rectangle is 42 m. The length is 1 m longer than the width. Find the dimensions of the rectangle. 44. In a right triangle, the measure of one acute angle is one-fourth the measure of the other. Find the measures of the acute angles. 45. A coin collection consists of 50¢ pieces and $1 coins. If there are 21 coins worth $15.50, how many 50¢ pieces and $1 coins are there? 46. Jacob has a piggy bank consisting of nickels and dimes. If there are 30 coins worth $1.90, how many nickels and dimes are in the bank? 47. One phone company charges $0.15 per minute for long-distance calls. A second company charges only $0.10 per minute for long-distance calls, but adds a monthly fee of $4.95. a. Write a linear function representing the cost for the first company for x minutes. b. Write a linear function representing the cost for the second company for x minutes. c. Find the number of minutes of long-distance calling for which the total bill from either company would be the same. 48. A rental car company rents a compact car for $20 a day, plus $0.25 per mile. A midsize car rents for $30 a day, plus $0.20 per mile. a. Write a linear function representing the cost to rent the compact car for x miles. b. Write a linear function representing the cost to rent a midsize car for x miles. c. Find the number of miles at which the cost to rent either car would be the same.

Linear Inequalities and Systems of Linear Inequalities in Two Variables

Section 3.5

1. Graphing Linear Inequalities in Two Variables

Concepts

A linear inequality in two variables x and y is an inequality that can be written in one of the following forms: ax ⫹ by 6 c, ax ⫹ by 7 c, ax ⫹ by ⱕ c, or ax ⫹ by ⱖ c, provided a and b are not both zero. A solution to a linear inequality in two variables is an ordered pair that makes the inequality true. For example, solutions to the inequality x ⫹ y 6 6 are ordered pairs (x, y) such that the sum of the x- and y-coordinates is less than 6. This inequality has an infinite number of solutions, and therefore it is convenient to express the solution set as a graph. To graph a linear inequality in two variables, we will follow these steps.

1. Graphing Linear Inequalities in Two Variables 2. Compound Linear Inequalities in Two Variables 3. Graphing a Feasible Region

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PROCEDURE Graphing a Linear Inequality in Two Variables Step 1 Solve for y, if possible. Step 2 Graph the related equation. Draw a dashed line if the inequality is strict, ⬍ or ⬎. Otherwise, draw a solid line. Step 3 Shade above or below the line as follows: • Shade above the line if the inequality is of the form y 7 ax ⫹ b or y ⱖ ax ⫹ b. • Shade below the line if the inequality is of the form y 6 ax ⫹ b or y ⱕ ax ⫹ b. Note: A dashed line indicates that the line is not included in the solution set. A solid line indicates that the line is included in the solution set. This process is demonstrated in Example 1.

Graphing a Linear Inequality in Two Variables

Example 1

Graph the solution set.

⫺3x ⫹ y ⱕ 1

Solution:

y

⫺3x ⫹ y ⱕ 1 y ⱕ 3x ⫹ 1

5

Solve for y.

Next graph the line defined by the related equation y ⫽ 3x ⫹ 1. Because the inequality is of the form y ⱕ ax ⫹ b, the solution to the inequality is the region below the line y ⫽ 3x ⫹ 1. See Figure 3-9.

4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

5

x

⫺4 ⫺5

Figure 3-9

Skill Practice Graph the solution set. 1. 2x ⫹ y ⱖ ⫺4 y 5 4 3 2

Test point 1 (0, 0) ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

5

x

⫺4 ⫺5

Figure 3-10

After graphing the solution to a linear inequality, we can verify that we have shaded the correct side of the line by using test points. In Example 1, we can pick an arbitrary ordered pair within the shaded region. Then substitute the x- and y-coordinates in the original inequality. If the result is a true statement, then that ordered pair is a solution to the inequality and suggests that other points from the same region are also solutions. For example, the point (0, 0) lies within the shaded region (Figure 3-10). ⫺3x ⫹ y ⱕ 1

Answer

⫺3102 ⫹ 102 ⱕ 1

Substitute (0, 0) in the original inequality.

?

y

1.

?

5

0 ⫹ 0 ⱕ 1 ✔ True

4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1

2 3 4

5

x

The point (0, 0) from the shaded region is a solution.

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In Example 2, we will graph the solution set to a strict inequality. A strict inequality uses the symbol ⬍ or ⬎. In such a case, the boundary line will be drawn as a dashed line. This indicates that the boundary itself is not part of the solution set.

Graphing a Linear Inequality in Two Variables

Example 2

⫺4y 6 5x

Graph the solution set.

Solution: ⫺4y 6 5x ⫺4y 5x 7 ⫺4 ⫺4

Solve for y. Because we divide both sides by a negative number, reverse the inequality sign.

5 y 7 ⫺ x 4 y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

⫺4 ⫺5

5

x

Graph the line defined by the related equation, y ⫽ ⫺54x. The boundary line is drawn as a dashed line because the inequality is strict. Also note that the line passes through the origin. Because the inequality is of the form y 7 ax ⫹ b, the solution to the inequality is the region above the line. See Figure 3-11.

Figure 3-11

Skill Practice Graph the solution set. 2. ⫺3y 6 x

In Example 2, we cannot use the origin as a test point, because the point (0, 0) is on the boundary line. Be sure to select a test point strictly within the shaded region. In this case, we choose (2, 1). See Figure 3-12. ⫺4y 6 5x ?

⫺4112 6 5122 ?

⫺4 6 10 ✔ True

Substitute (2, 1) in the original inequality. The point (2, 1) from the shaded region is a solution to the original inequality. y 5 4 3 Test point 2 (2, 1) 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

Answer 2. 5

x

⫺4 ⫺5

Figure 3-12

In Example 3, we encounter a situation in which we cannot solve for the y variable.

y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1

2 3 4

5

x

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Graphing a Linear Inequality

Example 3

4x ⱖ ⫺12

Graph the solution set. y

Solution:

5 4 3

4x ⱖ ⫺12

Test point (0, 0)

2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

In this inequality, there is no y variable. However, we can simplify the inequality by solving for x.

x ⱖ ⫺3 5

x

Graph the related equation x ⫽ ⫺3. This is a vertical line. The boundary is drawn as a solid line, because the inequality is not strict, ⱖ. To shade the appropriate region, refer to the inequality, x ⱖ ⫺3. The points for which x is greater than ⫺3 are to the right of x ⫽ ⫺3. Therefore, shade the region to the right of the line (Figure 3-13).

⫺4 ⫺5

Figure 3-13

Selecting a test point such as (0, 0) from the shaded region indicates that we have shaded the correct side of the line. 4x ⱖ ⫺12

Substitute x ⫽ 0.

?

4102 ⱖ ⫺12 ✔

True

Skill Practice Graph the solution set. 3. ⫺2x ⱖ 2

2. Compound Linear Inequalities in Two Variables Some applications require us to find the union or intersection of the solution sets of two or more linear inequalities.

Graphing a Compound Linear Inequality

Example 4

Graph the solution set of the compound inequality. y 7 12x ⫹ 1

and

x⫹y 6 1

Solution: Solve each inequality for y. First inequality

Second inequality

y 7 12x ⫹ 1

x⫹y 6 1 y 6 ⫺x ⫹ 1

The inequality is of the form y 7 ax ⫹ b. Shade above the boundary line. (See Figure 3-14.)

Answer

y

y

3.

The inequality is of the form y 6 ax ⫹ b. Shade below the boundary line. (See Figure 3-15.) y

5

5

5

4 3

4 3

4 3

2 1

2 1

2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1

2 3 4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

⫺4 ⫺5

Figure 3-14

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

⫺4 ⫺5

Figure 3-15

5

x

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The intersection is the solution set to the system of inequalities. See Figure 3-17.

The region bounded by the inequalities is the region above the line y  12x  1 and below the line y  x  1. This is the intersection or “overlap” of the two regions (shown in purple in Figure 3-16).

y

y 5

5 4 3

y  12 x  1

4 3 2 1

2 1 5 4 3 2 1 1 2 3

269

Linear Inequalities and Systems of Linear Inequalities in Two Variables

1

2 3 4

5

x

5 4 3 2 1 1 2 3

y  x  1

1

2 3 4

5

x

4 5

4 5

Figure 3-17

Figure 3-16

Skill Practice Graph the solution set. 4. x  3y 7 3 and y 6 2x  4

Example 5 demonstrates the union of the solution sets of two linear inequalities.

Graphing a Compound Linear Inequality

Example 5

Graph the solution set of the compound inequality. 3y  6

or

yx0

Solution: First inequality

Second inequality

3y  6

yx0

y2

yx

The graph of y  2 is the region on and below the horizontal line y  2. (See Figure 3-18.)

The inequality y  x is of the form y  ax  b. Graph a solid line and the region below the line. (See Figure 3-19.) y

y 5

5

4 3

4 3

2 1

2 1

5 4 3 2 1 1 2 3

1

2 3 4

4 5

Figure 3-18

5

x

5 4 3 2 1 1 2 3

1

2 3 4

4 5

Figure 3-19

5

x

Answer y

4. 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

1

2 3 4

5

x

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The solution to the compound inequality 3y ⱕ 6 or y ⫺ x ⱕ 0 is the union of these regions (Figure 3-20).

y 5 4 3

Skill Practice Graph the solution set.

2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

5

x

⫺4 ⫺5

5. 2y ⱕ 4

or

Example 6

yⱕx⫹1

Graphing a Compound Linear Inequality

Figure 3-20

Describe the region of the plane defined by the system of inequalities. xⱕ0

and

yⱖ0

Solution: xⱕ0 yⱖ0

y

x ⱕ 0 on the y-axis and in the second and third quadrants. y ⱖ 0 on the x-axis and in the first and second quadrants. The intersection of these regions is the set of points in the second quadrant (with the boundary included).

5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

5

x

⫺4 ⫺5

Skill Practice Graph the region defined by the system of inequalities. 6. x ⱕ 0

and

y ⱕ 0

3. Graphing a Feasible Region When two variables are related under certain constraints, a system of linear inequalities can be used to show a region of feasible values for the variables. The feasible region represents the ordered pairs that are true for each inequality in the system. Answers 5.

y

Example 7

5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1

2 3 4

5

x

Susan has two tests on Friday: one in chemistry and one in psychology. Because the two classes meet in consecutive hours, she has no study time between tests. Susan estimates that she has a maximum of 12 hr of study time before the tests, and she must divide her time between chemistry and psychology. Let x represent the number of hours Susan spends studying chemistry.

⫺4 ⫺5

6.

Graphing a Feasible Region

Let y represent the number of hours Susan spends studying psychology. y

a. Find a set of inequalities to describe the constraints on Susan’s study time.

5

b. Graph the constraints to find the feasible region defining Susan’s study time.

4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1

2 3 4

5

x

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Solution: a. Because Susan cannot study chemistry or psychology for a negative period of time, we have x  0 and y  0. Furthermore, her total time studying cannot exceed 12 hr: x  y  12. A system of inequalities that defines the constraints on Susan’s study time is: y y0 x  y  12 b. The first two conditions x  0 and y  0 represent the set of points in the first quadrant. The third condition x  y  12 represents the set of points below and including the line x  y  12 (Figure 3-21).

16 14

Hours (psychology)

x0

12 10

x  y  12

8 6 4 2

4 2 2 4

2 4 6

8 10 12 14 16

x

Hours (chemistry)

Figure 3-21

Discussion:

y 16

Hours (psychology)

1. Refer to the feasible region. Is the ordered pair (8, 5) part of the feasible region? No. The ordered pair (8, 5) indicates that Susan spent 8 hr studying chemistry and 5 hr studying psychology. This is a total of 13 hr, which exceeds the constraint that Susan only had 12 hr to study. The point (8, 5) lies outside the feasible region, above the line x  y  12 (Figure 3-22).

14 12 10 8

x  y  12

6

(8, 5)

4 2

4 2 2 4

(7, 3) 2 4 6

8 10 12 14 16

x

Hours (chemistry)

Figure 3-22 2. Is the ordered pair (7, 3) part of the feasible region? Yes. The ordered pair (7, 3) indicates that Susan spent 7 hr studying chemistry and 3 hr studying psychology. This point lies within the feasible region and satisfies all three constraints.

x0

70

True

y0

30

True

x  y  12

172  132  12

True

Notice that the ordered pair (7, 3) corresponds to a point where Susan is not making full use of the 12 hr of study time. y

1 This inequality may also be written as y  x. 2 Figure 3-23 shows the first quadrant with the 1 constraint y  x. 2

16

Hours (psychology)

3. Suppose there was one additional constraint imposed on Susan’s study time. She knows she needs to spend at least twice as much time studying chemistry as she does studying psychology. Because the time studying chemistry must be at least twice the time studying psychology, we have x  2y.

14 12 10 8

1

y  2x

6 4 2

4 2 2 4

2

4 6

8 10 12 14 16

Hours (chemistry)

Figure 3-23

x

271

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4. At what point in the feasible region is Susan making the most efficient use of her time for both classes? First and foremost, Susan must make use of all 12 hr. This occurs for points along the line x  y  12. Susan will also want to study for both classes with approximately twice as much time devoted to chemistry. Therefore, Susan will be deriving the maximum benefit at the point of intersection of the line 1 x  y  12 and the line y  x. 2 1 Using the substitution method, replace y  x into the equation 2 x  y  12. y

1 x  x  12 2

Hours (psychology)

2x  x  24

16

Clear fractions.

3x  24 x8

Solve for x.

y

To solve for y, substitute x  8 1 into the equation y  x. 2

182 2

y4

14 12

x + y = 12

10 8

1

y 2x (8, 4)

6 4 2

4 2 2 4

2

4 6

8 10 12 14 16

x

Hours (chemistry)

Therefore, Susan should spend 8 hr studying chemistry and 4 hr studying psychology. Answer

Skill Practice

7. a. x  0 and y  0 b. x  y  30 y c.

Number of dogs

35 30 25 20 15 10 5

105 5 10

x  y  30

5 10 15 20 25 30 35

x

7. A local pet rescue group has a total of 30 cages that can be used to hold cats and dogs. Let x represent the number of cages used for cats, and let y represent the number used for dogs. a. Write a set of inequalities to express the fact that the number of cat and dog cages cannot be negative. b. Write an inequality to describe the constraint on the total number of cages for cats and dogs. c. Graph the system of inequalities to find the feasible region describing the available cages.

Number of cats

Section 3.5 Practice Exercises Boost your GRADE at mathzone.com!

• Practice Problems • Self-Tests • NetTutor

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Study Skills Exercise 1. Define the key term linear inequality in two variables..

Review Exercises For Exercises 2–5, solve the inequalities. 2. 5 6 x  1

and 2x  6  6

4. 4  y 6 3y  12

or 21y  32  12

3. 5  x  4

and 6 7 3x  3

5. 2x 6 4

or 3x  1  13

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Concept 1: Graphing Linear Inequalities in Two Variables For Exercises 6–9, decide if the given point is a solution to the inequality. 6. 2x  y 7 8 a. 13, 52

7. 3y  x 6 5

b. 11, 102

c. 14, 22

a. 11, 72

d. 10, 02

8. y  2

a. 15, 32

c. 10, 02

d. 12, 32

9. x  5

b. 14, 22

c. 10, 02

b. 15, 02

d. 13, 22

a. 14, 52

b. 15, 12

c. (8, 8)

d. (0, 0)

10. When should you use a dashed line to graph the solution to a linear inequality? For Exercises 11–16, decide which inequality symbol should be used 16, 7, , 2 by looking at the graph. y

11.

12.

2 1 1

2 3

4

5

x

54 3 21 1 2 3 4 5

3 4 5

xy

y

2 y

14.

1 2 3 4 5

x

54 3 2 1 1 2 3

y

y

15.

1 2

4 5

x

5 4 3 2 1 1 2 3 4 5

x

3

0

x

4 y

16.

5 4 3 2 1

5 4 3 2 1

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

5

2x  3

5 4 3 2 1

x

y 5 4 3 2 1

5 4 3 2 1

4 3

5 4 3 21 1 2

13.

y

5

1 2 3 4 5

and

y

x

0

5 4 3 2 1 1 2 3 4 5

x

0

1 2 3 4 5

and

y

x

0

For Exercises 17–40, graph the solution set. (See Examples 1–3.) 17. x  2y 7 4

18. x  3y  6

4 5

y

y

y

3 2 1

4 3 2 1

5 4 3 2 1 5 4 3 2 1 1 2 3

19. 5x  2y 6 10

1 2

3 4

5

x

3 2 1 1

1 2

3 4

5 6 7

x

5 4 3 2 1 1 2

2 3

3

4 5

5

6

7

4 6

1 2 3 4

5

x

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Chapter 3 Systems of Linear Equations and Inequalities

20. x  3y 6 8

21. 2x  6y  12

y

22. 4x 6 3y  12 y

y

5 4 3 2 1

4 3 2 1

5 4 3 2 1

2 1 1 2

1 2 3 4

5 6 7

8

x

3 4

2 1 1 2 3

1

2

3 4

5 6 7 8

x

23. 2y  4x

25. y  2

y

y

y

5 4

5 4

3 2 1

3 2 1 1 2 3 4 5

x

26. y  5

5 4 3 2 1

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

x

1

2

3 4

5

2

2 29. y  x  4 5

2 3 4 5 6

2

3 4

5

x

1

2

3 4

5 6 7

x

5 4 3 2 1 1

2 3

2 3

4 5

4 5

5

1

2

3 4

5

1 2

3 4

5

x

y

5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

x

1 31. y  x  6 3

y

y

3 2 1 1

1

5 30. y   x  4 2

4 3 2 1

3 4

5 4 3 2 1

5 4 3 2 1 1

5 4 3 2 1 1

2

y

5 4 3 2 1

x

1

28. x  6 6 7 y

y

5 4 3 2 1 1 2 3 4 5

27. 4x 6 5 8 7 6 5 4 3 2 1

x

6

24. 6x 6 2y

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5 6

5

4 5

5

4 3 2 1 1 2 3 4

1

2

3 4

5

x

8 7 6 5 4 3 2 1 5 4 3 2 1 1 2

x

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1 32. y   x  2 4

33. y  5x 7 0

y

1 x 7 0 2 y 6 5 4 3 2 1

5 4 3 2 1 1

2

3 4

5 6 7 8

x

5 4 3 2 1 1 2 3

1

2

3 4

5

x

4 3 2 1 1

y x  6 1 5 4

36. x 

2

3 4

5 6

x

5 4 3 2 1 1

1

2

3 4

5

x

y

y

3 4

5

x

2 3 4 5

8

x

y

5 4 3 2 1 1 2

5 6 7

5 40. x   y 4

5 4 3 2 1 5 4 3 2 1 1

3 4

4

2 39. x  y 3

38. 0.3x  0.2y 6 0.6

1 2

2 3

4 5

4

x

6 5 4 3 2 1 2 1 1

2 3

2 3

5 6

y

5 4 3 2 1

1

3 4

37. 0.1x  0.2y  0.6 y

4 3 2 1 1

2

4

y 2 2

y 6 5 4 3 2 1

1

2 3

4 5

4 5

35.

34. y 

y

5 4 3 2 1 2 1 1 2 3

275

Linear Inequalities and Systems of Linear Inequalities in Two Variables

5 4 3 2 1

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

x

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

x

Concept 2: Compound Linear Inequalities in Two Variables For Exercises 41–55, graph the solution set of each compound inequality. (See Examples 4–6.) y 7 x  2

41. y 6 4 and y

2 3

x  2y 6 6

2

3 4

5 6

x

4 3 2 1 1 2 3 4

x3

6 5 4 3 2 1

6 5 4 3 2 1

1

43. 2x  y  5 or y

y

7 6 5 4 3 2 1 4 3 2 1 1

42. y 6 3 and

1 2

3 4

5 6

x

4 3 2 1 1 2 3 4

1 2 3 4 5 6

x

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44. x  3y  3 or

x  2

45. x  y 6 3 and 4x  y 6 6

1 2

3 4

5

x

2x  3y  6

1

3 4

5

x

4 3 2 1 1

51. x  2

y 7 4

xy3

or

49. x 7 4

1

1

2

3 4

5 6

x

2 1 1

4 5

2 3

6

4

52. x  0

y0

or

5 4 3 2 1 1 3 4

5 6 7

x

2 3

53. x 7 0 and

xy 6 6

y

54. x 6 0

1

2

3 4

2

3 4

5 6 7

x

5 6 7 8

y  3

or

1 2

3 4

5

x

7 6 5 4 3 2 1 1 2 3

2 3

4 5

4 5

6 7

xy 6 2

and

5 4 3 2 1 1 2 3 4 5

x

y

55. y  0

1

2

3

x

x  y  4

or y

y

1

x

3 2 1

5 4 3 2 1

5 4 3 2 1

7 6 5 4 3 2 1

5

y 6 2

and

y

2

3 4

6 5 4 3 2 1

2 3

5 4 3 2 1

1

2

y

y 7 6 5 4 3 2 1

2 3

4 3 2 1

x

y

1 2

and

3 2 1 1

5 6

4 3 2 1

4 5

3 2 1 1

3 4

54 3 2 1 1

48. 3x  2y  4

2 3

50. x 6 3

1 2

4

y

1

8 7 6 5

2 3

5 4 3 2 1 2

9

4 3 2 1 1

47. 2x  y  2 or

3

y

6 5 4 3 2 1

4 5

5 4

and 3x  y 6 9

y

y 5 4 3 2 1 5 4 3 2 1 1 2 3

46. x  y 6 4

1

2

3 4

5

x

5 4 3 2 1 1 2 3 4 5

1 2

3 4

5

x

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277

Linear Inequalities and Systems of Linear Inequalities in Two Variables

Concept 3: Graphing a Feasible Region For Exercises 56–59, graph the feasible regions. 56. x ⫹ y ⱕ 3 and

57. x ⫺ y ⱕ 2 and

x ⱖ 0, y ⱖ 0

x ⱖ 0, y ⱖ 0

y

y

5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

58. x ⱖ 0, y ⱖ 0

3 4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

x ⫹ y ⱕ 8 and

x ⫹ y ⱕ 5 and

3x ⫹ 5y ⱕ 30

x ⫹ 2y ⱕ 6 y

y

5 4 3 2 1 1 2

59. x ⱖ 0, y ⱖ 0

1

2

3 4

5

x

5 4

8 7 6 5 4 3 2 1 ⫺2 ⫺1 ⫺1 ⫺2

3 2 1 ⫺3 ⫺2 ⫺1 ⫺1

1

2 3 4

5 6 7

x

⫺2 1 2

3 4

25

Width

20 15 10 5

b. Sketch part of the solution set for this inequality that represents all possible values for the length and width of the dog run. (Hint: Note that both the length and the width must be positive.)

0

61. Suppose Rick has 40 ft of fencing with which he can build a rectangular garden. Let x represent the length of the garden and let y represent the width.

5

10 15 Length

20

25

5

10 15 Length

20

25

x

y

25 20 Width

a. Write an inequality representing the fact that the total perimeter of the garden is at most 40 ft.

⫺4 ⫺5 y

60. Suppose Sue has 50 ft of fencing with which she can build a rectangular dog run. Let x represent the length of the dog run and let y represent the width. a. Write an inequality representing the fact that the total perimeter of the dog run is at most 50 ft.

⫺3

x

5 6 7 8

b. Sketch part of the solution set for this inequality that represents all possible values for the length and width of the garden. (Hint: Note that both the length and the width must be positive.)

15 10 5 0

x

62. A manufacturer produces two models of desks. Model A requires 4 hr to stain and finish and 3 hr to assemble. Model B requires 3 hr to stain and finish and 1 hr to assemble. The total amount of time available for staining and finishing is 24 hr and for assembling is 12 hr. Let x represent the number of Model A desks produced, and let y represent the number of Model B desks produced.

b. Write an inequality in terms of the number of Model A and Model B desks that can be produced if the total time for staining and finishing is at most 24 hr. c. Write an inequality in terms of the number of Model A and Model B desks that can be produced if the total time for assembly is no more than 12 hr. d. Graph the feasible region formed by graphing the preceding inequalities.

Model B desks

y

a. Write two inequalities that express the fact that the number of desks to be produced cannot be negative.

8 7 6 5 4 3 2 1

⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1

2

3 4

5 6 7

Model A desks

e. Is the point (3, 1) in the feasible region? What does the point (3, 1) represent in the context of this problem? f. Is the point (5, 4) in the feasible region? What does the point (5, 4) represent in the context of this problem?

x

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63. In scheduling two drivers for delivering pizza, James needs to have at least 65 hr scheduled this week. His two drivers, Karen and Todd, are not allowed to get overtime, so each one can work at most 40 hr. Let x represent the number of hours that Karen can be scheduled, and let y represent the number of hours Todd can be scheduled. (See Example 7.) y

b. Write two inequalities that express the fact that neither Karen nor Todd is allowed overtime (i.e., each driver can have at most 40 hr). c. Write an inequality that expresses the fact that the total number of hours from both Karen and Todd needs to be at least 65 hr. d. Graph the feasible region formed by graphing the inequalities.

Hours (Todd)

a. Write two inequalities that express the fact that Karen and Todd cannot work a negative number of hours.

65 60 55 50 45 40 35 30 25 20 15 10 5 5 10 15 20 25 30 35 40 45 50 55 60 65 Hours (Karen)

x

e. Is the point (35, 40) in the feasible region? What does the point (35, 40) represent in the context of this problem? f. Is the point (20, 40) in the feasible region? What does the point (20, 40) represent in the context of this problem?

Section 3.6 Concepts 1. Solutions to Systems of Linear Equations in Three Variables 2. Solving Systems of Linear Equations in Three Variables 3. Applications of Linear Equations in Three Variables 4. Solving Dependent and Inconsistent Systems

y

Figure 3-24

1. Solutions to Systems of Linear Equations in Three Variables In Sections 3.1–3.3, we solved systems of linear equations in two variables. In this section, we will expand the discussion to solving systems involving three variables. A linear equation in three variables can be written in the form Ax  By  Cz  D, where A, B, and C are not all zero. For example, the equation 2x  3y  z  6 is a linear equation in three variables. Solutions to this equation are ordered triples of the form (x, y, z) that satisfy the equation. Some solutions to the equation 2x  3y  z  6 are Solution:

Check:

11, 1, 12

2112  3112  112 ⱨ 6 ✔ True

10, 1, 32

2102  3112  132 ⱨ 6 ✔ True

12, 0, 22

z

x

Systems of Linear Equations in Three Variables and Applications

2122  3102  122 ⱨ 6 ✔ True

Infinitely many ordered triples serve as solutions to the equation 2x  3y  z  6. The set of all ordered triples that are solutions to a linear equation in three variables may be represented graphically by a plane in space. Figure 3-24 shows a portion of the plane 2x  3y  z  6 in a 3-dimensional coordinate system. An example of a system of three linear equations in three variables is shown here. 2x  y  3z  7 3x  2y  z  11 2x  3y  2z  3

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Systems of Linear Equations in Three Variables and Applications

A solution to a system of linear equations in three variables is an ordered triple that satisfies each equation. Geometrically, a solution is a point of intersection of the planes represented by the equations in the system. A system of linear equations in three variables may have one unique solution, infinitely many solutions, or no solution (Table 3-2, Table 3-3, and Table 3-4). Table 3-2 One unique solution (planes intersect at one point) • The system is consistent. • The system is independent.

Table 3-3 No solution (the three planes do not all intersect) • The system is inconsistent. • The system is independent.

Table 3-4 Infinitely many solutions (planes intersect at infinitely many points) • The system is consistent. • The system is dependent.

279

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Chapter 3 Systems of Linear Equations and Inequalities

2. Solving Systems of Linear Equations in Three Variables To solve a system involving three variables, the goal is to eliminate one variable. This reduces the system to two equations in two variables. One strategy for eliminating a variable is to pair up the original equations two at a time.

PROCEDURE Solving a System of Three Linear Equations in Three Variables Step 1 Write each equation in standard form Ax ⫹ By ⫹ Cz ⫽ D. Step 2 Choose a pair of equations, and eliminate one of the variables by using the addition method. Step 3 Choose a different pair of equations and eliminate the same variable. Step 4 Once steps 2 and 3 are complete, you should have two equations in two variables. Solve this system by using the methods from Sections 3.2 and 3.3. Step 5 Substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable. Solve for the third variable. Step 6 Check the ordered triple in each of the original equations. Then write the solution as an ordered triple within set notation.

Example 1

Solving a System of Linear Equations in Three Variables 2x ⫹ y ⫺ 3z ⫽ ⫺7

Solve the system.

3x ⫺ 2y ⫹ z ⫽ 11 ⫺2x ⫺ 3y ⫺ 2z ⫽

3

Solution: A

2x ⫹ y ⫺ 3z ⫽ ⫺7

Step 1: The equations are already in standard form.

B

3x ⫺ 2y ⫹ z ⫽ 11

• It is often helpful to label the equations. • The y variable can be easily eliminated from equations A and B and from equations

C ⫺2x ⫺ 3y ⫺ 2z ⫽

3

A and C . This is accomplished by creating opposite coefficients for the y terms and then adding the equations.

TIP: It is important to

Step 2: Eliminate the y variable from equations A and B .

note that in steps 2 and 3, the same variable is eliminated.

A 2x ⫹ y ⫺ 3z ⫽ ⫺7 B 3x ⫺ 2y ⫹ z ⫽ 11

Multiply by 2.

4x ⫹ 2y ⫺ 6z ⫽ ⫺14 3x ⫺ 2y ⫹ z ⫽ 7x

11

⫺ 5z ⫽ ⫺3 D

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Systems of Linear Equations in Three Variables and Applications

Step 3: Eliminate the y variable again, this time from equations A and C . A

2x ⫹ y ⫺ 3z ⫽ ⫺7

C ⫺2x ⫺ 3y ⫺ 2z ⫽

Multiply by 3.

6x ⫹ 3y ⫺ 9z ⫽ ⫺21 ⫺2x ⫺ 3y ⫺ 2z ⫽

3

4x

3

⫺ 11z ⫽ ⫺18 E

Step 4: Now equations D and E can be paired up to form a linear system in two variables. Solve this system. D 7x ⫺ 5z ⫽ ⫺3 E 4x ⫺ 11z ⫽ ⫺18

Multiply by ⫺4.

⫺28x ⫹ 20z ⫽

12

28x ⫺ 77z ⫽ ⫺126

Multiply by 7.

⫺57z ⫽ ⫺114 z⫽2 Once one variable has been found, substitute this value into either equation in the two-variable system, that is, either equation D or E . D 7x ⫺ 5z ⫽ ⫺3 7x ⫺ 5122 ⫽ ⫺3

Substitute z ⫽ 2 into equation D .

7x ⫺ 10 ⫽ ⫺3

A

7x ⫽

7

x⫽

1

2x ⫹ y ⫺ 3z ⫽ ⫺7

Step 5:

2112 ⫹ y ⫺ 3122 ⫽ ⫺7 2 ⫹ y ⫺ 6 ⫽ ⫺7 y ⫺ 4 ⫽ ⫺7 y ⫽ ⫺3

The solution set is 5(1, ⫺3, 2)6. Check:

Now that two variables are known, substitute these values (x and z) into any of the original three equations to find the remaining variable y. Substitute x ⫽ 1 and z ⫽ 2 into equation A .

Step 6: Check the ordered triple in the three original equations.

2x ⫹ y ⫺ 3z ⫽ ⫺7 3x ⫺ 2y ⫹ z ⫽ 11 ⫺2x ⫺ 3y ⫺ 2z ⫽ 3

2112 ⫹ 1⫺32 ⫺ 3122 ⱨ ⫺7 ✔ True 3112 ⫺ 21⫺32 ⫹ 122 ⱨ 11 ✔ True

⫺2112 ⫺ 31⫺32 ⫺ 2122 ⱨ

3 ✔ True

Skill Practice Solve the system. 1. x ⫹ 2y ⫹ z ⫽ 1 3x ⫺ y ⫹ 2z ⫽ 13 2x ⫹ 3y ⫺ z ⫽ ⫺8

Answer

1. 5(1, ⫺2, 4)6

281

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Chapter 3 Systems of Linear Equations and Inequalities

3. Applications of Linear Equations in Three Variables Example 2

Applying Systems of Linear Equations to Geometry

In a triangle, the smallest angle measures 10⬚ more than one-half the measure of the largest angle. The middle angle measures 12⬚ more than the measure of the smallest angle. Find the measure of each angle.

Solution: Let x represent the measure of the smallest angle. Let y represent the measure of the middle angle. Let z represent the measure of the largest angle. x

To solve for three variables, we need to establish three independent relationships among x, y, and z. z ⫹ 10 2

A

x⫽

B

y ⫽ x ⫹ 12

y

z

The smallest angle measures 10° more than one-half the measure of the largest angle. The middle angle measures 12⬚ more than the measure of the smallest angle.

C x ⫹ y ⫹ z ⫽ 180

The sum of the angles inscribed in a triangle is 180°. Clear fractions and write each equation in standard form.

A

x y

B C

z ⫽ ⫹ 10 2

Standard Form Multiply by 2.

2x ⫽ z ⫹ 20

⫺ z ⫽ 20

⫺x ⫹ y

⫽ x ⫹ 12

x⫹y⫹z⫽

2x

⫽ 12

x ⫹ y ⫹ z ⫽ 180

180

Notice equation B is missing the z variable. Therefore, we can eliminate z again by pairing up equations A and C . A 2x C

⫺ z ⫽ 120

x ⫹ y ⫹ z ⫽ 180 3x ⫹ y

⫽ 200

B ⫺x ⫹ y ⫽ 12

D

Multiply by ⫺1.

D 3x ⫹ y ⫽ 200

x ⫺ y ⫽ ⫺12 3x ⫹ y ⫽ 200 4x

⫽ 188 x ⫽ 47

From equation B we have ⫺x ⫹ y ⫽ 12 From equation C we have

x ⫹ y ⫹ z ⫽ 180

Pair up equations B and D to form a system of two variables. Solve for x. ⫺47 ⫹ y ⫽ 12

y ⫽ 59

47 ⫹ 59 ⫹ z ⫽ 180

z ⫽ 74

The smallest angle measures 47⬚, the middle angle measures 59⬚, and the largest angle measures 74⬚.

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283

Skill Practice 2. The perimeter of a triangle is 30 in. The shortest side is 4 in. shorter than the longest side. The longest side is 6 in. less than the sum of the other two sides. Find the length of each side.

Example 3

Applying Systems of Linear Equations to Nutrition

Doctors have become increasingly concerned about the sodium intake in the U.S. diet. Recommendations by the American Medical Association indicate that most individuals should not exceed 2400 mg of sodium per day. Torie ate 1 slice of pizza, 1 serving of ice cream, and 1 glass of soda for a total of 1030 mg of sodium. David ate 3 slices of pizza, no ice cream, and 2 glasses of soda for a total of 2420 mg of sodium. Emilie ate 2 slices of pizza, 1 serving of ice cream, and 2 glasses of soda for a total of 1910 mg of sodium. How much sodium is in one serving of each item?

Solution: Let x represent the sodium content of 1 slice of pizza. Let y represent the sodium content of 1 serving of ice cream. Let z represent the sodium content of 1 glass of soda. x ⫹ y ⫹ z ⫽ 1030

From Torie’s meal we have:

A

From David’s meal we have:

B 3x

From Emilie’s meal we have:

C 2x ⫹ y ⫹ 2z ⫽ 1910

⫹ 2z ⫽ 2420

Equation B is missing the y variable. Eliminating y from equations A and C , we have A

x ⫹ y ⫹ z ⫽ 1030

Multiply by ⫺1.

⫺x ⫺ y ⫺ z ⫽ ⫺1030

C 2x ⫹ y ⫹ 2z ⫽ 1910

2x ⫹ y ⫹ 2z ⫽

1910

⫹ z⫽

880

D x

Solve the system formed by equations B and D . 3x ⫹ 2z ⫽

B 3x ⫹ 2z ⫽ 2420 D

x ⫹ z ⫽ 880

Multiply by ⫺2.

2420

⫺2x ⫺ 2z ⫽ ⫺1760 x

From equation D we have x ⫹ z ⫽ 880 From equation A we have x ⫹ y ⫹ z ⫽ 1030

⫽

660

660 ⫹ z ⫽ 880

z ⫽ 220

660 ⫹ y ⫹ 220 ⫽ 1030

y ⫽ 150

Therefore, 1 slice of pizza has 660 mg of sodium, 1 serving of ice cream has 150 mg of sodium, and 1 glass of soda has 220 mg of sodium. Skill Practice 3. Annette, Barb, and Carlita work in a clothing shop. One day the three had combined sales of $1480. Annette sold $120 more than Barb. Barb and Carlita combined sold $280 more than Annette. How much did each person sell? Answers 2. The sides are 8 in., 10 in., and 12 in. 3. Annette sold $600, Barb sold $480, and Carlita sold $400.

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Chapter 3 Systems of Linear Equations and Inequalities

4. Solving Dependent and Inconsistent Systems Example 4

Solving a Dependent System of Linear Equations

Solve the system. If there is not a unique solution, label the system as either dependent or inconsistent. A 3x ⫹ y ⫺ z ⫽ 8 B 2x ⫺ y ⫹ 2z ⫽ 3 C

x ⫹ 2y ⫺ 3z ⫽ 5

Solution: The first step is to make a decision regarding the variable to eliminate. The y variable is particularly easy to eliminate because the coefficients of y in equations A and B are already opposites. The y variable can be eliminated from equations B and C by multiplying equation B by 2. A 3x ⫹ y ⫺ z ⫽ 8

Pair up equations A and B to eliminate y.

B 2x ⫺ y ⫹ 2z ⫽ 3 5x

⫹ z ⫽ 11 D

B

2x ⫺ y ⫹ 2z ⫽ 3

C

x ⫹ 2y ⫺ 3z ⫽ 5

Multiply by 2.

4x ⫺ 2y ⫹ 4z ⫽ 6 x ⫹ 2y ⫺ 3z ⫽ 5 ⫹ z ⫽ 11 E

5x

Pair up equations B and C to eliminate y.

Because equations D and E are equivalent equations, it appears that this is a dependent system. By eliminating variables we obtain the identity 0 ⫽ 0. D

5x ⫹ z ⫽ 11

E

5x ⫹ z ⫽ 11

Multiply by ⫺1.

⫺5x ⫺ z ⫽ ⫺11 5x ⫹ z ⫽

11

0⫽

0

The result 0 ⫽ 0 indicates that there are infinitely many solutions and that the system is dependent. Skill Practice Solve the system. If the system does not have a unique solution, identify the system as dependent or inconsistent. 4.

Answer 4. Infinitely many solutions; dependent system

x⫹ y⫹ z⫽ 8 2x ⫺ y ⫹ z ⫽ 6 ⫺5x ⫺ 2y ⫺ 4z ⫽ ⫺30

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Section 3.6

Example 5

Systems of Linear Equations in Three Variables and Applications

285

Solving an Inconsistent System of Linear Equations

Solve the system. If there is not a unique solution, identify the system as either dependent or inconsistent. 2x ⫹ 3y ⫺ 7z ⫽ 4 ⫺4x ⫺ 6y ⫹ 14z ⫽ 1 5x ⫹ y ⫺ 3z ⫽ 6

Solution: We will eliminate the x variable from equations A and B . A

2x ⫹ 3y ⫺ 7z ⫽ 4

B

⫺4x ⫺ 6y ⫹ 14 z ⫽ 1

Multiply by 2.

4x ⫹ 6y ⫺ 14z ⫽ 8 ⫺4x ⫺ 6y ⫹ 14z ⫽ 1 0⫽9

(contradiction)

The result 0 ⫽ 9 is a contradiction, indicating that the system has no solution, 5 6. The system is inconsistent. Skill Practice 5. Solve the system. If the system does not have a unique solution, identify the system as dependent or inconsistent. x ⫺ 2y ⫹ z ⫽ 5 x ⫺ 3y ⫹ 2z ⫽ ⫺7 ⫺2x ⫹ 4y ⫺ 2z ⫽ 6

Answer 5. No solution; inconsistent system

Section 3.6 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Linear equation in three variables

b. Ordered triple

Review Exercises

2. Determine if the ordered pair 1⫺4, ⫺72 is a solution to the system.

⫺5x ⫹ 3y ⫽ ⫺1 4x ⫺ 2y ⫽ ⫺2

For Exercises 3–4, solve the systems by using two methods: (a) the substitution method and (b) the addition method. 3. 3x ⫹ y ⫽ 4 4x ⫹ y ⫽ 5

4.

2x ⫺ 5y ⫽ 3 ⫺4x ⫹ 10y ⫽ 3

5. Marge can ride her bike 24 mi in 113 hr riding with the wind. Riding against the wind she can ride 24 mi in 2 hr. Find the speed at which Marge can ride in still air and the speed of the wind.

Concept 1: Solutions to Systems of Linear Equations in Three Variables 6. How many solutions are possible when solving a system of three equations with three variables?

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7. Which of the following points are solutions to the system? 12, 1, 72, 13, ⫺10, ⫺62, 14, 0, 22

8. Which of the following points are solutions to the system? 11, 1, 32, 10, 0, 42, 14, 2, 12

2x ⫺ y ⫹ z ⫽ 10 4x ⫹ 2y ⫺ 3z ⫽ 10 x ⫺ 3y ⫹ 2z ⫽ 8

⫺3x ⫺ 3y ⫺ 6z ⫽ ⫺24 ⫺9x ⫺ 6y ⫹ 3z ⫽ ⫺45 9x ⫹ 3y ⫺ 9z ⫽ 33

9. Which of the following points are solutions to the system? 112, 2, ⫺22, 14, 2, 12, 11, 1, 12

10. Which of the following points are solutions to the system? 10, 4, 32, 13, 6, 102, 13, 3, 12 x ⫹ 2y ⫺ z ⫽ 5 x ⫺ 3y ⫹ z ⫽ ⫺5 ⫺2x ⫹ y ⫺ z ⫽ ⫺4

⫺x ⫺ y ⫺ 4z ⫽ ⫺6 x ⫺ 3y ⫹ z ⫽ ⫺1 4x ⫹ y ⫺ z ⫽ 4

Concept 2: Solving Systems of Linear Equations in Three Variables For Exercises 11–22, solve the system of equations. (See Example 1.) 11. 2x ⫹ y ⫺ 3z ⫽ ⫺12 3x ⫺ 2y ⫺ z ⫽ 3 ⫺x ⫹ 5y ⫹ 2z ⫽ ⫺3

12. ⫺3x ⫺ 2y ⫹ 4z ⫽ ⫺15 2x ⫹ 5y ⫺ 3z ⫽ 3 4x ⫺ y ⫹ 7z ⫽ 15

13.

16. y ⫽ 2x ⫹ z ⫹ 1 ⫺3x ⫺ 1 ⫽ ⫺2y ⫹ 2z 5x ⫹ 3z ⫽ 16 ⫺ 3y

14.

6x ⫺ 5y ⫹ z ⫽ 7 5x ⫹ 3y ⫹ 2z ⫽ 0 ⫺2x ⫹ y ⫺ 3z ⫽ 11

15. 4x ⫹ 2z ⫽ 12 ⫹ 3y 2y ⫽ 3x ⫹ 3z ⫺ 5 y ⫽ 2x ⫹ 7z ⫹ 8

17.

x⫹ y⫹z⫽ 6 ⫺x ⫹ y ⫺ z ⫽ ⫺2 2x ⫹ 3y ⫹ z ⫽ 11

18.

20.

x ⫹ y ⫹ 3z ⫽ 2 2x ⫺ 3z ⫽ 5 3y ⫹ 3z ⫽ 2

⫽ 8 21. 4x ⫹ 9y 8x ⫹ 6z ⫽ ⫺1 6y ⫹ 6z ⫽ ⫺1

x ⫺ y ⫺ z ⫽ ⫺11 x ⫹ y ⫺ z ⫽ 15 2x ⫺ y ⫹ z ⫽ ⫺9

x ⫺ 3y ⫺ 4z ⫽ ⫺7 5x ⫹ 2y ⫹ 2z ⫽ ⫺1 4x ⫺ y ⫺ 5z ⫽ ⫺6

19. 2x ⫺ 3y ⫹ 2z ⫽ ⫺1 x ⫹ 2y ⫽ ⫺4 x ⫹ z⫽ 1 ⫹ 2z ⫽ 11 y ⫺ 7z ⫽ 4 x ⫺ 6y ⫽ 1

22. 3x

Concept 3: Applications of Linear Equations in Three Variables 23. A triangle has one angle that measures 5⬚ more than twice the smallest angle, and the third angle measures 11⬚ less than 3 times the measure of the smallest angle. Find the measures of the three angles. (See Example 2.)

24. The largest angle of a triangle measures 4⬚ less than 5 times the measure of the smallest angle. The middle angle measures twice that of the smallest angle. Find the measures of the three angles.

25. The perimeter of a triangle is 55 cm. The measure of the shortest side is 8 cm less than the middle side. The measure of the longest side is 1 cm less than the sum of the other two sides. Find the lengths of the sides.

26. The perimeter of a triangle is 5 ft. The longest side of the triangle measures 20 in. more than the shortest side. The middle side is 3 times the measure of the shortest side. Find the lengths of the three sides in inches.

27. A movie theater charges $7 for adults, $5 for children under age 17, and $4 for seniors over age 60. For one showing of Batman the theater sold 222 tickets and took in $1383. If twice as many adult tickets were sold as the total of children and senior tickets, how many tickets of each kind were sold? (See Example 3.)

28. Baylor University in Waco, Texas, had twice as many students as Vanderbilt University in Nashville, Tennessee. Pace University in New York City had 2800 more students than Vanderbilt University. If the enrollment for all three schools totaled 27,200, find the enrollment for each school.

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29. Goofie Golf has 18 holes that are par 3, par 4, or par 5. Most of the holes are par 4. In fact, there are 3 times as many par 4’s as par 3’s. There are 3 more par 5’s than par 3s. How many of each type are there? 30. Combining peanuts, pecans, and cashews makes a party mixture of nuts. If the amount of peanuts equals the amount of pecans and cashews combined, and if there are twice as many cashews as pecans, how many ounces of each nut is used to make 48 oz of party mixture? 31. Souvenir hats, T-shirts, and jackets are sold at a rock concert. Three hats, two T-shirts, and one jacket cost $140. Two hats, two T-shirts, and two jackets cost $170. One hat, three T-shirts, and two jackets cost $180. Find the prices of the individual items. 32. Annie and Maria traveled overseas for 7 days and stayed in three different hotels in three different cities: Stockholm, Sweden; Oslo, Norway; and Paris, France. The total bill for all seven nights (not including tax) was $1040. The total tax was $106. The nightly cost (excluding tax) to stay at the hotel in Paris was $80 more than the nightly cost (excluding tax) to stay in Oslo. Find the cost per night for each hotel excluding tax.

Number of Nights

Cost/Night ($)

Tax Rate

Paris, France

1

x

8%

Stockholm, Sweden

4

y

11%

Oslo, Norway

2

z

10%

City

Concept 4: Solving Dependent and Inconsistent Systems (Mixed Exercises) For Exercises 33–44, solve the system. If there is not a unique solution, label the system as either dependent or inconsistent. (See Examples 1, 4, and 5.) 33.

2x ⫹ y ⫹ 3z ⫽ 2 x ⫺ y ⫹ 2z ⫽ ⫺4 ⫺2x ⫹ 2y ⫺ 4z ⫽ 8

34.

36.

3x ⫹ 2y ⫹ z ⫽ 3 x ⫺ 3y ⫹ z ⫽ 4 ⫺6x ⫺ 4y ⫺ 2z ⫽ 1

37. 12x ⫹ 23y 1 3y

2x ⫹ y ⫽ ⫺3 2y ⫹ 16z ⫽ ⫺10 ⫺7x ⫺ 3y ⫹ 4z ⫽ 8

⫽ ⫺ 12z ⫽

1 5x

39. ⫺3x ⫹ y ⫺ z ⫽ 8 ⫺4x ⫹ 2y ⫹ 3z ⫽ ⫺3 2x ⫹ 3y ⫺ 2z ⫽ ⫺1 42.

x⫹ y⫽ z 2x ⫹ 4y ⫺ 2z ⫽ 6 3x ⫹ 6y ⫺ 3z ⫽ 9

⫺ 14z ⫽

35.

5 2 ⫺103 3 4

6x ⫺ 2y ⫹ 2z ⫽ 2 4x ⫹ 8y ⫺ 2z ⫽ 5 ⫺2x ⫺ 4y ⫹ z ⫽ ⫺2

38. 12x ⫹ 14y ⫹ z ⫽ 3 ⫹ 14y ⫹ 14z ⫽ x ⫺ y ⫺ 23z ⫽

1 8x

9 8 1 3

40.

2x ⫹ 3y ⫹ 3z ⫽ 15 3x ⫺ 6y ⫺ 6z ⫽ ⫺23 ⫺9x ⫺ 3y ⫹ 6z ⫽ 8

41. 2x ⫹ y ⫽ 31z ⫺ 12 3x ⫺ 21y ⫺ 2z2 ⫽ 1 212x ⫺ 3z2 ⫽ ⫺6 ⫺ 2y

43.

⫺0.1y ⫹ 0.2z ⫽ 0.2 0.1x ⫹ 0.1y ⫹ 0.1z ⫽ 0.2 ⫺0.1x ⫺ 0.3z ⫽ 0.2

⫽ 0 44. ⫺0.4x ⫺ 0.3y 0.3y ⫹ 0.1z ⫽ ⫺0.1 0.4x ⫺ 0.1z ⫽ 1.2

Expanding Your Skills The systems in Exercises 45–48 are called homogeneous systems because each system has (0, 0, 0) as a solution. However, if a system is dependent, it will have infinitely many more solutions. For each system determine whether (0, 0, 0) is the only solution or if the system is dependent. 45. 2x ⫺ 4y ⫹ 8z ⫽ 0 ⫺x ⫺ 3y ⫹ z ⫽ 0 x ⫺ 2y ⫹ 5z ⫽ 0

46. 2x ⫺ 4y ⫹ z ⫽ 0 x ⫺ 3y ⫺ z ⫽ 0 3x ⫺ y ⫹ 2z ⫽ 0

47.

4x ⫺ 2y ⫺ 3z ⫽ 0 ⫺8x ⫺ y ⫹ z ⫽ 0 3 2x ⫺ y ⫺ z ⫽ 0 2

48. 5x ⫹ y ⫽0 4y ⫺ z ⫽ 0 5x ⫹ 5y ⫺ z ⫽ 0

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Section 3.7

Solving Systems of Linear Equations by Using Matrices

Concepts

1. Introduction to Matrices

1. Introduction to Matrices 2. Solving Systems of Linear Equations by Using the Gauss-Jordan Method

In Sections 3.2, 3.3, and 3.6, we solved systems of linear equations by using the substitution method and the addition method. We now present a third method called the Gauss-Jordan method that uses matrices to solve a linear system. A matrix is a rectangular array of numbers (the plural of matrix is matrices). The rows of a matrix are read horizontally, and the columns of a matrix are read vertically. Every number or entry within a matrix is called an element of the matrix. The order of a matrix is determined by the number of rows and number of columns. A matrix with m rows and n columns is an m ⫻ n (read as “m by n”) matrix. Notice that with the order of a matrix, the number of rows is given first, followed by the number of columns. Example 1

Determining the Order of a Matrix

Determine the order of each matrix. 2 a. c 5

⫺4 p

1 d 27

1.9 0 ¥ b. ≥ 7.2 ⫺6.1

1 c. £ 0 0

0 1 0

0 0§ 1

d. 3a

b

c4

Solution: a. This matrix has two rows and three columns. Therefore, it is a 2 ⫻ 3 matrix. b. This matrix has four rows and one column. Therefore, it is a 4 ⫻ 1 matrix. A matrix with one column is called a column matrix. c. This matrix has three rows and three columns.Therefore, it is a 3 ⫻ 3 matrix. A matrix with the same number of rows and columns is called a square matrix. d. This matrix has one row and three columns. Therefore, it is a 1 ⫻ 3 matrix. A matrix with one row is called a row matrix. Skill Practice Determine the order of the matrix. 1.

⫺5 £ 1 8

2 3§ 9

2. 34 ⫺8 4

5 £ 10 § 15

3.

4.

c

2 ⫺1

⫺0.5 d 6

A matrix can be used to represent a system of linear equations written in standard form. To do so, we extract the coefficients of the variable terms and the constants within the equation. For example, consider the system 2x ⫺ y ⫽

5

x ⫹ 2y ⫽ ⫺5 The matrix A is called the coefficient matrix. Answers 1. 3 ⫻ 2 3. 3 ⫻ 1

2. 1 ⫻ 2 4. 2 ⫻ 2

A⫽ c

2 1

⫺1 d 2

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If we extract both the coefficients and the constants from the equations, we can construct the augmented matrix of the system: c

⫺1 5 ` d 2 ⫺5

2 1

A vertical bar is inserted into an augmented matrix to designate the position of the equal signs.

Writing an Augmented Matrix for a System of Linear Equations

Example 2

Write the augmented matrix for each linear system. a. ⫺3x ⫺ 4y ⫽ 3

⫺ 3z ⫽ 14

b. 2x

2x ⫹ 4y ⫽ 2

2y ⫹ z ⫽ 2 x⫹ y

⫽ 4

Solution: 2 b. £ 0 1

⫺3 ⫺4 3 a. c ` d 2 4 2

⫺3 14 1 † 2§ 0 4

0 2 1

TIP: Notice that zeros are inserted to denote the coefficient of each missing term.

Skill Practice Write the augmented matrix for each system. 5. ⫺x ⫹ y ⫽ 4 2x ⫺ y ⫽ 1

6.

2x ⫺ y ⫹ z ⫽ 14 ⫺3x ⫹ 4y ⫽ 8 x ⫺ y ⫹ 5z ⫽ 0

Writing a Linear System from an Augmented Matrix

Example 3

Write a system of linear equations represented by each augmented matrix. a. c

2 4

2 b. £ 1 3

⫺5 ⫺8 ` d 1 6

⫺1 1 1

3 14 ⫺2 † ⫺5 § ⫺1 2

1 c. £ 0 0

0 1 0

0 4 0 † ⫺1 § 1 0

Solution: a. 2x ⫺ 5y ⫽ ⫺8 4x ⫹ y ⫽

b. 2x ⫺ y ⫹ 3z ⫽ 14 x ⫹ y ⫺ 2z ⫽ ⫺5

6

3x ⫹ y ⫺ z ⫽ c.

x ⫹ 0y ⫹ 0z ⫽

x⫽

4

0x ⫹ y ⫹ 0z ⫽ ⫺1 0x ⫹ 0y ⫹ z ⫽

or

Answers 5. c

4

y ⫽ ⫺1 z⫽

0

2

2 6. £⫺3 1

0

Skill Practice Write a system of linear equations represented by each augmented matrix. 7.

2 c ⫺1

3 5 ` d 8 1

8.

⫺3 £ 14 ⫺8

2 1 3

1 4 0 † 20 § 5 6

⫺1 2

9.

1 £0 0

0 1 0

0 ⫺8 0 † 2§ 1 15

1 4 ` d ⫺1 1 ⫺1 4 ⫺1

1 14 0 † 8§ 5 0

7. 2x ⫹ 3y ⫽ 5 ⫺x ⫹ 8y ⫽ 1 8. ⫺3x ⫹ 2y ⫹ z ⫽ 4 14x ⫹ y ⫽ 20 ⫺8x ⫹ 3y ⫹ 5z ⫽ 6 9. x ⫽ ⫺8, y ⫽ 2, z ⫽ 15

289

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2. Solving Systems of Linear Equations by Using the Gauss-Jordan Method We know that interchanging two equations results in an equivalent system of linear equations. Interchanging two rows in an augmented matrix results in an equivalent augmented matrix. Similarly, because each row in an augmented matrix represents a linear equation, we can perform the following elementary row operations that result in an equivalent augmented matrix.

PROPERTY Elementary Row Operations The following elementary row operations performed on an augmented matrix produce an equivalent augmented matrix: • Interchange two rows. • Multiply every element in a row by a nonzero real number. • Add a multiple of one row to another row.

When we are solving a system of linear equations by any method, the goal is to write a series of simpler but equivalent systems of equations until the solution is obvious. The Gauss-Jordan method uses a series of elementary row operations performed on the augmented matrix to produce a simpler augmented matrix. In particular, we want to produce an augmented matrix that has 1’s along the diagonal of the matrix of coefficients and 0’s for the remaining entries in the matrix of coefficients. A matrix written in this way is said to be written in reduced row echelon form. For example, the augmented matrix from Example 3(c) is written in reduced row echelon form. 1 £0 0

0 1 0

0 4 0 † 1 § 1 0

The solution to the corresponding system of equations is easily recognized as x  4, y  1, and z  0. Similarly, matrix B represents a solution of x  a and y  b. B c Example 4

1 0

0 a ` d 1 b

Solving a System by Using the Gauss-Jordan Method

Solve by using the Gauss-Jordan method. 2x  y 

5

x  2y  5

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Solution: c

2 1

⫺1 5 ` d 2 ⫺5

Set up the augmented matrix.

c

1 2

2 ⫺5 ` d ⫺1 5

Switch row 1 and row 2 to get a 1 in the upper left position.

⫺2R1 ⫹ R2 1 R2

c

1 0

2 ⫺5 ` d ⫺5 15

Multiply row 1 by ⫺2 and add the result to row 2. This produces an entry of 0 below the upper left position.

⫺15R2 1 R2

1 c 0

2 ⫺5 ` d 1 ⫺3

Multiply row 2 by ⫺15 to produce a 1 along the diagonal in the second row.

c

1 0

0 1 ` d 1 ⫺3

Multiply row 2 by ⫺2 and add the result to row 1. This produces a 0 in the first row, second column.

C⫽ c

1 0

0 1 ` d 1 ⫺3

R1 3 R2

⫺2R2 ⫹ R1 1 R1

The matrix C is in reduced row echelon form. From the augmented matrix, we have x ⫽ 1 and y ⫽ ⫺3. The solution set is {(1, ⫺3)}. Skill Practice 10. Solve by using the Gauss-Jordan method. x ⫺ 2y ⫽ ⫺21 2x ⫹ y ⫽ ⫺2

The order in which we manipulate the elements of an augmented matrix to produce reduced row echelon form was demonstrated in Example 4. In general, the order is as follows. • First produce a 1 in the first row, first column. Then use the first row to obtain 0’s in the first column below this element. • Next, if possible, produce a 1 in the second row, second column. Use the second row to obtain 0’s above and below this element. • Next, if possible, produce a 1 in the third row, third column. Use the third row to obtain 0’s above and below this element. • The process continues until reduced row echelon form is obtained.

Answer

10. 5 1⫺5, 826

291

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Example 5

Solving a System by Using the Gauss-Jordan Method

Solve by using the Gauss-Jordan method. ⫽ ⫺y ⫹ 5

x ⫺2x

⫹ 2z ⫽

y ⫺ 10

3x ⫹ 6y ⫹ 7z ⫽ 14

Solution: First write each equation in the system in standard form. x ⫺2x ⫹

⫽ ⫺y ⫹ 5

x⫹ y

2z ⫽ y ⫺ 10

3x ⫹ 6y ⫹ 7z ⫽ 1 £ ⫺2 3

⫺1R2 ⫹ R1 1 R1 ⫺3R2 ⫹ R3 1 R3

2R3 ⫹ R1 1 R1 ⫺2R3 ⫹ R2 1 R2

5

⫺2x ⫺ y ⫹ 2z ⫽ ⫺10

3x ⫹ 6y ⫹ 7z ⫽ 14

2R1 ⫹ R2 1 R2 ⫺3R1 ⫹ R3 1 R3

⫽

1 ⫺1 6

14

0 5 2 † ⫺10 § 7 14

Set up the augmented matrix.

1 £0 0

1 1 3

0 5 2 † 0§ 7 ⫺1

Multiply row 1 by 2 and add the result to row 2. Multiply row 1 by ⫺3 and add the result to row 3.

1 £0 0

0 1 0

⫺2 5 2 † 0§ 1 ⫺1

Multiply row 2 by ⫺1 and add the result to row 1. Multiply row 2 by ⫺3 and add the result to row 3.

1 £0 0

0 1 0

0 3 0 † 2§ 1 ⫺1

Multiply row 3 by 2 and add the result to row 1. Multiply row 3 by ⫺2 and add the result to row 2.

From the reduced row echelon form of the matrix, we have x ⫽ 3, y ⫽ 2, and z ⫽ ⫺1. The solution set is {(3, 2, ⫺1)}. Skill Practice Solve by using the Gauss-Jordan method. 11. x ⫹ y ⫹ z ⫽ 2 x⫺ y⫹ z⫽4 x ⫹ 4y ⫹ 2z ⫽ 1

Answer

11. 511, ⫺1, 22 6

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293

It is particularly easy to recognize a dependent or inconsistent system of equations from the reduced row echelon form of an augmented matrix. This is demonstrated in Examples 6 and 7. Example 6

Solving a Dependent System by Using the Gauss-Jordan Method

Solve by using the Gauss-Jordan method. x  3y  4 1 3 x y2 2 2

Solution: 1 3 4 ` d c1 32 2 2 c

12R1  R2 1 R2

1 0

3 4 ` d 0 0

Set up the augmented matrix.

Multiply row 1 by 12 and add the result to row 2.

The second row of the augmented matrix represents the equation 0  0. The system is dependent. The solution set is 51x, y2 0 x  3y  46. Skill Practice Solve by using the Gauss-Jordan method. 12. 4x  6y  16 6x  9y  24

Example 7

Solving an Inconsistent System by Using the Gauss-Jordan Method

Solve by using the Gauss-Jordan method. x  3y  2 3x  9y  1

Solution:

3R1  R2 1 R2

c

1 3 2 ` d 3 9 1

c

1 0

3 2 ` d 0 7

Set up the augmented matrix.

Multiply row 1 by 3 and add the result to row 2.

The second row of the augmented matrix represents the contradiction 0  7. The system is inconsistent. There is no solution, { }. Skill Practice Solve by using the Gauss-Jordan method. 13. 6x  10y  1 15x  25y  3

Answers 12. Infinitely many solutions; 5 1x, y2 0 4x  6y  166; dependent system 13. No solution; { }; inconsistent system

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Calculator Connections Topic: Entering a Matrix into a Calculator Many graphing calculators have a matrix editor in which the user defines the order of the matrix and then enters the elements of the matrix. For example, the 2  3 matrix D c

2 3

3 13 ` d 1 8

is entered as shown.

Once an augmented matrix has been entered into a graphing calculator, a rref function can be used to transform the matrix into reduced row echelon form.

Section 3.7

Practice Exercises

Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Matrix

b. Order of a matrix

c. Column matrix

d. Square matrix

e. Row matrix

f. Coefficient matrix

g. Augmented matrix

h. Reduced row echelon form

Review Exercises 2. How much 50% acid solution should be mixed with pure acid to obtain 20 L of a mixture that is 70% acid? For Exercises 3–5, solve the system by using any method. 3. x  6y  9

4. x  y  z  8

5. 2x  y  z  4

x  2y  13

x  2y  z  3

x  y  3z  7

x  3y  2z  7

x  3y  4z  22

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Concept 1: Introduction to Matrices For Exercises 6–14, (a) determine the order of each matrix and (b) determine if the matrix is a row matrix, a column matrix, a square matrix, or none of these. (See Example 1.) 4 5 6. ≥ ¥ 3 0 3 1

9 d 3

5 12. c 4.3

8.1 9

9. c

9 8. £ 1 5

5 7. £ 1 § 2 10. 34 4.2 0 d 18 3

13. c

11. 30

74 1 3

3 4

2

1

4 8 8

8

5 14. £ 1 0

6 d 78

3 4§ 7

11

54

1 2§ 7

For Exercises 15–18, set up the augmented matrix. (See Example 2.) 15.

x  2y  1

16.

2x  y  7 17.

x  2y

x  3y  3 2x  5y  4

5z

 17  2z

18. 5x

2x  6y  3z  2

8x

3x  y  2z 

8x  3y  12z 

1

 6z 

26  y 24

For Exercises 19–22, write a system of linear equations represented by the augmented matrix. (See Example 3.) 4 19. c 12

2 20. c 7

3 6 ` d 5 6

5 15 ` d 15 45

1 21. £ 0 0

0 1 0

0 4 0 † 1 § 1 7

1 22. £ 0 0

0 1 0

0 0.5 0 † 6.1 § 1 3.9

Concept 2: Solving Systems of Linear Equations by Using the Gauss-Jordan Method 23. Given the matrix E E c

3 9

24. Given the matrix F 2 8 ` d 1 7

F c

8 0 ` d 13 2

1 12

a. What is the element in the second row and third column?

a. What is the element in the second row and second column?

b. What is the element in the first row and second column?

b. What is the element in the first row and third column?

25. Given the matrix Z Z c

2 2

26. Given the matrix J 1 11 ` d 1 1

write the matrix obtained by multiplying the elements in the first row by 12. 27. Given the matrix K K c

5 1

J c

1 0

1 7 ` d 3 6

write the matrix obtained by multiplying the elements in the second row by 13. 28. Given the matrix L

2 1 ` d 4 3

write the matrix obtained by interchanging rows 1 and 2.

L c

9 7

6 13 ` d 2 19

write the matrix obtained by interchanging rows 1 and 2.

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Chapter 3 Systems of Linear Equations and Inequalities

29. Given the matrix M M⫽ c

30. Given the matrix N

1 5 2 ` d ⫺3 ⫺4 ⫺1

N⫽ c

write the matrix obtained by multiplying the first row by 3 and adding the result to row 2. 31. Given the matrix R

1 3 ⫺5 ` d ⫺2 2 12

write the matrix obtained by multiplying the first row by 2 and adding the result to row 2. 32. Given the matrix S

1 3 0 ⫺1 R ⫽ £ 4 1 ⫺5 † 6§ ⫺2 0 ⫺3 10

1 S⫽ £ 5 ⫺3

a. Write the matrix obtained by multiplying the first row by ⫺4 and adding the result to row 2.

2 1 4

0 10 ⫺4 † 3 § 5 2

a. Write the matrix obtained by multiplying the first row by ⫺5 and adding the result to row 2. b. Using the matrix obtained from part (a), write the matrix obtained by multiplying the first row by 3 and adding the result to row 3.

b. Using the matrix obtained from part (a), write the matrix obtained by multiplying the first row by 2 and adding the result to row 3.

For Exercises 33–48, solve the systems by using the Gauss-Jordan method. (See Examples 4–7.) 33.

x ⫺ 2y ⫽ ⫺1

34.

2x ⫹ y ⫽ ⫺7 37.

x ⫹ 3y ⫽ 3

38.

42.

⫺3x ⫺ 6y ⫽ 12 45. x ⫹ y ⫹ z ⫽ 6

2x ⫹ 5y ⫽

39.

1

⫺4x ⫺ 10y ⫽ ⫺2

x ⫹ 3y ⫽ ⫺1

x⫹ y⫽

⫺4x ⫺ 9y ⫽ 3

x ⫹ 2y ⫽ 13

x ⫹ 3y ⫹ 8z ⫽

x⫺y⫽4

40. 2x ⫺ y ⫽ 0

2x ⫹ y ⫽ 5

x⫹y⫽3

3x ⫹ y ⫽ ⫺4 ⫺6x ⫺ 2y ⫽

46. 2x ⫺ 3y ⫺ 2z ⫽ 11

x⫹y⫺z⫽0

36. 2x ⫺ 3y ⫽ ⫺2

43.

4

2x ⫺ 4y ⫽ ⫺4

x⫺y⫹z⫽2

x ⫹ 3y ⫽ 6

35.

2x ⫺ 5y ⫽ 4

4x ⫹ 12y ⫽ 12 41.

x ⫺ 3y ⫽ 3

1

3x ⫺ y ⫹ 14z ⫽ ⫺2

44. 2x ⫹ y ⫽ 4 6x ⫹ 3y ⫽ ⫺1

3

47. x ⫺ 2y ⫽ 5 ⫺ z

48. 5x ⫽ 10z ⫹ 15

2x ⫹ 6y ⫹ 3z ⫽ ⫺10

x ⫺ y ⫹ 6z ⫽ 23

3x ⫺ y ⫺ 2z ⫽

x ⫹ 3y ⫺ 12z ⫽ 13

5

For Exercises 49– 52, use the augmented matrices A, B, C, and D to answer true or false. A⫽ c

6 5

⫺4 2 ` d ⫺2 7

B⫽ c

5 6

⫺2 7 ` d ⫺4 2

C⫽ c

1 1 ⫺23 ` 3d 5 ⫺2 7

D⫽ c

5 ⫺12

⫺2 7 ` d 8 ⫺4

49. The matrix A is a 2 ⫻ 3 matrix.

50. Matrix B is equivalent to matrix A.

51. Matrix A is equivalent to matrix C.

52. Matrix B is equivalent to matrix D.

53. What does the notation R2 3 R1 mean when one is performing the Gauss-Jordan method?

54. What does the notation 2R3 1 R3 mean when one is performing the Gauss-Jordan method?

55. What does the notation ⫺3R1 ⫹ R2 1 R2 mean when one is performing the GaussJordan method?

56. What does the notation 4R2 ⫹ R3 1 R3 mean when one is performing the Gauss-Jordan method?

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Group Activity

Graphing Calculator Exercises For Exercises 57–62, use the matrix features on a graphing calculator to express each augmented matrix in reduced row echelon form. Compare your results to the solution you obtained in the indicated exercise. 57. c

1 2

⫺2 ⫺1 ` d 1 ⫺7

58. c

Compare with Exercise 33. 2 60. c 1

1 2

⫺3 3 ` d ⫺5 4

59. c

Compare with Exercise 34. 1 1 61. £ 1 ⫺1 1 1

⫺3 ⫺2 ` d 2 13

Compare with Exercise 36.

1 3 6 ` d ⫺4 ⫺9 3

Compare with Exercise 35.

1 6 1 † 2§ ⫺1 0

2 62. £ 1 3

Compare with Exercise 45.

⫺3 3 ⫺1

⫺2 11 8 † 1§ 14 ⫺2

Compare with Exercise 46.

Group Activity Creating a Quadratic Model of the Form y ⴝ at2 ⴙ bt ⴙ c Estimated time: 20 minutes Group Size: 3 Natalie Dalton was a player on the Daytona State College women’s fast-pitch softball team. She threw a ball 120 ft from right field to make a play at third base. A photographer used strobe photography to follow the path of the ball. The height of the ball (in ft) at 0.2-sec intervals was recorded in the table. Time (sec) t

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Height (ft) y

5

11

16

19

21

22

21

19

16

12

6

1. Plot the points 1t, y2 from the data.

them (t1, y1), (t2, y2), and (t3, y3).

Height (ft)

2. Select three data points and label

20 15 10

(t1, y1) ⫽

5

(t2, y2) ⫽

0

(t3, y3) ⫽

Height of Softball vs. Time

y 25

0

0.4

0.8

1.2 1.6 Time (sec)

2

t 2.4

3. Substitute the values of t1 and y1 for t and y in the quadratic model y ⫽ at2 ⫹ bt ⫹ c. The equation you are left with should only have the variables a, b, and c. Call this equation 1. Then repeat this step two more times, using the points (t2, y2) and (t3, y3). You should then be left with three equations involving a, b, and c only. Equation 1: Equation 2: Equation 3:

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Chapter 3 Systems of Linear Equations and Inequalities

4. Solve the system of equations from step 3 for the variables a, b, and c. 5. Replace the values of a, b, and c into the quadratic function y  at2  bt  c. 6. Use the function found in step 5 to approximate the height of the ball after 12 sec. 7. Use the function found in step 5 to approximate the height of the ball after 1.4 seconds. How well does this value match the observed value of 19 ft?

Chapter 3

Summary

Section 3.1

Solving Systems of Linear Equations by the Graphing Method

Key Concepts

Examples

A system of linear equations in two variables can be solved by graphing. A solution to a system of linear equations is an ordered pair that satisfies each equation in the system. Graphically, this represents a point of intersection of the lines. There may be one solution, infinitely many solutions, or no solution.

Example 1 Solve by graphing.

xy3 2x  y  0

Write each equation in slope-intercept form (y  mx  b) to graph the lines. y  x  3 y  2x y

xy3

One solution Consistent Independent

Infinitely many solutions Consistent Dependent

No solution Inconsistent Independent

A system of equations is consistent if there is at least one solution. A system is inconsistent if there is no solution. A linear system in x and y is dependent if two equations represent the same line. The solution set is the set of all points on the line. If two linear equations represent different lines, then the system of equations is independent.

5 4 3

2x  y  0

(1, 2)

2 1 5 4 3 2 1 1 2

1

2

3 4

5

x

3 4 5

The solution is the point of intersection, (1, 2). The solution set is {(1, 2)}.

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Summary

Section 3.2

Solving Systems of Linear Equations by the Substitution Method

Key Concepts

Examples

Substitution Method

Example 1 2y  6x  14 2x  y  5

Isolate a variable. y

 2x  5 v

1. Isolate one of the variables. 2. Substitute the quantity found in step 1 into the other equation. 3. Solve the resulting equation. 4. Substitute the value from step 3 back into the equation from step 1 to solve for the remaining variable. 5. Check the ordered pair in both equations, and write the answer as an ordered pair within set notation.

Substitute

212x  52  6x  14 4x  10  6x  14 2x  10  14 2x  4 x2 y  2x  5

Now solve for y.

y  2122  5 y1 The ordered pair (2, 1) checks in both equations. The solution set is 512, 126. A system is consistent if there is at least one solution. A system is inconsistent if there is no solution. An inconsistent system is detected by a contradiction (such as 0  52.

Example 2 y  2x  3 4x  2y  1

4x  212x  32  1 4x  4x  6  1 6  1

Contradiction. The system is inconsistent. There is no solution, 5 6. A system is independent if the two equations represent different lines. A system is dependent if the two equations represent the same line. This produces infinitely many solutions. A dependent system is detected by an identity (such as 0  02.

Example 3 x  3y  1 2x  6y  2

213y  12  6y  2 6y  2  6y  2 22

Identity. The system is dependent. There are infinitely many solutions. 51x, y2 0 x  3y  16

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Chapter 3 Systems of Linear Equations and Inequalities

Section 3.3

Solving Systems of Linear Equations by the Addition Method

Key Concepts

Examples

Addition Method

Example 1

1. Write both equations in standard form Ax  By  C. 2. Clear fractions or decimals (optional). 3. Multiply one or both equations by nonzero constants to create opposite coefficients for one of the variables. 4. Add the equations from step 3 to eliminate one variable. 5. Solve for the remaining variable. 6. Substitute the known value from step 5 back into one of the original equations to solve for the other variable. 7. Check the ordered pair in both equations and write the solution set.

3x  4y  18 5x  3y  1

Mult. by 3.

9x  12y  54

Mult. by 4.

20x  12y  4 29x

 58 x2

3122  4y  18 6  4y  18 4y  12 y  3 The ordered pair (2, 3) checks in both equations. The solution set is 512, 326 .

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Section 3.4

301

Applications of Systems of Linear Equations in Two Variables

Key Concepts

Examples

Solve application problems by using systems of linear equations in two variables.

Example 1 Mercedes invested $1500 more in a certificate of deposit that pays 6.5% simple interest than she did in a savings account that pays 4% simple interest. If her total interest at the end of 1 yr is $622.50, find the amount she invested in the 6.5% account.

• • • • •

Cost applications Mixture applications Principal and interest applications Uniform motion applications Geometry applications

Steps to Solve Applications: 1. 2. 3. 4. 5.

Label two variables. Construct two equations in words. Write two equations. Solve the system. Write the answer.

Let x represent the amount of money invested at 6.5%. Let y represent the amount of money invested at 4%. a

amount invested Amount invested b  $1500 ba at 4% at 6.5%

°

interest earned Interest earned from 6.5% ¢  ° from 4% ¢  $622.50 account account x  y  1500

0.065x  0.04y  622.50

Using substitution gives 0.0651y  15002  0.04y  622.50 0.065y  97.5  0.04y  622.50 0.105y  525 y  5000 x  150002  1500  6500 Mercedes invested $6500 at 6.5%.

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Chapter 3 Systems of Linear Equations and Inequalities

Section 3.5

Linear Inequalities and Systems of Linear Inequalities in Two Variables

Key Concepts

Examples

A linear inequality in two variables is an inequality of the form ax ⫹ by 6 c, ax ⫹ by 7 c, ax ⫹ by ⱕ c, or ax ⫹ by ⱖ c.

Example 1 Graph the solution to the inequality 2x ⫺ y 6 4. Solve for y: 2x ⫺ y 6 4

Graphing a Linear Inequality in Two Variables 1. Solve for y, if possible. 2. Graph the related equation. Draw a dashed line if the inequality is strict, < or >. Otherwise, draw a solid line. 3. Shade above or below the line according to the following convention. • Shade above the line if the inequality is of the form y 7 ax ⫹ b or y ⱖ ax ⫹ b. • Shade below the line if the inequality is of the form y 6 ax ⫹ b or y ⱕ ax ⫹ b. You can use test points to check that you have shaded the correct region. Select an ordered pair from the proposed solution set and substitute the values of x and y in the original inequality. If the test point produces a true statement, then you have shaded the correct region. The union or intersection of two or more linear inequalities is the union or intersection of the solution sets.

⫺y 6 ⫺2x ⫹ 4 y 7 2x ⫺ 4 Graph the related equation, y ⫽ 2x ⫺ 4, with a dashed line. Shade above the line.

y 5 4 3 2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1 2 3

4 5

x

⫺3 ⫺4 ⫺5

Example 2 Graph the solution set. x 6 0

and y 7 2

y 5 4 3 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2

1 2 3

4 5

x

⫺3 ⫺4 ⫺5

Example 3 Graph the solution set. x ⱕ 0 or

yⱖ2

y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

1 2 3 4 5

x

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Summary

Section 3.6

Systems of Linear Equations in Three Variables and Applications

Key Concepts

Examples

A linear equation in three variables can be written in the form Ax  By  Cz  D, where A, B, and C are not all zero. The graph of a linear equation in three variables is a plane in space.

Example 1 A

x  2y  z  4

B 3x  y  z  5 C 2x  3y  2z  7 A

x  2y  z  4

B 3x  y  z  5 A solution to a system of linear equations in three variables is an ordered triple that satisfies each equation. Graphically, a solution is a point of intersection among three planes. A system of linear equations in three variables may have one unique solution, infinitely many solutions (dependent system), or no solution (inconsistent system).

4x  y

9 D

2 ⴢ A 2x  4y  2z  8 C 2x  3y  2z  7 4x  7y

 15 E

D 4x  y  9

Multiply by 1.

E 4x  7y  15

4x  y  9 4x  7y  15 6y 

6

y

1

Substitute y  1 into either equation D or E . D 4x  112  9 4x  8 x2 Substitute x  2 and y  1 into equation A , B , or C . A 122  2112  z  4 z0

The solution set is 512, 1, 026.

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Chapter 3 Systems of Linear Equations and Inequalities

Section 3.7

Solving Systems of Linear Equations by Using Matrices

Key Concepts

Examples

A matrix is a rectangular array of numbers displayed in rows and columns. Every number or entry within a matrix is called an element of the matrix. The order of a matrix is determined by the number of rows and number of columns. A matrix with m rows and n columns is an m ⫻ n matrix.

Example 1

A system of equations written in standard form can be represented by an augmented matrix consisting of the coefficients of the terms of each equation in the system.

Example 2

[1 2 5] is a 1 ⫻ 3 matrix (a row matrix). c

⫺1 1

8 d is a 2 ⫻ 2 matrix (a square matrix). 5

4 c d is a 2 ⫻ 1 matrix (a column matrix). 1

The augmented matrix for 4x ⫹ y ⫽ ⫺12 x ⫺ 2y ⫽

The Gauss-Jordan method can be used to solve a system of equations by using the following elementary row operations on an augmented matrix. 1. Interchange two rows. 2. Multiply every element in a row by a nonzero real number. 3. Add a multiple of one row to another row. These operations are used to write the matrix in reduced row echelon form. 1 c 0

0 a ` d 1 b

This represents the solution, x ⫽ a and y ⫽ b.

6

c

is

4 1

1 ⫺12 ` d ⫺2 6

Example 3 Solve the system from Example 2 by using the GaussJordan method. R1 3 R2

c

1 4

⫺2 6 ` d 1 ⫺12

⫺4R1 ⫹ R2 1 R2

c

1 0

⫺2 6 ` d 9 ⫺36

1 9 R2

c

1 0

⫺2 6 ` d 1 ⫺4

c

1 0

0 ⫺2 ` d 1 ⫺4

1 R2

2R2 ⫹ R1 1 R1

From the reduced row echelon form of the matrix we have x ⫽ ⫺2 and y ⫽ ⫺4. The solution set is 51⫺2, ⫺426 .

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Review Exercises

Chapter 3

Review Exercises y

8. y  12x  2

Section 3.1

5 4 3

4x  8y  8

1. Determine if the ordered pair is a solution to the system.

2 1

5x  7y  4

5 4 3 2 1 1 2

1 y x1 2 a. (2, 2)

3 9. y  x  4 4

12. 6x  y  5

1 7 x y 4 4

5x  y  3

1

2

3 4

5

x

For Exercises 13–22, solve the systems by using the addition method.

3 4 5

13.

5 4 3

15. 3x  4y  1

2

3 4

5

x

3

17.

4 5

y

4 5

3x  4y  10

2

16. 3x  y  1

2x  5y  1

1 1 x  y   3 3

2y  3x  8

18. 3x  y  16 31x  y2  y  2x  2

2 1

3

14. 4x  3y  5

6x  4y  4

5 4 3

5 4 3 2 1 1 2

2 3 x y1 5 5 2 1 x y 3 3

y

2 1

3x  y  2

3 6 x 11 11

Section 3.3

2 1

5 4 3 2 1 1 2

y

11. 4x  y  7

y 5 4 3

5 4 3 2 1 1 2

10. 3x  11y  9

x  2y  6

For Exercises 5–8, solve the system by graphing.

7. 6x  2y  4

x

For Exercises 9–12, solve the systems by using the substitution method.

4. Lines with different slopes intersect in one point.

y  x  5

5

Section 3.2

3. Parallel lines form an inconsistent system.

6. y  2x  7

3 4

4 5

b. (2, 2)

2. An inconsistent system has one solution.

g 1x2  2x  4

2

3

For Exercises 2–4, answer true or false.

5. f1x2  x  1

1

1

2

3 4

5

x

19. 1y  4x2  2x  9 20. 14x  352  3y 2x  2y  10 21. 0.4x  0.3y 

1x  152  y

1.8 22. 0.02x  0.01y  0.11

0.6x  0.2y  1.2

0.01x  0.04y 

0.26

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Section 3.4 23. Melinda invested twice as much money in an account paying 5% simple interest as she did in an account paying 3.5% simple interest. If her total interest at the end of 1 yr is $303.75, find the amount she invested in the 5% account. 24. A school carnival sold tickets to ride on a Ferris wheel. The charge was $1.50 for adults and $1.00 for students. If 54 tickets were sold for a total of $70.50, how many of each type of ticket were sold?

b. Write a linear function describing the total cost for x min of long-distance calls from Company 2. c. How many minutes of long-distance calls would result in equal cost for both offers? 28. Two angles are complementary. One angle measures 6 more than 5 times the measure of the other. What are the measures of the two angles?

Section 3.5 For Exercises 29–34, graph the solution set. 29. 2x 7 y  5

30. 2x  8  3y y

y

3 2 1

7 6 5 4 3 2 1

5 4 3 2 1 1

1 2

3 4

5

5 4 3 2 1 1

1 2

3 4

5

x

5 4 3 2 1 1

2 3

2 3

4 5

4 5

1 33. x  y 2

34. x 6

a. Write a linear function describing the total cost for x min of long-distance calls from Company 1.

2

x

2 y 5

y

Company 2: $12.95 per month, plus $0.08 per minute for long-distance calls

5

y 5 4 3 2 1

5 4 3 2 1

y

5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

3 4

32. x  2

y

Company 1: $9.95 per month, plus $0.10 per minute for long-distance calls

1

x

6 7

31. x 7 3

27. Two phone companies offer discount rates to students.

5

4 5

x

2 3

26. It takes a pilot 134 hr to travel with the wind from Jacksonville, Florida, to Myrtle Beach, South Carolina. Her return trip takes 2 hr flying against the wind. What is the speed of the wind and the speed of the plane in still air if the distance between Jacksonville and Myrtle Beach is 280 mi?

3 4

2 3

5 4 3 2 1 1

25. How many liters of 20% saline solution must be mixed with 50% saline solution to produce 16 L of a 31.25% saline solution?

1 2

5 4 3 2 1 1 2 3 4 5

x

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

x

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For Exercises 35–38, graph the system of inequalities. 35. 2x  y 7 2

and

2x  y  2

y 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

36. 3x  y  6

1 2 3 4 5

3x  y 6 6

or

y

Let x represent the number of acres of orange trees. Let y represent the number of acres of grapefruit trees.

5 4 3 2 1 1 2 3 4 5

or

1 2 3 4 5

x

y  x y

5 4 3 2 1 1 2 3 4 5

y  0,

c. Write an inequality that expresses the fact that the farmer wants to plant at least 4 times as many orange trees as grapefruit trees.

1 2 3 4 5

x

d. Sketch the inequalities in parts (a)–(c) to find the feasible region for the farmer’s distribution of orange and grapefruit trees. y

and

2 y x4 3

y 6 5 4 3 2 1 3 2 1 1 2 3 4

a. Write two inequalities that express the fact that the farmer cannot use a negative number of acres to plant orange and grapefruit trees. b. Write an inequality that expresses the fact that the total number of acres used for growing orange and grapefruit trees is at most 100.

5 4 3 2 1

38. x  0,

39. Suppose a farmer has 100 acres of land on which to grow oranges and grapefruit. Furthermore, because of demand from his customers, he wants to plant at least 4 times as many acres of orange trees as grapefruit trees.

x

5 4 3 2 1

37. y  x

307

100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 60 70 80 90 100

1

2

3 4 5 6 7

x

x

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Chapter 3 Systems of Linear Equations and Inequalities

Section 3.6 For Exercises 40–43, solve the systems of equations. If a system does not have a unique solution, label the system as either dependent or inconsistent. 40.

42.

5x  5y  5z  30

41. 5x  3y  z  5

x  y  z  2

x  2y  z  6

10x  6y  2z  4

x  2y  z  8

x y z

4

53. c

2x  5y

8

1 D £ 4 3

1 4

3 1 ` d 1 6

2 1 2

0 3 1 † 0§ 2 5

a. Write the matrix obtained by multiplying the first row by 4 and adding the result to row 2. b. Using the matrix obtained in part (a), write the matrix obtained by multiplying the first row by 3 and adding the result to row 3. For Exercises 57–60, solve the system by using the Gauss-Jordan method. 57. x  y 

For Exercises 47–50, determine the order of each matrix.

13

0 5 0 † 2§ 1 8

56. Given the matrix D

Section 3.7

49. 30

0 1 0

b. Write the matrix obtained by multiplying the first row by 4 and adding the result to row 2.

46. The smallest angle in a triangle measures 9° less than the middle angle. The largest angle is 26° more than 3 times the measure of the smallest angle. Find the measure of the three angles.

4 0 6

1 54. £ 0 0

a. What is the element in the second row and first column?

45. Three pumps are working to drain a construction site. Working together, the pumps can drain 950 gal/hr of water. The slowest pump drains 150 gal/hr less than the fastest pump. The fastest pump drains 150 gal/hr less than the sum of the other two pumps. How many gallons can each pump drain per hour?

2 £ 5 1



C c

44. The perimeter of a right triangle is 30 ft. One leg is 2 ft longer than twice the shortest leg. The hypotenuse is 2 ft less than 3 times the shortest leg. Find the lengths of the sides of this triangle.

47.

9 0 ` d 1 3

55. Given the matrix C

2y  3z  2

2x  4y  6z  12

1 0

 4z  5

43. 3x

x  2y  3z  6

For Exercises 53–54, write a corresponding system of equations from the augmented matrix.

1 3 § 10

4

5 48. £ 9 0

164

6 2§ 3

For Exercises 51–52, set up the augmented matrix. 3 51. x  y   x  y  1

52.

x y z

4

2x  y  3z 

8

2x  2y  z  9

58.

x  y  1 59.

7 50. £ 12 § 4

3

4x  3y 

6

12x  5y  6

x  y  z  4 2x  y  2z 

9

x  2y  z 

5

60.

x y z

4

2x  y  3z 

8

2x  2y  z  9

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Test

Test

Chapter 3

1. Determine if the ordered pair 1 14, 22 is a solution to the system.

6. Solve the system by graphing. f1x2  x  3

4x  3y  5 12x  2y 

y 5 4 3

3 g1x2   x  2 2

7

2 1 5 4 3 2 1 1 2

For Exercises 2–4, match each figure with the appropriate description, a, b, or c.

2

3 4

5

x

3 4 5

a. The system is consistent and dependent. There are infinitely many solutions.

7. Solve the system by using the substitution method.

b. The system is consistent and independent. There is one solution.

3x  5y  13

c. The system is inconsistent and independent. There are no solutions. 2.

1

3.

yx9 8. Solve the system by using the addition method.

y

y

6x  8y  5 3x  2y  1 x

x

For Exercises 9–13, solve the system of equations. 9. 7y  5x  21 9y  2x  27

4.

y

11.

1 1 17 x y 5 2 5 1 1 1x  22   y 4 6

x

10.

3x  5y  7 18x  30y  42

12. 4x  5  2y y  2x  4

13. 0.03y  0.06x  0.3 0.4x  2  0.5y 14. Graph the solution set 2x  5y  10.

5. Solve the system by graphing. 4x  2y  4 3x  y 

7

y 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

1

2

3 4

5

x

y 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

x

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Chapter 3 Systems of Linear Equations and Inequalities

For Exercises 15–16, graph the solution set. 15. x ⫹ y 6 3 and 3x ⫺ 2y 7 ⫺6

For Exercises 18–19, solve the system. 18. 2x ⫹ 2y ⫹ 4z ⫽ ⫺6

y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

3x ⫹ y ⫹ 2z ⫽ 29 x ⫺ y ⫺ z ⫽ 44 x

1 2 3 4 5

2x ⫽ 11 ⫹ y ⫺ z x ⫹ 21y ⫹ z2 ⫽ 8 y

16. 5x ⱕ 5 or x ⫹ y ⱕ 0

20. How many liters of a 20% acid solution should be mixed with a 60% acid solution to produce 200 L of a 44% acid solution?

5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1 2

3 4

5

x

⫺2 ⫺3 ⫺4 ⫺5

17. After menopause, women are at higher risk for hip fractures as a result of low calcium. As early as their teen years, women need at least 1200 mg of calcium per day (the USDA recommended daily allowance). One 8-oz glass of skim milk contains 300 mg of calcium, and one antacid tablet (regular strength) contains 400 mg of calcium. Let x represent the number of 8-oz glasses of milk that a woman drinks per day. Let y represent the number of antacid tablets (regular strength) that a woman takes per day. a. Write two inequalities that express the fact that the number of glasses of milk and the number of antacid tablets taken each day cannot be negative. b. Write a linear inequality in terms of x and y for which the daily calcium intake is at least 1200 mg. c. Graph the inequalities. y 5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

19. 21x ⫹ z2 ⫽ 6 ⫹ x ⫺ 3y

21. Two angles are complementary. Two times the measure of one angle is 60° less than the measure of the other. Find the measure of each angle. 22. Working together, Joanne, Kent, and Geoff can process 504 orders per day for their business. Kent can process 20 more orders per day than Joanne can process. Geoff can process 104 fewer orders per day than Kent and Joanne combined. Find the number of orders that each person can process per day. 23. Write an example of a 3 ⫻ 2 matrix. 24. Given the matrix A 1 A⫽ £ 4 ⫺5

2 0 ⫺6

1 ⫺3 1 † ⫺2 § 3 0

a. Write the matrix obtained by multiplying the first row by ⫺4 and adding the result to row 2. b. Using the matrix obtained in part (a), write the matrix obtained by multiplying the first row by 5 and adding the result to row 3. For Exercises 25–26, solve by using the Gauss-Jordan method.

1 2 3 4 5

x

25. 5x ⫺ 4y ⫽ 34 x ⫺ 2y ⫽ 8

26.

x⫹ y⫹z⫽1 2x ⫹ y

⫽0

⫺2y ⫺ z ⫽ 5

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Cumulative Review Exercises

Cumulative Review Exercises

Chapters 1–3 1. Simplify.

2332  817  52 4

2. Simplify.

734  21w  52  312w  12 4  20

For Exercises 14–15, graph the equations. 1 14. y   x  4 3

15. x  2 y

y

For Exercises 3–5, solve the equation. 3. 512x  12  213x  12  7  218x  12 4.

1 3 1 1a  22  12a  12   2 4 6

For Exercises 6–11, solve the inequality. Write the answer in interval notation if possible. 6. 3y  21y  12 6 8

8. 4x 7 16 9. 0 

5 4 3

2 1

2 1

5 4 3 2 1 1 2

5. 4  0 2x  3 0  9

7. 4x 7 16 or

5 4 3

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

4 5

4 5

1

2

3 4

5

x

16. Find the slope of the line passing through the points 14, 102 and 16, 102. 17. Find an equation for the line that passes through the points 13, 82 and 12, 42. Write the answer in slope-intercept form.

6x  3  9

and 6x  3  9

18. Solve the system by using the addition method.

3x  9 5 6

2x  3y  6 1 3 x y1 2 4

10. 0x  4 0  1 6 11 11. 4 6 02x  4 0

19. Solve the system by using the substitution method. 2x  y  4

12. Graph the solution set. x  5y  5

y  3x  1

y 5 4 3 2 1 5 4 3 2 1 1 2 3

1 2

3 4

5

x

4 5

13. Identify the slope and the x- and y-intercepts of the line 5x  2y  15.

20. A child’s piggy bank contains 19 coins consisting of nickels, dimes, and quarters. The total amount of money in the bank is $3.05. If the number of quarters is 1 more than twice the number of nickels, find the number of each type of coin in the bank. 21. Two video clubs rent DVDs according to the following fee schedules: Club 1: $25 initiation fee plus $2.50 per DVD Club 2: $10 initiation fee plus $3.00 per DVD a. Write a linear function describing the total cost of renting x DVDs from club 1. b. Write a linear function describing the total cost of renting x DVDs from club 2. c. How many DVDs would have to be rented to make the cost for club 1 the same as the cost for club 2?

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24. Write an example of a 2 ⫻ 4 matrix.

22. Solve the system. 3x ⫽ 3 ⫺ 2y ⫺ 3z 4x ⫺ 5y ⫹ 7z ⫽ 1 2x ⫹ 3y ⫺ 2z ⫽ 6 23. Determine the order of the matrix. c

4 ⫺2

5 6

1 d 0

25. Solve the system by using the Gauss-Jordan method. 2x ⫺ 4y ⫽ ⫺2 4x ⫹ y ⫽

5

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Polynomials

CHAPTER OUTLINE 4.1 Properties of Integer Exponents and Scientific Notation 314 4.2 Addition and Subtraction of Polynomials and Polynomial Functions 323 4.3 Multiplication of Polynomials 334 4.4 Division of Polynomials 343 Problem Recognition Exercises: Operations on Polynomials

353

4.5 Greatest Common Factor and Factoring by Grouping 354 4.6 Factoring Trinomials 362 4.7 Factoring Binomials 376 Problem Recognition Exercises: Factoring Summary

385

4.8 Solving Equations by Using the Zero Product Rule 388 Group Activity: Investigating Pascal’s Triangle

402

1

3

2

Chapter 4 In this chapter, we study addition, subtraction, multiplication, and division of polynomials, along with an important operation called factoring. Are You Prepared? To prepare for this chapter, use the crossword puzzle to review the properties of real numbers and exponents from Chapter R. Across 1. Simplify and give the word name for 1⫺22 4. 6. a(b ⫹ c) ⫽ ab ⫹ ac illustrates the ________ property. 1 ⫺1 1 ⫺2 8. Simplify. a b ⫹ a b 6 3 9. Simplify and give the word name for y 0. 3r 4 2 10. Simplify. a 5 b ⴢ 12r2 3 r

4

5 7

6

8

9

Down 2. Simplify. 31x ⫹ 42 ⫺ 21x ⫹ 22 3. Simplify and give the word name for ⫺22. 4. True or false: 1x ⫹ 221x ⫹ 32 ⫽ 1x ⫹ 321x ⫹ 22 16d 4 5. Simplify. 2d 3 7. Simplify. ⫺415 ⫹ t2 ⫹ 51t ⫹ 32 ⫹ 10

10

313

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Chapter 4 Polynomials

Section 4.1

Properties of Integer Exponents and Scientific Notation

Concepts

1. Simplifying Expressions with Exponents

1. Simplifying Expressions with Exponents 2. Scientific Notation

In Section R.3, we learned that exponents are used to represent repeated multiplication. The following properties of exponents (Table 4-1) are often used to simplify algebraic expressions. Table 4-1

Properties of Exponents*

Description

Property

Example

b ⴢb b

b ⴢb b  b6

b ⴢ b  1b ⴢ b21b ⴢ b ⴢ b ⴢ b2  b6

b5  b52 b2  b3

b5 bⴢbⴢbⴢbⴢb  bⴢb b2  b3

1ab2 m  ambm

1ab2 3  a3b3

a m am a b  m b b

a 3 a3 a b  3 b b

1ab2 3  1ab21ab21ab2  1a ⴢ a ⴢ a21b ⴢ b ⴢ b2  a3b3

m

Multiplication of like bases

n

mn

bm  bmn bn

Division of like bases

1bm 2 n  bmⴢn

Power rule Power of a product Power of a quotient

2

4

24

1b4 2 2  b4ⴢ2  b8

Expanded Form

2

4

1b4 2 2  1b ⴢ b ⴢ b ⴢ b21b ⴢ b ⴢ b ⴢ b2  b8

a 3 a a a a b a ba ba b b b b b aⴢaⴢa a3   3 bⴢbⴢb b

*Assume that a and b are real numbers 1b  02 and that m and n represent integers.

In addition to the properties of exponents, two definitions are used to simplify algebraic expressions.

DEFINITION b 0 and b ⴚ n Let n be an integer, and b be a real number such that b  0. *1.

b0  1

Example:

50  1

n

1 1 1 3 1 1 2. bn  a b  n Example: 43  a b  3 or b b 4 64 4 3. From the definition of bn we also have: a n b n bn a b  a b  n for a  0, b  0. a b a 3 2 7 2 72 49 Example: a b  a b  2 or 7 3 9 3 *Note: The value of 00 is not defined by definition 1 because the base, b, must not equal 0.

The definition of b0 is consistent with the properties of exponents. For example, if b is a nonzero real number and n is an integer, then bn 1 bn The expression b0  1 n

b  bnn  b0 bn

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315

Properties of Integer Exponents and Scientific Notation

The definition of bn is also consistent with the properties of exponents. If b is a nonzero real number, then b3 bⴢbⴢb 1   2 bⴢbⴢbⴢbⴢb b5 b The expression b2 

1 b2

b3  b35  b2 b5

Simplifying Expressions with Exponents

Example 1

Simplify the expressions a. 122 4

c. 24

b. 24

d. 17x2 0

e. 7x0

4. 18y2 0

5. 8y 0

Solution:

a. 122 4  122122122122  16 b. 2  1 ⴢ 24 4

 1 ⴢ 12 ⴢ 2 ⴢ 2 ⴢ 22

 16

c. 24  1 ⴢ 124 2  1 ⴢ a  1 ⴢ 

1 2ⴢ2ⴢ2ⴢ2

1 16

d. 17x2 0  1 e.

1 b 24

because b0  1

7x0  7 ⴢ x0  7 ⴢ 1  7

Skill Practice Simplify the expressions. 1. 132 2

Example 2

3. 32

2. 32

Using Properties of Exponents

Simplify the expressions. Write the final answer with positive exponents only. a. x3x5x2

b.

y7 y4

c. 1b2 2 5 Answers 1. 9

2. 9

4. 1

5. 8

3. 

1 9

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Chapter 4 Polynomials

Solution: a. x3x5x2  x35 122

Multiply like bases by adding exponents.

 x6

Simplify.

 y74

Divide like bases by subtracting exponents.

y

Simplify.

7

b.

y

y4 3

c. 1b2 2 5

 b2152

Apply the power rule.

10

b 

Multiply the exponents.

1 b10

Write the answer with positive exponents.

Skill Practice Simplify the expressions. Write the final answer with positive exponents only. 6. w 7w 3w

7.

Example 3

t 11 t6

8. 1p 4 2 2

Simplifying an Expression with Exponents

Simplify the expression.

1 3 a b  122 2  30 5

Solution: 1 3 a b  122 2  30 5 1 2  53  a b  30 2

Simplify negative exponents.

1 1 4

Evaluate expressions with exponents.



500 1 4   4 4 4

Write the expressions with a common denominator.



503 4

Simplify.

 125 

Skill Practice Simplify the expression. 2 1 1 0 9. a b  41  a b 3 4

Answers 6. w 5

7. t 5

8.

1 p8

9.

3 4

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Section 4.1

Example 4

Properties of Integer Exponents and Scientific Notation

Simplifying an Expression with Exponents

Simplify the expression. Write the answer with positive exponents only. 12a7b⫺4 2 3 14a3b⫺2 2 2

Solution: 12a7b⫺4 2 3 14a3b⫺2 2 2 ⫽

23a21b⫺12 42a6b⫺4

Apply the power rule.

⫽

8a21b⫺12 16a6b⫺4

Simplify the coefficients.

⫽

8a21⫺6b⫺12⫺ 1⫺42 16

Divide like bases by subtracting exponents.

1

8a15b⫺8 ⫽ 16

Simplify.

2

⫽

15

a 2b8

Simplify negative exponents.

Skill Practice Simplify the expression. Write the final answer with positive exponents only. 10.

13x 3y⫺4 2 2 1x⫺2y2 ⫺4

Example 5

Simplifying an Expression with Exponents

Simplify the expression. Write the answer with positive exponents only. 4xy⫺3 a

8x2 ⫺2 b 3x5y2

Solution: 4xy⫺3 a

8x2 ⫺2 8 ⫺2 ⫺3 b ⫽ 4xy ⴢ a b 3x5y2 3x3y2

Simplify within parentheses.

⫽

4xy⫺3 8⫺2 ⴢ ⫺2 ⫺6 ⫺4 1 3 x y

Raise the expression in parentheses to the ⫺2 power.

⫽

4x 32x6y4 ⴢ y3 82

Simplify negative exponents.

⫽

4 ⴢ 9 ⴢ x ⴢ x6 ⴢ y4 3

64y

Multiply the fractions and simplify the expressions 32 and 82.

1

4 ⴢ 9 ⴢ x1⫹6y4⫺3 ⫽ 64

Add the exponents on x. Subtract the exponents on y.

16

9x7y ⫽ 16

Answer 10.

Simplify.

9 x 2y 4

317

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Chapter 4 Polynomials

Skill Practice Simplify the expression. Write the answer with positive exponents only. 11. 9m5t a

3m2t 3 3 b 4t 6

2. Scientific Notation Scientists in a variety of fields often work with very large or very small numbers. For instance, the distance between the Earth and the Sun is approximately 93,000,000 mi. The mass of an electron is 0.000 000 000 000 000 000 000 000 000 000 911 kg. Scientific notation was devised as a shortcut method of expressing very large and very small numbers. The principle behind scientific notation is to use a power of 10 to express the magnitude of the number. For example, a number such as 50,000 can be written as 5  10,000 or equivalently as 5.0  104. Similarly, the 1 number 0.0035 is equal to 3.5  1000 or, equivalently, 3.5  103.

DEFINITION Scientific Notation

A number expressed in the form a  10n, where 1  0 a 0 6 10 and n is an integer, is said to be written in scientific notation.

Consider the following numbers in scientific notation: The distance between the Sun and the Earth:

93,000,000 mi  9.3  107 mi 7 places

The mass of an electron:

0.000 000 000 000 000 000 000 000 000 000 911 kg  9.11  1031 kg

31 places

In each case, the power of 10 corresponds to the number of place positions that the decimal point is moved. The power of 10 is sometimes called the order of magnitude (or simply the magnitude) of the number. The order of magnitude of the distance between the Earth and Sun is 107 mi (tens of millions). The mass of an electron has an order of magnitude of 1031 kg. Example 6

Writing Numbers in Scientific Notation

Fill in the table by writing the numbers in scientific notation or standard notation as indicated.

Answer 11. 

64m11t 10 3

Quantity

Standard Notation

Number of NASCAR fans

75,000,000 people

Width of an influenza virus

0.000000001 m

Scientific Notation

Cost of hurricane Katrina

$1.25  1011

Probability of winning the Florida state lottery

4.35587878  108

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Properties of Integer Exponents and Scientific Notation

319

Solution: Quantity

Standard Notation

Scientific Notation

Number of NASCAR fans

75,000,000 people

7.5 ⫻ 107 people

Width of an influenza virus

0.000000001 m

1.0 ⫻ 10⫺9 m

Cost of hurricane Katrina

$125,000,000,000

$1.25 ⫻ 1011

Probability of winning the Florida state lottery

0.0000000435587878

4.35587878 ⫻ 10⫺8

Skill Practice Rewrite each number in either scientific notation or standard notation. 12. 2,600,000

13. 0.00088

14. ⫺5.7 ⫻ 10⫺8

15. 1.9 ⫻ 10 5

Calculator Connections Topic: Using Scientific Notation on a Calculator Calculators use scientific notation to display very large or very small numbers. To enter scientific notation in a calculator, try using the key or the key to express the power of 10. Notice that E in the display of the calculator introduces the power of 10.

Example 7

Applying Scientific Notation

a. During a recent economic crisis, the U.S. government passed a bill to “bail out” troubled financial institutions that had a large number of mortgage-related assets. The government committed an estimated $750,000,000,000 for the “bailout.” How much money does this represent per person if there were 300,000,000 people living in the United States at that time? b. The mean distance between the Earth and the Andromeda Galaxy is approximately 1.8 ⫻ 106 light-years. Assuming that 1 light-year is 6.0 ⫻ 1012 mi, what is the distance in miles to the Andromeda Galaxy?

Solution: a. Express each value in scientific notation. Then divide the total amount to be paid off by the number of people. 7.5 ⫻ 1011 3.0 ⫻ 108 ⫽a

7.5 1011 b⫻a 8b 3.0 10

Divide 7.5 by 3.0 and subtract the powers of 10.

⫽ 2.5 ⫻ 103 In standard notation, this amounts to $2500 per person.

Answers 12. 2.6 ⫻ 10 6 14. ⫺0.000000057

13. 8.8 ⫻ 10⫺ 4 15. 190,000

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Chapter 4 Polynomials

b. Multiply the number of light-years by the number of miles per light-year. 11.8 ⫻ 106 216.0 ⫻ 1012 2

⫽ 11.8216.02 ⫻ 1106 211012 2

Calculator Connections

⫽ 10.8 ⫻ 1018

Use a calculator to check the solutions to Example 7.

Multiply 1.8 and 6.0 and add the powers of 10. The number 10.8 ⫻ 1018 is not in “proper” scientific notation because 10.8 is not between 1 and 10.

⫽ 11.08 ⫻ 101 2 ⫻ 1018

Rewrite 10.8 as 1.08 ⫻ 101.

⫽ 1.08 ⫻ 1101 ⫻ 1018 2

Apply the associative property of multiplication.

⫽ 1.08 ⫻ 1019

The distance between the Earth and the Andromeda Galaxy is 1.08 ⫻ 1019 mi. Skill Practice

Answers 16. Approximately 2.5 ⫻ 105 pennies 17. 4.56 ⫻ 1013 mi

Section 4.1

16. The thickness of a penny is 6.0 ⫻ 10⫺2 in. How many pennies would have to be stacked to equal the height of the Empire State Building (1.5 ⫻ 104 in.)? 17. The distance from the Earth to the “nearby’’ star, Barnard’s Star, is 7.6 light-years (where 1 light-year ⫽ 6.0 ⫻ 1012 mi). How many miles away is Barnard’s Star?

Practice Exercises • Practice Problems • Self-Tests • NetTutor

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Study Skills Exercise 1. Define the key term scientific notation.

Concept 1: Simplifying Expressions with Exponents 2. Write the expressions in expanded form and simplify.

3.

b4 ⴢ b3 and 1b4 2 3

Write the expressions in expanded form and simplify. ab3 and (ab)3

For Exercises 4–9, write an example of each property. (Answers may vary.) 4. bn ⴢ bm ⫽ bn⫹m 7.

bn ⫽ bn⫺m bm

1b ⫽ 02

5. 1ab2 n ⫽ anbn a n an 8. a b ⫽ n b b

1b ⫽ 02

6. 1bn 2 m ⫽ bnm 0 9. b ⫽ 1

1b ⫽ 02

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321

Properties of Integer Exponents and Scientific Notation

For Exercises 10–28, simplify. (See Example 1.) 2 1 10. a b 3

1 1 11. a b 3

12. 31

13. 52

14. 82

15. 52

16. 82

17. 152 2

18. 182 2

1 3 19. a b 4

3 1 20. a b 8

3 4 21. a b 2

1 2 22. a b 9

2 3 23. a b 5

1 5 24. a b 2

25. 110ab2 0

26. 113x2 0

27. 10ab0

28. 13x0

For Exercises 29–80, simplify and write the answer with positive exponents only. (See Examples 2–5.) 138 136

57 53

29. y3 ⴢ y5

30. x4 ⴢ x8

31.

33. 1y2 2 4

34. 1z3 2 4

35. 13x2 2 4

36. 12y5 2 3

37. p3

38. q5

39. 710 ⴢ 713

40. 119 ⴢ 117 44. b1b8

41.

45.

49.

w3 w5 r r1 a3 b2

53. 24  22

32.

42.

t4 t8

43. a2a5

46.

s1 s

47.

50.

c4 d1

51. 16xyz2 2 0

52. 17ab3 2 0

55. 12  52

56. 42  22

54. 32  31

z6 z2

48.

2 2 1 2 1 0 1 1 2 0 1 2 4 1 3 2 2 0 57. a b  a b  a b 58. a b  a b  a b 59. a b  a b  a b 3 2 3 6 3 4 5 2 7 61.

p2q

62.

5 1

pq

m1n3 m4n2

63.

48ab10 32a4b3

w8 w3

4 0 2 2 9 1 60. a b  a b  a b 5 3 5 64.

25x2y12 10x5y7

65. 13x4y5z2 2 4

66. 16a2b3c2 2

67. 14m2n21m6n3 2

69. 1p2q2 3 12pq4 2 2

70. 1mn3 2 2 15m2n2 2

71. a

x2 3 2 b 15x y2 y

72. a

a 2 2 3 b 13a b 2 b2

75. a

2x6y5

76. a

6a2b3 2 b 5a1b

73.

18a2b2 2 4 116a3b7 2 2

77. a

2x3y0 6 5

4x y

2

b

74.

13x2y3 2 2 12xy4 2 3

78. a

a3b2c0 2 b a b c 1 2 3

3x2y4

79. 3xy5 a

3

b

2x4y 5 3

6x y

2

b

68. 16pq3 212p4q2

80. 7x3y4 a

3x1y5 3 2

9x y

b

3

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Concept 2: Scientific Notation 81. The European Organization for Nuclear Research (known by the acronym CERN) has built the world’s largest high-energy particle accelerator, called the Large Hadron Collider (LHC). Scientists hope the LHC will answer many open questions in physics. Write the following numbers in scientific notation. (See Example 6.) a. The LHC cost $8,000,000,000 to build. b. 3,000,000 DVDs worth of data will be produced each year. c. 14,000,000,000,000 electron volts (eV) of energy will be produced to smash the protons together. d. 1 eV is equivalent to 0.000 000 000 000 000 000 1602 joules (J). 82. Write the numbers in scientific notation. a. The estimated population of the United States in 2007 was approximately 292,600,000. b. The size of the smallest visible object in an optical microscope is 0.0000002 m. c. A trillion is defined as 1,000,000,000,000. 83. Write the numbers in standard notation. a. The Andromeda Galaxy contains at least 2 ⫻ 1011 stars. b. The diameter of a capillary is 4.0 ⫻ 10⫺6 m. c. The mean distance of Venus from the Sun is 1.082 ⫻ 1011 m. 84. Write the numbers in standard notation. a. At the end of a recent year, the Department of Energy’s inventory of high-level radioactive waste was approximately 3.784 ⫻ 105 m3. b. The diameter of a water molecule is 3 ⫻ 10⫺10 m. c. The distance a bullet will travel in one second when fired from a 0.22 caliber gun is 4.1 ⫻ 102 m. For Exercises 85–90, determine which numbers are in “proper” scientific notation. If the number is not in “proper” scientific notation, correct it. 85. 35 ⫻ 104

86. 0.469 ⫻ 10⫺7

87. 7.0 ⫻ 100

88. 8.12 ⫻ 101

89. 9 ⫻ 101

90. 6.9 ⫻ 100

For Exercises 91–98, perform the indicated operations and write the answer in scientific notation. 91. 16.5 ⫻ 103 215.2 ⫻ 10⫺8 2

92. 13.26 ⫻ 10⫺6 218.2 ⫻ 109 2

94. (3,400,000,000)(70,000,000,000,000)

95. 18.5 ⫻ 10⫺2 2 ⫼ 12.5 ⫻ 10⫺15 2

97. 1900,000,0002 ⫼ 1360,0002

98. 10.00000000022 ⫼ 18,000,0002

93. (0.0000024)(6,700,000,000) 96. 13 ⫻ 109 2 ⫼ 11.5 ⫻ 1013 2

99. If one H2O molecule contains 2 hydrogen atoms and 1 oxygen atom, and 10 H2O molecules contain 20 hydrogen atoms and 10 oxygen atoms, how many hydrogen atoms and oxygen atoms are contained in 6.02 ⫻ 1023 H2O molecules? (See Example 7.) 100. The star named Alpha Centauri is 4.3 light-years from the Earth. If 1 light-year is approximately 6 ⫻ 109 mi, how far is Alpha Centauri? 101. If the county of Queens, New York, has a population of approximately 2,200,000 and the area is 110 mi2, how many people are there per square mile? (See Example 7.)

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102. If the county of Catawba, North Carolina, has a population of approximately 150,000 and the area is 400 mi2, how many people are there per square mile? 103. According to the Federal Emergency Management Agency (FEMA), the annual loss due to earthquakes in California is approximately $3.5 ⫻ 109. If this is representative as a yearly average, find the loss over 15 yr. 104. Avogadro’s number Na ⫽ 6.02 ⫻ 1023 is the number of atoms in 1 mole of an element. a. How many atoms are in 5 moles of carbon-12? b. If 75 g of carbon-12 has 4.515 ⫻ 1025 atoms, how many moles is this?

Expanding Your Skills 105. A 20-yr-old starts a savings plan for her retirement. She will put $20 per month into a mutual fund that she hopes will average 6% growth annually. a. If she plans to retire at age 65, for how many months will she be depositing money? b. By age 65, how much money will she have deposited? c. The value of an account built in this fashion is given by A ⫽ P ⴢ c a1 ⫹

r N 12 b ⫺ 1 d ⴢ a1 ⫹ b r 12

where A is the final amount of money in the account, P is the amount of the monthly deposit, and N is the number of months. Use a calculator to find the total amount in the woman’s retirement account at age 65. For Exercises 106–111, simplify each expression. Assume that a and b represent positive integers and x and y are nonzero real numbers. 106. xa⫹1xa⫹5

109.

107. ya⫺5ya⫹7

x3a⫺3 xa⫹1

110.

108.

x3b⫺2yb⫹1

111.

x2b⫹1y2b⫹2

y2a⫹1 ya⫺1 x2a⫺2ya⫹3 xa⫹4ya⫺3

Addition and Subtraction of Polynomials and Polynomial Functions

Section 4.2

1. Polynomials: Basic Definitions

Concepts

One commonly used algebraic expression is called a polynomial. A polynomial in x is defined as a finite sum of terms of the form axn, where a is a real number and the exponent n is a whole number. For each term, a is called the coefficient, and n is called the degree of the term. For example:

1. Polynomials: Basic Definitions 2. Addition of Polynomials 3. Subtraction of Polynomials 4. Polynomial Functions

Term (Expressed in the Form axn) Coefficient 5

3x

Degree

3

5

14

1

14

7

rewrite as 7x0

7

0

1 p 2

rewrite as 12p1

1 2

1

x

14

rewrite as 1x

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If a polynomial has exactly one term, it is called a monomial. A two-term polynomial is called a binomial, and a three-term polynomial is called a trinomial. Usually the terms of a polynomial are written in descending order according to degree. In descending order, the highest-degree term is written first and is called the leading term. Its coefficient is called the leading coefficient. The degree of a polynomial is the greatest degree of all its terms. Thus, the leading term determines the degree of the polynomial.

Expression

Descending Order

2x9

Monomials

2

9

49

49

0

10y  7y

2

7y  10y

7

2

2 6 b 3

2  b6 3

2  3

1

w  2w3  9w6

9w6  2w3  w

2.5a4  a8  1.3a3

a8  2.5a4  1.3a3

2

Trinomials

Degree of Polynomial

2x9

49 Binomials

Leading Coefficient

9

6

1

8

Polynomials may have more than one variable. In such a case, the degree of a term is the sum of the exponents of the variables contained in the term. For example, the term 2x3y4z has degree 8 because the exponents applied to x, y, and z are 3, 4, and 1, respectively. The following polynomial has a degree of 12 because the highest degree of its terms is 12. 11x4y3z





5x3y2z7

2x2y



7

degree

degree

degree

degree

8

12

3

0

2. Addition of Polynomials To add or subtract two polynomials, we combine like terms. Recall that two terms are like terms if they each have the same variables and the corresponding variables are raised to the same powers.

Example 1

Adding Polynomials

Add the polynomials. a. 13t 3  2t 2  5t2  1t 3  6t2

2 1 4 1 b. a w2  w  b  a w2  8w  b 3 8 3 4

Solution:

a. 13t 3  2t 2  5t2  1t 3  6t2

 3t 3  t 3  2t 2  15t2  16t2

Group like terms.

 4t 3  2t 2  11t

Add like terms.

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2 1 4 1 b. a w2  w  b  a w2  8w  b 3 8 3 4 2 4 1 1  w2  w2  1w2  8w   a b 3 3 8 4

Group like terms.

6 1 2  w2  7w  a  b 3 8 8

Add fractions with common denominators.

 2w2  7w 

1 8

Simplify.

Skill Practice Add the polynomials.

1. 12x 2  5x  22  16x 2  8x  82 1 1 3 1 2. a m 2  2m  b  a m2  7m  b 4 3 4 12

Example 2

Adding Polynomials 1a2b  7ab  62  15a2b  2ab  72

Add the polynomials.

Solution: Polynomials can be added vertically. Be sure to line up the like terms. a2b  7ab  6  5a2b  2ab  7 6a2b  5ab  1

Add like terms.

Skill Practice Add the polynomials.

3. 15a 2b  6ab 2  22  12a2b  ab2  32

3. Subtraction of Polynomials Subtraction of two polynomials is similar to subtracting real numbers. Add the opposite of the second polynomial to the first polynomial. The opposite (or additive inverse) of a real number a is a. Similarly, if A is a polynomial, then A is its opposite.

Example 3

Finding the Opposite of a Polynomial

Find the opposite of the polynomials. a. 5a  2b  c

b. 5.5y4  2.4y3  1.1y

TIP: Notice that the sign

Solution: Expression

Opposite

Simplified Form

a. 5a  2b  c

15a  2b  c2

5a  2b  c

b. 5.5y4  2.4y3  1.1y

15.5y4  2.4y3  1.1y2

5.5y4  2.4y3  1.1y

Skill Practice Find the opposite of the polynomials. 4. 7z  6w

5. 2p  3q  r  1

of each term is changed when finding the opposite of a polynomial.

Answers 1. 8x 2  3x  10 1 1 2. m 2  5m  2 4 3. 3a 2b  5ab 2  1 4. 7z  6w 5. 2p  3q  r  1

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DEFINITION Subtraction of Polynomials

If A and B are polynomials, then A  B  A  1B2 .

Example 4

Subtracting Polynomials

Subtract the polynomials.

13x2  2x  52  14x2  7x  22

Solution:

13x2  2x  52  14x2  7x  22

 13x2  2x  52  14x2  7x  22

 3x2  14x2 2  2x  7x  152  122  x2  9x  7

Add the opposite of the second polynomial. Group like terms. Combine like terms.

Skill Practice Subtract the polynomials. 6. 16a 2  2a2  13a2  2a  32

Example 5

Subtracting Polynomials

Subtract the polynomials.

16x2y  2xy  52  1x2y  32

Solution:

16x2y  2xy  52  1x2y  32

Subtraction of polynomials can be performed vertically by vertically aligning like terms. Then add the opposite of the second polynomial. “Placeholders” (shown in red) may be used to help line up like terms. 6x2y  2xy  5 1x2y  0xy  32

Add the opposite.

6x2y  2xy  5  x2y  0xy  3 5x2y  2xy  8

Skill Practice Subtract the polynomials. 7. 17p 2q  62  12p2q  4pq  42

Example 6 Subtract

Subtracting Polynomials

1 4 3 2 1 x  x  2 4 5

from

3 4 1 2 x  x  4x 2 2

Solution: In general, to subtract a from b, we write b  a. Therefore, to subtract

Answers 6. 9a 2  4a  3 7. 5p 2q  4pq  10

1 4 3 2 1 x  x  2 4 5

from

3 4 1 2 x  x  4x 2 2

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we have 3 1 1 3 1 a x4 ⫹ x2 ⫺ 4xb ⫺ a x4 ⫺ x2 ⫹ b 2 2 2 4 5 3 1 1 3 1 ⫽ x4 ⫹ x2 ⫺ 4x ⫺ x4 ⫹ x2 ⫺ 2 2 2 4 5

Subtract the polynomials by adding the opposite of the second polynomial.

3 1 1 3 1 ⫽ x4 ⫺ x4 ⫹ x2 ⫹ x2 ⫺ 4x ⫺ 2 2 2 4 5

Group like terms.

3 3 1 2 1 ⫽ x4 ⫺ x4 ⫹ x2 ⫹ x2 ⫺ 4x ⫺ 2 2 4 4 5

Write like terms with a common denominator.

2 5 1 ⫽ x4 ⫹ x2 ⫺ 4x ⫺ 2 4 5

Combine like terms.

5 1 ⫽ x4 ⫹ x2 ⫺ 4x ⫺ 4 5

Simplify.

Skill Practice 1 1 1 1 3 8. Subtract p 3 ⫹ p 2 ⫹ p from p 3 ⫹ p 2 ⫺ p 2 3 2 3 4

4. Polynomial Functions A polynomial function is a function defined by a finite sum of terms of the form axn, where a is a real number and n is a whole number. For example, the functions defined here are polynomial functions: f1x2 ⫽ 3x ⫺ 8 g1x2 ⫽ 4x5 ⫺ 2x3 ⫹ 5x ⫺ 3 1 3 5 h1x2 ⫽ ⫺ x4 ⫹ x3 ⫺ 4x2 ⫹ x ⫺ 1 2 5 9 k1x2 ⫽ 7

17 ⫽ 7x0, which is of the form ax n, where n ⫽ 0 is a whole number2

The following functions are not polynomial functions: m1x2 ⫽

1 ⫺8 x

q1x2 ⫽ 0x 0 Example 7

1 a ⫽ x⫺1, the exponent ⫺1 is not a whole numberb x 1 0 x 0 is not of the form ax n 2

Evaluating a Polynomial Function

Given P1x2 ⫽ x3 ⫹ 2x2 ⫺ x ⫺ 2, find the function values. a. P1⫺32

b. P1⫺12

c. P102

d. P122

Answer 1 5 3 8. ⫺ p 3 ⫹ p 2 ⫺ p 6 12 2

327

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Solution: a.

P1x2  x3  2x2  x  2

P132  132 3  2132 2  132  2  27  2192  3  2  27  18  3  2  8

b. P112  112 3  2112 2  112  2  1  2112  1  2  1  2  1  2 0

c. P102  102 3  2102 2  102  2

P(x)

 2

d. P122  122 3  2122 2  122  2  8  2142  2  2 8822  12 The function values can be confirmed from the graph of y  P(x) (Figure 4-1).

(2, 12)

12 10 8 6 4 (1, 0) 2

x 5 4 3 2 1 1 2 3 4 5 2 (0, 2) 4 3 2 6 P(x)  x  2x  x  2 (3, 8) 8

Figure 4-1

Skill Practice Given: P 1x2  2x3  4x  6 9. Find P(0). 11. Find P(1).

Example 8

10. Find P(2). 12. Find P(2).

Using Polynomial Functions in an Application

The length of a rectangle is 4 m less than 3 times the width. Let x represent the width. Write a polynomial function P that represents the perimeter of the rectangle and simplify the result.

Solution: Let x represent the width. Then 3x  4 is the length. The perimeter of a rectangle is given by P  2L  2W. Thus, P1x2  213x  42  21x2

x

3x  4

 6x  8  2x  8x  8 Skill Practice Answers 9. P102  6 10. P122  30 11. P112  12 12. P122  18 13. P 1x2  x  14x  22  12x  32  7x  1

13. The longest side of a triangle is 2 ft less than 4 times the shortest side. The middle side is 3 ft more than twice the shortest side. Let x represent the shortest side. Find a polynomial function P that represents the perimeter of the triangle, and simplify the result.

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Using a Polynomial Function in an Application

The percent of male smokers in the United States has varied by year since 1990. The function defined by M(x)  0.003x2  0.257x  28.0 approximates the percent of male smokers, M(x), where x is the number of years since 1990. See Figure 4-2. a. Evaluate M(5) and interpret the meaning in the context of this problem.

Percent

Example 9

Addition and Subtraction of Polynomials and Polynomial Functions

30 25 20 15 10 5 0 0

Percent of Male Smokers, United States

M(x)  0.003x2  0.257x  28.0

5 10 20 15 Year (x  0 corresponds to 1990)

25

Source: U.S. National Center for Health Statistics.

Figure 4-2

b. Determine the percent of male smokers in the year 2008.

Solution: a. M152  0.003152 2  0.257152  28.0

Substitute x  5 into the function.

⬇ 26.6 M152 ⬇ 26.6 means that in the year 1995, 26.6% of males in the United States smoked. b. The year 2008 is 18 yr since 1990. Therefore, evaluate M(18). M1182  0.003 1182 2  0.2571182  28.0

Substitute x  18 into the function.

⬇ 22.4 M1182 ⬇ 22.4 means that in the year 2008, 22.4% of males in the United States smoked. Skill Practice The population of Kenya can be modeled by P(t)  0.017t 2  1.04t  38. The value t is the number of years since 2007 and P(t) is the population of Kenya (in millions). Source: Central Intelligence Agency 14. Evaluate P(3) and interpret its meaning. 15. If this trend continues, predict the population of Kenya for the year 2015.

Answers 14. M (3)  41.273 means that in the year 2010 (t  3 years since 2007), the population of Kenya will be 41.273 million. 15. M (8)  47.408; In the year 2015, the population of Kenya will be 47.408 million.

Section 4.2 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Polynomial

b. Coefficient

c. Degree of a term

d. Monomial

e. Binomial

f. Trinomial

g. Leading term

h. Leading coefficient

i. Degree of a polynomial

j. Like terms

k. Polynomial function

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Review Exercises For Exercises 2–6, simplify the expression. 2. a

6tv2 1 b 15t 3

1 2 2 0 4. a b  a b 4 3

3. 12ac2 215a1c4 2

5. 13.4  105 215.0  102 2

6.

6.2  103 3.1  105

Concept 1: Polynomials: Basic Definitions For Exercises 7–12, write the polynomial in descending order. Then identify the leading coefficient and the degree. 7. a2  6a3  a 10. 8  4y  y5  y2

8. 2b  b4  5b2

9. 6x2  x  3x4  1

11. 100  t 2

12. 51  s2

For Exercises 13–18, write a polynomial in one variable that is described by the following. (Answers may vary.) 13. A monomial of degree 5

14. A monomial of degree 4

15. A trinomial of degree 2

16. A trinomial of degree 3

17. A binomial of degree 4

18. A binomial of degree 2

Concept 2: Addition of Polynomials For Exercises 19–30, add the polynomials and simplify. (See Examples 1–2.) 19. 14m2  4m2  15m2  6m2

20. 13n3  5n2  12n3  2n2

21. 13x4  x3  x2 2  13x3  7x2  2x2

22. 16x3  2x2  122  1x2  3x  92

1 2 3 1 23. a w3  w2  1.8wb  a w3  w2  2.7wb 2 9 2 9

7 5 1 1 24. a2.9t4  t  b  a8.1t4  t  b 8 3 8 3

25. Add 19x2y  5xy  12 to 18x2y  xy  152 .

26. Add 1x3y2  5xy2 to 110x3y2  x2y  102 .

27. Add 16a2  7a  12 to 12a2  4a  82.

28. Add 18p3  12p  12 to 1p3  6p2  142.

29.

12x3  6x  8  13x3  5x2  4x 2

30.

8y4  8y3  6y2 9  14y4  5y3  10y  32

Concept 3: Subtraction of Polynomials For Exercises 31–36, write the opposite of the given polynomial. (See Example 3.) 31. 30y3

32. 2x2

33. 4p3  2p  12

34. 8t 2  4t  3

35. 11ab2  a2b

36. 23rs  4r  9s

For Exercises 37–46, subtract the polynomials and simplify. (See Examples 4–5.) 37. 113z5  z2 2  17z5  5z2 2

38. 18w4  3w2 2  112w4  w2 2

39. 13x3  3x2  x  62  1x3  x2  x  12

40. 18x3  6x  72  15x3  2x  42

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41. 1⫺3xy3 ⫹ 3x2y ⫺ x ⫹ 62 ⫺ 1⫺xy3 ⫺ xy ⫺ x ⫹ 12 43.

4t 3 ⫺ 6t 2 ⫺ 18 ⫺ 13t 3 ⫹ 7t 2 ⫹ 9t ⫺ 52

331

42. 1⫺8x2y2 ⫹ 6xy2 ⫹ 7xy2 ⫺ 15xy2 ⫺ 2xy ⫺ 42 44.

5w3 ⫺ 9w2 ⫹ 6w ⫹ 13 ⫺ 17w3 ⫺ 10w ⫺ 82

1 1 1 3 2 1 45. a a2 ⫺ ab ⫹ b2 ⫹ 3b ⫺ a⫺ a2 ⫹ ab ⫺ b2 ⫺ 5b 5 2 10 10 5 2 4 1 1 1 2 9 46. a a2 ⫺ ab ⫹ b2 ⫺ 7b ⫺ a a2 ⫺ ab ⫺ b2 ⫹ 1b 7 7 14 2 7 14 47. Subtract 19x2 ⫺ 5x ⫹ 12 from 18x2 ⫹ x ⫺ 152. (See Example 6.) 48. Subtract 1⫺x3 ⫹ 5x2 from 110x3 ⫹ x2 ⫺ 102. 49. Find the difference of 13x5 ⫺ 2x3 ⫹ 42 and 1x4 ⫹ 2x3 ⫺ 72. 50. Find the difference of 17x10 ⫺ 2x4 ⫺ 3x2 and 1⫺4x3 ⫺ 5x4 ⫹ x ⫹ 52.

Mixed Exercises For Exercises 51–74, add or subtract as indicated. Write the answers in descending order, if possible. 51. 18y2 ⫺ 4y3 2 ⫺ 13y2 ⫺ 8y3 2

52. 1⫺9y2 ⫺ 82 ⫺ 14y2 ⫹ 32

53. 1⫺2r ⫺ 6r4 2 ⫹ 1⫺r4 ⫺ 9r2

54. 1⫺8s9 ⫹ 7s 2 2 ⫹ 17s9 ⫺ s 2 2

55. 15xy ⫹ 13x2 ⫹ 3y2 ⫺ 14x2 ⫺ 8y2

56. 16p2q ⫺ 2q2 ⫺ 1⫺2p2q ⫹ 132

57. 111ab ⫺ 23b2 2 ⫹ 17ab ⫺ 19b2 2

58. 1⫺4x2y ⫹ 92 ⫹ 18x2y ⫺ 122

59. 冤2p ⫺ 13p ⫹ 52冥 ⫹ 14p ⫺ 62 ⫹ 2

60. ⫺1q ⫺ 22 ⫺ 冤4 ⫺ 12q ⫺ 32 ⫹ 5冥

61. 5 ⫺ 冤2m2 ⫺ 14m2 ⫹ 12冥

62. 冤4n3 ⫺ 1n3 ⫹ 42冥 ⫹ 3n3

63. 16x3 ⫺ 52 ⫺ 1⫺3x3 ⫹ 2x2 ⫺ 12x3 ⫺ 6x2

64. 19p4 ⫺ 22 ⫹ 17p4 ⫹ 12 ⫺ 18p4 ⫺ 102

65. 1⫺ab ⫹ 5a2b2 ⫺ 37ab2 ⫺ 2ab ⫺ 17a2b ⫹ 2ab2 2 4 66. 1m3n2 ⫹ 4m2n2 ⫺ 3⫺5m3n2 ⫺ 4mn ⫺ 17m2n ⫺ 6mn2 4 67. 18x3 ⫺ x2 ⫹ 32 ⫺ 35x2 ⫹ x ⫺ 14x3 ⫹ x ⫺ 22 4

68. 1y2 ⫹ 6y ⫺ 62 ⫺ 3 12y3 ⫺ 4y2 ⫺ 13y2 ⫹ y ⫹ 12 4

69.

12a2b ⫺ 4ab2 ⫺ ab ⫺14a2b ⫹ ab2 ⫺ 5ab2

70.

2x2 ⫺ 7xy ⫹ 3y2 ⫺ 19x2 ⫺ 10xy ⫺ y2 2

71.

⫺5x4 ⫺ 11x 2 ⫹6 4 3 ⫺ 1⫺5x ⫹ 3x ⫹ 5x2 ⫺ 10x ⫹ 52

72.

9z4 ⫹ 2z2 ⫹ 11 4 3 2 ⫺ 19z ⫺ 4z ⫹ 8z ⫺ 9z ⫺ 42

73.

⫺2.2p5 ⫺ 9.1p4 ⫹ 5.3p2 ⫺ 7.9p ⫹ ⫺ 6.4p4 ⫺ 8.5p3 ⫺ 10.3p2

74.

5.5w4 ⫹ 4.6w2 ⫺ 9.3w ⫺ 8.3 ⫹ 0.4w4 ⫺ 7.3w3 ⫺ 5.8w ⫹ 4.6

For Exercises 75–76, find the perimeter. 75.

2x3 ⫹ 6x

4x3 ⫺ 5x

6x3 ⫹ x

76.

5x ⫺ 2

2x ⫺ 6

3x ⫺ 1

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Concept 4: Polynomial Functions For Exercises 77–84, determine whether the given function is a polynomial function. If it is a polynomial function, state the degree. If not, state the reason why. 2 77. h1x2  x2  5 3

78. k1x2  7x4  0.3x  x3

79. p1x2  8x3  2x2 

80. q1x2  x2  4x3

81. g1x2  7

82. g1x2  4x

83. M1x2  0x 0  5x

84. N1x2  x2  0 x 0

3 x

85. Given P1x2  x4  2x  5, find the function values. (See Example 7.) a. P(2)

b. P(1)

c. P(0)

d. P(1)

86. Given N1x2  x2  5x, find the function values. a. N(1)

b. N(1)

c. N(2)

d. N(0)

87. Given H1x2  12x3  x  14, find the function values. a. H(0)

b. H(2)

c. H(2)

d. H(1)

88. Given K1x2  23x2  19, find the function values. a. K(0)

b. K(3)

c. K(3)

d. K(1)

89. A rectangular garden is designed to be 3 ft longer than it is wide. Let x represent the width of the garden. Find a function P that represents the perimeter in terms of x. (See Example 8.)

90. Pauline measures a rectangular conference room and finds that the length is 4 yd greater than twice the width. Let x represent the width. Find a function P that represents the perimeter in terms of x. 91. The cost in dollars of producing x toy cars is C1x2  2.2x  1. The revenue received is R1x2  5.98x. To calculate profit, subtract the cost from the revenue. a. Write and simplify a function P that represents profit in terms of x. b. Find the profit of producing 50 toy cars. 92. The cost in dollars of producing x lawn chairs is C1x2  4.5x  10.1. The revenue for selling x chairs is R1x2  12.99x. To calculate profit, subtract the cost from the revenue. a. Write and simplify a function P that represents profit in terms of x.

93. The function defined by D1x2  5.2x2  40.4x  1636 approximates the average yearly dormitory charge for 4-yr universities x years since 1990. D(x) is the cost in dollars, and x represents the number of years since 1990. (See Example 9.) a. Evaluate D(0) and D(18) and interpret their meaning in the context of this problem. b. If this trend continues, what will the annual dormitory charge be in the year 2012?

Cost ($)

b. Find the profit of producing 100 lawn chairs.

6,000 5,000 4,000 3,000 2,000 1,000 0

Yearly Dormitory Cost, Four-Year Universities D(x)  5.2x2  40.4x  1636

0

4 8 12 16 20 Year (x  0 corresponds to 1990)

24

Source: U.S. National Center for Education Statistics

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94. The population of Mexico can be modeled by P1t2  0.022t 2  2.012t  102, where t is the number of years since 2000 and P(t) is the number of people in millions. a. Evaluate P(0) and P(6), and interpret their meaning in the context of this problem. Round to 1 decimal place if necessary.

160 140 120 100 80 60 40 20 0 0

Population (millions)

Section 4.2

b. If this trend continues, what will the population of Mexico be in the year 2010? Round to 1 decimal place if necessary.

Population of Mexico P(t)  0.022t2  2.012t  102

2 4 6 8 10 12 Year (t  0 corresponds to 2000)

14

95. The number of women, W, to be paid child support in the United States can be approximated by W1t2  143t  6580 where t is the number of years since 2000, and W(t) is the yearly total measured in thousands. (Source: U.S. Bureau of the Census) a. Evaluate W(0), W(5), and W(10). b. Interpret the meaning of the function value W(10). 96. The total yearly amount of child support due (in billions of dollars) in the United States can be approximated by D1t2  0.925t  4.625 where t is the number of years since 2000, and D(t) is the amount due (in billions of dollars). a. Evaluate D(0), D(4), and D(8). b. Interpret the meaning of the function value of D(8).

Expanding Your Skills 97. A toy rocket is shot from ground level at an angle of 60° from the horizontal. See the figure. The x- and y-positions of the rocket (measured in feet) vary with time t according to x1t2  25t

a. Evaluate x(0) and y(0), and write the values as an ordered pair. Interpret the meaning of these function values in the context of this problem. Match the ordered pair with a point on the graph. b. Evaluate x(1) and y(1) and write the values as an ordered pair. Interpret the meaning of these function values in the context of this problem. Match the ordered pair with a point on the graph. c. Evaluate x(2) and y(2), and write the values as an ordered pair. Match the ordered pair with a point on the graph.

Vertical Distance (ft)

y1t2  16t2  43.3t y 35 30 25 20 15 10 5 00

Path of Rocket

10

20 30 40 50 Horizontal Distance (ft)

60

70

x

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Section 4.3

Multiplication of Polynomials

Concepts

1. Multiplying Polynomials

1. Multiplying Polynomials 2. Special Case Products: Difference of Squares and Perfect Square Trinomials 3. Translations Involving Polynomials 4. Applications Involving a Product of Polynomials

The properties of exponents covered in Section 4.1 can be used to simplify many algebraic expressions including the multiplication of monomials. To multiply monomials, first use the associative and commutative properties of multiplication to group coefficients and like bases. Then simplify the result by using the properties of exponents. Example 1

Multiplying Monomials

Multiply the monomials.

13x2y7 215x3y2

Solution:

13x2y7 215x3y2

 13 ⴢ 521x2 ⴢ x3 21y7 ⴢ y2

Group coefficients and like bases.

 15x y

Add exponents and simplify.

5 8

Skill Practice Multiply the monomials. 1. 18r 3s214r4s4 2

The distributive property is used to multiply polynomials: a1b  c2  ab  ac. Example 2

Multiplying a Polynomial by a Monomial

Multiply the polynomials. a. 5y3 12y2  7y  62

1 b. 4a3b7c a2ab2c4  a5bb 2

Solution: a. 5y3 12y2  7y  62

 15y3 212y2 2  15y3 217y2  15y3 2162  10y5  35y4  30y3

b. 4a3b7c a2ab2c4 

Simplify each term.

1 5 a bb 2

1  14a3b7c212ab2c4 2  14a3b7c2a a5bb 2

Apply the distributive property.

 8a4b9c5  2a8b8c

Simplify each term.

Skill Practice Multiply the polynomials. Answers 1. 32r s 2. 12b 4  18b 3  48b 2 3. 4s 3t 3  2s 2t 4 7 5

Apply the distributive property.

2. 6b2 12b2  3b  82

1 1 3. 8st 3a s 2  stb 2 4

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Section 4.3

Multiplication of Polynomials

Thus far, we have illustrated polynomial multiplication involving monomials. Next, the distributive property will be used to multiply polynomials with more than one term. For example:

e

1x ⫹ 321x ⫹ 52 ⫽ 1x ⫹ 32x ⫹ 1x ⫹ 325

Apply the distributive property.

⫽ 1x ⫹ 32x ⫹ 1x ⫹ 325

Apply the distributive property again.

⫽ x2 ⫹ 3x ⫹ 5x ⫹ 15 ⫽ x2 ⫹ 8x ⫹ 15 Note:

Combine like terms.

Using the distributive property results in multiplying each term of the first polynomial by each term of the second polynomial:

1x ⫹ 321x ⫹ 52 ⫽ x2 ⫹ 5x ⫹ 3x ⫹ 15 ⫽ x2 ⫹ 8x ⫹ 15

Example 3

Multiplying Polynomials

Multiply the polynomials.

12x2 ⫹ 4213x2 ⫺ x ⫹ 52

Solution: 12x2 ⫹ 4213x2 ⫺ x ⫹ 52 ⫽ 12x2 213x2 2 ⫹ 12x2 21⫺x2 ⫹ 12x2 2152 ⫹ 14213x2 2 ⫹ 1421⫺x2 ⫹ 142152

Multiply each term in the first polynomial by each term in the second. Apply the distributive property.

⫽ 6x4 ⫺ 2x3 ⫹ 10x2 ⫹ 12x2 ⫺ 4x ⫹ 20

Simplify each term.

⫽ 6x4 ⫺ 2x3 ⫹ 22x2 ⫺ 4x ⫹ 20

Combine like terms.

TIP: Multiplication of polynomials can be performed vertically by a process similar to column multiplication of real numbers. 12x2 ⫹ 4213x2 ⫺ x ⫹ 52

3x2 ⫺ x ⫹ 5 ⫻ 2x2 ⫹ 4 12x2 ⫺ 4x ⫹ 20 6x4 ⫺ 2x3 ⫹ 10x2 6x4 ⫺ 2x3 ⫹ 22x2 ⫺ 4x ⫹ 20

Note:

When multiplying by the column method, it is important to align like terms vertically before adding terms.

Skill Practice Multiply the polynomials. 4. 12y ⫺ 1213y 2 ⫺ 2y ⫺ 12

Answer 4. 6y 3 ⫺ 7y 2 ⫹ 1

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Example 4

Multiplying Polynomials 13y  2217y  62

Multiply the polynomials.

Solution: 13y  2217y  62

Multiply each term in the first polynomial by each term in the second.

 13y217y2  13y2162  12217y2  122162

Apply the distributive property.

 21y2  18y  14y  12

Simplify each term.

 21y2  4y  12

Combine like terms.

TIP: The acronym, FOIL (first outer inner last) can be used as a memory device to multiply the two binomials. First

Outer terms

Outer

Inner

Last

First terms

13y  2217y  62  13y217y2  13y2162  12217y2  122162 Inner terms

Last terms

 21y 2  18y  14y  12  21y 2  4y  12

Note: It is important to realize that the acronym FOIL may only be used when finding the product of two binomials.

Skill Practice Multiply the polynomials. 5. 14t  5212t  32

2. Special Case Products: Difference of Squares and Perfect Square Trinomials In some cases, the product of two binomials takes on a special pattern. I.

The first special case occurs when multiplying the sum and difference of the same two terms. For example:

12x  3212x  32

 4x2  6x  6x  9  4x2  9

w

Notice that the “middle terms” are opposites. This leaves only the difference between the square of the first term and the square of the second term. For this reason, the product is called a difference of squares.

Note: The binomials 2x  3 and 2x  3 are called conjugates. In one expression, 2x and 3 are added, and in the other, 2x and 3 are subtracted. In general, a  b and a  b are conjugates of each other. Answer 5. 8t 2  22t  15

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Section 4.3

II.

Multiplication of Polynomials

The second special case involves the square of a binomial. For example:

13x ⫹ 72 2

When squaring a binomial, the product will be a trinomial called a perfect square trinomial. The first and third terms are formed by squaring the terms of the binomial. The middle term is twice the product of the terms in the binomial.

⫽ 13x ⫹ 7213x ⫹ 72

⫽ 9x2 ⫹ 21x ⫹ 21x ⫹ 49 ⫽ 9x2 ⫹ 42x

⫹

w

49

⫽ 13x2 2 ⫹ 213x2172 ⫹ 172 2 Note: The expression 13x ⫺ 72 2 also results in a perfect square trinomial, but the middle term is negative. 13x ⫺ 7213x ⫺ 72 ⫽ 9x2 ⫺ 21x ⫺ 21x ⫹ 49 ⫽ 9x2 ⫺ 42x ⫹ 49

The following summarizes these special case products.

FORMULA Special Case Product Formulas 1. 1a ⫹ b21a ⫺ b2 ⫽ a2 ⫺ b2

The product is called a difference of squares.

2. 1a ⫹ b2 2 ⫽ a2 ⫹ 2ab ⫹ b2 1a ⫺ b2 2 ⫽ a2 ⫺ 2ab ⫹ b2

}

The product is called a perfect square trinomial.

It is advantageous for you to become familiar with these special case products because they will be presented again when we factor polynomials. Example 5

Using Special Products

Use the special product formulas to multiply the polynomials. a. 16c ⫺ 7d216c ⫹ 7d2

b. 15x ⫺ 22 2

c. 14x3 ⫹ 3y2 2 2

Solution:

a. 16c ⫺ 7d216c ⫹ 7d2

a ⫽ 6c, b ⫽ 7d

⫽ 16c2 2 ⫺ 17d2 2

Apply the formula a2 ⫺ b2.

⫽ 36c2 ⫺ 49d2

Simplify each term.

b. 15x ⫺ 22 2

⫽ 15x2 ⫺ 215x2122 ⫹ 122 2

a ⫽ 5x, b ⫽ 2 Apply the formula a2 ⫺ 2ab ⫹ b2.

2

⫽ 25x2 ⫺ 20x ⫹ 4

Simplify each term.

c. 14x ⫹ 3y 2 3

2 2

⫽ 14x3 2 2 ⫹ 214x3 213y2 2 ⫹ 13y2 2 2 ⫽ 16x6 ⫹ 24x3y2 ⫹ 9y4

a ⫽ 4x3, b ⫽ 3y2 Apply the formula a2 ⫹ 2ab ⫹ b2. Simplify each term.

Skill Practice Multiply the polynomials. 6. 15x ⫺ 4y215x ⫹ 4y2

7. 1c ⫺ 32 2

8. 17s2 ⫹ 2t2 2

Answers 6. 25x 2 ⫺ 16y 2 7. c 2 ⫺ 6c ⫹ 9 8. 49s 4 ⫹ 28s 2t ⫹ 4t 2

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The special case products can be used to simplify more complicated algebraic expressions.

Using Special Products

Example 6

3x  1y  z2 4 3x  1y  z2 4

Multiply.

Solution:

3x  1y  z2 4 3x  1y  z2 4

This product is in the form 1a  b2 1a  b2, where a  x and b  1y  z2.

 1x2 2  1y  z2 2

Apply the formula a2  b2.

 x2  y2  2yz  z2

Apply the distributive property.

 1x2 2  1y2  2yz  z2 2

Expand 1y  z2 2 by using the special case product formula.

Skill Practice Multiply.

9. 3 a  1c  52 4 3a  1c  52 4

Using Special Products

Example 7

1x  y2 3

Multiply.

Solution: 1x  y2 3

 1x  y2 2 1x  y2  1x2  2xy  y2 21x  y2

 1x2 21x2  1x2 21y2  12xy21x2  12xy21y2  1y2 21x2  1y2 21y2

Rewrite as the square of a binomial and another factor.

Expand 1x  y2 2 by using the special case product formula. Apply the distributive property.

 x3  x2y  2x2y  2xy2  xy2  y3

Simplify each term.

 x  3x y  3xy  y

Combine like terms.

3

2

2

3

Skill Practice Multiply. 10. 1t  22 3

3. Translations Involving Polynomials Example 8

Translating Between English Form and Algebraic Form

Complete the table.

English Form

Algebraic Form

The square of the sum of x and y

Answers 9. a 2  c 2  10c  25 10. t 3  6t 2  12t  8

x2  y2 The square of the product of 3 and x

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Section 4.3

Multiplication of Polynomials

Solution: English Form

Algebraic Form

The square of the sum of x and y

Notes The sum is squared, not the individual terms.

1x  y2 2

The sum of the squares of x and y The square of the product of 3 and x

x2  y2

The individual terms x and y are squared first. Then the sum is taken.

13x2 2

The product of 3 and x is taken. Then the result is squared.

Skill Practice Translate to algebraic form: 11. The square of the difference of a and b 12. The difference of the square of a and the square of b 13. Translate to English form: a  b 2.

4. Applications Involving a Product of Polynomials Example 9

Applying a Product of Polynomials

A box is created from a sheet of cardboard 20 in. on a side by cutting a square from each corner and folding up the sides (Figures 4-3 and 4-4). Let x represent the length of the sides of the squares removed from each corner. a. Write a function V that represents the volume of the box in terms of x. b. Find the volume if a 4-in. square is removed. x

20  2x

x

20

x

20  2x

20  2x

x x

20

20  2x

Figure 4-4

Figure 4-3

Solution: a. The volume of a rectangular box is given by the formula V  lwh. The length and width can both be expressed as 20  2x. The height of the box is x. The volume is given by Vlⴢwⴢh

V1x2  120  2x2120  2x2x  120  2x2 2x

 1400  80x  4x 2 2x  400x  80x 2  4x 3  4x  80x  400x 3

2

Answers

11. 1a  b2 2 12. a 2  b 2 13. The difference of a and the square of b

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b. If a 4-in. square is removed from the corners of the box, we have x  4. Thus, the volume is V1x2  4x3  80x2  400x V142  4142 3  80142 2  400142  41642  801162  400142  256  1280  1600  576 x

The volume is 576 in.3 Skill Practice A rectangular photograph is mounted on a square piece of cardboard whose sides have length x. The border that surrounds the photo is 3 in. on each side and 4 in. on both top and bottom.

4 in.

x

x

3 in.

3 in.

14. Write a function A for the area of the photograph in terms of x. 15. Determine the area of the photograph if x is 12.

Answers

14. A 1x2  1x  821x  62; A 1x2  x 2  14x  48 15. 24 in.2

4 in. x

Section 4.3 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Difference of squares

b. Conjugates

c. Perfect square trinomial

Review Exercises

2. Simplify. 14x2y  2xy  3xy2 2  12x2y  4xy2 2  16x2y  5xy2 3. Simplify. 12  3x2  35  16x2  4x  12 4

4. Given f 1x2  4x3  5, find the function values. a. f 132

b. f 102

c. f 122

5. Given g1x2  x4  x2  3, find the function values. a. g112

b. g122

c. g102

6. Write the distributive property of multiplication over addition. Give an example of the distributive property. (Answers may vary.)

Concept 1: Multiplying Polynomials For Exercises 7–40, multiply the polynomials. (See Examples 1–4.) 7. 17x4y216xy5 2

8. 14a3b7 212ab3 2

10. 18.5c4d5e216cd2e2

11.

1 12a  32 5

13. 2m3n2 1m2n3  3mn2  4n2

14. 3p2q 1p3q3  pq2  4p2

9. 12.2a6b4c7 215ab4c3 2 12.

1 16b  42 3

1 2 15. 6xy2 a x  xyb 2 3

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341

5 1 16. 12ab a a  ab2 b 6 4

17. 1x  y21x  2y2

18. 13a  521a  22

19. 16x  1215  2x2

20. 17  3x21x  82

21. 1y2  12212y2  32

22. 14p2  1212p2  52

23. 15s  3t215s  2t2

24. 14a  3b214a  b2

25. 1n2  10215n  32

26. 1m2  8213m  72

27. 11.3a  4b212.5a  7b2

28. 12.1x  3.5y214.7x  2y2

29. 12x  y213x2  2xy  y2 2

30. 1h  5k21h2  2hk  3k2 2

31. 1x  721x2  7x  492

32. 1x  321x2  3x  92

33. 14a  b21a3  4a2b  ab2  b3 2

34. 13m  2n21m3  2m2n  mn2  2n3 2 1 36. a a2  2ab  b2 b12a  b2 2

1 35. a a  2b  cb 1a  6b  c2 2

37. 1x2  2x  1213x  52

1 1 39. a y  10b a y  15b 5 2

38. 1x  y  2z215x  y  z2

1 2 40. a x  6b a x  9b 3 2

Concept 2: Special Case Products: Difference of Squares and Perfect Square Trinomials For Exercises 41–60, multiply by using the special case products. (See Example 5.) 41. 1a  821a  82

42. 1b  221b  22

43. 13p  1213p  12

44. 15q  3215q  32

1 1 45. ax  b ax  b 3 3

1 1 1 1 46. a x  b a x  b 2 3 2 3

47. 13h  k213h  k2

48. 1x  7y21x  7y2

49. 13h  k2 2

50. 1x  7y2 2

51. 1t  72 2

52. 1w  92 2

53. 1u  3v2 2

54. 1a  4b2 2

1 2 55. ah  kb 6

2 2 56. a x  1b 5

57. 12z2  w3 212z2  w3 2

58. 1a4  2b3 21a4  2b3 2

59. 15x2  3y2 2

60. 14p3  2m2 2

61. Multiply the expressions. Explain their similarities.

62. Multiply the expressions. Explain their similarities. a. 1A  B21A  B2

a. 1A  B21A  B2

b. 3A  13h  k2 4 3A  13h  k2 4

b. 3 1x  y2  B 4 3 1x  y2  B4 For Exercises 63–68, multiply the expressions. (See Example 6.) 63. 3 1w  v2  24 3 1w  v2  2 4

64. 3 1x  y2  64 3 1x  y2  64

65. 32  1x  y2 4 3 2  1x  y2 4

66. 3a  1b  12 4 3 a  1b  12 4

67. 3 13a  42  b4 3 13a  42  b4

68. 3 15p  72  q4 3 15p  72  q4

69. Explain how to multiply 1x  y2 3.

70. Explain how to multiply 1a  b2 3.

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For Exercises 71–74, multiply the expressions. (See Example 7.) 71. 12x ⫹ y2 3

72. 1x ⫺ 5y2 3

75. Explain how you would multiply the binomials 1x ⫺ 221x ⫹ 6212x ⫹ 12

For Exercises 77–84, simplify the expressions. 77. 2a2 1a ⫹ 5213a ⫹ 12

78. ⫺5y12y ⫺ 321y ⫹ 32

81. ⫺312x ⫹ 72 ⫺ 14x ⫺ 12 2 82. 1 p ⫹ 102 2 ⫺ 41 p ⫹ 62 2

73. 14a ⫺ b2 3

74. 13a ⫹ 4b2 3

76. Explain how you would multiply the binomials

1a ⫹ b21a ⫺ b212a ⫹ b212a ⫺ b2

79. 1x ⫹ 321x ⫺ 321x ⫹ 52 83. 1y ⫹ 12 2 ⫺ 12y ⫹ 32 2

80. 1t ⫹ 221t ⫺ 321t ⫹ 12

84. 1b ⫺ 32 2 ⫺ 13b ⫺ 12 2

Concept 3: Translations Involving Polynomials For Exercises 85–88, translate from English form to algebraic form. (See Example 8.) 85. The square of the sum of r and t

86. The square of a plus the cube of b

87. The difference of x squared and y cubed

88. The square of the product of 3 and a

For Exercises 89–92, translate from algebraic form to English form. (See Example 8.) 89. p3 ⫹ q2

90. a3 ⫺ b3

91. xy2

92. 1c ⫹ d2 3

Concept 4: Applications Involving a Product of Polynomials 93. A rectangular garden has a walk around it of width x. The garden is 20 ft by 15 ft. Write a function representing the combined area A(x) of the garden and walk. Simplify the result.

94. An 8-in. by 10-in. photograph is in a frame of width x. Write a function that represents the area A(x) of the frame alone. Simplify the result.

8 in. x

x

15 ft

10 in.

20 ft

95. A box is created from a square piece of cardboard 8 in. on a side by cutting a square from each corner and folding up the sides. Let x represent the length of the sides of the squares removed from each corner. (See Example 9.) a. Write a function representing the volume of the box.

x

b. Find the volume if 1-in. squares are removed from the corners.

x 8 in.

96. A box is created from a rectangular piece of metal with dimensions 12 in. by 9 in. by removing a square from each corner of the metal sheet and folding up the sides. Let x represent the length of the sides of the squares removed from each corner. a. Write a function representing the volume of the box. b. Find the volume if 2-in. squares are removed from the corners.

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Division of Polynomials

343

For Exercises 97–102, write an expression for the area and simplify your answer. 97. Square

98. Square

99. Rectangle x⫺2 x⫹2

x⫺2

x⫹3

100. Rectangle

101. Triangle

102. Triangle

2x ⫺ 3 2x ⫹ 3

4x

x⫹3

x⫺1

2x ⫺ 6

For Exercises 103–104, write an expression for the volume and simplify your answer. 103.

104.

2x 3x  10 x7 3x

x3

x

Expanding Your Skills

105. Explain how to multiply 1x ⫹ 22 4.

106. Explain how to multiply 1y ⫺ 32 4.

107. 12x ⫺ 32 multiplied by what binomial will result 108. 14x ⫹ 12 multiplied by what binomial will result in the trinomial 12x2 ⫺ 5x ⫺ 2? Check your answer by in the trinomial 10x2 ⫺ 27x ⫹ 18? Check your multiplying the binomials. answer by multiplying the binomials. 109. 14y ⫹ 32 multiplied by what binomial will result 110. 13y ⫺ 22 multiplied by what binomial will result in in the trinomial 8y2 ⫹ 2y ⫺ 3? Check your the trinomial 3y2 ⫺ 17y ⫹ 10? Check your answer answer by multiplying the binomials. by multiplying the binomials.

Division of Polynomials

Section 4.4

1. Division by a Monomial

Concepts

Division of polynomials is presented in this section as two separate cases. The first case illustrates division by a monomial divisor. The second case illustrates division by a polynomial with two or more terms. To divide a polynomial by a monomial, divide each individual term in the polynomial by the divisor and simplify the result.

1. Division by a Monomial 2. Long Division 3. Synthetic Division

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PROCEDURE Dividing a Polynomial by a Monomial If a, b, and c are polynomials such that c  0, then ab a b   c c c

Example 1

Similarly,

ab a b   c c c

Dividing a Polynomial by a Monomial

Divide the polynomials. a.

3x4  6x3  9x 3x

b. 110c 3d  15c 2d 2  2cd 3 2  15c 2d 2 2

Solution: a.

3x4  6x3  9x 3x 

3x4 6x3 9x   3x 3x 3x

Divide each term in the numerator by 3x.

 x3  2x2  3

Simplify each term, using the properties of exponents.

b. 110c 3d  15c 2d 2  2cd 3 2  15c 2d 2 2 

10c 3d  15c 2d 2  2cd 3 5c 2d 2



10c 3d 15c 2d 2 2cd 3   5c 2d 2 5c 2d 2 5c 2d 2



2c 2d 3 d 5c

Divide each term in the numerator by 5c 2d 2. Simplify each term.

Skill Practice Divide. 1.

18y 3  6y 2  12y 6y

2. 124a 3b 2  16a 2b 3  8ab2  18ab2

2. Long Division If the divisor has two or more terms, a long division process similar to the division of whole numbers is used. Example 2

Using Long Division to Divide Polynomials

Divide the polynomials by using long division.

13x2  14x  102  1x  22

Answers 1. 3y 2  y  2 2. 3a 2b  2ab 2  1

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Solution: x  2 冄3x2  14x  10

Division of Polynomials

Divide the leading term in the dividend by the leading term in the divisor. 3x2  3x. This is the first term in the x quotient.

3x x  2 冄 3x2  14x  10 3x2  6x

Multiply 3x by the divisor and record the result: 3x1x  22  3x2  6x.

3x x  2 冄3x  14x  10 3x2  6x 8x 2

3x  8 x  2 冄3x2  14x  10 3x2  6x 8x  10 8x  16 3x  8 x  2 冄3x2  14x  10 3x2  6x 8x  10 8x  16  26

Next, subtract the quantity 3x2  6x. To do this, add its opposite.

Bring down the next column and repeat the process. 8x Divide the leading term by x:  8 x Multiply the divisor by 8 and record the result: 81x  22  8x  16.

TIP: Take a minute to compare the long division process with numbers to the process of dividing polynomials. 148 31冄 4602 31 150 124 262 248 14

Subtract the quantity 8x  16 by adding its opposite. The remainder is 26. We do not continue because the degree of the remainder is less than the degree of the divisor.

Summary: The quotient is

3x  8

The remainder is

26

The divisor is

x2

The dividend is

3x2  14x  10

The solution to a long division problem is often written in the form: Quotient 

remainder . divisor

This answer is 3x  8 

26 . x2

Skill Practice Divide.

3. 14x 2  6x  82  1x  32 Answer 3. 4x  6 

10 x3

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TIP: The division of polynomials can be checked in the same fashion as the division of real numbers. To check, we know that Dividend  (divisor)(quotient)  remainder 3x2  14x  10 ⱨ 1x  22 13x  82  1262 ⱨ 3x2  8x  6x  16  1262 ⱨ 3x2  14x  10 ✔

Example 3

True

Using Long Division to Divide Polynomials

Divide the polynomials by using long division: 12x3  10x2  562  12x  42

Solution: First note that the dividend has a missing power of x and can be written as 2x3  10x2  0x  56. The term 0x is a placeholder for the missing term. It is helpful to use the placeholder to keep the powers of x lined up.

TIP: To get the first term in the quotient, you can also ask, “What can be multiplied by 2x to get 2x 3?”

x2 2x  4 冄 2x3  10x2  0x  56 2x3  4x2

x2  2x  4 冄2x  10x2  2x3  4x2 14x2  14x 2  3

7x 0x  56 0x 28x

x2  7x  14 2x  4 冄2x3  10x2  0x  56 2x3  4x2 14x2  0x 14x2  28x 28x  56 28x  56

Leave space for the missing power of x. 2x3 Divide  x2 to get the first 2x term of the quotient.

Subtract by adding the opposite. Bring down the next column. 14x2  7x to get the next 2x term in the quotient. Divide

Subtract by adding the opposite. Bring down the next column. 28x  14 to get the next 2x term in the quotient. Divide

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x2  7x  14 2x  4 冄2x3  10x2  0x  56 2x3  4x2 14x2  0x 14x2  28x 28x  56 28x  56 0

Division of Polynomials

347

Subtract by adding the opposite.

The remainder is 0.

The answer is x2  7x  14. Skill Practice Divide. 4.

4y 3  2y  2 2y  2

In Example 3, the quotient is x2  7x  14 and the remainder is 0. Because the remainder is zero, 2x  4 divides evenly into 2x3  10x2  56. For this reason, the divisor and quotient are factors of 2x3  10x2  56. To check, we have Dividend  (divisor) (quotient)  remainder 2x3  10x2  56 ⱨ 12x  421x2  7x  142  0

ⱨ 2x3  14x2  28x  4x2  28x  56 ⱨ 2x3  10x2  56 ✔ True

Example 4 Divide.

Using Long Division to Divide Polynomials

115x3  4x  9x4  52  13x2  42

Solution: Write the dividend in descending powers of x. The dividend has a missing power of x2.

9x4  15x3  4x  5 9x4  15x3  0x2  4x  5

The divisor has a missing power of x and can be written as 3x2  0x  4. 3x2  5x  4 3x2  0x  4 冄9x4  15x3  0x2  4x  5 19x4  0x3  12x2 2 15x3  12x2  4x 115x3  0x2  20x2 12x2  16x  5 112x2  0x  162 16x  11

The answer is 3x2  5x  4 

TIP: Both the divisor and dividend must be written in descending order before performing polynomial division. The remainder is 16x  11. The degree of the remainder is less than the degree of the divisor.

16x  11 . 3x 2  4 Answers

Skill Practice Divide.

5. 1x 3  1  2x 2 2  1x 2  12

4. 2y 2  2y  1 x  1 5. x  2  2 x 1

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3. Synthetic Division In this section, we introduced the process of long division to divide two polynomials. Next, we will learn another technique, called synthetic division, to divide two polynomials. Synthetic division can be used when dividing a polynomial by a first-degree divisor of the form x  r, where r is a constant. Synthetic division is considered a “shortcut” because it uses the coefficients of the divisor and dividend without writing the variables. Consider dividing the polynomials 13x2  14x  102  1x  22 . 3x  8 x  2 冄 3x2  14x  10 13x2  6x2 8x  10 18x  162 26 First note that the divisor x  2 is in the form x  r, where r  2. Therefore, synthetic division can be used to find the quotient and remainder. Step 1: Write the value of r in a box.

14

2 3

10

3

Step 3: Skip a line and draw a horizontal line below the list of coefficients.

Step 4: Bring down the leading coefficient from the dividend and write it below the line. 2

Step 5: Multiply the value of r by the number below the line 12  3  62 . Write the result in the next column above the line.

Step 2: Write the coefficients of the dividend to the right of the box.

3

14

10

6 3

8

Step 6: Add the numbers in the column above the line 114  62 , and write the result below the line.

Repeat steps 5 and 6 until all columns have been completed. Step 7: To get the final result, we use the numbers below the line. The number in the last column is the remainder. The other numbers are the coefficients of the quotient.

2 3 14 10 6 16 3

8

26

Quotient: 3x  8,

A box is usually drawn around the remainder.

remainder: 26

The degree of the quotient will always be 1 less than that of the dividend. Because the dividend is a second-degree polynomial, the quotient will be a firstdegree polynomial. In this case, the quotient is 3x  8 and the remainder is 26.

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Section 4.4

Example 5

Division of Polynomials

Using Synthetic Division to Divide Polynomials

Divide the polynomials 15x  4x3  6  x4 2  1x  32 by using synthetic division.

Solution: As with long division, the terms of the dividend and divisor should be written in descending order. Furthermore, missing powers must be accounted for by using placeholders (shown here in red). 5x  4x3  6  x4  x4  4x3  0x 2  5x  6 To use synthetic division, the divisor must be in the form (x  r). The divisor x  3 can be written as x  (3). Hence, r  3. Step 1: Write the value of r in a box.

3

1 4

0

5

6

1

Step 2: Write the coefficients of the dividend to the right of the box.

Step 3: Skip a line and draw a horizontal line below the list of coefficients.

Step 4: Bring down the leading coefficient from the dividend and write it below the line.

Step 5: Multiply the 3 1 4 0 5 6 value of r by the 3 number below 1 1 the line (3  1  3). Write the result in the next column above the line.

Step 6: Add the numbers in the column above the line: 4  (3)  1.

Repeat steps 5 and 6:

3

1 4

0

5

6

3 3 9 42 1 1 3 14 48

remainder

The quotient is

constant

x  x  3x  14.

x-term coefficient

The remainder is 48.

x2-term coefficient

3

2

x3-term coefficient

The answer is x3  x2  3x  14 

48 . x3

Skill Practice Divide the polynomials by using synthetic division. 6. 15y2  4y  2y3  52  1y  32

Answer 6. 2y 2  y  1 

2 y3

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TIP: It is interesting to compare the long division process to the synthetic division process. For Example 5, long division is shown on the left, and synthetic division is shown on the right. Notice that the same pattern of coefficients used in long division appears in the synthetic division process. x3  x2  3x  14 x  3 冄 x  4x3  0x2  5x  6 1x 4  3x 3 2 x3  0x2 1x3  3x2 2 3x2  5x 13x2  9x2 14x  6 4

3

x3

114x  422 48

Example 6

4 0 5 6 3 3 9 42 1 1 3 14 48 1

x2

x

constant remainder

Quotient: x3  x2  3x  14 Remainder: 48

Using Synthetic Division to Divide Polynomials

Divide the polynomials by using synthetic division.

1p4  812  1p  32

Solution: 1p4  812  1p  32

1p4  0p3  0p2  0p  812  1p  32 3

1 1

0 3 3

0 9 9

0 27 27

Insert placeholders (red) for missing powers of p.

81 81 0

Quotient: p3  3p2  9p  27 Remainder: 0 The answer is p3  3p2  9p  27. Skill Practice Divide the polynomials by using synthetic division. Answer 7. x  x  1 2

7. 1x3  12  1x  12

Section 4.4 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key term synthetic division.

Review Exercises

2. a. Add 13x  12  12x  52 .

b. Multiply 13x  1212x  52 .

3. a. Subtract 1a  10b2  15a  b2 . b. Multiply 1a  10b215a  b2 .

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4. a. Subtract 12y2  12  1y2  5y  12 . b. Multiply 12y2  121y2  5y  12 .

Division of Polynomials

5. a. Add 1x2  x2  16x2  x  22 .

b. Multiply 1x2  x216x2  x  22 .

For Exercises 6–8, answers may vary. 6. Write an example of a product of two binomials and simplify. 7. Write an example of the square of a binomial and simplify. 8. Write an example of the product of conjugates and simplify.

Concept 1: Division by a Monomial For Exercises 9–24, divide the polynomials. Check your answer by multiplication. (See Example 1.) 9.

16t 4  4t 2  20t 4t

10.

2x3  8x2  2x 2x

11. 136y  24y2  6y3 2  13y2

12. 16p2  18p4  30p5 2  16p2

13. 14x3y  12x2y2  4xy3 2  14xy2

14. 125m5n  10m4n  m3n2  15m3n2

17. 13p4  6p3  2p2  p2  16p2

18. 14q3  8q2  q2  112q2

15. 18y4  12y3  32y2 2  14y2 2 19. 1a3  5a2  a  52  1a2

16. 112y5  8y6  16y4  10y3 2  12y3 2 20. 12m5  3m4  m3  m2  9m2  1m2 2

6s3t5  8s2t4  10st2 21. 2st4

8r4w2  4r3w  2w3 22. 4r3w

23. 18p4q7  9p5q6  11p3q  42  1p2q2

24. 120a5b5  20a3b2  5a2b  62  1a2b2

Concept 2: Long Division

25. a. Divide 12x3  7x2  5x  12  1x  22 , and identify the divisor, quotient, and remainder. b. Explain how to check by using multiplication. 26. a. Divide 1x3  4x2  7x  32  1x  32 , and identify the divisor, quotient, and remainder. b. Explain how to check by using multiplication.

For Exercises 27– 46, divide the polynomials by using long division. Check your answer by multiplication. (See Examples 2–4.)

27. 1x2  11x  192  1x  42

3 2 28. 1x  7x  13x  32  1x  22

29. 13y3  7y2  4y  32  1y  32

30. 1z3  2z2  2z  52  1z  42

31. 112a2  77a  1212  13a  112

32. 128x2  29x  62  14x  32

33. 19y  18y2  202  13y  42

34. 12y  3y2  12  1y  12

35. 118x3  7x  122  13x  22

36. 18x3  6x  222  12x  12

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37. 18a3  12  12a  12

38. 181x4  12  13x  12

39. 1x4  x3  x2  4x  22  1x2  x  12

40. 12a5  7a4  11a3  22a2  29a  102  12a2  5a  22

41. 12x3  10x  x4  252  1x2  52

42. 15x3  x4  4  10x2  1x2  22

43. 1x4  3x2  102  1x2  22

44. 13y4  25y2  182  1y2  32

45. 1n4  162  1n  22

46. 1m3  272  1m  32

Concept 3: Synthetic Division 47. Explain the conditions under which you may use synthetic division to divide polynomials. 48. Can synthetic division be used directly to divide 14x4  3x3  7x  92 by 12x  52 ? Explain why or why not. 49. Can synthetic division be used to divide 16x 5  3x 2  2x  142 by 1x 2  32 ? Explain why or why not. 50. Can synthetic division be used to divide 13x4  x  12 by 1x  52? Explain why or why not. 51. The following table represents the result of a synthetic division. 5

1 1

2 5 3

4 15 11

3 55 58

52. The following table represents the result of a synthetic division. 2

2 2

3 4 1

0 2 2

1 4 5

6 10 16

Use x as the variable.

Use x as the variable.

a. Identify the divisor.

a. Identify the divisor.

b. Identify the quotient.

b. Identify the quotient.

c. Identify the remainder.

c. Identify the remainder.

For Exercises 53–68, divide by using synthetic division. Check your answer by multiplication. (See Examples 5–6.) 53. 1x2  2x  482  1x  82

54. 1x2  4x  122  1x  62

55. 1t2  3t  42  1t  12

56. 1h2  7h  122  1h  32

57. 15y2  5y  12  1y  12

58. 13w2  w  52  1w  22

59. 13  7y2  4y  3y3 2  1y  32

60. 12z  2z2  z3  52  1z  32

61. 1x3  3x2  42  1x  22

65. 1x3  2162  1x  62

66. 1y4  162  1y  22

62. 13y4  25y2  182  1y  32

63. 1a5  322  1a  22

64. 1b3  272  1b  32

1 3 67. 14w4  w2  6w  32  aw  b 68. 112y4  5y3  y2  y  32  ay  b 2 4

Mixed Exercises For Exercises 69–80, divide the polynomials by using an appropriate method. 69. 1x3  8x2  3x  22  1x  42 71. 122x2  11x  332  111x2

70. 18xy2  9x2y  6x2y2 2  1x2y2 2

72. 12m3  4m2  5m  332  1m  32

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Problem Recognition Exercises

73. 112y3  17y2  30y  102  13y2  2y  52

74. 190h12  63h9  45h8  36h7 2  19h9 2

75. 14x4  6x3  3x  12  12x2  12

76. 1y4  3y3  5y2  2y  52  1y  22

79. 15x3  9x2  10x2  15x2 2

80. 115k4  3k3  4k2  42  13k2  12

77. 116k11  32k10  8k8  40k4 2  18k8 2

353

78. 14m3  18m2  22m  102  12m2  4m  32

Expanding Your Skills 81. Given P1x2  4x3  10x2  8x  20, a. Evaluate P142.

b. Divide. 14x3  10x2  8x  202  1x  42 c. Compare the value found in part (a) to the remainder found in part (b). 82. Given P1x2  3x3  12x2  5x  8, a. Evaluate P162.

b. Divide. 13x3  12x2  5x  82  1x  62 c. Compare the value found in part (a) to the remainder found in part (b). 83. Based on your solutions to Exercises 81–82, make a conjecture about the relationship between the value of a polynomial function P(x) at x  r, and the value of the remainder of P1x2  1x  r2. 84. a. Use synthetic division to divide. 17x2  16x  92  1x  12 b. Based on your solution to part (a), is x  1 a factor of 7x2  16x  9? 85. a. Use synthetic division to divide. 18x2  13x  52  1x  12 b. Based on your solution to part (a), is x  1 a factor of 8x2  13x  5?

Problem Recognition Exercises Operations on Polynomials Perform the indicated operations. 1. a. 13x  12 2

b. 13x  1213x  12

c. 13x  12  13x  12

2. a. 19m  52  19m  52 b. 19m  5219m  52 c. 19m  52 2

3. a.

4x2  8x  10 2x

4. a.

3y2  15y  4 3y

b.

4x2  8x  10 2x  1

b.

3y2  15y  4 3y  6

c. 14x2  8x  102  1x  12

c. 13y2  15y  42  1y  62

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5. a. 1p ⫺ 521p ⫹ 52 ⫺ 1 p2 ⫹ 52

b. 1p ⫺ 521p ⫹ 52 ⫺ 1p ⫹ 52 2

6. a. 1x ⫹ 421x ⫺ 42 ⫺ 1x ⫹ 42 2 b. 1x ⫹ 421x ⫺ 42 ⫺ 1x2 ⫹ 42

c. 1p ⫺ 521p ⫹ 52 ⫺ 1p2 ⫺ 252 7. 15t 2 ⫺ 6t ⫹ 22 ⫺ 13t 2 ⫺ 7t ⫹ 32 9. 16z ⫹ 5216z ⫺ 52

c. 1x ⫹ 421x ⫺ 42 ⫺ 1x2 ⫺ 162 8. ⫺5x2 13x2 ⫹ x ⫺ 22 10. 16y3 ⫹ 2y2 ⫹ y ⫺ 22 ⫹ 13y3 ⫺ 4y ⫹ 32

11. 13b ⫺ 4212b ⫺ 12

12. 15a ⫹ 2212a2 ⫹ 3a ⫹ 12

13. 1t 3 ⫺ 4t 2 ⫹ t ⫺ 92 ⫹ 1t ⫹ 122 ⫺ 12t 2 ⫺ 6t2

14. 12b3 ⫺ 3b ⫺ 102 ⫼ 1b ⫺ 22

15. 1k ⫹ 42 2 ⫹ 1⫺4k ⫹ 92

16. 13x4 ⫺ 11x3 ⫺ 4x2 ⫺ 5x ⫹ 202 ⫼ 1x ⫺ 42

17. ⫺2t1t 2 ⫹ 6t ⫺ 32 ⫹ t13t ⫹ 2213t ⫺ 22

18.

1 1 2 1 1 19. a p3 ⫺ p2 ⫹ 5b ⫺ a⫺ p3 ⫹ p2 ⫺ pb 4 6 3 3 5

20. ⫺6w3 11.2w ⫺ 2.6w2 ⫹ 5.1w3 2

21. 16a2 ⫺ 4b2 2

1 1 1 1 22. a z2 ⫺ b a z2 ⫹ b 2 3 2 3

23. 1m ⫺ 32 2 ⫺ 21m ⫹ 82

24. 12x ⫺ 521x ⫹ 12 ⫺ 1x ⫺ 32 2

25. 1m2 ⫺ 6m ⫹ 7212m2 ⫹ 4m ⫺ 32 27. 35 ⫺ 1a ⫹ b2 4 2

7x2y3 ⫺ 14xy2 ⫺ x2 ⫺7xy

26. 1x3 ⫺ 642 ⫼ 1x ⫺ 42 28. 3a ⫺ 1x ⫺ y2 4 3a ⫹ 1x ⫺ y2 4

29. 1x ⫹ y2 2 ⫺ 1x ⫺ y2 2

30. 1a ⫺ 42 3

1 1 1 1 31. a⫺ x ⫹ b a x ⫺ b 2 3 4 2

1 32. ⫺3x2y3z4 a x4yzw3 b 6

Section 4.5

Greatest Common Factor and Factoring by Grouping

Concepts

1. Factoring Out the Greatest Common Factor

1. Factoring Out the Greatest Common Factor 2. Factoring Out a Negative Factor 3. Factoring Out a Binomial Factor 4. Factoring by Grouping

Sections 4.5 through 4.7 are devoted to a mathematical operation called factoring. To factor an integer means to write the integer as a product of two or more integers. To factor a polynomial means to express the polynomial as a product of two or more polynomials. For example, in the product 5 ⴢ 7 ⫽ 35, the numbers 5 and 7 are factors of 35. In the product 12x ⫹ 121x ⫺ 62 ⫽ 2x2 ⫺ 11x ⫺ 6, the quantities 12x ⫹ 12 and 1x ⫺ 62 are factors of 2x2 ⫺ 11x ⫺ 6.

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Section 4.5

Greatest Common Factor and Factoring by Grouping

355

The greatest common factor (GCF) of a polynomial is the greatest factor that divides each term of the polynomial evenly. For example, the greatest common factor of 9x4 ⫹ 18x3 ⫺ 6x2 is 3x2. To factor out the greatest common factor from a polynomial, follow these steps:

PROCEDURE Factoring Out the Greatest Common Factor Step 1 Identify the greatest common factor of all terms of the polynomial. Step 2 Write each term as the product of the GCF and another factor. Step 3 Use the distributive property to factor out the greatest common factor. Note: To check the factorization, multiply the polynomials.

Example 1

Factoring Out the Greatest Common Factor

Factor out the greatest common factor. a. 12x3 ⫹ 30x2

b. 12c 2d 3 ⫺ 30c 3d 2 ⫺ 3cd

Solution: a. 12x3 ⫹ 30x2

⫽ 6x2 12x2 ⫹ 6x2 152

⫽ 6x2 12x ⫹ 52

The GCF is 6x2. Write each term as the product of the GCF and another factor. Factor out 6x2 by using the distributive property.

TIP: Any factoring problem can be checked by multiplying the factors. Check:

6x2 12x ⫹ 52 ⫽ 12x3 ⫹ 30x2 ✔

b. 12c2d3 ⫺ 30c3d2 ⫺ 3cd

⫽ 3cd 14cd2 2 ⫺ 3cd 110c2d2 ⫺ 3cd 112 ⫽ 3cd14cd 2 ⫺ 10c 2d ⫺ 12

Check:

Avoiding Mistakes The GCF is 3cd. Write each term as the product of the GCF and another factor. Factor out 3cd by using the distributive property.

In Example 1(b), the GCF of 3cd is equal to one of the terms of the polynomial. In such a case, you must leave a 1 in place of that term after the GCF is factored out. 3cd 14cd 2 ⫺ 10c 2d ⫺ 12

3cd14cd2 ⫺ 10c2d ⫺ 12 ⫽ 12c2d3 ⫺ 30c3d2 ⫺ 3cd ✔

Skill Practice Factor out the greatest common factor. 1. 45y5 ⫺ 15y 2 ⫹ 30y

2. 16a 2b 5 ⫹ 12a 3b 3 ⫹ 4a 3b 2

2. Factoring Out a Negative Factor Sometimes it is advantageous to factor out the opposite of the GCF, particularly when the leading coefficient of the polynomial is negative. This is demonstrated in Example 2. Notice that this changes the signs of the remaining terms inside the parentheses. Answers

1. 15y 13y 4 ⫺ y ⫹ 22 2. 4a 2b 2 14b 3 ⫹ 3ab ⫹ a2

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Example 2

Factoring Out a Negative Factor

Factor out the quantity ⫺5a2b from the polynomial ⫺5a4b ⫺ 10a3b2 ⫹ 15a2b3.

Solution: ⫺5a4b ⫺ 10a3b2 ⫹ 15a2b3

The GCF is 5a2b. However, in this case we will factor out the opposite of the GCF, ⫺5a2b.

⫽ ⫺5a2b1a2 2 ⫹ ⫺5a2b12ab2 ⫹ ⫺5a2b1⫺3b2 2 ⫽ ⫺5a2b1a2 ⫹ 2ab ⫺ 3b2 2

Write each term as the product of ⫺5a2b and another factor.

Factor out ⫺5a2b by using the distributive property.

Skill Practice Factor out the quantity ⫺6xy from the polynomial. 3. 24x 4y 3 ⫺ 12x 2y ⫹ 18xy 2

3. Factoring Out a Binomial Factor The distributive property may also be used to factor out a common factor that consists of more than one term. This is shown in Example 3. Example 3

Factoring Out a Binomial Factor

Factor out the greatest common factor.

x3 1x ⫹ 22 ⫺ x1x ⫹ 22 ⫺ 91x ⫹ 22

Solution:

x3 1x ⫹ 22 ⫺ x1x ⫹ 22 ⫺ 91x ⫹ 22 ⫽ 1x ⫹ 221x3 2 ⫺ 1x ⫹ 221x2 ⫺ 1x ⫹ 22192 ⫽ 1x ⫹ 22 1x3 ⫺ x ⫺ 92

The GCF is the quantity 1x ⫹ 22. Write each term as the product of 1x ⫹ 22 and another factor.

Factor out 1x ⫹ 22 by using the distributive property.

Skill Practice Factor out the greatest common factor. 4. a 2 1b ⫹ 22 ⫹ 51b ⫹ 22

Answers

3. ⫺6xy 1⫺4x 3y 2 ⫹ 2x ⫺ 3y2 4. 1b ⫹ 221a 2 ⫹ 52

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Greatest Common Factor and Factoring by Grouping

357

4. Factoring by Grouping When two binomials are multiplied, the product before simplifying contains four terms. For example: 13a ⫹ 22 12b ⫺ 72 ⫽ 13a ⫹ 22 12b2 ⫹ 13a ⫹ 22 1⫺72 ⫽ 13a ⫹ 22 12b2 ⫹ 13a ⫹ 22 1⫺72 ⫽ 6ab ⫹ 4b ⫺ 21a ⫺ 14 In Example 4, we learn how to reverse this process. That is, given a four-term polynomial, we will factor it as a product of two binomials. The process is called factoring by grouping.

PROCEDURE Factoring by Grouping To factor a four-term polynomial by grouping: Step 1 Identify and factor out the GCF from all four terms. Step 2 Factor out the GCF from the first pair of terms. Factor out the GCF from the second pair of terms. (Sometimes it is necessary to factor out the opposite of the GCF.) Step 3 If the two terms share a common binomial factor, factor out the binomial factor.

Example 4

Factoring by Grouping

Factor by grouping. 6ab ⫺ 21a ⫹ 4b ⫺ 14

Solution: 6ab ⫺ 21a ⫹ 4b ⫺ 14

⫽ 6ab ⫺ 21a

Step 1:

⫹ 4b ⫺ 14

⫽ 3a12b ⫺ 72 ⫹ 212b ⫺ 72

Group the first pair of terms and the second pair of terms. Step 2:

Factor out the GCF from each pair of terms. Note: The two terms now share a common binomial factor of 12b ⫺ 72.

⫽ 12b ⫺ 7213a ⫹ 22 Check:

Identify and factor out the GCF from all four terms. In this case the GCF is 1.

Step 3:

Factor out the common binomial factor.

12b ⫺ 72 13a ⫹ 22 ⫽ 2b13a2 ⫹ 2b122 ⫺ 713a2 ⫺ 7122

Avoiding Mistakes In step 2, the expression 3a 12b ⫺ 72 ⫹ 212b ⫺ 72 is not yet factored because it is a sum, not a product. To factor the expression, you must carry it one step further. 3a 12b ⫺ 72 ⫹ 212b ⫺ 72 ⫽ 12b ⫺ 72 13a ⫹ 22 The factored form must be represented as a product.

⫽ 6ab ⫹ 4b ⫺ 21a ⫺ 14 ✔ Skill Practice Factor by grouping. 5. 7c 2 ⫹ cd ⫹ 14c ⫹ 2d

Answer

5. 17c ⫹ d 21c ⫹ 22

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Example 5

Factoring by Grouping

Factor by grouping. x3 ⫹ 3x2 ⫺ 3x ⫺ 9

Solution: x3 ⫹ 3x2 ⫺ 3x ⫺ 9

⫽ x3 ⫹ 3x2

Step 1:

⫺ 3x ⫺ 9

⫽ x2 1x ⫹ 32 ⫺ 31x ⫹ 32

Identify and factor out the GCF from all four terms. In this case the GCF is 1. Group the first pair of terms and the second pair of terms.

Step 2:

Factor out x2 from the first pair of terms. Factor out ⫺3 from the second pair of terms (this causes the signs to change in the second parentheses). The terms now contain a common binomial factor.

⫽ 1x ⫹ 32 1x2 ⫺ 32

Step 3:

Factor out the common binomial 1x ⫹ 32.

TIP: One frequent question is, can the order be switched between factors? The answer is yes. Because multiplication is commutative, the order in which two or more factors are written does not matter. Thus, the following factorizations are equivalent: 1x ⫹ 32 1x2 ⫺ 32 ⫽ 1x2 ⫺ 321x ⫹ 32

Skill Practice Factor by grouping. 6. a3 ⫺ 4a2 ⫺ 3a ⫹ 12

Example 6

Factoring by Grouping

Factor by grouping. 24p2q2 ⫺ 18p2q ⫹ 60pq2 ⫺ 45pq

Solution: 24p2q2 ⫺ 18p2q ⫹ 60pq2 ⫺ 45pq ⫽ 3pq18pq ⫺ 6p ⫹ 20q ⫺ 152 ⫽ 3pq18pq ⫺ 6p

Answer

6. 1a 2 ⫺ 321a ⫺ 42

⫹ 20q ⫺ 152

Step 1:

Remove the GCF 3pq from all four terms. Group the first pair of terms and the second pair of terms.

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Section 4.5

⫽ 3pq 32p14q ⫺ 32 ⫹ 514q ⫺ 32 4

⫽ 3pq14q ⫺ 32 12p ⫹ 52

Greatest Common Factor and Factoring by Grouping

Step 2:

Factor out the GCF from each pair of terms. The terms share the binomial factor 14q ⫺ 32.

Step 3:

Factor out the common binomial 14q ⫺ 32.

359

Skill Practice Factor the polynomial. 7. 24x 2y ⫺ 12x 2 ⫹ 20xy ⫺ 10x

Notice that in step 3 of factoring by grouping, a common binomial is factored from the two terms. These binomials must be exactly the same in each term. If the two binomial factors differ, try rearranging the original four terms.

Example 7

Factoring by Grouping Where Rearranging Terms Is Necessary

Factor the polynomial. 4x ⫹ 6pa ⫺ 8a ⫺ 3px

Solution: 4x ⫹ 6pa ⫺ 8a ⫺ 3px

⫽ 4x ⫹ 6pa

Identify and factor out the GCF from all four terms. In this case the GCF is 1.

Step 2:

The binomial factors in each term are different.

⫺ 8a ⫺ 3px

⫽ 212x ⫹ 3pa2 ⫺ 118a ⫹ 3px2 ⫽ 4x ⫺ 8a

Step 1:

⫺ 3px ⫹ 6pa

⫽ 41x ⫺ 2a2 ⫺ 3p1x ⫺ 2a2

⫽ 1x ⫺ 2a2 14 ⫺ 3p2

Avoiding Mistakes

Try rearranging the original four terms in such a way that the first pair of coefficients is in the same ratio as the second pair of coefficients. Notice that the ratio 4 to ⫺8 is the same as the ratio ⫺3 to 6. Step 2:

Remember that when factoring by grouping, the binomial factors must be exactly the same.

Factor out 4 from the first pair of terms. Factor out ⫺3p from the second pair of terms.

Step 3:

Factor out the common binomial factor.

Skill Practice Factor the polynomial. 8. 3ry ⫹ 2s ⫹ sy ⫹ 6r

Answers 7. 2x16x ⫹ 5212y ⫺ 12 8. 13r ⫹ s212 ⫹ y2

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Section 4.5 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Greatest common factor (GCF)

b. Factoring by grouping

Review Exercises For Exercises 2–8, perform the indicated operation. 2. 1⫺4a3b5c21⫺2a7c2 2

3. 17t 4 ⫹ 5t 3 ⫺ 9t2 ⫺ 1⫺2t 4 ⫹ 6t 2 ⫺ 3t2

4. 15x3 ⫺ 9x ⫹ 52 ⫹ 14x3 ⫹ 3x2 ⫺ 2x ⫹ 12 ⫺ 16x3 ⫺ 3x2 ⫹ x ⫹ 12 6. 1a ⫹ 6b2 2

5. 15y2 ⫺ 321y2 ⫹ y ⫹ 22 7.

6v3 ⫺ 12v2 ⫹ 2v ⫺2v

8.

3x3 ⫹ 2x2 ⫺ 4 x⫹2

Concept 1: Factoring Out the Greatest Common Factor For Exercises 9–24, factor out the greatest common factor. (See Example 1.) 9. 3x ⫹ 12

10. 15x ⫺ 10

11. 6z2 ⫹ 4z

12. 49y3 ⫺ 35y2

13. 4p6 ⫺ 4p

14. 5q2 ⫺ 5q

15. 12x4 ⫺ 36x2

16. 51w4 ⫺ 34w3

17. 9st 2 ⫹ 27t

18. 8a2b3 ⫹ 12a2b

19. 9a4b3 ⫹ 27a3b4 ⫺ 18a2b5 20. 3x5y4 ⫺ 15x4y5 ⫹ 9x2y7

21. 10x2y ⫹ 15xy2 ⫺ 5xy

22. 12c3d ⫺ 15c2d ⫹ 3cd

23. 13b2 ⫺ 11a2b ⫺ 12ab

24. 6a3 ⫺ 2a2b ⫹ 5a2

Concept 2: Factoring Out a Negative Factor For Exercises 25–30, factor out the indicated quantity. (See Example 2.) 25. ⫺x2 ⫺ 10x ⫹ 7: Factor out ⫺1. 26. ⫺5y2 ⫹ 10y ⫹ 3: Factor out ⫺1. 27. ⫺12x3y ⫺ 6x2y ⫺ 3xy: Factor out ⫺3xy. 28. ⫺32a4b2 ⫹ 24a3b ⫹ 16a2b: Factor out ⫺8a2b. 29. ⫺2t 3 ⫹ 11t 2 ⫺ 3t: Factor out ⫺t. 30. ⫺7y2z ⫺ 5yz ⫺ z: Factor out ⫺z.

Concept 3: Factoring Out a Binomial Factor For Exercises 31–38, factor out the GCF. (See Example 3.) 31. 2a13z ⫺ 2b2 ⫺ 513z ⫺ 2b2

32. 5x13x ⫹ 42 ⫹ 213x ⫹ 42

33. 2x2 12x ⫺ 32 ⫹ 12x ⫺ 32

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Greatest Common Factor and Factoring by Grouping

34. z1w ⫺ 92 ⫹ 1w ⫺ 92

35. y12x ⫹ 12 2 ⫺ 312x ⫹ 12 2

37. 3y1x ⫺ 22 2 ⫹ 61x ⫺ 22 2

38. 10z1z ⫹ 32 2 ⫺ 21z ⫹ 32 2

361

36. a1b ⫺ 72 2 ⫹ 51b ⫺ 72 2

39. Construct a polynomial that has a greatest common factor of 3x2. (Answers may vary.) 40. Construct two different trinomials that have a greatest common factor of 5x2y3. (Answers may vary.) 41. Construct a binomial that has a greatest common factor of 1c ⫹ d2. (Answers may vary.)

Concept 4: Factoring by Grouping 42. If a polynomial has four terms, what technique would you use to factor it? 43. Factor the polynomials by grouping. a. 2ax ⫺ ay ⫹ 6bx ⫺ 3by b. 10w2 ⫺ 5w ⫺ 6bw ⫹ 3b c. Explain why you factored out 3b from the second pair of terms in part (a) but factored out the quantity ⫺3b from the second pair of terms in part (b). 44. Factor the polynomials by grouping. a. 3xy ⫹ 2bx ⫹ 6by ⫹ 4b2 b. 15ac ⫹ 10ab ⫺ 6bc ⫺ 4b2 c. Explain why you factored out 2b from the second pair of terms in part (a) but factored out the quantity ⫺2b from the second pair of terms in part (b). For Exercises 45–64, factor each polynomial by grouping (if possible). (See Examples 4–7.) 45. y3 ⫹ 4y2 ⫹ 3y ⫹ 12

46. ab ⫹ b ⫹ 2a ⫹ 2

47. 6p ⫺ 42 ⫹ pq ⫺ 7q

48. 2t ⫺ 8 ⫹ st ⫺ 4s

49. 2mx ⫹ 2nx ⫹ 3my ⫹ 3ny

50. 4x2 ⫹ 6xy ⫺ 2xy ⫺ 3y2

51. 10ax ⫺ 15ay ⫺ 8bx ⫹ 12by

52. 35a2 ⫺ 15a ⫹ 14a ⫺ 6

53. x3 ⫺ x2 ⫺ 3x ⫹ 3

54. 2rs ⫹ 4s ⫺ r ⫺ 2

55. 6p2q ⫹ 18pq ⫺ 30p2 ⫺ 90p

56. 5s2t ⫹ 20st ⫺ 15s2 ⫺ 60s

57. 100x3 ⫺ 300x2 ⫹ 200x ⫺ 600

58. 2x5 ⫺ 10x4 ⫹ 6x3 ⫺ 30x2

59. 6ax ⫺ by ⫹ 2bx ⫺ 3ay

60. 5pq ⫺ 12 ⫺ 4q ⫹ 15p

61. 4a ⫺ 3b ⫺ ab ⫹ 12

62. x2y ⫹ 6x ⫺ 3x3 ⫺ 2y

63. 7y3 ⫺ 21y2 ⫹ 5y ⫺ 10

64. 5ax ⫹ 10bx ⫺ 2ac ⫹ 4bc

65. Explain why the grouping method failed for Exercise 63. 66. Explain why the grouping method failed for Exercise 64.

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Mixed Exercises 67. Solve the equation U ⫽ Av ⫹ Acw for A by first factoring out A.

68. Solve the equation S ⫽ rt ⫹ wt for t by first factoring out t.

69. Solve the equation ay ⫹ bx ⫽ cy for y.

70. Solve the equation cd ⫹ 2x ⫽ ac for c.

71. The area of a rectangle of width w is given by A ⫽ 2w2 ⫹ w. Factor the right-hand side of the equation to find an expression for the length of the rectangle. 72. The amount in a savings account bearing simple interest at an annual interest rate r for t years is given by A ⫽ P ⫹ Prt where P is the principal amount invested. a. Solve the equation for P. b. Compute the amount of principal originally invested if the account is worth $12,705 after 3 yr at a 7% interest rate.

Expanding Your Skills For Exercises 73–80, factor out the greatest common factor and simplify. 73. 1a ⫹ 32 4 ⫹ 61a ⫹ 32 5

74. 14 ⫺ b2 4 ⫺ 214 ⫺ b2 3

75. 2413x ⫹ 52 3 ⫺ 3013x ⫹ 52 2

76. 1012y ⫹ 32 2 ⫹ 1512y ⫹ 32 3

77. 1t ⫹ 42 2 ⫺ 1t ⫹ 42

78. 1p ⫹ 62 2 ⫺ 1p ⫹ 62

79. 15w2 12w ⫺ 12 3 ⫹ 5w3 12w ⫺ 12 2

80. 8z4 13z ⫺ 22 2 ⫹ 12z3 13z ⫺ 22 3

Section 4.6

Factoring Trinomials

Concepts

1. Factoring Trinomials: AC-Method

1. Factoring Trinomials: AC-Method 2. Factoring Trinomials: Trialand-Error Method 3. Factoring Perfect Square Trinomials 4. Factoring by Using Substitution

In Section 4.5, we learned how to factor out the greatest common factor from a polynomial and how to factor a four-term polynomial by grouping. In this section, we present two methods to factor trinomials. The first method is called the ac-method. The second method is called the trial-and-error method. The product of two binomials results in a four-term expression that can sometimes be simplified to a trinomial. To factor the trinomial, we want to reverse the process. Multiply:

12x ⫹ 321x ⫹ 22 ⫽ ⫽

Factor:

2x2 ⫹ 7x ⫹ 6 ⫽ ⫽

Multiply the binomials. Add the middle terms.

Rewrite the middle term as a sum or difference of terms. Factor by grouping.

2x2 ⫹ 4x ⫹ 3x ⫹ 6 2x2 ⫹ 7x ⫹ 6

2x2 ⫹ 4x ⫹ 3x ⫹ 6 12x ⫹ 321x ⫹ 22

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Section 4.6

Factoring Trinomials

To factor a trinomial ax2 ⫹ bx ⫹ c by the ac-method, we rewrite the middle term bx as a sum or difference of terms. The goal is to produce a four-term polynomial that can be factored by grouping. The process is outlined as follows.

PROCEDURE The AC-Method to Factor ax 2 ⴙ bx ⴙ c (a ⴝ 0) Step 1 After factoring out the GCF, multiply the coefficients of the first and last terms, ac. Step 2 Find two integers whose product is ac and whose sum is b. (If no pair of integers can be found, then the trinomial cannot be factored further and is called a prime polynomial.) Step 3 Rewrite the middle term bx as the sum of two terms whose coefficients are the integers found in step 2. Step 4 Factor by grouping.

The ac-method for factoring trinomials is illustrated in Example 1. Before we begin, however, keep these two important guidelines in mind. • •

For any factoring problem you encounter, always factor out the GCF from all terms first. To factor a trinomial, write the trinomial in the form ax2 ⫹ bx ⫹ c.

Example 1

Factoring a Trinomial by the AC-Method

Factor by using the ac-method.

12x2 ⫺ 5x ⫺ 2

Solution: 12x2 ⫺ 5x ⫺ 2 a ⫽ 12

The GCF is 1.

b ⫽ ⫺5

c ⫽ ⫺2

Factors of –24

Factors of –24

1221⫺122

1⫺221122

1121⫺242

(3)(⫺8) 1421⫺62

1⫺121242 1⫺32182 1⫺42162

12x2 ⫺ 5x ⫺ 2 ⫽ 12x2 ⫹ 3x ⫺ 8x ⫺ 2 ⫽ 12x2 ⫹ 3x

⫺ 8x ⫺ 2

Step 1: The expression is written in the form ax2 ⫹ bx ⫹ c. Find the product ac ⫽ 121⫺22 ⫽ ⫺24. Step 2: List all the factors of ⫺24, and find the pair whose sum equals ⫺5. The numbers 3 and ⫺8 produce a product of ⫺24 and a sum of ⫺5. Step 3: Write the middle term of the trinomial as two terms whose coefficients are the selected numbers 3 and ⫺8. Step 4: Factor by grouping.

⫽ 3x14x ⫹ 12 ⫺ 214x ⫹ 12

⫽ 14x ⫹ 1213x ⫺ 22

The check is left for the reader.

Skill Practice 1. Factor by using the ac-method.

10x 2 ⫹ x ⫺ 3.

Answer

1. 15x ⫹ 3212x ⫺ 12

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TIP: One frequently asked question is whether the order matters when we rewrite the middle term of the trinomial as two terms (step 3). The answer is no. From Example 1, the two middle terms in step 3 could have been reversed. 12x2 ⫺ 5x ⫺ 2 ⫽ 12x2 ⫺ 8x ⫹ 3x ⫺ 2 ⫽ 4x13x ⫺ 22 ⫹ 113x ⫺ 22 ⫽ 13x ⫺ 2214x ⫹ 12 This example also shows that the order in which two factors are written does not matter. The expression 13x ⫺ 22 14x ⫹ 12 is equivalent to 14x ⫹ 12 13x ⫺ 22 because multiplication is a commutative operation.

Example 2

Factoring a Trinomial by the AC-Method

Factor the trinomial by using the ac-method.

⫺20c3 ⫹ 34c2d ⫺ 6cd2

Solution: ⫺20c3 ⫹ 34c2d ⫺ 6cd2

⫽ ⫺2c110c2 ⫺ 17cd ⫹ 3d2 2

Factor out ⫺2c. Step 1: Find the product a ⴢ c ⫽ 1102132 ⫽ 30

Factors of 30

Factors of 30 1⫺121⫺302

1 ⴢ 30

Step 2: The numbers ⫺2 and ⫺15 form a product of 30 and a sum of ⫺17.

1⫺221⫺152

2 ⴢ 15

1⫺321⫺102

3 ⴢ 10

1⫺521⫺62

5ⴢ6

⫽ ⫺2c110c ⫺ 17cd ⫹ 3d2 2 2

⫽ ⫺2c110c2 ⫺ 2cd

⫺ 15cd ⫹ 3d2 2

⫽ ⫺2c32c15c ⫺ d2 ⫺ 3d15c ⫺ d2 4

Step 3: Write the middle term of the trinomial as two terms whose coefficients are ⫺2 and ⫺15. Step 4: Factor by grouping.

⫽ ⫺2c15c ⫺ d212c ⫺ 3d2 Skill Practice 2. Factor by using the ac-method.

⫺4wz 3 ⫺ 2w 2z 2 ⫹ 20w 3z

TIP: In Example 2, removing the GCF from the original trinomial produced a new trinomial with smaller coefficients. This makes the factoring process simpler because the product ac is smaller. Original trinomial ⫺20c ⫹ 34c d ⫺ 6cd 3

2

With the GCF factored out 2

ac ⫽ 1⫺2021⫺62 ⫽ 120

Answer

2. ⫺2wz 12z ⫹ 5w21z ⫺ 2w2

⫺2c110c2 ⫺ 17cd ⫹ 3d 2 2 ac ⫽ 1102132 ⫽ 30

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Section 4.6

Factoring Trinomials

2. Factoring Trinomials: Trial-and-Error Method Another method that is widely used to factor trinomials of the form ax2  bx  c is the trial-and-error method. To understand how the trial-and-error method works, first consider the multiplication of two binomials: Product of 3 ⴢ 2

Product of 2 ⴢ 1

⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭

12x  3211x  22  2x2  4x  3x  6  2x2  7x  6 sum of products of inner terms and outer terms

In Example 3, we will factor this trinomial by reversing this process.

Example 3

Factoring a Trinomial by the Trial-and-Error Method 2x2  7x  6

Factor by the trial-and-error method.

Solution: To factor by the trial-and-error method, we must fill in the blanks to create the correct product. Factors of 2

2x2  7x  6  1ⵧx

ⵧ21ⵧx

ⵧ2

Factors of 6

• •

•

The first terms in the binomials must be 2x and x. This creates a product of 2x2, which is the first term in the trinomial. The second terms in the binomials must form a product of 6. This means that the factors must both be positive or both be negative. Because the middle term of the trinomial is positive, we will consider only positive factors of 6. The options are 1 ⴢ 6, 2 ⴢ 3, 6 ⴢ 1, and 3 ⴢ 2. Test each combination of factors until the correct product of binomials is found.

12x  121x  62  2x2  12x  1x  6  2x2  13x  6 12x  221x  32  2x2  6x  2x  6  2x2  8x  6 12x  621x  12  2x  2x  6x  6  2x2  8x  6

Incorrect.

Wrong middle term.

Incorrect.

Wrong middle term.

Incorrect.

Wrong middle term.

2

12x  321x  22  2x2  4x  3x  6  2x2  7x  6

Correct.

The factored form of 2x  7x  6 is 12x  321x  22 . 2

Skill Practice Factor by the trial-and-error method. 3. 5y2  9y  4

Answer

3. 15y  421y  12

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When applying the trial-and-error method, sometimes it is not necessary to test all possible combinations of factors. For the trinomial, 2x2 ⫹ 7x ⫹ 6, the GCF is 1. Therefore, any binomial factor that shares a common factor greater than 1 will not work and does not need to be tested. For example, the following binomials cannot work: 12x ⫹ 621x ⫹ 12

⎫ ⎪ ⎬ ⎪ ⎭

⎫ ⎪ ⎬ ⎪ ⎭

12x ⫹ 221x ⫹ 32

Common factor of 2

Common factor of 2

Although the trial-and-error method is tedious, its principle is generally easy to remember. We reverse the process of multiplying binomials.

PROCEDURE The Trial-and-Error Method to Factor ax 2 ⴙ bx ⴙ c Step 1 Factor out the greatest common factor. Step 2 List all pairs of positive factors of a and pairs of positive factors of c. Consider the reverse order for either list of factors. Step 3 Construct two binomials of the form Factors of a

1ⵧ x

ⵧ21ⵧ x

ⵧ2

Factors of c

Step 4 Test each combination of factors and signs until the correct product is found. Step 5 If no combination of factors produces the correct product, the trinomial cannot be factored further and is a prime polynomial.

Example 4

Factoring a Trinomial by the Trial-and-Error Method

Factor by the trial-and-error method.

13y ⫺ 6 ⫹ 8y2

Solution: 8y2 ⫹ 13y ⫺ 6 1ⵧy

Write in the form ax2 ⫹ bx ⫹ c.

ⵧ21ⵧy

ⵧ2

Step 1: The GCF is 1.

Factors of 8

Factors of 6

1ⴢ8

1ⴢ6

2ⴢ4

2ⴢ3 3ⴢ2 6ⴢ1

12y 12y 12y 12y 11y

11y

1214y

62

2214y

32

3214y

22

6214y

12

1218y

62

3218y

22

Step 2: List the positive factors of 8 and positive factors of 6. Consider the reverse order in one list of factors.

¶ (reverse order)

Step 3: Construct all possible binomial factors by using different combinations of the factors of 8 and 6. y

Without regard to signs, these factorizations cannot work because the terms in the binomial share a common factor greater than 1.

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Section 4.6

Factoring Trinomials

Test the remaining factorizations. Keep in mind that to produce a product of 6, the signs within the parentheses must be opposite (one positive and one negative). Also, the sum of the products of the inner terms and outer terms must be combined to form 13y. 11y 6218y

11y 2218y

12

Incorrect. Wrong middle term. Regardless of signs, the product of inner terms 48y and the product of outer terms 1y cannot be combined to form the middle term 13y.

32

Correct.

The terms 16y and 3y can be combined to form the middle term 13y, provided the signs are applied correctly. We require 16y and 3y.

The correct factorization of 8y 2  13y  6 is 1y  2218y  32 . Skill Practice Factor by the trial-and-error method. 4. 5t  6  4t 2

In Example 4, the factors of 6 must have opposite signs to produce a negative product. Therefore, one binomial factor is a sum and one is a difference. Determining the correct signs is an important aspect of factoring trinomials. We suggest the following guidelines:

TIP: Given the trinomial ax2  bx  c 1a 7 02, the signs can be determined as follows: 1. If c is positive, then the signs in the binomials must be the same (either both positive or both negative). The correct choice is determined by the middle term. If the middle term is positive, then both signs must be positive. If the middle term is negative, then both signs must be negative. c is positive.

Example:

20x 2  43x  21 14x  32 15x  72

c is positive.

Example:

same signs

20x 2  43x  21 14x  3215x  72 same signs

2. If c is negative, then the signs in the binomials must be different. The middle term in the trinomial determines which factor gets the positive sign and which factor gets the negative sign. c is negative.

Example:

x 2  3x  28 1x  72 1x  42 different signs

c is negative.

Example:

x 2  3x  28 1x  721x  42 different signs

Answer

4. 14t  321t  22

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Example 5

Factoring a Trinomial by the Trial-and-Error Method

Factor by the trial-and-error method. ⫺80x3y ⫹ 208x2y2 ⫺ 20xy3

Solution: ⫺80x3y ⫹ 208x2y2 ⫺ 20xy3

⫽ ⫺4xy120x2 ⫺ 52xy ⫹ 5y2 2

Step 1: Factor out ⫺4xy.

⫽ ⫺4xy1ⵧx ⵧy21ⵧx ⵧy2

Factors of 20

Factors of 5

1 ⴢ 20

1ⴢ5

2 ⴢ 10

5ⴢ1

Step 2: List the positive factors of 20 and positive factors of 5. Consider the reverse order in one list of factors.

4ⴢ5 Step 3: Construct all possible binomial factors by using different combinations of the factors of 20 and factors of 5. The signs in the parentheses must both be negative. ⫺4xy11x ⫺ 1y2120x ⫺ 5y2 ⫺4xy14x ⫺ 1y215x ⫺ 5y2

d

⫺4xy12x ⫺ 1y2110x ⫺ 5y2

Incorrect. These binomials contain a common factor.

⫺4xy11x ⫺ 5y2120x ⫺ 1y2

Incorrect.

Wrong middle term. ⫺4xy1x ⫺ 5y2120x ⫺ 1y2 ⫽ ⫺4xy120x2 ⫺ 101xy ⫹ 5y2 2

⫺4xy14x ⫺ 5y215x ⫺ 1y2

Incorrect.

Wrong middle term. ⫺4xy14x ⫺ 5y215x ⫺ 1y2 ⫽ ⫺4xy120x2 ⫺ 29x ⫹ 5y2 2

⫺4xy12x ⫺ 5y2 110x ⫺ 1y2

Correct.

⫺4xy12x ⫺ 5y2 110x ⫺ 1y2 ⫽ ⴚ4xy120x 2 ⴚ 52xy ⴙ 5y2 2 ⫽ ⫺80x3y ⫹ 208x2y2 ⫺ 20xy3

The correct factorization of ⫺80x 3y ⫹ 208x 2y 2 ⫺ 20xy 3 is ⫺4xy12x ⫺ 5y2110x ⫺ y2. Skill Practice Factor by the trial-and-error method. 5. ⫺4z3 ⫺ 22z 2 ⫺ 30z

Answer

5. ⫺2z 12z ⫹ 521z ⫹ 32

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Section 4.6

Factoring Trinomials

Factoring a Trinomial by the Trial-and-Error Method

Example 6

Factor completely. 2x2  9x  14

Solution: 2x2  9x  14

The GCF is 1 and the trinomial is written in the form ax2  bx  c.

12x  1421x  12

Incorrect. (2x  14) contains a common factor of 2.

12x  221x  72 12x  121x  142  2x2  15x  14 12x  721x  22  2x  11x  14 2

Incorrect. (2x  2) contains a common factor of 2. Incorrect. Wrong middle term. Incorrect. Wrong middle term.

No combination of factors results in the correct product. Therefore, the trinomial is prime (cannot be factored). Skill Practice Factor completely. 6. 6r 2  13r  10

If a trinomial has a leading coefficient of 1, the factoring process simplifies significantly. Consider the trinomial x2  bx  c. To produce a leading term of x2, we can construct binomials of the form 1x  ⵧ21x  ⵧ2 . The remaining terms may be satisfied by two numbers p and q whose product is c and whose sum is b: Factors of c

Sum  b

冎

冧

1x  p21x  q2  x2  qx  px  pq  x2  1p  q2x  pq

Product  c

This process is demonstrated in Example 7.

Example 7

Factoring a Trinomial with a Leading Coefficient of 1

Factor completely. x2  10x  16

Solution: x2  10x  16  1x

ⵧ21x ⵧ2

Factor out the GCF from all terms. In this case, the GCF is 1. The trinomial is written in the form x2  bx  c. To form the product x2, use the factors x and x.

Answer 6. Prime

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Next, look for two numbers whose product is 16 and whose sum is 10. Because the middle term is negative, we will consider only the negative factors of 16. Factors of 16 11162 2182 4142

Sum

1  1162  17

2  182  10 4  142  8

The numbers are 2 and 8.

Therefore, x2  10x  16  1x  221x  82 . Skill Practice Factor completely. 7. c 2  6c  27

3. Factoring Perfect Square Trinomials Recall from Section 4.3 that the square of a binomial always results in a perfect square trinomial. 1a  b2 2  1a  b21a  b2  a2  ab  ab  b2  a2  2ab  b2 1a  b2 2  1a  b21a  b2  a2  ab  ab  b2  a2  2ab  b2

For example, 12x  72 2  12x2 2  212x2172  172 2  4x2  28x  49 a  2x

b7

a2  2ab  b2

To factor the trinomial 4x2  28x  49, the ac-method or the trial-and-error method can be used. However, recognizing that the trinomial is a perfect square trinomial, we can use one of the following patterns to reach a quick solution.

FORMULA Factored Form of a Perfect Square Trinomial TIP: The following are perfect squares. 12  1 22  4 32  9 42  16 o

(x1)2  x2 (x2)2  x4 (x3)2  x6 (x4)2  x8 o

Any expression raised to an even power (multiple of 2) is a perfect square.

Answer

7. 1c  921c  32

a2  2ab  b2  1a  b2 2 a2  2ab  b2  1a  b2 2

TIP: To determine if a trinomial is a perfect square trinomial, follow these steps: 1. Check if the first and third terms are both perfect squares with positive coefficients. 2. If this is the case, identify a and b, and determine if the middle term equals 2ab or 2ab.

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Section 4.6

Example 8

Factoring Trinomials

Factoring a Perfect Square Trinomial x2  12x  36

Factor completely.

Solution: x2  12x  36

The GCF is 1. • The first and third terms are positive. • The first term is a perfect square: x2  1x2 2 • The third term is a perfect square: 36  162 2 • The middle term is twice the product of x and 6:

Perfect squares

 x2  12x  36

12x  21x2162

 1x2 2  21x2162  162 2

The trinomial is in the form a2  2ab  b2, where a  x and b  6.

 1x  62 2

Factor as 1a  b2 2.

Skill Practice Factor completely. 8. x 2  2x  1

Example 9

Factoring a Perfect Square Trinomial

Factor completely.

4x2  36xy  81y2

Solution: 4x2  36xy  81y2

The GCF is 1. • The first and third terms are positive. • The first term is a perfect square: 4x2  12x2 2. • The third term is a perfect square: 81y2  19y2 2. • The middle term:

Perfect squares

 4x2  36xy  81y2

 12x2  212x219y2  19y2 2

36xy  212x219y2 2

 12x  9y2 2

The trinomial is in the form a2  2ab  b2, where a  2x and b  9y. Factor as 1a  b2 2.

Skill Practice Factor completely. 9. 9y2  12yz  4z 2

Answers

8. 1x  12 2 9. 13y  2z2 2

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4. Factoring by Using Substitution Sometimes it is convenient to use substitution to convert a polynomial into a simpler form before factoring. Example 10

Using Substitution to Factor a Polynomial

Factor by using substitution.

12x  72 2  312x  72  40

Solution:

12x  72 2  312x  72  40  u 2  3u  40

Substitute u  2x  7. The trinomial is simpler in form.

 1u  821u  52

 312x  72  84 312x  72  54

 12x  7  8212x  7  52

 12x  15212x  22

 12x  1521221x  12

Factor the trinomial. Reverse substitute. Replace u by 2x  7. Simplify. The second binomial has a GCF of 2. Factor out the GCF from the second binomial.

 212x  1521x  12 Skill Practice Factor by using substitution. 10. 13x  12 2  213x  12  15

Example 11

Using Substitution to Factor a Polynomial

Factor by using substitution.

TIP: The ac-method or trial-and-error method can also be used for Example 11 without using substitution.

6y6  5y3  4

Solution: 6y6  5y3  4

 61y3 2 2  51y3 2  4

Let u  y3.

 6u2  5u  4

Substitute u for y3 in the trinomial.

 12u  1213u  42

 12y  1213y  42 3

3

Factor the trinomial. Reverse substitute. Replace u with y3.

Skill Practice Factor by using substitution. 11. 2x4  7x2  3

Answers 10. 313x  221x  22 11. 12x 2  12 1x 2  32

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Section 4.6

Factoring Trinomials

As you work through the exercises in this section, keep these guidelines in mind to factor trinomials.

PROCEDURE Factoring Trinomials of the Form ax 2 ⴙ bx ⴙ c (a ⴝ 0) When factoring trinomials, the following guidelines should be considered: Step 1 Factor out the greatest common factor. Step 2 Check to see if the trinomial is a perfect square trinomial. If so, factor it as either 1a  b2 2 or 1a  b2 2. (With a perfect square trinomial, you do not need to use the ac-method or trial-and-error method.) Step 3 If the trinomial is not a perfect square, use either the ac-method or the trial-and-error method to factor. Step 4 Check the factorization by multiplication. Note: Consider using substitution if a trinomial is in the form au2  bu  c, where u is an algebraic expression.

Section 4.6 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms: a. Prime polynomial

b. Perfect square trinomial

Review Exercises 2. Explain how to check a factoring problem. For Exercises 3–8, factor the polynomial completely. 3. 36c2d7e11  12c3d5e15  6c2d4e7

4. 5x3y3  15x4y2  35x2y4

5. 2x13a  b2  13a  b2

6. 61v  82  3u1v  82

7. wz2  2wz  33az  66a

8. 3a2x  9ab  abx  3b2

Concepts 1–2: Factoring Trinomials In Exercises 9–46, factor the trinomial completely by using any method. Remember to look for a common factor first. (See Examples 1–7.) 9. b2  12b  32

10. a2  12a  27

11. y2  10y  24

12. w2  3w  54

13. x2  13x  30

14. t 2  9t  8

15. c2  6c  16

16. z2  3z  28

17. 2x2  7x  15

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18. 2y2 ⫺ 13y ⫹ 15

19. a ⫹ 6a2 ⫺ 5

20. 10b2 ⫺ 3 ⫺ 29b

21. s2 ⫹ st ⫺ 6t2

22. p2 ⫺ pq ⫺ 20q2

23. 3x2 ⫺ 60x ⫹ 108

24. 4c2 ⫹ 12c ⫺ 72

25. 2c2 ⫺ 2c ⫺ 24

26. 3x2 ⫹ 12x ⫺ 15

27. 2x2 ⫹ 8xy ⫺ 10y2

28. 20z2 ⫹ 26zw ⫺ 28w2

29. 33t2 ⫺ 18t ⫹ 2

30. 5p2 ⫺ 10p ⫹ 7

31. 3x2 ⫹ 14xy ⫹ 15y2

32. 2a2 ⫹ 15ab ⫺ 27b2

33. 5u3v ⫺ 30u2v2 ⫹ 45uv3

34. 3a3 ⫹ 30a2b ⫹ 75ab2

35. x3 ⫺ 5x2 ⫺ 14x

36. p3 ⫹ 2p2 ⫺ 24p

37. ⫺23z ⫺ 5 ⫹ 10z2

38. 3 ⫹ 16y2 ⫹ 14y

39. b2 ⫹ 2b ⫹ 15

40. x2 ⫺ x ⫺ 1

41. ⫺2t 2 ⫹ 12t ⫹ 80

42. ⫺3c2 ⫹ 33c ⫺ 72

43. 14a2 ⫹ 13a ⫺ 12

44. 12x2 ⫺ 16x ⫹ 5

45. 6a2b ⫹ 22ab ⫹ 12b

46. 6cd2 ⫹ 9cd ⫺ 42c

Concept 3: Factoring Perfect Square Trinomials 47. a. Multiply the binomials 1x ⫹ 521x ⫹ 52 .

48. a. Multiply the binomials 12w ⫺ 5212w ⫺ 52 .

b. Factor x2 ⫹ 10x ⫹ 25.

b. Factor 4w2 ⫺ 20w ⫹ 25.

49. a. Multiply the binomials 13x ⫺ 2y2 2 .

50. a. Multiply the binomials 1x ⫹ 7y2 2 .

b. Factor 9x2 ⫺ 12xy ⫹ 4y2.

b. Factor x2 ⫹ 14xy ⫹ 49y2.

For Exercises 51–54, fill in the blank to make the trinomial a perfect square trinomial. 51. 9x2 ⫹ 1 53. 64z4 ⫹ 1

2 ⫹ 25

52. 16x4 ⫺ 1

2⫹1

2 ⫹ t2

54. 9m4 ⫺ 1

2 ⫹ 49n2

For Exercises 55–66, factor out the greatest common factor, if necessary. Then determine if the polynomial is a perfect square trinomial. If it is, factor it. (See Examples 8–9.) 55. y2 ⫺ 8y ⫹ 16

56. x2 ⫹ 10x ⫹ 25

57. 64m2 ⫹ 80m ⫹ 25

58. 100c2 – 140c + 49

59. w2 ⫺ 5w ⫹ 9

60. 2a2 ⫹ 14a ⫹ 98

61. 9a2 ⫺ 30ab ⫹ 25b2

62. 16x4 ⫺ 48x2y ⫹ 9y2

63. 16t 2 ⫺ 80tv ⫹ 20v2

64. 12x2 ⫺ 12xy ⫹ 3y2

65. 5b4 ⫺ 20b2 ⫹ 20

66. a4 ⫹ 12a2 ⫹ 36

Concept 4: Factoring by Using Substitution For Exercises 67–70, factor the polynomial in part (a). Then use substitution to help factor the polynomials in parts (b) and (c). 67. a. u2 ⫺ 10u ⫹ 25

68. a. u2 ⫹ 12u ⫹ 36

b. x4 ⫺ 10x2 ⫹ 25

b. y4 ⫹ 12y2 ⫹ 36

c. 1a ⫹ 12 2 ⫺ 101a ⫹ 12 ⫹ 25

c. 1b ⫺ 22 2 ⫹ 121b ⫺ 22 ⫹ 36

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Section 4.6

69. a. u2  11u  26

Factoring Trinomials

375

70. a. u2  17u  30

b. w6  11w3  26

b. z6  17z3  30

c. 1y  42 2  111y  42  26

c. 1x  32 2  171x  32  30

For Exercises 71–82, factor by using substitution. (See Examples 10–11.) 71. 13x  12 2  13x  12  6

72. 12x  52 2  12x  52  12

73. 21x  52 2  91x  52  4

74. 41x  32 2  71x  32  3

75. 31y  42 2  51y  42  2

76. 13t  22 2  13t  22  20

77. 3y6  11y3  6

78. 3x4  5x2  12

79. 4p4  5p2  1

80. t 4  3t 2  2

81. x4  15x2  36

82. t6  16t 3  63

Mixed Exercises 83. A student factored 4y2  10y  4 as 12y  1212y  42 on her factoring test. Why did her professor deduct several points, even though 12y  1212y  42 does multiply out to 4y2  10y  4?

84. A student factored 9w2  36w  36 as 13w  62 2 on his factoring test. Why did his instructor deduct several points, even though 13w  62 2 does multiply out to 9w2  36w  36?

For Exercises 85–105, factor completely by using an appropriate method. (Be sure to note the number of terms in the polynomial.) 85. w4  12w 2  36

86. 9  6t 2  t 4

87. 81w2  90w  25

88. 49a2  28ab  4b2

89. 3x1a  b2  61a  b2

90. 4p1t  82  21t  82

91. 12a2bc2  4ab2c2  6abc3

92. 18x2z  6xyz  30xz2

93. 20x3  74x2  60x

94. 24y3  90y2  75y

95. 2y2  9y  4

96. 3w2  12w  4

97. 21w2  52 2  1w2  52  15

98. 51t2  32 2  211t2  32  4

99. 1  4d  3d2

100. 2  5a  2a2

101. ax  5a2  2bx  10ab

102. my  y2  3xm  3xy

103. 8z2  24zw  224w2

104. 9x2  18xy  135y2

105. ay  ax  5cy  5cx

For Exercises 106–114, factor the expression that defines each function. 106. f1x2  2x2  13x  7

107. g1x2  3x2  14x  8

108. m1t2  t2  22t  121

109. n1t2  t2  20t  100

110. P1x2  x3  4x2  3x

111. Q1x2  x4  6x3  8x2

112. h1a2  a3  5a2  6a  30

113. k1a2  a3  4a2  2a  8

114. f 1x2  3x 3  9x2  5x  15

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Section 4.7

Factoring Binomials

Concepts

1. Difference of Squares

1. Difference of Squares 2. Using a Difference of Squares in Grouping 3. Sum and Difference of Cubes 4. Summary of Factoring Binomials 5. Factoring Binomials of the Form x 6 ⴚ y 6

Up to this point we have learned how to • • • •

Factor out the greatest common factor from a polynomial. Factor a four-term polynomial by grouping. Recognize and factor perfect square trinomials. Factor trinomials by the ac-method and by the trial-and-error method.

Next, we will learn how to factor binomials that fit the pattern of a difference of squares. Recall from Section 4.3 that the product of two conjugates results in a difference of squares 1a  b21a  b2  a2  b2 Therefore, to factor a difference of squares, the process is reversed. Identify a and b and construct the conjugate factors.

FORMULA Factored Form of a Difference of Squares a2  b2  1a  b21a  b2

Example 1

Factoring a Difference of Squares

Factor the binomial completely.

16x2  9

Solution: 16x2  9  14x2 2  132 2

 14x  3214x  32

The GCF is 1. The binomial is a difference of squares. Write in the form a2  b2, where a  4x and b  3. Factor as 1a  b21a  b2 .

Skill Practice Factor completely. 1. 4z 2  1

Example 2

Factoring a Difference of Squares

Factor the binomial completely.

98c2d  50d3

Solution: 98c2d  50d3

 2d149c 2  25d 2 2  2d3 17c2 2  15d2 2 4  2d17c  5d217c  5d2

Answers

1. 12z  12 12z  12 2. 7yz 1y  3z2 1y  3z2

The GCF is 2d. The resulting binomial is a difference of squares. Write in the form a2  b2, where a  7c and b  5d. Factor as 1a  b21a  b2.

Skill Practice Factor completely. 2. 7y3z  63yz3

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Section 4.7

Example 3

Factoring Binomials

Factoring a Difference of Squares

Factor the binomial completely.

z4  81

Solution: z4  81

The GCF is 1. The binomial is a difference of squares.

 1z2 2 2  192 2

Write in the form a2  b2, where a  z2 and b  9.

 1z2  921z2  92

Factor as 1a  b21a  b2.

z2  9 is also a difference of squares.

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

 1z2  921z  321z  32 Skill Practice Factor completely. 3. b4  16 The difference of squares a2  b2 factors as 1a  b21a  b2. However, the sum of squares is not factorable.

PROPERTY Sum of Squares Suppose a and b have no common factors. Then the sum of squares a2  b2 is not factorable over the real numbers. That is, a2  b2 is prime over the real numbers.

To see why a2  b2 is not factorable, consider the product of binomials: 1a

b21a

b2 ⱨ a2  b2

If all possible combinations of signs are considered, none produces the correct product. 1a  b21a  b2  a2  b2

1a  b21a  b2  a  2ab  b 2

2

1a  b21a  b2  a2  2ab  b2

Wrong sign Wrong middle term Wrong middle term

After exhausting all possibilities, we see that if a and b share no common factors, then the sum of squares a2  b2 is a prime polynomial.

Answer

3. 1b 2  421b  221b  22

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2. Using a Difference of Squares in Grouping Sometimes a difference of squares can be used along with other factoring techniques.

Using a Difference of Squares in Grouping

Example 4

y3 ⫺ 6y2 ⫺ 4y ⫹ 24

Factor completely.

Solution: y3 ⫺ 6y2 ⫺ 4y ⫹ 24 ⫽ y ⫺ 6y 3

⫺ 4y ⫹ 24

2

⫽ y2 1y ⫺ 62 ⫺ 41y ⫺ 62 ⫽ 1y ⫺ 621y2 ⫺ 42

The GCF is 1. The polynomial has four terms. Factor by grouping. y2 ⫺ 4 is a difference of squares.

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

⫽ 1y ⫺ 621y ⫹ 221y ⫺ 22 Skill Practice Factor completely. 4. a 3 ⫹ 5a 2 ⫺ 9a ⫺ 45

Factoring a Four-Term Polynomial by Grouping Three Terms

Example 5 Factor completely.

x2 ⫺ y2 ⫺ 6y ⫺ 9

Solution: Grouping “2 by 2” will not work to factor this polynomial. However, if we factor out ⫺1 from the last three terms, the resulting trinomial will be a perfect square trinomial. x2

⫺ y2 ⫺ 6y ⫺ 9 ⫽ x ⫺ 11y ⫹ 6y ⫹ 92 2

Avoiding Mistakes When factoring the expression x 2 ⫺ 1y ⫹ 32 2 as a difference of squares, be sure to use parentheses around the quantity 1y ⫹ 32. This will help you remember to "distribute the negative” in the expression 3 x ⫺ 1y ⫹ 32 4 . 冤 x ⫺ 1y ⫹ 32 冥 ⫽ 1x ⫺ y ⫺ 32

2

⫽ x2 ⫺ 1y ⫹ 32 2

⫽ 冤x ⫺ 1y ⫹ 32冥 冤x ⫹ 1y ⫹ 32 冥

⫽ 1x ⫺ y ⫺ 321x ⫹ y ⫹ 32

Skill Practice Factor completely. 5. x 2 ⫹ 10x ⫹ 25 ⫺ y 2 Answers

4. 1a ⫹ 52 1a ⫺ 321a ⫹ 32 5. 1x ⫹ 5 ⫺ y2 1x ⫹ 5 ⫹ y2

Group the last three terms. Factor out ⫺1 from the last three terms. Factor the perfect square trinomial y2 ⫹ 6y ⫹ 9 as 1y ⫹ 32 2.

The quantity x2 ⫺ 1y ⫹ 32 2 is a difference of squares, a2 ⫺ b2, where a ⫽ x and b ⫽ 1y ⫹ 32. Factor as a2 ⫺ b2 ⫽ 1a ⫹ b21a ⫺ b2.

Apply the distributive property to clear the inner parentheses.

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Section 4.7

379

Factoring Binomials

TIP: From Example 5, the expression x2  1 y  32 2 can also be factored by using substitution. Let u  y  3. x2  1 y  32 2  x2  u2

Substitution u  y  3.

 1x  u2 1x  u2

Factor as a difference of squares.

 冤x  1 y  32冥 冤x  1 y  32冥

 1x  y  321x  y  32

Substitute back. Apply the distributive property.

3. Sum and Difference of Cubes For binomials that represent the sum or difference of cubes, factor by using the following formulas.

TIP: The following are

FORMULA Factored Form of a Sum and Difference of Cubes Sum of cubes:

a  b  1a  b21a  ab  b 2 3

3

2

2

Difference of cubes: a3  b3  1a  b21a2  ab  b2 2

perfect cubes. 13  1 23  8 33  27 43  64 o

Multiplication can be used to confirm the formulas for factoring a sum or difference of cubes. 1a  b21a2  ab  b2 2  a3  a2b  ab2  a2b  ab2  b3  a3  b3 ✔

(x1)3  x3 (x2)3  x6 (x3)3  x9 (x4)3  x12 o

Any expression raised to a multiple of 3 is a perfect cube.

1a  b21a2  ab  b2 2  a3  a2b  ab2  a2b  ab2  b3  a3  b3 ✔

To help you remember the formulas for factoring a sum or difference of cubes, keep the following guidelines in mind. • The factored form is the product of a binomial and a trinomial. • The first and third terms in the trinomial are the squares of the terms within the binomial factor. Therefore, these terms are always positive. • Without regard to sign, the middle term in the trinomial is the product of terms in the binomial factor. Square the first term of the binomial.

Product of terms in the binomial

x3  8  1x2 3  122 3  1x  22 3 1x2 2  1x2122  122 2 4 Square the last term of the binomial.

• The sign within the binomial factor is the same as the sign of the original binomial. • The first and third terms in the trinomial are always positive. • The sign of the middle term in the trinomial is opposite the sign within the binomial. Same sign

Positive

x3  8  1x2 3  122 3  1x  22 3 1x2 2  1x2122  122 2 4 Opposite signs

TIP: To help remember the placement of the signs in factoring the sum or difference of cubes, remember SOAP: Same sign, Opposite signs, Always Positive.

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Example 6 Factor.

Factoring a Difference of Cubes

8x3  27

Solution: 8x3  27  12x2 3  132 3 a3  b3  1a  b21a2  ab  b2 2

12x2 3  132 3  12x  32 3 12x2 2  12x2132  132 2 4  12x  3214x2  6x  92

8x3 and 27 are perfect cubes. Write as a3  b3, where a  2x and b  3.

Apply the difference of cubes formula. Simplify.

Skill Practice Factor completely. 6. 125p 3  8

Example 7 Factor.

Factoring a Sum of Cubes

125t3  64z6

Solution: 125t3  64z6  15t2 3  14z2 2 3 a3  b3  1a  b21a2  ab  b2 2 15t2 3  14z2 2 3  3 15t2  14z2 2 4 3 15t2 2  15t214z2 2  14z2 2 2 4  15t  4z2 2125t2  20tz2  16z4 2

125t 3 and 64z6 are perfect cubes. Write as a3  b3, where a  5t and b  4z2. Apply the sum of cubes formula.

Simplify.

Skill Practice Factor completely. 7. x 3  1000y6

4. Summary of Factoring Binomials After factoring out the greatest common factor, the next step in any factoring problem is to recognize what type of pattern it follows. Exponents that are divisible by 2 are perfect squares, and those divisible by 3 are perfect cubes. The formulas for factoring binomials are summarized here.

Answers

6. 15p  22 125p 2  10p  42 7. 1x  10y 2 2 1x 2  10xy 2  100y 4 2

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Section 4.7

Factoring Binomials

SUMMARY Factoring Binomials • Difference of squares: • Difference of cubes: • Sum of cubes:

Example 8

a2  b2  1a  b21a  b2

a3  b3  1a  b21a2  ab  b2 2

a3  b3  1a  b21a2  ab  b2 2

Review of Factoring Binomials

Factor the binomials. a. m3 

1 8

b. 9k2  24m2

c. 128y6  54x3

d. 50y6  8x2

Solution: a. m3 

1 8

1 3  1m2 3  a b 2 1 1 1  am  b am2  m  b 2 2 4 b. 9k2  24m2

 313k2  8m2 2

c. 128y6  54x3

 2164y6  27x3 2  23 14y2 2 3  13x2 3 4

 214y2  3x2116y4  12xy2  9x2 2 d. 50y6  8x2

 2125y  4x 2 6

2

 23 15y3 2 2  12x2 2 4  215y3  2x215y3  2x2

m3 is a perfect cube: m3  1m2 3. 1 1 3 1 8 is a perfect cube: 8  1 2 2 . This is a difference of cubes, where a  m and b  12:

a3  b3  1a  b21a2  ab  b2 2. Factor out the GCF. The resulting binomial is not a difference of squares or a sum or difference of cubes. It cannot be factored further over the real numbers. Factor out the GCF. Both 64 and 27 are perfect cubes, and the exponents of both x and y are multiples of 3. This is a sum of cubes, where a  4y2 and b  3x.

a3  b3  1a  b21a2  ab  b2 2. Factor out the GCF.

Both 25 and 4 are perfect squares. The exponents of both x and y are multiples of 2. This is a difference of squares, where a  5y3 and b  2x. a2  b2  1a  b21a  b2.

Skill Practice Factor the binomials. 8. x2 

1 25

9. 16y3  4y

10. 24a7  3a

11. 18p4  50t 2

Answers 1 1 8. ax  bax  b 5 5 9. 4y 14y 2  12 10. 3a 12a 2  1214a 4  2a 2  12 11. 213p 2  5t 213p 2  5t 2

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5. Factoring Binomials of the Form x6 ⴚ y6 Example 9

Factoring Binomials

Factor the binomial x6 ⫺ y6 as a. A difference of cubes b. A difference of squares

Solution: Notice that the expressions x6 and y6 are both perfect squares and perfect cubes because the exponents are both multiples of 2 and of 3. Consequently, x6 ⫺ y6 can be interpreted initially as either a difference of cubes or a difference of squares. a. x6 ⫺ y6 Difference of cubes

⫽ 1x2 2 3 ⫺ 1y2 2 3 ⫽ 1x2 ⫺ y2 2 3 1x2 2 2 ⫹ 1x2 21y2 2 ⫹ 1y2 2 2 4

Apply the formula a3 ⫺ b3 ⫽ 1a ⫺ b21a2 ⫹ ab ⫹ b2 2. Factor x2 ⫺ y2 as a difference of squares.

⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

⫽ 1x2 ⫺ y2 21x4 ⫹ x2y2 ⫹ y4 2

Write as a3 ⫺ b3, where a ⫽ x2 and b ⫽ y2.

⫽ 1x ⫹ y21x ⫺ y21x4 ⫹ x2y2 ⫹ y4 2

The expression x4 ⫹ x2y2 ⫹ y4 cannot be factored by using the skills learned thus far.

b. x6 ⫺ y6 Difference of squares

⫽ 1x3 2 2 ⫺ 1y3 2 2

Write as a2 ⫺ b2, where a ⫽ x3 and b ⫽ y3.

⫽ 1x3 ⫹ y3 21x3 ⫺ y3 2 Difference of cubes

Factor x3 ⫹ y3 as a sum of cubes. Factor x3 ⫺ y3 as a difference of cubes. ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Sum of cubes

Apply the formula a2 ⫺ b2 ⫽ 1a ⫹ b21a ⫺ b2.

⫽ 1x ⫹ y21x2 ⫺ xy ⫹ y2 21x ⫺ y21x2 ⫹ xy ⫹ y2 2

TIP: If given a choice between factoring a binomial as a difference of squares or as a difference of cubes, it is recommended that you factor initially as a difference of squares. As Example 9 illustrates, factoring as a difference of squares leads to a more complete factorization. a6 ⫺ b6 ⫽ 1a ⫺ b21a2 ⫹ ab ⫹ b2 21a ⫹ b21a2 ⫺ ab ⫹ b2 2

Answer

12. 1a ⫺ 221a ⫹ 221a 2 ⫹ 2a ⫹ 42 1a 2 ⫺ 2a ⫹ 42

Skill Practice Factor completely. 12. a6 ⫺ 64

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Section 4.7

Section 4.7

Factoring Binomials

383

Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercises 1. Multiplying polynomials and factoring polynomials are inverse operations. That is, to check a factoring problem you can multiply, and to check a multiplication problem you can factor. To practice both operations, write a factored polynomial on one side of a 3 ⫻ 5 card with the directions, Multiply. On the other side of the card, write the expanded form of the polynomial with the directions, Factor. Now you can mix up the cards and get a good sense of what is meant by the directions: Factor and Multiply. 2. Define the key terms. a. Difference of squares

b. Sum of squares

c. Sum of cubes

d. Difference of cubes

Review Exercises For Exercises 3–8, factor completely. 3. 4x2 ⫺ 20x ⫹ 25

4. 9t2 ⫺ 42t ⫹ 49

5. 10x ⫹ 6xy ⫹ 5 ⫹ 3y

6. 21a ⫹ 7ab ⫺ 3b ⫺ b2

7. 32p2 ⫺ 28p ⫺ 4

8. 6q2 ⫹ 37q ⫺ 35

Concept 1: Difference of Squares 9. Explain how to identify and factor a difference of squares.

10. Can you factor 25x2 ⫹ 4?

For Exercises 11–22, factor the binomials. Identify the binomials that are prime. (See Examples 1–3.) 11. x2 ⫺ 9

12. y2 ⫺ 25

13. 16 ⫺ 49w2

14. 81 ⫺ 64b2

15. 8a2 ⫺ 162b2

16. 50c2 ⫺ 72d2

17. 25u2 ⫹ 1

18. w2 ⫹ 4

19. 2a4 ⫺ 32

20. 5y4 ⫺ 5

21. 49 ⫺ k6

22. 4 ⫺ h6

Concept 2: Using a Difference of Squares in Grouping For Exercises 23–36, use the difference of squares along with factoring by grouping. (See Examples 4–5.) 23. x3 ⫺ x2 ⫺ 16x ⫹ 16

24. x3 ⫹ 5x2 ⫺ x ⫺ 5

25. 4x3 ⫹ 12x2 ⫺ x ⫺ 3

26. 5x3 ⫺ x2 ⫺ 45x ⫹ 9

27. 9y3 ⫹ 7y2 ⫺ 36y ⫺ 28

28. 9z3 ⫺ 5z2 ⫺ 36z ⫹ 20

29. 49x2 ⫹ 28x ⫹ 4 ⫺ y2

30. 100y2 ⫹ 140y ⫹ 49 ⫺ z2

31. w2 ⫺ 9n2 ⫹ 6n ⫺ 1

32. m2 ⫺ 25c2 ⫹ 20c ⫺ 4

33. p4 ⫺ 10p2 ⫹ 25 ⫺ t4

34. m4 ⫺ 14m2 ⫹ 49 ⫺ z4

35. 9u4 ⫺ 4v4 ⫹ 20v2 ⫺ 25

36. x4 ⫺ 9y4 ⫺ 42y2 ⫺ 49

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Concept 3: Sum and Difference of Cubes 37. Explain how to identify and factor a sum of cubes. 38. Explain how to identify and factor a difference of cubes. For Exercises 39–52, factor the sum or difference of cubes. (See Examples 6–7.) 39. 8x3 ⫺ 1 (Check by multiplying.)

40. y3 ⫹ 64 (Check by multiplying.)

41. 125c3 ⫹ 27

42. 216u3 ⫺ v3

43. x3 ⫺ 1000

44. y3 ⫺ 27

45. 64t 6 ⫹ 1

46. 125r6 ⫹ 1

47. 2000y6 ⫹ 2x3

48. 3a6 ⫹ 24b3

49. 16z4 ⫺ 54z

50. x5 ⫺ 64x2

51. p12 ⫺ 125

52. t9 ⫺ 8

55. 18d12 ⫺ 32

56. 3z8 ⫺ 12

Concept 4: Summary of Factoring Binomials For Exercises 53–80, factor completely. (See Example 8.) 53. 36y2 ⫺

1 25

54. 16p2 ⫺

1 9

57. 242v2 ⫹ 32

58. 8p2 ⫹ 200

59. 4x2 ⫺ 16

60. 9m2 ⫺ 81n2

61. 25 ⫺ 49q2

62. 1 ⫺ 25p2

63. 1t ⫹ 2s2 2 ⫺ 36

64. 15x ⫹ 42 2 ⫺ y2

65. 27 ⫺ t3

66. 8 ⫹ y3

67. 27a3 ⫹

69. 2m3 ⫹ 16

70. 3x3 ⫺ 375

71. x4 ⫺ y4

73. a9 ⫹ b9

74. 27m9 ⫺ 8n9

75.

1 3 1 p ⫺ 8 125

76. 1 ⫺

77. 4w2 ⫹ 25

78. 64 ⫹ a2

79.

1 2 1 2 x ⫺ y 25 4

80.

1 8

68. b3 ⫹

27 125

72. 81u4 ⫺ 16v4 1 3 d 27

1 2 4 a ⫺ b2 100 49

Concept 5: Factoring Binomials of the Form x 6 ⴚ y 6 For Exercises 81–88, factor completely. (See Example 9.) 81. a6 ⫺ b6 (Hint: First factor as a difference of squares.) 82. 64x6 ⫺ y6

83. 64 ⫺ y6

84. 1 ⫺ p6

86. 27q6 ⫹ 125p6

87. 8x6 ⫹ 125

88. t6 ⫹ 1

85. h6 ⫹ k6 (Hint: Factor as a sum of cubes.)

Mixed Exercises 89. Find a difference of squares that has 12x ⫹ 32 as one of its factors. 91. Find a difference of cubes that has 14a2 ⫹ 6a ⫹ 92 as its trinomial factor.

90. Find a difference of squares that has 14 ⫺ p2 as one of its factors. 92. Find a sum of cubes that has 125c2 ⫺ 10cd ⫹ 4d 2 2 as its trinomial factor.

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Problem Recognition Exercises

93. Find a sum of cubes that has 14x2  y2 as its binomial factor.

94. Find a difference of cubes that has 13t  r 2 2 as its binomial factor.

95. Consider the shaded region:

96. A manufacturer needs to know the area of a metal washer. The outer radius of the washer is R and the inner radius is r.

a. Find an expression that represents the area of the shaded region. b. Factor the expression found in part (a).

a. Find an expression that represents the area of the washer.

c. Find the area of the shaded region if x  6 in. and y  4 in.

b. Factor the expression found in part (a).

x

c. Find the area of the washer if R  12 in. and r  14 in. (Round to the nearest 0.01 in.2)

y

R y

x

r

Expanding Your Skills For Exercises 97–100, factor the polynomials by using the difference of squares, sum of cubes, or difference of cubes with grouping. 97. x2  y2  x  y

98. 64m2  25n2  8m  5n

99. x3  y3  x  y

100. 4pu3  4pv3  7yu3  7yv3

Problem Recognition Exercises Factoring Summary We now review the techniques of factoring presented thus far along with a general strategy for factoring polynomials.

PROCEDURE Factoring Strategy Step 1 Factor out the greatest common factor (Section 4.5). Step 2 Identify whether the polynomial has two terms, three terms, or more than three terms. Step 3 If the polynomial has more than three terms, try factoring by grouping (Section 4.5 and Section 4.7). Step 4 If the polynomial has three terms, check first for a perfect square trinomial. Otherwise, factor the trinomial with the ac-method or the trial-and-error method (Section 4.6). Step 5 If the polynomial has two terms, determine if it fits the pattern for a difference of squares, difference of cubes, or sum of cubes. Remember, a sum of squares is not factorable over the real numbers (Section 4.7). Step 6 Be sure to factor the polynomial completely. Step 7 Check by multiplying.

385

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Chapter 4 Polynomials

1. What is meant by a prime factor? 2. What is the first step in factoring any polynomial? 3. When factoring a binomial, what patterns do you look for? 4. When factoring a trinomial, what pattern do you look for first? 5. What do you look for when factoring a four-term polynomial? 6. How would you use substitution to factor 314x2 ⫹ 12 2 ⫹ 2014x2 ⫹ 12 ⫹ 12. For Exercises 7–66, a. Identify the category in which the polynomial best fits (you may need to factor out the GCF first). Choose from • difference of squares

• sum of squares

• difference of cubes

• sum of cubes

• perfect square trinomial

• trinomial (ac-method or trial-and-error)

• four terms—grouping

• none of these

b. Factor the polynomial completely. 7. 6x2 ⫺ 21x ⫺ 45

8. 8m3 ⫺ 10m2 ⫺ 3m

9. 8a2 ⫺ 50

10. ab ⫹ ay ⫺ b2 ⫺ by

11. 14u2 ⫺ 11uv ⫹ 2v2

12. 9p2 ⫺ 12pq ⫹ 4q2

13. 16x3 ⫺ 2

14. 9m2 ⫹ 16n2

15. 27y3 ⫹ 125

16. 3x2 ⫺ 16

17. 128p6 ⫹ 54q3

18. 5b2 ⫺ 30b ⫹ 45

19. 16a4 ⫺ 1

20. 81u2 ⫺ 90uv ⫹ 25v2

21. p2 ⫺ 12p ⫹ 36 ⫺ c2

22. 4x2 ⫹ 16

23. 12ax ⫺ 6ay ⫹ 4bx ⫺ 2by

24. 125y3 ⫺ 8

25. 5y2 ⫹ 14y ⫺ 3

26. 2m4 ⫺ 128

27. t 2 ⫺ 100

28. 4m2 ⫺ 49n2

29. y3 ⫹ 27

30. x3 ⫹ 1

31. d 2 ⫹ 3d ⫺ 28

32. c2 ⫹ 5c ⫺ 24

33. x2 ⫺ 12x ⫹ 36

34. p2 ⫹ 16p ⫹ 64

35. 2ax2 ⫺ 5ax ⫹ 2bx ⫺ 5b

36. 8x2 ⫺ 4bx ⫹ 2ax ⫺ ab

37. 10y2 ⫹ 3y ⫺ 4

38. 12z2 ⫹ 11z ⫹ 2

39. 10p2 ⫺ 640

40. 50a2 ⫺ 72

41. z4 ⫺ 64z

42. t4 ⫺ 8t

43. b3 ⫺ 4b2 ⫺ 45b

44. y3 ⫺ 14y2 ⫹ 40y

45. 9w2 ⫹ 24wx ⫹ 16x2

46. 4k2 ⫺ 20kp ⫹ 25p2

47. 60x2 ⫺ 20x ⫹ 30ax ⫺ 10a

48. 50x2 ⫺ 200x ⫹ 10cx ⫺ 40c

49. w4 ⫺ 16

50. k4 ⫺ 81

51. t 6 ⫺ 8

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52. p6 ⫹ 27

53. 8p2 ⫺ 22p ⫹ 5

54. 9m2 ⫺ 3m ⫺ 20

55. 36y2 ⫺ 12y ⫹ 1

56. 9a2 ⫹ 42a ⫹ 49

57. 2x2 ⫹ 50

58. 4y2 ⫹ 64

59. 12r 2s2 ⫹ 7rs2 ⫺ 10s2

60. 7z2w2 ⫺ 10zw2 ⫺ 8w2

61. x2 ⫹ 8xy ⫺ 33y2

62. s2 ⫺ 9st ⫺ 36t 2

63. m6 ⫹ n3

64. a3 ⫺ b6

65. x2 ⫺ 4x

66. y2 ⫺ 9y

For Exercises 67–101, factor completely using the strategy found on page 385. 67. x2 1x ⫹ y2 ⫺ y2 1x ⫹ y2

68. u2 1u ⫺ v2 ⫺ v2 1u ⫺ v2

69. 1a ⫹ 32 4 ⫹ 61a ⫹ 32 5

70. 14 ⫺ b2 4 ⫺ 214 ⫺ b2 3

71. 2413x ⫹ 52 3 ⫺ 3013x ⫹ 52 2

72. 1012y ⫹ 32 2 ⫹ 1512y ⫹ 32 3

73.

1 2 1 1 x ⫹ x⫹ 100 35 49

76. 1x3 ⫹ 42 2 ⫺ 101x3 ⫹ 42 ⫹ 24 79. y3 ⫹

1 64

74.

1 2 1 1 a ⫹ a⫹ 25 15 36

77. 16p4 ⫺ q4 80. z3 ⫹

1 125

12 1 1 t ⫹ t⫹ 9 6 16

75. 15x2 ⫺ 12 2 ⫺ 415x2 ⫺ 12 ⫺ 5 78. s4t4 ⫺ 81 81. 6a3 ⫹ a2b ⫺ 6ab2 ⫺ b3 1 2 1 1 y ⫹ y⫹ 25 5 4

82. 4p3 ⫹ 12p2q ⫺ pq2 ⫺ 3q3

83.

85. x2 ⫹ 12x ⫹ 36 ⫺ a2

86. a2 ⫹ 10a ⫹ 25 ⫺ b2

87. p2 ⫹ 2pq ⫹ q2 ⫺ 81

88. m2 ⫺ 2mn ⫹ n2 ⫺ 9

89. b2 ⫺ 1x2 ⫹ 4x ⫹ 42

90. p2 ⫺ 1y2 ⫺ 6y ⫹ 92

91. 4 ⫺ u2 ⫹ 2uv ⫺ v2

92. 25 ⫺ a2 ⫺ 2ab ⫺ b2

93. 6ax ⫺ by ⫹ 2bx ⫺ 3ay

94. 5pq ⫺ 12 ⫺ 4q ⫹ 15p

95. u6 ⫺ 64 [Hint: Factor first as a difference of squares, 1u3 2 2 ⫺ 182 2.]

96. 1 ⫺ v6

97. x8 ⫺ 1

99. a2 ⫺ b2 ⫹ a ⫹ b

100. 25c2 ⫺ 9d 2 ⫹ 5c ⫺ 3d

84.

98. y8 ⫺ 256 101. 5wx3 ⫹ 5wy3 ⫺ 2zx3 ⫺ 2zy3

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Chapter 4 Polynomials

Section 4.8

Solving Equations by Using the Zero Product Rule

Concepts

1. Solving Equations by Using the Zero Product Rule

1. Solving Equations by Using the Zero Product Rule 2. Applications of Quadratic Equations 3. Definition of a Quadratic Function 4. Applications of Quadratic Functions

In Section 1.1, we defined a linear equation in one variable as an equation of the form ax  b  0 1a  02 . A linear equation in one variable is sometimes called a first-degree polynomial equation because the highest degree of all its terms is 1. A second-degree polynomial equation is called a quadratic equation.

DEFINITION Quadratic Equation in One Variable If a, b, and c are real numbers such that a  0, then a quadratic equation is an equation that can be written in the form ax2  bx  c  0

The following equations are quadratic because they can each be written in the form ax2  bx  c  0 1a  02 . 4x2  4x  1

x1x  22  3

1x  421x  42  9

4x2  4x  1  0

x2  2x  3

x2  16  9

x2  2x  3  0

x2  25  0 x2  0x  25  0

One method to solve a quadratic equation is to factor and apply the zero product rule. The zero product rule states that if the product of two factors is zero, then one or both of its factors is equal to zero.

PROPERTY The Zero Product Rule If ab  0, then a  0 or b  0. For example, the quadratic equation x2  x  12  0 can be written in factored form as 1x  421x  32  0. By the zero product rule, one or both factors must be zero: x  4  0 or x  3  0. Therefore, to solve the quadratic equation, set each factor to zero and solve for x. 1x  421x  32  0

x40

or

x4

or

x30 x  3

Apply the zero product rule. Set each factor to zero. Solve each equation for x.

Quadratic equations, like linear equations, arise in many applications of mathematics, science, and business. The following steps summarize the factoring method to solve a quadratic equation.

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Section 4.8

Solving Equations by Using the Zero Product Rule

PROCEDURE Solving a Quadratic Equation by Factoring Step 1 Write the equation in the form ax2  bx  c  0. Step 2 Factor completely. Step 3 Apply the zero product rule. That is, set each factor equal to zero and solve the resulting equations.* *The solution(s) found in step 3 may be checked by substitution in the original equation.

Solving a Quadratic Equation

Example 1 Solve.

2x2  5x  12

Solution: 2x2  5x  12 2x2  5x  12  0

Write the equation in the form ax2  bx  c  0 .

12x  321x  42  0

2x  3  0

Factor completely.

or

x40

2x  3

or

x4

3 2

or

x4

x

Check: x  

3 2

Set each factor equal to zero. Solve each equation.

Check: x  4

2x2  5x  12

2x2  5x  12

3 2 3 2a b  5a b ⱨ 12 2 2

2142 2  5142 ⱨ 12

9 15 ⱨ 2a b  12 4 2

21162  20 ⱨ 12

9 15 ⱨ  12 2 2

32  20 ⱨ 12 

24 ⱨ 12  2 3 The solution set is e  , 4 f . 2 Skill Practice Solve. 1. y 2  2y  35

Answer

1. 57, 56

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Chapter 4 Polynomials

Example 2 Solve.

Solving a Quadratic Equation

6x2 ⫹ 8x ⫽ 0

Solution: 6x2 ⫹ 8x ⫽ 0 2x13x ⫹ 42 ⫽ 0 2x ⫽ 0

Factor completely.

3x ⫹ 4 ⫽ 0

or

x⫽0

Set each factor equal to zero.

3x ⫽ ⫺4 x⫽⫺

Solve each equation for x.

4 3

4 The solution set is e 0, ⫺ f . 3

The solutions check.

Skill Practice Solve. 2. 9x 2 ⫽ 21x

Example 3 Solve.

Solving a Quadratic Equation

9x14x ⫹ 22 ⫺ 10x ⫽ 8x ⫹ 25

Solution: 9x14x ⫹ 22 ⫺ 10x ⫽ 8x ⫹ 25 36x2 ⫹ 18x ⫺ 10x ⫽ 8x ⫹ 25

Clear parentheses.

36x ⫹ 8x ⫽ 8x ⫹ 25

Combine like terms.

36x ⫺ 25 ⫽ 0

Make one side of the equation equal to zero. The equation is in the form ax2 ⫹ bx ⫹ c ⫽ 0. (Note: b ⫽ 0.)

2

2

16x ⫺ 5216x ⫹ 52 ⫽ 0

Factor completely.

6x ⫺ 5 ⫽ 0

or

6x ⫽ 5

or

6x ⫽ ⫺5

5 6

or

x⫽⫺

x⫽

6x ⫹ 5 ⫽ 0

Set each factor equal to zero.

5 6

5 5 The solution set is e , ⫺ f . 6 6 Skill Practice Solve. 3. 5a 12a ⫺ 32 ⫹ 41a ⫹ 12 ⫽ 3a 13a ⫺ 22 Answers 7 2. e 0, f 3 3. 54, 16

Solve each equation. The check is left to the reader.

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Example 4 Solve.

Solving Equations by Using the Zero Product Rule

Solving an Equation

2x1x  52  3  2x2  5x  1

Solution: 2x1x  52  3  2x2  5x  1 2x2  10x  3  2x2  5x  1 15x  2  0

Make one side of the equation equal to zero. The equation is not quadratic. It is in the form ax  b  0, which is linear. Solve by using the method for linear equations.

15x  2 x The solution set is e 

Clear parentheses.

2 15

2 f. 15

The check is left to the reader.

Skill Practice Solve. 4. t 2  3t  1  t 2  2t  11

The zero product rule can be used to solve higher-degree polynomial equations provided one side of the equation is zero and the other is written in factored form.

Example 5 Solve.

Solving a Higher-Degree Polynomial Equation

21y  721y  12110y  32  0

Solution: 21y  721y  12110y  32  0 One side of the equation is zero, and the other side is already factored. 2  0 or

No solution

y70

y  7

or y  1  0

or

y1

or

10y  3  0

or

y

3 10

Set each factor equal to zero. Solve each equation for y.

Notice that when the constant factor is set to zero, the result is the contradiction 2 = 0. The constant factor does not produce a solution to the equation. Therefore, the only solutions are 7, 1, and 103 . Each solution can be checked in the original equation. The solution set is e 7, 1, 

3 f. 10

Skill Practice Solve. 5. 31w  2212w  121w  82  0 Answers 4. 526

1 5. e 2,  , 8 f 2

391

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Chapter 4 Polynomials

Solving a Higher-Degree Polynomial Equation

Example 6

z3  3z2  4z  12  0

Solve.

Solution: z3  3z2  4z  12  0 z3  3z2

This is a higher-degree polynomial equation.

 4z  12  0

One side of the equation is zero. Now factor. Because there are four terms, try factoring by grouping.

z2 1z  32  41z  32  0 1z  321z2  42  0

z2  4 can be factored further as a difference of squares.

1z  321z  221z  22  0 z30 z  3

or

z20

or

or

z2

or

z20 z  2

The solution set is 53, 2, 26 .

Set each factor equal to zero. Solve each equation.

Skill Practice Solve. 6. x 3  x 2  9x  9  0

2. Applications of Quadratic Equations Solving an Application of a Quadratic Equation

Example 7

The product of two consecutive odd integers is 35. Find the integers.

Solution: Let x represent the smaller odd integer and x  2 represent the next consecutive odd integer. a

First odd next odd bⴢa b  35 integer integer x ⴢ 1x  22  35 x2  2x  35 x  2x  35  0 2

1x  721x  52  0 x70

or

x50

x  7 or

x5

Verbal model Mathematical equation Clear parentheses. Set the equation equal to zero. Factor. Set each factor equal to zero. Solve each equation.

If x  7 then the next odd integer is x  2  5. If x  5 then the next odd integer is x  2  7. There are two pairs of odd integers that are solutions, 7, 5 and 5, 7. Answers 6. {1, 3, 3} 7. The integers are 8 and 6 or 6 and 8.

Skill Practice 7. The product of two consecutive even integers is 48. Find the integers.

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Solving Equations by Using the Zero Product Rule

393

Solving an Application of a Quadratic Equation 2w  6

The length of a basketball court is 6 ft less than 2 times the width. If the total area is 4700 ft2, find the dimensions of the court. w

Solution: If the width of the court is represented by w, then the length can be represented by 2w  6 (Figure 4-5). A  1length21width2

4700  12w  62w

Figure 4-5

Area of a rectangle Mathematical equation

4700  2w2  6w 2w2  6w  4700  0

Set the equation equal to zero and factor.

21w2  3w  23502  0

Factor out the GCF.

21w  5021w  472  0

Factor the trinomial.

20

w  50  0

or

or

w  47  0

Set each factor equal to zero.

contradiction

w  50

or

w  47

A negative width is not possible.

The width is 50 ft. The length is 2w  6  21502  6  94 ft. Skill Practice 8. The width of a rectangle is 5 in. less than 3 times the length. The area is 2 in.2 Find the length and width.

A right triangle is a triangle that contains a 90° angle. Furthermore, the sum of the squares of the two legs (the shorter sides) of a right triangle equals the square of the hypotenuse (the longest side). This important fact is known as the Pythagorean theorem. For the right triangle shown in Figure 4-6, the Pythagorean theorem is stated as

TIP: When applying the (leg) a

(hypotenuse) c

b (leg)

a2  b2  c2

Figure 4-6

Pythagorean theorem, it does not matter which leg you label a and which you label b. Since the lengths of the legs are interchangeable you can also write the Pythagorean theorem as leg2  leg2  hyp2.

In this formula, a and b are the legs and c is the hypotenuse. Notice that the hypotenuse is the longest side and is opposite the right angle. The triangle given in Figure 4-7 is a right triangle. We have a2





b2

c = 13 ft

c2

15 ft2 2  112 ft2 2  113 ft2 2 25 ft2  144 ft2  169 ft2 169 ft  169 ft  2

b = 12 ft

2

a = 5 ft

Figure 4-7

Answer 8. The width is 1 in., and the length is 2 in.

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Application of a Quadratic Equation

Example 9

A region of coastline off Biscayne Bay is approximately in the shape of a right angle. The corresponding triangular area has sandbars and is marked off on navigational charts as being shallow water. If one leg of the triangle is 0.5 mi shorter than the other leg, and the hypotenuse is 2.5 mi, find the lengths of the legs of the triangle (Figure 4-8).

x

x  0.5

Shallow 2.5

Figure 4-8

Solution: Let x represent the longer leg. Then x  0.5 represents the shorter leg. a2  b2  c2

x  1x  0.52  12.52 2

2

Pythagorean theorem 2

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

x2  1x2 2  21x210.52  10.52 2  6.25 x2  x2  x  0.25  6.25

TIP: Recall that the square of a binomial results in a perfect square trinomial.

1a  b2 2  a2  2ab  b2

1x0.52 2  1x2 2  21x210.52  10.52 2  x2  x  0.25

2x2  x  6  0 12x  321x  22  0 2x  3  0 x

3 2

or

x20

or

x2

Write the equation in the form ax2  bx  c  0. Factor. Set both factors to zero. Solve both equations for x.

The side of a triangle cannot be negative, so we reject the solution x  32. Therefore, one leg of the triangle is 2 mi. The other leg is x  0.5  2  0.5  1.5 mi. Skill Practice 9. The longer leg of a right triangle measures 7 ft more than the shorter leg. The hypotenuse is 8 ft longer than the shorter leg. Find the lengths of the sides of the triangle.

3. Definition of a Quadratic Function In Section 2.7, we graphed several basic functions by plotting points, including f 1x2  x2. This function is called a quadratic function, and its graph is in the shape of a parabola. In general, any second-degree polynomial function is a quadratic function.

DEFINITION Quadratic Function Answer 9. The sides are 5 ft, 12 ft, and 13 ft.

Let a, b, and c represent real numbers such that a  0. Then a function defined by f 1x2  ax2  bx  c is called a quadratic function.

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The graph of a quadratic function is a parabola that opens upward or downward. The leading coefficient a determines the direction of the parabola. For the quadratic function defined by f 1x2  ax2  bx  c: If a 7 0, the parabola opens upward. For example, f 1x2  x2

y

x

If a 6 0, the parabola opens downward. For example, g1x2  x2

y

x

Recall from Section 2.7 that the x-intercepts of a function y  f 1x2 are the real solutions to the equation f 1x2  0. The y-intercept is found by evaluating f(0). Example 10

Finding the x- and y-Intercepts of a Quadratic Function

Find the x- and y-intercepts.

f 1x2  x2  x  12

Solution:

To find the x-intercept, substitute f 1x2  0. f 1x2  x2  x  12 0  x2  x  12

Substitute 0 for f(x). The result is a quadratic equation.

0  1x  421x  32 x4 0

or

x 4

or

Factor.

x30 x  3

Set each factor equal to zero. Solve each equation.

The x-intercepts are (4, 0) and (3, 0). To find the y-intercept, find f 102 . f 1x2  x2  x  12

f 102  102 2  102  12

Substitute x  0.

 12 The y-intercept is (0, 12). Skill Practice 10. Find the x- and y-intercepts of the function defined by f 1x2  x 2  8x  12. Answer

10. x-intercepts: 16, 02 and 12, 02 ; y-intercept: 10, 122

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Chapter 4 Polynomials

Calculator Connections The graph of f 1x2  x2  x  12 supports the solution to Example 10. Use the zero feature to confirm that the graph crosses the x-axis at 3 and 4.The y-intercept is given as (0, 12).

4. Applications of Quadratic Functions Example 11

Application of a Quadratic Function

A model rocket is shot vertically upward with an initial velocity of 288 ft/sec. The function given by h1t2  16t2  288t relates the rocket’s height h(t) (in feet) to the time t after launch (in seconds). a. Find h(0), h(5), h(10), and h(15), and interpret the meaning of these function values in the context of the rocket’s height and time after launch. b. Find the t-intercepts of the function, and interpret their meaning in the context of the rocket’s height and time after launch. c. Find the time(s) at which the rocket is at a height of 1152 ft.

Solution: a.

h1t2  16t2  288t h102  16102 2  288102  0 h152  16152 2  288152  1040 h1102  161102 2  2881102  1280 h1152  161152 2  2881152  720 h102  0 means that at t  0 sec, the height of the rocket is 0 ft. h152  1040 means that 5 sec after launch, the height is 1040 ft. h1102  1280 means that 10 sec after launch, the height is 1280 ft. h1152  720 means that 15 sec after launch, the height is 720 ft.

b. The t-intercepts of the function are represented by the real solutions of the equation h1t2  0. 16t2  288t  0

Set h1t2  0.

16t1t  182  0

Factor.

16t  0

or

t0

or

t  18  0

Apply the zero product rule.

t  18

The rocket is at ground level initially (at t  0 sec) and then again after 18 sec when it hits the ground.

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c. Set h1t2  1152 and solve for t. h1t2  16t2  288t 1152  16t2  288t 16t2  288t  1152  0

Substitute 1152 for h(t). Set the equation equal to zero.

161t  18t  722  0

Factor out the GCF.

161t  621t  122  0

Factor.

2

or

t  12

The rocket will reach a height of 1152 ft after 6 sec (on the way up) and after 12 sec (on the way down). (See Figure 4-9.)

h(t) 1500 1250 1000 750 500 250 0 0 250

Height of Rocket Versus Time After Launch h(t)  16t2  288t

(6, 1152)

(12, 1152)

Height (ft)

t6

3

6

15

9 12 Time (sec)

18

t

Figure 4-9

Skill Practice An object is dropped from the top of a building that is 144 ft high. The function given by h 1t2 16t 2 144 relates the height h (t) of the object (in feet) to the time t in seconds after it is dropped. 11. a. Find h (0) and interpret the meaning of the function value in the context of this problem. b. Find the t-intercept(s) and interpret the meaning in the context of this problem. c. When will the object be 128 ft above ground level.

Answer 11a. h (0)  144, which is the initial height of the object (after 0 sec). b. The t-intercept is (3, 0), which means the object is at ground level (0 ft high) after 3 sec. The intercept (3, 0) does not make sense for this problem since time cannot be negative. c. One second after release, the object will be 128 ft above ground level.

Section 4.8 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Quadratic equation

b. Zero product rule

c. Quadratic function

d. Parabola

Review Exercises 2. Write the factored form for each binomial, if possible. a. x2  y2

b. x2  y2

c. x3  y3

d. x3  y3

For Exercises 3–8, factor completely. 3. 10x2  3x

4. 7x2  28

5. 2p2  9p  5

6. 3q2  4q  4

7. t3  1

8. z2  11z  30

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Chapter 4 Polynomials

Concept 1: Solving Equations by Using the Zero Product Rule 9. What conditions are necessary to solve an equation by using the zero product rule?

10. State the zero product rule.

For Exercises 11–16, determine which of the equations are written in the correct form to apply the zero product rule directly. If an equation is not in the correct form, explain what is wrong. 11. 2x1x  32  0

12. 1u  121u  32  10

13. 3p2  7p  4  0

14. t2  t  12  0

15. a1a  32 2  5

2 1 16. a x  5bax  b  0 3 2

For Exercises 17–52, solve the equation. (See Examples 1–6.) 17. 1x  321x  52  0

18. 1x  721x  42  0

19. 12w  9215w  12  0

21. x1x  42110x  32  0

22. t1t  6213t  112  0

23. 0  51y  0.421y  2.12

24. 0  41z  7.521z  9.32

25. x2  6x  27  0

26. 2x2  x  15  0

27. 2x2  5x  3

28. 11x  3x2  4

29. 10x2  15x

30. 5x2  7x

31. 61y  22  31y  12  8

32. 4x  31x  92  6x  1

33. 9  y1y  62

34. 62  t1t  162  2

35. 9p2  15p  6  0

36. 6y2  2y  48

37. 1x  1212x  121x  32  0

38. 2x1x  42 2 14x  32  0

39. 1y  321y  42  8

40. 1t  1021t  52  6

41. 12a  121a  12  6

42. w16w  12  2

43. p2  1p  72 2  169

44. x2  1x  22 2  100

45. 3t1t  52  t2  2t2  4t  1

46. a2  4a  2  1a  321a  52

47. 2x3  8x2  24x  0

48. 2p3  20p2  42p  0

49. w3  16w

50. 12x3  27x

51. 0  2x3  5x2  18x  45

52. 0  3y3  y2  48y  16

20. 13a  1214a  52  0

Concept 2: Applications of Quadratic Equations 53. If 5 is added to the square of a number, the result is 30. Find all such numbers. 54. Four less than the square of a number is 77. Find all such numbers. 55. The square of a number is equal to 12 more than the number. Find all such numbers. 56. The square of a number is equal to 20 more than the number. Find all such numbers. 57. The product of two consecutive integers is 42. Find the integers. 58. The product of two consecutive integers is 110. Find the integers. 59. The product of two consecutive odd integers is 63. Find the integers. (See Example 7.) 60. The product of two consecutive even integers is 120. Find the integers.

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61. A rectangular pen is to contain 35 ft2 of area. If the width is 2 ft less than the length, find the dimensions of the pen. (See Example 8.) 62. The length of a rectangular photograph is 7 in. more than the width. If the area is 78 in.2, what are the dimensions of the photograph? 63. The length of a rectangular room is 5 yd more than the width. If the area is 300 yd2, find the length and the width of the room. 64. The top of a rectangular dining room table is twice as long as it is wide. Find the dimensions of the table if the area is 18 ft2. 65. The height of a triangle is 1 in. more than the base. If the height is increased by 2 in. while the base remains the same, the new area becomes 20 in.2 a. Find the base and height of the original triangle. b. Find the area of the original triangle. 66. The base of a triangle is 2 cm more than the height. If the base is increased by 4 cm while the height remains the same, the new area is 56 cm2. a. Find the base and height of the original triangle. b. Find the area of the original triangle. 67. The area of a triangular garden is 25 ft2. The base is twice the height. Find the base and the height of the triangle. 68. The height of a triangle is 1 in. more than twice the base. If the area is 18 in.2, find the base and height of the triangle. 69. The sum of the squares of two consecutive positive integers is 41. Find the integers. 70. The sum of the squares of two consecutive, positive even integers is 164. Find the integers. Clayton

71. Justin must travel from Summersville to Clayton. He can drive 10 mi through the mountains at 40 mph. Or he can drive east and then north on superhighways at 60 mph. The alternative route forms a right angle as shown in the diagram. The eastern leg is 2 mi less than the northern leg. (See Example 9.) a. Find the total distance Justin would travel in going the alternative route. b. If Justin wants to minimize the time of the trip, which route should he take? 10 mi

x

72. A 17-ft ladder is standing up against a wall. The distance between the base of the ladder and the wall is 7 ft less than the distance between the top of the ladder and the base of the wall. Find the distance between the base of the ladder and the wall. x2

73. A right triangle has side lengths represented by three consecutive even integers. Find the lengths of the three sides, measured in meters.

Summersville

74. The hypotenuse of a right triangle is 3 m more than twice the short leg. The longer leg is 2 m more than twice the shorter leg. Find the lengths of the sides. 75. Determine the length of the radius of a circle whose area is numerically equal to its circumference. 76. Determine the length of the radius of a circle whose area is numerically twice its circumference.

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Concept 3: Definition of a Quadratic Function

For Exercises 77–80, a. Find the values of x for which f 1x2 ⫽ 0. b. Find f 102 . 77. f 1x2 ⫽ x2 ⫺ 3x

78. f 1x2 ⫽ 4x2 ⫹ 2x

79. f 1x2 ⫽ x2 ⫺ 6x ⫺ 7

80. f 1x2 ⫽ 2x2 ⫹ 11x ⫹ 5

For Exercises 81–84, find the x- and y-intercepts for the functions defined by y ⫽ f(x). (See Example 10.) 81. f 1x2 ⫽

1 1x ⫺ 221x ⫹ 1212x2 2

82. f 1x2 ⫽ 1x ⫹ 121x ⫺ 221x ⫹ 32 2

83. f 1x2 ⫽ x2 ⫺ 2x ⫹ 1

84. f 1x2 ⫽ x2 ⫹ 4x ⫹ 4

For Exercises 85– 88, find the x-intercepts of each function and use that information to match the function with its graph. 85. g1x2 ⫽ x2 ⫺ 9

87. f 1x2 ⫽ 41x ⫹ 12

86. h1x2 ⫽ x1x ⫺ 221x ⫹ 42

88. k1x2 ⫽ 1x ⫹ 121x ⫹ 321x ⫺ 221x ⫺ 12 a.

b.

y

c.

y 15 12

21

6 4 2

9 6 3

15

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2

4 6

8 10

x

⫺5⫺4 ⫺3 ⫺2 ⫺1 ⫺3 ⫺6 ⫺9 ⫺12 ⫺15

d.

y

10 8

18

6 4 2

12

1

2 3 4 5

9 6

x

3 ⫺5⫺4 ⫺3 ⫺2 ⫺1 ⫺3 ⫺6 ⫺9

y 10 8

1

2 3 4 5

x

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2

4 6

8 10

Concept 4: Applications of Quadratic Functions 89. A rocket is fired upward from ground level with an initial velocity of 490 m/sec. The height of the rocket s(t) in meters is a function of the time t in seconds after launch. (See Example 11.) s1t2 ⫽ ⫺4.9t2 ⫹ 490t a. What characteristics of s indicate that it is a quadratic function? b. Find the t-intercepts of the function. c. What do the t-intercepts mean in the context of this problem? d. At what times is the rocket at a height of 485.1 m? 90. A certain company makes water purification systems. The factory can produce x water systems per year. The profit P(x) the company makes is a function of the number of systems x it produces. P1x2 ⫽ ⫺2x2 ⫹ 1000x a. Is this function linear or quadratic? b. Find the number of water systems x that would produce a zero profit. c. What points on the graph do the answers in part (b) represent? d. Find the number of systems for which the profit is $80,000.

x

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For Exercises 91–94, factor the expressions represented by f(x). Explain how the factored form relates to the graph of the function. Can the graph of the function help you determine the factors? 91. f 1x2 ⫽ x2 ⫺ 7x ⫹ 10

92. f 1x2 ⫽ x2 ⫺ 2x ⫺ 3

y 5 4

y 5 4 3

3

2 1

2 1 ⫺3 ⫺2 ⫺1 ⫺1

1 2 3

4 5

x

6 7

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

2 3

4 5

1 2 3

4 5

6 7

x

⫺2 ⫺3

⫺2 ⫺3 ⫺4 ⫺5

93. f 1x2 ⫽ x2 ⫹ 2x ⫹ 1

1

⫺4 ⫺5

94. f 1x2 ⫽ x2 ⫺ 8x ⫹ 16

y 5 4

y 5 4

3

3

2 1

2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2 3

4 5

x

⫺3 ⫺2 ⫺1 ⫺1

⫺2 ⫺3

⫺2 ⫺3

⫺4 ⫺5

⫺4 ⫺5

x

Expanding Your Skills 95. The surface area of a right circular cylinder is represented by SA ⫽ 2pr2 ⫹ 2prh. If the surface area is 156p ft2 and the height is 7 ft, determine the radius of the cylinder. 96. Determine the length and width of a rectangle with a perimeter of 20 yd and an area of 16 yd2. 97. Determine the length and width of a rectangle with a perimeter of 28 ft and an area of 48 ft2. For Exercises 98–101, find an equation that has the given solutions. For example, 2 and ⫺1 are solutions to 1x ⫺ 221x ⫹ 12 ⫽ 0 or x2 ⫺ x ⫺ 2 ⫽ 0. In general, x1 and x2 are solutions to the equation a1x ⫺ x1 21x ⫺ x2 2 ⫽ 0, where a can be any nonzero real number. For each problem, there is more than one correct answer depending on your choice of a. 98. x ⫽ ⫺3 and x ⫽ 1

99. x ⫽ 2 and x ⫽ ⫺2

100. x ⫽ 0 and x ⫽ ⫺5

101. x ⫽ 0 and x ⫽ ⫺3

Graphing Calculator Exercises For Exercises 102–105, graph Y1. Use the Zero feature to approximate the x-intercepts. Then solve Y1 ⫽ 0 and compare the solutions to the x-intercepts. 102. Y1 ⫽ ⫺x2 ⫹ x ⫹ 2

103. Y1 ⫽ ⫺x2 ⫺ x ⫹ 20

104. Y1 ⫽ x2 ⫺ 6x ⫹ 9

105. Y1 ⫽ x2 ⫹ 4x ⫹ 4

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Chapter 4 Polynomials

Group Activity Investigating Pascal’s Triangle Estimated time: 15 minutes Group Size: 2 or 3

1. Determine the value of 1a  b2 0. Write your answer here: __________________ 2. Write the binomials in expanded form. [For example, for (a  b)2, multiply the binomials (a  b)(a  b) and write the result in the space provided.] Check your answers with another group. a. (a  b)1 ____________________________ b. (a  b)2 ____________________________ c. (a  b)3 ____________________________ 3. a. How many terms were in the expansion of (a  b)2? ________ b. How many terms were in the expansion of (a  b)3? ________ c. How many terms would you expect in the expansion of (a  b)4? ________ d. How many terms would you expect in the expansion of (a  b)5? ________ e. How many terms would you expect in the expansion of (a  b)n? ________ 4. Based on your answers from question 2, do you see a pattern regarding the exponents on the factors of a and b in each term? Explain. 5. The coefficients of each expansion also follow a pattern. The triangle of numbers shown here is called Pascal’s triangle. Pascal’s triangle gives the coefficients of a binomial expansion of the form (a  b)n. Work with your group members to see if you can figure out the pattern for the coefficients of (a  b)4, (a  b)5, (a  b)6, and (a  b)7. (a  b)n (a  b)

Coefficients of the Expansion

0

1

(a  b)1 (a  b)

(a  b)3 (a  b)

1

2

1 1

1 2

3

1 3

1

4

(a  b)5 (a  b)6 6. Using the expected number of terms found in question 3, the pattern of the exponents found in question 4, and the pattern for the coefficients in question 5, write the expansions for the following. a. (a  b)4  ______________________________________________________ b. (a  b)5  ______________________________________________________ c. (a  b)6  ______________________________________________________

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Chapter 4

Summary

Section 4.1

Properties of Integer Exponents and Scientific Notation

Key Concepts

Let a and b 1b  02 represent real numbers and m and n represent positive integers. m

b  bmn bn

bm ⴢ bn  bmn 1b 2  b

1ab2  a b

a m am a b  m b b

b0  1

m n

mn

403

m

m m

Examples Example 1 a

2x2y z1 a 

1 n 1 bn  a b  n b b



A number expressed in the form a  10n, where 1  0a 0 6 10 and n is an integer, is written in scientific notation.

3

b 1x4y0 2 23x6y3 z3

b1x 4 ⴢ 12

23x10y3 z3 1 2x yz

3 10 3 3

or

1 8x y z

10 3 3

Example 2 0.0000002  35,000

 12.0  107 213.5  104 2  7.0  103 or 0.007

Section 4.2

Addition and Subtraction of Polynomials and Polynomial Functions

Key Concepts

Examples

A polynomial in x is defined by a sum of terms of the form axn, where a is a real number and n is a whole number.

Example 1

• a is the coefficient of the term. • n is the degree of the term. The degree of a polynomial is the greatest degree of its terms. The term of a polynomial with the greatest degree is the leading term. Its coefficient is the leading coefficient. A one-term polynomial is a monomial. A two-term polynomial is a binomial. A three-term polynomial is a trinomial.

To add or subtract polynomials, add or subtract like terms.

7y4  2y2  3y  8 is a polynomial with leading coefficient 7 and degree 4. Example 2 f1x2  4x3  6x  11 f is a polynomial function with leading term 4x3 and leading coefficient 4. The degree of f is 3. Example 3 For

f 1x2  4x3  6x  11, find f 112. f 112  4112 3  6112  11  9

Example 4 14x3y  3x2y2 2  17x3y  5x2y2 2  4x3y  3x2y2  7x3y  5x2y2  11x3y  8x2y2

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Chapter 4 Polynomials

Section 4.3

Multiplication of Polynomials

Key Concepts

Examples

To multiply polynomials, multiply each term in the first polynomial by each term in the second polynomial.

Example 1

Special Products

1x  2213x2  4x  112  3x3  4x2  11x  6x2  8x  22  3x3  10x2  19x  22

1. Multiplication of conjugates 1x  y21x  y2  x2  y2

Example 2

The product is called a difference of squares.

2. Square of a binomial

Example 3

1x  y2 2  x2  2xy  y2

1x  y2 2  x2  2xy  y2

The product is called a perfect square trinomial.

Section 4.4

13x  5213x  52  13x2 2  152 2  9x2  25 14y  32 2  14y2 2  12214y2132  132 2  16y2  24y  9

Division of Polynomials

Key Concepts

Examples

Division of polynomials:

Example 1

1. For division by a monomial, use the properties ab a b   c c c

and

ab a b   c c c

for c  0.

12a2  6a  9 3a 

12a2 6a 9   3a 3a 3a

 4a  2  2. If the divisor has more than one term, use long division. • First write each polynomial in descending order. • Insert placeholders for missing powers.

3 a

Example 2 13x2  5x  12  1x  22 3x  11 x  2 冄 3x2  5x  1 13x2  6x2 11x  1 111x  222 23 Answer:

3x  11 

23 x2

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3. Synthetic division may be used to divide a polynomial by a binomial in the form x  r, where r is a constant.

405

Example 3 13x2  5x  12  1x  22

2 3

5 1 6 22 3 11 23

Answer:

Section 4.5

3x  11 

23 x2

Greatest Common Factor and Factoring by Grouping

Key Concepts

Examples

The greatest common factor (GCF) is the largest factor common to all terms of a polynomial. To factor out the GCF from a polynomial, use the distributive property. A four-term polynomial may be factored by grouping.

Example 1 3x2 1a  b2  6x1a  b2  3x1a  b2x  3x1a  b2122  3x1a  b21x  22

Steps to Factor by Grouping 1. Identify and factor out the GCF from all four terms. 2. Factor out the GCF from the first pair of terms. Factor out the GCF from the second pair of terms. (Sometimes it is necessary to factor out the opposite of the GCF.) 3. If the two pairs of terms share a common binomial factor, factor out the binomial factor.

Section 4.6

Example 2 60xa  30xb  80ya  40yb  1036xa  3xb  8ya  4yb4  1033x12a  b2  4y12a  b2 4

 1012a  b213x  4y2

Factoring Trinomials

Key Concepts

Examples

AC-Method

Example 1

To factor trinomials of the form ax  bx  c: 2

1. Factor out the GCF. Find the product ac. 2. Find two integers whose product is ac and whose sum is b. (If no pair of numbers can be found, then the trinomial is prime.) 3. Rewrite the middle term bx as the sum of two terms whose coefficients are the numbers found in step 2. 4. Factor the polynomial by grouping.

10y2  35y  20  512y2  7y  42 ac  122142  8

Find two integers whose product is 8 and whose sum is 7. The numbers are 8 and 1 . 532y2  8y  1y  44

 532y1y  42  11 y  42 4

 51y  4212y  12

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Chapter 4 Polynomials

Trial-and-Error Method

Example 2

To factor trinomials in the form ax  bx  c: 2

1. Factor out the GCF. 2. List the pairs of factors of a and the pairs of factors of c. Consider the reverse order in either list. 3. Construct two binomials of the form

10y 2  35y  20  512y 2  7y  42 The pairs of factors of 2 are 2 ⴢ 1. The pairs of factors of 4 are

Factors of a

1ⵧx ⵧ21ⵧx ⵧ2 Factors of c

4. Test each combination of factors until the product of the outer terms and the product of inner terms add to the middle term. 5. If no combination of factors works, the polynomial is prime.

1 ⴢ 142 2 ⴢ 122 4 ⴢ 112

1 ⴢ 4 2 ⴢ 2 4 ⴢ 1

12y  221y  22  2y2  2y  4 12y  421y  12  2y  2y  4 2

12y  121y  42  2y2  7y  4 12y  221y  22  2y  2y  4 2

12y  421y  12  2y2  2y  4 12y  121y  42  2y  7y  4 2

No No No No No Yes

Therefore, 10y2  35y  20 factors as 512y  121y  42. The factored form of a perfect square trinomial is the square of a binomial: a  2ab  b  1a  b2 2

2

a  2ab  b  1a  b2 2

2

Example 3 9w2  30wz  25z2

 13w2 2  213w215z2  15z2 2

2 2

 13w  5z2 2

Sometimes it is easier to factor a polynomial after making a substitution.

Example 4 17v2  12 2  17v2  12  12  u2  u  12

 1u  321u  42

 17v2  1  3217v2  1  42  17v  2217v  52 2

Section 4.7

2

Factoring Binomials

Key Concepts

Examples

Factoring Binomials: Summary

Example 1

Difference of squares:

25u2  9v4  15u  3v2 215u  3v2 2

a2  b2  1a  b21a  b2 Sum of squares: If a and b share no common factors, then a2  b2 is prime.

Let u  17v2  12. Substitute. Factor. Back substitute. Simplify.

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Summary

Difference of cubes:

Example 2

a  b  1a  b21a  ab  b 2

8c3  d 6  12c  d 2 214c 2  2cd 2  d 4 2

Sum of cubes:

Example 3

3

3

2

2

a3  b3  1a  b21a2  ab  b2 2

27w9  64x3

Sometimes it is necessary to group three terms with one term.

Example 4

 13w3  4x219w6  12w3x  16x2 2

4a2  12ab  9b2  c2  4a2  12ab  9b2

 c2

 12a  3b2  c 2

Group 3 by 1.

2

Perfect square trinomial.

 12a  3b  c212a  3b  c2

Section 4.8

Difference of squares.

Solving Equations by Using the Zero Product Rule

Key Concepts

Examples

An equation of the form ax  bx  c  0, where a  0, is a quadratic equation. The zero product rule states that if ab  0 , then a  0 or b  0 . The zero product rule can be used to solve a quadratic equation or higher-degree polynomial equation that is factored and equal to zero. 2

Example 1 0  x12x  321x  42 x0

or

2x  3  0

or

3 2

or

x

3 The solution set is e 0, , 4 f . 2 f1x2  ax2  bx  c 1a  02 defines a quadratic function. The x-intercepts of a function defined by y  f1x2 are determined by finding the real solutions to the equation f1x2  0 . The y-intercept of y  f 1x2 is at f 102. y 7 6 5 4 3 2

2 3

Find the x-intercepts. f1x2  3x2  8x  5 0  3x2  8x  5

0  13x  521x  12

3x  5  0

or

x10

5 3

or

x1

x

1 3 2 1 1

Example 2

1

2

3 4

5

6 7

x

The x-intercepts are 1 53, 02 and (1, 0). Find the y-intercept. f 1x2  3x2  8x  5

f102  3102 2  8102  5 f 102  5

The y-intercept is 10, 52.

x40 x  4

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Chapter 4 Polynomials

Chapter 4

Review Exercises

Section 4.1 For Exercises 1–8, simplify the expression and write the answer with positive exponents. 1. 13x2 3 13x2 2 3.

2.

24x5y3

4.

8x y 4

5. 12a b 2

2 5 3

7. a

4x4y2 5x1y4

b

6.

4

8.

16x 4 213x 8 2

Section 4.2 For Exercises 17–18, identify the polynomial as a monomial, binomial, or trinomial; then give the degree of the polynomial. 17. 6x4  10x  1

18x2y3

18. 18

5 5

12x y

14a b 2

2 3 2

a

25x2y3 5x4y2

19. Given the polynomial function defined by g1x2  4x  7, find the function values. a. g102

5

b

b. A nanometer is 0.000001 of a millimeter.

c. g132

20. Given the polynomial function defined by p1x2  x4  x  12, find the function values.

9. Write the numbers in scientific notation. a. The population of Asia was 3,686,600,000 in 2000.

b. g142

a. p102

b. p112

c. p122

21. The amount, A(x), of bottled water consumed per capita in the United States can be approximated by A(x)  0.047x2  1.46x  16.8

10. Write the numbers in scientific notation. a. A millimeter is 0.001 of a meter.

where x is the number of years since the year 2000 and A(x) is measured in gallons.

b. The population of Asia is predicted to be 5,155,700,000 by 2040.

a. Evaluate A(5) and interpret the meaning in the context of the problem.

11. Write the numbers in standard form.

b. Interpret the meaning of A(15).

a. A micrometer is 1  103 of a millimeter. b. A nanometer is 1  109 of a meter. 50.0 Gallons

12. Write the numbers in standard form. a. The total square footage of shopping centers in the United States is approximately 5.23  109 ft2.

13.

2,500,000 0.0004

14.

15. 13.6  108 219.0  102 2 16. 17.0  1012 215.2  103 2

0.0005 25,000

A(x)  0.047x2  1.46x  16.8

40.0 30.0 20.0 10.0 0.0

b. The total yearly sales of those shopping centers is $1.091  1012. (Source: International Council of Shopping Centers.) For Exercises 13–16, perform the indicated operations. Write the answer in scientific notation.

Per Capita Consumption of Bottled Water

0

2 4 6 8 10 Year (x  0 represents to 2000)

12

For Exercises 22–33, add or subtract the polynomials as indicated. 22. 1x2  2x  3xy  72  13x2  x  2xy  62 23.

7xy  3xz  5yz 13xy  15xz  8yz

24.

4a3  8a2  3a 17a3  3a2  9a2

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Review Exercises

25. 13a2  2a  a3 2  15a2  a3  8a2 5 1 1 3 3 1 26. a x4  x2  b  a x4  x2  b 8 4 2 8 4 2 1 1 1 1 1 5 27. a x4  x2  b  a x4  x2  b 6 2 3 6 4 3 28. 17x  y2  312x  y2  13x  6y2 4

409

52. A square garden is surrounded by a walkway of uniform width x. If the sides of the garden are given by the expression 2x  3, find and simplify a polynomial for the following. a. The area of the garden. b. The area of the walkway and garden. c. The area of the walkway only.

29. 14x  4y2  3 14x  2y2  13x  7y2 4 30. Add 4x  6 to 7x  5. 2x  3

31. Add 2x2  4x to 2x2  7x. x

32. Subtract 4x  6 from 7x  5.

2x  3 x

33. Subtract 2x  4x from 2x  7x. 2

2

Section 4.3 For Exercises 34–51, multiply the polynomials. 34. 2x1x2  7x  42

35. 3x16x2  5x  42

36. 1x  621x  72

37. 1x  221x  92

1 1 38. a x  1b a x  5b 2 2

1 1 39. a  2yb a  yb 5 5

40. 13x  5219x2  15x  252 41. 1x  y21x2  xy  y2 2 42. 12x  52 2

2 1 43. a x  4b 2

44. 13y  11213y  112

45. 16w  1216w  12

2 2 46. a t  4b a t  4b 3 3

1 1 47. az  b az  b 4 4

48. 3 1x  22  b 4 3 1x  22  b4 49. 3c  1w  32 4 3c  1w  32 4 50. 12x  12 3

51. 1y2  32 3

53. The length of a rectangle is 2 ft more than 3 times the width. Let x represent the width of the rectangle. a. Write a function P that represents the perimeter of the rectangle. b. Write a function A that represents the area of the rectangle. 54. In parts (a) and (b), one of the statements is true and the other is false. Identify the true statement and explain why the false statement is incorrect. a. 2x2  5x  7x3

12x2 215x2  10x3

b. 4x  7x  3x

4x  7x  3

Section 4.4 For Exercises 55–56, divide the polynomials. 55. 16x3y  12x2y2  9xy3 2  13xy2

56. 110x4  15x3  20x2 2  15x2 2 57. a. Divide 19y4  14y2  82  13y  22. b. Identify the quotient and the remainder. c. Explain how you can check your answer. For Exercises 58–61, divide the polynomials by using long division. 58. 1x2  7x  102  1x  52 59. 1x2  8x  162  1x  42

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Chapter 4 Polynomials

60. 12x5  4x4  2x3  42  1x2  3x2

75. 4ax2  2bx2  6ax  3xb

61. 12x5  3x3  x2  42  1x2  x2

76. y3  6y2  y  6

62. Explain the conditions under which you may use synthetic division. 63. The following table is the result of a synthetic division. 3

5

2

6

1

6

33

93

297

11

31

99

298

2

2

Use x as the variable. a. Identify the divisor. b. Identify the quotient. c. Identify the remainder. For Exercises 64–68, divide the polynomials by using synthetic division. 64. 1t3  3t2  8t  122  1t  22 65. 1x2  7x  142  1x  52 66. 1x2  8x  202  1x  42 67. 1w3  6w2  82  1w  32 68. 1p4  162  1p  22

Section 4.5 For Exercises 69–72, factor by removing the greatest common factor. 69. x3  4x2  11x 70. 21w3  7w  14 71. 5x1x  72  21x  72 72. 3t1t  42  51t  42 For Exercises 73–76, factor by grouping (remember to take out the GCF first). 73. m  8m  m  8 3

2

74. 24x3  36x2  72x  108

Section 4.6 77. What characteristics determine a perfect square trinomial? For Exercises 78–87, factor the polynomials by using any method. 78. 18x2  27xy  10y2

79. 3m2  mt  10t 2

80. 60a2  65a3  20a4

81. 2k2  7k3  6k4

82. 49x2  36  84x

83. 80z  32  50z2

84. 19w  22 2  419w  22  5

85. 14x  32 2  1214x  32  36 86. 18a4  39a2  15

87. 3w4  2w2  5

Section 4.7 For Exercises 88–94, factor the binomials. 1 88. 25  y2 89. x3  27 90. b2  64 91. h3  9h

92. a3  64

93. k4  16

94. 9y3  4y

For Exercises 95–98, factor by grouping and by using the difference of squares. 95. x2  8xy  16y2  9 (Hint: Group three terms that make up a perfect square trinomial, then factor as a difference of squares.) 96. a2  12a  36  b2 97. t2  16t  64  25c2 98. y2  6y  9  16x2

Section 4.8 99. How do you determine if an equation is quadratic? 100. What shape is the graph of a quadratic function?

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Review Exercises

For Exercises 101–104, label the equation as quadratic or linear. 101. x2  6x  7

102. 1x  321x  42  9

103. 2x  5  3

104. x  3  5x2

105. x2  2x  15  0

107. 2t1t  52  1  3t  3  t2 108. 31x  121x  5212x  92  0

109. f1x2  4x2  4 110. g1x2  2x2  2

a. Complete the table to determine the height of the missile for the given values of t. Height h(t) (ft)

1 3 10

b. y

20

y

3 2

5 4

1

3 2 1

x

4 3 2

c.

42

2 3

1 2 3

4

x

Height h(t)

8

30

d. y

y

50 40

5 4

30

3

20 10

2 1 4

x

4 3 2 1 1 2 3

(ft)

a.

2 3

where h(t) is the height in feet and t is the time in seconds.

0

1 1 112. k1x2   x2  8 2

1

Width

Time t (sec)

111. h1x2  5x3  10x2  20x  40

4 3 2 1 10 20 30

Length

h1t2  16t2  672t  1280

For Exercises 109–112, find the x- and y-intercepts of the function. Then match the function with its graph.

4 6

10 ft

114. A missile is shot upward from a submarine 1280 ft below sea level. The initial velocity of the missile is 672 ft/sec. A function that approximates the height of the missile (relative to sea level) is given by

106. 8x2  59x  21

2

113. A moving van has the capacity to hold 1200 ft3 in volume. If the van is 10 ft high and the length is 1 ft less than twice the width, find the dimensions of the van. You-Haul

For Exercises 105–108, use the zero product rule to solve the equations.

8 6 4 2 1 2 3 4 5

411

0

Time t (sec)

1280 ft 1

2 3

4

x

b. Interpret the meaning of a negative value of h(t). c. Factor the function to find the time required for the missile to emerge from the water and the time required for the missile to reenter the water. (Hint: The height of the missile will be zero at sea level.)

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Chapter 4 Polynomials

Chapter 4

Test

For Exercises 1–4, simplify the expression, and write the answer with positive exponents only. 1.

20a7 4a6

3. a

x6x3 x2

2.

3x6 2 b 5y7

5. Multiply.

18. Explain the process to solve a polynomial equation by the zero product rule.

121xy2 2 3 1x4y2 1x0y5 2 1

4.

18.0  10 217.1  10 2 6

17. Explain the strategy for factoring a polynomial expression.

For Exercises 19–32, factor completely. 19. 3a2  27ab  54b2

20. c4  1

21. xy  7x  3y  21

22. 49  p2

23. 10u2  30u  20

24. 12t2  75

25. 5y2  50y  125

26. 21q2  14q

27. 2x3  x2  8x  4

3 28. y  125

29. x2  8x  16  y2

30. r6  256r2

5

6. Divide. (Write the answer in scientific notation.) 19,200,0002  10.0042 7. For the function defined by F1x2  5x3  2x2  8, find the function values F112 , F122 , and F102 . 8. Perform the indicated operations. Write the answer in descending order. 15x2  7x  32  1x2  5x  252  14x2  4x  202

31. 1x2  12 2  31x2  12  2 32. 12a  6ac  2b  bc

For Exercises 9–11, multiply the polynomials. 9. 12a  521a2  4a  92

For Exercises 33–38, solve the equation. 33. 12x  321x  52  0

1 3 10. a x  b 16x  42 3 2

34. w2  7w  0

11. 15x  4y2 215x  4y2 2 12. Explain why 15x  72 2  25x2  49. 13. Write and simplify an expression that describes the area of the square.

35. y2  6y  16 36. x15x  42  1 37. 4p  64p3  0 1 1 38. t2  t  0 2 16

7x  4

14. Divide the polynomials.

12x y  5x y  6xy  xy2  12xy2 3 4

2 2

3

15. Divide the polynomials.

110p3  13p2  p  32  12p  32

16. Divide the polynomials by using synthetic division. 1y4  2y  52  1y  22

39. A child launches a toy rocket from the ground. The height, h(x), of the rocket can be determined by its horizontal distance, x, from the launch pad by h1x2  

x2 x 256

where x and h are in feet and x 0 and h 0. How many feet from the launch pad will the rocket hit the ground?

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Cumulative Review Exercises

a.

40. The population of Japan, P(t) (in millions), can be approximated by:

b. y

y 50 40 30 20 10

25 20

P(t)  0.01t 2  0.062t  127.7,

15 10 5

where t  0 represents the year 2000. a. Evaluate p142 and interpret in the context of the problem.

5 4 3 2 1 5 10 15 20

b. Approximate the number of people in Japan in the year 2006. c. If the trend continues, predict the population of Japan in the year 2015.

1

2 3

4 5

x

25

c.

5 4 3 2 1 10 20 30 40 50

41. f1x2  x  6x  8 2

43. p1x2  2x2  8x  6

4 5

2

4 6

8 10

x

y

10 8

10 8

6

6

4 2

4 2

108 6 4 2 2 4 6 8 10

42. k1x2  x3  4x2  9x  36

2 3

d. y

For Exercises 41–44, find the x- and y-intercepts of the function. Then match the function with its graph.

1

2

4 6

8 10

x

108 6 4 2

x

4 6 8 10

44. q1x2  x3  x2  12x

Cumulative Review Exercises

Chapters 1–4

1. Graph the inequality and express the set in interval notation: All real numbers at least 5, but not more than 12

5. The recent population of Mexico was approximately 9.85  107. At the current growth rate of 1.7%, this number is expected to double after 42 yr. How many people does this represent? Express the answer in scientific notation.

2. Simplify the expression 3x2  5x  2  41x2  32 . 3. Graph from memory.

b. y  0 x 0

a. y  x2 y

y

5 4 3 2 1 54 3 2 1 1 2 3 4 5

6. In the 2006 Orange Bowl football championship, Penn State scored 3 points more than Florida State in a three-overtime thriller. The total number of points scored was 49. Find the number of points scored by each team.

1 2 3 4 5

x

5 4 3 2 1 5 4 3 2 1 1 2 3

1

4 5

1 2 1 3 4. Simplify the expression a b  a b . 3 2

2

3 4

5

x

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Chapter 4 Polynomials

7. Find the value of each angle in the triangle.

14. A telephone pole is leaning after a storm (see figure). What is the slope of the pole?

(2x) (2x  5)

x

8. Divide 1x3  642  1x  42.

14 ft

9. Determine the slope and y-intercept of the line 4x  3y  9, and graph the line. y

15. Given P1x2  16x2  x  5, find the function value P162.

5 4 3 2 1 5 4 3 2 1 1 2 3

1

2

3 4

5

x

16. Solve.

1 1 1 x   1x  32 3 6 2

17. Given 3x  2y  5, solve for y.

4 5

18. A student scores 76, 85, and 92 on her first three algebra tests.

10. Write an equation of the line that is perpendicular to the y-axis and passes through the point 12, 52. 11. Simplify the expression.

a. Is it possible for her to score high enough on the fourth test to bring her test average up to 90? Assume that each test is weighted equally and that the maximum score on a test is 100 points. b. What is the range of values required for the fourth test so that the student’s test average will be between 80 and 89, inclusive?

36a2b4 3 a b 18b6 12. Solve the system. 2x  y  2z 

1

3x  5y  2z  11 x  y  2z  1 13. Determine whether the relation is a function. a. 512, 12, 13, 12, 18, 12, 15, 126 b.

2 ft

y

19. How many liters of a 40% acid solution and how many liters of a 15% acid solution must be mixed to obtain 25 L of a 30% acid solution? 20. Multiply the polynomials 14b  3212b2  12 . 21. Add the polynomials.

15a2  3a  12  13a3  5a  62

22. Divide the polynomials 16w3  5w2  2w2  12w2 2. For Exercises 23–24, solve the equations. 23. y2  5y  14

x

24. a3  9a2  20a  0 25. Solve the inequality and write the solution set in interval notation. 5 6 8  0 2x  5 0

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5

CHAPTER OUTLINE 5.1 Rational Expressions and Rational Functions 416 5.2 Multiplication and Division of Rational Expressions 426 5.3 Addition and Subtraction of Rational Expressions 431 5.4 Complex Fractions 441 Problem Recognition Exercises: Operations on Rational Expressions 449

5.5 Solving Rational Equations 449 Problem Recognition Exercises: Rational Equations vs. Expressions 457

5.6 Applications of Rational Equations and Proportions 458 5.7 Variation 469 Group Activity: Computing the Future Value of an Investment

478

Chapter 5 In this chapter, we define a rational expression as a ratio of two polynomials. The operations on rational expressions mirror the operations on fractions. Are You Prepared? To prepare for this chapter, work Exercises 1–8 to practice simplifying expressions containing fractions and solving equations containing fractions. Fill in the words Solution Word corresponding to the solutions to Exercises 1–8. 4 15 1. ⴢ 9 40

6 2 2. ⫼ 35 7

8 7 3. ⫺ 3 6

6 2 4. ⫺ ⫹ 5 3

16 5. ⫺ ⫼6 3

2 6. ⫺4 ⫼ a⫺ b 3

x⫺5 x ⫽ 2 3

To

8.

a 4

8

3

{15}

denominator

{5}

rational

⫺

1 2 1 x⫺ ⫽1⫹ x 2 3 6

or 2

expressions

1

6

8 9

common

8 15

subtract

⫺

Solve the equations. 7.

3 2

. 5

7

6

need

3 5

add

1 6

we

415

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416

Chapter 5 Rational Expressions and Rational Equations

Section 5.1

Rational Expressions and Rational Functions

Concepts

1. Rational Functions

1. Rational Functions 2. Simplifying Rational Expressions 3. Simplifying Ratios of ⴚ1

In Chapter 4, we introduced polynomials and polynomial functions. The ratio of two polynomials defines a rational expression. This leads to the following definition of a rational function.

DEFINITION Rational Function

A function is a rational function if it can be written in the form f 1x2 ⫽

f(x)

where p and q are polynomial functions and q1x2 ⫽ 0.

5 4 3

f(x) ⫽

2 1 ⫺5 ⫺4

⫺1 ⫺1 ⫺2

1

2

⫺3 ⫺4 ⫺5

Figure 5-1

1 x

3 4

p 1x2 q 1x2

,

For example, the functions f, g, h, and k are rational functions. 5

1 f1x2 ⫽ , x

x

g1x2 ⫽

2 , x⫺3

h1a2 ⫽

a⫹7 , a2 ⫺ 5

k1t2 ⫽

t⫹4 2t ⫺ 11t ⫹ 5 2

In Section 2.7, we introduced the rational function defined by f 1x2 ⫽ 1x . Recall that f 1x2 ⫽ 1x has a restriction on its domain that x ⫽ 0 and the graph of f 1x2 ⫽ 1x has a vertical asymptote at x ⫽ 0 (Figure 5-1). The domain of a function is the set of values that when substituted into the function produce a real number. For a rational function, we must exclude values that make the denominator zero. To find the domain of a rational function, we offer these guidelines.

PROCEDURE Finding the Domain of a Rational Function Step 1 Set the denominator equal to zero and solve the resulting equation. Step 2 The domain is the set of all real numbers excluding the values found in step 1.

Example 1 Given g1x2 ⫽

Evaluating a Rational Function

2 x⫺3

a. Determine the function values (if they exist). g(0), g(1), g(2), g(2.5), g(2.9), g(3), g(3.1), g(3.5), g(4), and g(5). b. Write the domain of the function.

Solution: a.

g102 ⫽

2 2 ⫽⫺ 102 ⫺ 3 3

g132 ⫽

2 2 ⫽ 132 ⫺ 3 0

g112 ⫽

2 ⫽ ⫺1 112 ⫺ 3

g13.12 ⫽

2 2 ⫽ ⫽ 20 13.12 ⫺ 3 0.1

g122 ⫽

2 ⫽ ⫺2 122 ⫺ 3

g13.52 ⫽

2 2 ⫽ ⫽4 13.52 ⫺ 3 0.5

g12.52 ⫽

2 2 ⫽ ⫽ ⫺4 12.52 ⫺ 3 ⫺0.5

g142 ⫽

2 ⫽2 142 ⫺ 3

g12.92 ⫽

2 2 ⫽ ⫽ ⫺20 12.92 ⫺ 3 ⫺0.1

g152 ⫽

2 ⫽1 152 ⫺ 3

(undefined)

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Section 5.1

b.

g1x2 

2 x3

417

Rational Expressions and Rational Functions

The value of the function is undefined when the denominator equals zero.

x30

Set the denominator equal to zero and solve for x.

x3

The value x  3 must be excluded from the domain.

The domain can be written in set-builder notation or in interval notation. Set-builder notation: 5x 0x is a real number, x  36 Interval notation: 1, 32 ´ 13, 2

(

4 3 2 1

(

0

1

2

3

4

5

Skill Practice Determine the function values. x2 h 1x2  x6 1. h(0) 2. h(2) 3. h(2) 5. Write the domain in set-builder notation. 6. Write the domain in interval notation.

g(x)  x 2 3

g(x) 5 4 3

4. h(6)

2 1 5 4 3 2 1 1 2

1

2

4 5

3 4 5

The graph of g1x2  x 2 3 is shown in Figure 5-2. Notice that the function has a vertical asymptote at x  3 where the function is undefined (Figure 5-2).

Figure 5-2

Calculator Connections Topic: Using Dot Mode on a Graphing Calculator A Table feature can be used to check the function values in Example 1, g1x2  x 2 3 .

The graph of a rational function may be misleading on a graphing calculator. For example, g1x2  2 1x  32 has a vertical asymptote at x  3, but the vertical asymptote is not part of the graph (Figure 5-3). A graphing calculator may try to “connect the dots” between two consecutive points, one to the left of x  3 and one to the right of x  3. This creates a line that is nearly vertical and appears to be part of the graph.

Figure 5-3

To show that this line is not part of the graph, we can graph the function in a Dot Mode (see the owner’s manual for your calculator). The graph of g1x2  2 1x  32 in dot mode indicates that the line x  3 is not part of the function (Figure 5-4).

Answers 1 1 2. 3 2 3. 0 4. Undefined 5. 5x 0 x is a real number, x  66 6. 1, 62 ´ 16,  2 1.

Figure 5-4

x

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Chapter 5 Rational Expressions and Rational Equations

Example 2

Finding the Domain of a Rational Function

Write the domain in set-builder notation and in interval notation. a. f 1x2 

x4 2x2  11x  5

b. g1x2 

x x2  4

Solution: a. f 1x2 

x4 2x2  11x  5

2x2  11x  5  0

12x  121x  52  0

Set the denominator equal to zero and solve for x. This is a factorable quadratic equation.

2x  1  0

or

x50

1 2

or

x5

x

The domain written in set-builder notation is 5x 0 x is a real number and x  12, x  56.

The domain written in interval notation is 1, 12 2 ´ 1 12, 52 ´ 15, 2 . b. g1x2 

x x 4 2

Because the quantity x2 is nonnegative for any real number x, the denominator x2  4 cannot equal zero; therefore, no real numbers are excluded from the domain. The domain written in set-builder notation is 5x 0 x is a real number}.

TIP: The domain of a rational function excludes values for which the denominator is zero. The values for which the numerator is zero do not affect the domain of the function.

The domain written in interval notation is 1, 2.

Skill Practice Write the domain in set-builder notation and in interval notation. 7. h1x2 

x  10 2x 2  x  1

8. g1t2 

t t2  1

2. Simplifying Rational Expressions p

A rational expression is an expression in the form q, where p and q are polynomials and q  0. As with fractions, it is often advantageous to simplify rational expressions to lowest terms. The method for simplifying rational expressions mirrors the process to simplify fractions. In each case, factor the numerator and denominator. Common factors in the numerator and denominator form a ratio of 1 and can be simplified.

Answers

7. 5x 0 x is a real number and x  12 , x  16 ; 1, 12 2 ´ 112 , 12 ´ 11,  2 8. 5t 0 t is a real number6 ; 1,  2

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Section 5.1

15 35

factor

x2  x  12 x2  16

factor

Simplifying a fraction: Simplifying a rational expression:

Rational Expressions and Rational Functions

419

3ⴢ5 3 5 3 3  ⴢ  ⴢ1 7ⴢ5 7 5 7 7 1x  321x  42

1x  421x  42

  

1x  32

1x  42

ⴢ

1x  42

1x  32

1x  42

ⴢ1

1x  42

1x  32 1x  42

This process is stated formally as the fundamental principle of rational expressions.

PROPERTY Fundamental Principle of Rational Expressions Let p, q, and r represent polynomials. Then p r p p pr  ⴢ  ⴢ1 qr q r q q

for q  0 and r  0

A rational expression is not defined for the values of the variable that make the denominator equal to zero. We refer to these values as restrictions on the variable. Example 3

Simplifying a Rational Expression

Given the expression

2x3  12x2  16x 6x  24

a. Factor the numerator and denominator. b. Determine the restrictions on x. c. Simplify the expression.

Solution: a.

2x3  12x2  16x 6x  24  

2x1x2  6x  82 61x  42

Factor the numerator and denominator.

2x1x  421x  22 61x  42

b. The expression is not defined for all values of x for which the denominator is equal to zero. 61x  42  0

Solve the equation.

x  4 The restriction on x is that x  4.

Avoiding Mistakes Always determine the restrictions on the variable before simplifying an expression.

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Chapter 5 Rational Expressions and Rational Equations

c.

2x1x  421x  22 2 ⴢ 31x  42

 

x1x  22 3

1

ⴢ

21x  42 21x  42

x1x  22

Simplify the ratio of common factors. provided x  4

3 x2  3x  28 2x  14

Skill Practice Given

9. a. Factor the numerator and denominator. b. Determine the restrictions on x. c. Simplify the expression.

It is important to note that the expressions 2x3  12x2  16x 6x  24

and

x1x  22 3

are equal for all values of x that make each expression a real number. Therefore, x1x  22 2x3  12x2  16x  6x  24 3 for all values of x except x  4. (At x  4, the original expression is undefined.) From this point forward, when we simplify rational expressions we will not explicitly write the restrictions by the simplified form. These will be implied from the original expression. Example 4 involves simplifying a quotient of two monomials. This was also covered in Section 4.1.

Simplifying a Rational Expression

Example 4 2x2y8

Simplify.

8x4y3

Solution: 2x2y8

This expression has the restriction that x  0 and y  0.

8x4y3 

2x2y8

Factor the denominator.

23x4y3 1

  Answers 9. a.

1x  721x  42

b. x  7

21x  72 x4 c. provided x  7 2 2 b 10. 2a 3

y5 22x2

ⴢ

2x2y3 2x2y3

Simplify common factors whose ratio is 1.

y5 4x2

Skill Practice Simplify. 10.

9a 5b 3 18a 8b

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Section 5.1

421

Simplifying a Rational Expression

Example 5 Simplify.

Rational Expressions and Rational Functions

t3 ⫹ 8 t ⫹ 4t ⫹ 4 2

Solution:

1t ⫹ 221t2 ⫺ 2t ⫹ 42 t3 ⫹ 8 ⫽ t2 ⫹ 4t ⫹ 4 1t ⫹ 22 2

Factor the numerator and denominator. The numerator is a sum of cubes. The denominator is a perfect square trinomial. The restriction on t is t ⫽ ⫺2.

1t ⫹ 22 1t2 ⫺ 2t ⫹ 42 1

⫽

⫽

Simplify common factors whose ratio is 1.

1t ⫹ 221t ⫹ 22

TIP: Recall the formula to factor a sum of cubes. a3 ⫹ b3 ⫽ 1a ⫹ b21a2 ⫺ ab ⫹ b2 2 Recall the formula to factor a perfect square trinomial. a2 ⫹ 2ab ⫹ b2 ⫽ 1a ⫹ b2 2

t2 ⫺ 2t ⫹ 4 t⫹2

Skill Practice Simplify. 11.

p3 ⫺ 27 p2 ⫺ 6p ⫹ 9

Avoiding Mistakes Because the fundamental property of rational expressions is based on the identity property of multiplication, reducing applies only to factors (remember that factors are multiplied). Therefore, terms that are added or subtracted cannot be reduced. For example: 1

3x 3x x x ⫽ ⫽ 112 ⴢ ⫽ 3y 3y y y

However,

Reduce common factor.

x⫹3 cannot be simplified. y⫹3

Cannot reduce common terms.

3. Simplifying Ratios of ⴚ1 When two factors are identical in the numerator and denominator, they form a ratio of 1 and can be simplified. Sometimes we encounter two factors that are opposites and form a ratio equal to ⫺1. For example: Simplified Form

Details/Notes

⫺5 ⫽ ⫺1 5

The ratio of a number and its opposite is ⫺1.

100 ⫽ ⫺1 ⫺100

The ratio of a number and its opposite is ⫺1.

x⫹7 ⫽ ⫺1 ⫺x ⫺ 7

x⫹7 x⫹7 x⫹7 1 ⫽ ⫽ ⫽ ⫽ ⫺1 ⫺x ⫺ 7 ⫺11x ⫹ 72 ⫺11x ⫹ 72 ⫺1

1

Factor out ⫺1.

Answer 1

2⫺x ⫽ ⫺1 x⫺2

⫺11⫺2 ⫹ x2 ⫺11x ⫺ 22 2⫺x ⫺1 ⫽ ⫽ ⫽ ⫽ ⫺1 x⫺2 x⫺2 x⫺2 1

11.

p 2 ⫹ 3p ⫹ 9 p⫺3

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Chapter 5 Rational Expressions and Rational Equations

Recognizing factors that are opposites is useful when simplifying rational expressions. For example, a ⫺ b and b ⫺ a are opposites because the opposite of a ⫺ b can be b written ⫺1a ⫺ b2 ⫽ ⫺a ⫹ b ⫽ b ⫺ a. Therefore, in general, ab ⫺ ⫺ a ⫽ ⫺1.

Example 6

Simplifying a Rational Expression

Simplify the rational expression to lowest terms.

x⫺5 25 ⫺ x2

Solution: x⫺5 25 ⫺ x2 ⫽

x⫺5 15 ⫺ x215 ⫹ x2

Factor. Notice that x ⫺ 5 and 5 ⫺ x are opposites and form a ratio of ⫺1.

⫺1

x⫺5 ⫽ 15 ⫺ x215 ⫹ x2 ⫽ 1⫺12a ⫽⫺

In general,

a⫺b ⫽ ⫺1. b⫺a

1 b 5⫹x

1 1 or ⫺ 5⫹x x⫹5

TIP: The factor of ⫺1 may be applied in front of the rational expression, or it may be applied to the numerator or to the denominator. Therefore, the final answer may be written in different forms. ⫺

1 x⫹5

or

⫺1 x⫹5

or

Skill Practice Simplify the expression. Answer

12.

5 12. ⫺ x⫹3

Section 5.1

20 ⫺ 5x x 2 ⫺ x ⫺ 12

Practice Exercises

Boost your GRADE at mathzone.com!

• Practice Problems • Self-Tests • NetTutor

Study Skills Exercise 1. Define the key terms. a. Rational expression

b. Rational function

• e-Professors • Videos

1 ⫺1x ⫹ 52

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Rational Expressions and Rational Functions

Concept 1: Rational Functions For Exercises 2–6, determine the function values, if possible. (See Example 1.) 2. f 1x2 

5 ; f 102, f 122, f 112, f 162 x1

3. k1x2 

3 ; k102, k112, k122, k142 x4

4. m1x2 

x4 ; m162, m142, m102, m142 x6

5. n1a2 

3a  1 1 ; n112, n102, na b, n112 3 a2  1

6. f 1t2 

2t  8 ; f 142, f 142, f 132, f 132 t2  9

For Exercises 7–20, a. Write the domain in set-builder notation. b. Write the domain in interval notation. (See Example 2.) 7. f 1x2 

9 x

9. h1v2 

v1 v7

10. p1t2 

t9 t3

11. k1x2 

3x  1 2x  5

12. n1t2 

6t  5 3t  8

14. k1a2 

a2 2a  3a  5 x x  16

13. f 1q2 

8. g1a2  

q1 q  6q  27 2

10 a

2

15. h1c2 

c c  25

16. m1x2 

17. f 1x2 

x5 x2  25

18. g1t2 

t4 t 2  16

19. p1x2 

x5 3

20. r1x2 

x2 8

2

2

423

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Chapter 5 Rational Expressions and Rational Equations

For Exercises 21–24, write the domain of each function in interval notation and use that information to match the function with its graph. 21. m 1x2 

22. n 1x2 

1 x4

y

a.

y

b.

5

x

y

d.

4 3 2 1

2 1

2 1

2 1 2 3 4

y

c.

1 x1 5

5

1

24. p 1x2 

1 x4 5 4 3

5 4 3

5 4 3 2 1 1 2

23. q 1x2 

1 x1

5 4 3 2 1 1 2

1 2 3 4

5

x

5 4 3 2 1 1 2

4 3

1 2 3 4

5

x

5 4 3 2 1 1 2

3

3

3

3

4 5

4 5

4 5

4 5

Concept 2: Simplifying Rational Expressions For Exercises 25–26, simplify the expression if possible. 25. a.

8x 4y

b.

8x 4y

26. a.

a  21 14  b

b.

21a 14b

For Exercises 27–30, a. Factor the numerator and denominator. b. Determine the restrictions on x. c. Simplify the expression. (See Example 3.) 27.

x2  6x  8 x2  3x  4

28.

x2  6x 2x  11x  6

29.

x2  18x  81 x2  81

30.

x2  14x  49 x2  49

2

For Exercises 31–52, simplify the rational expressions. (See Examples 4–5.) 31.

100x3y5 36xy8

32.

35.

3m4n 12m6n4

36.

39.

x5 x2  25

40.

43.

2t 2  7t  4 2t 2  5t  3

44.

48ab3c2 6a7bc0 5x3y2 4 2

20x y

3z  6 3z2  12 y2  8y  9 y2  5y  4

33.

7w11z6 14w3z3

34.

12r9s3 24r8s4

37.

6a  18 9a  27

38.

5y  15 3y  9

41.

7c 21c2  35c

42.

45.

1p  1212p  12 4

1p  12 2 12p  12 2

46.

2p  3 2p  7p  6 2

r 1r  32 5

r3 1r  32 2

1 2 3

4 5

x

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Section 5.1

47.

9 ⫺ z2 2z2 ⫹ z ⫺ 15

51.

10x3 ⫺ 25x 2 ⫹ 4x ⫺ 10 ⫺4 ⫺ 10x2

48.

2c 2 ⫹ 2c ⫺ 12 ⫺8 ⫹ 2c ⫹ c 2

Rational Expressions and Rational Functions

49.

2z3 ⫹ 128 16 ⫹ 8z ⫹ z 2

52.

8x3 ⫺ 12x2 ⫹ 6x ⫺ 9 16x4 ⫺ 9

50.

p3 ⫺ 1 5 ⫺ 10p ⫹ 5p2

Concept 3: Simplifying Ratios of ⴚ1 For Exercises 53–64, simplify the rational expressions. (See Example 6.) 53.

r⫹6 6⫹r

54.

a⫹2 2⫹a

55.

b⫹8 ⫺b ⫺ 8

56.

7⫹w ⫺7 ⫺ w

57.

10 ⫺ x x ⫺ 10

58.

y ⫺ 14 14 ⫺ y

59.

2t ⫺ 2 1⫺t

60.

5p ⫺ 10 2⫺p

61.

c⫹4 c⫺4

62.

b⫹2 b⫺2

63.

64.

4w2 ⫺ 49z2 14z ⫺ 4w

y⫺x 12x2 ⫺ 12y2

Mixed Exercises For Exercises 65–84, simplify the rational expression. 65.

t2 ⫺ 1 t2 ⫹ 7t ⫹ 6

66.

69.

⫺16a2bc4 8ab2c4

70.

73.

b⫹4 2b ⫹ 5b ⫺ 12

74.

77.

81.

2

1a ⫺ 22 2 1a ⫺ 52 3 1a ⫺ 22 1a ⫺ 52 3

78.

x2 ⫹ 4x ⫹ 4 x2 ⫺ 4 ⫺9x3yz2 27x4yz c⫺6 3c ⫺ 17c ⫺ 6 2

t2 1t ⫺ 112 4

t 1t ⫺ 112 5

2

4y3 ⫹ 12y2 ⫺ y ⫺ 3 x3 ⫺ 2x2 ⫺ 25x ⫹ 50 82. x3 ⫹ 5x2 ⫺ 4x ⫺ 20 2y3 ⫹ y2 ⫺ 18y ⫺ 9

67.

8p ⫹ 8 2p ⫺ 4p ⫺ 6 2

68.

15y ⫺ 15 3y ⫹ 9y ⫺ 12 2

71.

x2 ⫺ y2 8y ⫺ 8x

72.

p2 ⫺ 49 14 ⫺ 2p

75.

⫺2x ⫹ 34 ⫺4x ⫹ 6

76.

⫺9w ⫺ 3 3w ⫹ 12

79.

4x ⫺ 2x2 5x ⫺ 10

80.

83.

t3 ⫹ 8 3t2 ⫹ t ⫺ 10

84.

2y ⫺ 6 3y2 ⫺ y3 w3 ⫺ 27 4w2 ⫺ 5w ⫺ 21

Expanding Your Skills 85. Write a rational expression whose domain is 1⫺⬁, 22 ´ 12, ⬁2 . (Answers may vary.)

86. Write a rational expression whose domain is 1⫺⬁, 32 ´ 13, ⬁ 2 . (Answers may vary.)

87. Write a rational function whose domain is 1⫺⬁, ⫺52 ´ 1⫺5, ⬁2 . (Answers may vary.)

88. Write a rational function whose domain is 1⫺⬁, ⫺62 ´ 1⫺6, ⬁2 . (Answers may vary.)

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Section 5.2

Multiplication and Division of Rational Expressions

Concepts

1. Multiplication of Rational Expressions

1. Multiplication of Rational Expressions 2. Division of Rational Expressions

Recall that to multiply fractions, we multiply the numerators and multiply the denominators. The same is true for multiplying rational expressions.

PROPERTY Multiplication of Rational Expressions Let p, q, r, and s represent polynomials, such that q ⫽ 0 and s ⫽ 0. Then p r pr ⴢ ⫽ q s qs

For example: Multiply the Fractions

Multiply the Rational Expressions

2 5 10 ⴢ ⫽ 3 7 21

2x 5z 10xz ⴢ ⫽ 3y 7 21y

Sometimes it is possible to simplify a ratio of common factors to 1 before multiplying. To do so, we must first factor the numerators and denominators of each fraction. 1

1

1

3ⴢ5 7ⴢ3ⴢ5 1 7 ⴢ ⫽ ⫽ 2ⴢ5 3ⴢ7 2ⴢ5ⴢ3ⴢ7 2

Factor.

7 15 ⴢ 10 21

The same process is also used to multiply rational expressions.

PROCEDURE Multiplying Rational Expressions Step 1 Factor the numerator and denominator of each expression. Step 2 Multiply the numerators, and multiply the denominators. Step 3 Reduce the ratios of common factors to 1 or ⫺1 and simplify.

Multiplying Rational Expressions

Example 1 Multiply.

5a ⫺ 5b 2 ⴢ 2 10 a ⫺ b2

Solution: 5a ⫺ 5b 2 ⴢ 2 10 a ⫺ b2 ⫽ ⫽

Avoiding Mistakes If all factors in the numerator simplify to 1, do not forget to write the factor of 1 in the numerator.

51a ⫺ b2 2 ⴢ 5ⴢ2 1a ⫺ b21a ⫹ b2 51a ⫺ b2 ⴢ 2

5 ⴢ 2 ⴢ 1a ⫺ b21a ⫹ b2 5 1a ⫺ b2 ⴢ 2 1

⫽

1

Factor numerator and denominator.

Multiply.

1

5 ⴢ 2 ⴢ 1a ⫺ b21a ⫹ b2

⫽

1 a⫹b

Reduce common factors and simplify.

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Section 5.2

Multiplication and Division of Rational Expressions

Skill Practice Multiply. 3y ⫺ 6 y2 ⫹ 3y ⫹ 2 1. ⴢ 6y y2 ⫺ 4

Multiplying Rational Expressions

Example 2

4w ⫺ 20p

Multiply.

2w2 ⫺ 50p2

ⴢ

2w2 ⫹ 7wp ⫺ 15p2 3w ⫹ 9p

Solution: 4w ⫺ 20p 2w2 ⫹ 7wp ⫺ 15p2 ⴢ 3w ⫹ 9p 2w2 ⫺ 50p2 ⫽ ⫽ ⫽

41w ⫺ 5p2

21w ⫺ 25p 2 2

⫽

ⴢ

12w ⫺ 3p21w ⫹ 5p2 31w ⫹ 3p2

2 ⴢ 21w ⫺ 5p2

21w ⫺ 5p21w ⫹ 5p2

ⴢ

Factor numerator and denominator.

12w ⫺ 3p21w ⫹ 5p2

Factor further.

31w ⫹ 3p2

2 ⴢ 21w ⫺ 5p212w ⫺ 3p21w ⫹ 5p2

Multiply.

21w ⫺ 5p21w ⫹ 5p2 ⴢ 31w ⫹ 3p2 2 ⴢ 21w ⫺ 5p2 12w ⫺ 3p21w ⫹ 5p2 1

⫽

2

1

1

21w ⫺ 5p2 1w ⫹ 5p2 ⴢ 31w ⫹ 3p2

Simplify common factors.

212w ⫺ 3p2 31w ⫹ 3p2

Notice that the expression is left in factored form to show that it has been simplified to lowest terms. Skill Practice Multiply. 2.

p 2 ⫹ 8p ⫹ 16 2p ⫹ 6 ⴢ 2 10p ⫹ 10 p ⫹ 7p ⫹ 12

2. Division of Rational Expressions Recall that to divide fractions, multiply the first fraction by the reciprocal of the second fraction. Divide:

15 10 ⫼ 14 49

Multiply by the reciprocal of the second fraction.

1

1

3ⴢ5ⴢ7ⴢ7 21 15 49 ⫽ ⫽ ⴢ 2ⴢ7ⴢ2ⴢ5 4 14 10

The same process is used for dividing rational expressions.

PROPERTY Division of Rational Expressions Let p, q, r, and s represent polynomials, such that q ⫽ 0, r ⫽ 0, s ⫽ 0. Then p p s ps r ⫼ ⫽ ⴢ ⫽ q s q r qr

Answers 1.

y⫹1 2y

2.

p⫹4 51p ⫹ 12

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Example 3 Divide.

Dividing Rational Expressions

8t 3 ⫹ 27 4t 2 ⫺ 6t ⫹ 9 ⫼ 9 ⫺ 4t 2 2t 2 ⫺ t ⫺ 3

Solution: 8t 3 ⫹ 27 4t 2 ⫺ 6t ⫹ 9 ⫼ 2 2 9 ⫺ 4t 2t ⫺ t ⫺ 3 ⫽

Avoiding Mistakes When dividing rational expressions, your first step should be to take the reciprocal of the second fraction (divisor). Do this first so that you do not forget.

⫽

8t 3 ⫹ 27 2t 2 ⫺ t ⫺ 3 ⴢ 9 ⫺ 4t 2 4t 2 ⫺ 6t ⫹ 9

12t ⫹ 3214t 2 ⫺ 6t ⫹ 92 12t ⫺ 321t ⫹ 12 ⴢ 13 ⫺ 2t213 ⫹ 2t2 4t 2 ⫺ 6t ⫹ 9

12t ⫹ 32 14t 2 ⫺ 6t ⫹ 92 12t ⫺ 32 1t ⫹ 12 1

⫽

Multiply the first fraction by the reciprocal of the second.

⫺1

1

13 ⫺ 2t2 13 ⫹ 2t214t 2 ⫺ 6t ⫹ 92

⫽ 1⫺12

Factor numerator and denominator. Notice 8t 3 ⫹ 27 is a sum of cubes. Furthermore, 4t 2 ⫺ 6t ⫹ 9 does not factor over the real numbers. Simplify to lowest terms.

1t ⫹ 12 1

⫽ ⫺1t ⫹ 12

⫺t ⫺ 1

or

Skill Practice Divide. 3.

10x2 ⫹ 12x ⫹ 2 x2 ⫹ x ⫼ 5x3 ⫺ x2 25x2 ⫺ 1

TIP: In Example 3, the factors (2t ⫺ 3) and (3 ⫺ 2t) are opposites and form a ratio of ⫺1. The factors (2t ⫹ 3) and (3 ⫹ 2t) are equal and form a ratio of 1.

Answer

2t ⫺ 3 ⫽ ⫺1 3 ⫺ 2t

1 3. 2x

whereas

2t ⫹ 3 ⫽1 3 ⫹ 2t

Section 5.2 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Write an example of how to multiply two fractions, divide two fractions, add two fractions, and subtract two fractions. For example: a.

5 9 ⴢ 12 10

b.

3 15 ⫼ 4 8

c.

5 7 ⫹ 6 9

d.

5 2 ⫺ 4 7

As you learn about rational expressions, compare the operations on rational expressions to those on fractions. This is a great place to use 3 ⫻ 5 cards again. Write an example of an operation with fractions on one side and the same operation with rational expressions on the other side.

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429

Review Exercises For Exercises 2–9, simplify the rational expression to lowest terms. 2.

3w2 ⫺ 12 w ⫹ 5w ⫹ 6

3.

t2 ⫺ 5t ⫺ 6 t2 ⫺ 7t ⫹ 6

4.

6.

5x2yz3 20xyz

7.

7x ⫹ 14 7x ⫺ 7x ⫺ 42

8.

2

2

5y ⫹ 15

5.

5y2 ⫹ 16y ⫹ 3 25 ⫺ x2 x ⫺ 10x ⫹ 25

9.

2

2⫺p p2 ⫺ p ⫺ 2 a3b2c5 2a3bc2

Concept 1: Multiplication of Rational Expressions For Exercises 10–21, multiply the rational expressions. (See Examples 1–2.) 10.

8w2 3 ⴢ 9 2w4

11.

16 z4 ⴢ z7 8

12.

13.

27r 5 28rs3 ⴢ 32 7s 9r s

14.

3z ⫹ 12 16z3 ⴢ 3 9z ⫹ 36 8z

15.

16.

3y2 ⫹ 18y ⫹ 15 y ⫺ 5 ⴢ 2 6y ⫹ 6 y ⫺ 25

17.

10w ⫺ 8 3w2 ⫺ w ⫺ 14 ⴢ w⫹2 25w2 ⫺ 16

18.

19.

3x ⫺ 15 10x ⫺ 20x2 ⴢ 5⫺x 4x2 ⫺ 2x

20. x1x ⫹ 52 2 ⴢ

2 x ⫺ 25

5p2q4 12pq3

ⴢ

6p2 20q2

x2y x ⫺ 4x ⫺ 5 2

x ⫺ 5y x2 ⫹ xy

ⴢ

2x2 ⫺ 13x ⫹ 15 xy3

y2 ⫺ x2 10y ⫺ 2x

21. y1y2 ⫺ 42 ⴢ

2

ⴢ

y y⫹2

Concept 2: Division of Rational Expressions For Exercises 22–35, divide the rational expressions. (See Example 3.) 22.

5x 10x2 ⫼ 7 21

23.

24.

6x2y2 3xy2 ⫼ 1x ⫺ 22 1x ⫺ 22 2

25.

26.

t 2 ⫹ 5t ⫼ 1t ⫹ 52 t⫹1

27.

6p ⫹ 7 ⫼ 136p2 ⫺ 492 p⫹2

28.

a a3 ⫹ 6a2 ⫺ 40a ⫼ a ⫺ 10 a2 ⫺ 100

29.

b2 ⫺ 6b ⫹ 9 b2 ⫺ 9 ⫼ 4 b2 ⫺ b ⫺ 6

31.

6s2 ⫹ st ⫺ 2t 2 3s2 ⫹ 17st ⫹ 10t 2 ⫼ 6s2 ⫺ 5st ⫹ t 2 6s2 ⫹ 13st ⫺ 5t 2

33.

a3 ⫹ a ⫹ a2 ⫹ 1 a3 ⫹ a ⫹ a2b ⫹ b ⫼ 2 2 2 2 a ⫹ a ⫹ ab ⫹ b 2a ⫹ 2ab ⫹ ab2 ⫹ b3

35.

8x ⫺ 4x2 3x ⫹ 6 ⫼ xy ⫺ 2y ⫹ 3x ⫺ 6 y⫹3

30.

32.

34.

2x2 ⫹ 5xy ⫹ 2y2 4x2 ⫺ y2

⫼

x2 ⫹ xy ⫺ 2y2 2x2 ⫹ xy ⫺ y2

x4 ⫺ x3 ⫹ x2 ⫺ x x3 ⫺ 4x2 ⫹ x ⫺ 4 ⫼ 3 3 2 2x ⫹ 2x ⫹ x ⫹ 1 2x ⫺ 8x2 ⫹ x ⫺ 4 3y ⫺ y2 y3 ⫺ 27

⫼

y y2 ⫹ 3y ⫹ 9

2a 10a5 ⫼ 3 77 7b 1r ⫹ 32 2 3

4r s

⫼

r⫹3 rs

3

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Mixed Exercises For Exercises 36–55, perform the indicated operations. 36.

40.

8a4b3 a7b2 ⫼ 3c 9c

37.

10x2 ⫺ 13xy ⫺ 3y2 8x2 ⫺ 10xy ⫺ 3y2

42. 13m2 ⫺ 12m2 ⫼

ⴢ

4x3y 3x5 ⫼ 2 7 2x y 6y6

2y ⫹ 8x 2x2 ⫹ 2y2

m2 ⫺ 4m m2 ⫺ 6m ⫹ 8

2 5x 2 ⴢ ⫼ 15x 25x2 12

41.

2a3 ⫹ 4a2b 6a2 ⫹ ab ⫺ b2 ⴢ 10a2 ⫹ 5ab 3a2 ⫹ 5ab ⫺ 2b2

1a ⫹ b2 2 a3 ⫺ b3 a2 ⫹ ab ⫹ b2 ⴢ 2 ⫼ a⫺b a ⫺ b2 1a ⫺ b2 2

45.

46.

x2 ⫺ 4y2 2y ⫼ 1x ⫹ 2y2 ⴢ x ⫹ 2y x ⫺ 2y

47.

39.

x4 ⫺ 16 x2 ⫹ x ⫺ 6

43. 12x2 ⫹ 82 ⫼

44.

m2 ⫺ n2 m2 ⫺ 2mn ⫹ n2 1m ⫺ n2 4 ⫼ ⴢ 1m ⫺ n2 2 m2 ⫺ mn ⫹ n2 m3 ⫹ n3 x2 ⫺ 6xy ⫹ 9y2 x2 ⫺ 4y2

ⴢ

x2 ⫺ 5xy ⫹ 6y2 x2 ⫺ 9y2 ⫼ 3y ⫺ x x ⫹ 2y

8x2 ⫹ 12xy ⫹ 18y2 2x ⫹ 3y

49.

25m2 ⫺ 1 5m ⫹ 1 ⫼ 125m3 ⫺ 1 25m2 ⫹ 5m ⫹ 1

50.

m3 ⫹ 2m2 ⫺ mn2 ⫺ 2n2 m3 ⫺ 25m ⴢ m3 ⫺ m2 ⫺ 20m m3 ⫹ m2n ⫺ 4m ⫺ 4n

51.

2a2 ⫹ ab ⫺ 8a ⫺ 4b a2 ⫺ 6a ⫹ 9 ⴢ 2a2 ⫺ 6a ⫹ ab ⫺ 3b a2 ⫺ 16

52.

7 14 ⴢ 1x ⫹ 52 ⫼ 3x ⫹ 15 9x ⫺ 27

53.

45 27 ⴢ 18x ⫹ 42 ⫼ 2x ⫹ 1 4x ⫹ 2

55.

2x2 ⫺ 11x ⫺ 6 2x2 ⫺ 5x ⫺ 3 ⫼ 2 3x ⫺ 2 3x ⫺ 7x ⫺ 6

48.

54.

8x3 ⫺ 27y3

y2 3 4y ⫼ ⴢ 7 14 y

38.

4x ⫺ 9y 2

2

⫼

12y ⫹ 3 6y2 ⫺ y ⫺ 12

⫼

4y2 ⫺ 19y ⫺ 5 2y2 ⫺ y ⫺ 3

Expanding Your Skills For Exercises 56–59, write an expression for the area of the figure and simplify. 56.

57. b2 in. 5a

8 hk

k2 cm 2h2

4a2 in. b

58.

cm

59. 5x ⫺ 15 ft 4x

12x m x2 ⫺ 4

x⫹2m 6

x2 ft x⫺3

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Section 5.3

Addition and Subtraction of Rational Expressions

Addition and Subtraction of Rational Expressions

Section 5.3

1. Addition and Subtraction of Rational Expressions with Like Denominators

Concepts

To add or subtract rational expressions, the expressions must have the same denominator. As with fractions, we add or subtract rational expressions with the same denominator by combining the terms in the numerator and then writing the result over the common denominator. Then, if possible, we simplify the expression to lowest terms.

PROPERTY Addition and Subtraction of Rational Expressions Let p, q, and r represent polynomials where q  0. Then 1.

p pr r   q q q

2.

pr p r   q q q

Adding and Subtracting Rational Expressions with Like Denominators

Example 1

Add or subtract as indicated. a.

1 3  8 8

b.

5x 3  2x  1 2x  1

Solution: 1 3 13 a.   8 8 8 

4 8



1 2

c.

x2 x  12  x4 x4

Add the terms in the numerator.

Simplify the fraction.

b.

5x 3 5x  3   2x  1 2x  1 2x  1

Add the terms in the numerator. The answer is already in lowest terms.

c.

x2 x  12  x4 x4

Combine the terms in the numerator. Use parentheses to group the terms in the numerator that follow the subtraction sign. This will help you remember to apply the distributive property.



x2  1x  122 x4



x2  x  12 x4

Apply the distributive property.



1x  421x  32

Factor the numerator and denominator.



1x  42 1x  32

Simplify the rational expression.

1x  42

1

1x  42

x3

431

1. Addition and Subtraction of Rational Expressions with Like Denominators 2. Least Common Denominator 3. Equivalent Rational Expressions 4. Addition and Subtraction of Rational Expressions with Unlike Denominators

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Skill Practice Add or subtract as indicated. 1.

5 1  12 12

2.

4c  3 8  c2 c2

3.

t2 5t  14  t7 t7

2. Least Common Denominator If two rational expressions have different denominators, each expression must be rewritten with a common denominator before adding or subtracting the expressions. The least common denominator (LCD) of two or more rational expressions is defined as the least common multiple of the denominators. For example, consider the fractions 201 and 81 . By inspection, the least common denominator is 40. To understand why, find the prime factorization of both denominators. 20  22 ⴢ 5

and

8  23

A common multiple of 20 and 8 must be a multiple of 5, a multiple of 22, and a multiple of 23. However, any number that is a multiple of 23  8 is automatically a multiple of 22  4. Therefore, it is sufficient to construct the least common denominator as the product of unique prime factors, where each factor is raised to its highest power. The LCD of

1 1 and 3 is 23 ⴢ 5  40. 2 ⴢ5 2 2

PROCEDURE Steps to Find the LCD of Two or More Rational Expressions Step 1 Factor all denominators completely. Step 2 The LCD is the product of unique prime factors from the denominators, where each factor is raised to the highest power to which it appears in any denominator.

Example 2

Finding the LCD of Rational Expressions

Find the LCD of the rational expressions. a.

1 5 7 , , 12 18 30

b.

1 5 , 3 2x y 16xy2z

Solution: a.

1 5 7 , , 12 18 30 1 5 7 , , 2 2 ⴢ3 2ⴢ3 2ⴢ3ⴢ5 2

LCD  22 ⴢ 32 ⴢ 5  180

Answers 1.

1 3

2.

4c  5 c2

3. t  2

Factor the denominators completely. The LCD is the product of the factors 2, 3, and 5. Each factor is raised to its highest power.

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Section 5.3

b.

Addition and Subtraction of Rational Expressions

1 5 , 3 2x y 16xy 2z 1 5 , 4 2 3 2x y 2 xy z

Factor the denominators completely.

LCD ⫽ 2 4x3y 2z

The LCD is the product of the factors 2, x, y, and z. Each factor is raised to its highest power.

⫽ 16x3y2z

Skill Practice Find the LCD of the rational expressions. 4.

7 1 5 , , 40 15 6

Example 3

5.

1 5 , 3 2 9a b 18a 4b

Finding the LCD of Rational Expressions

Find the LCD of the rational expressions. a.

x2 ⫹ 3 6 , x ⫹ 9x ⫹ 20 x2 ⫹ 8x ⫹ 16

b.

2

x⫹4 1 , x⫺3 3⫺x

Solution: a.

x2 ⫹ 3 6 , 2 2 x ⫹ 9x ⫹ 20 x ⫹ 8x ⫹ 16 x2 ⫹ 3 6 , 1x ⫹ 421x ⫹ 52 1x ⫹ 42 2

LCD ⫽ 1x ⫹ 521x ⫹ 42 2

b.

x⫹4 1 , x⫺3 3⫺x

Factor the denominators completely. The LCD is the product of the factors 1x ⫹ 52 and 1x ⫹ 42. Each factor is raised to its highest power. The denominators are already factored.

Notice that x ⫺ 3 and 3 ⫺ x are opposite factors. If ⫺1 is factored from either expression, the binomial factors will be the same. x⫹4 1 x⫹4 1 , , x ⫺ 3 ⫺11⫺3 ⫹ x2 ⫺11⫺x ⫹ 32 3 ⫺ x Factor out ⫺1. same binomial factors

Factor out ⫺1. same binomial factors

LCD ⫽ 1x ⫺ 321⫺12

LCD ⫽ 1⫺1213 ⫺ x2

⫽ ⫺x ⫹ 3

⫽ ⫺3 ⫹ x

⫽3⫺x

⫽x⫺3

The LCD can be expressed as either 13 ⫺ x2 or 1x ⫺ 32. Skill Practice Find the LCD of the rational expressions. 6.

5 x⫹1 , 2 x ⫹ 4x ⫹ 4 x ⫺ x ⫺ 6 2

7.

6 1 , z⫺7 7⫺z

Answers 4. 120 5. 18a 4b 2 6. (x ⫹ 2)2(x ⫺ 3) 7. z ⫺ 7 or 7 ⫺ z

433

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Chapter 5 Rational Expressions and Rational Equations

3. Equivalent Rational Expressions Rational expressions can be added if they have common denominators. Once the LCD has been determined, each rational expression must be converted to an equivalent rational expression with the indicated denominator. Using the identity property of multiplication, we know that for q  0 and r  0, p p p r pr  ⴢ1 ⴢ  q q q r qr This principle is used to convert a rational expression to an equivalent expression with a different denominator. For example, 12 can be converted to an equivalent expression with a denominator of 12 as follows: 1 1 1 6 1ⴢ6 6  ⴢ1 ⴢ   2 2 2 6 2ⴢ6 12 In this example, we multiplied 12 by a convenient form of 1. The ratio 66 was chosen so that the product produced a new denominator of 12. Notice that multiplying 12 by 66 is equivalent to multiplying the numerator and denominator of the original expression by 6.

Example 4

Creating Equivalent Rational Expressions

Convert each expression to an equivalent rational expression with the indicated denominator. a.

7  5p2 20p6

b.

w  2 w5 w  3w  10

Solution: a.

7  5p2 20p6 5p2 must be multiplied by 4p4 to create 20p6. Multiply numerator and denominator by 4p4.

28p4 7 4p4 ⴢ  5p2 4p4 20p6 b.

w  2 w5 w  3w  10 

w w w2  ⴢ w5 w5 w2

TIP: Notice that in Example 4 we multiplied the polynomials in the numerator but left the denominator in factored form. This convention is followed because when we add and subtract rational expressions, the terms in the numerators must be combined.

Answers 8.

2x 2y 16x y

3 2

9.

5b 2  15b b2  9

1w  521w  22



Factor the denominator. Multiply the numerator and denominator by the missing factor 1w  22.

w2  2w 1w  521w  22

Skill Practice Convert each expression to an equivalent rational expression with the indicated denominator. 8.

1  8xy 16x 3y 2

9.

5b  2 b3 b 9

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Section 5.3

Addition and Subtraction of Rational Expressions

4. Addition and Subtraction of Rational Expressions with Unlike Denominators To add or subtract rational expressions with unlike denominators, we must convert each expression to an equivalent expression with the same denominator. For example, consider adding the expressions x 3 2  x 5 1 . The LCD is 1x  221x  12. For each expression, identify the factors from the LCD that are missing in the denominator. Then multiply the numerator and denominator of the expression by the missing factor(s): 132

1x  22

ⴢ

1x  12 1x  12



152

1x  12

ⴢ

1x  22 1x  22

The rational expressions now have the same denominator and can be added.



31x  12  51x  22 1x  221x  12

Combine terms in the numerator.



3x  3  5x  10 1x  221x  12

Clear parentheses and simplify.



8x  7 1x  221x  12

PROCEDURE Steps to Add or Subtract Rational Expressions Step 1 Factor the denominator of each rational expression. Step 2 Identify the LCD. Step 3 Rewrite each rational expression as an equivalent expression with the LCD as its denominator. Step 4 Add or subtract the numerators, and write the result over the common denominator. Step 5 Simplify, if possible.

Example 5

Adding Rational Expressions with Unlike Denominators

3 4  2 7b b

Add.

Solution: 3 4  2 7b b

The denominators are already factored.

Step 2:

The LCD is 7b2.



3 b 4 7 ⴢ  2ⴢ 7b b b 7

Step 3:

Write each expression with the LCD.



3b 28  2 2 7b 7b

Step 4:

Add the numerators, and write the result over the LCD.



3b  28 7b2

Step 5: Simplify.

Skill Practice Add. 10.

Step 1:

1 4  3 5y 3y

Answer 10.

12y 2  5 15y 3

435

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Chapter 5 Rational Expressions and Rational Equations

Example 6

Subtracting Rational Expressions with Unlike Denominators

3t  2 5  2t  12 t  4t  12

Subtract.

2

Solution: 3t  2 5  2t  12 t 2  4t  12 



3t  2 5  1t  621t  22 21t  62 122

ⴢ

13t  22

122 1t  621t  22



Step 1:

Factor the denominators.

Step 2:

The LCD is 21t  621t  22.

1t  22 5 ⴢ 21t  62 1t  22

Step 3:

Write each expression with the LCD.



213t  22  51t  22 21t  621t  22

Step 4: Add the numerators and write the result over the LCD.



6t  4  5t  10 21t  621t  22

Step 5: Simplify.



t6 21t  621t  22

Combine like terms.

1

t6  2 1t  621t  22 

Simplify.

1 21t  22

Skill Practice Subtract. 11.

2x  3 5  3x 3 x x2 2

Example 7

Adding and Subtracting Rational Expressions with Unlike Denominators

Add and subtract as indicated.

2 x 3x  18   2 x x3 x  3x

Solution: 2 x 3x  18   2 x x3 x  3x 

Answer 1 11. 31x  22



2 x 3x  18   x x3 x1x  32

x x 3x  18 2 1x  32 ⴢ  ⴢ  x 1x  32 1x  32 x x1x  32

Step 1:

Factor the denominators.

Step 2:

The LCD is x1x  32.

Step 3: Write each expression with the LCD.

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Section 5.3



21x  32  x2  13x  182 x1x  32

Step 4:



2x  6  x  3x  18 x1x  32

Step 5:



x2  x  12 x1x  32



437

Addition and Subtraction of Rational Expressions

Add the numerators, and write the result over the LCD.

2

Simplify.

Avoiding Mistakes It is important to insert parentheses around the quantity being subtracted.

Combine like terms.

1x  421x  32 x1x  32

Factor the numerator.

1x  421x  3 2 1





Simplify.

x1x  32

x4 x

Skill Practice Add. 12.

a 2  a  24 5  2 a3 a 9

Example 8

Subtracting Rational Expressions with Unlike Denominators

4 6  w w

Subtract.

Solution: 4 6  w w



112 6 4  ⴢ w w 112



6 4  w w

 

6  142 w

10 w

Step 1:

The denominators are already factored.

Step 2:

The denominators are opposites and differ by a factor of 1. The LCD can either be taken as w or w. We will use an LCD of w.

Step 3:

Write each expression with the LCD. Note that 1w2112  w.

Step 4:

Subtract the numerators, and write the result over the LCD.

Step 5:

Simplify.

Skill Practice Subtract. 13.

3 5  y y

Answers 12.

a3 a3

13. 

8 y

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Adding Rational Expressions with Unlike Denominators

Example 9

y2 x2  xy yx

Add.

Solution: y2 x2  xy yx



112 y2 x2  ⴢ 1x  y2 1y  x2 112



y2 x2  xy xy



x2  y2 xy 1x  y21x  y2

Step 1:

The denominators are already factored.

Step 2:

The denominators are opposites and differ by a factor of 1. The LCD can be taken as either 1x  y2 or 1y  x2. We will use an LCD of 1x  y2.

Step 3:

Write each expression with the LCD. Note that 1y  x2112  y  x  x  y.

Step 4:

Combine the numerators, and write the result over the LCD.

1



xy

Step 5: Factor and simplify to lowest terms.

xy Skill Practice Add. Answer

14.

3a 15  a5 5a

14. 3

Section 5.3 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Least common denominator (LCD)

b. Equivalent rational expressions

Review Exercises For Exercises 2–6, perform the indicated operation. 2.

x2 x  xy yx

3.

9b  9 2b  4 ⴢ 4b  8 3b  3

5.

15  a2 2 25a2  1 ⴢ 2 10a  2 a  10a  25

6.

x2  z2 x2  2xz  z2  14x2z4 3xz3

4.

8a2 4a  21b 7b3

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Addition and Subtraction of Rational Expressions

439

Concept 1: Addition and Subtraction of Rational Expressions with Like Denominators For Exercises 7–14, add or subtract as indicated and simplify if possible. (See Example 1.) 7.

3 7  5x 5x

9.

x 3  2 x  2x  3 x  2x  3

10.

x 6  2 x  4x  12 x  4x  12

11.

5x  1 3x  6  12x  921x  62 12x  921x  62

12.

4x 5x  6  8x  1 8x  1

13.

x2 x  12  x5 x5

14.

2x  1 x5  x2 x2

8.

2

1 5  2 2x2 2x

2

Concept 2: Least Common Denominator For Exercises 15–26, find the least common denominator (LCD). (See Examples 2–3.) y2 35

16.

y ; 15a

18.

13 ; 12cd5

20.

x ; 12x  121x  72

2 12x  121x  12

22.

14 ; 1x  22 2 1x  92

41 x1x  221x  92

x5 x  8x  12

24.

7a ; a4

a  12 a2  16

5 4a

26.

10 ; x6

x1 6x

15.

5 ; 8

3 20x

17.

5 ; 6m4

19.

6 ; 1x  421x  22

21.

3 ; x1x  121x  72 2

23.

5 ; x6

25.

3a ; a4

1 15mn7 8 1x  421x  62 1 x2 1x  72

2

9 8c3

Concept 3: Equivalent Rational Expressions For Exercises 27–32, fill in the blank to make an equivalent fraction with the given denominator. (See Example 4.) 27.

5  2 3x 9x y

28.

5  2 3 xy 4x y

29.

2x  x1 x1x  121x  22

30.

5x  2x  5 12x  521x  82

31.

y  2 y6 y  5y  6

32.

t2  2 t8 t  6t  16

Concept 4: Addition and Subtraction of Rational Expressions with Unlike Denominators For Exercises 33–58, add or subtract as indicated. (See Examples 5–9.) 33.

4 5  2 3p 2p

34.

6 1  2 10ab 5a b

35.

t1 s1  s t

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Chapter 5 Rational Expressions and Rational Equations

36.

y⫺2 x⫹2 ⫺ x y

37.

4a ⫺ 2 a⫺2 ⫺ 3a ⫹ 12 a⫹4

38.

6y ⫹ 5 y⫹2 ⫺ 5y ⫺ 25 y⫺5

39.

10 2 ⫹ b1b ⫹ 52 b

40.

6 3 ⫹ w w1w ⫺ 22

41.

x⫺2 x⫹2 ⫺ x⫺6 6⫺x

42.

x ⫺ 10 x ⫹ 10 ⫺ x⫺8 8⫺x

43.

6b 1 ⫺ b⫺4 b⫹1

44.

a 5 ⫺ a⫺3 a⫹6

45.

2 4 ⫹ 2x ⫹ 1 x⫺2

46.

3 1 ⫹ y⫹6 3y ⫹ 1

47.

y⫺2 2y2 ⫺ 15y ⫹ 12 ⫹ y⫺4 y2 ⫺ 16

48.

x⫹1 x2 ⫹ 13x ⫹ 18 ⫹ 2 x⫹3 x ⫺9

49.

x⫹2 x ⫺ 2 2 x ⫺ 36 x ⫹ 9x ⫹ 18

50.

7 x ⫹ 2 x ⫺x⫺2 x ⫹ 4x ⫹ 3

51.

5 8 ⫹ w ⫺w

52.

5 4 ⫹ y ⫺y

53.

n 2n ⫺ 5 ⫹ 5⫺n n⫺5

54.

2c ⫺ 7 c ⫹ 7⫺c c⫺7

55.

x 2 ⫹ 3x ⫺ 15 25 ⫺ x2

56.

5 4 ⫺ 2 2 9⫺x x ⫹ 4x ⫹ 3

57.

m 4 t 2 ⫺ ⫺ 58. 2 2 2 20 ⫹ 9m ⫹ m 12 ⫹ 7m ⫹ m 6 ⫹ 5t ⫹ t 2 ⫹ 3t ⫹ t 2

2

Mixed Exercises For Exercises 59–78, simplify. 59.

x⫹3 x⫹5 ⫹ 2 2x x

60.

x⫹2 x⫹4 ⫹ 2 15x 5x

61. w ⫹ 2 ⫹

1 h⫹3

63.

9 x⫺3 ⫺ 2 x2 ⫺ 2x ⫹ 1 x ⫺x

64.

t⫺2 6 t⫹1 ⫺ ⫹ 2 t⫹3 t⫺3 t ⫺9

66.

y⫺3 y⫹1 ⫺4y ⫹ 7 ⫺ ⫹ 2 y ⫺ 2 2y ⫺ 5 2y ⫺ 9y ⫹ 10

67. 1x ⫺ 12 ⴢ c

62. h ⫺ 3 ⫹

65.

68. 13x ⫺ 22 ⴢ c 71.

2x 1 1 ⫺ ⫹ x⫺y y⫺x x ⫺ y2 2

74. 1y ⫹ 82 ⴢ c 77.

x 2 3z z ⫹ d 69. ⫺ x ⫹ 1 z ⫺ 3 z ⫹ 4 3x ⫹ x ⫺ 2 2

y 4 ⫺ 2 d 2y ⫹ 1 2y ⫹ 17y ⫹ 8

⫺10 15 ⫹ 2 z2 ⫺ 6z ⫹ 5 z ⫺ 4z ⫺ 5

70.

1 w⫺2

2 z⫹1 ⫺ 2 4z2 ⫺ 12z ⫹ 9 2z ⫺ 3z 3 x d ⫹ 2x ⫺ 2 x ⫺1

2p p ⫺ p⫺5 p⫹6

72.

3w ⫺ 1 2⫺w w ⫺ ⫺ w⫺1 1⫺w 2w ⫹ w ⫺ 3

73. 12p ⫹ 12 ⴢ c

75.

3 2 ⫹ y y⫺6

76.

78.

⫺4 3 ⫹ 2 n2 ⫹ 6n ⫹ 5 n ⫹ 7n ⫹ 10

2

2

2p 1 ⫺ d 6p ⫹ 3 p⫹4

p ⫺8 ⫹ p p⫹5

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For Exercises 79–82, write an expression that represents the perimeter of the figure and simplify. 79.

80. x  1 yd x

x  1 cm x

2 cm 3x

2 yd x2

x  4 yd 3

6 cm x2

81.

82. 3 ft x2

5 m x3

x ft x1

2x m x5

Complex Fractions

Section 5.4

1. Simplifying Complex Fractions by Method I

Concepts

A complex fraction is a fraction whose numerator or denominator contains one or more fractions. For example:

1. Simplifying Complex Fractions by Method I 2. Simplifying Complex Fractions by Method II 3. Using Complex Fractions in Applications

5x2 y 10x y2

1 1  2 3 3 1  4 6

2 and

are complex fractions. Two methods will be presented to simplify complex fractions. The first method (Method I) follows the order of operations to simplify the numerator and denominator separately before dividing. The process is summarized as follows.

PROCEDURE Simplifying a Complex Fraction—Method I Step 1 Add or subtract expressions in the numerator to form a single fraction. Add or subtract expressions in the denominator to form a single fraction. Step 2 Divide the rational expressions from step 1. Step 3 Simplify to lowest terms, if possible.

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Chapter 5 Rational Expressions and Rational Equations

Example 1

Simplifying a Complex Fraction by Method I

Simplify the expression.

5x2 y 10x y2

Solution: 5x2 y 10x y2

Step 1: The numerator and denominator of the complex fraction are already single fractions.

⫽

5x2 y2 ⴢ y 10x

⫽

5x2y2 10xy

⫽

1 2⫺1 2⫺1 x y 2

1 ⫽ xy 2

or

Step 2: Multiply the numerator of the complex fraction by the reciprocal of the denominator.

Step 3: Simplify.

xy 2

Skill Practice Simplify the expression. 18a 3 b2 1. 6a 2 b

Sometimes it is necessary to simplify the numerator and denominator of a complex fraction before the division is performed. This is illustrated in Example 2.

Example 2

Simplifying a Complex Fraction by Method I

Simplify the expression.

1 3 ⫺ 4x 2 1 3⫺ 2x

Solution: 1 3 ⫺ 4x 2 1 3⫺ 2x

Answer 1.

3a b

1 2x 3 ⫺ ⴢ 4x 2x 2 ⫽ 2x 3 1 ⴢ ⫺ 2x 1 2x

Step 1:

Combine fractions in the numerator and denominator separately.

The LCD in the numerator is 4x. The LCD in the denominator is 2x.

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Section 5.4

1 ⫺ 6x 4x ⫽ 6x ⫺ 1 2x ⫺1

Step 2:

Complex Fractions

Divide the expression in the numerator of the complex fraction by the expression in the denominator.

1

1 ⫺ 6x 2x ⫽ ⴢ 4x 6x ⫺ 1

Multiply by the reciprocal of the divisor.

2

1 ⫽⫺ 2

Step 3:

Simplify to lowest terms.

Skill Practice Simplify the expression. 1 4 ⫺ 9m 3 2. 1 4⫺ 3m

2. Simplifying Complex Fractions by Method II We will now use a second method to simplify complex fractions—Method II. Recall that multiplying the numerator and denominator of a rational expression by the same quantity does not change the value of the expression. This is the basis for Method II.

PROCEDURE Simplifying a Complex Fraction—Method II Step 1 Multiply the numerator and denominator of the complex fraction by the LCD of all individual fractions within the expression. Step 2 Apply the distributive property, and simplify the numerator and denominator. Step 3 Simplify to lowest terms, if possible.

Example 3

Simplifying a Complex Fraction by Method II 6 x 2 3 ⫺ 2 x x 4⫺

Simplify by using Method II.

Solution: 6 x 2 3 ⫺ 2 x x 4⫺

6 x2 ⴢ a4 ⫺ b x ⫽ 2 3 x2 ⴢ a ⫺ 2 b x x

The LCD of all individual terms is x2.

Step 1:

Multiply the numerator and denominator of the complex fraction by the LCD of x2.

Answer 2. ⫺

1 3

443

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Chapter 5 Rational Expressions and Rational Equations

6 x2 ⴢ 142 ⫺ x2 ⴢ a b x ⫽ 2 3 x2 ⴢ a b ⫺ x2 ⴢ a 2 b x x ⫽

Step 2:

Apply the distributive property.

Step 3:

Factor and simplify.

4x2 ⫺ 6x 2x ⫺ 3 2x12x ⫺ 3 2 1

⫽

2x ⫺ 3

⫽ 2x Skill Practice Simplify by using Method II. 1 y 3. 1 1⫺ 2 y y⫺

Example 4

Simplifying a Complex Fraction by Method II

Simplify by using Method II.

x⫺1 ⫺ x⫺2 1 ⫹ 2x⫺1 ⫺ 3x⫺2

Solution: x⫺1 ⫺ x⫺2 1 ⫹ 2x⫺1 ⫺ 3x⫺2

TIP: When writing 2 x⫺1 with positive exponents recall that 2x⫺1 ⫽ 2 ⴢ ⫽

1 x

2 1 2 ⴢ ⫽ 1 x x

1 1 ⫺ 2 x x ⫽ 2 3 1⫹ ⫺ 2 x x

1 1 x2 ⴢ a ⫺ 2 b x x ⫽ 2 3 x2 ⴢ a1 ⫹ ⫺ 2 b x x

1 1 x2 a b ⫺ x2 a 2 b x x ⫽ 2 3 x2 112 ⫹ x2 a b ⫺ x2 a 2 b x x ⫽

Answer 3. y

x⫺1 x ⫹ 2x ⫺ 3 2

Rewrite the expression with positive exponents. The LCD of all individual terms is x2.

Step 1:

Multiply the numerator and denominator of the complex fraction by the LCD x2.

Step 2:

Apply the distributive property.

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Section 5.4

⫽

x⫺1 1x ⫹ 321x ⫺ 12

Step 3:

Complex Fractions

Factor and simplify to lowest terms.

1

x⫺1 ⫽ 1x ⫹ 321x ⫺ 12 ⫽

1 x⫹3

Skill Practice Simplify by using Method II. 4.

c ⫺1b ⫺ b ⫺1c b ⫺1 ⫹ c ⫺1

Example 5

Simplifying a Complex Fraction by Method II

Simplify the expression by Method II.

1 1 ⫺ w⫹3 w⫺3 9 1⫹ 2 w ⫺9

Solution: 1 1 ⫺ w⫹3 w⫺3 9 1⫹ 2 w ⫺9 1 1 ⫺ w⫹3 w⫺3 ⫽ 9 1⫹ 1w ⫹ 321w ⫺ 32

Factor all denominators to find the LCD.

1 1 9 1 The LCD of , , , and is 1w ⫹ 321w ⫺ 32. 1 w⫹3 w⫺3 1w ⫹ 321w ⫺ 32 1w ⫹ 321w ⫺ 32a

1 1 ⫺ b w⫹3 w⫺3 ⫽ 9 1w ⫹ 321w ⫺ 32 c 1 ⫹ d 1w ⫹ 321w ⫺ 32

Step 1: Multiply the numerator and denominator of the complex fraction by 1w ⫹ 32 1w ⫺ 32.

1w ⫹ 321w ⫺ 32a

1 1 b ⫺ 1w ⫹ 321w ⫺ 32 a b w⫹3 w⫺3 ⫽ 9 1w ⫹ 321w ⫺ 321 ⫹ 1w ⫹ 321w ⫺ 32 c d 1w ⫹ 321w ⫺ 32 ⫽

1w ⫺ 32 ⫺ 1w ⫹ 32 1w ⫹ 321w ⫺ 32 ⫹ 9

⫽

w⫺3⫺w⫺3 w2 ⫺ 9 ⫹ 9

⫽

⫺6 w2

⫽⫺

6 w2

Step 3:

Step 2: Distributive property.

Simplify.

Apply the distributive property.

Answer 4. b ⫺ c

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Chapter 5 Rational Expressions and Rational Equations

Skill Practice Simplify by using Method II. 2 1 ⫺ x⫹1 x⫺1 5. x 1 ⫺ x⫺1 x⫹1

3. Using Complex Fractions in Applications Finding the Slope of a Line

Example 6

Find the slope of the line that passes through the given points. 1 5 a , b 2 6

and

9 5 a , b 8 4

Solution: 1 5 9 5 Let 1x1, y1 2 ⫽ a , b and 1x2, y2 2 ⫽ a , b. 2 6 8 4 Slope ⫽

Label the points.

y2 ⫺ y1 x2 ⫺ x1

5 5 ⫺ 4 6 ⫽ 9 1 ⫺ 8 2

Apply the slope formula.

Using Method II to simplify the complex fractions, we have 5 24 a ⫺ 4 m⫽ 9 24 a ⫺ 8

5 b 6 1 b 2

6 4 5 5 24 a b ⫺ 24 a b 4 6 ⫽ 3 12 1 9 24 a b ⫺ 24 a b 8 2

⫽

30 ⫺ 20 27 ⫺ 12

⫽

10 2 ⫽ 15 3

Multiply numerator and denominator by the LCD, 24.

Apply the distributive property.

Simplify.

2 The slope is . 3 Skill Practice 3 1 7 5 6. Find the slope of the line that contains the points a⫺ , b and a , ⫺ b. 5 4 10 2 Answers 5.

x⫺3 x2 ⫹ 1

6. m ⫽ ⫺

55 26

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Section 5.4 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercise 1. Define the key term complex fraction.

Review Exercises For Exercises 2–4, simplify to lowest terms. 2.

x3 ⫹ y3 5x ⫹ 5y

3.

25a3b3c 15a4bc

4.

6t2 ⫺ 27t ⫹ 30 12t ⫺ 30

For Exercises 5–8, perform the indicated operations. 5.

5 3 ⫹ 2x x2

6.

2y ⫺ 4 y2 ⫹ 3y ⫹ 2 ⴢ y⫹1 y2 ⫺ 4

7.

3 1 ⫺ a⫺5 a⫹1

8.

7 14b ⫼ 2 12 ⫺ 6b b ⫹b⫺6

Concept 1: Simplifying Complex Fractions by Method I For Exercises 9–16, simplify the complex fractions by using Method I. (See Examples 1–2.) 5x2 9y2 9. 3x y2x 2 1 ⫹ 3 6 13. 1 1 ⫺ 2 4

3w2 4rs 10. 15wr s2

x⫺6 3x 11. 3x ⫺ 18 9

7 3 ⫹ 8 4 14. 1 5 ⫺ 3 6

8⫺ 15.

5 2x

5 ⫺2 8x

a⫹4 6 12. 16 ⫺ a2 3 3 5x

10 ⫺ 16.

3 ⫺5 10x

Concept 2: Simplifying Complex Fractions by Method II For Exercises 17–48, simplify the complex fractions by using Method II. (See Examples 3–5.) 7y y⫹3 17. 1 4y ⫹ 12

6x x⫺5 18. 1 4x ⫺ 20

19.

3q ⫺q p 21. q q⫺ p

b ⫹ 3b a 22. 2b b⫹ a

2 3 ⫹ 2 a a 23. 4 9 ⫺ 2 a a

25.

t⫺1 ⫺ 1 1 ⫺ t⫺2

26.

d⫺2 ⫺ c⫺2 c⫺1 ⫺ d⫺1

1⫹

27.

1 3

5 ⫺1 6

⫺8 6w ⫺4 w⫺1

4 5 20. 3 ⫺1 ⫹ 10 2⫹

2 1 ⫹ 2 y y 24. 4 1 ⫺ 2 y y 6

28. 2z ⫺

10 z⫺4

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y y⫹3 29. y ⫹y y⫹3

4 w⫺4 30. 4 ⫺1 w⫺4

2 t⫹1 2 2⫹ t

2⫺ 33.

1 ⫹ y⫹2 y 37. 2 ⫺ y⫺3 y 41.

3 p⫺1 3 3⫺ p

3⫹ 34.

4 ⫺3 7 ⫹2

1 ⫹ t⫺4 t 38. 6 ⫹ t⫹5 t

x⫺2 x ⫹ 3x⫺1

42.

1 1 ⫺ 4⫹h 4 45. h

1 ⫹5 2 ⫺4

x⫺1 ⫹ x⫺2 5x⫺2

1 1 ⫺ 3 ⫹ 3h 3 46. h

1 6 ⫺ 2 x x 31. 3 4 1⫺ ⫹ 2 x x

12 1 ⫺ 2 x x 32. 9 3 ⫹ ⫺2 x x2

3 2 ⫺ a a⫹1 35. 2 3 ⫺ a a⫹1

5 4 ⫹ b b⫹1 36. 4 5 ⫺ b b⫹1

2 2 ⫺ x x⫹h 39. h

1 1 ⫺ 2x ⫹ 2h 2x 40. h

1⫺

43.

1⫹

2a⫺1 ⫹ 3b⫺2 a⫺1 ⫺ b⫺1

44.

2m⫺1 ⫹ n⫺1 m⫺2 ⫺ 4n⫺1

⫺3 3 ⫹ x x⫹h 48. h

6 6 ⫺ x x⫹h 47. h

Concept 3: Using Complex Fractions in Applications 49. The slope formula is used to find the slope of the line passing through the points 1x1, y1 2 and 1x2, y2 2. Write the slope formula from memory. For Exercises 50–53, find the slope of the line that passes through the given points. (See Example 6.) 1 2 1 50. a1 , b, a , ⫺2b 2 5 4

3 3 51. a⫺ , b, 1⫺1, ⫺32 7 5

5 9 1 1 52. a , b, a⫺ , ⫺ b 8 10 16 5

1 1 1 1 53. a , b, a , b 4 3 8 6

Expanding Your Skills 54. Show that 1x ⫹ x ⫺1 2 ⫺1 ⫽ simplifying.

x by writing the expression on the left without negative exponents and x ⫹1

55. Show that 1x⫺1 ⫹ y⫺1 2 ⫺1 ⫽

2

xy by writing the expression on the left without negative exponents and x⫹y

simplifying. 56. Simplify.

x 1 1 ⫺ a1 ⫹ b x

⫺1

57. Simplify.

x 1 ⫺1 1 ⫺ a1 ⫺ b x

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Solving Rational Equations

449

Problem Recognition Exercises Operations on Rational Expressions For Exercises 1–24, identify the operation (addition, subtraction, multiplication, or division), then simplify the expression. In each case, be sure to ask yourself if you need a common denominator. 3.

16x2 ⫺ 9 x 5x2 ⫺ 6x ⫹ 1 ⫼ ⫺ 2 2 x ⫺1 4x ⫹ 7x ⫹ 3 4x ⫺ 3

y⫹2 4 3 ⫹ 2 ⫺ y⫹1 y⫺1 y ⫺1

6.

8w2 4w ⫺ 2 w ⫺ 16w w ⫺ 4w

t ⫺9 t⫹3 ⫼ t t⫹2

1 a 9. 1 4⫺ 2 a

1.

2 3 ⫺ ⫹1 2y ⫺ 3 2y

2. 1x ⫹ 52 ⫹ a

4.

a2 ⫺ 25 a2 ⫹ 4a ⫹ ab ⫹ 4b ⴢ 2 3a ⫹ 3ab a2 ⫹ 9a ⫹ 20

5.

7.

a ⫺ 16 x ⫹ 3 ⴢ 2x ⫹ 6 a ⫺ 4

8.

2

6x2y 5 10. 3x y

13.

3 x⫺2 ⫺ x⫺2 6 3y ⫹ 6

27 ⫼ 16. 2 y⫺5 y ⫺ 3y ⫺ 10

7 b x⫺4

2⫹

2

11.

14.

6xy x ⫺y 2

2

⫹

x⫹y y⫺x

12. 1x2 ⫺ 6x ⫹ 82 ⴢ a

5 x⫹7 ⫹ x⫹7 10 2 ⫺3 y 2 1⫺ y

19.

4x2 ⫹ 22x ⫹ 24 6x ⫹ 6 ⴢ 2 4x ⫹ 4 4x ⫺ 9

20.

22.

2x ⫺1 ⫹ 3x ⫺2 x ⫺2 ⫺ 5x ⫺1

23. 1y ⫹ 22 ⴢ

12x3y5z 5x

4

⫼

16xy7 2

10z

2y ⫹ 1 y2 ⫺ 4

⫺

y⫺2 y⫹3

3 b x⫺2

15.

1 w⫹2 ⫺ w⫺1 3w ⫺ 3

18.

3 2 ⫺ ⫹5 t⫺3 t⫹2

21.

3x ⫺ 1 7 ⫹ 4 6x ⫺ 2

24.

a2 100 ⫺ 20a ⫺ a ⫺ 10 10 ⫺ a

y⫹ 17.

3

Solving Rational Equations

Section 5.5

1. Solving Rational Equations

Concepts

Thus far, we have studied two types of equations in one variable: linear equations and quadratic equations. In this section, we will study another type of equation called a rational equation.

1. Solving Rational Equations 2. Formulas Involving Rational Equations

DEFINITION Rational Equation An equation with one or more rational expressions is called a rational equation.

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Chapter 5 Rational Expressions and Rational Equations

The following equations are rational equations: 1 1 1 x⫹ ⫽ x 2 3 4

3 1 2 ⫹ ⫽ x 5 3

3⫺

6w 6 ⫽ w⫹1 w⫹1

To understand the process of solving a rational equation, first review the procedure of clearing fractions from Section 1.1.

Solving a Rational Equation

Example 1

Solve the equation.

1 1 1 x⫹ ⫽ x 2 3 4

Solution: 1 1 1 x⫹ ⫽ x 2 3 4

The LCD of all terms in the equation is 12.

1 1 1 12 a x ⫹ b ⫽ 12 a xb 2 3 4

Multiply both sides by 12 to clear fractions.

1 1 1 12 ⴢ x ⫹ 12 ⴢ ⫽ 12 ⴢ x 2 3 4 6x ⫹ 4 ⫽ 3x

Apply the distributive property. Solve the resulting equation.

3x ⫽ ⫺4 x⫽⫺

4 3 Check:

1 1 1 x⫹ ⫽ x 2 3 4 1 4 1 1 4 a⫺ b ⫹ ⱨ a⫺ b 2 3 3 4 3 2 1 1 ⫺ ⫹ ⱨ ⫺ 3 3 3 1 1 ⫺ ⱨ⫺ ✔ 3 3

4 The solution set is e ⫺ f . 3 Skill Practice Solve the equation. 1.

21 1 1 x⫹ ⫽ x 2 20 5

The same process of clearing fractions is used to solve rational equations when variables are present in the denominator.

Answer 7 1. e ⫺ f 2

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Section 5.5

Example 2

Solving Rational Equations

Solving a Rational Equation 3 1 2 ⫹ ⫽ x 5 3

Solve the equation.

Solution: The LCD of all terms in the equation is 15x. Note that in this equation there is a restriction that x ⫽ 0.

2 3 1 ⫹ ⫽ x 5 3

15x ⴢ

3 1 2 15x a ⫹ b ⫽ 15x a b x 5 3

Multiply by 15x to clear fractions.

2 3 1 ⫹ 15x ⴢ ⫽ 15x ⴢ x 5 3

Apply the distributive property.

9x ⫹ 15 ⫽ 10x

Solve the resulting equation.

15 ⫽ x Check: x ⫽ 15

1 2 3 ⫹ ⫽ x 5 3 3 1 ⱨ2 ⫹ 5 1152 3 9 1 ⱨ2 ⫹ 15 15 3 10 ⱨ 2 ✔ 15 3

The solution set is 5156. Skill Practice Solve the equation. 2.

4 3 ⫹ ⫽ ⫺1 y 3

Example 3

Solving a Rational Equation 3⫺

Solve the equation.

6w 6 ⫽ w⫹1 w⫹1

Solution: 3⫺

6w 6 ⫽ w⫹1 w⫹1

The LCD of all terms in the equation is w ⫹ 1. Note that in this equation there is a restriction that w ⫽ ⫺1.

1w ⫹ 12 132 ⫺ 1w ⫹ 12a

6w 6 b ⫽ 1w ⫹ 12a b w⫹1 w⫹1

Multiply by 1w ⫹ 12 on both sides to clear fractions.

1w ⫹ 12132 ⫺ 1w ⫹ 12a

6w 6 b ⫽ 1w ⫹ 12a b w⫹1 w⫹1

Apply the distributive property.

3w ⫹ 3 ⫺ 6w ⫽ 6 ⫺3w ⫽ 3 w ⫽ ⫺1

Solve the resulting equation.

Answer 9 2. e ⫺ f 7

451

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Check:

3⫺ 3⫺

6w 6 ⫽ w⫹1 w⫹1

61⫺12 6 ⱨ 1⫺12 ⫹ 1 1⫺12 ⫹ 1 The denominator is 0 for the value of w ⫽ ⫺1.

The value ⫺1 is one of the restrictions on w found in the first step. As expected, the value w ⫽ ⫺1 does not check. Since no other potential solution exists, the equation has no solution. The solution set is the empty set, 5 6. Skill Practice Solve the equation. 3. 5 ⫺

4x 8 ⫽ x⫹2 x⫹2

Examples 1–3 show that the steps to solve a rational equation mirror the process of clearing fractions from Section 1.1. However, we must check whether the potential solutions are defined in each expression in the original equation. A potential solution that does not check is called an extraneous solution. The steps for solving a rational equation are summarized as follows.

PROCEDURE Solving a Rational Equation Step 1 Factor the denominators of all rational expressions. Identify any values of the variable for which any expression is undefined. Step 2 Identify the LCD of all terms in the equation. Step 3 Multiply both sides of the equation by the LCD. Step 4 Solve the resulting equation. Step 5 Check the potential solutions in the original equation. Note that any value from step 1 for which the equation is undefined cannot be a solution to the equation.

Solving a Rational Equation

Example 4

1⫹

Solve the equation.

3 28 ⫽ 2 x x

Solution: 1⫹

3 28 ⫽ 2 x x

3 28 x 2 a1 ⫹ b ⫽ x 2 a 2 b x x x2 ⴢ 1 ⫹ x2 ⴢ

3 28 ⫽ x2 ⴢ 2 x x

x2 ⫹ 3x ⫽ 28 x2 ⫹ 3x ⫺ 28 ⫽ 0

Answer

3. 5 6 (The value ⫺2 does not check.)

or

Multiply both sides by x2 to clear fractions. Apply the distributive property. The resulting equation is quadratic. Set the equation equal to zero and factor.

1x ⫹ 721x ⫺ 42 ⫽ 0 x ⫽ ⫺7

The LCD of all terms in the equation is x2. Expressions will be undefined for x ⫽ 0.

x⫽4

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x ⫽ ⫺7

Check:

1⫹

1⫹

3 28 ⫽ 2 x x

3 ⱨ 28 ⫺7 1⫺72 2

1⫹

3 ⱨ 28 4 142 2

49 21 ⱨ 28 ⫺ 49 49 49

The solution set is 5⫺7, 46.

Check: x ⫽ 4

3 28 ⫽ 2 x x

1⫹

28 ⱨ 28 ✔ 49 49

Solving Rational Equations

16 12 ⱨ 28 ⫹ 16 16 16 28 ⱨ 28 ✔ 16 16

Skill Practice Solve the equation. 4. 1 ⫹

6 16 ⫽ 2 x x

Solving a Rational Equation

Example 5

2p 36 ⫽ ⫺1 p⫹3 p ⫺9

Solve.

2

Solution: 2p 36 ⫽ ⫺1 p⫹3 p ⫺9 2

2p 36 ⫽ ⫺1 1p ⫹ 321p ⫺ 32 p⫹3

The LCD is 1p ⫹ 321p ⫺ 32 . Expressions will be undefined for p = 3 and p = –3. Multiply both sides by the LCD to clear fractions.

1p ⫹ 321p ⫺ 32 c

2p 36 d ⫽ 1p ⫹ 321p ⫺ 32a b ⫺ 1p ⫹ 321p ⫺ 321 1p ⫹ 321p ⫺ 32 p⫹3

1p ⫹ 321p ⫺ 32 c

2p 36 d ⫽ 1p ⫹ 321p ⫺ 32a b ⫺ 1p ⫹ 321p ⫺ 321 1p ⫹ 321p ⫺ 32 p⫹3

36 ⫽ 2p1p ⫺ 32 ⫺ 1p ⫹ 321p ⫺ 32 36 ⫽ 2p2 ⫺ 6p ⫺ 1p2 ⫺ 92

Solve the resulting equation. The equation is quadratic.

36 ⫽ 2p ⫺ 6p ⫺ p ⫹ 9 2

2

36 ⫽ p2 ⫺ 6p ⫹ 9 0 ⫽ p2 ⫺ 6p ⫺ 27

0 ⫽ 1 p ⫺ 921p ⫹ 32 p⫽9

or

Set the equation equal to zero and factor.

p ⫽ ⫺3

Answer

4. 5⫺8, 26

453

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Check: p ⫽ 9

Check: p ⫽ ⫺3

2p 36 ⫺1 ⫽ p⫹3 p ⫺9

2p 36 ⫺1 ⫽ p⫹3 p ⫺9

2192 36 ⱨ ⫺1 192 ⫹ 3 192 ⫺ 9

21⫺32 36 ⱨ ⫺1 2 1⫺32 ⫹ 3 1⫺32 ⫺ 9

2

2

2

36 ⱨ 18 ⫺1 72 12

Denominator is zero.

1ⱨ3 ⫺1 2 2 1ⱨ1 ✔ 2 2 Here the value ⫺3 is not a solution to the original equation because it is restricted in the original equation. However, 9 checks in the original equation. The solution set is 596.

Skill Practice Solve. 5.

6 20x x ⫺ 2 ⫽ x⫹2 x⫹2 x ⫺x⫺6

2. Formulas Involving Rational Equations Solving a Literal Equation Involving Rational Expressions

Example 6

Solve for the indicated variable.

Avoiding Mistakes Variables in algebra are casesensitive. For example, M and m are different variables.

V⫽

mv m⫹M

for m

Solution: V⫽

mv m⫹M

V1m ⫹ M2 ⫽ a

for m

mv b1m ⫹ M2 m⫹M

Multiply by the LCD and clear fractions.

V1m ⫹ M2 ⫽ mv Vm ⫹ VM ⫽ mv Vm ⫺ mv ⫽ ⫺VM

Collect all m terms on one side.

m1V ⫺ v2 ⫽ ⫺VM

Factor out m.

m1V ⫺ v2 ⫺VM ⫽ 1V ⫺ v2 1V ⫺ v2

Divide by 1V ⫺ v2 .

m⫽

Answer

5. 5⫺96 (The value ⫺2 does not check.)

Use the distributive property to clear parentheses.

⫺VM V⫺v

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TIP: The factor of ⫺1 that appears in the numerator may be written in the denominator or out in front of the expression. The following expressions are equivalent: ⫺VM V⫺v

or

VM VM ⫽ ⫺1V ⫺ v2 v⫺V

or

m⫽

⫺

VM V⫺v

Skill Practice Solve the equation for x. 6. y ⫽

ax ⫹ b x⫹d

Answer 6. x ⫽

b ⫺ yd y⫺a

or

x⫽

yd ⫺ b a⫺y

Section 5.5 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Rational equation

b. Extraneous solution

Review Exercises For Exercises 2–7, perform the indicated operations. 2.

1 1 ⫹ 2 x ⫺ 16 x ⫹ 8x ⫹ 16

3.

3 2 ⫺ 2 y ⫺1 y ⫺ 2y ⫹ 1

4.

5.

2t 2 ⫹ 7t ⫹ 3 ⫼ 1t ⫹ 32 4t 2 ⫺ 1

6.

1 ⫹ x⫺1 1 ⫺ x⫺2

7.

2

2

m2 ⫺ 9 ⫼ 1m2 ⫺ m ⫺ 122 m2 ⫺ 3m x⫹y x⫺1 ⫹ y⫺1

Concept 1: Solving Rational Equations 8. Why is it important to check your answer when solving a rational equation? For Exercises 9–42, solve the rational equations. (See Examples 1–5.) x⫹2 x⫺4 1 ⫺ ⫽ 3 4 2

10.

x⫹6 x⫹8 ⫺ ⫽0 3 5

11.

3y 5y ⫺2⫽ 4 6

12.

2w 4w ⫺8⫽ 5 2

13.

5 7 ⫺ ⫹3⫽0 4p 6

14.

7 3 ⫺ ⫺2⫽0 15w 10

15.

1 3 4 5 ⫺ ⫽ ⫺ x 2 2x 12

16.

2 1 11 1 ⫹ ⫽ ⫺ 3x 4 6x 3

17.

3 5 ⫹2⫽ x⫺4 x⫺4

18.

5 7 ⫺2⫽ x⫹3 x⫹3

19.

1 2 ⫹ ⫽1 3 w⫺3

20.

3 7 ⫹ ⫽2 5 p⫹2

9.

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21.

12 12 2 ⫺ ⫽ x x x⫺5

22.

25 25 2 ⫺ ⫽ y y y⫺2

23.

3 4 ⫺ ⫽ ⫺1 2 a a

24.

3 1 ⫽2⫹ 2 w w

25.

1 a ⫺ 4a⫺1 ⫽ 0 4

26.

1 t ⫺ 12t⫺1 ⫽ 0 3

27.

y 2 6 ⫹ 2 ⫽ y y⫹3 y ⫹ 3y

28.

⫺8 t 1 ⫹ ⫽ t⫺6 t t ⫺ 6t

29.

4 8 ⫺ 2 ⫽ ⫺2 t⫺2 t ⫺ 2t

30.

x 72 ⫽ 2 ⫹4 x⫹6 x ⫺ 36

31.

6 1 4 ⫺ ⫽ 2 5y ⫹ 10 y⫺5 y ⫺ 3y ⫺ 10

32.

⫺3 2 10 ⫺ 2 ⫽ 2 x ⫺ 7x ⫹ 12 x ⫹ x ⫺ 12 x ⫺ 16

33.

x 1 5 ⫹ ⫽ x⫺5 5 x⫺5

34.

x 2 2 ⫹ ⫽ x⫺2 3 x⫺2

35.

6 1 1 ⫺ ⫽ x⫺3 4x ⫺ 4 x ⫺ 4x ⫹ 3

36.

1 5 2 ⫺ ⫹ ⫽0 x⫹3 x⫺3 4x ⫺ 36

37.

4 k2 1 ⫺ ⫺ ⫽0 k⫹2 k⫺2 4 ⫺ k2

38.

h 4 h ⫺ ⫽⫺ 2 h⫺4 h⫺4

39.

5 5 2 ⫹ ⫽ x⫺3 x⫺4 x ⫺ 7x ⫹ 12

40.

9 3 5 ⫺ ⫽ x⫹2 x⫹5 x ⫹ 7x ⫹ 10

41.

4 7 1 ⫺ 2 ⫽ 2 x ⫹ 7x ⫹ 12 x ⫹ 8x ⫹ 15 x ⫹ 9x ⫹ 20

42.

5 2 8 ⫺ 2 ⫽ 2 x ⫺ 6x ⫹ 8 x ⫹ 3x ⫺ 10 x ⫹ x ⫺ 20

2

2

2

2

2

2

2

2

Concept 2: Formulas Involving Rational Equations For Exercises 43–60, solve the formula for the indicated variable. (See Example 6.) 43. K ⫽

ma F

for m

44. K ⫽

46. K ⫽

IR E

for R

47. I ⫽

49. h ⫽

2A B⫹b

52.

T ⫹ mf ⫽g m

55. a ⫹ b ⫽

58.

2A h

b⫹a 1 ⫽ ab f

for B

for m

ma F

for a

45. K ⫽

E R⫹r

for R

48. I ⫽

E R⫹r

for h

51. x ⫽

at ⫹ b t

50.

V ⫽ r2 ph

53.

x⫺y ⫽z xy

for x

1 1 1 ⫽ ⫹ R R1 R2

for P

57.

s2 ⫺ s1 t2 ⫺ t1

for t2

60. a ⫽

56. 1 ⫹ rt ⫽

for b

59. v ⫽

for E

54. w ⫺ n ⫽ P wn

A P

for h

IR E

v2 ⫺ v1 t2 ⫺ t1

for r

for t

for w

for R

for v1

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457

Mixed Exercises For Exercises 61–74, solve the equations. 61.

3 2 ⫺4 ⫹ ⫽ 2 x x⫹2 x ⫹ 2x

62.

63. 4c1c ⫹ 12 ⫽ 31c2 ⫹ 42 65.

1 8 2 ⫹ ⫽ y y⫺3 y ⫺ 3y 2

64. 3t12t ⫺ 22 ⫽ 51t 2 ⫺ 12

4 3 2 ⫺ ⫽ 2 v⫺1 v⫹5 v ⫹ 4v ⫺ 5

66.

67. 51x ⫺ 92 ⫽ 31x ⫹ 42 ⫺ 214x ⫹ 12

⫺2 3 6 ⫺ ⫽ 2 a⫹4 a⫺5 a ⫺ a ⫺ 20

68. 4z ⫺ 315z ⫺ 32 ⫽ z ⫺ 12

69.

3y y 5 ⫺ ⫽ 10 2y 5

70.

2h 8 h ⫺ ⫽ 3 3h 2

71.

1 2 5 14d ⫺ 12 ⫹ 12d ⫹ 22 ⫽ 14d ⫹ 12 2 3 6

72.

2 1 1 12b ⫹ 52 ⫹ 17b ⫺ 102 ⫽ 13b ⫹ 22 5 10 2

73. 8t⫺1 ⫹ 2 ⫽ 3t⫺1

74. 6z⫺2 ⫺ 5z⫺1 ⫽ 0

Expanding Your Skills 75. If 5 is added to the reciprocal of a number, the result is 163 . Find the number. 76. If 23 is added to the reciprocal of a number, the result is 173 . Find the number. 77. If 7 is decreased by the reciprocal of a number, the result is 92 . Find the number. 78. If a number is added to its reciprocal, the result is 136. Find the number.

Problem Recognition Exercises Rational Equations vs. Expressions 1. a. Simplify.

3 10 1 ⫹ 2 ⫺ w⫺5 w⫹5 w ⫺ 25

2. a. Simplify.

b. Solve.

3 10 1 ⫹ 2 ⫽0 ⫺ w⫺5 w⫹5 w ⫺ 25

b. Solve.

c. What is the difference in the type of problem given in parts (a) and (b)?

x 2 ⫹ ⫺1 2x ⫹ 4 3x ⫹ 6 x 2 ⫹ ⫽1 2x ⫹ 4 3x ⫹ 6

c. What is the difference in the type of problem given in parts (a) and (b)?

For Exercises 3–20, first ask yourself whether the problem is an expression to simplify or an equation to solve. Then simplify or solve as indicated. 3.

2 1 ⫹ a ⫹ 3 a ⫹ 4a ⫹ 3

6.

x 12 1 5 3 ⫺ 2 ⫺ ⫺ ⫽ 0 7. x⫺1 b ⫹ 2 b ⫺ 1 b2 ⫹ b ⫺ 2 x ⫺x

2

4.

1 4 ⫹ 2 c⫹6 c ⫹ 8c ⫹ 12

5.

7 1 3 ⫹ ⫺ ⫽0 y ⫹ 1 y ⫺ 2 y ⫺y⫺2

8.

4 3 ⫹ 5t ⫺ 20 t⫺4

2

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9.

3 7 ⫺5⫽ ⫺1 w w

10.

⫺3 1 ⫺ ⫽ ⫺2 y y2

11.

4p ⫹ 1 p⫺3 ⫹ 8p ⫺ 12 2p ⫺ 3

12.

x⫹1 x2 ⫺ 2x ⫹ 4 x⫹2

13.

1 1 ⫹ 6x 2x2

14.

5 1 ⫹ 2 4a 6a

15.

3 2 ⫺1 ⫹ 2⫽ 2t t 3t

16.

1 1 ⫺3 ⫹ ⫽ 2 5b 2b b

17.

3 2 ⫺ 2 c ⫹ 4c ⫹ 3 c ⫹ 6c ⫹ 9

18.

1 3 ⫺ 2 y ⫺ 10y ⫹ 25 y ⫺ 7y ⫹ 10

19.

36 4 3 ⫺ ⫽ w ⫺ 4 2w2 ⫺ 7w ⫺ 4 2w ⫹ 1

20.

5 25 2 ⫺ ⫽ 2 x⫺3 x⫹2 x ⫺x⫺6

2

Section 5.6

2

Applications of Rational Equations and Proportions

Concepts

1. Solving Proportions

1. 2. 3. 4.

A proportion is a rational equation that equates two ratios.

Solving Proportions Applications of Proportions Similar Triangles Applications of Rational Equations

DEFINITION Ratio and Proportion 1. The ratio of a to b is

a b

1b ⫽ 02 and can also be expressed as

a : b or a ⫼ b. 2. An equation that equates two ratios or rates is called a proportion. a c Therefore, if b ⫽ 0 and d ⫽ 0, then ⫽ is a proportion. b d The process for solving rational equations can be used to solve proportions. Example 1

Solving a Proportion

Solve the proportion.

5 95 ⫽ y 19

Solution: 5 95 ⫽ y 19

The LCD is 19y. Note that y ⫽ 0.

19y a

5 95 b ⫽ 19y a b y 19

Multiply both sides by the LCD.

19y a

5 95 b ⫽ 19y a b y 19

Clear fractions.

5y ⫽ 1805

Solve the resulting equation.

5y 1805 ⫽ 5 5 y ⫽ 361

The solution set is 53616.

The solution 361 checks in the original equation.

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459

Skill Practice Solve the proportion. 1.

8 12 ⫽ x 5

2. Applications of Proportions Example 2

Solving a Proportion

The recommended ratio of total cholesterol to HDL cholesterol is 7 to 2. If Rich’s blood test revealed that he has a total cholesterol level of 210 mg/dL (milligrams per deciliter), what should his HDL level be to fit within the recommendations?

Solution: One method of solving this problem is to set up a proportion. Write two equivalent ratios depicting the amount of total cholesterol to HDL cholesterol. Let x represent the unknown amount of HDL cholesterol. Given ratio

7 210 ⫽ x 2

Amount of total cholesterol Amount of HDL cholesterol

7 210 2x a b ⫽ 2x a b x 2 7x ⫽ 420

Multiply both sides by the LCD 2x. Clear fractions.

x ⫽ 60 Rich’s HDL cholesterol level should be 60 mg/dL to fit within the recommended level. Skill Practice 2. The ratio of cats to dogs at an animal rescue facility is 8 to 5. How many dogs are in the facility if there are 400 cats?

Example 3

Solving a Proportion

The ratio of male to female police officers in a certain town is 11 : 3. If the total number of officers is 112, how many are men and how many are women?

Solution: Let x represent the number of male police officers. Then 112 ⫺ x represents the number of female police officers.

Answers 1. e

15 f 2

2. There are 250 dogs.

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11 x ⫽ 3 112 ⫺ x

Male Female 31112 ⫺ x2a

Number of males Number of females

11 x b ⫽ 31112 ⫺ x2a b 3 112 ⫺ x

111112 ⫺ x2 ⫽ 3x

Multiply both sides by 31112 ⫺ x2 . The resulting equation is linear.

1232 ⫺ 11x ⫽ 3x 1232 ⫽ 14x 1232 14x ⫽ 14 14 x ⫽ 88 Then 112 ⫺ x ⫽ 112 ⫺ 88 ⫽ 24 There are 88 male police officers and 24 female officers. Skill Practice 3. Professor Wolfe has a ratio of passing students to failing students of 5 to 4. One semester he had a total of 207 students. How many students passed and how many failed?

3. Similar Triangles Proportions are used in geometry with similar triangles. Two triangles are similar if their corresponding angles are equal. In such a case, the lengths of the corresponding sides are proportional. The triangles in Figure 5-5 are similar. Therefore, the following ratios are equivalent. a b c ⫽ ⫽ z x y

c

a

z

x

b

y

Figure 5-5

Example 4

Using Similar Triangles in an Application

The shadow cast by a yardstick is 2 ft long. The shadow cast by a tree is 11 ft long. Find the height of the tree.

Solution: Let x represent the height of the tree.

Answer 3. 115 passed and 92 failed.

We will assume that the measurements were taken at the same time of day. Therefore, the angle of the sun is the same on both objects, and we can set up similar triangles (Figure 5-6).

x 1 yd ⫽ 3 ft 2 ft

11 ft

Figure 5-6

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Height of yardstick Height of tree

Applications of Rational Equations and Proportions

Length of yardstick’s shadow Length of tree’s shadow

3 ft 2 ⫽ x ft 11 ft 2 3 ⫽ x 11

Write an equation.

3 2 11x ⴢ a b ⫽ 11x ⴢ a b x 11 33 ⫽ 2x

Multiply by the LCD. Solve the equation.

16.5 ⫽ x

Interpret the results.

The tree is 16.5 ft high. Skill Practice 4. Solve for x, given that the triangles are similar.

9m 3m x

4m

4. Applications of Rational Equations Example 5

Solving an Application Involving Distance, Rate, and Time

An athlete’s average speed on her bike is 14 mph faster than her average speed running. She can bike 31.5 mi in the same time that it takes her to run 10.5 mi. Find her speed running and her speed biking.

Solution: Because the speed biking is given in terms of the speed running, let x represent the running speed. Let x represent the speed running. Then x ⫹ 14 represents the speed biking. Organize the given information in a chart. Distance

Rate

Time

Running

10.5

x

10.5 x

Biking

31.5

x ⫹ 14

31.5 x ⫹ 14

Because d ⫽ rt, then t ⫽

d r

Answer 4. x ⫽ 12 m

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The time required to run 10.5 mi is the same as the time required to bike 31.5 mi, so we can equate the two expressions for time: 10.5 31.5 ⫽ x x ⫹ 14 x1x ⫹ 142a

The LCD is x1x ⫹ 142. Multiply by x1x ⫹ 142 to clear fractions.

10.5 31.5 b ⫽ x1x ⫹ 142a b x x ⫹ 14

10.51x ⫹ 142 ⫽ 31.5x

The resulting equation is linear.

10.5x ⫹ 147 ⫽ 31.5x

Solve for x.

⫺21x ⫽ ⫺147 x⫽7 Then x ⫹ 14 ⫽ 7 ⫹ 14 ⫽ 21. The athlete runs 7 mph and bikes 21 mph. Skill Practice 5. Devon can cross-country ski 5 km/hr faster than his sister, Shanelle. Devon skis 45 km in the same amount of time that Shanelle skis 30 km. Find their speeds.

Example 6

Solving an Application Involving Distance, Rate, and Time

Valentina travels 70 km to Rome by train, and then takes a bus 30 km to the Coliseum. The bus travels 24 km/hr slower than the train. If the total time traveling on the bus and train is 2 hr, find the average speed of the train and the average speed of the bus.

Solution: Because the speed of the bus is given in terms of the speed of the train, let x represent the speed of the train. Let x represent the speed of the train. Let x ⫺ 24 represent the speed of the bus. Organize the given information in a chart. Distance

Rate

Time

Train

70

x

70 x

Bus

30

x ⫺ 24

30 x ⫺ 24

d Fill in the last column with t ⫽ . r

Answer 5. Shanelle skis 10 km/hr and Devon skis 15 km/hr.

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In this problem, we are given that the total time is 2 hr. So we add the two times to equal 2. 30 70 ⫹ ⫽2 x x ⫺ 24 x1x ⫺ 242 a

70 30 b ⫽ x1x ⫺ 242 ⫽ x1x ⫺ 242122 x x ⫺ 24

The LCD is x1x ⫺ 242. Multiply by the LCD to clear the fractions.

Avoiding Mistakes Remember to multiply all terms on both sides of the equation by the LCD.

701x ⫺ 242 ⫹ 30x ⫽ 2x1x ⫺ 242 70x ⫺ 1680 ⫹ 30x ⫽ 2x2 ⫺ 48x

x ⫺ 60 ⫽ 0 x ⫽ 60

The resulting equation is quadratic.

0 ⫽ 2x2 ⫺ 148x ⫹ 1680

Set the equation equal to 0.

0 ⫽ 21x2 ⫺ 74x ⫹ 8402

Factor.

0 ⫽ 21x ⫺ 6021x ⫺ 142

Solve for x.

or x ⫺ 14 ⫽ 0

Set each factor equal to 0.

or x ⫽ 14

If x ⫽ 14, then the rate of the bus would be 14 ⫺ 24 ⫽ ⫺10. Because this is not reasonable, the solution is x ⫽ 60. That is, the average rate of the train is 60 km/hr and the average rate of the bus is 60 ⫺ 24 or 36 km/hr. Skill Practice 6. Jason drives 50 mi to a train station and then continues his trip with a 210-mi train ride. The car travels 20 mph slower than the train. If the total travel time is 4 hr, find the average speed of the car and the average speed of the train.

Example 7

Solving an Application Involving “Work’’

JoAn can wallpaper a bathroom in 3 hr. Bonnie can wallpaper the same bathroom in 5 hr. How long would it take them if they worked together?

Solution: Let x represent the amount of time required for both people working together to complete the job.

Answer 6. The car’s average speed is 50 mph and the train’s average speed is 70 mph.

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One method to approach this problem is to add the rates of speed at which each person works. a

JoAn’s Bonnie’s speed working b⫹a b⫽a b speed speed together 1 job 3 hr

⫹

1 job 5 hr

1 job x hr

⫽

1 1 1 ⫹ ⫽ x 3 5 1 1 1 15x ⴢ a ⫹ b ⫽ 15x ⴢ a b x 3 5 5x ⫹ 3x ⫽ 15 8x ⫽ 15 x⫽

15 8

or

x⫽1

7 8

7 JoAn and Bonnie can wallpaper the bathroom in 1 hr. 8 Skill Practice 7. Antonio can install a new roof in 4 days. Carlos can install the same size roof in 6 days. How long will it take them to install a roof if they work together?

TIP: An alternative approach to solving a “work” problem is to determine the portion of the job that each person can complete in 1 hr. Let x represent the amount of time required to complete the job working together. Then • JoAn completes 31 of the job in 1 hr, and 31 x jobs in x hours. 1

1

• Bonnie completes 5 of the job in 1 hr and 5 x jobs in x hours. a

Portion of the job portion of the job 1 whole b⫹a b⫽a b completed by JoAn completed by Bonnie job 1 x 3

1 x 5

⫹

⫽

1

1 1 15 ⴢ a x ⫹ xb ⫽ 15 ⴢ 112 3 5 5x ⫹ 3x ⫽ 15 8x ⫽ 15 x⫽ The time working together is 1

Answer 7. It will take them

12 2 days or 2 days. 5 5

7 hr. 8

15 8

or x ⫽ 1

7 8

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Applications of Rational Equations and Proportions

465

Section 5.6 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Ratio

b. Proportion

c. Similar triangles

Review Exercises For Exercises 2–10, perform the indicated operation and simplify, or solve the equation. 2. 3 ⫺

6 ⫽x⫹8 x

6 ⫽x⫹7 x

4.

5 3 ⫺ 3x ⫺ 6 4x ⫺ 8

2 5 ⫺1 ⫺ ⫽ y⫺1 4 y⫹1

7.

5 10 ⫽7⫺ w⫺2 w⫹2

3. 2 ⫹

5.

4 1 ⫹ 5t ⫺ 1 10t ⫺ 2

6.

8.

ab a2 a ⫹ 1 ⫼ ⴢ 6 12 ab ⫹ b

9.

8p2 ⫺ 32 p ⫺ 4p ⫹ 4 2

ⴢ

3p2 ⫺ 3p ⫺ 6 2p ⫹ 20p ⫹ 32 2

10.

3t 2t ⫽ t⫹ t⫹6 t⫹6

Concept 1: Solving Proportions For Exercises 11–26, solve the proportions. (See Example 1.) 11.

y 20 ⫽ 6 15

12.

12 14 ⫽ x 18

13.

9 m ⫽ 75 50

14.

n 12 ⫽ 15 45

15.

p⫺1 p⫹3 ⫽ 4 3

16.

q⫺5 q⫹2 ⫽ 2 3

17.

x⫹1 4 ⫽ 5 15

18.

t⫺1 2 ⫽ 7 21

19.

5 ⫺ 2x 1 ⫽ x 4

20.

2y ⫹ 3 3 ⫽ y 2

21.

y⫺3 2 ⫽ y⫺1 4

22.

1 x⫺3 ⫽ x⫺5 3

23.

1 w ⫽ 49w 9

24.

1 z ⫽ 4z 25

25.

x⫹3 2 ⫽ 5x ⫹ 26 x⫹4

26.

⫺2 x⫺3 ⫽ x⫺2 8x ⫹ 11

Concept 2: Applications of Proportions 27. A preschool advertises that it has a 3-to-1 ratio of children to adults. If 18 children are enrolled, how many adults must be on the staff? (See Example 2.) 28. An after-school care facility tries to maintain a 4-to-1 ratio of children to adults. If the facility hired five adults, what is the maximum number of children that can enroll? 29. A 3.5-oz box of candy has a total of 21.0 g of fat. How many grams of fat would a 14-oz box of candy contain? 30. A 6-oz box of candy has 350 calories. How many calories would a 10-oz box contain? 31. A fisherman in the North Atlantic catches eight swordfish for a total of 1840 lb. How many swordfish were caught if a commercial fishing boat arrives in port with 230,000 lb of swordfish? 32. If a 64-oz bottle of laundry detergent costs $12.00, how much would an 80-oz bottle cost?

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33. Pam drives her Toyota Prius 243 mi in city driving on 4.5 gal of gas. At this rate how many gallons of gas are required to drive 621 mi? 34. On a map, the distance from Sacramento, California, to San Francisco, California, is 8 cm. The legend gives the actual distance as 96 mi. On the same map, Fatima measured 7 cm from Sacramento to Modesto, California. What is the actual distance? 35. Yellowstone National Park in Wyoming has the largest population of freeroaming bison. To approximate the number of bison, 200 are captured and tagged and then left free to roam. Later, a sample of 120 bison is observed and 6 have tags. Approximate the population of bison in the park. 36. Laws have been instituted in Florida to help save the manatee. To establish the number of manatees in Florida, 150 manatees were tagged. A new sample was taken later, and among the 40 manatees in the sample, 3 were tagged. Approximate the number of manatees in Florida. 37. The ratio of men to women enrolled in Math for Elementary Teachers at Indiana University-Purdue University Indianapolis is 1 to 5. If the total enrollment in these classes is approximately 186 students per semester, how many men are enrolled? (See Example 3.) 38. The ratio of Hank’s income spent on rent to his income spent on car payments is 3 to 1. If he spends a total of $1640 per month on the rent and car payment, how much does he spend on each item? 39. The ratio of single men in their 20s to single women in their 20s is 119 to 100 (Source: U.S. Census). In a random group of 1095 single college students in their 20s, how many are men and how many are women? 40. A chemist mixes water and alcohol in a 7 to 8 ratio. If she makes a 450-L solution, how much is water and how much is alcohol?

Concept 3: Similar Triangles For Exercises 41–44, the triangles shown here are similar. Find the lengths of the missing sides. (See Example 4.) 41.

42. b

a

6 ft

11.2 ft y 14 ft

10 ft

8 cm

43.

z

x

18 cm

x

13 cm

12 cm

44.

1.75 in. 5 in.

y

4.55 in.

12 m a

c

b

3.75 m

5m

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Concept 4: Applications of Rational Equations For Exercises 45–46, use the fact that distance ⫽ (rate)(time). 45. A truck travels 7 mph faster than a car. Let x represent the speed of the car. a. Write an expression for the speed of the truck. b. Write an expression for the time it takes the car to travel 48 mi. c. Write an expression for the time it takes the truck to travel 83 mi. 46. A car travels 4 mph slower than a motorcycle. Let x represent the speed of the motorcycle. a. Write an expression for the speed of the car. b. Write an expression for the time it takes the motorcycle to travel 50 mi. c. Write an expression for the time it takes the car to travel 145 mi. 47. A motorist travels 80 mi while driving in a bad rainstorm. In sunny weather, the motorist drives 20 mph faster and covers 120 mi in the same amount of time. Find the speed of the motorist in the rainstorm and the speed in sunny weather. (See Example 5.) 48. Brooke walks 2 km/hr slower than her older sister Adrianna. If Brooke can walk 12 km in the same amount of time that Adrianna can walk 18 km, find their speeds. 49. Two out-of-town firefighting crews have been called to a wildfire in the mountains. The Wescott Fire Station is 96 mi from the fire, and the Broadmoor Fire Station is 88 mi from the fire. The fire truck from the Wescott Fire Station travels 6.4 mph faster than the Broadmoor fire truck. If it takes the trucks the same amount of time to reach the fire, what is the average speed of each truck? 50. Kathy can run 3 mi to the beach in the same time Dennis can ride his bike 7 mi to work. Kathy runs 8 mph slower than Dennis rides his bike. Find their speeds. 51. A bicyclist rides 30 mi against a wind and returns 30 mi with the wind. His average speed for the return trip is 5 mph faster. How fast did the cyclist ride against the wind if the total time of the trip was 5 hr? (See Example 6.) 52. A boat travels 60 mi to an island and 60 mi back again. Changes in the wind and tide made the average speed on the return trip 3 mph slower than the speed on the way out. If the total time of the trip took 9 hr, find the speed going to the island and the speed of the return trip. 53. Celeste walked 140 ft on a moving walkway at the airport. Then she walked on the ground for 100 ft. She travels 2 ft/sec faster on the walkway than she does on the ground. If the time it takes her to travel the total distance of 240 ft is 40 sec, how fast does Celeste travel on and off the moving walkway? 54. Julio rides his bike for 6 mi and gets a flat tire. Then he has to walk with the bike for another mile. His speed walking is 6 mph less than his speed riding the bike. If the total time is 1 hr, find his speed riding the bike and his speed walking. 55. Beatrice does professional triathlons. She runs 2 mph faster than her friend Joe, a weekend athlete. If they each run 12 mi, Beatrice finishes 30 min (12 hr) ahead of Joe. Determine how fast each person runs.

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56. A bus leaves a terminal at 9:00. A car leaves 1 hr later and averages 10 mph faster than the bus. If the car overtakes the bus after 200 mi, find the average speed of the bus and the average speed of the car. 57. One painter can paint a room in 6 hr. Another painter can paint the same room in 8 hr. How long would it take them working together? (See Example 7.) 58. Karen can wax her SUV in 2 hr. Clarann can wax the same SUV in 3 hr. If they work together, how long will it take them to wax the SUV? 59. A new housing development offers fenced-in yards that all have the same dimensions. Joel can fence a yard in 12 hr, and Michael can fence a yard in 15 hr. How long will it take if they work together? 60. Ted can change an advertisement on a billboard in 4 hr. Marie can do the same job in 5 hr. How long would it take them if they worked together? 61. A swimming pool takes 30 hr to fill using an old pump. When a new pump was installed, it took only 12 hr to fill the pool with both pumps. However, the old pump had to be repaired. a. Determine how long it would take for the new pump to fill the pool alone. b. If the new pump begins filling the empty pool at 4 P.M. on Thursday, when should the technician return to stop the pump? 62. One carpenter can complete a kitchen in 8 days. With the help of another carpenter, they can do the job together in 4 days. How long would it take the second carpenter if he worked alone? 63. Gus works twice as fast as Sid. Together they can dig a garden in 4 hr. How long would it take each person working alone? 64. It takes a child 3 times longer to vacuum a house than an adult. If it takes 1 hr for one adult and one child working together to vacuum a house, how long would it take each person working alone?

Expanding Your Skills 65. Find the value of y so that the slope of the line between the points (3, 1) and (11, y) is 12. 66. Find the value of x so that the slope of the line between the points 1⫺2, ⫺52 and (x, 10) is 3. 67. Find the value of x so that the slope of the line between the points 14, ⫺22 and (x, 2) is 4. 68. Find the value of y so that the slope of the line between the points (3, 2) and 1⫺1, y2 is ⫺34.

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Section 5.7

Variation

Variation

Section 5.7

1. Definition of Direct and Inverse Variation

Concepts

In this section, we introduce the concept of variation. Direct and inverse variation models can show how one quantity varies in proportion to another.

1. Definition of Direct and Inverse Variation 2. Translations Involving Variation 3. Applications of Variation

DEFINITION Direct and Inverse Variation Let k be a nonzero constant real number. Then the following statements are equivalent: 1. y varies directly as x. y is directly proportional to x.

f

y ⫽ kx

2. y varies inversely as x. y is inversely proportional to x.

f

y⫽

k x

Note: The value of k is called the constant of variation. For a car traveling at 30 mph, the equation d ⫽ 30t indicates that the distance traveled is directly proportional to the time of travel. For positive values of k, when two variables are directly related, as one variable increases, the other variable will also increase. Likewise, if one variable decreases, the other will decrease. In the equation d ⫽ 30t, the longer the time of the trip, the greater the distance traveled. The shorter the time of the trip, the shorter the distance traveled. For positive values of k, when two variables are inversely related, as one variable increases, the other will decrease, and vice versa. Consider a car traveling between Toronto and Montreal, a distance of 500 km. The time required to make the trip is inversely proportional to the speed of travel: t ⫽ 500 Ⲑr. As the rate of speed r increases, the quotient 500 Ⲑr will decrease. Hence, the time will decrease. Similarly, as the rate of speed decreases, the trip will take longer.

2. Translations Involving Variation The first step in using a variation model is to translate an English phrase to an equivalent mathematical equation.

Example 1

Translating to a Variation Model

Translate each expression to an equivalent mathematical model. a. The circumference of a circle varies directly as the radius. b. At a constant temperature, the volume of a gas varies inversely as the pressure. c. The length of time of a meeting is directly proportional to the square of the number of people present.

469

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Solution: a. Let C represent circumference and r represent radius. The variables are directly related, so use the model C ⫽ kr. b. Let V represent volume and P represent pressure. Because the variables k are inversely related, use the model V ⫽ . P c. Let t represent time and let N be the number of people present at a meeting. Because t is directly related to N2, use the model t ⫽ kN2. Skill Practice Translate to a variation model. 1. The time t it takes to drive a particular distance is inversely proportional to the speed s. 2. The amount of your paycheck P varies directly with the number of hours h that you work. 3. q varies inversely as the square of t.

Sometimes a variable varies directly as the product of two or more other variables. In this case, we have joint variation.

DEFINITION Joint Variation Let k be a nonzero constant real number. Then the following statements are equivalent: y varies jointly as w and z. f y is jointly proportional to w and z.

Example 2

y ⫽ kwz

Translating to a Variation Model

Translate each expression into an equivalent mathematical model. a. y varies jointly as u and the square root of v. b. The gravitational force of attraction between two planets varies jointly as the product of their masses and inversely as the square of the distance between them.

Solution: a. y ⫽ ku 1v b. Let m1 and m2 represent the masses of the two planets. Let F represent the gravitational force of attraction and d represent the distance between km1m2 the planets. The variation model is F ⫽ . d2 Answers 1. t ⫽ 3. q ⫽ 5. x ⫽

k s k

Skill Practice Translate to a variation model. 2. P ⫽ kh

t2 k1y z

4. a ⫽ kbc

4. a varies jointly as b and c. 5. x varies directly as the square root of y and inversely as z.

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Variation

3. Applications of Variation Consider the variation models y ⫽ kx and y ⫽ k Ⲑx. In either case, if values for x and y are known, we can solve for k. Once k is known, we can use the variation equation to find y if x is known, or to find x if y is known. This concept is the basis for solving many problems involving variation.

PROCEDURE Steps to Find a Variation Model Step 1 Write a general variation model that relates the variables given in the problem. Let k represent the constant of variation. Step 2 Solve for k by substituting known values of the variables into the model from step 1. Step 3 Substitute the value of k into the original variation model from step 1.

Example 3

Solving an Application Involving Direct Variation

The variable z varies directly as w. When w is 16, z is 56. a. Write a variation model for this situation. Use k as the constant of variation. b. Solve for the constant of variation. c. Find the value of z when w is 84.

Solution: a. z ⫽ kw b. z ⫽ kw 56 ⫽ k1162 k1162 56 ⫽ 16 16 7 ⫽k 2

Substitute known values for z and w. Then solve for the unknown value of k. To isolate k, divide both sides by 16. Simplify

56 7 to . 16 2

7 c. With the value of k known, the variation model can now be written as z ⫽ w. 2 7 z ⫽ 1842 To find z when w ⫽ 84, substitute w ⫽ 84 into the equation. 2 z ⫽ 294 Skill Practice The variable q varies directly as the square of v. When v is 2, q is 40. 6. Write a variation model for this relationship. 7. Solve for the constant of variation. 8. Find q when v ⫽ 7. Answers 6. q ⫽ kv 2 8. q ⫽ 490

7. k ⫽ 10

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Example 4

Solving an Application Involving Direct Variation

The speed of a racing canoe in still water varies directly as the square root of the length of the canoe. a. If a 16-ft canoe can travel 6.2 mph in still water, find a variation model that relates the speed of a canoe to its length. b. Find the speed of a 25-ft canoe.

Solution: a. Let s represent the speed of the canoe and L represent the length. The general variation model is s ⫽ k2L. To solve for k, substitute the known values for s and L. s ⫽ k1L 6.2 ⫽ k116

Substitute s ⫽ 6.2 mph and L ⫽ 16 ft.

6.2 ⫽ k ⴢ 4 6.2 4k ⫽ 4 4

Solve for k.

k ⫽ 1.55 s ⫽ 1.55 1L

Substitute k ⫽ 1.55 into the model s ⫽ k2L.

b. s ⫽ 1.551L ⫽ 1.55125

Find the speed when L ⫽ 25 ft.

⫽ 7.75 mph

The speed of a 25-ft canoe is 7.75 mph.

Skill Practice The amount of water needed by a mountain hiker varies directly as the time spent hiking. The hiker needs 2.4 L for a 3-hr hike. 9. Write a model that relates the amount of water needed to the time of the hike. 10. How much water will be needed for a 5-hr hike?

Example 5

Solving an Application Involving Inverse Variation

The loudness of sound measured in decibels (dB) varies inversely as the square of the distance between the listener and the source of the sound. If the loudness of sound is 17.92 dB at a distance of 10 ft from a stereo speaker, what is the decibel level 20 ft from the speaker?

Solution: Let L represent the loudness of sound in decibels and d represent the distance in feet.The inverse relationship between decibel level and the square of the distance is modeled by L⫽

Answers 9. w ⫽ 0.8t

10. 4 L

k d2

17.92 ⫽

k 1102 2

17.92 ⫽

k 100

Substitute L ⫽ 17.92 dB and d ⫽ 10 ft.

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Section 5.7

117.922100 ⫽

k ⴢ 100 100

Variation

Solve for k (clear fractions).

k ⫽ 1792 L⫽

1792 d2

Substitute k ⫽ 1792 into the original model L ⫽

k . d2

With the value of k known, we can find L for any value of d. L⫽

1792 1202 2

Find the loudness when d ⫽ 20 ft.

⫽ 4.48 dB Notice that the loudness of sound is 17.92 dB at a distance 10 ft from the speaker. When the distance from the speaker is increased to 20 ft, the decibel level decreases to 4.48 dB.This is consistent with an inverse relationship. For k ⬎ 0, as one variable is increased, the other is decreased. It also seems reasonable that the farther one moves away from the source of a sound, the softer the sound becomes. Skill Practice 11. The yield on a bond varies inversely as the price. The yield on a particular bond is 4% when the price is $100. Find the yield when the price is $80.

Example 6

Solving an Application Involving Joint Variation

In the early morning hours of August 29, 2005, Hurricane Katrina plowed into the Gulf Coast of the United States, bringing unprecedented destruction to southern Louisiana, Mississippi, and Alabama. The winds of a hurricane are strong enough to send a piece of plywood through a tree. The kinetic energy of an object varies jointly as the weight of the object at sea level and as the square of its velocity. During a hurricane, a 0.5-lb stone traveling at 60 mph has 81 joules (J) of kinetic energy. Suppose the wind speed doubles to 120 mph. Find the kinetic energy.

Solution: Let E represent the kinetic energy, let w represent the weight, and let v represent the velocity of the stone. The variation model is E ⫽ kwv2 81 ⫽ k10.521602 2

Substitute E ⫽ 81 J, w ⫽ 0.5 lb, and v ⫽ 60 mph.

81 ⫽ k10.52136002

Simplify exponents.

81 ⫽ k118002 k118002 81 ⫽ 1800 1800

Divide by 1800.

0.045 ⫽ k

Solve for k. Answer 11. 5%

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With the value of k known, the model E ⫽ kwv2 can be written as E ⫽ 0.045wv2. We now find the kinetic energy of a 0.5-lb stone traveling at 120 mph. E ⫽ 0.04510.5211202 2 ⫽ 324 The kinetic energy of a 0.5-lb stone traveling at 120 mph is 324 J. Skill Practice 12. The amount of simple interest earned in an account varies jointly as the interest rate and time of the investment. An account earns $40 in 2 yr at 4% interest. How much interest would be earned in 3 yr at a rate of 5%?

In Example 6, when the velocity increased by 2 times, the kinetic energy increased by 4 times (note that 324 J ⫽ 4ⴢ 81 J2. This factor of 4 occurs because the kinetic energy is proportional to the square of the velocity. When the velocity increased by 2 times, the kinetic energy increased by 22 times.

Answer 12. $75

Section 5.7

Practice Exercises

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Study Skills Exercises 1. It is not too early to think about your final exam. Write the page number of the cumulative review for Chapters 1–5. Make this exercise set part of your homework this week. 2. Define the key terms. a. Direct variation

b. Inverse variation

c. Joint variation

Review Exercises For Exercises 3–8, solve the proportion. 3.

x 13 ⫽ 4 10

4.

8 6 ⫽ y 11

5.

3 w⫹2 ⫽ 8 6

6.

2 x⫺4 ⫽ 3 2

7.

p⫺5 2 ⫽ p 7

8.

k 1 ⫽ k⫹1 9

Concept 1: Definition of Direct and Inverse Variation 9. In the equation r ⫽ kt, does r vary directly or inversely with t?

10. In the equation w ⫽ kv , does w vary directly or inversely with v?

Concept 2: Translations Involving Variation For Exercises 11–22, write a variation model. Use k as the constant of variation. (See Examples 1–2.) 11. T varies directly as q.

12. W varies directly as z.

13. b varies inversely as c.

14. m varies inversely as t.

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Section 5.7

Variation

15. Q is directly proportional to x and inversely proportional to y.

16. d is directly proportional to p and inversely proportional to n.

17. c varies jointly as s and t.

18. w varies jointly as p and f.

19. L varies jointly as w and the square root of v.

20. q varies jointly as v and the square root of w.

21. x varies directly as the square of y and inversely as z.

22. a varies directly as n and inversely as the square of d.

475

Concept 3: Applications of Variation For Exercises 23–28, find the constant of variation k. (See Example 3.) 23. y varies directly as x, and when x is 4, y is 18.

24. m varies directly as x and when x is 8, m is 22.

25. p is inversely proportional to q and when q is 16, p is 32.

26. T is inversely proportional to x and when x is 40, T is 200.

27. y varies jointly as w and v. When w is 50 and v is 0.1, y is 8.75.

28. N varies jointly as t and p. When t is 1 and p is 7.5, N is 330.

Solve Exercises 29–40 by using the steps found on page 471. (See Examples 3–4.) 29. x varies directly as p. If x ⫽ 50 when p ⫽ 10, find x when p is 14.

30. y is directly proportional to z. If y ⫽ 12 when z ⫽ 36, find y when z is 21.

31. b is inversely proportional to c. If b is 4 when c is 3, find b when c ⫽ 2.

32. q varies inversely as w. If q is 8 when w is 50, find q when w is 125.

33. Z varies directly as the square of w, and Z ⫽ 14 when w ⫽ 4. Find Z when w ⫽ 8.

34. m varies directly as the square of x. If m = 200 when x ⫽ 20, find m when x is 32.

35. Q varies inversely as the square of p, and Q ⫽ 4 when p ⫽ 3. Find Q when p ⫽ 2.

36. z is inversely proportional to the square of t. If z ⫽ 15 when t ⫽ 4, find z when t ⫽ 10.

37. L varies jointly as a and the square root of b, and L ⫽ 72 when a ⫽ 8 and b ⫽ 9. Find L when a ⫽ 12 and b ⫽ 36.

38. Y varies jointly as the cube of x and the square root of w, and Y ⫽ 128 when x ⫽ 2 and w ⫽ 16. Find Y when x ⫽ 12 and w ⫽ 64.

39. B varies directly as m and inversely as n, and B ⫽ 20 when m ⫽ 10 and n ⫽ 3. Find B when m ⫽ 15 and n ⫽ 12.

40. R varies directly as s and inversely as t, and R ⫽ 14 when s ⫽ 2 and t ⫽ 9. Find R when s ⫽ 4 and t ⫽ 3.

For Exercises 41–58, use a variation model to solve for the unknown value. 41. The amount of medicine that a physician prescribes for a patient varies directly as the weight of the patient. A physician prescribes 3 grams(g) of a medicine for a 150-lb person. a. How many grams should be prescribed for a 180-lb person? b. How many grams should be prescribed for a 225-lb person? c. How many grams should be prescribed for a 120-lb person?

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42. The number of people that a turkey can serve varies directly as the weight of the turkey. A 15-lb turkey can serve 10 people. a. How many people will a 12-lb turkey serve? b. How many people will an 18-lb turkey serve? c. How many people will a 21-lb turkey serve? 43. The unit cost of producing CDs is inversely proportional to the number of CDs produced. If 5000 CDs are produced, the cost per CD is $0.48. a. What would be the unit cost if 6000 CDs were produced? b. What would be the unit cost if 8000 CDs were produced? c. What would be the unit cost if 2400 CDs were produced? 44. An author self-publishes a book and finds that the number of books she can sell per month varies inversely as the price of the book. The author can sell 1500 books per month when the price is set at $8 per book. a. How many books would she expect to sell if the price were $12? b. How many books would she expect to sell if the price were $15? c. How many books would she expect to sell if the price were $6? 45. The amount of pollution entering the atmosphere varies directly as the number of people living in an area. If 80,000 people cause 56,800 tons of pollutants, how many tons enter the atmosphere in a city with a population of 500,000? 46. The area of a picture projected on a wall varies directly as the square of the distance from the projector to the wall. If a 10-ft distance produces a 16-ft2 picture, what is the area of a picture produced when the projection unit is moved to a distance 20 ft from the wall? 47. The intensity of a light source varies inversely as the square of the distance from the source. If the intensity is 48 lumens (lm) at a distance of 5 ft, what is the intensity when the distance is 8 ft? (See Example 5.) 48. The frequency of a vibrating string is inversely proportional to its length. A 24-in. piano string vibrates at 252 cycles/sec. What is the frequency of an 18-in. piano string? 49. The current in a wire varies directly as the voltage and inversely as the resistance. If the current is 9 amperes (A) when the voltage is 90 volts (V) and the resistance is 10 ohms (⍀), find the current when the voltage is 185 V and the resistance is 10 ⍀. 50. The resistance of a wire varies directly as its length and inversely as the square of its diameter. A 40-ft wire with 0.1-in. diameter has a resistance of 4 ⍀. What is the resistance of a 50-ft wire with a diameter of 0.20 in.? 51. The amount of simple interest earned in an account varies jointly as the amount of principal invested and the amount of time the money is invested. If $2500 in principal earns $500 in interest after 4 yr, then how much interest will be earned on $7000 invested for 10 yr? (See Example 6.) 52. The amount of simple interest earned in an account varies jointly as the amount of principal invested and the amount of time the money is invested. If $6000 in principal earns $840 in interest after 2 yr, then how much interest will be earned on $4500 invested for 8 yr? 53. The stopping distance of a car is directly proportional to the square of the speed of the car. If a car traveling at 40 mph has a stopping distance of 109 ft, find the stopping distance of a car that is traveling at 25 mph. (Round your answer to one decimal place.)

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Section 5.7

Variation

477

54. The weight of a medicine ball varies directly as the cube of its radius. A ball with a radius of 3 in. weighs 4.32 lb. How much would a medicine ball weigh if its radius were 5 in.? 55. The surface area of a cube varies directly as the square of the length of an edge. The surface area is 24 ft2 when the length of an edge is 2 ft. Find the surface area of a cube with an edge that is 5 ft. 56. The period of a pendulum is the length of time required to complete one swing back and forth. The period varies directly as the square root of the length of the pendulum. If it takes 1.8 sec for a 0.81-m pendulum to complete one period, what is the period of a 1-m pendulum? 57. The power in an electric circuit varies jointly as the current and the square of the resistance. If the power is 144 watts (W) when the current is 4 A and the resistance is 6 ⍀, find the power when the current is 3 A and the resistance is 10 ⍀. 58. The strength of a wooden beam varies jointly as the width of the beam and the square of the thickness of the beam and inversely as the length of the beam. A beam that is 48 in. long, 6 in. wide, and 2 in. thick can support a load of 417 lb. Find the maximum load that can be safely supported by a board that is 12 in. wide, 72 in. long, and 4 in. thick.

L ⫽ length

T ⫽ thickness W ⫽ width

Expanding Your Skills 59. The area A of a square varies directly as the square of the length l of its sides. a. Write a general variation model with k as the constant of variation. b. If the length of the sides is doubled, what effect will that have on the area? c. If the length of the sides is tripled, what effect will that have on the area? 60. In a physics laboratory, a spring is fixed to the ceiling. With no weight attached to the end of the spring, the spring is said to be in its equilibrium position. As weights are applied to the end of the spring, the force stretches the spring a distance d from its equilibrium position. A student in the laboratory collects the following data: Force F (lb)

2

4

6

Distance d (cm)

2.5

5.0

7.5

8 10.0

10 12.5

a. Based on the data, do you suspect a direct relationship between force and distance or an inverse relationship? b. Find a variation model that describes the relationship between force and distance.

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Group Activity Computing the Future Value of an Investment Materials: Calculator Estimated time: 15 minutes Group Size: 3 Suppose you are able to save $100 per month. If you invest the money in an account that pays 6% annual interest, how much money would you have at the end of 10 yr? This question can be answered by using the following formula. S ⫽ Rc

11 ⫹ i2 n ⫺ 1 i

d

S is the future value of the investment. R is the amount saved per period. i is the interest rate per period. n is the total number of periods.

where

In this example, R ⫽ $100 i⫽

(amount invested per month)

0.06 ⫽ 0.005 12

(annual interest rate divided by 12 months)

n ⫽ 11221102 ⫽ 120 Therefore, S ⫽ $100 c

(12 months per year times 10 years)

11 ⫹ 0.0052 120 ⫺ 1 0.005

d

S ⫽ $16,387.93 1. Compute the future value of an account if you save $150 per month for 30 yr at an annual interest rate of 6%. R⫽ i⫽

S⫽

n⫽ 2. Compute the future value of an account given the monthly saving, interest rate, and time period given in the table. With your group members, discuss how the value of the account varies with the interest rate and time the money is invested. Monthly Savings

Annual interest rate

Years

$100

6%

20

$150

6%

20

$200

6%

20

Monthly Savings

Annual interest rate

Years

$200

2.4%

20

$200

3.6%

20

$200

4.8%

20

Future Value

Future Value

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Summary

Monthly Savings

Annual interest rate

Years

$200

6%

25

$200

6%

30

$200

6%

35

479

Future Value

3. Suppose you need to accumulate $50,000 in 15 yr to pay for your child’s college tuition. You decide to invest the money in a mutual fund and the average rate of return is 9% per year. How much should you save each month to reach your goal? Round up to the nearest dollar. 4. Suppose you invest $5000 in an account on July 1 each year for 20 yr. If the annual growth on your investment is 8%, how much will the account be worth in 20 yr? (Hint: In this example, the savings is yearly, not monthly.) R⫽ i⫽

S⫽

n⫽

Chapter 5

Summary

Section 5.1

Rational Expressions and Rational Functions

Key Concepts

p A rational expression is in the form where p and q q are polynomials and q ⫽ 0. A rational function is a function of the form p1x2 f1x2 ⫽ , where p1x2 and q1x2 are polynomial funcq1x2 tions and q1x2 ⫽ 0. The domain of a rational function excludes the values for which the denominator is zero. To simplify a rational expression to lowest terms, factor the numerator and denominator completely. Then simplify factors whose ratio is 1 or ⫺1. A rational expression written in lowest terms will still have the same restrictions on the domain as the original expression.

Examples Example 1 Find the domain of the function. f 1x2 ⫽

x⫺3 1x ⫹ 4212x ⫺ 12

1 Domain: e x 0 x is a real number and x ⫽ ⫺4, x ⫽ f 2 or

1⫺⬁, ⫺42 ´ 1⫺4, 12 2 ´ 1 12, ⬁2

Example 2 Simplify to lowest terms. t 2 ⫺ 6t ⫺ 16 5t ⫹ 10

1t ⫺ 821t ⫹ 22 51t ⫹ 22

⫽

t⫺8 5

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Section 5.2

Multiplication and Division of Rational Expressions

Key Concepts

Examples

To multiply rational expressions, factor the numerators and denominators completely. Then simplify factors whose ratio is 1 or ⫺1.

Example 1 b2 ⫺ a2 a2 ⫺ 3ab ⫹ 2b2 ⴢ 2 2a ⫹ 2b a ⫺ 2ab ⫹ b 2

⫺1

⫽

1

1a ⫺ b21a ⫺ b2 ⴢ 21a ⫹ b2

⫽⫺ To divide rational expressions, multiply by the reciprocal of the divisor.

1

1b ⫺ a21b ⫹ a2 ⴢ 1a ⫺ 2b21a ⫺ b2 a ⫺ 2b 2

or

2b ⫺ a 2

Factor. Simplify.

Example 2 3x ⫹ 1 9x ⫹ 3 ⫼ 4x ⫹ 8 x2 ⫺ 4 1

313x ⫹ 12

1

41x ⫹ 22 ⫽ ⴢ 1x ⫺ 221x ⫹ 22 3x ⫹ 1 ⫽

12 x⫺2

Factor. Multiply by the reciprocal. Simplify.

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Summary

Section 5.3

481

Addition and Subtraction of Rational Expressions

Key Concepts

Examples

To add or subtract rational expressions, the expressions must have the same denominator. The least common denominator (LCD) is the product of unique factors from the denominators, in which each factor is raised to its highest power.

Example 1

Steps to Add or Subtract Rational Expressions

Example 2

1. Factor the denominator of each rational expression. 2. Identify the LCD. 3. Rewrite each rational expression as an equivalent expression with the LCD as its denominator. [This is accomplished by multiplying the numerator and denominator of each rational expression by the missing factor(s) from the LCD.] 4. Add or subtract the numerators, and write the result over the common denominator. 5. Simplify, if possible.

For

1 31x ⫺ 12 3 1x ⫹ 22

and

⫺5 61x ⫺ 121x ⫹ 72 2

LCD ⫽ 61x ⫺ 12 3 1x ⫹ 221x ⫹ 72 2

c 1 ⫺ 2c ⫺ 8 c2 ⫺ c ⫺ 12 ⫽

c 1 ⫺ 1c ⫺ 421c ⫹ 32 21c ⫺ 42

Factor the denominators.

The LCD is 21c ⫺ 421c ⫹ 32 1c ⫹ 32 2 c 1 ⴢ ⫺ ⴢ 2 1c ⫺ 421c ⫹ 32 21c ⫺ 42 1c ⫹ 32

⫽

2c ⫺ 1c ⫹ 32 21c ⫺ 421c ⫹ 32

⫽

2c ⫺ c ⫺ 3 21c ⫺ 421c ⫹ 32

⫽

c⫺3 21c ⫺ 421c ⫹ 32

Write equivalent fractions with LCD. Subtract.

Simplify.

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Section 5.4

Complex Fractions

Key Concepts

Examples

Complex fractions can be simplified by using Method I or Method II. Method I uses the order of operations to simplify the numerator and denominator separately before multiplying by the reciprocal of the denominator of the complex fraction.

Example 1 Simplify by using Method I. 5 x 10 x2



5 x2  ⴢ x 10  To use Method II, multiply the numerator and denominator of the complex fraction by the LCD of all the individual fractions. Then simplify the result.

x 2

Example 2 Simplify by using Method II. 1

4 w2

The LCD is w2.

6 1 1  2 w w w2 a1 

4 b w2  1 6 w2 a1   2 b w w 

w2  4 w2  w  6

1w  22 1w  22  1w  32 1w  22 1



w2 w3

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Summary

Section 5.5

Solving Rational Equations

Key Concepts

Examples

Steps to Solve a Rational Equation

Example 1

1. Factor the denominators of all rational expressions. Identify any restrictions on the variable. 2. Identify the LCD of all expressions in the equation. 3. Multiply both sides of the equation by the LCD. 4. Solve the resulting equation. 5. Check each potential solution.

1 1 ⫺2w ⫺ ⫽ w 2w ⫺ 1 2w ⫺ 1

The LCD is w 12w ⫺ 12. w12w ⫺ 12

1 1 ⫺ w12w ⫺ 12 ⴢ w 2w ⫺ 1 ⫽ w12w ⫺ 12 ⴢ

12w ⫺ 121 ⫺ w112 ⫽ w1⫺2w2 2w ⫺ 1 ⫺ w ⫽ ⫺2w2 2w2 ⫹ w ⫺ 1 ⫽ 0

⫺2w 2w ⫺ 1

1quadratic equation2

12w ⫺ 12 1w ⫹ 12 ⫽ 0 w⫽

1 2

or

w ⫽ ⫺1

The solution set is {⫺1}. (The value

Section 5.6

1 does not check.) 2

Applications of Rational Equations and Proportions

Key Concepts

Examples

An equation that equates two ratios is called a proportion.

Example 1 An old water pump can fill a tank in 6 hr, and a new pump can fill the tank in 4 hr. How long will it take to fill the tank if both pumps are working?

a c ⫽ provided b ⫽ 0, d ⫽ 0 b d

Let x represent the time working together. Example 1 A sample of 85 g of a particular ice cream contains 17 g of fat. How much fat does 324 g of the same ice cream contain?

fat 1g2 ice cream 1g2 185 ⴢ 3242 ⴢ 1

17 x ⫽ 85 324

1 x 17 ⫽ 185 ⴢ 3242 ⴢ 85 324

fat 1g2 ice cream 1g2

Multiply by the LCD.

5508 ⫽ 85x x ⫽ 64.8 g There would be 64.8 g of fat in 324 g of ice cream.

a

Speed of speed of speed working b⫹a b⫽a b old pump new pump together 1 1 1 ⫹ ⫽ x 6 4 1 1 1 12x ⴢ a ⫹ b ⫽ 12x ⴢ a b x 6 4 2x ⫹ 3x ⫽ 12 5x ⫽ 12 x⫽

12 5

It will take

12 hr. 5

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Section 5.7

Variation

Key Concepts

Examples

Direct Variation

Example 1 t varies directly as the square root of x.

y varies directly as x. f y ⫽ kx y is directly proportional to x.

Inverse Variation k y varies inversely as x. f y⫽ y is inversely proportional to x. x

Joint Variation y varies jointly as w and z. f y ⫽ kwz y is jointly proportional to w and z.

Steps to Find a Variation Model 1. Write a general variation model that relates the variables given in the problem. Let k represent the constant of variation. 2. Solve for k by substituting known values of the variables into the model from step 1. 3. Substitute the value of k into the original variation model from step 1.

Chapter 5

Example 2 W is inversely proportional to the cube of x. W⫽

k x3

Example 3 y is jointly proportional to x and to the square of z. y ⫽ kxz2

Example 4 C varies directly as the square root of d and inversely as t. If C ⫽ 12 when d is 9 and t is 6, find C if d is 16 and t is 12. Step 1: C ⫽

k1d t

Step 2: 12 ⫽

k19 kⴢ3 1 12 ⫽ 1 k ⫽ 24 6 6

Step 3: C ⫽

24116 241d 1C⫽ 1C⫽8 t 12

Review Exercises 2. Let h 1x2 ⫽

Section 5.1 1. Let k 1y2 ⫽

t ⫽ k1x

y y2 ⫺ 1

.

a. Find the function values (if they exist): k 122, 1 k102, k 112, k 1⫺12, k a b. 2 b. Identify the domain for k. Write the answer in interval notation.

x . x2 ⫹ 1

a. Find the function values (if they exist): h 112, 1 h 102, h 1⫺12, h 1⫺32, h a b. 2 b. Identify the domain for h. Write the answer in interval notation.

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Review Exercises

For Exercises 3–10, simplify the rational expressions. 2

3 3

3.

5.

7.

9.

28a b 14a2b3

4.

x2  4x  3 x3

6.

x3  27 9  x2

8.

2t2  3t  5 7  6t  t2

10.

3

20.

ab  2a  b  2 ab  3b  4a  12 ⴢ ab  3b  2a  6 ab  b  4a  4

21.

x2  8x  20 x2  6x  40  2 2 x  6x  16 x  3x  40

22.

2b  b2 b2  2b  4 ⴢ b3  8 b2

23.

3w2 4 2w  ⴢ w 21 7

25x yz 125xyz

k2  3k  10 k2  5k  6 a4  81 3a y3  4y y2  5y  6

24. For Exercises 11–14, write the domain of each function in interval notation. Use that information to match the function with its graph. 11. f 1x2 

1 x3

12. m 1x2 

1 x2

13. k 1x2 

6 x2  3x

14. p 1x2 

2 x2  4

a.

b.

y

8 6 4

8 6

4 2

4 2 x

2 4 6 8

2 4

8 6 4 2 2 4

6

c.

d.

y

2

4 2

1

8 6 4 2 2 4

x

2 4 6 8

8 6 4 2

2 4 6

8

2

4 6

8

x

2

For Exercises 15–26, multiply or divide as indicated.

19.

3a  9 a3 ⴢ 6a  18 a2 x  4y x  xy 2



k5 b 3k  5

27.

1 1 1  2 3 x x x

28.

1 5  x2 x2

29.

y 3  2y  1 1  2y

30.

a2 3  2a  6 a3

31.

4k 3  2 k2  2k  1 k 1

x

Section 5.2

17.

x2  x  20 x2  x  6 2x  10 ⴢ  2 2 10  5x x  4x  4 12  x  x

32. 4x  3 

2x  1 x4

1

6 8

15.

y3  y

For Exercises 27–38, add or subtract as indicated.

y

8 6

7y2  14y

Section 5.3

6 8

8

y3  8



26. 19k2  252 ⴢ a

y

8 6

25.

5y2  20

16.

20y  5x x y 2

2

4y 2y  8  5 15

18. 1x2  5x  242 a

7k  28 k2  2k  8 ⴢ 2k  4 k2  2k  8

x8 b x3

33.

2a2  2a 2  2 a3 a  2a  15

34.

6 7  2 x2  4x  3 x  5x  6

35.

2 8 3x  5

36.

7 1  2 4k  k  3 4k  7k  3 2

485

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7 13 54. 1 ⫹ m⫺1 ⫽ m⫺1 6 6

37.

6a 2 3a ⫺ ⫹ 3a ⫺ 1 a ⫺2 3a ⫺ 7a ⫹ 2

53. 5y⫺2 ⫹ 1 ⫽ 6y⫺1

38.

y 2y ⫺ 5 ⫺ ⫺4 y⫺3 y⫹2

55. Solve for x.

c⫽

56. Solve for P.

A P ⫽P⫹ rt rt

2

Section 5.4 For Exercises 39–46, simplify the complex fraction. 2x 2 3x ⫺ 3 39. 4x 6x ⫺ 6

k⫹2 3 40. 5 k⫺2

2 1 ⫹ x xy 41. 4 x2

4 ⫺1 y 42. 1 4 ⫺ 2 y y

1 ⫹1 a⫺1 43. 1 ⫺1 a⫹1

1 3 ⫺ x⫺1 1⫺x 44. 2 2 ⫺ x x⫺1

45.

1 ⫹ xy⫺1 x2y⫺2 ⫺ 1

46.

5a⫺1 ⫹ 1ab2 ⫺1 3a⫺2

For Exercises 47–48, find the slope of the line containing the two points. 2 7 13 5 47. a , ⫺ b and a , ⫺ b 3 4 6 3

Section 5.5 For Exercises 49–54, solve the equation. 49.

x⫹3 8 ⫺ 2 ⫽0 2 x ⫺x x ⫺1

50.

y 3 18 ⫹ ⫽ 2 y⫹3 y⫺3 y ⫺9 72 x⫺8

Section 5.6 For Exercises 57–60, solve the proportions. 57.

5 x ⫽ 4 6

58.

x 6 ⫽ 36 7

59.

51x ⫹ 12 x⫹2 ⫽ 3 4

60.

x ⫺3 ⫽ x⫹2 5

61. In one game Peyton Manning completed 34 passes for 357 yd. At this rate how many yards would be gained for 22 passes? 62. Erik bought $108 Canadian with $100 American. At this rate, how many Canadian dollars can he buy with $235 American? 63. Tony rode 175 mi on a 2-day bicycle ride to benefit the Multiple Sclerosis Foundation. The second day he rode 5 mph slower than the first day because of a strong headwind. If Tony rode 100 mi on the first day and 75 mi on the second day in a total time of 10 hr, how fast did he ride each day? 64. Stephen drove his car 45 mi. He ran out of gas and had to walk 3 mi to a gas station. His speed driving is 15 times his speed walking. If the total time for the drive and walk was 112 hr, what was his speed driving?

8 1 13 9 48. a , ⫺ b and a , b 15 3 10 5

51. x ⫺ 9 ⫽

ax ⫹ b x

52.

3x ⫹ 1 x⫺1 ⫽ ⫹2 x⫹5 x⫹1

65. Doug and Jean work as phone solicitors. They work in batches of 400 calls. Doug can finish a batch in an average of 8 hr, and Jean can finish a batch in 10 hr. How long would it take them to finish a batch if they worked together? 66. Two pipes can fill a tank in 6 hr. The larger pipe works twice as fast as the smaller pipe. How long would it take each pipe to fill the tank if they worked separately?

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Test

Section 5.7 67. The force applied to a spring varies directly with the distance that the spring is stretched. a. Write a variation model using k as the constant of variation.

487

70. The distance d that one can see to the horizon varies directly as the square root of the height above sea level. If a person 25 m above sea level can see 30 km, how far can a person see if she is 64 m above sea level?

b. When 6 lb of force is applied, the spring stretches 2 ft. Find k. c. How much force is required to stretch the spring 4.2 ft? 68. Suppose y varies inversely with x and y ⫽ 32 when x ⫽ 2. Find y when x ⫽ 4. 69. Suppose y varies jointly with x and the square root of z, and y ⫽ 3 when x ⫽ 3 and z ⫽ 4. Find y when x ⫽ 8 and z ⫽ 9.

Chapter 5

Test

1. For the function h1x2 ⫽

2x ⫺ 14 x2 ⫺ 49

a. Evaluate h 102, h 152, h 172, and h 1⫺72, if possible.

8.

x2 8x ⫺ 16 ⫺ x⫺4 x⫺4

9.

4x 2 ⫹x⫹ x⫹1 x⫹1

b. Write the domain of h in interval notation. 5x ⫺ 3 2. Write the domain of k1x2 ⫽ in interval 7 notation. 3. For the function f(x) ⫽

2x ⫹ 6 x2 ⫺ x ⫺ 12

3 k 10. 4 4⫹ k 3⫹

11.

2u⫺1 ⫹ 2v⫺1 4u⫺3 ⫹ 4v⫺3

12.

ax ⫹ bx ⫹ 2a ⫹ 2b x ⫺ 3 x⫹2 ⴢ ⫼ ax ⫺ 3a ⫹ bx ⫺ 3b x ⫺ 5 ax ⫺ 5a

13.

3 1 1 ⫺ 2 ⫺ 2 x ⫹ 8x ⫹ 15 x ⫹ 7x ⫹ 12 x ⫹ 9x ⫹ 20

a. Write the domain in set-builder notation. b. Simplify to lowest terms.

2

For Exercises 4–5, simplify to lowest terms. 4.

12m3n7 18mn8

5.

9x2 ⫺ 9 3x2 ⫹ 2x ⫺ 5

6. Find the slope of the line containing the points 1 3 5 8 a , ⫺ b and a , ⫺ b. 12 4 6 3 For Exercises 7–13, simplify. 7.

2x ⫺ 5 ⴢ 12x2 ⫺ x ⫺ 152 25 ⫺ 4x2

For Exercises 14–16, solve the equation. 14.

7 z⫺5 6 ⫺ 2 ⫽ z z⫹1 z ⫺1

16.

4x 16 ⫽3⫹ x⫺4 x⫺4

17. Solve for T.

15.

3 4 ⫹ ⫽1 y⫹3 y ⫺9

1 ⫹ Tv ⫽p T

2

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Chapter 5 Rational Expressions and Rational Equations

F

18. Solve for m1.

Gm1m2 r2

19. If the reciprocal of a number is added to 3 times the number, the result is 132. Find the number. 20. The following triangles are similar. Find the lengths of the missing sides.

12 m

10 m

18 m

b

a

21. On a certain map, the distance between New York and Los Angeles is 8.2 in., and the actual distance is 2820 mi. What is the distance between two cities that are 5.7 in. apart on the same map? Round to the nearest mile.

Number

ⴚ22

25. The period of a pendulum varies directly as the square root of the length of the pendulum. If the period of the pendulum is 2.2 sec when the length is 4 ft, find the period when the length is 9 ft.

Cumulative Review Exercises

1. Check the sets to which each number belongs. Set

23. Barbara can type a chapter in a book in 4 hr. Jack can type a chapter in a book in 10 hr. How long would it take them to type a chapter if they worked together? 24. Write a variation model using k as the constant of variation. The variable x varies directly as y and inversely as the square of t.

21 m

Chapters 1–5

22. Lance can ride 48 mi on his bike against the wind. With the wind at his back, he rides 4 mph faster and can ride 60 mi in the same amount of time. Find his speed riding against the wind and his speed riding with the wind.

P

6

ⴚ12

Real numbers

6. Solve the system of equations. 3x  3y  z  13 2x  y  2z  4 x  2y  3z  15 7. Find an equation of the line through 13, 52 that is perpendicular to the line y  3x. Write the answer in slope-intercept form.

Irrational numbers Rational numbers Integers Whole numbers

8. Graph the line 2x  y  3.

Natural numbers

y

2. Solve the system of equations. 3x  4y  4 2x  8y  8 3. Perform the indicated operations. 12x  321x  42  1x  52 2 4. The area of a trapezoid is given by A  12 h1b1  b2 2. a. Solve for b1. b. Find b1 when h  4 cm, b2  6 cm, and A  32 cm2. 5. The dimensions of a rectangular swimming pool are such that the length is 10 m less than twice the width. If the perimeter is 160 m, find the length and width.

5 4 3 2 1 5 4 3 2 1 1 2

1

2

3 4

5

x

3 4 5

9. The speed of a car varies inversely as the time to travel a fixed distance. A car traveling the speed limit of 60 mph travels between two points in 10 sec. How fast is a car moving if it takes only 8 sec to cover the same distance? 10. Find the x-intercepts. f 1x2  12x3  17x2  6x

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Cumulative Review Exercises

11. Factor 64y3  8z6 completely over the real numbers. 12. Factor.

10x2  x  2

13. Perform the indicated operations. 2x2  11x  21 2x2  98  4x2  10x  6 x2  x  xa  a 14. Perform the indicated operations. x 1 1  2  2 x  5x  50 x  7x  10 x  8x  20 2

15. Simplify the complex fraction. 1

49 c2

7 1 c 16. Solve.

y3  5y2  y  5

489

17. Solve the equation. 4y y 9   2 y2 y1 y y2 18. Max knows that the distance between Roanoke, Virginia, and Washington, D.C., is 195 mi. On a certain map, the distance between the two cities is 6.5 in. On the same map, the distance between Roanoke and Cincinnati, Ohio, is 9.25 in. Find the distance in miles between Roanoke and Cincinnati. Round to the nearest mile. 19. Determine whether the equation represents a horizontal or vertical line. Then identify the slope of the line. a. x  5

b. 2y  8

20. Simple interest varies jointly as the interest rate and as the time the money is invested. If an investment yields $1120 interest at 8% for 2 yr, how much interest will the investment yield at 10% for 5 yr?

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6

Radicals and Complex Numbers

CHAPTER OUTLINE 6.1 Definition of an nth root 492 6.2 Rational Exponents 503 6.3 Simplifying Radical Expressions 510 6.4 Addition and Subtraction of Radicals

517

6.5 Multiplication of Radicals 522 Problem Recognition Exercises: Simplifying Radical Expressions

530

6.6 Division of Radicals and Rationalization 530 6.7 Solving Radical Equations 540 6.8 Complex Numbers 550 Group Activity: Margin of Error of Survey Results

559

Chapter 6 In this chapter, we study radical expressions. This includes operations on square roots, cube roots, fourth roots, and so on. We also revisit the Pythagorean theorem and its applications as well as introduce the set of complex numbers. Are You Prepared? Take a minute to review your knowledge of square roots and operations on polynomials. These skills will help you be successful in this chapter. For Exercises 1–10, perform the indicated operations. Then write the letter corresponding to each answer in the space at the bottom of the page. 1. 14x ⫺ 321x ⫹ 52 2. 4x2 13x2 ⫹ 5x ⫹ 12 5. 24 6. 42 4 2 9. 1⫺3x ⫹ 9x ⫺ 15x2 ⫺ 1⫺3x4 ⫺ 7x2 ⫹ 9x ⫹ 92 N 16x 2 ⫺ 9

R

16

Y

16x ⫺ 24x ⫺ 9

A

G

4x ⫹ 17x ⫺ 15

B 12x 4

2

2

not a real number

3. 14x ⫺ 3214x ⫹ 32 7. 2⫺4 10. 12x216x3 2

4. 14x ⫺ 32 2 8. 2102 ⫺ 62

I 16x 2 ⫺ 24x ⫹ 9 E 12x 4 ⫹ 20x 3 ⫹ 4x 2

M U

8 2

In this chapter, we also introduce a new type of number called an ___ ___ ___ ___ ___ ___ ___ ___ ___ 4 8 7 1 4 3 7 6 9

___ ___ ___ ___ ___ ___. 3 5 8 10 2 6

491

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Chapter 6 Radicals and Complex Numbers

Section 6.1

Definition of an nth Root

Concepts

1. Definition of a Square Root

1. Definition of a Square Root 2. Definition of an n th Root 3. Roots of Variable Expressions 4. Pythagorean Theorem 5. Radical Functions

The reverse operation to squaring a number is to find its square roots. For example, finding a square root of 36 is equivalent to asking, “what number when squared equals 36?” One obvious answer to this question is 6 because 162 2  36, but 6 will also work, because 162 2  36.

DEFINITION Square Root b is a square root of a if b2  a.

Example 1

Identifying Square Roots

Identify the square roots of the real numbers. a. 25

b. 49

c. 0

d. 9

Solution:

a. 5 is a square root of 25 because 152 2  25.

5 is a square root of 25 because 152 2  25.

b. 7 is a square root of 49 because 172 2  49.

7 is a square root of 49 because 172 2  49.

c. 0 is a square root of 0 because 102 2  0.

d. There are no real numbers that when squared will equal a negative number; therefore, there are no real-valued square roots of 9. Skill Practice Identify the square roots of the real numbers. 1. 64

2. 16

3. 1

4. 100

TIP: • All positive real numbers have two real-valued square roots (one positive and one negative). • Zero has only one square root, which is zero itself. • A negative number has no real-valued square roots.

Recall from Section R.3 that the positive square root of a real number can be denoted with a radical sign 1 .

DEFINITION Positive and Negative Square Roots Let a represent a positive real number. Then Answers 1. 2. 3. 4.

8 and 8 4 and 4 1 and 1 No real-valued square roots

1. 1a is the positive square root of a. The positive square root is also called the principal square root. 2. 1a is the negative square root of a. 3. 10  0

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Definition of an n th Root

493

Simplifying Square Roots

Example 2

Simplify the square roots. a. 136

b.

4 A9

c. 10.04

Solution: a. 136 denotes the positive square root of 36. 136 ⫽ 6

b.

4 4 denotes the positive square root of . A9 9 4 2 ⫽ A9 3

c. 10.04 denotes the positive square root of 0.04. Note: 10.2210.22 ⫽ 0.04

10.04 ⫽ 0.2

Skill Practice Simplify the square roots. 5. 181

6.

36 B 49

7. 10.09

The numbers 36, 49 , and 0.04 are perfect squares because their square roots are rational numbers. Radicals that cannot be simplified to rational numbers are irrational numbers. Recall that an irrational number cannot be written as a terminating or repeating decimal. For example, the symbol 113 is used to represent the exact value of the square root of 13. The symbol 142 is used to represent the exact value of the square root of 42. These values can be approximated by a rational number by using a calculator. 113 ⬇ 3.605551275

Calculator Connections Use a calculator to approximate the values of 113 and 142.

142 ⬇ 6.480740698

TIP: Before using a calculator to evaluate a square root, try estimating the value first. 113 must be a number between 3 and 4 because 19 6 113 6 116. 142 must be a number between 6 and 7 because 136 6 142 6 149.

A negative number cannot have a real number as a square root because no real number when squared is negative. For example, 1⫺25 is not a real number because there is no real number b for which 1b2 2 ⫽ ⫺25.

Answers 5. 9

6.

6 7

7. 0.3

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Chapter 6 Radicals and Complex Numbers

Simplifying Square Roots

Example 3

Simplify the square roots, if possible. a. 1⫺144

b. ⫺1144

c. 1⫺0.01

1 d. ⫺ A9

Solution: a. 1⫺144 is not a real number. b. ⫺1144 ⫽ ⫺1 ⴢ 1144 ⫽ ⫺1 ⴢ

12

TIP: For the expression ⫺1144, the factor of ⫺1 is outside the radical.

⫽ ⫺12 c. 1⫺0.01 is not a real number. 1 B9

d. ⫺

⫽ ⫺1 ⴢ

1 B9

⫽ ⫺1 ⴢ

1 3

⫽⫺

1 3

Skill Practice Simplify the square roots, if possible. 8. 1⫺81

9. ⫺164

10. ⫺10.25

11.

B

⫺

1 4

2. Definition of an nth Root Finding a square root of a number is the reverse process of squaring a number. This concept can be extended to finding a third root (called a cube root), a fourth root, and in general an nth root.

DEFINITION n th Root b is an nth root of a if bn ⫽ a. Example: 2 is a fourth root of 16 because 24 ⫽ 16.

The radical sign 1 is used to denote the principal square root of a number. The n symbol 1 is used to denote the principal nth root of a number. In the expression n 1a, n is called the index of the radical, and a is called the radicand. For a square 2 root, the index is 2, but it is usually not written 1 1 a is denoted simply as 1a2 . A rad3 ical with an index of 3 is called a cube root, denoted by 1 a. Answers 8. Not a real number 9. ⫺8 10. ⫺0.5 11. Not a real number

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Section 6.1

495

Definition of an n th Root

n

PROPERTY 1 a n

1. If n 7 1 is an even integer and a 7 0, then 1 a is the principal (positive) nth root of a. n 2. If n 7 1 is an odd integer, then 1 a is the nth root of a. n 3. If n 7 1 is an integer, then 1 0  0.

For the purpose of simplifying radicals, it is helpful to know the following powers. Perfect Squares

Perfect Cubes

1 1

1 1

1 1

15  1

22  4

23  8

24  16

25  32

32  9

33  27

34  81

35  243

4  16

4  64

4  256

45  1024

52  25

53  125

54  625

55  3125

2

Perfect Fifth Powers

4

3

2

Example 4

Perfect Fourth Powers

3

4

Identifying the n th Root of a Real Number

Simplify the expressions, if possible. a. 14

3 b. 164

5 c. 1 32

6 e. 1 1,000,000

f. 1100

4 g. 1 16

Solution:

4 d. 181

because 122 2  4

a. 14  2

Calculator Connections

because 142 3  64

3 b. 1 64  4

because 122 5  32

5 c. 1 32  2

Topic: Approximating n th Roots

because 132 4  81

4 d. 1 81  3

e. 1 1,000,000  10

because 1102  1,000,000

f. 1100 is not a real number.

No real number when squared equals 100.

6

6

4

g. 116 is not a real number.

No real number when raised to the fourth power equals 16.

A calculator can be used to approximate nth roots by using the function. On most calculators, the index is entered first.

Skill Practice Simplify if possible. 4 12. 1 16

3 13. 1 1000

5 14. 1 1

5 16. 1 100,000

17. 136

3 18. 1 27

5 15. 1 32

Examples 4(f) and 4(g) illustrate that an nth root of a negative quantity is not a real number if the index is even. This is because no real number raised to an even power is negative.

3. Roots of Variable Expressions Finding an nth root of a variable expression is similar to finding an nth root of a numerical expression. For roots with an even index, however, particular care must be taken to obtain a nonnegative result.

Answers 12. 2 13. 10 15. 2 16. 10 17. Not a real number

14. 1 18. 3

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Chapter 6 Radicals and Complex Numbers

DEFINITION

n

1a n

n 1. If n is a positive odd integer, then 1a n  a.

n 2. If n is a positive even integer, then 1a n  0 a 0.

The absolute value bars are necessary for roots with an even index because the variable a may represent a positive quantity or a negative quantity. By using absolute n value bars, 2an  0 a 0 is nonnegative and represents the principal nth root of an. n

Simplifying Expressions of the Form 1a n

Example 5

Simplify the expressions. 4 a. 2132 4

5 b. 2 132 5

c. 21x  22 2

3 d. 2 1a  b2 3

e. 2y4

Solution:

4 a. 2132 4  0 3 0  3

Because this is an even-indexed root, absolute value bars are necessary to make the answer positive.

5 b. 2 132 5  3

This is an odd-indexed root, so absolute value bars are not necessary.

c. 21x  22 2  0 x  2 0

Because this is an even-indexed root, absolute value bars are necessary. The sign of the quantity x  2 is unknown; however, 0x  2 0  0 regardless of the value of x.

3 d. 2 1a  b2 3  a  b

This is an odd-indexed root, so absolute value bars are not necessary.

e. 2y4  21y2 2 2  0 y2 0

Because this is an even-indexed root, use absolute value bars.

 y2

However, because y2 is nonnegative, the absolute value bars are not necessary.

Skill Practice Simplify the expressions. 19. 2142 2

3 142 3 20. 2

21. 2 1y  92 2

3 1t  12 3 22. 2

4 8 v 23. 2

If n is an even integer, then 2an  0 a 0 ; however, if the variable a is assumed n to be nonnegative, then the absolute value bars may be dropped. That is, 2an  a provided a  0. In many examples and exercises, we will make the assumption that the variables within a radical expression are positive real numbers. In such a case, n the absolute value bars are not needed to evaluate 2an. In Chapter 4 you became familiar with the patterns associated with perfect squares and perfect cubes. In general, any expression raised to an even power is a perfect square. An expression raised to a power that is a multiple of three is a perfect cube. n

Perfect Squares 1x 2  x 1 2

2

1x2 2 2  x4 Answers 19. 4 22. t  1

20. 4 23. v 2

21. 0 y  9 0

1x3 2 2  x6 1x4 2 2  x8

Perfect Cubes 1x1 2 3  x3 1x2 2 3  x6 1x3 2 3  x9

1x4 2 3  x12

These patterns may be extended to higher powers.

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497

Definition of an n th Root

Simplifying n th Roots

Example 6

Simplify the expressions. Assume that all variables are positive real numbers. 3 b. 2 27a3

a. 2y8

Solution:

a. 2y8 ⫽ 21y4 2 2 ⫽ y4 c.

a5 B b5 5

c.

d. ⫺ 4 B

81x4y8 16

TIP: In Example 6, the variables are assumed to represent positive numbers. Therefore, absolute value bars are not necessary in the simplified form.

3 3 b. 2 27a3 ⫽ 2 13a2 3 ⫽ 3a

81x4y8 3xy2 4 3xy2 ⫽⫺4a b ⫽⫺ B 16 B 2 2

5 a5 a 5 a ⫽ a b ⫽ 5 Bb B b b 5

d. ⫺

4

Skill Practice Simplify the expressions. Assume all variables represent positive real numbers. 3 25. 2 64y12

24. 2t 6

26.

x4 B y4 4

32a5 B b10

27. ⫺

5

4. Pythagorean Theorem In Section 4.8, we used the Pythagorean theorem in several applications. For the triangle shown in Figure 6-1, the Pythagorean theorem may be stated as a2 ⫹ b2 ⫽ c2. In this formula, a and b are the legs of the right triangle, and c is the hypotenuse. Example 7

a (leg)

b (leg)

Applying the Pythagorean Theorem

Use the Pythagorean theorem and the definition of the principal square root to find the length of the unknown side.

c (hypotenuse)

Figure 6-1 5 ft

13 ft

Solution: ?

Label the sides of the triangle. a2 ⫹ b2 ⫽ c2

152 ⫹ b ⫽ 1132 2

2

2

25 ⫹ b ⫽ 169 2

Apply the Pythagorean theorem.

a ⫽ 5 ft

c ⫽ 13 ft

Simplify.

b2 ⫽ 169 ⫺ 25

Isolate b2.

b ⫽ 144

By definition, b must be one of the square roots of 144. Because b represents the length of a side of a triangle, choose the positive square root of 144.

2

b ⫽ 12

b⫽?

The third side is 12 ft long. Skill Practice Use the Pythagorean theorem and the definition of the principal square root to find the length of the unknown side of the right triangle. 28. 12 cm

15 cm

Answers 24. t 3

?

27. ⫺

25. 4y 4 2a b2

28. 9 cm

26.

x y

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Chapter 6 Radicals and Complex Numbers

Example 8

Applying the Pythagorean Theorem

Two boats leave a dock at 12:00 noon. One travels due north at 6 mph, and the other travels due east at 8 mph (Figure 6-2). How far apart are the two boats after 2 hr?

Solution: a  12 miles Dock

c?

The boat traveling north travels a distance of 16 mph212 hr2  12 mi. The boat traveling east travels a distance of 18 mph212 hr2  16 mi. The course of the boats forms a right triangle where the hypotenuse represents the distance between them. a2  b2  c2

b  16 miles

1122 2  1162 2  c2

Figure 6-2

Apply the Pythagorean theorem.

144  256  c2

Simplify.

400  c

2

1400  c

By definition, c must be one of the square roots of 400. Choose the positive square root of 400 to represent the distance between the two boats.

20  c The boats are 20 mi apart. Skill Practice

29. Two cars leave from the same place at the same time. One travels west at 40 mph, and the other travels north at 30 mph. How far apart are they after 2 hr?

5. Radical Functions

n If n is an integer greater than 1, then a function written in the form f 1x2  1 x is called a radical function. Note that if n is an even integer, then the function will be a real number only if the radicand is nonnegative.Therefore, the domain is restricted to nonnegative real numbers, or equivalently, 30, 2 . If n is an odd integer, then the domain is all real numbers.

Example 9

Determining the Domain of Radical Functions

For each function, write the domain in interval notation. 4 a. g1t2  1 t2

3 b. h1a2  1 a3

c. k1x2  13  5x  2

Solution: a.

4 g1t2  1 t2

t20 t2

The index is even. The radicand must be nonnegative. Set the radicand greater than or equal to zero. Solve for t.

The domain is 32, 2. b. h1a2  13 a  3

The index is odd; therefore, the domain is all real numbers.

The domain is 1, 2. Answer 29. 100 mi

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Section 6.1

c.

k1x2 ⫽ 13 ⫺ 5x ⫹ 2

The index is even; therefore, the radicand must be nonnegative.

3 ⫺ 5x ⱖ 0

Set the radicand greater than or equal to zero.

⫺5x ⱖ ⫺3

Solve for x.

⫺5x ⫺3 ⱕ ⫺5 ⫺5

Reverse the inequality sign.

xⱕ

Definition of an n th Root

3 5

The domain is 1⫺⬁, 35 4 . Skill Practice For each function, write the domain in interval notation. 30. f 1x2 ⫽ 1x ⫹ 5

3 31. g1t 2 ⫽ 1 t⫺9

32. h1a2 ⫽ 11 ⫺ 2a

Calculator Connections We can graph the functions defined in Example 9. The graphs support the answers we obtained for the domain of each function. 4 g1t2 ⫽ 1 t⫺2

3 h1a2 ⫽ 1 a⫺3

k1x2 ⫽ 13 ⫺ 5x ⫹ 2

Answers 30. [⫺5, ⬁) 32. (⫺⬁, 12 ]

Section 6.1

31. (⫺⬁, ⬁)

Practice Exercises

Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Square root

b. Radical sign

c. Principal square root

d. Perfect square

e. nth root

f. Index

g. Radicand

h. Cube root

i. Pythagorean theorem

j. Radical function

Concept 1: Definition of a Square Root 3 2. Simplify the expression 1 8. Explain how you can check your answer.

499

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Chapter 6 Radicals and Complex Numbers

3. a. Find the square roots of 64. (See Example 1.)

4. a. Find the square roots of 121.

b. Find 164.

b. Find 1121.

c. Explain the difference between the answers in part (a) and part (b).

c. Explain the difference between the answers in part (a) and part (b).

5. a. What is the principal square root of 81?

6. a. What is the principal square root of 100?

b. What is the negative square root of 81?

b. What is the negative square root of 100?

7. Using the definition of a square root, explain why 136 is not a real number. For Exercises 8–19, evaluate the roots without using a calculator. Identify those that are not real numbers. (See Examples 2–3.)

8. 125

9. 149

10. 125

11. 149

12. 125

13. 149

14.

16. 10.64

17. 10.81

18. 10.0144

100 B 121

15.

64 B9

19. 20.16

Concept 2: Definition of an n th Root 4 20. Using the definition of an nth root, explain why 1 16 is not a real number.

For Exercises 21–38, evaluate the roots without using a calculator. Identify those that are not real numbers. (See Example 4.)

21. a. 164 3

d. 264 22. a. 116 4

d. 216

3 b. 2 64

c. 164

e. 164

3 f. 2 64

4 b. 2 16

c. 116

e. 116

4 f. 2 16

1 B8 3

26.

1 B 32

30.

8 3  B 27

3 23. 2 27

3 24. 1125

25.

5 27. 2 32

4 28. 1 1

29.

4 31. 11

6 32. 21

6 33. 21,000,000

4 34. 110,000

3 35. 10.008

4 36. 10.0016

4 0.0625 37. 1

3 38. 1 0.064

3

B



125 64

5

Concept 3: Roots of Variable Expressions For Exercises 39–58, simplify the radical expressions. (See Example 5.) 39. 2a2

4 4 40. 2 a

3 3 41. 2 a

5 5 42. 2 a

6 6 43. 2 a

7 7 44. 2 a

45. 21x  12 2

3 46. 2 1y  32 3

47. 2x2y4

3 48. 2 1u  v2 3

49. 

x3 , B y3 3

y0

50.

a4 , B b8 4

b0

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51.

2 4

2x

4

,

x0

55. 2122 10 10

501

Definition of an n th Root

52. 2152 2

3 53. 2 1922 3

6 54. 2 1502 6

5 56. 2 122 5

7 57. 2 19232 7

6 58. 2 14172 6

For Exercises 59–74, simplify the expressions. Assume all variables are positive real numbers. (See Example 6.) 59. 2x2y4

60. 216p2 p6 B 81

25 B q2

63. 

67.

64. 

h2k4 B 16

68.

5 71. 2 32 y10

4x2 B y8

61.

a6 B b2

62.

65. 29x2y4z2

66. 24a4b2c6

t3 B 27

69. 

3 72. 2 64x6y3

w2 B z4

3

70.

6 73. 2 64p12q18

4 16 B w4

4 74. 2 16r12s8

Concept 4: Pythagorean Theorem For Exercises 75–78, find the length of the third side of each triangle by using the Pythagorean theorem. (See Example 7.) 75.

76.

12 cm

77.

15 cm

12 ft

5 ft ?

78. 10 m

?

8 in.

26 m ?

? 6 in.

For Exercises 79–82, use the Pythagorean theorem. 79. Roberto and Sherona began running from the same place at the same time. They ran along two different paths that formed right angles with each other. Roberto ran 4 mi and stopped, while Sherona ran 3 mi and stopped. How far apart were they when they stopped? (See Example 8.)

80. Leine and Laura began hiking from their campground. Laura headed south while Leine headed east. Laura walked 12 mi and Leine walked 5 mi. How far apart were they when they stopped walking?

81. Two mountain bikers take off from the same place at the same time. One travels north at 4 mph, and the other travels east at 3 mph. How far apart are they after 5 hr?

82. Professor Ortiz leaves campus on her bike, heading west at 12 ft/sec. Professor Wilson leaves campus at the same time and walks south at 5 ft/sec. How far apart are they after 40 sec?

Concept 5: Radical Functions For Exercises 83–86, evaluate the function for the given values of x. Then write the domain of the function in interval notation. (See Example 9.) 83. h1x2  1x  2

84. k1x2  1x  1

a. h102

a. k132

b. h112

b. k122

c. h122

c. k112

d. h132

d. k102

e. h162

e. k132

3 85. g1x2  1x  2

a. g 162 b. g 112 c. g 122

d. g 132

3 86. f 1x2  1x  1

a. f 192 b. f 122 c. f 102 d. f 172

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For each function defined in Exercises 87–94, write the domain in interval notation. (See Example 9.) 87. f1x2  15  2x

88. g1x2  13  4x

3 89. k1x2  1 4x  7

3 90. R1x2  1 x1

91. M1x2  1x  5  3

92. N1x2  1x  3  1

3 93. F1x2  1 x72

3 94. G1x2  1 x  10  4

For Exercises 95–98, find the domain of the function. Then match the function with its graph. 95. p1x2  11  x a.

96. q1x2  1x  5 b.

y

7 6 5 4 3

1

2 1

6 5 4 3 2 1 1 2 3

1 2

3 4

x

c.

y

7 6 5 4 3

5 4 3 2 1 1 2 3

3 98. R1x2  1 x1

97. T1x2  1x  10

d.

y

10

8 6

8 6 4

4 2 108 6 4 1 2

3 4

5

2 4

x

y

10

2 2

4

6 8 10

x

4 2 2 4

2 4

6 8 10 12 14 16

x

6 8 10

6 8 10

Mixed Exercises For Exercises 99–102, write the English phrase as an algebraic expression. 99. The sum of q and the square of p

100. The product of 11 and the cube root of x

101. The quotient of 6 and the cube root of x

102. The difference of y and the principal square root of x

103. If a square has an area of 64 in.2, then what are the lengths of the sides?

104. If a square has an area of 121 m2, then what are the lengths of the sides?

s?

s?

A  64 in.2 s  ?

A  121 m2 s  ?

Graphing Calculator Exercises For Exercises 105–112, use a calculator to evaluate the expressions to four decimal places. 105. 169

106. 15798

3 107. 2  1 5

4 109. 72 25

3 110. 32 9

111.

3  119 11

4 108. 3  21 10

112.

113. Graph h1x2  1x  2. Use the graph to confirm the domain found in Exercise 83. 114. Graph k1x2  1x  1. Use the graph to confirm the domain found in Exercise 84. 3 115. Graph g1x2  1 x  2. Use the graph to confirm the domain found in Exercise 85. 3 116. Graph f 1x2  1 x  1. Use the graph to confirm the domain found in Exercise 86.

5  2115 12

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Rational Exponents

Rational Exponents

Section 6.2

1. Definition of a1/n and am/n

Concepts

In Section 4.1, the properties for simplifying expressions with integer exponents were presented. In this section, the properties are expanded to include expressions with rational exponents. We begin by defining expressions of the form a1Ⲑn.

DEFINITION a1/ n n Let a be a real number, and let n be an integer such that n 7 1. If 1 a is a real number, then

1. Definition of a1/n and am /n 2. Converting Between Rational Exponents and Radical Notation 3. Properties of Rational Exponents 4. Applications Involving Rational Exponents

a1 Ⲑ n ⫽ 2 a n

Evaluating Expressions of the Form a1/n

Example 1

Convert the expression to radical form and simplify, if possible. a. 1⫺82 1Ⲑ 3

b. 811Ⲑ 4

c. ⫺1001 Ⲑ 2

d. 1⫺1002 1Ⲑ 2

e. 16⫺1Ⲑ 2

Solution:

3 a. 1⫺82 1Ⲑ3 ⫽ 1 ⫺8 ⫽ ⫺2 4 b. 811Ⲑ4 ⫽ 1 81 ⫽ 3

c. ⫺1001Ⲑ2 ⫽ ⫺1 ⴢ 1001Ⲑ2

The exponent applies only to the base of 100.

⫽ ⫺12100 d. 1⫺1002

⫽ ⫺10 1Ⲑ2

is not a real number because 1⫺100 is not a real number.

e. 16⫺1Ⲑ2 ⫽

1 161Ⲑ2

⫽

1 116

⫽

1 4

Write the expression with a positive exponent. 1 Recall that b⫺n ⫽ n . b

Skill Practice Convert the expression to radical form and simplify, if possible. 1. 1⫺642 1Ⲑ 3

2. 161Ⲑ4

3. ⫺361Ⲑ2

4. 1⫺362 1Ⲑ2

5. 64⫺1Ⲑ3

n If 1 a is a real number, then we can define an expression of the form amⲐn in such a way that the multiplication property of exponents still holds true. For example:

163Ⲑ4

4 1161Ⲑ4 2 3 ⫽ 1 1 162 3 ⫽ 122 3 ⫽ 8 4 4 1163 2 1Ⲑ4 ⫽ 2163 ⫽ 24096 ⫽ 8

Answers 1. ⫺4

2. 2

4. Not a real number

3. ⫺6 5.

1 4

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Chapter 6 Radicals and Complex Numbers

DEFINITION a m/n Let a be a real number, and let m and n be positive integers such that m and n n share no common factors and n 7 1. If 1 a is a real number, then amⲐn ⫽ 1a1Ⲑn 2 m ⫽ 1 1a2 m n

and

amⲐn ⫽ 1am 2 1Ⲑn ⫽ 1am n

The rational exponent in the expression amⲐn is essentially performing two operations. The numerator of the exponent raises the base to the mth power. The denominator takes the nth root. Example 2

Evaluating Expressions of the Form a m/n

Convert each expression to radical form and simplify. a. 82Ⲑ3

b. 1005Ⲑ2

c. a

1 3Ⲑ2 b 25

d. 4⫺3Ⲑ2

e. 1⫺812 3/4

Solution:

Calculator Connections A calculator can be used to confirm the results of Example 2(a)–2(c).

3 a. 82Ⲑ3 ⫽ 1 1 82 2

⫽ 122 2

Take the cube root of 8 and square the result. Simplify.

⫽4 b. 1005Ⲑ2 ⫽ 1 11002 5 ⫽ 1102 5

Take the square root of 100 and raise the result to the fifth power. Simplify.

⫽ 100,000 c. a

1 3Ⲑ2 1 3 b ⫽a b 25 A 25 1 3 ⫽a b 5 ⫽

Take the square root of

1 and cube the result. 25

Simplify.

1 125

1 3Ⲑ2 1 d. 4⫺3Ⲑ2 ⫽ a b ⫽ 3 2 4 4Ⲑ

Write the expression with positive exponents.

⫽

1 1 142 3

Take the square root of 4 and cube the result.

⫽

1 23

Simplify.

⫽

1 8

4 e. 1⫺812 3/4 is not a real number because 1⫺81 is not a real number.

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Rational Exponents

Skill Practice Convert each expression to radical form and simplify. 6. 93Ⲑ2

8. a

7. 85Ⲑ3

1 4Ⲑ 3 b 27

9. 32⫺4Ⲑ5

10. 1⫺42 3/2

2. Converting Between Rational Exponents and Radical Notation Example 3

Using Radical Notation and Rational Exponents

Convert each expression to radical notation. Assume all variables represent positive real numbers. b. 15x2 2 1Ⲑ3

a. a3Ⲑ5

c. 3y1Ⲑ4

d. z⫺3Ⲑ4

Solution: 5 3 5 a. a3Ⲑ5 ⫽ 2 a or Q 2 a R3 3 b. 15x2 2 1Ⲑ3 ⫽ 2 5x2 4 c. 3y1Ⲑ4 ⫽ 31 y

d. z⫺3Ⲑ4 ⫽

Note that the coefficient 3 is not raised to the 14 power.

1 1 ⫽ 4 3/4 z 2z3

Skill Practice Convert each expression to radical notation. Assume all variables represent positive real numbers. 11. t 4Ⲑ5

12. 12y 3 2 1Ⲑ4

Example 4

13. 10p1Ⲑ2

14. q⫺2Ⲑ3

Using Radical Notation and Rational Exponents

Convert each expression to an equivalent expression by using rational exponents. Assume that all variables represent positive real numbers. 4 3 a. 2 b

Solution: 4 3 a. 2 b ⫽ b3Ⲑ4

b. 17a

c. 71a

b. 17a ⫽ 17a2 1Ⲑ2

c. 71a ⫽ 7a1Ⲑ 2

Skill Practice Convert to an equivalent expression using rational exponents. Assume all variables represent positive real numbers. 3 2 15. 2 x

16. 15y

17. 51y

3. Properties of Rational Exponents In Section 4.1, several properties and definitions were introduced to simplify expressions with integer exponents. These properties also apply to rational exponents.

Answers 6. 27 9.

1 16

5 11. 2t 4

14.

1 3

2q 2 17. 5y 1Ⲑ 2

7. 32

8.

1 81

10. Not a real number 4 12. 2 2y 3

13. 10 2p

15. x 2 Ⲑ 3

16. 15y2 1Ⲑ 2

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Chapter 6 Radicals and Complex Numbers

SUMMARY

Definitions and Properties of Exponents

Let a and b be nonzero real numbers. Let m and n be rational numbers such that am, an, and bm are real numbers. Description

Property

1. Multiplying like bases 2. Dividing like bases

a a ⫽a m n

4. Power of a product 5. Power of a quotient Description

⫽ x5Ⲑ3

x3Ⲑ5 ⫽ x2Ⲑ5 x1Ⲑ5

121Ⲑ3 2 1Ⲑ2 ⫽ 21Ⲑ6

1ab2 m ⫽ ambm

19y2 1/2 ⫽ 91/2y1/2 ⫽ 3y1/2

Definition

Example

a m am a b ⫽ m b b

a

4 1Ⲑ2 41Ⲑ2 2 b ⫽ 1Ⲑ2 ⫽ 25 5 25

1 1 a⫺m ⫽ a b ⫽ m a a

1 1Ⲑ3 1 182 ⫺1Ⲑ3 ⫽ a b ⫽ 8 2

a0 ⫽ 1

50 ⫽ 1

m

1. Negative exponents

x x

am ⫽ am⫺n an

1am 2 n ⫽ amn

3. The power rule

Example 1Ⲑ3 4Ⲑ3

m⫹n

2. Zero exponent

Simplifying Expressions with Rational Exponents

Example 5

Use the properties of exponents to simplify the expressions. Assume all variables represent positive real numbers. a. y2Ⲑ5y3Ⲑ5

b.

6a⫺1/2 a3/2

c. a

s1/2t1/3 4 b w3/4

Solution: a. y2Ⲑ5y3Ⲑ5 ⫽ y12Ⲑ52 ⫹ 13Ⲑ52

Multiply like bases by adding exponents.

5Ⲑ5

⫽y

Simplify.

⫽y ⫺1/2

b.

6a a3/2

Divide like bases by subracting exponents.

⫽ 6a1⫺1/22 ⫺ 13/22

4 Simplify: ⫺ ⫽ ⫺2 2

⫽ 6a⫺2 ⫽ c. a

6 a2

Simplify the negative exponent.

s1/2 t 1/3 4 s11/22 ⴢ4t 11/32 ⴢ4 b ⫽ w3/4 w13/42 ⴢ4 ⫽

s2t 4/3 w3

Apply the power rule. Multiply exponents. Simplify.

Skill Practice Use the properties of exponents to simplify the expressions. Assume all variables represent positive real numbers. Answers 18. x 5/4

19.

4 k

20.

a 2b 3 c15/4

18. x 1Ⲑ 2 ⴢ x 3Ⲑ4

19.

4k⫺2/3 k1/3

20. a

a1/3b1/2 6 b c 5/8

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507

4. Applications Involving Rational Exponents Example 6

Applying Rational Exponents

Suppose P dollars in principal is invested in an account that earns interest annually. If after t years the investment grows to A dollars, then the annual rate of return r on the investment is given by A 1Ⲑt r⫽a b ⫺1 P Find the annual rate of return on $5000 which grew to $6894.21 after 6 yr.

Solution: A 1Ⲑt r⫽a b ⫺1 P ⫽a

6894.21 1Ⲑ6 b ⫺1 5000

where A ⫽ $6894.21, P ⫽ $5000, and t ⫽ 6

⬇ 0.055 or 5.5% The annual rate of return is 5.5%. Skill Practice 21. The formula for the radius of a sphere is r⫽a

3V 1Ⲑ 3 b 4p

where V is the volume. Find the radius of a sphere whose volume is 113.04 in.3 (Use 3.14 for p.)

Answer 21. 3 in.

Section 6.2 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

For the exercises in this set, assume that all variables represent positive real numbers unless otherwise stated.

Study Skills Exercises 1. Before you do your homework for this section, go back to Section 4.1 and review the properties of exponents. Do several problems from the Section 4.1 exercises. This will help you with the concepts in Section 6.2. 2. Define the key terms. a. a1 /n

b. a m/n

Review Exercises 3 3. Given: 1 27

4. Given: 118

a. Identify the index.

a. Identify the index.

b. Identify the radicand.

b. Identify the radicand.

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For Exercises 5–8, evaluate the radicals. 5. 125

3 6. 1 8

4 7. 1 81

4 8. 1 1 162 3

Concept 1: Definition of a1/n and a m/n For Exercises 9–20, convert the expressions to radical form and simplify. (See Example 1.) 9. 1441Ⲑ 2

10. 161Ⲑ4

11. ⫺1441Ⲑ 2

12. ⫺161Ⲑ4

13. 1⫺1442 1Ⲑ 2

14. 1⫺162 1Ⲑ4

15. 1⫺642 1Ⲑ 3

16. 1⫺322 1Ⲑ5

17. 1252 ⫺1Ⲑ 2

18. 1272 ⫺1Ⲑ 3

19. ⫺49⫺1/2

20. ⫺64⫺1/2

21. Explain how to interpret the expression amⲐn as a radical. 3 3 4 22. Explain why 1 1 82 4 is easier to evaluate than 2 8.

For Exercises 23–26, simplify the expression, if possible. (See Example 2.) 23. a. 163Ⲑ4

24. a. 813Ⲑ4

b. ⫺163Ⲑ4

25. a. 253Ⲑ 2

b. ⫺813Ⲑ4

26. a. 43Ⲑ 2

b. ⫺253Ⲑ 2

b. ⫺43Ⲑ 2

c. 1⫺162 3Ⲑ4

c. 1⫺812 3Ⲑ4

c. 1⫺252 3Ⲑ 2

c. 1⫺42 3Ⲑ 2

e. ⫺16⫺3Ⲑ4

e. ⫺81⫺3Ⲑ4

e. ⫺25⫺3Ⲑ 2

e. ⫺4⫺3Ⲑ 2

d. 16⫺3Ⲑ4

d. 81⫺3Ⲑ4

f. 1⫺162 ⫺3Ⲑ4

d. 25⫺3Ⲑ 2

f. 1⫺812 ⫺3Ⲑ4

d. 4⫺3Ⲑ 2

f. 1⫺252 ⫺3Ⲑ 2

f. 1⫺42 ⫺3Ⲑ 2

For Exercises 27–50, simplify the expression. (See Example 2.) 27. 64⫺3Ⲑ 2

28. 81⫺3Ⲑ 2

29. 2433Ⲑ5

31. ⫺27⫺4Ⲑ3

32. ⫺16⫺5Ⲑ4

33. a

35. 1⫺42 ⫺3Ⲑ 2

36. 1⫺492 ⫺3Ⲑ 2

37. 1⫺82 1Ⲑ3

39. ⫺81Ⲑ3

40. ⫺91Ⲑ2

41.

43.

1 1000⫺1Ⲑ3

47. a

1 ⫺3Ⲑ4 1 ⫺1Ⲑ 2 b ⫺a b 16 49

44.

1 81⫺3Ⲑ4

48. a

1 1Ⲑ4 1 1Ⲑ 2 b ⫺a b 16 49

100 ⫺3Ⲑ 2 b 9

1 36⫺1Ⲑ2

30. 15Ⲑ 3 34. a

49 ⫺1Ⲑ 2 b 100

38. 1⫺92 1Ⲑ2 42.

1 16⫺1Ⲑ2

1 2Ⲑ3 1 1Ⲑ2 45. a b ⫹ a b 8 4

1 ⫺2Ⲑ3 1 ⫺1Ⲑ2 ⫹a b 46. a b 8 4

1 1Ⲑ2 1 ⫺1Ⲑ3 49. a b ⫹ a b 4 64

50. a

1 1Ⲑ2 1 ⫺5Ⲑ6 b ⫹a b 36 64

Concept 2: Converting Between Rational Exponents and Radical Notation For Exercises 51–58, convert each expression to radical notation. (See Example 3.) 51. q2Ⲑ 3

52. t 3Ⲑ5

53. 6y3Ⲑ4

54. 8b4 Ⲑ 9

55. 1x 2y2 1Ⲑ3

56. 1c 2d2 1Ⲑ6

57. 1qr2 ⫺1 Ⲑ5

58. 17x2 ⫺1 Ⲑ4

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For Exercises 59–66, write each expression by using rational exponents rather than radical notation.

(See Example 4.)

Section 6.2

3

4

3

59. 1x

60. 1a

61. 101b

62. ⫺21t

3 2 63. 2 y

6 64. 2z5

4 65. 2a2b3

66. 1abc

Concept 3: Properties of Rational Exponents For Exercises 67–90, simplify the expressions by using the properties of rational exponents. Write the final answer using positive exponents only. (See Example 5.) p5Ⲑ3

q5Ⲑ4

67. x1Ⲑ4x⫺5Ⲑ4

68. 22Ⲑ3 2⫺5Ⲑ3

69.

71. 1y1Ⲑ5 2 10

72. 1x1Ⲑ2 2 8

73. 6⫺1Ⲑ563Ⲑ5

74. a⫺1Ⲑ3a2Ⲑ3

77. 1a1Ⲑ3a1Ⲑ4 2 12

78. 1x2Ⲑ3x1Ⲑ2 2 6

75.

4t⫺1Ⲑ3 t4Ⲑ3

79. 15a2c⫺1Ⲑ2d1Ⲑ 2 2 2 83. a

16w⫺2z 1Ⲑ3 b 2wz⫺8

87. 1x2y⫺1Ⲑ 3 2 6 1x1Ⲑ 2yz2Ⲑ 3 2 2

76.

5s⫺1Ⲑ3 s5Ⲑ3

80. 12x⫺1Ⲑ3y2z5Ⲑ3 2 3 84. a

50p⫺1q 1Ⲑ2 b 2pq⫺3

88. 1a⫺1Ⲑ 3b1Ⲑ 2 2 4 1a⫺1Ⲑ2b3Ⲑ5 2 10

p2Ⲑ3

81. a

x⫺2Ⲑ3 12 b y⫺3Ⲑ4

85. 125x2y4z6 2 1Ⲑ2 89. a

x3my2m 1Ⲑm b z5m

70.

q1Ⲑ4

82. a

m⫺1Ⲑ4 ⫺4 b n⫺1Ⲑ2

86. 18a6b3c9 2 2Ⲑ3 90. a

a4nb3n 1Ⲑn b cn

Concept 4: Applications Involving Rational Exponents 91. If P dollars in principal grows to A dollars after t years with annual interest, then the interest rate is given A 1/t by r ⫽ a b ⫺ 1. (See Example 6.) P a. In one account, $10,000 grows to $16,802 after 5 yr. Compute the interest rate. Round your answer to a tenth of a percent. b. In another account $10,000 grows to $18,000 after 7 yr. Compute the interest rate. Round your answer to a tenth of a percent. c. Which account produced a higher average yearly return? 92. If the area A of a square is known, then the length of its sides, s, can be computed by the formula s ⫽ A1Ⲑ2. a. Compute the length of the sides of a square having an area of 100 in.2 b. Compute the length of the sides of a square having an area of 72 in.2 Round your answer to the nearest 0.1 in. 93. The radius r of a sphere of volume V is given by r ⫽ a

3V 1Ⲑ3 b . Find the radius of a sphere having a volume 4p

of 85 in.3 Round your answer to the nearest 0.1 in. 94. Is 1a ⫹ b2 1Ⲑ2 the same as a1Ⲑ2 ⫹ b1Ⲑ2? If not, give a counterexample.

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Chapter 6 Radicals and Complex Numbers

Expanding Your Skills For Exercises 95–106, write the expression using rational exponents. Then simplify and convert back to radical notation. Assume that all variables represent positive real numbers. 15

Example: 2x10

Rational exponents

6 3 95. 2 y 9 99. 2 x6

103. 216x8y6

x10/15

Simplify

3 2 2 x

12

4 96. 2 w2 12

Radical notation

x2/3

18

97. 2z3

98. 2t 3

6

102. 2m2p8

3

106. 264m5n9p

100. 2p9

101. 2x3y6

104. 281a12b20

105. 2 8x3y7z8

8

3

For Exercises 107–110, write the expression as a single radical. 3 107. 21x

3 108. 2 1x

5 3 109. 2 1w

3 4 110. 2 1w

For Exercises 111–118, use a calculator to approximate the expressions. Round to four decimal places, if necessary. 111. 91Ⲑ2 3

115. 252

112. 125⫺1Ⲑ3 4

116. 263

Section 6.3

113. 50⫺1Ⲑ4

114. 11722 3Ⲑ5

117. 2103

118. 216

3

Simplifying Radical Expressions

Concepts

1. Multiplication Property of Radicals

1. Multiplication Property of Radicals 2. Simplifying Radicals by Using the Multiplication Property of Radicals 3. Simplifying Radicals by Using the Order of Operations

You may have already noticed certain properties of radicals involving a product or quotient.

PROPERTY Multiplication Property of Radicals n n Let a and b represent real numbers such that 1 a and 1 b are both real. Then n n n 1 ab ⫽ 1 aⴢ 1 b

The multiplication property of radicals follows from the property of rational exponents. 1ab ⫽ 1ab2 1Ⲑn ⫽ a1Ⲑnb1Ⲑn ⫽ 1a ⴢ 1b n

n

n

The multiplication property of radicals indicates that a product within a radicand can be written as a product of radicals, provided the roots are real numbers. For example: 1144 ⫽ 116 ⴢ 19 The reverse process is also true. A product of radicals can be written as a single radical provided the roots are real numbers and they have the same indices. 13 ⴢ 112 ⫽ 136

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Section 6.3

Simplifying Radical Expressions

2. Simplifying Radicals by Using the Multiplication Property of Radicals In algebra, it is customary to simplify radical expressions.

DEFINITION Simplified Form of a Radical Consider any radical expression where the radicand is written as a product of prime factors. The expression is in simplified form if all the following conditions are met: 1. The radicand has no factor raised to a power greater than or equal to the index. 2. The radicand does not contain a fraction. 3. There are no radicals in the denominator of a fraction.

For example, the following radicals are not simplified. 3 5 1. The expression 2 x fails condition 1. 1 2. The expression fails condition 2. B4 1 3. The expression 3 fails condition 3. 28

The expressions 2x2, 2x4, 2x6, and 2x8 are not simplified because they fail condition 1 (the exponents are not less than the index). However, each radicand is a perfect square and is easily simplified for x ⱖ 0. 2x2 ⫽ x 2x4 ⫽ x2 2x6 ⫽ x3 2x8 ⫽ x4 However, how is an expression such as 2x9 simplified? This and many other radical expressions are simplified by using the multiplication property of radicals. We demonstrate the process in Examples 1–4. Example 1

Using the Multiplication Property to Simplify a Radical Expression

Simplify the expression for x ⱖ 0.

2x9

Solution: The expression 2x9 is equivalent to 2x8 ⴢ x. 2x9 ⫽ 2x8 ⴢ x ⫽ 2x8 ⴢ 2x ⫽ x4 1x

Apply the multiplication property of radicals.

Note that x8 is a perfect square because x8 ⫽ 1x4 2 2. Simplify.

Skill Practice Simplify the expression. Assume a ⱖ 0. 1. 2a11

Answer 1. a 5 1a

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In Example 1, the expression x9 is not a perfect square. Therefore, to simplify 2x , it was necessary to write the expression as the product of the largest perfect square and a remaining or “left-over” factor: 2x9 ⫽ 2x8 ⴢ x. This process also applies to simplifying nth roots, as shown in Example 2. 9

Using the Multiplication Property to Simplify Radical Expressions

Example 2

Simplify each expression. Assume all variables represent positive real numbers. 4

3 b. 2 w7z9

a. 2b7

Solution: The goal is to rewrite each radicand as the product of the greatest perfect square (perfect cube, perfect fourth power, and so on) and a leftover factor. 4

4

a. 2b7 ⫽ 2b4 ⴢ b3 4

b4 is the greatest perfect fourth power in the radicand.

4

⫽ 2 b4 ⴢ 2 b3

Apply the multiplication property of radicals.

4

⫽ b2b3

Simplify.

3 3 b. 2 w7z9 ⫽ 2 1w6z9 2 ⴢ 1w2

w6z9 is the greatest perfect cube in the radicand.

3 3 ⫽ 2 w6z9 ⴢ 2 w

Apply the multiplication property of radicals.

3 ⫽ w2z3 1 w

Simplify.

Skill Practice Simplify the expressions. Assume all variables represent positive real numbers. 4 25 2. 2 v

Calculator Connections A calculator can be used to support the solution to Example 3. The decimal approximation for 156 and 2114 agree for the first 10 digits. This in itself does not make 156 ⫽ 2114. It is the multiplication property of radicals that guarantees that the expressions are equal.

3 17 10 3. 2 p q

Each expression in Example 2 involves a radicand that is a product of variable factors. If a numerical factor is present, sometimes it is necessary to factor the coefficient before simplifying the radical.

Using the Multiplication Property to Simplify a Radical

Example 3

Simplify the expression.

Solution: 256 ⫽ 223 ⴢ 7

⫽ 212 2 ⴢ 12 ⴢ 72 2

4. 224 4 2. v 6 1 v

3 2 3. p 5q 3 2 p q

4. 2 16

Factor the radicand. 2

2 is the greatest perfect square in the radicand.

⫽ 222 ⴢ 22 ⴢ 7

Apply the multiplication property of radicals.

⫽ 2114

Simplify.

Skill Practice Simplify. Answers

156

2 56 2 28 2 14 7

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Section 6.3

Simplifying Radical Expressions

Using the Multiplication Property to Simplify Radicals

Example 4

Simplify the expressions.Assume that all variables represent positive real numbers. 3 b. 2 40x3y5z7

a. 6250

Solution: a. 6250 ⫽ 6252 ⴢ 2

3

Factor the radicand.

⫽ 6 ⴢ 252 ⴢ 22

Apply the multiplication property of radicals.

⫽ 6 ⴢ 5 ⴢ 22

Simplify.

⫽ 3022

Simplify.

b. 240x y z

3 5 7

3

⫽ 22 5x y z 3

3 5 7

⫽ 2 12 x y z 2 ⴢ 15y z2 3

3 3 3 6

2

3 3 3 3 6 3 ⫽ 2 2xyz ⴢ 2 5y2z

⫽

3 2xyz2 2 5y2z

Factor the radicand. 23x3y3z6 is the greatest perfect cube.

2 40 2 20 2 10 5

Apply the multiplication property of radicals. Simplify.

Skill Practice Simplify. Assume that a ⬎ 0 and b ⬎ 0. 4 6. 2 32a10b19

5. 5118

TIP: To simplify the numerical factor within a radical, we can also write the factor as the product of the greatest perfect square (cube, etc.) and a “leftover” factor.

Example: 150 ⫽ 125 ⴢ 2 ⫽ 5 12 Greatest perfect square factor of 50

3 3 3 Example: 2 40 ⫽ 2 8 ⴢ 5 ⫽ 22 5

Greatest perfect cube factor of 40

3. Simplifying Radicals by Using the Order of Operations Often a radical can be simplified by applying the order of operations. In Examples 5 and 6, the first step will be to simplify the expression within the radicand.

Using the Order of Operations to Simplify Radicals

Example 5

Use the order of operations to simplify the expressions. Assume a ⬎ 0. a.

a7 B a3

b.

3 3 B 81

Solution: a.

a7 B a3

The radicand contains a fraction. However, the fraction can be reduced to lowest terms.

⫽ 2a4 ⫽ a2

Simplify the radical.

Answers 5. 15 12

4 6. 2a 2b4 2 2a 2b 3

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b.

3 3 B 81

The radical contains a fraction that can be simplified.

⫽

1 B27

Reduce to lowest terms.

⫽

1 3

Simplify.

3

Skill Practice Use the order of operations to simplify the expressions. Assume v ⬎ 0. 7.

v 21 B v5

8.

Example 6

64 B 2 5

Applying the Order of Operations

Use the order of operations to simplify.

7 150 10

Solution: 7150 7125 ⴢ 2 ⫽ 10 10 ⫽

7 ⴢ 512 10

25 is the greatest perfect square in the radicand. Simplify the radical.

1

Avoiding Mistakes

7 ⴢ 512 ⫽ 10

The expression 7 12 2 cannot be simplified further because one factor of 2 is in the radicand and the other is outside the radical.

Simplify the fraction to lowest terms.

2

⫽

712 2

Skill Practice Simplify. Answers 7. v 8

8. 2

9.

2 13 3

9.

21300 30

Section 6.3 Practice Exercises Boost your GRADE at mathzone.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

For the exercises in this set, assume that all variables represent positive real numbers unless otherwise stated.

Study Skills Exercise 1. The final exam is just around the corner. Your old tests and quizzes provide good material to study for the final exam. Use your old tests to make a list of the concepts on which you need to concentrate. Ask your professor for help if there are still concepts that you do not understand.

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Section 6.3

Simplifying Radical Expressions

Review Exercises For Exercises 2–4, simplify the expression. Write the answer with positive exponents only. 2. 1a2b⫺4 2 1Ⲑ 2 a

a b b⫺3

3. a

p4 q

⫺6

b

⫺1Ⲑ2

1p3q⫺2 2

4. 1x1Ⲑ3y5Ⲑ6 2 ⫺6

5. Write x4Ⲑ 7 in radical notation.

6. Write y2Ⲑ5 in radical notation.

7. Write 2y9 by using rational exponents.

3 2 8. Write 2 x by using rational exponents.

Concept 2: Simplifying Radicals by Using the Multiplication Property of Radicals For Exercises 9–32, simplify the radicals. (See Examples 1–4.) 9. 2x5

10. 2p15

3 7 11. 2 q

3 17 12. 2 r

13. 2a5b4

14. 2c9d6

4 8 13 15. ⫺2 xy

4 16 17 16. ⫺2 p q

17. 128

18. 163

19. 120

20. 150

21. 5118

22. 2124

3 23. 2 54

3 24. 2 250

25. 225ab3

26. 264m5n20

3 27. 2 40x7

3 28. 2 81y17

3 29. 2 ⫺16x6yz3

3 30. 2 ⫺192a6bc2

4 31. 2 80w4z7

4 32. 2 32p8qr5

Concept 3: Simplifying Radicals by Using the Order of Operations For Exercises 33–44, simplify the radicals. (See Examples 5–6.) 33.

x3 Bx

34.

y5 By

35.

37.

50 B2

38.

98 B2

39.

3 3 B 24

40.

3 2 B 250

41.

3 51 16 6

42.

7118 9

43.

3 51 72 12

44.

3 31 250 10

p7

36.

B p3

q11 B q5

Mixed Exercises For Exercises 45–72, simplify the radicals. 45. 180

46. 1108

47. ⫺6175

48. ⫺818

49. 225x4y3

50. 2125p3q2

3 51. 2 27x2y3z4

3 52. 2 108a3bc2

3y3

53.

12w5 B 3w

54.

64x9 B 4x3

55.

57.

16a2b B 2a2b4

58.

⫺27a4 B 8a

59. 223a14b8c31d22

60. 275u12v20w65x80

3

3

B 300y

15

56.

4h B 100h5

61. 218a6b3

62. 272m5n2

63. ⫺5a212a3b4c

64. ⫺7y275xy5z6

65. 27x5y

66. 210pq7

67. 254a4b2

68. 248r6s2

69.

2127 3

70.

7124 2

71.

31125 20

72.

10163 12

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For Exercises 73–76, write a mathematical expression for the English phrase and simplify. 73. The quotient of 1 and the cube root of w6

74. The principal square root of the quotient of h2 and 49

75. The principal square root of the quantity k raised to the third power

76. The cube root of 2x4

For Exercises 77–80, determine the length of the third side of the right triangle. Write the answer as a simplified radical. 77.

78. 10 ft

?

?

4 in.

12 in. 8 ft

79.

?

80.

18 m

3 cm

12 m

7 cm ?

81. On a baseball diamond, the bases are 90 ft apart. Find the exact distance from home plate to second base. Then round to the nearest tenth of a foot.

82. Linda is at the beach flying a kite. The kite is directly over a sand castle 60 ft away from Linda. If 100 ft of kite string is out (ignoring any sag in the string), how high is the kite? (Assume that Linda is 5 ft tall.) See figure.

2nd base 100 ft

90 ft

60 ft

5 ft

90 ft Home plate

Expanding Your Skills 83. Tom has to travel from town A to town C across a small mountain range. He can travel one of two routes. He can travel on a four-lane highway from A to B B and then from B to C at an average speed of 55 mph. Or he can travel on a two-lane road directly from town A to town C, but 40 mi his average speed will 50 mi be only 35 mph. If Tom is in a hurry, which route will take him to town C faster? A

C

84. One side of a rectangular pasture is 80 ft in length. The diagonal distance is 110 yd. If fencing costs $3.29 per foot, how much will it cost to fence the pasture?

110 yd

80 ft

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Section 6.4

Addition and Subtraction of Radicals

517

Addition and Subtraction of Radicals

Section 6.4

1. Addition and Subtraction of Radicals

Concept

DEFINITION Like Radicals

1. Addition and Subtraction of Radicals

Two radical terms are called like radicals if they have the same index and same radicand.

Like radicals can be added or subtracted by using the distributive property. Same index

⎫ ⎬ ⎭

Distributive property

31x ⫹ 71x ⫽ 13 ⫹ 72 1x ⫽ 101x Same radicand

Example 1

Adding and Subtracting Radicals

Add or subtract as indicated. a. 6111 ⫺ 2111

b. 13 ⫹ 13

Solution: a. 6111 ⫺ 2111

⫽ 16 ⫺ 22 111

Apply the distributive property.

⫽ 4111

Simplify.

b. 13 ⫹ 13 ⫽ 113 ⫹ 113

Note that 13 ⫽ 113.

⫽ 11 ⫹ 12 13

Apply the distributive property.

⫽ 213

Simplify.

Avoiding Mistakes The process of adding like radicals with the distributive property is similar to adding like terms. The end result is that the numerical coefficients are added and the radical factor is unchanged. 13 ⫹ 13 ⫽ 113 ⫹ 113 ⫽ 213 Be careful: 13 ⫹ 13 ⫽ 16 In general: 1x ⫹ 1y ⫽ 1x ⫹ y

Skill Practice Add or subtract as indicated. 1. 516 ⫺ 816

2. 110 ⫹ 110

Answers 1. ⫺3 16

2. 2110

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Example 2

Adding and Subtracting Radicals

Add or subtract as indicated. 3 3 3 a. ⫺21 ab ⫹ 71 ab ⫺ 1 ab

b.

1 3 x13y ⫺ x13y 4 2

Solution: 3 3 3 a. ⫺21 ab ⫹ 71 ab ⫺ 1 ab 3 ⫽ 1⫺2 ⫹ 7 ⫺ 12 1 ab

Apply the distributive property.

⫽ 41ab

Simplify.

3

b.

1 3 x13y ⫺ x13y 4 2 1 3 ⫽ a ⫺ b x13y 4 2

Apply the distributive property.

6 1 ⫽ a ⫺ b x13y 4 4

Get a common denominator.

5 ⫽ ⫺ x13y 4

Simplify.

Skill Practice Add or subtract as indicated. 3 3 3 3. 51xy ⫺ 31xy ⫹ 71xy 1 5 4. y12 ⫹ y12 6 4

Example 3 shows that it is often necessary to simplify radicals before adding or subtracting. Example 3

Adding and Subtracting Radicals

Simplify the radicals and add or subtract as indicated.Assume all variables represent positive real numbers. a. 318 ⫹ 12

b. 82x3y2 ⫺ 3y2x3

c. 250x2y5 ⫺ 13y22x2y3 ⫹ xy298y3

Solution: a. 318 ⫹ 12 ⫽ 3 ⴢ 212 ⫹ 12

The radicands are different. Try simplifying the radicals first. Simplify: 28 ⫽ 223 ⫽ 222

⫽ 612 ⫹ 12

Answers 3 3. 91 xy

4.

13 y 12 12

⫽ 16 ⫹ 12 12

Apply the distributive property.

⫽ 712

Simplify.

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Section 6.4

b. 82x3y2 ⫺ 3y2x3

Addition and Subtraction of Radicals

The radicands are different. Simplify the radicals first. Simplify 2x3y2 ⫽ xy 1x and 2x3 ⫽ x1x.

⫽ 8xy1x ⫺ 3xy 1x ⫽ 18 ⫺ 32xy1x

Apply the distributive property.

⫽ 5xy1x

Simplify.

c. 250x2y5 ⫺ 13y22x2y3 ⫹ xy 298y3

Simplify each radical. 250x2y5 ⫽ 225 ⴢ 2x2y5 ⫽ 5xy2 12y

⫽ 5xy2 12y ⫺ 13xy2 12y ⫹ 7xy2 12y

⫽ 15 ⫺ 13 ⫹ 72xy2 12y

e ⫺13y22x2y3 ⫽ ⫺13xy2 12y xy298y3 ⫽ xy249 ⴢ 2y3 ⫽ 7xy2 12y Apply the distributive property.

⫽ ⫺xy 12y 2

Skill Practice Simplify the radicals and add or subtract as indicated. Assume all variables represent positive real numbers. 5. 175 ⫹ 213 6. 42a 2b ⫺ 6a1b 7. ⫺322y3 ⫹ 5y118y ⫺ 2250y3

In some cases, when two radicals are added, the resulting sum must be written in factored form. This is demonstrated in Example 4. Example 4

Adding Radical Expressions

Add the radicals. Assume that x ⱖ 0.

322x2 ⫹ 28

Solution: 322x2 ⫹ 28 ⫽ 3x12 ⫹ 212 ⫽ 13x ⫹ 22 12

Simplify each radical. Notice that the radicands are the same, but the terms are not like terms. The first term has a factor of x and the second does not. Apply the distributive property. The expression cannot be simplified further because 3x and 2 are not like terms.

Skill Practice Add the radicals. Assume that y ⱖ 0. 8. 4245 ⫺ 25y4

Answers 5. 713 7. 2y 12y

6. ⫺2a1b 8. 112 ⫺ y 2 2 15

519

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Section 6.4 Practice Exercises Boost your GRADE at mathzone.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

For the exercises in this set, assume that all variables represent positive real numbers, unless otherwise stated.

Study Skills Exercise 1. Define the key term like radicals.

Review Exercises For Exercises 2–5, simplify the radicals. 3 2. 2 ⫺16s4t9

4 7 4 3. ⫺2 xy

6. Write 14x2 2 1Ⲑ3 in radical notation. 8. Simplify.

4. 236a2b3

5.

7b8 B 56b2 3

4 3 7. Write 2 x y using rational exponents.

32⫺1Ⲑ5

For Exercises 9–10, simplify the expressions. Write the answer with positive exponents only. 10. 1x1Ⲑ2y⫺3Ⲑ4 2 ⫺4

9. y2Ⲑ3y1Ⲑ4

Concept 1: Addition and Subtraction of Radicals For Exercises 11–12, determine if the radical terms are like. 3 11. a. 12 and 2 2

3 3 12. a. 72x and 2x

b. 12 and 312

3 4 b. 2 x and 2 x

c. 12 and 15

4 4 c. 22 x and x2 2

13. Explain the similarities between the pairs of expressions. a. 715 ⫹ 415

and

b. ⫺216 ⫺ 913

and

7x ⫹ 4x ⫺2x ⫺ 9y

14. Explain the similarities between the pairs of expressions. a. ⫺413 ⫹ 513 b. 1317 ⫺ 18

and and

⫺4z ⫹ 5z 13a ⫺ 18

For Exercises 15–32, add or subtract the radical expressions, if possible. (See Examples 1–2.) 15. 315 ⫹ 615

16. 51a ⫹ 31a

3 3 3 17. 31 tw ⫺ 21 tw ⫹ 1 tw

3 3 3 18. 61 7 ⫺ 21 7⫹ 1 7

19. 6110 ⫺ 110

20. 13111 ⫺ 111

4 4 4 21. 1 3 ⫹ 71 3⫺ 1 14

22. 2111 ⫹ 3113 ⫹ 5111

23. 81x ⫹ 21y ⫺ 61x

24. 10110 ⫺ 8110 ⫹ 12

3 3 25. 1 ab ⫹ a1 b

4 4 26. x1 y ⫺ y1 x

3 27. 12t ⫹ 1 2t

4 3 28. 1 5c ⫹ 1 5c

29.

5 3 7 3 z16 ⫹ z1 6 6 9

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Section 6.4

30.

3 4 1 4 a1b ⫹ a1 b 4 6

Addition and Subtraction of Radicals

31. 0.81x1y ⫺ 0.11x1y

33. Explain the process for adding the two radicals. Then find the sum. 34. Explain the process for subtracting two radicals. Then find the sum.

521

32. 7.51pq ⫺ 6.31pq 312 ⫹ 7150 112x ⫺ 175x

For Exercises 35–64, add or subtract the radical expressions as indicated. (See Examples 3–4.) 35. 136 ⫹ 181

36. 3180 ⫺ 5145

37. 2112 ⫹ 148

38. 5132 ⫹ 2150

39. 417 ⫹ 163 ⫺ 2128 40. 813 ⫺ 2127 ⫹ 175

41. 5118 ⫹ 132 ⫺ 4150 42. 7172 ⫺ 18 ⫹ 4150

3 3 43. 2 81 ⫺ 2 24

3 3 44. 172 81 ⫺ 22 24

45. 312a ⫺ 18a ⫺ 172a 46. 112t ⫺ 127t ⫹ 513t

3 2 6 3 47. 2s2 2 s t ⫹ 3t2 2 8s8

3 4 3 48. 42 x ⫺ 2x2 x

3 4 3 49. 72 x ⫺ x2 x

3 3 50. 62y10 ⫺ 3y2 2y4

51. 5p220p2 ⫹ p2 180

52. 2q248q2 ⫺ 227q4

3 2 3 53. 2 ab⫺ 2 8a2b

54. w180 ⫺ 32125w2

55. 5x1x ⫹ 61x

56. 9y2 12 ⫹ 412

57. 250x2 ⫺ 328

58. 29x3 ⫺ 225x

3 3 3 59. 112 54cd3 ⫺ 22 2cd3 ⫹ d2 16c

61.

3 3 2 2 3 8 2 60. x2 64x5y2 ⫺ x2 2 x y ⫹ 52 xy

3 4 ab224a3 ⫹ 254a5b2 ⫺ a2b1150a 2 3

3 3 3 63. x2 16 ⫺ 22 27x ⫹ 2 54x3

2 5 62. mn 172n ⫹ n28m2n ⫺ 250m2n3 3 6 4 5 4 4 64. 52 y ⫺ 2y2 y⫹ 2 16y7

Mixed Exercises For Exercises 65–70, answer true or false. If an answer is false, explain why or give a counterexample. 65. 1x ⫹ 1y ⫽ 1x ⫹ y

66. 1x ⫹ 1x ⫽ 2 1x

3 3 3 67. 51 x ⫹ 21 x ⫽ 71 x

3 68. 61x ⫹ 5 1 x ⫽ 11 1x

69. 1y ⫹ 1y ⫽ 12y

70. 2c2 ⫹ d 2 ⫽ c ⫹ d

For Exercises 71–74, write the English phrase as an algebraic expression. Simplify each expression. 71. The sum of the principal square root of 48 and the principal square root of 12

72. The sum of the cube root of 16 and the cube root of 2

73. The difference of 5 times the cube root of x6 and the square of x

74. The sum of the cube of y and the principal fourth root of y12

For Exercises 75–78, write an English phrase that translates the mathematical expression. (Answers may vary.) 3 76. 43 ⫺ 1 4

75. 118 ⫺ 52

4 77. 1 x ⫹ y3

78. a4 ⫹ 1a

For Exercises 79–80, find the exact value of the perimeter, and then approximate the value to one decimal place. 79.

80.

75 ft

3 ft

2 6 cm 2 24 cm

54 cm

27 ft

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81. The figure has perimeter 1412 ft. Find the value of x.

82. The figure has perimeter 1217 cm. Find the value of x.

50 ft

112 cm x x

Expanding Your Skills

y

83. a. An irregularly shaped garden is shown in the figure. All distances are expressed in yards. Find the perimeter. Hint: Use the Pythagorean theorem to find the length of each side. Write the final answer in radical form.

(6, 9) (7, 7)

(0, 6)

b. Approximate your answer to two decimal places. c. If edging costs $1.49 per foot and sales tax is 6%, find the total cost of edging the garden.

Section 6.5

(2, 2)

(4, 1) x

Multiplication of Radicals

Concepts

1. Multiplication Property of Radicals

1. Multiplication Property of Radicals 2. Expressions of the Form n a )n (1 3. Special Case Products 4. Multiplying Radicals with Different Indices

In this section, we will learn how to multiply radicals by using the multiplication property of radicals first introduced in Section 6.3.

PROPERTY The Multiplication Property of Radicals n n Let a and b represent real numbers such that 1 a and 1 b are both real. Then n n n 1 ab ⫽ 1 aⴢ 1 b

To multiply two radical expressions, we use the multiplication property of radicals along with the commutative and associative properties of multiplication. Example 1

Multiplying Radical Expressions

Multiply each expression and simplify the result. Assume all variables represent positive real numbers. 1 3 2 3 a. 1312215162 b. 12x1y21⫺7 1xy2 c. 115c 2 cd2a 2 cd b 3

Solution:

a. 1312215162

⫽ 13 ⴢ 521 12 ⴢ 162 ⫽ 15112

Commutative and associative properties of multiplication Multiplication property of radicals

⫽ 1522 3 2

⫽ 15 ⴢ 213 ⫽ 3013

Simplify the radical.

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Section 6.5

Multiplication of Radicals

b. 12x1y21⫺71xy2

⫽ 12x21⫺721 1y ⴢ 1xy2

Commutative and associative properties of multiplication

⫽ ⫺14x2xy2

Multiplication property of radicals

⫽ ⫺14xy1x

Simplify the radical.

1 3 3 c. 115c2cd2a 2cd2 b 3 1 3 3 ⫽ a15c ⴢ b 1 2 cd ⴢ 2 cd 2 2 3 3 2 3 ⫽ 5c2 cd 3

⫽ 5cd 2c

Commutative and associative properties of multiplication Multiplication property of radicals

2

Simplify the radical.

Skill Practice Multiply the expressions and simplify the results. Assume all variables represent positive real numbers. 1. 141621⫺2132 2. 13ab1b21⫺21ab2 3 3 3. 12 1 4ab2152 2a 2b2

When multiplying radical expressions with more than one term, we use the distributive property.

Example 2 Multiply.

Multiplying Radical Expressions

311112 ⫹ 1112

Solution: 311112 ⫹ 1112

⫽ 3111 ⴢ 122 ⫹ 3111 ⴢ 111

Apply the distributive property.

⫽ 6211 ⫹ 32112

Multiplication property of radicals

⫽ 6111 ⫹ 3 ⴢ 11

Simplify the radical.

⫽ 6111 ⫹ 33 Skill Practice Multiply. 4. 51512 15 ⫺ 22

Answers 1. ⫺24 12 3 2 3. 20a 2 b

2. ⫺6ab 2 1a 4. 50 ⫺ 1015

523

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Example 3

Multiplying Radical Expressions

Multiply.

a. 1 15 ⫹ 31221215 ⫺ 122

b. 1⫺10a 1b ⫹ 7b21a1b ⫹ 2b2

Solution: a. 1 15 ⫹ 31221215 ⫺ 122 ⫽ 2252 ⫺ 210 ⫹ 6210 ⫺ 3222

Apply the distributive property.

⫽ 2 ⴢ 5 ⫹ 5110 ⫺ 3 ⴢ 2

Simplify radicals and combine like radicals.

⫽ 10 ⫹ 5110 ⫺ 6 ⫽ 4 ⫹ 5110

Combine like terms.

b. 1⫺10a1b ⫹ 7b21a1b ⫹ 2b2 ⫽ ⫺10a2 2b2 ⫺ 20ab2b ⫹ 7ab2b ⫹ 14b2

Apply the distributive property.

⫽ ⫺10a2b ⫺ 13ab1b ⫹ 14b2

Simplify and combine like terms.

Skill Practice Multiply.

5. 12 13 ⫺ 311021 13 ⫹ 21102

Example 4 Multiply.

6. 1x 1y ⫹ y213x 1y ⫺ 2y2

Multiplying Radical Expressions

121x ⫹ 1y216 ⫺ 1x ⫹ 81y2

Solution: 12 1x ⫹ 1y216 ⫺ 1x ⫹ 81y2 ⫽ 122x ⫺ 22x2 ⫹ 162xy ⫹ 62y ⫺ 2xy ⫹ 82y2

Apply the distributive property.

⫽ 122x ⫺ 2x ⫹ 16 2xy ⫹ 62y ⫺ 2xy ⫹ 8y

Simplify the radicals.

⫽ 122x ⫺ 2x ⫹ 15 2xy ⫹ 62y ⫹ 8y

Combine like terms.

Skill Practice Multiply.

7. 1 1t ⫹ 31w214 ⫺ 1t ⫺ 1w2

Answers 5. ⫺54 ⫹ 130 6. 3x 2y ⫹ xy 1y ⫺ 2y 2 7. 41t ⫺ t ⫺ 41tw ⫹ 12 1w ⫺ 3w

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525

n

2. Expressions of the Form ( 1 a )n The multiplication property of radicals can be used to simplify an expression of the form 1 1a2 2, where a ⱖ 0. 1 2a2 2 ⫽ 2a ⴢ 2a ⫽ 2a2 ⫽ a

where a ⱖ 0

n n This logic can be applied to nth roots. If 1a is a real number, then 1 1a2 n ⫽ a.

Example 5

Simplifying Radical Expressions

Simplify the expressions. Assume all variables represent positive real numbers. a. 1 1112 2

3 c. 1 1 pq2 3

5 b. 1 1 z2 5

Solution:

a. 1 1112 2 ⫽ 11

5 b. 1 1 z2 5 ⫽ z

Skill Practice Simplify. 8. 1 1142 2

7 9. 1 1 q2 7

3 c. 1 1 pq2 3 ⫽ pq

5 10. 1 1 3z2 5

3. Special Case Products From Examples 2–4, you may have noticed a similarity between multiplying radical expressions and multiplying polynomials. In Section 4.3 we learned that the square of a binomial results in a perfect square trinomial: 1a ⫹ b2 2 ⫽ a2 ⫹ 2ab ⫹ b2 1a ⫺ b2 2 ⫽ a2 ⫺ 2ab ⫹ b2

The same patterns occur when squaring a radical expression with two terms. Example 6

Squaring a Two-Term Radical Expression

Square the radical expressions. Assume all variables represent positive real numbers. a. 1 1d ⫹ 32 2

b. 15 1y ⫺ 122 2

Solution:

a. 1 1d ⫹ 32 2 a2 ⫹ 2ab ⫹ b2

⫽ 1 1d2 2 ⫹ 21 1d2132 ⫹ 132 2 ⫽ d ⫹ 61d ⫹ 9

This expression is in the form 1a ⫹ b2 2, where a ⫽ 1d and b ⫽ 3. Apply the formula 1a ⫹ b2 2 ⫽ a2 ⫹ 2ab ⫹ b2.

TIP: The product

1 1d ⫹ 32 2 can also be found by using the distributive property. 11d ⫹ 32 2 ⫽ 1 1d ⫹ 32 11d ⫹ 32

Simplify.

Answers 8. 14

9. q

10. 3z

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b. 151y ⫺ 122 2 a2 ⫺ 2ab ⫹ b2

⫽ 151y2 2 ⫺ 2151y21 122 ⫹ 1 122 2 ⫽ 25y ⫺ 10 12y ⫹ 2

This expression is in the form 1a ⫺ b2 2, where a ⫽ 51y and b ⫽ 12. Apply the formula 1a ⫺ b2 2 ⫽ a2 ⫺ 2ab ⫹ b2. Simplify.

Skill Practice Square the radical expressions. Assume all variables represent positive real numbers. 11. 1 1a ⫺ 52 2

12. 141x ⫹ 132 2

Recall from Section 4.3 that the product of two conjugate binomials results in a difference of squares. 1a ⫹ b21a ⫺ b2 ⫽ a2 ⫺ b2

The same pattern occurs when multiplying two conjugate radical expressions. Example 7

Multiplying Conjugate Radical Expressions

Multiply the radical expressions. Assume all variables represent positive real numbers. a. 1 13 ⫹ 221 13 ⫺ 22

Solution:

TIP: The product

1 13 ⫹ 22 1 13 ⫺ 22 can also be found by using the distributive property. 1 13 ⫹ 22 1 13 ⫺ 22

a. 1 13 ⫹ 221 13 ⫺ 22 a2 ⫺ b2

1 3 1 3 b. a 1s ⫺ 1tb a 1s ⫹ 1tb 3 4 3 4 The expression is in the form 1a ⫹ b21a ⫺ b2, where a ⫽ 13 and b ⫽ 2.

⫽ 1 132 2 ⫺ 122 2

Apply the formula 1a ⫹ b21a ⫺ b2 ⫽ a2 ⫺ b2.

⫽3⫺4

Simplify.

⫽ ⫺1 1 3 1 3 b. a 1s ⫺ 1tb a 1s ⫹ 1tb 3 4 3 4 a2 ⫺ b2

Answers 11. a ⫺ 10 1a ⫹ 25 12. 16x ⫹ 813x ⫹ 3 1 4 13. ⫺4 14. a ⫺ b 4 25

This expression is in the form 1a ⫺ b21a ⫹ b2, where a ⫽ 13 1s and b ⫽ 34 1t.

2 2 1 3 ⫽ a 1sb ⫺ a 1tb 3 4

Apply the formula 1a ⫹ b21a ⫺ b2 ⫽ a2 ⫺ b2.

1 9 ⫽ s⫺ t 9 16

Simplify.

Skill Practice Multiply the conjugates. Assume all variables represent positive real numbers. 1 2 2 1 13. 1 15 ⫹ 321 15 ⫺ 32 14. a 1a ⫹ 1bb ⴢ a 1a ⫺ 1bb 2 5 2 5

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Multiplication of Radicals

4. Multiplying Radicals with Different Indices The product of two radicals can be simplified provided the radicals have the same index. If the radicals have different indices, then we can use the properties of rational exponents to obtain a common index. Example 8

Multiplying Radicals with Different Indices

Multiply the expressions. Write the answers in radical form. 3 4 a. 1 5ⴢ 1 5

3 b. 1 7 ⴢ 12

Solution: 3 4 a. 1 5ⴢ 1 5

⫽ 51Ⲑ351Ⲑ4 ⫽5

Rewrite each expression with rational exponents.

11Ⲑ32 ⫹ 11Ⲑ42

Because the bases are equal, we can add the exponents.

⫽ 514Ⲑ122 ⫹ 13Ⲑ122

Write the fractions with a common denominator.

7Ⲑ12

⫽5

Simplify the exponent.

12

⫽ 2 57

Rewrite the expression as a radical.

3

b. 17 ⴢ 12 ⫽ 71Ⲑ321Ⲑ2

Rewrite each expression with rational exponents.

2Ⲑ6 3Ⲑ6

⫽7 2

Write the rational exponents with a common denominator.

⫽ 17223 2 1Ⲑ6 6

Apply the power rule of exponents.

⫽ 27 2

Rewrite the expression as a single radical.

6 ⫽1 392

Simplify.

2 3

Skill Practice Multiply the expressions. Write the answers in radical form. Assume all variables represent positive real numbers. 3 15. 1x ⴢ 1 x

4 3 3 16. 2 a ⴢ 1 b

Answers 6 5 15. 2 x

12

16. 2a 9b 4

Section 6.5 Practice Exercises Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

For the exercises in this set, assume that all variables represent positive real numbers, unless otherwise stated.

Review Exercises For Exercises 1–3, simplify the radicals. 3 1. 2 ⫺16x5y6z7

2. ⫺220a2b3c

3.

8y3z5 B y

For Exercises 4–6, simplify the expressions. Write the answer with positive exponents only. 4. x1/3y1/4x⫺1/6y1/3

5. p1/8q1/2p⫺1/4q3/2

6.

b1/4 b3/2

527

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For Exercises 7–8, add or subtract as indicated. 3 3 7. ⫺21 7 ⫹ 41 7

8. 428x3 ⫺ x250x

Concept 1: Multiplication Property of Radicals For Exercises 9–44, multiply the radical expressions. (See Examples 1–4.) 3 3 9. 27 ⴢ 23

4 4 10. 1 6ⴢ 1 2

11. 12 ⴢ 110

3 3 12. 2 4ⴢ 2 12

4 4 13. 1 16 ⴢ 1 64

14. 25x3 ⴢ 210x4

3 3 15. 14 1 42121 52

16. 1215213172

17. 18a 1b21⫺3 1ab2

4 3 4 18. 1p2 q 21 2 pq2

19. 130 ⴢ 112

20. 120 ⴢ 154

21. 16x ⴢ 112x

22. 1 23ab2 21 221a2b2

23. 25a3b2 ⴢ 220a3b3

3 3 24. 2 m2n2 ⴢ 2 48m4n2

25. 14 23xy3 21⫺2 26x3y2 2

4 4 26. 121 3x2142 27x6 2

3 3 3 27. 1 2 4a2b21 2 2ab3 21 2 54a2b2

3 3 3 28. 1 2 9x3y21 2 6xy21 2 8x2y5 2

29. 131413 ⫺ 62

30. 3 151215 ⫹ 42

31. 121 16 ⫺ 132

32. 151 13 ⫹ 172

1 33. ⫺ 1x161x ⫹ 72 3

1 34. ⫺ 1y18 ⫺ 31y2 2

35. 1 13 ⫹ 211021413 ⫺ 1102

36. 18 17 ⫺ 1521 17 ⫹ 3152

37. 1 1x ⫹ 421 1x ⫺ 92

38. 1 1w ⫺ 221 1w ⫺ 92

3 3 39. 1 1y ⫹ 221 1y ⫺ 32

5 5 40. 14 ⫹ 1p215 ⫹ 1p2

41. 1 1a ⫺ 31b2191a ⫺ 1b2

42. 1111m ⫹ 41n21 1m ⫹ 1n2

43. 1 1p ⫹ 21q218 ⫹ 31p ⫺ 1q2

44. 151s ⫺ 1t21 1s ⫹ 5 ⫹ 61t2

n

Concept 2: Expressions of the Form ( 1a)n For Exercises 45–52, simplify the expressions. Assume all variables represent positive real numbers. (See Example 5.) 45. 1 1152 2

46. 1 1582 2

47. 1 13y2 2

48. 1 119yz2 2

3 49. 1 262 3

5 50. 1 2 242 5

51. 1709 ⴢ 1709

52. 1401 ⴢ 1401

Concept 3: Special Case Products 53. a. Write the formula for the product of two conjugates. 1x ⫹ y21x ⫺ y2 ⫽ b. Multiply 1x ⫹ 521x ⫺ 52 .

54. a. Write the formula for squaring a binomial. 1x ⫹ y2 2 ⫽ b. Multiply 1x ⫹ 52 2.

For Exercises 55– 66, multiply the special products. (See Examples 6–7.) 55. 1 113 ⫹ 42 2

56. 16 ⫺ 1112 2

57. 1 1p ⫺ 172 2

58. 1 1q ⫹ 122 2

59. 1 12a ⫺ 31b2 2

60. 1 13w ⫹ 41z2 2

61. 1 13 ⫹ x21 13 ⫺ x2

62. 1y ⫹ 1621y ⫺ 162

63. 1 16 ⫹ 1221 16 ⫺ 122

64. 1 115 ⫹ 1521 115 ⫺ 152

2 1 2 1 65. a 1x ⫹ 1yba 1x ⫺ 1yb 3 2 3 2

1 1 1 1 66. a 1s ⫹ 1tba 1s ⫺ 1tb 4 5 4 5

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Section 6.5

For Exercises 67– 68, multiply the expressions.

Multiplication of Radicals

67. a. 1 13 ⫹ 1x21 13 ⫺ 1x2

68. a. 1 15 ⫹ 1y21 15 ⫺ 1y2

c. 1 13 ⫺ 1x21 13 ⫺ 1x2

c. 1 15 ⫺ 1y21 15 ⫺ 1y2

b. 1 13 ⫹ 1x21 13 ⫹ 1x2

529

b. 1 15 ⫹ 1y21 15 ⫹ 1y2

Mixed Exercises For Exercises 69–76, identify each statement as true or false. If an answer is false, explain why. 69. 13 ⴢ 12 ⫽ 16

3 70. 15 ⴢ 12 ⫽ 110

71. 1x ⫺ 152 2 ⫽ x ⫺ 5

72. 31215x2 ⫽ 615x

73. 51314x2 ⫽ 15120x

74.

75.

15x ⫽ 1x 5

76. 1 1t ⫺ 121 1t ⫹ 12 ⫽ t ⫺ 1

31x ⫽ 1x 3

For Exercises 77–84, perform the indicated operations. 77. 1⫺16x2 2

78. 1⫺18a2 2

79. 1 13x ⫹ 12 2

80. 1 1x ⫺ 12 2

81. 1 1x ⫹ 3 ⫺ 42 2

82. 1 1x ⫹ 1 ⫹ 32 2

83. 1 12t ⫺ 3 ⫹ 52 2

84. 1 13w ⫺ 2 ⫺ 42 2

87.

88.

For Exercises 85–88, find the exact area. 85.

86. 6 2m

40 ft

2 18 yd

3 5 in.

7 6 yd 6 12 in. 10 12 m 3 2 ft

Concept 4: Multiplying Radicals with Different Indices For Exercises 89–100, multiply or divide the radicals with different indices. Write the answers in radical form and simplify. (See Example 8.) 4 89. 1x ⴢ 1 x

3 90. 1 y ⴢ 1y

5 3 91. 1 2z ⴢ 1 2z

3 2 93. 2 p ⴢ 2p3

4 3 3 2 94. 2 q ⴢ 2 q

95.

3 6 97. 1 xⴢ 1 y

6 98. 1a ⴢ 1 b

4 99. 1 8 ⴢ 13

2u3 3 2 u

3 4 92. 1 5w ⴢ 1 5w

96.

2v5 4 2 v

6 100. 111 ⴢ 1 2

Expanding Your Skills For Exercises 101–106, multiply. 3 4 101. 2 2xy ⴢ 2 5xy

3 102. 26ab ⴢ 2 7ab

3 103. 2 4m2n ⴢ 26mn

4 3 104. 2 5xy3 ⴢ 2 10x2y

3 3 3 2 3 3 2 105. 12 a⫹ 2 b21 2 a ⫺2 ab ⫹ 2 b2

3 3 3 2 3 3 2 106. 12 a⫺2 b212 a ⫹2 ab ⫹ 2 b2

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Problem Recognition Exercises Simplifying Radical Expressions For Exercises 1–20, simplify the expressions. 1. a. 224

2. a. 254

3. a. 2200y6

4. a. 232z15

3 b. 224

3 b. 254

3 b. 2200y6

3 b. 232z15

5. a. 280

6. a. 248

7. a. 2x5y6

8. a. 2a10b12

3 b. 280

3 b. 248

3 b. 2x5y6

3 b. 2a10b12

4 c. 280

4 c. 248

4 c. 2x5y6

4 c. 2a10b12

3 9. a. 232s5t6

3 10. a. 296v7w20

4

4

b. 232s5t6 5

b. 296v7w20 5

c. 232s t

7

b. 210 ⴢ 210

20

14. a. 327 ⫺ 1027

b. 226 ⴢ 526 17. a. 5218 ⫺ 428

b. 25 ⴢ 25

12. a. 210 ⫹ 210

c. 296v w

5 6

13. a. 226 ⫺ 526

11. a. 25 ⫹ 25

b. 327 ⴢ 1027 18. a. 250 ⫺ 272

15. a. 28 ⫹ 22

16. a. 212 ⫹ 23

b. 28 ⴢ 22

b. 212 ⴢ 23

3 3 19. a. 42 24 ⫹ 62 3

3 3 20. a. 22 54 ⫺ 52 2

3 3 b. 42 24 ⴢ 62 3

3 3 b. 22 54 ⴢ 52 2

b. 5218 ⴢ 428

b. 250 ⴢ 272

Section 6.6

Division of Radicals and Rationalization

Concepts

1. Simplified Form of a Radical

1. Simplified Form of a Radical 2. Division Property of Radicals 3. Rationalizing the Denominator—One Term 4. Rationalizing the Denominator—Two Terms

Recall that for a radical expression to be in simplified form the following three conditions must be met.

DEFINITION Simplified Form of a Radical Consider any radical expression in which the radicand is written as a product of prime factors. The expression is in simplified form if all the following conditions are met: 1. The radicand has no factor raised to a power greater than or equal to the index. 2. The radicand does not contain a fraction. 3. No radicals are in the denominator of a fraction.

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Division of Radicals and Rationalization

In the previous sections, we have concentrated on the first condition in the simplification process. Next, we will demonstrate how to satisfy the second and third conditions involving radicals and fractions.

2. Division Property of Radicals The multiplication property of radicals makes it possible to write a product within a radical to be separated as a product of radicals. We now state a similar property for radicals involving quotients.

PROPERTY

Division Property of Radicals n

n

Let a and b represent real numbers such that 1 a and 1 b are both real. Then, n

a 2a ⫽ n Bb 2b

b⫽0

n

The division property of radicals indicates that a quotient within a radicand can be written as a quotient of radicals provided the roots are real numbers. For example: 25 125 ⫽ B 36 136 The reverse process is also true. A quotient of radicals can be written as a single radical provided that the roots are real numbers and that they have the same index. 3

Same index

1 125 3

18

⫽

125 B 8 3

In Examples 1 and 2, we will apply the division property of radicals to simplify radical expressions.

Example 1

Using the Division Property to Simplify Radicals

Use the division property of radicals to simplify the expressions. Assume the variables represent positive real numbers. a.

a6 B b4

b.

3

81y5

B x3

Solution: a.

a6 B b4 ⫽ ⫽

The radicand contains an irreducible fraction. 2a6 2b4 a3 b2

Apply the division property to rewrite as a quotient of radicals. Simplify the radicals.

531

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b.

3

81y5

The radicand contains an irreducible fraction.

B x3 ⫽

3 2 81y5 3

Apply the division property to rewrite as a quotient of radicals.

2x

3

3

⫽ ⫽

2 34 ⴢ y5

Factor the radicand in the numerator to simplify the radical.

3

2 x3 3 3y2 3y2 x

Simplify the radicals in the numerator and the denominator. The expression is simplified because it now satisfies all conditions.

Skill Practice Simplify the expressions. 1.

x4 B y10

2.

Example 2

w7 B 64 3

Using the Division Property to Simplify a Radical

Use the division property of radicals to simplify the expressions. Assume the variables represent positive real numbers. 4 2 8p7 4 3 2 p

Solution: 4 2 8p7

There is a radical in the denominator of the fraction.

4 3 2 p

⫽

4

8p7

B p

3

Apply the division property to write the quotient under a single radical.

4 ⫽2 8p4

Simplify the fraction.

4 ⫽ p2 8

Simplify the radical.

Skill Practice Simplify the expression. 3.

3 2 16y 4 3 2 y

3. Rationalizing the Denominator—One Term The third condition restricts radicals from the denominator of an expression. The process to remove a radical from the denominator is called rationalizing the denominator. In many cases, rationalizing the denominator creates an expression that is computationally simpler. For example, 6 ⫽ 213 13 Answers 1.

x2 y5

2.

3 w21 w 4

3 3. 2y 1 2

and

⫺2 ⫽ 2 ⫺ 16 2 ⫹ 16

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Section 6.6

Division of Radicals and Rationalization

We will demonstrate the process to rationalize the denominator as two separate cases: • Rationalizing the denominator (one term) • Rationalizing the denominator (two terms involving square roots) To begin the first case, recall that the nth root of a perfect nth power simplifies completely. 2x2 ⫽ x 3

xⱖ0

2x ⫽ x 3

4 4 2 x ⫽x 5

xⱖ0

2x ⫽ x # # # 5

Therefore, to rationalize a radical expression, use the multiplication property of radicals to create an nth root of an nth power.

Example 3

Rationalizing Radical Expressions

Fill in the missing radicand to rationalize the radical expressions. Assume all variables represent positive real numbers. a. 2a ⴢ 2? ⫽ 2a2 ⫽ a

3 3 3 3 b. 2 yⴢ 2 ?⫽ 2 y ⫽y

4 4 4 4 4 c. 2 2z3 ⴢ 2 ?⫽ 2 2 z ⫽ 2z

Solution: a. 2a ⴢ 2? ⫽ 2a2 ⫽ a

What multiplied by 1a will equal 2a2?

2a ⴢ 2a ⫽ 2a2 ⫽ a 3 3 3 3 b. 2 yⴢ 2 ?⫽ 2 y ⫽y

3 3 3 What multiplied by 1 y will equal 2 y?

3 3 2 3 3 2 yⴢ 2 y ⫽ 2 y ⫽y

c.

4 4 4 4 4 2 2z3 ⴢ 2 ?⫽ 2 2 z ⫽ 2z 4 4 3 4 4 4 2 2z3 ⴢ 2 2z⫽ 2 2 z ⫽ 2z

4 What multiplied by 2 2z3 will equal 4 4 4 22 z ?

Skill Practice Fill in the missing radicand to rationalize the radical expression. 4. 27 ⴢ 2?

5 5 5. 2 t 2 ⴢ 2 ?

3 3 6. 2 5x 2 ⴢ 2 ?

To rationalize the denominator of a radical expression, multiply the numerator and denominator by an appropriate expression to create an nth root of an nth power in the denominator.

Answers 4. 7

5. t 3

6. 52x

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Chapter 6 Radicals and Complex Numbers

Example 4

Rationalizing the Denominator—One Term 5

Simplify the expression.

3

2a

a⫽0

Solution: To remove the radical from the denominator, a cube root of a perfect cube is 3 2 needed in the denominator. Multiply numerator and denominator by 2 a 3 3 2 3 3 because 2a ⴢ 2a ⫽ 2a ⫽ a. 5 3 2 a

⫽ ⫽ ⫽

5 3 2 a

ⴢ

3 2 2 a 3 2 2 a

3 52a2 3 3 2 a 3 2 52 a a

Multiply the radicals. Simplify.

Skill Practice Simplify the expression. Assume y 7 0. 7.

2 4

2y

Note that for a ⫽ 0, the expression expression

5 3

2a

3 2 2 a 3 2 2 a

⫽ 1. In Example 4, multiplying the

by this ratio does not change its value.

Example 5

Rationalizing the Denominator—One Term

Simplify the expression.

y5 B7

Solution: y5 B7 ⫽ ⫽ ⫽

The radical contains an irreducible fraction. 2y5

Apply the division property of radicals.

17 y2 1y

Simplify the radical in the numerator.

17 y2 1y 17

ⴢ

17 17

Rationalize the denominator. Note: 27 ⴢ 27 ⫽ 272 ⫽ 7 .

y 17y 2

⫽

Answer 4

7.

22 y 3 y

Avoiding Mistakes

272

y2 17y ⫽ 7

Simplify.

A factor within a radicand cannot be simplified with a factor outside the radicand. For example, 17y cannot be simplified. 7

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535

Skill Practice Simplify the expression. 8.

8 B3

Rationalizing the Denominator—One Term

Example 6

Simplify the expression.

15 3

225s

Solution: 15 3

225s ⫽ ⫽ ⫽

15 3

252s

ⴢ

3 1525s2

3 2 5s2 3

25s2

Because 25 ⫽ 52 , one additional factor of 5 is needed to form a perfect cube. Two additional factors of s are needed to make a perfect cube. 3 Multiply numerator and denominator by 2 5s2 .

3 3 3 2 5s 3 152 5s2 5s

Simplify the perfect cube.

TIP: In the expression

3

3 15 2 5s2 ⫽ 5s

Reduce to lowest terms.

1

3

⫽

325s2 s

3

15 25s2 , the factor of 15 5s and the factor of 5 may be reduced because both are outside the radical. 3 3 15 2 5s2 15 2 5s2 ⫽ ⴢ 5s 5 s

Skill Practice Simplify the expression. 9.

18

⫽

3

23y 2

3 32 5s2 s

4. Rationalizing the Denominator—Two Terms Example 7 demonstrates how to rationalize a two-term denominator involving square roots. First recall from the multiplication of polynomials that the product of two conjugates results in a difference of squares. 1a ⫹ b21a ⫺ b2 ⫽ a2 ⫺ b2 If either a or b has a square root factor, the expression will simplify without a radical. That is, the expression is rationalized. For example: 12 ⫹ 16212 ⫺ 162 ⫽ 122 2 ⫺ 1 162 2 ⫽4⫺6 ⫽ ⫺2

Answers 8.

216 3

9.

613 9y y

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Chapter 6 Radicals and Complex Numbers

Rationalizing the Denominator—Two Terms

Example 7

Simplify the expression by rationalizing the denominator.

⫺2 2 ⫹ 16

Solution: ⫺2 2 ⫹ 16 ⫽

Avoiding Mistakes When constructing the conjugate of an expression, change only the sign between the terms (not the sign of the leading term).

1⫺22

12 ⫹ 162

ⴢ

12 ⫺ 162

Multiply the numerator and denominator by the conjugate of the denominator.

12 ⫺ 162

conjugates

⫺212 ⫺ 162

122 2 ⫺ 1 162 2

In the denominator, apply the formula 1a ⫹ b21a ⫺ b2 ⫽ a2 ⫺ b2.

⫽

⫺212 ⫺ 162 4⫺6

Simplify.

⫽

⫺212 ⫺ 162 ⫺2

⫽

⫺212 ⫺ 162 ⫺2

⫽

⫽ 2 ⫺ 16 Skill Practice Simplify by rationalizing the denominator. 10.

8 3 ⫹ 15

Example 8

Rationalizing the Denominator—Two Terms

Rationalize the denominator.

Solution:

14 ⫹ 1x2 1 1x ⫺ 72

⫽

Answers 10. 6 ⫺ 215 y ⫹ 41y ⫺ 32 11. 64 ⫺ y

ⴢ

4 ⫹ 1x 1x ⫺ 7

1 1x ⫹ 72 1 1x ⫹ 72

1421 1x2 ⫹ 142172 ⫹ 1 1x21 1x2 ⫹ 1 1x2172 1 1x2 ⫺ 172 2

⫽

41x ⫹ 28 ⫹ x ⫹ 71x x ⫺ 49

⫽

x ⫹ 111x ⫹ 28 x ⫺ 49

2

Skill Practice Rationalize the denominator. 1y ⫺ 4 11. 8 ⫺ 1y

Multiply numerator and denominator by the conjugate of the denominator.

Apply the distributive property. Multiply the radicals. Simplify.

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Section 6.6

312 ⫹ 215 12 ⫺ 415

Rationalize the denominator.

13 12 ⫹ 2152 1 12 ⫺ 4152

⫽

ⴢ

537

Rationalizing the Denominator—Two Terms

Example 9

Solution:

Division of Radicals and Rationalization

1 12 ⫹ 4152

Multiply numerator and denominator by the conjugate of the denominator.

1 12 ⫹ 4152

13122 ⴢ 1 122 ⫹ 1312214152 ⫹ 121521 122 ⫹ 1215214152 1 122 ⫺ 14152 2

⫽

6 ⫹ 12110 ⫹ 2110 ⫹ 40 2 ⫺ 80

⫽

46 ⫹ 14110 ⫺78

or

⫺

2

Apply the distributive property.

Multiply the radicals.

46 ⫹ 14110 78

2 123 ⫹ 71102 1

⫽⫺

Factor and simplify.

78 39

⫽⫺

23 ⫹ 7110 39

Skill Practice Rationalize the denominator. 512 ⫺ 215 12. 12 ⫺ 15

Answer 12. ⫺ 110

Section 6.6 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

For the exercises in this set, assume that all variables represent positive real numbers unless otherwise stated.

Study Skills Exercise 1. Define the key term rationalizing the denominator.

Review Exercises

2. Simplify. 14x2y4 2 1/2 164y⫺3 2 1/3

For Exercises 3–10, perform the indicated operations. 3. 2y245 ⫹ 3220y2

4. 3x272x ⫺ 9250x3

5. 1⫺61y ⫹ 32131y ⫹ 12

6. 1 1w ⫹ 122121w ⫺ 42

7. 18 ⫺ 1t2 2

8. 1 1p ⫹ 42 2

9. 1 12 ⫹ 1721 12 ⫺ 172 10. 1 13 ⫹ 521 13 ⫺ 52

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538

Chapter 6 Radicals and Complex Numbers

Concept 2: Division Property of Radicals For Exercises 11–22, simplify using the division property of radicals. Assume all variables represent positive real numbers. (See Examples 1–2.) 11.

15.

19.

49x4 B y6 3

⫺16 j 3

B

k3

4 2 3b3 4 2 48b11

12.

16.

20.

100p2

13.

B q

8

32x B y10 5

17.

3 2 128wz8

21.

3 2 2wz2

8a2 B x6

14.

272ab5

18.

28ab 23yz2

22.

2w4

4w3 B 25y4 26x3 224x 250x3z 29y4

Concept 3: Rationalizing the Denominator—One Term The radical expressions in Exercises 23–30 have radicals in the denominator. Fill in the missing radicands to rationalize the denominators. (See Example 3.) 23.

x x 1? ⫽ ⴢ 15 15 1?

24.

27.

8 8 1?? ⫽ ⴢ 13z 13z 1??

28.

2 2 1? ⫽ ⴢ 1x 1x 1? 10 27w

⫽

10 27w

ⴢ

25. 2?? 2??

29.

7 3

1x

⫽

1 4 2 8a2

7 3

1x ⫽

ⴢ

3 1 ?

26.

3

1?

1 4 2 8a2

ⴢ

4 2 ?? 4 2 ??

30.

5 4

1y

⫽

1 3 2 9b2

5 4

1y ⫽

For Exercises 31–58, rationalize the denominator. (See Examples 4–6.) 31.

1 13

32.

1 17

33.

1 Bx

34.

1 Bz

35.

6 12y

36.

9 13t

37.

a3 B2

38.

b3 B3

39.

6 18

40.

2 148

41.

43.

47.

51.

55.

⫺6

44.

4

1x 4 B w2 3

48.

2 3

24x

2

1 2x

7

52.

56.

⫺2

45.

5

1y 5 B z2 3

6 3

23y

2

1 2y

5

3

42.

3

22 7

46.

3

14

2 3

27 1 3

19

49.

16 B3

50.

81 B8

53.

8 7124

54.

5 3150

57.

4

2 28x

5

58.

4

6 227t7

Concept 4: Rationalizing the Denominator—Two Terms 59. What is the conjugate of 12 ⫺ 16?

60. What is the conjugate of 111 ⫹ 15?

61. What is the conjugate of 1x ⫹ 23?

62. What is the conjugate of 17 ⫺ 1y?

ⴢ

4 1 ? 4 1 ?

1 3 2 9b2

ⴢ

3 2 ?? 3 2 ??

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Section 6.6

Division of Radicals and Rationalization

539

For Exercises 63–82, rationalize the denominators. (See Examples 7–9.) 63.

4 12 ⫹ 3

64.

6 4 ⫺ 13

65.

8 16 ⫺ 2

66.

⫺12 15 ⫺ 3

67.

17 13 ⫹ 2

68.

18 13 ⫹ 1

69.

⫺1 1p ⫹ 1q

70.

6 1a ⫺ 1b

71.

x⫺5 1x ⫹ 15

72.

73.

1w ⫹ 2 9 ⫺ 1w

74.

10 ⫺ 1t 1t ⫺ 6

75.

31x ⫺ 1y 1y ⫹ 1x

76.

21a ⫹ 1b 1b ⫺ 1a

77.

3110 2 ⫹ 110

78.

417 3 ⫹ 17

79.

213 ⫹ 17 313 ⫺ 17

80.

512 ⫺ 15 512 ⫹ 15

81.

15 ⫹ 4 2 ⫺ 15

82.

3 ⫹ 12 12 ⫺ 5

y⫺2 1y ⫺ 12

Mixed Exercises For Exercises 83–86, write the English phrase as an algebraic expression. Then simplify the expression. 83. 16 divided by the cube root of 4 84. 21 divided by the principal fourth root of 27 85. 4 divided by the difference of x and the principal square root of 2 86. 8 divided by the sum of y and the principal square root of 3 87. The time T1x2 (in seconds) for a pendulum to make one complete swing back and forth is approximated by x A 32

T1x2 ⫽ 2p

where x is the length of the pendulum in feet. Determine the exact time required for one swing for a pendulum that is 1 ft long. Then approximate the time to the nearest hundredth of a second. 88. An object is dropped off a building x meters tall. The time T1x2 (in seconds) required for the object to hit the ground is given by T1x2 ⫽

10x A 49

Find the exact time required for the object to hit the ground if it is dropped off the First National Plaza Building in Chicago, a height of 230 m. Then round the time to the nearest hundredth of a second.

Expanding Your Skills For Exercises 89–94, simplify each term of the expression. Then add or subtract as indicated. 89.

16 1 ⫹ 2 16

90.

92.

6 2 ⫺ 212 ⫹ B2 B6

3 93. 1 25 ⫹

1 ⫹ 17 17 3 3 1 5

91. 115 ⫺

94.

1 3 1 4

3 5 ⫹ A5 A3

3 ⫹ 1 54

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540

Chapter 6 Radicals and Complex Numbers

For Exercises 95–102, rationalize the numerator by multiplying both numerator and denominator by the conjugate of the numerator. 95.

13 ⫹ 6 2

99.

15 ⫹ 3h ⫺ 15 h

96.

100.

Section 6.7

17 ⫺ 2 5

97.

17 ⫹ 2h ⫺ 17 h

101.

1a ⫺ 1b 1a ⫹ 1b

98.

14 ⫹ 5h ⫺ 2 h

102.

1p ⫹ 1q 1p ⫺ 1q 19 ⫹ 4h ⫺ 3 h

Solving Radical Equations

Concepts

1. Solutions to Radical Equations

1. Solutions to Radical Equations 2. Solving Radical Equations Involving One Radical 3. Solving Radical Equations Involving More than One Radical 4. Applications of Radical Equations and Functions

An equation with one or more radicals containing a variable is called a radical 3 n equation. For example, 1 x ⫽ 5 is a radical equation. Recall that 1 1 a2 n ⫽ a, provided n 1a is a real number. The basis of solving a radical equation is to eliminate the radical by raising both sides of the equation to a power equal to the index of the radical. 3 To solve the equation 1 x ⫽ 5, cube both sides of the equation. 3 1 x⫽5

3 11 x2 3 ⫽ 152 3

x ⫽ 125 By raising each side of a radical equation to a power equal to the index of the radical, a new equation is produced. Note, however, that some of (or all) the solutions to the new equation may not be solutions to the original radical equation. For this reason, it is necessary to check all potential solutions in the original equation. For example, consider the equation x ⫽ 4. By squaring both sides we produce a quadratic equation. x⫽4 1x2 2 ⫽ 142 2

Square both sides.

x2 ⫽ 16 x2 ⫺ 16 ⫽ 0

1x ⫺ 421x ⫹ 42 ⫽ 0 x⫽4

or

x ⫽ ⫺4

Squaring both sides produces a quadratic equation. Solving this equation, we find two solutions. However, the solution ⫺4 does not check in the original equation. (⫺4 does not check and is called an extraneous solution.)

PROCEDURE Solving a Radical Equation Step 1 Isolate the radical. If an equation has more than one radical, choose one of the radicals to isolate. Step 2 Raise each side of the equation to a power equal to the index of the radical. Step 3 Solve the resulting equation. If the equation still has a radical, repeat steps 1 and 2. *Step 4 Check the potential solutions in the original equation. *Extraneous solutions can arise only when both sides of the equation are raised to an even power. Therefore, an equation with odd-index roots will not have an extraneous solution. However, it is still recommended that you check all potential solutions regardless of the type of root.

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Section 6.7

Solving Radical Equations

2. Solving Radical Equations Involving One Radical Solving an Equation Containing One Radical

Example 1

Solve the equation.

1p ⫹ 5 ⫽ 9

Solution: 1p ⫹ 5 ⫽ 9 1p ⫽ 4

Isolate the radical.

1 1p2 ⫽ 4 2

2

Because the index is 2, square both sides.

p ⫽ 16 Check: p ⫽ 16

Check p ⫽ 16 as a potential solution.

1p ⫹ 5 ⫽ 9 116 ⫹ 5 ⱨ 9 4⫹5ⱨ9✔

True, 16 is a solution to the original equation.

The solution set is 5166.

Skill Practice Solve. 1. 1x ⫺ 3 ⫽ 2

Solving an Equation Containing One Radical

Example 2

Solve the equation.

3 1 w⫺1⫺2⫽2

Solution: 3 1 w⫺1⫺2⫽2 3 1 w⫺1⫽4

1 1w ⫺ 12 ⫽ 142 3

3

Isolate the radical. 3

w ⫺ 1 ⫽ 64

Because the index is 3, cube both sides. Simplify.

w ⫽ 65 Check: w ⫽ 65 3 1 65 ⫺ 1 ⫺ 2 ⱨ 2

Check w ⫽ 65 as a potential solution.

164 ⫺ 2 ⱨ 2 3

4⫺2ⱨ2✔

True, 65 is a solution to the original equation.

The solution set is 5656.

Skill Practice Solve. 3 2. 1 t⫹2⫹5⫽3 Answers 1. 5256

2. 5⫺106

541

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Chapter 6 Radicals and Complex Numbers

Solving an Equation Containing One Radical

Example 3

7 ⫽ 1x ⫹ 32 1/4 ⫹ 9

Solve the equation.

Solution:

7 ⫽ 1x ⫹ 32 1/4 ⫹ 9

4 Note that 1x ⫹ 32 1/4 ⫽ 1 x ⫹ 3.

4 7⫽ 1 x⫹3⫹9

TIP: After isolating the radical in Example 3, the equation shows a fourth root equated to a negative number: 4 ⫺2 ⫽ 1 x⫹3

By definition, a principal fourth root of any real number must be nonnegative. Therefore, there can be no real solution to this equation.

4 ⫺2 ⫽ 1 x⫹3

Isolate the radical.

1⫺22 4 ⫽ 1 1x ⫹ 32 4 4

Because the index is 4, raise both sides to the fourth power.

16 ⫽ x ⫹ 3 x ⫽ 13

Solve for x.

Check: x ⫽ 13 4 7⫽ 1 x⫹3⫹9

4 7ⱨ 1 1132 ⫹ 3 ⫹ 9 4 7ⱨ 1 16 ⫹ 9

7 ⱨ 2 ⫹ 9 (false)

13 is not a solution to the original equation.

4 The equation 7 ⫽ 1x ⫹ 3 ⫹ 9 has no solution.

The solution set is the empty set, 5 6. Skill Practice Solve. 3. 3 ⫽ 6 ⫹ 1x ⫺ 12 1/4

Solving an Equation Containing One Radical

Example 4

y ⫹ 214y ⫺ 3 ⫽ 3

Solve the equation.

Solution: y ⫹ 214y ⫺ 3 ⫽ 3 214y ⫺ 3 ⫽ 3 ⫺ y

1214y ⫺ 32 ⫽ 13 ⫺ y2 2

Isolate the radical term. 2

Because the index is 2, square both sides.

414y ⫺ 32 ⫽ 9 ⫺ 6y ⫹ y2

Note that 1214y ⫺ 32 2 ⫽ 22 1 14y ⫺ 32 2 and 13 ⫺ y2 2 ⫽ 13 ⫺ y213 ⫺ y2 ⫽ 9 ⫺ 6y ⫹ y2.

16y ⫺ 12 ⫽ 9 ⫺ 6y ⫹ y2 0 ⫽ y2 ⫺ 22y ⫹ 21 0 ⫽ 1y ⫺ 2121y ⫺ 12 y ⫺ 21 ⫽ 0 Answer

3. 5 6 (The value 82 does not check.)

y ⫽ 21

or

y⫺1⫽0

or

y⫽1

The equation is quadratic. Set one side equal to zero. Write the other side in descending order. Factor. Set each factor equal to zero. Solve.

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Section 6.7

Solving Radical Equations

Check: y ⫽ 1

Check: y ⫽ 21 y ⫹ 214y ⫺ 3 ⫽ 3

y ⫹ 214y ⫺ 3 ⫽ 3

21 ⫹ 2141212 ⫺ 3 ⱨ 3

1 ⫹ 214112 ⫺ 3 ⱨ 3

21 ⫹ 2181 ⱨ 3

1 ⫹ 211 ⱨ 3

21 ⫹ 18 ⱨ 3

1⫹2ⱨ3 3ⱨ3✔

39 ⱨ 3 False

The solution set is 516 (the value 21 does not check). Skill Practice Solve. 4. 21m ⫹ 3 ⫺ m ⫽ 3

3. Solving Radical Equations Involving More than One Radical Example 5

Solving an Equation with Two Radicals

Solve the equation. 3 3 1 2x ⫺ 4 ⫽ 1 1 ⫺ 8x

Solution: 3 3 1 2x ⫺ 4 ⫽ 1 1 ⫺ 8x

3 3 11 2x ⫺ 42 3 ⫽ 1 1 1 ⫺ 8x2 3

2x ⫺ 4 ⫽ 1 ⫺ 8x 10x ⫺ 4 ⫽ 1

Because the index is 3, cube both sides. Simplify. Solve the resulting equation.

10x ⫽ 5 x⫽ Check: x ⫽

1 2

Solve for x.

1 2

3 3 1 2x ⫺ 4 ⫽ 1 1 ⫺ 8x 3

B

1 1 2a b ⫺ 4 ⱨ 3 1 ⫺ 8a b 2 B 2 3 3 1 1⫺4ⱨ 1 1⫺4

1⫺3 ⱨ 1⫺3 ✔ (True) 3

3

Therefore, 12 is a solution to the original equation.

1 The solution set is e f . 2 Skill Practice Solve. 5 5 5. 1 2y ⫺ 1 ⫽ 1 10y ⫹ 3

Answers 4. 5⫺3, 16

1 5. e ⫺ f 2

543

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544

Chapter 6 Radicals and Complex Numbers

Solving an Equation with Two Radicals

Example 6

Solve the equation.

13m ⫹ 1 ⫺ 1m ⫹ 4 ⫽ 1

Solution: 13m ⫹ 1 ⫺ 1m ⫹ 4 ⫽ 1 13m ⫹ 1 ⫽ 1m ⫹ 4 ⫹ 1

1 13m ⫹ 12 ⫽ 1 1m ⫹ 4 ⫹ 12 2

Isolate one of the radicals. 2

Square both sides.

3m ⫹ 1 ⫽ m ⫹ 4 ⫹ 21m ⫹ 4 ⫹ 1 Note: 1 1m ⫹ 4 ⫹ 12 2

⫽ 1 1m ⫹ 42 2 ⫹ 2112 1m ⫹ 4 ⫹ 112 2

⫽ m ⫹ 4 ⫹ 21m ⫹ 4 ⫹ 1

TIP: In Example 6, we divided the equation by 2 because all coefficients were divisible by 2. This made the coefficients smaller before we squared both sides of the equation a second time.

3m ⫹ 1 ⫽ m ⫹ 5 ⫹ 21m ⫹ 4

Combine like terms.

2m ⫺ 4 ⫽ 21m ⫹ 4

Isolate the radical again.

m ⫺ 2 ⫽ 1m ⫹ 4

1m ⫺ 22 ⫽ 1 1m ⫹ 42 2

Divide both sides by 2. 2

Square both sides again.

m2 ⫺ 4m ⫹ 4 ⫽ m ⫹ 4

The resulting equation is quadratic.

m2 ⫺ 5m ⫽ 0

Set one side equal to zero.

m1m ⫺ 52 ⫽ 0 m⫽0 Check:

or

Factor.

m⫽5

m⫽0

Check: m ⫽ 5

13102 ⫹ 1 ⫺ 1102 ⫹ 4 ⱨ 1

13152 ⫹ 1 ⫺ 1152 ⫹ 4 ⫽ 1

11 ⫺ 14 ⱨ 1

116 ⫺ 19 ⱨ 1

1⫺2ⱨ1

(False)

The solution set is 556 (the value 0 does not check). Skill Practice Solve. 6. 13c ⫹ 1 ⫺ 1c ⫺ 1 ⫽ 2

Answer 6. 51, 56

4⫺3ⱨ1 ✔

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Section 6.7

Solving Radical Equations

4. Applications of Radical Equations and Functions Example 7

Applying a Radical Equation in Geometry

For a pyramid with a square base, the length of a side of the base b is given by b⫽

3V A h

where V is the volume and h is the height. The Pyramid of the Pharoah Khufu (known as the Great Pyramid) at Giza has a square base (Figure 6-3). If the distance around the bottom of the pyramid is 921.6 m and the height is 146.6 m, what is the volume of the pyramid?

h b b

Figure 6-3

Solution: The length of a side b (in meters) is given by b⫽ 230.4 ⫽

3V B h 3V B 146.6

1230.42 2 ⫽ a 53,084.16 ⫽ 153,084.1621146.62 ⫽

921.6 ⫽ 230.4 m. 4

3V 2 b B 146.6

Substitute b ⫽ 230.4 and h ⫽ 146.6. Because the index is 2, square both sides.

3V 146.6

Simplify.

3V 1146.62 146.6

Multiply both sides by 146.6.

153,084.1621146.62 ⫽ 3V

153,084.1621146.62 3V ⫽ 3 3

Divide both sides by 3.

2,594,046 ⬇ V The volume of the Great Pyramid at Giza is approximately 2,594,046 m3. Skill Practice 7. The length of the legs, s, of an isosceles right triangle is s ⫽ 12A, where A is the area. If the legs of the triangle are 9 in., find the area. Answer 7. 40.5 in.2

545

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Chapter 6 Radicals and Complex Numbers

Example 8

Applying a Radical Function

On a certain surface, the speed s(x) (in miles per hour) of a car before the brakes were applied can be approximated from the length of its skid marks x (in feet) by s1x2 ⫽ 3.81x

s(x)

xⱖ0

See Figure 6-4.

Speed of Car Based on Length of Skid Marks

Speed (mph)

80 60 40 20 0 0

50

100 150 200 250 300 350 400 Length of Skid Marks (ft)

x

Figure 6-4

a. Find the speed of a car before the brakes were applied if its skid marks are 361 ft long. b. How long would you expect the skid marks to be if the car had been traveling the speed limit of 50 mph? (Round to the nearest foot.)

Solution: s1x2 ⫽ 3.81x

a.

s13612 ⫽ 3.8 1361

Substitute x ⫽ 361.

⫽ 3.81192 ⫽ 72.2 If the skid marks are 361 ft, the car was traveling approximately 72.2 mph before the brakes were applied. s1x2 ⫽ 3.81x

b.

50 ⫽ 3.81x 50 ⫽ 1x 3.8 a

Substitute s1x2 ⫽ 50 and solve for x. Isolate the radical.

50 2 b ⫽x 3.8 x ⬇ 173

If the car had been going the speed limit (50 mph), then the length of the skid marks would have been approximately 173 ft. Skill Practice When an object is dropped from a height of 64 ft, the time t (x) (in seconds) it takes to reach a height x (in feet) is given by 1 t 1x2 ⫽ 164 ⫺ x 4 Answers 8.

3 sec 2

9. 48 ft

8. Find the time to reach a height of 28 ft from the ground. 9. What is the height after 1 sec?

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Section 6.7

Section 6.7

Solving Radical Equations

Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Radical equation

b. Extraneous solution

Review Exercises 2. Identify the equation as linear or quadratic. Then solve the equation. a. 2x ⫹ 3 ⫽ 23

b. 2x2 ⫺ 9x ⫽ 5

For Exercises 3–6, simplify the radical expressions. Assume all variables represent positive real numbers. 3.

9w3 B 16

4.

a2 B3

3 5. 2 54 c4

6.

49 B 5t 3

For Exercises 7–10, simplify each expression. Assume all radicands represent positive real numbers. 7. 1 14x ⫺ 62 2

8. 1 15y ⫹ 22 2

3 9. 1 1 9p ⫹ 72 3

3 10. 1 1 4t ⫹ 132 3

Concept 2: Solving Radical Equations Involving One Radical For Exercises 11–26, solve the equations. (See Examples 1–3.) 11. 1x ⫹ 4 ⫽ 6

12. 1x ⫹ 2 ⫽ 8

13. 15y ⫹ 1 ⫽ 4

14. 19z ⫺ 5 ⫺ 2 ⫽ 9

15. 6 ⫽ 12z ⫺ 3 ⫺ 3

16. 4 ⫽ 18 ⫹ 3a ⫺ 1

3 17. 2 x⫺2⫺3⫽0

3 18. 2 2x ⫺ 5 ⫺ 2 ⫽ 0

19. 115 ⫺ w2 1Ⲑ3 ⫹ 5 ⫽ 0

20. 1k ⫹ 182 1Ⲑ3 ⫹ 2 ⫽ 0

21. 3 ⫹ 1x ⫺ 16 ⫽ 0

22. 12 ⫹ 12x ⫹ 1 ⫽ 0

23. 216a ⫹ 7 ⫺ 2a ⫽ 0

24. 213 ⫺ w ⫺ w ⫽ 0

25. 12x ⫺ 52 1/4 ⫽ ⫺1

26. 1x ⫹ 162 1/4 ⫽ ⫺4

For Exercises 27–30, assume all variables represent positive real numbers. 3V A 4p 3

27. Solve for V:

r⫽

29. Solve for h2:

r ⫽ p 2r 2 ⫹ h2

28. Solve for V: r ⫽ 30. Solve for d:

V A hp

s ⫽ 1.31d

For Exercises 31–36, square the expression as indicated. 31. 1a ⫹ 52 2

32. 1b ⫹ 72 2

33. 1 15a ⫺ 32 2

34. 12 ⫹ 1b2 2

35. 1 1r ⫺ 3 ⫹ 52 2

36. 12 ⫺ 12t ⫺ 42 2

For Exercises 37–42, solve the radical equations, if possible. (See Example 4.) 37. 2a2 ⫹ 2a ⫹ 1 ⫽ a ⫹ 5

38. 2b2 ⫺ 5b ⫺ 8 ⫽ b ⫹ 7

39. 225w2 ⫺ 2w ⫺ 3 ⫽ 5w ⫺ 4

40. 24p2 ⫺ 2p ⫹ 1 ⫽ 2p ⫺ 3

41. 4 1p ⫺ 2 ⫺ 2 ⫽ ⫺p

42. x ⫺ 31x ⫺ 5 ⫽ 5

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Concept 3: Solving Radical Equations Involving More than One Radical For Exercises 43–66, solve the radical equations, if possible. (See Examples 5–6.) 4 4 43. 1 h⫹4⫽ 1 2h ⫺ 5

4 4 44. 2 3b ⫹ 6 ⫽ 1 7b ⫺ 6

3 3 45. 2 5a ⫹ 3 ⫺ 2 a ⫺ 13 ⫽ 0

3 3 46. 2 k⫺8⫺ 2 4k ⫹ 1 ⫽ 0

47. 15a ⫺ 9 ⫽ 15a ⫺ 3

48. 18 ⫹ b ⫽ 2 ⫹ 1b

49. 12h ⫹ 5 ⫺ 12h ⫽ 1

50. 13k ⫺ 5 ⫺ 13k ⫽ ⫺1

51. 1t ⫺ 9 ⫺ 1t ⫽ 3

52. 1y ⫺ 16 ⫺ 1y ⫽ 4

53. 6 ⫽ 2x2 ⫹ 3 ⫺ x

54. 2 ⫽ 2y2 ⫹ 5 ⫺ y

55. 13t ⫺ 7 ⫽ 2 ⫺ 13t ⫹ 1

56. 1p ⫺ 6 ⫽ 1p ⫹ 2 ⫺ 4

57. 1z ⫹ 1 ⫹ 12z ⫹ 3 ⫽ 1

58. 12y ⫹ 6 ⫽ 17 ⫺ 2y ⫹ 1

59. 16m ⫹ 7 ⫺ 13m ⫹ 3 ⫽ 1

60. 15w ⫹ 1 ⫺ 13w ⫽ 1

61. 2 ⫹ 212t ⫹ 3 ⫹ 213t ⫺ 5 ⫽ 0

62. 6 ⫹ 313x ⫹ 1 ⫹ 31x ⫺ 1 ⫽ 0

63. 31y ⫺ 3 ⫽ 14y ⫹ 3

64. 15x ⫺ 8 ⫽ 21x ⫺ 1

65. 1p ⫹ 7 ⫽ 12p ⫹ 1

66. 1t ⫽ 1t ⫺ 12 ⫹ 2

Concept 4: Applications of Radical Equations and Functions 67. If an object is dropped from an initial height h, its velocity at impact with the ground is given by v ⫽ 12gh where g is the acceleration due to gravity and h is the initial height. (See Example 7.) a. Find the initial height (in feet) of an object if its velocity at impact is 44 ft/sec. (Assume that the acceleration due to gravity is g ⫽ 32 ft/sec2.2 b. Find the initial height (in meters) of an object if its velocity at impact is 26 m/sec. (Assume that the acceleration due to gravity is g ⫽ 9.8 m/sec2.2 Round to the nearest tenth of a meter. 68. The time T (in seconds) required for a pendulum to make one complete swing back and forth is approximated by L Ag

T ⫽ 2p

where g is the acceleration due to gravity and L is the length of the pendulum (in feet). a. Find the length of a pendulum that requires 1.36 sec to make one complete swing back and forth. (Assume that the acceleration due to gravity is g ⫽ 32 ft/sec2.) Round to the nearest tenth of a foot. b. Find the time required for a pendulum to complete one swing back and forth if the length of the pendulum is 4 ft. (Assume that the acceleration due to gravity is g ⫽ 32 ft/sec2.) Round to the nearest tenth of a second. 69. The airline cost for x thousand passengers to travel round trip from New York to Atlanta is given by C1x2 ⫽ 10.3x ⫹ 1 where C(x) is measured in millions of dollars and x ⱖ 0. (See Example 8.)

a. Find the airline’s cost for 10,000 passengers 1x ⫽ 102 to travel from New York to Atlanta. b. If the airline charges $320 per passenger, find the profit made by the airline for flying 10,000 passengers from New York to Atlanta. c. Approximate the number of passengers who traveled from New York to Atlanta if the total cost for the airline was $4 million.

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Solving Radical Equations

549

70. The time t(d) in seconds it takes an object to drop d meters is given by t 1d2 ⫽

d B 4.9

a. Approximate the height of the Texas Commerce Tower in Houston if it takes an object 7.89 sec to drop from the top. Round to the nearest meter. b. Approximate the height of the Sears Tower in Chicago if it takes an object 9.51 sec to drop from the top. Round to the nearest meter.

Expanding Your Skills 71. The number of hours needed to cook a turkey that weighs x pounds can be approximated by 5 3 t 1x2 ⫽ 0.902 x

where t(x) is the time in hours and x is the weight of the turkey in pounds. a. Find the weight of a turkey that cooked for 4 hr. Round to the nearest pound. b. Find t(18) and interpret the result. Round to the nearest tenth of an hour. For Exercises 72–75, use the Pythagorean theorem to find a, b, or c. 72. Find b when a ⫽ 2 and c ⫽ y.

73. Find b when a ⫽ h and c ⫽ 5.

74. Find a when b ⫽ x and c ⫽ 8.

75. Find a when b ⫽ 14 and c ⫽ k.

Graphing Calculator Exercises 76. Graph Y1 and Y2 on a viewing window defined by ⫺10 ⱕ x ⱕ 40 and ⫺5 ⱕ y ⱕ 10. Y1 ⫽ 12x

and

Y2 ⫽ 8

Use an Intersect feature to approximate the x-coordinate of the point of intersection of the two graphs. How does the point of intersection relate to the solution to the equation 12x ⫽ 8? 78. Refer to Exercise 44. Graph Y1 and Y2 on a viewing window defined by ⫺5 ⱕ x ⱕ 20 and ⫺1 ⱕ y ⱕ 4. 4 Y1 ⫽ 1 3x ⫹ 6

and

4 Y2 ⫽ 1 7x ⫺ 6

Use an Intersect feature to approximate the x-coordinate of the point of intersection of the two graphs to support your solution to Exercise 44.

77. Graph Y1 and Y2 on a viewing window defined by ⫺10 ⱕ x ⱕ 20 and ⫺5 ⱕ y ⱕ 10. Y1 ⫽ 14x

and

Y2 ⫽ 6

Use an Intersect feature to approximate the x-coordinate of the point of intersection of the two graphs. How does the point of intersection relate to the solution to the equation 14x ⫽ 6? 79. Refer to Exercise 43. Graph Y1 and Y2 on a viewing window defined by ⫺5 ⱕ x ⱕ 20 and ⫺1 ⱕ y ⱕ 4. 4 Y1 ⫽ 1 x⫹4

and

4 Y2 ⫽ 1 2x ⫺ 5

Use an Intersect feature to approximate the x-coordinate of the point of intersection of the two graphs to support your solution to Exercise 43.

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Chapter 6 Radicals and Complex Numbers

Section 6.8

Complex Numbers

Concepts

1. Definition of i

1. Definition of i 2. Powers of i 3. Definition of a Complex Number 4. Addition, Subtraction, and Multiplication of Complex Numbers 5. Division and Simplification of Complex Numbers

In Section 6.1, we learned that there are no real-valued square roots of a negative number. For example, 19 is not a real number because no real number when squared equals 9. However, the square roots of a negative number are defined over another set of numbers called the imaginary numbers. The foundation of the set of imaginary numbers is the definition of the imaginary number i.

DEFINITION i i  11 Note: From the definition of i, it follows that i2  1. Using the imaginary number i, we can define the square root of any negative real number.

DEFINITION

1ⴚb for b ⬎ 0

Let b be a real number such that b 7 0. Then 1b  i 1b.

Simplifying Expressions in Terms of i

Example 1

Simplify the expressions in terms of i. a. 164

b. 150

c. 14

d. 129

Solution: a. 164  i 164  8i b. 150  i 150  i 252 ⴢ 2

Avoiding Mistakes

 5i 12

In an expression such as i 129, the i is often written in front of the square root. The expression 129 i is also correct, but may be misinterpreted as 129i (with i incorrectly placed under the radical).

c. 14  1 ⴢ 14  1 ⴢ 2i  2i d. 129  i 129

Skill Practice Simplify the expressions in terms of i. 1. 181

2. 120

3. 136

4. 17

The multiplication and division properties of radicals were presented in Sections 6.3 and 6.5 as follows: n n If a and b represent real numbers such that 1 a and 1 b are both real, then n

Answers 1. 9i

2. 2i15

n

3. 6i

4. i 17

n

n

1ab  1a ⴢ 1b

and

1a a  n Ab 1b n

b0

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Complex Numbers

n n The conditions that 1 a and 1 b must both be real numbers prevent us from applying the multiplication and division properties of radicals for square roots with a negative radicand. Therefore, to multiply or divide radicals with a negative radicand, first write the radical in terms of the imaginary number i. This is demonstrated in Example 2.

Example 2

Simplifying a Product or Quotient in Terms of i

Simplify the expressions. a.

1100 125

b. 125 ⴢ 19

c. 15 ⴢ 15

Solution: a.

1100 125 

10i 5i

Simplify each radical in terms of i before dividing.

2 b. 125 # 19  5i # 3i

Simplify each radical in terms of i first before multiplying.

 15i

Multiply.

 15112

Recall that i2  1.

 15

Simplify.

2

c. 15 ⴢ 15  i15 ⴢ i15  i2 ⴢ 1 152 2

 1 ⴢ 5  5 Skill Practice Simplify the expressions. 5.

136 19

6. 116 ⴢ 149

7. 12 ⴢ 12

Avoiding Mistakes In Example 2, we wrote the radical expressions in terms of i first, before multiplying or dividing. If we had mistakenly applied the multiplication or division property first, we would have obtained an incorrect answer. Correct: 125 ⴢ 19

 15i 2 13i 2  15i 2

 15112  15 correct

Be careful: 125 ⴢ 19 Z 1225  15 (incorrect answer)

125 and 19 are not real numbers. Therefore, the multiplication property of radicals cannot be applied.

Answers 5. 2

6. 28

7. 2

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2. Powers of i From the definition of i  11, it follows that i i i2  1

because i3  i2 ⴢ i  112i  i

i3  i

because i4  i2 ⴢ i2  112112  1

i4  1

because i5  i4 ⴢ i  112i  i

i5  i

because i6  i4 ⴢ i2  112112  1

i6  1

This pattern of values i, 1, i, 1, i, 1, i, 1, . . . continues for all subsequent powers of i. Table 6-1 lists several powers of i. Table 6-1

Powers of i

i1  i

i5  i

i9  i

i2  1

i6  1

i10  1

i3  i i4  1

i7  i i8  1

i11  i i12  1

To simplify higher powers of i, we can decompose the expression into multiples of i4 1i4  12 and write the remaining factors as i, i2, or i3.

Simplifying Powers of i

Example 3

Simplify the powers of i. a. i13

b. i18

c. i107

d. i32

Solution:

a. i13  1i12 2 ⴢ 1i2  1i4 2 3 ⴢ 1i2

Write the exponent as a multiple of 4 and a remainder.

 112 3 1i2

Recall that i4  1.

i

Simplify.

b. i18  1i16 2 ⴢ 1i2 2  1i 2 ⴢ 1i 2 4 4

Write the exponent as a multiple of 4 and a remainder.

2

 112 4 ⴢ 112

i4  1 and i2  1

 1

Simplify.

c. i107  1i104 2 ⴢ 1i3 2  1i4 2 26 1i3 2

Write the exponent as a multiple of 4 and a remainder.

 112 26 1i2

i4  1 and i3  i

 i

Simplify.

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Section 6.8

Complex Numbers

d. i32 ⫽ 1i4 2 8 ⫽ 112 8

i4 ⫽ 1

⫽1

Simplify.

Skill Practice Simplify the powers of i. 8. i 45

9. i 22

10. i 31

11. i 80

3. Definition of a Complex Number We have already learned the definitions of the integers, rational numbers, irrational numbers, and real numbers. In this section, we define the complex numbers.

DEFINITION Complex Number A complex number is a number of the form a ⫹ bi, where a and b are real numbers and i ⫽ 1⫺1. Notes: • If b ⫽ 0, then the complex number a ⫹ bi is a real number. • If b ⫽ 0, then we say that a ⫹ bi is an imaginary number. • The complex number a ⫹ bi is said to be written in standard form. The quantities a and b are called the real and imaginary parts (respectively) of the complex number. • The complex numbers a ⫺ bi and a ⫹ bi are called conjugates.

From the definition of a complex number, it follows that all real numbers are complex numbers and all imaginary numbers are complex numbers. Figure 6-5 illustrates the relationship among the sets of numbers we have learned so far. Complex Numbers Imaginary Numbers

Real Numbers Rational Numbers

5i

⫺ 27

Integers . . . , ⫺3, ⫺2, ⫺1

⫺2 ⫹ 4i

0.3

2 17

i 3 4 ⫹ 12 i

0.25

Irrational Numbers

Whole Numbers

␲

0 Natural Numbers 1, 2, 3, . . .

Figure 6-5

Answers 8. i 10. ⫺i

9. ⫺1 11. 1

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Example 4

Identifying the Real and Imaginary Parts of a Complex Number

Identify the real and imaginary parts of the complex numbers. a. 8  2i

b.

3 2

c. 1.75i

Solution: a. 8  2i b.

8 is the real part, and 2 is the imaginary part. Rewrite 32 in the form a  bi. 3 2 is the real part, and 0 is the imaginary part.

3 3   0i 2 2

c. 1.75i  0  1.75i

Rewrite 1.75i in the form a  bi. 0 is the real part, and 1.75 is the imaginary part.

Skill Practice Identify the real and imaginary parts of the complex numbers. 12. 22  14i

13. 50

14. 15i

TIP: Example 4(b) illustrates that a real number is also a complex number. 3 3   0i 2 2

Example 4(c) illustrates that an imaginary number is also a complex number. 1.75i  0  1.75i

4. Addition, Subtraction, and Multiplication of Complex Numbers The operations of addition, subtraction, and multiplication of real numbers also apply to imaginary numbers. To add or subtract complex numbers, combine the real parts and combine the imaginary parts. The commutative, associative, and distributive properties that apply to real numbers also apply to complex numbers. Example 5

Adding and Subtracting Complex Numbers

Add or subtract as indicated. Write the answers in the form a  bi. a. 11  5i2  13  7i2

1 3 1 1 b. a  ib  a  ib 4 5 2 10

Solution: real parts

a. 11  5i2  13  7i2  11  32  15  72i imaginary parts

 2  2i Answers 12. real: 22; imaginary: 14 13. real: 50; imaginary: 0 14. real: 0; imaginary: 15

Add the real parts. Add the imaginary parts. Simplify.

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Section 6.8

1 3 1 1 1 3 1 1 b. a  ib  a  ib    i   i 4 5 2 10 4 5 2 10

Complex Numbers

Apply the distributive property.

1 1 3 1  a  b  a  b i 4 2 5 10

Add real parts. Add imaginary parts.

2 6 1 1  a  b  a  b i 4 4 10 10

Get common denominators.

3 7   i 4 10

Simplify.

Skill Practice Perform the indicated operations. 1 1 3 2 15. a  ib  a  ib 2 4 5 3

Example 6

16. 16  11i 2  19  12i 2

Multiplying Complex Numbers

Multiply the complex numbers. Write the answers in the form a  bi. a. 110  5i2 12  3i 2

b. 11.2  0.5i2 11.2  0.5i2

Solution: a. 110  5i2 12  3i2  1102 122  110 2 13i2  15i2 122  15i2 13i2

Apply the distributive property.

 20  30i  10i  15i2  20  20i  1152 112

Recall i2  1.

 20  20i  15  35  20i

Write in the form a  bi.

b. 11.2  0.5i2 11.2  0.5i2

The expressions 11.2  0.5i2 and 11.2  0.5i2 are conjugates. The product is a difference of squares. 1a  b21a  b2  a2  b2

11.2  0.5i211.2  0.5i2  11.22 2  10.5i2 2  1.44  0.25i2

 1.44  0.2511 2

Apply the formula, where a  1.2 and b  0.5i. Recall i2  1.

 1.44  0.25  1.69  0i Skill Practice Multiply. 17. 14  6i 212  3i 2

18. 11.5  0.8i 211.5  0.8i 2

Answers 11 5  i 10 12 17. 10  24i 15.

16. 3  23i 18. 2.89  0i

555

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Chapter 6 Radicals and Complex Numbers

5. Division and Simplification of Complex Numbers The product of a complex number and its conjugate is a real number. For example: 15  3i215  3i2  25  9i2

 25  9112  25  9  34 To divide by a complex number, multiply the numerator and denominator by the conjugate of the denominator. This produces a real number in the denominator so that the resulting expression can be written in the form a  bi. Example 7

Dividing by a Complex Number

Divide the complex numbers and write the answer in the form a  bi. 4  3i 5  2i

Solution: 4  3i 5  2i

Multiply the numerator and denominator by the conjugate of the denominator:

14  3i2 15  2i2 142152  14212i2  13i2152  13i212i2 ⴢ  15  2i2 15  2i2 152 2  12i2 2

TIP: In Example 7, we asked for the answer in the form a  bi. However, if the second term is negative, we often leave an answer in terms of subtraction: 14 23 29  29 i. This is the same as 14 23 29  129 i2 .



20  8i  15i  6i2 25  4i2

Simplify numerator and denominator.



20  23i  6112 25  4112

Recall i2  1.



20  23i  6 25  4



14  23i 29

Simplify.



23 14  i 29 29

Write in the form a  bi.

Skill Practice Divide the complex numbers. Write the answer in the form a  bi. 19.

Answer 19.

4 7  i 13 13

2i 3  2i

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Section 6.8

Complex Numbers

Simplifying a Complex Number

Example 8

Simplify the complex number.

6 ⫹ 1⫺18 9

Solution: 6 ⫹ 1⫺18 6 ⫹ i 118 ⫽ 9 9

Write the radical in terms of i.

⫽

6 ⫹ 3i12 9

Simplify 118 ⫽ 312.

⫽

312 ⫹ i122 9

Factor the numerator.

3 12 ⫹ i122 1

⫽

Simplify.

9

3

⫽

2 ⫹ i12 3

or

2 12 ⫹ i 3 3

Write in the form a ⫹ bi.

Skill Practice Simplify the complex number. 20.

Answer

8 ⫺ 1⫺24 6

20.

4 ⫺ i 16 4 16 or ⫺ i 3 3 3

Section 6.8 Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Imaginary numbers

c. Complex number

b. i

d. Conjugate

Review Exercises For Exercises 2–4, perform the indicated operations. 2. ⫺215 ⫺ 3150 ⫹ 1125

3. 13 ⫺ 1x213 ⫹ 1x2

4. 1 15 ⫹ 122 2

6. 29t ⫹ 10 ⫹ 18 ⫽ 10

7. 14a ⫹ 29 ⫽ 21a ⫹ 5

For Exercises 5–7, solve the equations. 3 3 5. 13p ⫹ 7 ⫺ 12p ⫺ 1 ⫽ 0

8. Rationalize the denominator.

2 1x ⫺ 3

Concept 1: Definition of i 9. Simplify the expressions 1⫺1 and ⫺11.

10. Simplify i2.

557

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For Exercises 11–30, simplify the expressions. (See Examples 1–2.) 11. 1144

12. 181

13. 13

14. 117

15. 120

16. 175

17. 12 125213142

18. 1419213112

19. 7163  4128

20. 713  4127

21. 17 ⴢ 17

22. 111 ⴢ 111

23. 19 ⴢ 116

24. 125 ⴢ 136

25. 115 ⴢ 16

26. 112 ⴢ 150

27.

150 125

28.

127 19

29.

190 110

30.

1125 145

Concept 2: Powers of i For Exercises 31–42, simplify the powers of i. (See Example 3.) 31. i7

32. i38

33. i64

34. i75

35. i41

36. i25

37. i52

38. i0

39. i23

40. i103

41. i6

42. i82

Concept 3: Definition of a Complex Number 43. What is the conjugate of a complex number a  bi? 44. True or false? a. Every real number is a complex number.

b. Every complex number is a real number.

For Exercises 45–52, identify the real and imaginary parts of the complex number. (See Example 4.) 45. 5  12i

46. 22  16i

47. 6i

49. 35

50. 1

51.

48. 10i

3 i 5

1 1 52.   i 2 4

Concept 4: Addition, Subtraction, and Multiplication of Complex Numbers For Exercises 53–76, perform the indicated operations. Write the answer in the form a  bi. (See Examples 5–6.) 53. 12  i2  15  7i2

54. 15  2i2  13  4i2

1 2 1 5 55. a  ib  a  ib 2 3 5 6

56. a

57. 11  3i2  14  3i2

58. 12  i2  11  i2

11 7 2 3  ib  a  ib 10 5 5 5

59. 12  3i2  11  4i2  12  3i2

60. 12  5i2  17  2i2  13  4i2

61. 18i213i2

62. 12i214i2

63. 6i11  3i2

64. i13  4i2

65. 12  10i213  2i2

66. 14  7i212  3i2

67. 15  2i215  2i2

68. 14  11i214  11i2

69. 14  5i2 2

70. 13  2i2 2

71. 12  i213  2i214  3i2

72. 13  i213  i214  i2

73. 14  6i2 2

74. 13  5i2 2

1 3 1 3 1 2 1 2 75. a  ib a  ib 76. a  ib a  ib 2 4 2 4 3 6 3 6

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Group Activity

559

Concept 5: Division and Simplification of Complex Numbers For Exercises 77–90, divide the complex numbers. Write the answer in the form a  bi.

(See Example 7.)

77.

2 1  3i

78.

2 3i

79.

i 4  3i

80.

3  3i 1i

81.

5  2i 5  2i

82.

7  3i 4  2i

83.

3  7i 2  4i

84.

2  9i 1  4i

85.

13i 5  i

86.

15i 2  i

87.

2  3i 6i

88.

4i 2i

89.

10  i i

90.

6  i i

(Hint: The denominator can be written as 0  6i. Therefore, the conjugate is 0  6i, or simply 6i.)

For Exercises 91–98, simplify the complex numbers. Write the answer in the form a  bi. (See Example 8.) 91.

2  116 8

92.

6  14 4

93.

6  172 6

94.

20  1500 10

95.

8  148 4

96.

18  172 3

97.

5  150 10

98.

14  198 7

Expanding Your Skills For Exercises 99–102, determine if the complex number is a solution to the equation. 99. x2  4x  5  0; 101. x2  12  0;

2i

2i13

100. x2  6x  25  0; 102. x2  18  0;

3  4i

3i 12

Group Activity Margin of Error of Survey Results Materials: Calculator Estimated time: 20 minutes Group Size: 3 The members of the class will conduct a survey to estimate the percentage of the college population that answers “yes” to the several survey questions asked. 1. The members of the class will decide on three “Yes or No” questions that they will use to perform a survey. For example, here are some possible questions. “Do you study more than 10 hr a week?” “Do you work more than 25 hr a week?” “Are you taking more than 10 credit hours?”

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List the questions here: a. b. c.

2. Each student in the class will survey 20 people and record the number of “yes” responses in the table. Question

Number of “Yes” Responses

1 2 3

3. The instructor will then pool the results obtained from each student to get the total number of people surveyed by the whole class. Record the number of people surveyed (sample size) and the number of “yes” responses for each question in the table. Once the data are entered into the table, compute the percent, p, of “yes” responses in the fourth column. Write this value in decimal form. Round the values of p to three decimal places. (Note: For now, leave the last column blank.) Question

Sample Size

Number of “Yes” Responses

Percent “Yes” Responses, p

Margin of Error, E

1 2 3

4. The value of p represents the percent (written in decimal form) of people in the sample who answered “yes.” However, the percentage of people in the entire college population who would answer “yes” is unknown. The value of p is only an estimate. Statisticians often compute a margin of error associated with such an estimate by using the following formula. E  1.96

p11  p2 B

n

where E is the margin of error. n is the sample size. p is the percent (in decimal form) of “yes” responses.

Compute the margin of error for each of the three questions. Round the values of E to three decimal places. Record the results in the table.

5. The margin of error computed in step 4 is associated with what statisticians call a 95% level of confidence. To interpret the results, a statistician would say “With 95% confidence, the researcher estimates the percentage of “yes” responses from the entire college population to be between p  E.” Interpret the results from each of the three questions.

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Summary

Chapter 6

Summary

Section 6.1

Definition of an nth root

Key Concepts

Examples

b is an nth root of a if b ⫽ a. n

Example 1 2 is a square root of 4. ⫺2 is a square root of 4. ⫺3 is a cube root of ⫺27.

The expression 1a represents the principal square root of a. n The expression 1 a represents the principal nth root of a.

Example 2 136 ⫽ 6

3 1 ⫺64 ⫽ ⫺4

Example 3 1an ⫽ 0a 0 if n is even.

4 2 1x ⫹ 32 4 ⫽ 0 x ⫹ 3 0

n n

1an ⫽ a if n is odd.

Example 4

n

1a is not a real number if a 6 0 and n is even. f 1x2 ⫽ 1x defines a radical function. n

The Pythagorean Theorem

4 1 ⫺16 is not a real number.

Example 5

For g 1x2 ⫽ 1x the domain is 30, ⬁2 .

3 For h 1x2 ⫽ 1 x the domain is 1⫺⬁, ⬁2 .

Example 6

a2 ⫹ b2 ⫽ c2 x a

5 2 1x ⫹ 32 5 ⫽ x ⫹ 3

c

12

x2 ⫹ 122 ⫽ 132 b

13

x2 ⫹ 144 ⫽ 169 x2 ⫽ 25 x ⫽ 125 x⫽5

561

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Chapter 6 Radicals and Complex Numbers

Section 6.2

Rational Exponents

Key Concepts

Examples

Let a be a real number and n be an integer such that n n 7 1. If 1 a exists, then

Example 1

a

n

1 /n

⫽ 2a

m/n

⫽ Q 2a R ⫽ 2am

a

n

m

Example 2

n

3 272Ⲑ3 ⫽ Q 1 27 R ⫽ 132 2 ⫽ 9 2

All properties of integer exponents hold for rational exponents, provided the roots are real-valued.

Section 6.3

Example 3 p1/3 ⴢ p1/4

x4/3 x1/3

1y⫺1/2 2 6

⫽ p1/3⫹1/4

⫽ x4/3⫺1/3

⫽ y1⫺1/22162

⫽ p4/12⫹3/12

⫽ x3/3

⫽ y⫺3

⫽ p7/12

⫽x

⫽

Simplifying Radical Expressions

Key Concepts

Examples n

Let a and b represent real numbers such that 1a and n 1 b are both real. Then n n n 1 ab ⫽ 1 aⴢ 1 b

1211Ⲑ2 ⫽ 1121 ⫽ 11

Multiplication property

Example 1 112 ⫽ 14 ⴢ 3 ⫽ 14 ⴢ 13 ⫽ 213

A radical expression whose radicand is written as a product of prime factors is in simplified form if all the following conditions are met: 1. The radicand has no factor raised to a power greater than or equal to the index. 2. The radicand does not contain a fraction. 3. No radicals are in the denominator of a fraction.

Example 2 3 2 16x5y7 3 4 5 7 ⫽ 2 2xy 3 3 3 6 ⫽ 2 2 x y ⴢ 2x2y 3 3 3 6 3 ⫽ 2 2xy ⴢ 2 2x2y 3 ⫽ 2xy2 2 2x2y

1 y3

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Summary

Section 6.4

Addition and Subtraction of Radicals

Key Concepts

Examples

Like radicals have radical factors with the same index and the same radicand.

Example 1

Use the distributive property to add and subtract like radicals.

2x17  5x17  x17  12  5  12x17  2x 17 Example 2 4 4 5 x2 16x  32 x 4 4  2x1 x  3x1 x

4  12  32 x1 x 4  x1 x

Section 6.5

Multiplication of Radicals

Key Concepts

Examples

The Multiplication Property of Radicals

Example 1

n

n

If 1a and 1b are real numbers, then n

n

n

1a ⴢ 1b  1ab

3121 12  517  162  314  15114  3112  3 ⴢ 2  15114  3 ⴢ 213  6  15 114  613

To multiply or divide radicals with different indices, convert to rational exponents and use the properties of exponents.

Example 2 5 2 1p ⴢ 2 p

 p1/2 ⴢ p2/5  p5/10 ⴢ p4/10  p9/10 10

 2p9

563

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Chapter 6 Radicals and Complex Numbers

Section 6.6

Division of Radicals and Rationalization

Key Concepts

Examples

The Division Property of Radicals

Example 1

n

n

If 1a and 1b are real numbers, then n

a 1a  n Bb 1b n

b0

Simplify. 4x5 B y4  

The process of removing a radical from the denominator of an expression is called rationalizing the denominator. • Rationalizing a denominator with one term

24x5 2y4 2x 2 1x y2

Example 2 Rationalize the denominator. 4 3 2 t

  • Rationalizing a denominator with two terms involving square roots

4 3 2 t

ⴢ

3 2 42 t 3 3 2 t

3 2 2 t 3 2 2 t



3 2 42 t t

Example 3 Rationalize the denominator. 12  31x 1x  12 

1 12  31x2 1 1x  122

ⴢ

1 1x  122 1 1x  122



12 ⴢ 1x  22 ⴢ 12  31x ⴢ 1x  31x ⴢ 12 11x2 2  1 122 2



12x  2  3x  312x x2



3x  212x  2 x2

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Summary

Section 6.7

Solving Radical Equations

Key Concepts

Examples

Steps to Solve a Radical Equation

Example 1

1. Isolate the radical. If an equation has more than one radical, choose one of the radicals to isolate. 2. Raise each side of the equation to a power equal to the index of the radical. 3. Solve the resulting equation. If the equation still has a radical, repeat steps 1 and 2. 4. Check the potential solutions in the original equation.

Solve: 1b  5  1b  3  2 1b  5  1b  3  2

1 1b  52 2  1 1b  3  22 2 b  5  b  3  41b  3  4 b  5  b  7  41b  3 12  41b  3 3  1b  3

132 2  1 1b  32 2 9b3 6b Check: 16  5  16  3 ⱨ 2 11  19 ⱨ 2 13ⱨ2

No solution, 5 6

1false2

565

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Chapter 6 Radicals and Complex Numbers

Section 6.8

Complex Numbers

Key Concepts i  11

and

Examples i  1 2

For a real number b 7 0, 1b  i1b A complex number is in the form a  bi, where a and b are real numbers. The value a is called the real part, and b is called the imaginary part. To add or subtract complex numbers, combine the real parts and combine the imaginary parts.

Example 1 14 ⴢ 19  12i213i2  6i2  6 Example 2 13  5i2  12  i2  13  2i2  3  5i  2  i  3  2i  4  8i

Multiply complex numbers by using the distributive property.

Example 3 11  6i212  4i2  2  4i  12i  24i2  2  16i  24112  22  16i

Divide complex numbers by multiplying the numerator and denominator by the conjugate of the denominator.

Example 4 3 2  5i 

12  5i2 3  12  5i2 12  5i2



6  15i 4  25i2



6  15i 4  25



6  15i 29



6 15  i 29 29

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Review Exercises

Chapter 6

Review Exercises

For the exercises in this set, assume that all variables represent positive real numbers unless otherwise stated.

12. Use the Pythagorean theorem to find the length of the third side of the triangle. 17 cm

Section 6.1

15 cm

1. True or false? a. The principal nth root of an even-indexed root is always positive. b. The principal nth root of an odd-indexed root is always positive. 2. Explain why 21⫺ 32 2 ⫽ ⫺3.

a. 2a ⫹ b ⫽ a ⫹ b 2

b. 21a ⫹ b2 2 ⫽ a ⫹ b

50 A 32

4 5. 1625

6. 21 ⫺62 2

7. Evaluate the function values for f 1x2 ⫽ 1x ⫺ 1. a. f 1102

b. f 112

c. f 182

d. Write the domain of f in interval notation. 8. Evaluate the function values for g1t2 ⫽ 15 ⫹ t. a. g1⫺52

b. g1⫺42

c. g142

d. Write the domain of g in interval notation. 9. Write the English expression as an algebraic expression: Four more than the quotient of the cube root of 2x and the principal fourth root of 2x. For Exercises 10–11, simplify the expression. Assume that x and y represent any real number. 10. a. 2x2 4 4 c. 2 x

11. a. 24y2 100

c. 2y100

3 3 b. 2 x

5 d. 2 1x ⫹ 12 5 3 b. 2 27y3 101

d. 2y101

13. Are the properties of exponents the same for rational exponents and integer exponents? Give an example. (Answers may vary.)

15. Explain the process of eliminating a negative exponent from an algebraic expression. For Exercises 16–21, simplify the expressions. Write the answer with positive exponents only.

For Exercises 4–6, simplify the radicals. 4.

Section 6.2

14. In the expression xmⲐn what does n represent?

3. For a 7 0 and b 7 0, are the following statements true or false? 2

567

16. 1⫺1252 1Ⲑ3

17. 16⫺1Ⲑ4

18. a

1 ⫺3Ⲑ4 1 ⫺2Ⲑ3 b ⫺a b 16 8

19. 1b1Ⲑ2 ⴢ b1Ⲑ3 2 12

20. a

x⫺1Ⲑ4y⫺1Ⲑ3z3Ⲑ4

21. a

1Ⲑ3 ⫺1Ⲑ3 2Ⲑ3

2 x

y

b

⫺12

a12b⫺4c7 1/3 b a3b2c4

For Exercises 22–23, rewrite the expressions by using rational exponents. 4 3 22. 2 x

3 23. 2 2y2

For Exercises 24–26, use a calculator to approximate the expressions to four decimal places. 24. 101Ⲑ3

25. 17.82Ⲑ3

5 26. 2 1474

Section 6.3 27. List the criteria for a radical expression to be simplified. For Exercises 28–31, simplify the radicals. 28. 1108

4 5 29. 2 x yz4

3 30. ⫺22 250a3b10

31.

⫺16a4 B 2ab3 3

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Chapter 6 Radicals and Complex Numbers

32. Write an English phrase that describes the following mathematical expressions: (Answers may vary.) a.

b. 1x  12 3

2 Ax

33. An engineering firm made a mistake when building a 14 -mi bridge in the Florida Keys. The bridge was made without adequate expansion joints to prevent buckling during the heat of summer. During mid-June, the bridge expanded 1.5 ft, causing a vertical bulge in the middle. Calculate the height of the bulge h in feet. (Note: 1 mi  5280 ft.) Round to the nearest foot. 1 4

mile

Section 6.5 For Exercises 45–56, multiply the radicals and simplify the answer. 45. 13 ⴢ 112

4 4 46. 1 4ⴢ 1 8

47. 2131 17  31112

48. 3151213  152

49. 121x  32121x  32 50. 1 1y  421 1y  42 51. 1 17y  13x2 2

53. 11z  162121z  7162 54. 131a  1521 1a  2152 3 55. 2 u ⴢ 2u5

h

52. 1213w  52 2

4 56. 12 ⴢ 2 w3

Section 6.6 For Exercises 57–60, simplify the radicals. 57. 59.

3y5 B 25x6 2324w7 24w3

58. 60.

3

16x7y6

B

z9

3 2 3t14 3 2 192t2

Section 6.4 34. Complete the following statement: Radicals may be added or subtracted if . . . For Exercises 35–38, determine whether the radicals may be combined, and explain your answer. 3 35. 1 2x  212x

36. 2  1x

4 4 37. 1 3xy  21 3xy

38. 4132  7150

For Exercises 61–68, rationalize the denominator. 61.

63.

65.

For Exercises 39–42, add or subtract as indicated. 67.

39. 417  217  317 3 3 40. 21 64  31 54  16

3

62.

4 3

29p

2

5 215  210 t3 1t  13

3

5

66.

68.

2 3 1 2x

6 27  25 w7 1w  17

12 x2

42. x216x  422x  5x254x 2

64.

5 A 3w

69. Write the mathematical expression as an English phrase. (Answers may vary.)

41. 150  712  18 3

7 A 2y

2

For Exercises 43–44, answer true or false. If an answer is false, explain why. Assume all variables represent positive real numbers.

Section 6.7 Solve the radical equations in Exercises 70–77, if possible.

43. 5  3 1x  81x

70. 12y  7

44. 1y  1y  12y

3 72. 12w  3  5  2

71. 1a  6  5  0

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Review Exercises

4 4 73. 1 p  12  1 5p  16  0

569

For Exercises 83–86, rewrite the expressions in terms of i.

74. 1t  1t  5  5 75. 18x  1   1x  13

83. 116

84. 15

85. 175 ⴢ 13

86.

76. 22m2  4  29m  0 77. 2x  2  1  22x  5

124 16

For Exercises 87–90, simplify the powers of i.

78. A tower is supported by stabilizing wires. Find the exact length of each wire, and then round to the nearest tenth of a meter.

87. i38

88. i101

89. i19

90. i1000  i1002

For Exercises 91–94, perform the indicated operations. Write the final answer in the form a  bi. 12 m

91. 13  i2  12  4i2

92. 11  6i213  i2

93. 14  3i214  3i2

94. 15  i2 2

For Exercises 95–96, write the expressions in the form a + bi, and determine the real and imaginary parts.

6m

79. The velocity, v(d), of an ocean wave depends on the water depth d as the wave approaches land. v1d2  132d where v(d) is in feet per second and d is in feet.

95.

17  4i 4

96.

16  8i 8

For Exercises 97–100, divide and simplify. Write the final answer in the form a  bi.

a. Find v(20) and interpret its value. Round to one decimal place.

97.

2i 3  2i

98.

b. Find the depth of the water at a point where a wave is traveling at 16 ft/sec.

99.

5  3i 2i

100.

10  5i 2i 4i 4i

Section 6.8 For Exercises 101–102, simplify the expression.

80. Define a complex number. 81. Define an imaginary number. 82. Consider the following expressions. 3 4  6i

and

3 14  16

Compare the process of dividing by a complex number to the process of rationalizing the denominator.

101.

8  240 12

102.

6  2144 3

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Chapter 6 Radicals and Complex Numbers

Chapter 6

Test

1. a. What is the principal square root of 36?

18. Add or subtract as indicated.

b. What is the negative square root of 36?

315  415  2120

2. Which of the following are real numbers?

For Exercises 19–20, multiply the radicals.

a. 1100 3

c. 11000

19. 31x1 12  152

b. 1100

20. 1215  31x21415  1x2

3

d. 11000

3. Simplify. 3 3 a. 2 y

For Exercises 21–22, rationalize the denominator. Assume x 7 0.

4 4 b. 2 y

For Exercises 4–11, simplify the radicals. Assume that all variables represent positive numbers. 16 A 9

4 4. 1 81

5.

3 6. 1 32

7. 2a4b3c5

8. 218x5y3z4

9.

32w6 B 2w

x6 B 125y3

11.

2272 8

10.

3

2

22.

3 1 x

1x  2 3  1x

23. Rewrite the expressions in terms of i. a. 18

b. 2116

c.

2  18 4

For Exercises 24–30, perform the indicated operations and simplify completely. Write the final answer in the form a  bi.

12. a. Evaluate the function values f 182, f 162, f 142, and f 122 for f 1x2  12x  4. b. Write the domain of f in interval notation. 13. Use a calculator to evaluate

21.

3  25 to 17

24. 13  5i2  12  6i2

25. 14  i218  2i2

26. 116 ⴢ 149

27. 14  7i2 2

28. 12  10i212  10i2

29.

30.

3  2i 3  4i

6i 3  5i

four decimal places. For Exercises 14–15, simplify the expressions. Assume that all variables represent positive numbers. 14. 271 3

15. 82/3 ⴢ a

25x4y6 z2

1/2

b

For Exercises 16–17, use rational exponents to multiply or divide. Write the final answer in radical form. 6 16. 1 7 ⴢ 1y

17.

3 1 10 4 1 10

31. If the volume V of a sphere is known, the radius of the sphere can be computed by 3V r1V2  3 . A 4p Find r(10) to two decimal places. Interpret the meaning in the context of the problem.

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571

Cumulative Review Exercises

32. A patio 20 ft wide has a slanted roof, as shown in the picture. Find the length of the roof if there is an 8-in. overhang. Round the answer to the nearest foot.

For Exercises 33–35, solve the radical equation. 3 33. 12x  5  3

34. 15x  8  15x  1  1 ?

35. 1t  7  12t  3  2

8 ft

10 ft

8 in.

20 ft

Chapters 1–6

Cumulative Review Exercises

1. Simplify the expression. 62  235  813  12  4  24 2. Simplify the expression. 3x  312x  52  4y  213x  52  y 3. Solve.

912y  82  20  1y  52

For Exercises 4–6, solve the inequalities. Write the answers in interval notation, if possible. 4. 2a  4 6 14

5.

11  x  2 7 5

6. 09  4x 0  4 6 2

11. Bennette and Pepe are landscapers and have to dig a ditch in which to put underground piping for a sprinkler system. Bennette can dig 40 yd of ditch in 3 hr and Pepe can dig 40 yd of ditch in 5 hr. How long will it take them to dig 40 yd together? 12. Given the function defined by f 1x2  4x  2. a. Find f 122, f 102, f 142, and f 1 12 2 .

b. Write the ordered pairs that correspond to the function values in part (a). c. Graph y  f 1x2 .

y

7. Write an equation of the line that is parallel to the line 2x  y  9 and passes through the point 13, 12 . Write the answer in slope-intercept form. 8. Write an equation of a line parallel to the y-axis passing through the point (5, 8).

5 4 3 2 1 5 4 3 2 1 1

4x  3y  1 10. Determine if 12, 2, 12 2 is a solution to the system. 2x  y  4z  0 x  y  2z  5 3x  2y  2z  4

3 4

5

x

2 3 4 5

9. Solve the system of equations by using the addition method. 2x  3y  0

1 2

13. Simplify the expression. Write the final answer with positive exponents only. a

a3 2b14c1 3 12 b ab54c0

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Chapter 6 Radicals and Complex Numbers

14. Multiply or divide as indicated, and write the answer in scientific notation. a. 13.5  107 214  1012 2

15. Multiply the polynomials 12x  521x  32 . What is the degree of the product? 16. Perform the indicated operations and simplify. 13 1 15  16  132 17. Divide. 1x2  x  122 1x  32

19. Simplify. 20. Add. 21. Divide. a  bi.

3 2  2 x  5x x  25

23. Add.

6.28  105 b. 2.0  104

18. Simplify and subtract.

3 6 5   2 y2 y4 y  6y  8

22. Solve.

8 4 1  3 B 16 B 27

54c4 B cd3

13i Write the answer in the form 3  2i

a  10 a2  12a  20

6a 2a2  11a  6

24. Divide.

25. Perform the indicated operations. 15x2  4x  82  13x  52 2 26. Simplify.

27. Divide.

3

4245b3  5b280b

2

4 13  15 4 3  5i

28. Solve.

12x2  4x  21  0

29. Factor.

x2  6x  9  y2

30. Factor.

x6  8

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7

Quadratic Equations and Functions

CHAPTER OUTLINE 7.1 Square Root Property and Completing the Square 574 7.2 Quadratic Formula 583 7.3 Equations in Quadratic Form 598 Problem Recognition Exercises: Quadratic and Quadratic Type Equations

603

7.4 Graphs of Quadratic Functions 604 7.5 Vertex of a Parabola: Applications and Modeling 618 7.6 Nonlinear Inequalities 628 Problem Recognition Exercises: Recognizing Equations and Inequalities Group Activity: Creating a Quadratic Model of the Form y  a(x  h)  k 2

640 641

Chapter 7 In Chapter 7, we revisit quadratic equations. We have already learned how to solve quadratic equations by factoring and applying the zero product rule. Now we present two additional techniques that can be used to solve a quadratic equation even if the quadratic polynomial is not factorable. Are You Prepared? To prepare for this chapter, we recommend that you review how to solve quadratic equations by using the zero product rule and how to simplify radical expressions. Match the answers to the exercises below to the letter on the right. Then write the letter in the appropriate space at the bottom of the page to define an important vocabulary word. Solve.

Answers

1. x  9  0 2. x2  6x  8  0 3. 3x(x  1)  60

D 1  3i 1 23 A  i 2 2 C 54, 56 U 53, 36 1  22 T 2 Q 52, 46 R (0, 5) I (1, 0) and (5, 0)

2

Simplify. 5  250 7  21i 2  5. 6. 10 7 7. Determine the x-intercept(s). f(x)  x2  6x  8. Determine the y-intercept. f(x)  x2  6x  4.

212 4 5 5

A function defined by f(x)  ax2  bx  c 1a  02 is called a ___ ___ ___ ___ ___ ___ ___ ___ ___ function. 2 1 6 5 8 6 4 7 3

573

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574

Chapter 7 Quadratic Equations and Functions

Section 7.1 Concepts 1. Solving Quadratic Equations by Using the Square Root Property 2. Solving Quadratic Equations by Completing the Square 3. Literal Equations

Square Root Property and Completing the Square 1. Solving Quadratic Equations by Using the Square Root Property In Section 4.8 we learned how to solve a quadratic equation by factoring and applying the zero product rule. For example: x2  81 x2  81  0

Set one side equal to zero.

1x  921x  92  0 x90

or

x9

or

Factor.

x90

Set each factor equal to zero.

x  9

The solution set is 59, 96. It is important to note that the zero product rule can only be used if the equation is factorable. In this section and Section 7.2, we will learn how to solve quadratic equations, factorable and nonfactorable. The first technique utilizes the square root property.

PROPERTY The Square Root Property For any real number, k, if x2  k, then x  1k or x   1k. Note: The solution may also be written as  1k, read “plus or minus the square root of k.”

Example 1

Solving a Quadratic Equation by Using the Square Root Property

Use the square root property to solve the equation. x2  81

Solution: x2  81

The equation is in the form x2  k.

x   181

Apply the square root property.

x  9

The solution set is 59, 96.

Notice that the solutions are the same as those found by applying the zero product rule.

Skill Practice Solve by using the square root property. 1. a 2  49

For a quadratic equation, ax 2  bx  c  0, if b  0, then the equation is easily solved by using the square root property. This is demonstrated in Example 2. Answer 1. 576

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Section 7.1

Example 2

Square Root Property and Completing the Square

575

Solving a Quadratic Equation by Using the Square Root Property

Use the square root property to solve the equation. 3x2  75  0

Solution: 3x2  75  0

Rewrite the equation to fit the form x2  k.

3x2  75 x2  25 x   125

The equation is now in the form x2  k. Apply the square root property.

 5i Check: x  5i 3x2  75  0

3x2  75  0

315i2 2  75 ⱨ 0

315i2 2  75 ⱨ 0

31252  75 ⱨ 0

31252  75 ⱨ 0

3125i2 2  75 ⱨ 0

The solution set is 55i6.

Check: x  5i

Avoiding Mistakes A common mistake is to forget the  symbol when solving the equation x 2  k : x   1k

3125i2 2  75 ⱨ 0

75  75 ⱨ 0 

75  75 ⱨ 0 

Skill Practice Solve using the square root property. 2. 8x2  72  0

Example 3

Solving a Quadratic Equation by Using the Square Root Property

Use the square root property to solve the equation. 1w  32 2  20

Solution:

1w  32 2  20

The equation is in the form x2  k, where x  1w  32.

w  3   120

Apply the square root property.

w  3   222 ⴢ 5

Simplify the radical.

w  3  215 w  3  215

Solve for w.

The solution set is 53  2156. Skill Practice Solve using the square root property. 3. 1t  52 2  18

Answers 2. 53i 6

3. 55  3126

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Chapter 7 Quadratic Equations and Functions

2. Solving Quadratic Equations by Completing the Square In Example 3 we used the square root property to solve an equation where the square of a binomial was equal to a constant. 1w ⫹ 32 2 ⫽ 20

}

576

Square of a binomial

Constant

The square of a binomial is the factored form of a perfect square trinomial. For example: Perfect Square Trinomial

Factored Form

x ⫹ 10x ⫹ 25

1x ⫹ 52 2

2

1t ⫺ 32 2

t 2 ⫺ 6t ⫹ 9

1p ⫺ 72 2

p 2 ⫺ 14p ⫹ 49

For a perfect square trinomial with a leading coefficient of 1, the constant term is the square of one-half the linear term coefficient. For example: x2 ⫹ 10x ⫹ 25 3 12 1102 4 2 In general an expression of the form x2 ⫹ bx ⫹ n is a perfect square trinomial if n ⫽ 1 12b2 2. The process to create a perfect square trinomial is called completing the square.

Completing the Square

Example 4

Determine the value of n that makes the polynomial a perfect square trinomial. Then factor the expression as the square of a binomial. a. x2 ⫹ 12x ⫹ n

b. x2 ⫺ 26x ⫹ n

c. x2 ⫹ 11x ⫹ n

4 d. x2 ⫺ x ⫹ n 7

Solution: The expressions are in the form x2 ⫹ bx ⫹ n. The value of n equals the square of one-half the linear term coefficient 1 12 b2 2. a. x2 ⫹ 12x ⫹ n x2 ⫹ 12x ⫹ 36 1x ⫹ 62 2

n ⫽ 3 12 1122 4 2 ⫽ 162 2 ⫽ 36 Factored form

b. x ⫺ 26x ⫹ n 2

x2 ⫺ 26x ⫹ 169 1x ⫺ 132 2

n ⫽ 3 12 1⫺262 4 2 ⫽ 1⫺132 2 ⫽ 169 Factored form

c. x2 ⫹ 11x ⫹ n x2 ⫹ 11x ⫹ ax ⫹

11 2 b 2

121 4

n ⫽ 3 12 1112 4 2 ⫽ 1 112 2 2 ⫽ 121 4 Factored form

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d. x2  x2 

Square Root Property and Completing the Square

4 xn 7 n  3 12 147 2 4 2  127 2 2 

4 4 x 7 49

2 2 ax  b 7

4 49

Factored form

Skill Practice Determine the value of n that makes the polynomial a perfect square trinomial. Then factor. 4. x2  20x  n

5. y2  16y  n 7 7. w2  w  n 3

6. a2  5a  n

The process of completing the square can be used to write a quadratic equation ax2  bx  c  0 1a  02 in the form 1x  h2 2  k. Then the square root property can be used to solve the equation. The following steps outline the procedure.

PROCEDURE Solving a Quadratic Equation ax 2 ⫹ bx ⫹ c ⫽ 0 by Completing the Square and Applying the Square Root Property Step 1 Divide both sides by a to make the leading coefficient 1. Step 2 Isolate the variable terms on one side of the equation. Step 3 Complete the square. • Add the square of one-half the linear term coefficient to both sides, 1 12b2 2. • Factor the resulting perfect square trinomial. Step 4 Apply the square root property and solve for x.

Example 5

Solving a Quadratic Equation by Completing the Square and Applying the Square Root Property

Solve by completing the square and applying the square root property. x2  6x  13  0

Solution: x2  6x  13  0

Step 1: Since the leading coefficient a is equal to 1, we do not have to divide by a. We can proceed to step 2.

x2  6x  13

Step 2: Isolate the variable terms on one side.

x2  6x  9  13  9

Step 3: To complete the square, add 3 12 162 4 2  9 to both sides of the equation.

1x  32 2  4

Factor the perfect square trinomial.

Answers

4. n  100; 1x  102 2 5. n  64; 1y  82 2 25 5 2 6. n  ; aa  b 4 2 49 7 2 7. n  ; aw  b 36 6

577

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x ⫺ 3 ⫽ ⫾ 1⫺4

Step 4: Apply the square root property.

x ⫺ 3 ⫽ ⫾2i

Simplify the radical.

x ⫽ 3 ⫾ 2i

Solve for x.

The solutions are imaginary numbers and can be written as 3 ⫹ 2i and 3 ⫺ 2i. The solution set is 53 ⫾ 2i6.

Skill Practice Solve by completing the square and applying the square root property. 8. z2 ⫺ 4z ⫹ 26 ⫽ 0

Example 6

Solving a Quadratic Equation by Completing the Square and Applying the Square Root Property

Solve by completing the square and applying the square root property. 2m2 ⫹ 10m ⫽ 3

Solution: 2m2 ⫹ 10m ⫽ 3

The variable terms are already isolated on one side of the equation.

2m2 10m 3 ⫹ ⫽ 2 2 2 m2 ⫹ 5m ⫽ m2 ⫹ 5m ⫹

Divide by the leading coefficient, 2.

3 2 Add 3 12 152 4 2 ⫽ 1 52 2 2 ⫽

25 3 25 ⫽ ⫹ 4 2 4

5 2 6 25 am ⫹ b ⫽ ⫹ 2 4 4

25 4

to both sides.

Factor the left side and write the terms on the right with a common denominator.

5 2 31 am ⫹ b ⫽ 2 4 m⫹

5 31 ⫽⫾ 2 B 4 m⫽⫺

TIP: The solutions to Example 6 can also be written as: ⫺5 ⫾ 231 2

Answers

8. 52 ⫾ i 2226 114 3 f 9. e ⫺ ⫾ 2 2

5 231 ⫾ 2 2

The solution set is e ⫺

Apply the square root property. Subtract 52 from both sides and simplify the radical.

5 231 ⫾ f. 2 2

The solutions are irrational numbers.

Skill Practice Solve by completing the square and applying the square root property. 9. 4x2 ⫹ 12x ⫽ 5

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Square Root Property and Completing the Square

Solving a Quadratic Equation by Completing the Square and Applying the Square Root Property

Example 7

Solve by completing the square and applying the square root property. 2x12x  102  30  6x

Solution: 2x12x  102  30  6x 4x2  20x  30  6x

Clear parentheses.

4x2  26x  30  0

Write the equation in the form ax2  bx  c  0.

4x2 26x 30 0    4 4 4 4

Step 1: Divide both sides by the leading coefficient 4.

x2 

TIP: In general, if the solutions to a quadratic equation are rational numbers, the equation can be solved by factoring and using the zero product rule. Consider the equation from Example 7.

13 15 x 0 2 2 x2 

13 15 x 2 2

13 169 15 169 x2  x    2 16 2 16

Step 2: Isolate the variable terms on one side. Step 3: Add 3 12 1132 2 4 2  1134 2 2  to both sides.

13 2 120 169 ax  b    4 16 16 ax 

Factor the perfect square trinomial. Rewrite the right-hand side with a common denominator.

13 2 49 b  4 16

x

13 49  4 B 16

x

13 7  4 4

x

169 16

2x12x  102  30  6x 4x2  20x  30  6x 4x2  26x  30  0 212x2  13x  152  0 21x  5212x  3 2  0 x5

x

3 2

Step 4: Apply the square root property. Simplify the radical. x

13 7 20   5 4 4 4

x

13 7 6 3    4 4 4 2

Avoiding Mistakes If the terms are like terms, be sure to combine them.

13 7  4 4

3 The solution set is e 5, f . 2

or

The solutions are rational numbers.

Skill Practice Solve by completing the square and applying the square root property. 10. 2y 1y  12  3  y

Answer 3 10. e , 1 f 2

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3. Literal Equations Example 8

Solving a Literal Equation

Ignoring air resistance, the distance d (in meters) that an object falls in t sec is given by the equation d ⫽ 4.9t 2

where t ⱖ 0

a. Solve the equation for t. Do not rationalize the denominator. b. Using the equation from part (a), determine the amount of time required for an object to fall 500 m. Round to the nearest second.

Solution: a.

d ⫽ 4.9t 2 d ⫽ t2 4.9 t⫽⫾ ⫽

Isolate the quadratic term. The equation is in the form t 2 ⫽ k. d B 4.9

d B 4.9

b. t ⫽

d B 4.9

⫽

500 B 4.9

Apply the square root property. Because t represents time, t ⱖ 0. We reject the negative solution.

Substitute d ⫽ 500.

t ⬇ 10.1 The object will require approximately 10.1 sec to fall 500 m. Skill Practice The formula for the area of a circle is A ⫽ pr 2, where r is the radius. 11. Solve for r. (Do not rationalize the denominator.) 12. Use the equation from Skill Practice exercise 11 to find the radius for an area of 15.7 cm2. (Use 3.14 for p and round to 2 decimal places.)

Answers A Bp 12. The radius is 15 ⬇ 2.24 cm. 11. r ⫽

Section 7.1

Practice Exercises

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Study Skills Exercise 1. Define the key terms. a. Square root property

b. Completing the square

Concept 1: Solving Quadratic Equations by Using the Square Root Property For Exercises 2–17, solve the equations by using the square root property. (See Examples 1–3.) 2. x2 ⫽ 100

3. y2 ⫽ 4

4. a2 ⫽ 5

5. k2 ⫺ 7 ⫽ 0

6. 3v2 ⫹ 33 ⫽ 0

7. ⫺2m2 ⫽ 50

8. 1p ⫺ 52 2 ⫽ 9

9. 1q ⫹ 32 2 ⫽ 4

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10. 13x ⫺ 22 2 ⫺ 5 ⫽ 0

11. 12y ⫹ 32 2 ⫺ 7 ⫽ 0

12. 1h ⫺ 42 2 ⫽ ⫺8

13. 1t ⫹ 52 2 ⫽ ⫺18

14. 6p 2 ⫺ 3 ⫽ 2

15. 15 ⫽ 4 ⫹ 3w2

3 2 7 16. ax ⫺ b ⫹ ⫽ 0 2 4

4 2 3 17. am ⫹ b ⫹ ⫽0 5 25

581

18. Given the equation x2 ⫽ k, match the following statements. a. If k 7 0, then ______

i. there will be one real solution.

b. If k 6 0, then ______

ii. there will be two real solutions.

c. If k ⫽ 0, then ______

iii. there will be two imaginary solutions.

19. State two methods that can be used to solve the equation x2 ⫺ 36 ⫽ 0. Then solve the equation by using both methods. 20. Explain the difference between solving the equations: x ⫽ 116 and x2 ⫽ 16.

Concept 2: Solving Quadratic Equations by Completing the Square For Exercises 21–32, find the value of n so that the expression is a perfect square trinomial. Then factor the trinomial. (See Example 4.)

21. x2 ⫺ 6x ⫹ n

22. x2 ⫹ 12x ⫹ n

23. t2 ⫹ 8t ⫹ n

24. v2 ⫺ 18v ⫹ n

25. c2 ⫺ c ⫹ n

26. x2 ⫹ 9x ⫹ n

27. y2 ⫹ 5y ⫹ n

28. a2 ⫺ 7a ⫹ n

2 29. b2 ⫹ b ⫹ n 5

2 30. m2 ⫺ m ⫹ n 7

2 31. p 2 ⫺ p ⫹ n 3

3 32. w 2 ⫹ w ⫹ n 4

33. Summarize the steps used in solving a quadratic equation by completing the square and applying the square root property. 34. What types of quadratic equations can be solved by completing the square and applying the square root property? For Exercises 35–54, solve the quadratic equation by completing the square and applying the square root property. (See Examples 5–7.)

35. t 2 ⫹ 8t ⫹ 15 ⫽ 0

36. m2 ⫹ 6m ⫹ 8 ⫽ 0

37. x2 ⫹ 6x ⫽ ⫺16

38. x2 ⫺ 4x ⫽ ⫺15

39. p2 ⫹ 4p ⫹ 6 ⫽ 0

40. q2 ⫹ 2q ⫹ 2 ⫽ 0

41. ⫺3y ⫺ 10 ⫽ ⫺y2

42. ⫺24 ⫽ ⫺2y2 ⫹ 2y

43. 2a2 ⫹ 4a ⫹ 5 ⫽ 0

44. 3a2 ⫹ 6a ⫺ 7 ⫽ 0

45. 9x2 ⫺ 36x ⫹ 40 ⫽ 0

46. 9y2 ⫺ 12y ⫹ 5 ⫽ 0

47. 25p 2 ⫺ 10p ⫽ 2

48. 9n2 ⫺ 6n ⫽ 1

49. 12w ⫹ 521w ⫺ 12 ⫽ 2

50. 13p ⫺ 521p ⫹ 12 ⫽ ⫺3

51. n1n ⫺ 42 ⫽ 7

52. m1m ⫹ 102 ⫽ 2

53. 2x1x ⫹ 62 ⫽ 14

54. 3x1x ⫺ 22 ⫽ 24

Concept 3: Literal Equations 55. The distance (in feet) that an object falls in t sec is given by the equation d ⫽ 16t 2, where t ⱖ 0. a. Solve the equation for t. (See Example 8.) b. Using the equation from part (a), determine the amount of time required for an object to fall 1024 ft.

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56. The volume of a can that is 4 in. tall is given by the equation V  4pr 2, where r is the radius of the can, measured in inches. a. Solve the equation for r. Do not rationalize the denominator. b. Using the equation from part (a), determine the radius of a can with volume of 12.56 in.3 Use 3.14 for p. For Exercises 57–62, solve for the indicated variable. 57. A  pr2

for r

1r 7 02

58. E  mc2 1a 7 02

59. a2  b2  c2  d2

for a

1 61. V  pr2h 3

1r 7 02

for r

60. a2  b2  c2 1 62. V  s2h 3

63. A corner shelf is to be made from a triangular piece of plywood, as shown in the diagram. Find the distance x that the shelf will extend along the walls. Assume that the walls are at right angles. Round the answer to a tenth of a foot.

x

for c

1c 7 02 1b 7 02

for b for s

1s 7 02

64. The volume of a box with a square bottom and a height of 4 in. is given by V1x2  4x2, where x is the length (in inches) of the sides of the bottom of the box. a. If the volume of the box is 289 in.3, find the dimensions of the box. b. Are there two possible answers to part (a)? Why or why not?

x 6 ft

4 in.

x x

65. A square has an area of 50 in.2 What are the lengths of the sides? (Round to one decimal place.) 66. The amount of money A in an account with an interest rate r compounded annually is given by A  P11  r2 t where P is the initial principal and t is the number of years the money is invested. a. If a $10,000 investment grows to $11,664 after 2 yr, find the interest rate. b. If a $6000 investment grows to $7392.60 after 2 yr, find the interest rate. c. Jamal wants to invest $5000. He wants the money to grow to at least $6500 in 2 yr to cover the cost of his son’s first year at college. What interest rate does Jamal need for his investment to grow to $6500 in 2 yr? Round to the nearest hundredth of a percent.

a. Approximate the number of books required to make a profit of $20,000. [Hint: Let P1x2  20. Then complete the square to solve for x.] Round to one decimal place. b. Why are there two answers to part (a)?

P(x)

Profit P(x) ($1000)

67. A textbook company has discovered that the profit for selling its books is given by 1 P1x2   x2  5x 8 where x is the number of textbooks produced (in thousands) and P(x) is the corresponding profit (in thousands of dollars). The graph of the function is shown at right.

50 45 40 35 30 25 20 15 10 5 5 10 15 20 25 30 35 40 45 Number of Textbooks (thousands)

x

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68. If we ignore air resistance, the distance (in feet) that an object travels in free fall can be approximated by d1t2  16t 2, where t is the time in seconds after the object was dropped. a. If the CN Tower in Toronto is 1815 ft high, how long will it take an object to fall from the top of the building? Round to one decimal place. b. If the Renaissance Tower in Dallas is 886 ft high, how long will it take an object to fall from the top of the building? Round to one decimal place.

Quadratic Formula

Section 7.2

1. Derivation of the Quadratic Formula

If we solve a general quadratic equation ax  bx  c  0 1a  02 by completing the square and using the square root property, the result is a formula that gives the solutions for x in terms of a, b, and c. 2

ax2  bx  c  0

Begin with a quadratic equation in standard form.

ax2 bx c 0    a a a a

Divide by the leading coefficient.

c b x2  x   0 a a c b x2  x   a a x2 

Add the square of 12 the linear term coefficient to both sides of the equation.

ax 

b 2 b2 c b  2 a 2a 4a

Factor the left side as a perfect square.

ax 

b 2 b2  4ac b  2a 4a2

Combine fractions on the right side by getting a common denominator.

x

b2  4ac b  2a B 4a2

Apply the square root property.

x

b 2b2  4ac  2a 2a

Simplify the denominator.



1. Derivation of the Quadratic Formula 2. Solving Quadratic Equations by Using the Quadratic Formula 3. Using the Quadratic Formula in Applications 4. Discriminant 5. Mixed Review: Methods to Solve a Quadratic Equation

Isolate the terms containing x.

b 1 b 2 1 b 2 c xa ⴢ b a ⴢ b  a a 2 a 2 a

x

Concepts

b 2b2  4ac  2a 2a

b  2b  4ac 2a

Subtract

b from both sides. 2a

2

Combine fractions.

The solutions to the equation ax2  bx  c  0 for x in terms of the coefficients a, b, and c are given by the quadratic formula.

TIP: When applying the quadratic formula, note that a, b, and c are constants. The variable is x.

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FORMULA The Quadratic Formula

For a quadratic equation of the form ax2 ⫹ bx ⫹ c ⫽ 0 1a ⫽ 02 the solutions are x⫽

⫺b ⫾2b2 ⫺ 4ac 2a

The quadratic formula gives us another technique to solve a quadratic equation. This method will work regardless of whether the equation is factorable or not factorable.

2. Solving Quadratic Equations by Using the Quadratic Formula Solving a Quadratic Equation by Using the Quadratic Formula

Example 1

Solve the quadratic equation by using the quadratic formula. 2x2 ⫺ 3x ⫽ 5

Solution: 2x2 ⫺ 3x ⫽ 5 2x2 ⫺ 3x ⫺ 5 ⫽ 0

Avoiding Mistakes • The term, ⫺b, represents the opposite of b. • Remember to write the entire numerator over 2a.

a ⫽ 2, x⫽ ⫽

b ⫽ ⫺3,

Write the equation in the form ax2 ⫹ bx ⫹ c ⫽ 0. c ⫽ ⫺5

Identify a, b, and c.

⫺b ⫾ 2b ⫺ 4ac 2a 2

Apply the quadratic formula.

⫺1⫺32 ⫾ 21⫺32 2 ⫺ 41221⫺52 2122

⫽

3 ⫾ 19 ⫹ 40 4

⫽

3 ⫾ 149 4

3 ⫾ 7 ⫽ 4

Substitute a ⫽ 2, b ⫽ ⫺3, and c ⫽ ⫺5. Simplify.

x⫽

3⫹7 10 5 ⫽ ⫽ 4 4 2

x⫽

3⫺7 ⫺4 ⫽ ⫽ ⫺1 4 4

5 The solution set is e , ⫺1 f . Both solutions check in the original equation. 2 Skill Practice Solve the equation by using the quadratic formula. 1. 6x2 ⫺ 5x ⫽ 4

Answer 1 4 1. e ⫺ , f 2 3

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Example 2

Quadratic Formula

585

Solving a Quadratic Equation by Using the Quadratic Formula

Solve the quadratic equation by using the quadratic formula. ⫺x1x ⫺ 62 ⫽ 11

Solution: ⫺x1x ⫺ 62 ⫽ 11 ⫺x2 ⫹ 6x ⫺ 11 ⫽ 0

Write the equation in the form ax2 ⫹ bx ⫹ c ⫽ 0.

⫺11⫺x2 ⫹ 6x ⫺ 112 ⫽ ⫺1102

If the leading coefficient of the quadratic polynomial is negative, we suggest multiplying both sides of the equation by ⫺1. Although this is not mandatory, it is generally easier to simplify the quadratic formula when the value of a is positive.

x2 ⫺ 6x ⫹ 11 ⫽ 0

a ⫽ 1, b ⫽ ⫺6, and c ⫽ 11 x⫽ ⫽

⫺b ⫾2b2 ⫺ 4ac 2a

2112 6 ⫾ 136 ⫺ 44 2

⫽

6 ⫾ 1⫺8 2

6 ⫾ 2i12 ⫽ 2

⫽

Apply the quadratic formula.

⫺1⫺62 ⫾ 21⫺62 2 ⫺ 41121112

⫽

⫽

Identify a, b, and c.

213 ⫾ i122 2 2 13 ⫾ i122 2

Substitute a ⫽ 1, b ⫽ ⫺6, and c ⫽ 11. Simplify.

Avoiding Mistakes When identifying a, b, and c, use the coefficients only, not the variable.

Avoiding Mistakes Simplify the radical.

Always simplify the radical completely before trying to reduce the fraction to lowest terms.

Factor the numerator.

Simplify the fraction to lowest terms. x ⫽ 3 ⫹ i12

⫽ 3 ⫾ i12

The solutions are imaginary numbers.

x ⫽ 3 ⫺ i 12 The solution set is 53 ⫾ i226. Skill Practice Solve the equation by using the quadratic formula. 2. y 1y ⫹ 42 ⫽ ⫺12

Answer

2. 5⫺2 ⫾ 2i 226

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3. Using the Quadratic Formula in Applications Example 3

Using the Quadratic Formula in an Application

A delivery truck travels south from Hartselle, Alabama, to Birmingham, Alabama, along Interstate 65.The truck then heads east to Atlanta, Georgia, along Interstate 20.The distance from Birmingham to Atlanta is 8 mi less than twice the distance from Hartselle to Birmingham. If the straight-line distance from Hartselle to Atlanta is 165 mi, find the distance from Hartselle to Birmingham and from Birmingham to Atlanta. (Round the answers to the nearest mile.)

Solution: The motorist travels due south and then due east. Therefore, the three cities form the vertices of a right triangle (Figure 7-1). Let x represent the distance between Hartselle and Birmingham. Then 2x ⫺ 8 represents the distance between Birmingham and Atlanta.

x

I-65

Hartselle

Use the Pythagorean theorem to establish a relationship among the three sides of the triangle.

165 mi

1x2 2 ⫹ 12x ⫺ 82 2 ⫽ 11652 2

I-20 Birmingham

2x ⫺ 8

Atlanta

Figure 7-1

x2 ⫹ 4x2 ⫺ 32x ⫹ 64 ⫽ 27,225 5x2 ⫺ 32x ⫺ 27,161 ⫽ 0

Write the equation in the form ax2 ⫹ bx ⫹ c ⫽ 0.

a⫽5

Identify a, b, and c.

x⫽ ⫽

⫽

b ⫽ ⫺32

c ⫽ ⫺27,161

⫺1⫺322 ⫾ 21⫺322 2 ⫺ 41521⫺27,1612 2152

Apply the quadratic formula.

32 ⫾ 21024 ⫹ 543,220 10

Simplify.

32 ⫾ 1544,244 10

x⫽

32 ⫹ 1544,244 ⬇ 76.97 10

x⫽

32 ⫺ 1544,244 ⬇ ⫺70.57 10

We reject the negative solution because distance cannot be negative. Rounding to the nearest whole unit, we have x ⫽ 77. Therefore, 2x ⫺ 8 ⫽ 2(77) ⫺ 8 ⫽ 146. The distance between Hartselle and Birmingham is 77 mi, and the distance between Birmingham and Atlanta is 146 mi. Skill Practice 3. Steve and Tammy leave a campground, hiking on two different trails. Steve heads south and Tammy heads east. By lunchtime they are 9 mi apart. Steve walked 3 mi more than twice as many miles as Tammy. Find the distance each person hiked. (Round to the nearest tenth of a mile.)

Answer 3. Tammy hiked 2.8 mi, and Steve hiked 8.6 mi.

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Section 7.2

Example 4

Quadratic Formula

587

Analyzing a Quadratic Function

A model rocket is launched straight upward from the side of a 144-ft cliff (Figure 7-2). The initial velocity is 112 ft/sec. The height of the rocket h(t) is given by

h(t)

h1t2 ⫽ ⫺16t 2 ⫹ 112t ⫹ 144 where h(t) is measured in feet and t is the time in seconds. Find the time(s) at which the rocket is 208 ft above the ground.

208 144

Solution: h1t2 ⫽ ⫺16t 2 ⫹ 112t ⫹ 144 208 ⫽ ⫺16t 2 ⫹ 112t ⫹ 144 16t 2 ⫺ 112t ⫹ 64 ⫽ 0

Figure 7-2

Divide by 16. This makes the coefficients smaller, and it is less cumbersome to solve.

t 2 ⫺ 7t ⫹ 4 ⫽ 0

⫽

t

0

Write the equation in the form at 2 ⫹ bt ⫹ c ⫽ 0.

16t 2 112t 64 0 ⫺ ⫹ ⫽ 16 16 16 16

t⫽

Substitute 208 for h(t).

The equation is not factorable. Apply the quadratic formula.

⫺1⫺72 ⫾ 21⫺72 2 ⫺ 4112 142 2112

Let a ⫽ 1, b ⫽ ⫺7, and c ⫽ 4. t⫽

7 ⫹ 233 ⬇ 6.37 2

t⫽

7 ⫺ 233 ⬇ 0.63 2

7 ⫾ 233 2

The rocket will reach a height of 208 ft after approximately 0.63 sec (on the way up) and after 6.37 sec (on the way down). Skill Practice 4. A rocket is launched from the top of a 96-ft building with an initial velocity of 64 ft/sec. The height h 1t 2 of the rocket is given by h 1t 2 ⫽ ⫺16t 2 ⫹ 64t ⫹ 96. Find the time it takes for the rocket to hit the ground. [Hint: h 1t 2 ⫽ 0 when the object hits the ground.]

4. Discriminant The radicand within the quadratic formula is the expression b2 ⫺ 4ac. This is called the discriminant. The discriminant can be used to determine the number of solutions to a quadratic equation as well as whether the solutions are rational, irrational, or imaginary numbers.

Answer 4. 2 ⫹ 110 ⬇ 5.16 sec

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PROCEDURE Using the Discriminant to Determine the Number and Type of Solutions to a Quadratic Equation Consider the equation ax2  bx  c  0, where a, b, and c are rational numbers and a  0. The expression b2  4ac is called the discriminant. Furthermore, • If b2  4ac 7 0, then there will be two real solutions. a. If b2  4ac is a perfect square, the solutions will be rational numbers. b. If b2  4ac is not a perfect square, the solutions will be irrational numbers. • If b2  4ac 6 0, then there will be two imaginary solutions. • If b2  4ac  0, then there will be one rational solution.

Example 5

Using the Discriminant

Use the discriminant to determine the type and number of solutions for each equation. a. 2x2  5x  9  0

b. 3x2  x  2

c. 2x12x  32  1

d. 3.6x2  1.2x  0.1

Solution: For each equation, first write the equation in standard form ax2  bx  c  0. Then determine the discriminant. Equation a. 2x2  5x  9  0

Discriminant

Solution Type and Number

b2  4ac  152 2  4122192  25  72  47

Because 47 6 0, there will be two imaginary solutions.

b2  4ac  112 2  4132122  1  1242  25

25 7 0 and 25 is a perfect square. There will be two rational solutions.

b2  4ac  162 2  4142112  36  1162  52

52 7 0, but 52 is not a perfect square. There will be two irrational solutions.

b2  4ac  11.22 2  413.6210.12  1.44  1.44 0

Because the discriminant equals 0, there will be only one rational solution.

b. 3x2  x  2 3x2  x  2  0

c.

2x12x  32  1 4x2  6x  1 4x2  6x  1  0

d. 3.6x2  1.2x  0.1 3.6x2  1.2x  0.1  0

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Quadratic Formula

Skill Practice Use the discriminant to determine the type and number of solutions for the equation. 5. 3y2  y  3  0

6. 4t 2  6t  2 2 2 1 8. x 2  x   0 3 3 6

7. 3t 1t  12  9

With the discriminant we can determine the number of real-valued solutions to the equation ax2  bx  c  0, and thus the number of x-intercepts to the function f 1x2  ax2  bx  c. The following illustrations show the graphical interpretation of the three cases of the discriminant. f1x2  x2  4x  3 Use x2  4x  3  0 to find the value of the discriminant.

y 5 4 3 2 1 5 4 3 2 1 1 2

b2  4ac  142 2  4112132 4

1 2 3

4

5

x

Since the discriminant is positive, there are two real solutions to the quadratic equation. Therefore, there are two x-intercepts to the corresponding quadratic function, (1, 0) and (3, 0).

3 4 5

f 1x2  x2  x  1 Use x2  x  1  0 to find the value of the discriminant.

y 5 4 3

b2  4ac  112 2  4112112  3

2 1 5 4 3 2 1 1 2

1

2 3

4

5

x

Since the discriminant is negative, there are no real solutions to the quadratic equation. Therefore, there are no x-intercepts to the corresponding quadratic function.

3 4 5

f 1x2  x2  2x  1 Use x2  2x  1  0 to find the value of the discriminant.

y 5 4 3

b2  4ac  122 2  4112112 0

2 1 5 4 3 2 1 1 2 3 4 5

1

2 3

4

5

x

Since the discriminant is zero, there is one real solution to the quadratic equation. Therefore, there is one x-intercept to the corresponding quadratic function, (1, 0).

Answers 5. 6. 7. 8.

35; two imaginary solutions 4; two rational solutions 117; two irrational solutions 0; one rational solution

589

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Example 6

Finding x- and y-Intercepts of a Quadratic Function

Given f 1x2  x2  3x  1 a. Find the discriminant and use it to determine if there are any x-intercepts. b. Find the x-intercept(s), if they exist. c. Find the y-intercept.

Solution: a. a  1, b  3, and c  1.

The discriminant is b2  4ac  132 2  4112112 94 5 Since 5  0, there are two x-intercepts.

b. The x-intercepts are given by the real solutions to the equation f(x)  0. In this case, we have f 1x2  x2  3x  1  0

x2  3x  1  0

x



3  2 132 2  142112112

Apply the quadratic formula.

2112

x

3  15 2

⬇ 2.62

x

3  15 2

⬇ 0.38

3  15 2

The solutions are a

The equation is in the form ax2  bx  c  0.

3  15 3  15 and . Therefore, the x-intercepts are 2 2

3  15 3  15 , 0b and a , 0b. 2 2

c. To find the y-intercept, evaluate f (0). f 102  102  3102  1  1 2

The y-intercept is located at (0, 1). The parabola is shown in the graph with the x- and y-intercepts labeled.

y 5 2

f(x)  x  3x  1

4 3 2 1

5 4 3 2 1 1 ⎧3v5 ⎫ ⎩ 2 , 0⎭ 2

≈ (0.4, 0)3

(0, 1) 1

2 3

4 5

⎧3v5 ⎫ ⎩ 2 , 0⎭

≈ (2.6, 0)

4 5

Skill Practice Given f 1x2  x2  5x  2, Answers 9. 17, there are two x-intercepts. 5  117 10. x-intercepts: a , 0b, 2 5  117 a , 0b 2 y-intercept: (0, 2)

9. Find the discriminant and use it to determine if there are any x-intercepts. 10. Find the x- and y-intercepts.

x

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Example 7

Quadratic Formula

Finding x- and y-Intercepts of a Quadratic Function

Given f 1x2  x 2  x  2, a. Find the discriminant and use it to determine if there are any x-intercepts. b. Find the x-intercept(s), if they exist. c. Find the y-intercept.

Solution: a. a  1, b  1, and c  2.

The discriminant is b2  4ac  112 2  4112122 18  7 Since 7  0, there are no x-intercepts.

y 5

b. There are no x-intercepts.

4 3

c. f 102  102  102  2 2

2 The y-intercept is located at (0, 2).

2 1 5 4 3 2 1 1 f(x)  x2  x  2 2

(0, 2) 1

2 3

4

5

x

3

The parabola is shown. Note: the graph does not intersect the x-axis.

4 5

Skill Practice Given f 1x 2  2x2  3x  5, 11. Find the discriminant and use it to determine if there are any x-intercepts. 12. Find the x- and y-intercepts.

5. Mixed Review: Methods to Solve a Quadratic Equation Three methods have been presented to solve quadratic equations.

SUMMARY Methods to Solve a Quadratic Equation Factor and use the zero product rule (Section 4.8). • This method works well if you can factor the equation easily. Use the square root property. Complete the square if necessary (Section 7.1). This method is particularly good if • the equation is of the form ax2  c  0 or • the equation is of the form x2  bx  c  0, where b is even. Quadratic Formula (Section 7.2) • This method works in all cases—just be sure to write the equation in the form ax2  bx  c  0 before applying the formula.

Example: x2  8x  15  0 factors as 1x  321x  52  0

Examples: 4x2  9  0 x2  10x  3  0 Example: 7x2  3x  11  0

Answers 11. 31, there are no x-intercepts. 12. (0, 5)

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Before solving a quadratic equation, take a minute to analyze it first. Each problem must be evaluated individually before choosing the most efficient method to find its solutions.

Solving a Quadratic Equation by Using Any Method

Example 8

Solve the equation by using any method.

1x ⫹ 32 2 ⫹ x2 ⫺ 9x ⫽ 8

Solution: 1x ⫹ 32 2 ⫹ x2 ⫺ 9x ⫽ 8 x2 ⫹ 6x ⫹ 9 ⫹ x2 ⫺ 9x ⫺ 8 ⫽ 0

2x2 ⫺ 3x ⫹ 1 ⫽ 0

12x ⫺ 121x ⫺ 12 ⫽ 0 2x ⫺ 1 ⫽ 0 x⫽

1 2

or

x⫺1⫽0

or

x⫽1

The solution set is 5 12, 16.

Clear parentheses and write the equation in the form ax2 ⫹ bx ⫹ c ⫽ 0. This equation is factorable. Factor. Apply the zero product rule. Solve for x.

This equation could have been solved by using any of the three methods, but factoring was the most efficient method. Skill Practice Solve using any method. 13. 2t 1t ⫺ 12 ⫹ t2 ⫽ 5

Solving a Quadratic Equation by Using Any Method

Example 9

Solve the equation by using any method. x 2 ⫹ 5 ⫽ ⫺2x

Solution: x 2 ⫹ 2x ⫹ 5 ⫽ 0 x 2 ⫹ 2x ⫽ ⫺5 x 2 ⫹ 2x ⫹ 1 ⫽ ⫺5 ⫹ 1 1x ⫹ 12 2 ⫽ ⫺4

x ⫹ 1 ⫽ ⫾ 1⫺4 x ⫽ ⫺1 ⫾ 2i

The solution set is 5⫺1 ⫾ 2i6.

The equation does not factor. Because a ⫽ 1 and b is even, we can easily complete the square.

Add 3 12 122 4 2 ⫽ 12 ⫽ 1 to both sides. Apply the square root property. Solve for x.

This equation could also have been solved by using the quadratic formula. Answers 5 13. e ⫺1, f 3 14. 52 ⫾ i136

Skill Practice Solve using any method. 14. x2 ⫺ 4x ⫽ ⫺7

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Section 7.2

Example 10

Quadratic Formula

Solving a Quadratic Equation by Using Any Method 1 2 1 1 x ⫺ x⫹ ⫽0 4 2 3

Solve the equation by using any method.

Solution: 1 1 1 12 ⴢ a x2 ⫺ x ⫹ b ⫽ 12 ⴢ 102 4 2 3 3x2 ⫺ 6x ⫹ 4 ⫽ 0

Clear fractions. The equation is in the form ax2 ⫹ bx ⫹ c ⫽ 0. The left-hand side does not factor.

a ⫽ 3, b ⫽ ⫺6, and c ⫽ 4 ⫺1⫺62 ⫾ 21⫺62 2 ⫺ 4132142

x⫽

2132

Apply the quadratic formula.

⫽

6 ⫾ 2⫺12 6

Simplify.

⫽

6 ⫾ 2i23 6

Simplify the radical.

2 13 ⫾ i232

1

⫽

Factor and simplify.

6

3

⫽

3 ⫾ i23 3

⫽

3 23 ⫾ i 3 3

⫽1⫾

Write in the form a ⫾ bi.

23 i 3

The solution set is e 1 ⫾

Simplify. 23 if. 3

Skill Practice Solve using any method. 15.

1 2 4 1 x ⫺ x⫹ ⫽0 5 5 2

Answer 15. e

4 ⫾ 26 f 2

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Solving a Quadratic Equation by Using Any Method

Example 11

Solve the equation by using any method.

9p2  11  0

Solution: 9p2  11  0

Because b  0, use the square root property.

9p  11 2

p2 

Isolate the variable term.

11 9

p

11 B 9

Apply the square root property.

p

211 3

Simplify the radical.

The solution set is e 

16. 4y2  13  0

213 f 2

Section 7.2

211 f. 3

Skill Practice Solve using any method.

Answer 16. e 

The equation is in the form x 2  k.

Practice Exercises

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Study Skills Exercise 1. Define the key terms. a. Quadratic formula

b. Discriminant

Review Exercises 2. Use substitution to determine if x  3  15 is a solution to x2  6x  4  0. For Exercises 3–6, simplify the expression. 3.

16  1320 4

4.

18  1180 3

5.

14  1147 7

6.

10  1175 5

Concept 2: Solving Quadratic Equations by Using the Quadratic Formula 7. What form should a quadratic equation be in so that the quadratic formula can be applied? 8. Write the quadratic formula from memory.

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For Exercises 9–34, solve the equation by using the quadratic formula. (See Examples 1–2.) 9. x2 ⫹ 11x ⫺ 12 ⫽ 0

10. 5x2 ⫺ 14x ⫺ 3 ⫽ 0

11. 9y2 ⫺ 2y ⫹ 5 ⫽ 0

12. 2t2 ⫹ 3t ⫺ 7 ⫽ 0

13. 12p 2 ⫺ 4p ⫹ 5 ⫽ 0

14. ⫺5n2 ⫹ 4n ⫺ 6 ⫽ 0

15. ⫺z2 ⫽ ⫺2z ⫺ 35

16. 12x2 ⫺ 5x ⫽ 2

17. y 2 ⫹ 3y ⫽ 8

18. k2 ⫹ 4 ⫽ 6k

19. 25x2 ⫺ 20x ⫹ 4 ⫽ 0

20. 9y2 ⫽ ⫺12y ⫺ 4

21. w1w ⫺ 62 ⫽ ⫺14

22. m1m ⫹ 62 ⫽ ⫺11

23. 1x ⫹ 221x ⫺ 32 ⫽ 1

24. 3y1y ⫹ 12 ⫺ 7y1y ⫹ 22 ⫽ 6

25. ⫺4a2 ⫺ 2a ⫹ 3 ⫽ 0

26. ⫺2m2 ⫺ 5m ⫹ 3 ⫽ 0

27.

30.

1 2 2 2 y ⫹ ⫽⫺ y 2 3 3 (Hint: Clear fractions first.) 1 2 7 w ⫹ w⫹1⫽0 4 4

33. 0.3t2 ⫹ 0.7t ⫺ 0.5 ⫽ 0

28.

2 2 1 1 p ⫺ p⫹ ⫽0 3 6 2

29.

31. 0.01x2 ⫹ 0.06x ⫹ 0.08 ⫽ 0 (Hint: Clear decimals first.)

1 2 3 h ⫹h⫹ ⫽0 5 5

32. 0.5y2 ⫺ 0.7y ⫹ 0.2 ⫽ 0

34. 0.01x2 ⫹ 0.04x ⫺ 0.07 ⫽ 0

For Exercises 35–38, factor the expression. Then use the zero product rule and the quadratic formula to solve the equation. There should be three solutions to each equation. 35. a. Factor. b. Solve. 37. a. Factor.

x3 ⫺ 27 x3 ⫺ 27 ⫽ 0 3x3 ⫺ 6x2 ⫹ 6x

b. Solve. 3x 3 ⫺ 6x 2 ⫹ 6x ⫽ 0 39. The volume of a cube is 27 ft3. Find the lengths of the sides.

36. a. Factor.

64x3 ⫹ 1

b. Solve. 64x3 ⫹ 1 ⫽ 0 38. a. Factor. b. Solve.

5x3 ⫹ 5x2 ⫹ 10x 5x 3 ⫹ 5x 2 ⫹ 10x ⫽ 0

40. The volume of a rectangular box is 64 ft3. If the width is 3 times longer than the height, and the length is 9 times longer than the height, find the dimensions of the box.

Concept 3: Using the Quadratic Formula in Applications 41. The hypotenuse of a right triangle measures 4 in. One leg of the triangle is 2 in. longer than the other leg. Find the lengths of the legs of the triangle. Round to one decimal place. (See Example 3.)

42. The length of one leg of a right triangle is 1 cm more than twice the length of the other leg. The hypotenuse measures 6 cm. Find the lengths of the legs. Round to one decimal place.

43. The hypotenuse of a right triangle is 10.2 m long. One leg is 2.1 m shorter than the other leg. Find the lengths of the legs. Round to one decimal place.

44. The hypotenuse of a right triangle is 17 ft long. One leg is 3.4 ft longer than the other leg. Find the lengths of the legs.

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45. The fatality rate (in fatalities per 100 million vehicle miles driven) can be approximated for drivers x years old according to the function, F1x2 ⫽ 0.0036x2 ⫺ 0.35x ⫹ 9.2. (Source: U.S. Department of Transportation)

46. The braking distance (in feet) of a car going v mph is given by d 1v2 ⫽

v2 ⫹v 20

vⱖ0

a. Approximate the fatality rate for drivers 16 yr old.

a. Find the speed for a braking distance of 150 ft. Round to the nearest mile per hour.

b. Approximate the fatality rate for drivers 40 yr old.

b. Find the speed for a braking distance of 100 ft. Round to the nearest mile per hour.

c. Approximate the fatality rate for drivers 80 yr old. d. For what age(s) is the fatality rate approximately 2.5? 47. Mitch throws a baseball straight up in the air from a cliff that is 48 ft high. The initial velocity is 48 ft/sec. The height (in feet) of the object after t sec is given by h1t2 ⫽ ⫺16t2 ⫹ 48t ⫹ 48. Find the time at which the height of the object is 64 ft. (See Example 4.)

48. An astronaut on the moon throws a rock into the air from the deck of a spacecraft that is 8 m high. The initial velocity of the rock is 2.4 m/sec. The height (in meters) of the rock after t sec is given by h1t2 ⫽ ⫺0.8t2 ⫹ 2.4t ⫹ 8. Find the time at which the height of the rock is 6 m.

Concept 4: Discriminant For Exercises 49–56, a. Write the equation in the form ax2 ⫹ bx ⫹ c ⫽ 0, a 7 0. b. Find the value of the discriminant. c. Use the discriminant to determine the number and type of solutions. (See Example 5.) 49. x2 ⫹ 2x ⫽ ⫺1

50. 12y ⫺ 9 ⫽ 4y2

51. 19m2 ⫽ 8m

52. 2n ⫺ 5n2 ⫽ 0

53. 5p2 ⫺ 21 ⫽ 0

54. 3k2 ⫽ 7

55. 4n1n ⫺ 22 ⫺ 5n1n ⫺ 12 ⫽ 4

56. 12x ⫹ 121x ⫺ 32 ⫽ ⫺9

For Exercises 57–62, determine the discriminant. Then use the discriminant to determine the number of x-intercepts for the function. 57. f 1x2 ⫽ x2 ⫺ 6x ⫹ 5

58. g1x2 ⫽ ⫺x2 ⫺ 4x ⫺ 3

59. h1x2 ⫽ 4x2 ⫹ 12x ⫹ 9

60. k1x2 ⫽ 25x2 ⫺ 10x ⫹ 1

61. p1x2 ⫽ 2x2 ⫹ 3x ⫹ 6

62. m1x2 ⫽ 3x2 ⫹ 4x ⫹ 7

For Exercises 63–68, find the x- and y-intercepts of the quadratic function. (See Examples 6–7.) 63. f 1x2 ⫽ x2 ⫺ 5x ⫹ 3 64. g1x2 ⫽ 2x2 ⫹ 7x ⫹ 2 65. g1x2 ⫽ ⫺x2 ⫹ x ⫺ 1 66. f 1x2 ⫽ 2x2 ⫹ x ⫹ 5

67. p1x2 ⫽ 2x2 ⫹ 5x ⫺ 2

68. h1x2 ⫽ 3x2 ⫹ 2x ⫺ 2

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Concept 5: Mixed Review: Methods to Solve a Quadratic Equation For Exercises 69–86, solve the quadratic equation by using any method. (See Examples 8–11.) 69. a2 ⫹ 3a ⫹ 4 ⫽ 0

70. 4z2 ⫹ 7z ⫽ 0

71. 1x ⫺ 22 2 ⫹ 2x2 ⫺ 13x ⫽ 10

72. 1x ⫺ 32 2 ⫹ 3x2 ⫺ 5x ⫽ 12

73. 4y2 ⫹ 8y ⫺ 5 ⫽ 0

74. k2 ⫺ 4k ⫽ ⫺13

1 2 75. ax ⫹ b ⫹ 4 ⫽ 0 2

76. 12y ⫹ 32 2 ⫽ 9

77. 2y1y ⫺ 32 ⫽ ⫺1

78. w1w ⫺ 52 ⫽ 4

79. 12t ⫹ 521t ⫺ 12 ⫽ 1t ⫺ 321t ⫹ 82

80. 1b ⫺ 121b ⫹ 42 ⫽ 13b ⫹ 221b ⫹ 12

81.

1 2 1 1 x ⫺ x⫹ ⫽0 8 2 4

82.

1 2 1 1 x ⫺ x⫹ ⫽0 6 2 4

83. 32z2 ⫺ 20z ⫺ 3 ⫽ 0

84. 8k2 ⫺ 14k ⫹ 3 ⫽ 0

85. 3p2 ⫺ 27 ⫽ 0

86. 5h2 ⫺ 120 ⫽ 0

Sometimes students shy away from completing the square and using the square root property to solve a quadratic equation. However, sometimes this process leads to a simple solution. For Exercises 87–88, solve the equations two ways. a. Solve the equation by completing the square and applying the square root property. b. Solve the equation by applying the quadratic formula. c. Which technique was easier for you? 87. x2 ⫹ 6x ⫽ 5

88. x2 ⫺ 10x ⫽ ⫺27

Graphing Calculator Exercises 89. Graph Y1 ⫽ x3 ⫺ 27. Compare the x-intercepts with the solutions to the equation x3 ⫺ 27 ⫽ 0 found in Exercise 35.

90. Graph Y1 ⫽ 64x3 ⫹ 1. Compare the x-intercepts with the solutions to the equation 64x3 ⫹ 1 ⫽ 0 found in Exercise 36.

91. Graph Y1 ⫽ 3x3 ⫺ 6x2 ⫹ 6x. Compare the x-intercepts with the solutions to the equation 3x3 ⫺ 6x2 ⫹ 6x ⫽ 0 found in Exercise 37.

92. Graph Y1 ⫽ 5x3 ⫹ 5x2 ⫹ 10x. Compare the x-intercepts with the solutions to the equation 5x3 ⫹ 5x2 ⫹ 10x ⫽ 0 found in Exercise 38.

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Section 7.3

Equations in Quadratic Form

Concepts

1. Solving Equations by Using Substitution

1. Solving Equations by Using Substitution 2. Solving Equations Reducible to a Quadratic

We have learned to solve a variety of different types of equations, including linear, radical, rational, and polynomial equations. Sometimes, however, it is necessary to use a quadratic equation as a tool to solve other types of equations. In Example 1, we will solve the equation 12x2  52 2  1612x2  52  39  0. Notice that the terms in the equation are written in descending order by degree. Furthermore, the first two terms have the same base, 2x2  5, and the exponent on the first term is exactly double the exponent on the second term. The third term is a constant. An equation in this pattern is called quadratic in form. exponent is double

third term is constant

12x2  52 2  1612x2  52 1  39  0. To solve this equation we will use substitution as demonstrated in Example 1.

Example 1

Solving an Equation in Quadratic Form

Solve the equation.

12x 2  52 2  1612x 2  52  39  0

Solution:

12x2  52 2  1612x 2  52  39  0 Substitute u  12x 2  52 .

u2



 39  0

16u

The equation is in the form au2  bu  c  0.

1u  1321u  32  0

Avoiding Mistakes When using substitution, it is critical to reverse substitute to solve the equation in terms of the original variable.

u  13

Factor.

u3

or

Apply the zero product rule.

Reverse substitute.

2x2  5  13

or

2x2  5  3

2x 2  18

or

2x 2  8

x2  9

or

x2  4

x   19

or

x   14

 3

or

 2

The solution set is 53, 3, 2, 26. Skill Practice Solve the equation. Answer 1. {1, 1, 2, 2}

Once the substitution is made, the equation becomes quadratic in the variable u.

1. 13t 2  102 2  513t 2  102  14  0

Reverse substitute.

Write the equations in the form x2  k. Apply the square root property.

All solutions check in the original equation.

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Section 7.3

Equations in Quadratic Form

For an equation written in descending order, notice that u was set equal to the variable factor on the middle term. This is generally the case.

Solving an Equation in Quadratic Form

Example 2

p2Ⲑ3 ⫺ 2p1Ⲑ3 ⫽ 8

Solve the equation.

Solution: p2Ⲑ3 ⫺ 2p1Ⲑ3 ⫽ 8 p2Ⲑ3 ⫺ 2p1Ⲑ3 ⫺ 8 ⫽ 0

Set the equation equal to zero.

1 p 2 ⫺ 21p 2 ⫺ 8 ⫽ 0 1Ⲑ3 2

1Ⲑ3 1

u2 ⫺ 2u

Make the substitution u ⫽ p1Ⲑ3.

Substitute u ⫽ p1Ⲑ3.

Then the equation is in the form au2 ⫹ bu ⫹ c ⫽ 0.

⫺8⫽0

1u ⫺ 42 1u ⫹ 22 ⫽ 0

u⫽4 1Ⲑ3

Factor. u ⫽ ⫺2

or Reverse substitute.

⫽4

or

p1Ⲑ3 ⫽ ⫺2

3 1 p⫽4

or

3 1 p ⫽ ⫺2

3 11 p2 3 ⫽ 142 3

or

3 11 p2 3 ⫽ 1⫺22 3

p

p ⫽ 64

Apply the zero product rule.

The equations are radical equations. Cube both sides.

p ⫽ ⫺8

or

The solution set is 564, ⫺86.

All solutions check in the original equation.

Skill Practice Solve the equation. 2. y 2Ⲑ 3 ⫺ y 1Ⲑ 3 ⫽ 12

Example 3

Solving an Equation in Quadratic Form

Solve the equation.

x ⫺ 1x ⫺ 12 ⫽ 0

Solution: The equation can be solved by first isolating the radical and then squaring both sides (this is left as an exercise—see Exercise 24). However, this equation is also quadratic in form. By writing 1x as x1/2, we see that the exponent on the first term is exactly double the exponent on the middle term. x1 ⫺ x1/2 ⫺ 12 ⫽ 0

1x1/2 2 2 ⫺ 1x1/2 2 1 ⫺ 12 ⫽ 0

Let u ⫽ x1/2.

u2 ⫺ u ⫺ 12 ⫽ 0

1u ⫺ 421u ⫹ 32 ⫽ 0 u ⫽ 4 or

u ⫽ ⫺3

x1/2 ⫽ 4 or x1/2 ⫽ ⫺3

Factor. Solve for u. Reverse substitute. Answer 2. {64, ⫺27}

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Chapter 7 Quadratic Equations and Functions

1x  4 or 1x  3

Avoiding Mistakes Recall from Section 6.7 that when each side of an equation is raised to an even power, we must check the potential solutions.

x  16

The solution set is 5166.

Solve each equation for x. Recall that the principal square root of a number cannot equal 3. The value 16 checks in the original equation.

Skill Practice Solve the equation. 3. z  1z  2  0

2. Solving Equations Reducible to a Quadratic Some equations are reducible to a quadratic equation. In Example 4, we solve a polynomial equation by factoring. The resulting factors are quadratic. Example 4

Solving a Polynomial Equation 4x4  7x2  2  0

Solve the equation.

Solution: 4x4  7x2  2  0

14x2  121x2  22  0 4x2  1  0 or x2  2  0 x2 

1 4

x

or x2  2

This is a polynomial equation. Factor. Set each factor equal to zero. Notice that the two equations are quadratic. Each can be solved by the square root property.

1 B4

or

x  12

Apply the square root property.

1 2

or

x   i 12

Simplify the radicals.

x

1 1 The solution set is e ,  , i 12, i 12 f . 2 2 Skill Practice Solve the equation. 4. 9x4  35x2  4  0

Example 5

Solve the equation.

3y 2  1 y2 y1

Solution: 3y 2  1 y2 y1

This is a rational equation. The LCD is 1y  221y  12. Also note that the restrictions on y are y  2 and y  1.

a Answers

3. 546 (The value 1 does not check.) 1 4. e  , 2i f 3

Solving a Rational Equation

3y 2  b ⴢ 1y  221y  12  1 ⴢ 1y  221y  12 y2 y1

Multiply both sides by the LCD.

3y 2 ⴢ 1y  221y  12  ⴢ 1y  221y  12  1 ⴢ 1y  221y  12 y2 y1

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Section 7.3

3y1y  12  21y  22  1y  221y  12

Clear fractions.

3y2  3y  2y  4  y2  y  2y  2

Apply the distributive property.

3y  5y  4  y  y  2

The equation is quadratic.

2y2  6y  2  0

Write the equation in descending order.

2y2 6y 2 0    2 2 2 2

Each coefficient in the equation is divisible by 2. Therefore, if we divide both sides by 2, the coefficients in the equation are smaller. This will make it easier to apply the quadratic formula.

2

2

y2  3y  1  0

y 



Equations in Quadratic Form

132  1132 2  4112112

Apply the quadratic formula.

2112 3  19  4 2 y

3  113 2

The solution set is e

3  113 2

3  113 y 2

Both solutions check.

3  113 3  113 , f. 2 2

Skill Practice Solve the equation. 5.

t 1  1 2t  1 t4

Section 7.3 Boost your GRADE at ALEKS.com!

Answer 5. e

5  315 f 2

Practice Exercises • Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. What is meant by an equation in quadratic form?

Review Exercises For Exercises 2–7, solve the quadratic equations. 2. 16  12x  32 2

3 2 7 3. ax  b  2 4

4. n 1n  62  13

5. x 1x  82  16

6. 6k2  7k  6

7. 2x2  8x  44  0

Concept 1: Solving Equations by Using Substitution 8. a. Solve the quadratic equation by factoring. b. Solve the equation by using substitution.

u2  2u  24  0

1x  52 2  21x  52  24  0

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Chapter 7 Quadratic Equations and Functions

9. a. Solve the quadratic equation by factoring. b. Solve the equation by using substitution. 10. a. Solve the quadratic equation by factoring. b. Solve the equation by using substitution.

u2 ⫹ 10u ⫹ 24 ⫽ 0

1y2 ⫹ 5y2 2 ⫹ 101y2 ⫹ 5y2 ⫹ 24 ⫽ 0 u2 ⫺ 2u ⫺ 35 ⫽ 0

1w2 ⫺ 6w2 2 ⫺ 21w2 ⫺ 6w2 ⫺ 35 ⫽ 0

For Exercises 11–24, solve the equation by using substitution. (See Examples 1–3.) 11. 1x2 ⫺ 2x2 2 ⫹ 21x2 ⫺ 2x2 ⫽ 3

12. 1x2 ⫹ x2 2 ⫺ 81x2 ⫹ x2 ⫽ ⫺12

13. 1y2 ⫺ 4y2 2 ⫺ 1y2 ⫺ 4y2 ⫽ 20

14. 1w2 ⫺ 2w2 2 ⫺ 111w2 ⫺ 2w2 ⫽ ⫺24

15. m2Ⲑ3 ⫺ m1Ⲑ3 ⫺ 6 ⫽ 0

16. 2n2Ⲑ3 ⫹ 7n1Ⲑ3 ⫺ 15 ⫽ 0

17. 2t2Ⲑ5 ⫹ 7t1Ⲑ5 ⫹ 3 ⫽ 0

18. p2Ⲑ5 ⫹ p1Ⲑ5 ⫺ 2 ⫽ 0

19. y ⫹ 61y ⫽ 16

20. p ⫺ 81p ⫽ ⫺15

21. 2x ⫹ 3 1x ⫺ 2 ⫽ 0

22. 3t ⫹ 51t ⫺ 2 ⫽ 0

23. 16 a

x⫹6 2 x⫹6 b ⫹ 8a b⫹1⫽0 4 4

24. 9 a

x⫹3 2 x⫹3 b ⫺ 6a b⫹1⫽0 2 2

25. In Example 3, we solved the equation x ⫺ 1x ⫺ 12 ⫽ 0 by using substitution. Now solve this equation by first isolating the radical and then squaring both sides. Don’t forget to check the potential solutions in the original equation. Do you obtain the same solution as in Example 3?

Concept 2: Solving Equations Reducible to a Quadratic For Exercises 26–36, solve the equations. (See Examples 4–5.)

28. x2 19x2 ⫹ 72 ⫽ 2

26. t 4 ⫹ t 2 ⫺ 12 ⫽ 0

27. w4 ⫹ 4w2 ⫺ 45 ⫽ 0

29. y2 14y2 ⫹ 172 ⫽ 15

30.

y 12 ⫺1⫽⫺ 10 5y

31. 1 ⫹

34.

32.

x⫹5 x x ⫹ 19 ⫹ ⫽ x 2 4x

33.

3x 2 ⫺ ⫽1 x⫹1 x⫺3

35.

x 1 ⫽ 2x ⫺ 1 x⫺2

36.

z 2 ⫽ 3z ⫹ 2 z⫹1

5 3 ⫽⫺ 2 x x

2t 1 ⫺ ⫽1 t⫺3 t⫹4

Mixed Exercises For Exercises 37–60, solve the equations. 37. x4 ⫺ 16 ⫽ 0

38. t4 ⫺ 625 ⫽ 0

39. 14x ⫹ 52 2 ⫹ 314x ⫹ 52 ⫹ 2 ⫽ 0

40. 215x ⫹ 32 2 ⫺ 15x ⫹ 32 ⫺ 28 ⫽ 0

41. 4m4 ⫺ 9m2 ⫹ 2 ⫽ 0

42. x4 ⫺ 7x2 ⫹ 12 ⫽ 0

43. x6 ⫺ 9x3 ⫹ 8 ⫽ 0

44. x6 ⫺ 26x3 ⫺ 27 ⫽ 0

45. 2x2 ⫹ 20 ⫽ 32x

46. 2x2 ⫹ 60 ⫽ 41x

47. 14t ⫹ 1 ⫽ t ⫹ 1

48. 1t ⫹ 10 ⫽ t ⫹ 4

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Problem Recognition Exercises

x1 2 x1 50. a b  3a b  10  0 5 5

51. x2/3  x1/3  20

52. x2/5  3x1/5  2

53. m4  2m2  8  0

54. 2c4  c2  1  0

55. a3  16a  a2  16  0 (Hint: Factor by grouping first.)

56. b3  9b  b2  9  0

57. x3  5x  4x2  20  0

49. 2a

t4 2 t4 b a b30 3 3

2 2 2 b  8a b  12  0 x3 x3

59. a

58. y3  8y  3y2  24  0

5 2 5 60. a b  6a b  16  0 x1 x1

Graphing Calculator Exercises 61. a. Solve the equation x4  4x2  4  0. b. How many solutions are real and how many solutions are imaginary? c. How many x-intercepts do you anticipate for the function defined by y  x4  4x2  4? d. Graph Y1  x4  4x2  4 on a standard viewing window. 62. a. Solve the equation x4  2x2  1  0. b. How many solutions are real and how many solutions are imaginary? c. How many x-intercepts do you anticipate for the function defined by y  x4  2x2  1? d. Graph Y1  x4  2x2  1 on a standard viewing window. 63. a. Solve the equation x4  x3  6x2  0. b. How many solutions are real and how many solutions are imaginary? c. How many x-intercepts do you anticipate for the function defined by y  x4  x3  6x2? d. Graph Y1  x4  x3  6x2 on a standard viewing window. 64. a. Solve the equation x4  10x2  9  0. b. How many solutions are real and how many solutions are imaginary? c. How many x-intercepts do you anticipate for the function defined by y  x4  10x2  9? d. Graph Y1  x4  10x2  9 on a standard viewing window.

Problem Recognition Exercises Quadratic and Quadratic Type Equations For Exercises 1–4, solve each equation by a. Completing the square and applying the square root property. b. Using the quadratic formula. 1. x2  10x  3  0

2. v2  16v  5  0

3. 3t 2  t  4  0

603

4. 4y2  3y  5  0

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Chapter 7 Quadratic Equations and Functions

In Exercises 5–24, we have presented all types of equations that you have learned up to this point. For each equation, a. First determine the type of equation that is presented. Choose from: linear equation, quadratic equation, quadratic in form, rational equation, or radical equation. b. Solve the equation by using a suitable method. 5. t2  5t  14  0

6. a2  9a  20  0

8. x4  3x2  4  0

9. x  3x1/2  4  0

7. a4  10a2  9  0 10. x2  9x  2  0

11. 8b1b  12  213b  42  4b12b  32

12. 6x1x  12  31x  42  3x12x  52

13. 5a1a  62  1013a  12

14. 4x1x  32  612x  42

15.

17. c2  20c  1  0

18. d 2  18d  4  0

19. 2u1u  32  412  u2

20. 3y1y  22  91y  12

21. 12b  3  b

22. 15t  6  t

23. x2/3  2x1/3  15  0

24. y2/3  5y1/3  4  0

16.

5 v 12   2 v4 v3 v  7v  12

Section 7.4 Concepts 1. Quadratic Functions of the Form f ( x) ⫽ x 2 ⫹ k 2. Quadratic Functions of the Form f ( x) ⫽ (x ⫺ h) 2 3. Quadratic Functions of the Form f ( x) ⫽ ax 2 4. Quadratic Functions of the Form f (x) ⫽ a(x ⫺ h)2 ⫹ k

t 3 17   2 t5 t4 t  t  20

Graphs of Quadratic Functions In Section 4.8, we defined a quadratic function as a function of the form f 1x2  ax2  bx  c 1a  02. We also learned that the graph of a quadratic function is a parabola. In this section, we will learn how to graph parabolas. A parabola opens upward if a 7 0 (Figure 7-3) and opens downward if a 6 0 (Figure 7-4). If a parabola opens upward, the vertex is the lowest point on the graph. If a parabola opens downward, the vertex is the highest point on the graph. The axis of symmetry is the vertical line that passes through the vertex. a0

a0

y

y 10

10 8 6 4 2 108 6 4 2 2 4

x  2

h(x)  x2  4x  5 2

4 6

8 10

x

108 6 4 2 2 4

6

6

8 10

8 10

Figure 7-3

x1

8 6 4 2

g(x)  x2  2x  1 2

4 6

Figure 7-4

8 10

x

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Section 7.4

Graphs of Quadratic Functions

1. Quadratic Functions of the Form f ( x ) ⴝ x 2 ⴙ k One technique for graphing a function is to plot a sufficient number of points on the graph until the general shape and defining characteristics can be determined. Then sketch a curve through the points.

Graphing Quadratic Functions of the Form f (x ) ⫽ x 2 ⫹ k

Example 1

Graph the functions f, g, and h on the same coordinate system. f 1x2  x2

g1x2  x2  1

h1x2  x2  2

Solution: Several function values for f, g, and h are shown in Table 7-1 for selected values of x. The corresponding graphs are pictured in Figure 7-5. Table 7-1 x

f(x) ⴝ x 2

g(x) ⴝ x 2 ⴙ 1

h(x) ⴝ x 2 ⴚ 2

3

9

10

7

2

4

5

2

1

1

2

1

0

0

1

2

1

1

2

1

2

4

5

2

3

9

10

7

10

g(x)  x2  1 f(x)  x2 y h(x)  x2  2

8 6 4 2 108 6 4 2 2 4 6

2

4 6

8 10

x

8 10

Figure 7-5

Skill Practice Refer to the graph of f 1x 2  x 2  k to determine the value of k. 1.

y 10 8 6 4 2 108 6 4 2 2 4 6

2

4

6 8 10

x

8 10

Notice that the graphs of g 1x2  x2  1 and h 1x2  x2  2 take on the same shape as f 1x2  x2. However, the y values of g are 1 greater than the y values of f. Hence, the graph of g 1x2  x2  1 is the same as the graph of f 1x2  x2 shifted up 1 unit. Likewise the y-values of h are 2 less than those of f. The graph of h 1x2  x2  2 is the same as the graph of f 1x2  x2 shifted down 2 units. The functions in Example 1 illustrate the following properties of quadratic functions of the form f 1x2  x2  k.

Answer 1. k  8

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PROPERTY Graphs of f ( x ) ⴝ x 2 ⴙ k

If k 7 0, then the graph of f 1x2  x2  k is the same as the graph of y  x2 shifted up k units. If k 6 0, then the graph of f 1x2  x2  k is the same as the graph of y  x2 shifted down 0k 0 units.

Calculator Connections Try experimenting with a graphing calculator by graphing functions of the form y  x2  k for several values of k.

Graphing Quadratic Functions of the Form f (x ) ⫽ x 2 ⫹ k

Example 2

Sketch the functions defined by

TIP: For more accuracy in the graph, plot one or two points near the vertex and use the symmetry of the curve to find additional points on the graph.

a. m1x2  x2  4

7 b. n1x2  x2  2

m(x) 10 8 6 4 2

10 8 6 4 2 2 4

Solution: a. m1x2  x2  4

m1x2  x  142 2

Because k  4, the graph is obtained by shifting the graph of y  x2 down 04 0 units (Figure 7-6). b. n1x2  x2 

7 2

Because k  72, the graph is obtained by shifting the graph of y  x2 up 72 units (Figure 7-7).

2. y 10 8 6 4 2 108 6 4 2 2 4 6 8 10

g(x)  x2  3 f(x)  x2

Figure 7-6 n(x) 8 7 6 5 4

6 8 10

n(x)  x2 

x

7 2

3 2 1

Skill Practice

4

x

8 10

8 10

1

2 3

Figure 7-7

h(x)  x2  5 2

2 4 6

6

5 4 3 2 1 1 2

Answer

m(x)  x2  4

2. Graph the functions f, g, and h on the same coordinate system. f 1 x 2  x2 g 1 x 2  x2  3 h 1x2  x2  5

4

5

x

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Section 7.4

Graphs of Quadratic Functions

2. Quadratic Functions of the Form f ( x)  ( x  h)2

The graph of f 1x2  x2  k represents a vertical shift (up or down) of the function y  x2. Example 3 shows that functions of the form f 1x2  1x  h2 2 represent a horizontal shift (left or right) of the function y  x2.

Graphing Quadratic Functions of the Form f (x ) ⫽ (x ⫺ h ) 2

Example 3

Graph the functions f, g, and h on the same coordinate system. f 1x2  x2

g1x2  1x  12 2

h1x2  1x  22 2

Solution: Several function values for f, g, and h are shown in Table 7-2 for selected values of x. The corresponding graphs are pictured in Figure 7-8. Table 7-2 x

f(x)  x 2 g(x)  (x  1)2 h(x)  (x  2)2

4

16

9

36

3

9

4

25

2

4

1

16

1

1

0

9

0

0

1

4

1

1

4

1

2

4

9

0

3

9

16

1

4

16

25

4

5

25

36

9

g(x)  (x  1)2 f(x)  x2 y h(x)  (x  2)2

10 8 6

4 2 10 8 6 4 2 2 4 6

2

4 6

8 10

x

8 10

Figure 7-8

Skill Practice Refer to the graph of f 1x 2  1x  h2 2 to determine the value of h. 3.

y 10 8 6 4 2 108 6 4 2 2 4 6

2

4

6 8 10

x

8 10

Example 3 illustrates the following properties of quadratic functions of the form f 1x2  1x  h2 2.

Answer 3. h  4

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PROPERTY Graphs of f ( x ) ⴝ ( x ⴚ h )2

If h 7 0, then the graph of f 1x2  1x  h2 2 is the same as the graph of y  x2 shifted h units to the right. If h 6 0, then the graph of f 1x2  1x  h2 2 is the same as the graph of y  x2 shifted 0 h 0 units to the left. From Example 3 we have

h1x2  1x  22 2

and

g1x2  1x  12 2

y  x2 shifted 2 units to the right

Example 4

g1x2  3x  112 4 2 y  x2 shifted 0 1 0 unit to the left

Graphing Functions of the Form f ( x ) ⫽ (x ⫺ h )2

Sketch the functions p and q. a. p1x2  1x  72 2

b. q1x2  1x  1.62 2

Solution:

a. p1x2  1x  72 2 Because h  7 7 0, shift the graph of y  x2 to the right 7 units (Figure 7-9).

p(x)

p(x)  (x  7)2

10 8 6 4 2 10 8 64 2 2 4

2

4 6

8 10

x

6 8 10

Figure 7-9

b. q1x2  1x  1.62 2

q1x2  3x  11.62 4 2

Because h  1.6 6 0, shift the graph of y  x2 to the left 1.6 units (Figure 7-10).

q(x) 10 8 6 4 2

q(x)  (x  1.6)2

10 8 6 4 2 2 4

2

4 6

6

Answer

8 10

4. g(x)  (x  3)2

f(x)  x2

y

108 6 4 2 2 4 6 8 10

Figure 7-10

h(x)  (x  6)2

10 8 6 4 2

Skill Practice

2

4

6 8 10

x

4. Graph the functions f, g, and h on the same coordinate system. f 1x2  x 2 g 1x2  1x  32 2 h 1x2  1x  62 2

8 10

x

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609

Graphs of Quadratic Functions

3. Quadratic Functions of the Form f ( x)  ax2 Examples 5 and 6 investigate functions of the form f 1x2  ax2 Example 5

1a  02 .

Graphing Functions of the Form f ( x ) ⫽ ax 2

Graph the functions f, g, and h on the same coordinate system. f 1x2  x2

g 1x2  2x2

h 1x2 

1 2 x 2

Solution: Several function values for f, g, and h are shown in Table 7-3 for selected values of x. The corresponding graphs are pictured in Figure 7-11.

Table 7-3 x

f(x)  x 2

g(x)  2x 2

h(x)  12x 2

3

9

18

9 2

2

4

8

2

1

1

2

1 2

0

0

0

0

1

1

2

1 2

2

4

8

2

3

9

18

9 2

g(x)  2x2 f(x)  x2 1 y h(x)  2 x2

10 8 6 4 2

10 8 6 4 2 2 4

2

4 6

8 10

x

6 8 10

Figure 7-11

Skill Practice 5. Graph the functions f, g, and h on the same coordinate system. f 1x2  x2 g 1x2  3x2 1 h 1x2  x2 3

In Example 5, the function values defined by g 1x2  2x2 are twice those of f 1x2  x2. The graph of g 1x2  2x2 is the same as the graph of f 1x2  x2 stretched vertically by a factor of 2 (the graph appears narrower than f 1x2  x2). In Example 5, the function values defined by h1x2  12 x2 are one-half those of f 1x2  x2. The graph of h1x2  12 x2 is the same as the graph of f 1x2  x2 shrunk vertically by a factor of 21 (the graph appears wider than f 1x2  x2). Example 6

Graphing Functions of the Form f ( x ) ⫽ ax 2

Graph the functions f, g, and h on the same coordinate system. f 1x2  x2

g 1x2  3x2

1 h 1x2   x2 3

Answer 5.

g(x)  3x2 f(x)  x2 y 1 h(x)  3 x2

10 8 6 4 2

108 6 4 2 2 4 6 8 10

2

4

6 8 10

x

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Chapter 7 Quadratic Equations and Functions

Solution: Several function values for f, g, and h are shown in Table 7-4 for selected values of x. The corresponding graphs are pictured in Figure 7-12. Table 7-4 y

x

f(x)  x 2

3

9

g(x)  3x 2

h(x)  13x 2

27

3

2

4

12

43

1

1

3

13

2 1 −5 −4 −3 −2 −1

−1 −2 −3 −4 −5

1 2 3

0

0

0

0

1

1

3

13

−7 −8

g(x) = 3x2

2

4

12

43

3

9

27

3

4

5

x

h(x) = 13 x2

−6

f(x) = x2

Figure 7-12

Skill Practice 6. Graph the functions f, g, and h on the same coordinate system. 1 h 1x2   x2 f 1x2  x2 g 1x2  2x2 2

Example 6 illustrates that if the coefficient of the square term is negative, the parabola opens downward. The graph of g 1x2  3x2 is the same as the graph of f 1x2  x2 with a vertical stretch by a factor of 03 0 . The graph of h1x2  13x2 is the same as the graph of f 1x2  x2 with a vertical shrink by a factor of 013 0 .

PROPERTY Graphs of f ( x )  ax 2 1. If a 7 0, the parabola opens upward. Furthermore, • If 0 6 a 6 1, then the graph of f 1 x2  ax2 is the same as the graph of y  x2 with a vertical shrink by a factor of a. • If a 7 1, then the graph of f 1 x2  ax2 is the same as the graph of y  x2 with a vertical stretch by a factor of a. 2. If a 6 0, the parabola opens downward. Furthermore, • If 0 6 0a 0 6 1, then the graph of f 1x2  ax2 is the same as the graph of y  x2 with a vertical shrink by a factor of 0a 0 . • If 0a 0 7 1, then the graph of f 1x2  ax2 is the same as the graph of y  x2 with a vertical stretch by a factor of 0 a 0 . Answer 6.

y

4. Quadratic Functions of the Form f ( x)  a( x  h)2  k

10 8 6 4 2 108 6 4 2 2 4 6 8 10

f(x) = x2 g(x) = 2x2

2

4

6 8 10

h(x) = 1 x2 2

x

We can summarize our findings from Examples 1–6 by graphing functions of the form f 1x2  a1x  h2 2  k 1a  02 . The graph of y  x2 has its vertex at the origin 10, 02 . The graph of f 1x2  a1x  h2 2  k is the same as the graph of y  x2 shifted to the right or left h units and shifted up or down k units. Therefore, the vertex is shifted from (0, 0) to (h, k). The axis of symmetry is the vertical line through the vertex. Therefore, the axis of symmetry must be the line x  h.

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Section 7.4

611

Graphs of Quadratic Functions

PROPERTY Graphs of f ( x )  a( x  h )2  k

1. The vertex is located at 1h, k2 . 2. The axis of symmetry is the line x  h. 3. If a 7 0, the parabola opens upward, and k is the minimum value of the function. 4. If a 6 0, the parabola opens downward, and k is the maximum value of the function. f(x)

Minimum value

f(x)

k

(h, k)

Maximum k value

Maximum (h, k) point

a0 Minimum point

x

x

a0 xh

Example 7

xh

Graphing a Function of the Form f (x ) ⫽ a (x ⫺ h )2 ⫹ k

Given the function defined by

f 1x2  21x  32 2  4

a. Identify the vertex. b. Sketch the function. c. Identify the axis of symmetry. d. Identify the maximum or minimum value of the function.

Solution:

a. The function is in the form f 1x2  a1x  h2 2  k, where a  2, h  3, and k  4. Therefore, the vertex is 13, 42 .

b. The graph of f is the same as the graph of y  x2 shifted to the right 3 units, shifted up 4 units, and stretched vertically by a factor of 2 (Figure 7-13). c. The axis of symmetry is the line x  3. d. Because a 7 0, the function opens upward. Therefore, the minimum function value is 4. Notice that the minimum value is the minimum y-value on the graph. Skill Practice

f(x) 10 8 6

f(x)  2(x  3)2  4

4 2 10 8 6 4 2 2 4

(3, 4) 2

4 6

6 8 10

8 10

x

Answers x3

Figure 7-13

7. Given the function defined by g 1x2  31x  12 2  3 a. Identify the vertex. b. Sketch the graph. c. Identify the axis of symmetry. d. Identify the maximum or minimum value of the function.

7. a. Vertex: (1, 3) b. y 10 8 6 4 2 108 6 4 2 2 4 6

2

4

6 8 10

8 10

c. Axis of symmetry: x  1 d. Minimum value: 3

x

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Chapter 7 Quadratic Equations and Functions

Calculator Connections Topic: Using the Maximum and Minimum Features Some graphing calculators have Minimum and Maximum features that enable the user to approximate the minimum and maximum values of a function.

Example 8

Graphing a Function of the Form f ( x ) ⫽ a (x ⫺ h ) 2 ⫹ k

Given the function defined by g 1x2  1x  22 2 

7 4

a. Identify the vertex. b. Sketch the function. c. Identify the axis of symmetry. d. Identify the maximum or minimum value of the function.

Solution: a. g 1x2  1x  22 2 

7 4

7  13 x  122 4 2  a b 4 The function is in the form g 1x2  a1x  h2 2  k, where a  1, h  2, and k  74 . Therefore, the vertex is 12, 74 2 . b. The graph of g is the same as the graph of y  x2 shifted to the left 2 units, shifted down 74 units, and opening downward (Figure 7-14).

g(x) 10 8 6 7

(2,  4 ) 10 8 6 4

8. a. Vertex: (4, 2) b.

d. The parabola opens downward, so the maximum function value is 74.

y

10 8 6 4 2 108 6 4 2 2 4 6

Skill Practice 8. Given the function defined by 2

4

6 8 10

8 10

c. Axis of symmetry: x  4 d. Maximum value: 2

x

a. b. c. d.

2 4 6

c. The axis of symmetry is the line x  2.

Answers

4 2 2

4 6

8 10

x

g(x)  (x  2)2 

8 10

x  2

Figure 7-14

1 h 1x2   1x  42 2  2 2

Identify the vertex. Sketch the graph. Identify the axis of symmetry. Identify the maximum or minimum value of the function.

7 4

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Section 7.4

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Graphs of Quadratic Functions

Practice Exercises • Practice Problems • Self-Tests • NetTutor

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• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Parabola

b. Vertex

c. Axis of symmetry

d. Maximum value

e. Minimum value

Review Exercises For Exercises 2–8, solve the equations. 2. x2  x  5  0

3. 1y  32 2  4

4. 12a  2  a  1

6. 2z2  3z  9  0

7. x2/3  5x1/3  6  0

8. m2 1m2  62  27

5. 5t1t  22  3

Concept 1: Quadratic Functions of the Form f (x ) ⴝ x 2 ⴙ k

9. Describe how the value of k affects the graph of a function defined by f 1x2  x2  k.

For Exercises 10–17, graph the functions. (See Examples 1–2.) 11. f 1x2  x2  2

10. g1x2  x2  1

y

y

1

2

3 4

5

x

14. T(x)  x2 

3 4

4 5

1

2

3 4

5

x

5 4 3 2 1 1 2 3

3 2

2

3 4

5

x

5 4 3 2 1 1 2 3 4 5

3 4

5

x

5 4

17. n1x2  x2 

y

2

3 4

5

x

5 4 3 2 1 1 2 3 4 5

Concept 2: Quadratic Functions of the Form f (x ) ⴝ (x ⴚ h )2

3 4

5

x

1 3

y

5 4 3 2 1 1

1 2

5 4 3 2 1 1 2 3 4 5

16. M1x2  x2 

5 4 3 2 1 1

1 2

4 5

y

y

5 4 3 2 1 1 2 3

5 4 3 2 1 1 2 3

15. S1x2  x2 

5 4 3 2 1

y 5 4 3 2 1

5 4 3 2 1

4 5

4 5

13. q1x2  x2  4

y

5 4 3 2 1

5 4 3 2 1 5 4 3 2 1 1 2 3

12. p1x2  x2  3

5 4 3 2 1

1 2 3 4 5

x

18. Describe how the value of h affects the graph of a function defined by f 1x2  1x  h2 2.

5 4 3 2 1 1 2 3 4 5

1 2

3 4

5

x

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For Exercises 19–26, graph the functions. (See Examples 3–4.) 19. r1x2  1x  12 2

20. h1x2  1x  22 2

y

y

7 6 5 4 3 2 1

21. k1x2  1x  32 2 y

1 2 3 4 5

x

1 2

6 5 4 3 2 1 1

3 4

x

3 2 1 1 2 3

2 3

3 2 23. A1x2  ax  b 4

3 2 24. r1x2  ax  b 2

1 2

1 2

3 4

5

x

6 5 4 3 2 1 1 2 3

3 4

5 6 7

x

2 1 1

1 2

3 4

y

y

3 4

x

x

26. V1x2  1x  2.52 2 7 6 5 4 3 2 1

7 6 5 4 3 2 1 1 2

5 6 7 8

2 3

25. W1x2  1x  1.252 2

7 6 5 4 3 2 1

5 4 3 2 1 1 2 3

7 6 5 4 3 2 1

y

y 7 6 5 4 3 2 1

y

7 6 5 4 3 2 1

7 6 5 4 3 2 1

54 3 2 1 1 2 3

22. L1x2  1x  42 2

1 2

5 4 3 2 1 1

3 4

5

x

1 2

5 4 3 2 1 1

3 4

5

x

2 3

2 3

Concept 3: Quadratic Functions of the Form f (x ) ⴝ ax 2

27. Describe how the value of a affects the graph of a function defined by f 1x2  ax2, where a  0. 28. How do you determine whether the graph of a function defined by h1x2  ax2  bx  c 1a  02 opens upward or downward?

For Exercises 29–36, graph the functions. (See Examples 5–6.) 29. f1x2  4x2

y

y 8 7 6 4 3

5 4 3 2 1 1 2

2 1 5 4 3 2 1 1 2

1 2 3

4

5

x

33. c1x2  x2

3 4 5

4 5

4 5

4

5

1 2 3

4

5

x

2 3 4 5 6 7 8

1

2

3 4

5

x

5 4 3 2 1 1 2 3

1 2

3 4

5

x

2

3 4

5

1 2

3 4

5

x

1 36. f 1x2   x2 4 y

5

3

4 3

2 1

2 1

5 4 3 2 1 1

5 4 3 2 1 1

1

4 5

y

2 1 5 4 3 2 1 1

y 5 4 3 2 1

1 35. v1x2   x2 5

y

y

3

x

5 4 3 2 1 1 2 3

1 2 3

34. g1x2  4x2

5 4 3 2 1 5 4 3 2 1 1 2

y

5 4 3 2 1

5 4 3 2 1

5

1 32. f 1x2  x2 5

1 31. h1x2  x2 4

30. g1x2  3x2

1 2

3 4

5

x

2 3

2 3

4 5

4 5

6 7

x

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Graphs of Quadratic Functions

Concept 4: Quadratic Functions of the Form f (x )  a (x  h )2  k For Exercises 37–44, match the function with its graph. 1 37. f 1x2   x2 4

38. g1x2  1x  32 2

39. k1x2  1x  32 2

1 40. h 1x2  x2 4

41. t1x2  x2  2

42. m1x2  x2  4

43. p1x2  1x  12 2  3

44. n1x2  1x  22 2  3

a.

b.

y

c.

y

d.

y

y

5

5

5

5

4 3

4 3

4 3

4 3

2 1

2 1

2 1

2 1

5 4 3 2 1 1 2

1 2 3

4

5

x

5 4 3 2 1 1 2

1 2 3

4 5

x

5 4 3 2 1 1 2

1

2 3

4

5

x

5 4 3 2 1 1 2

3

3

3

3

4 5

4 5

4 5

4 5

e.

y

g.

y

f.

y

h.

5

4 3

4 3

4 3

4 3

2 1

2 1

2 1

2 1

5 4 3 2 1 1 2

1

2 3

4

5

x

5 4 3 2 1 1 2

1

2 3

4

5

x

5 4 3 2 1 1 2

1 2 3

4

5

x

2 3

4

5

1

2 3

4

5

x

y

5

5

1

5

5 4 3 2 1 1 2

3

3

3

3

4 5

4 5

4 5

4 5

x

For Exercises 45–64, graph the parabola and the axis of symmetry. Label the coordinates of the vertex, and write the equation of the axis of symmetry. (See Examples 7–8.) 45. y  1x  32 2  2

46. y  1x  22 2  3

y

y

8 7 6 5 4 3 2 1 3 2 1 1

1 2

3 4

5 6 7

x

4 5 6 7 8

2

3 4

5 6 7

x

3 4

5 6 7 8

x

3 2 1 1 2 3 4 5 6 7 8

1 2

3 4

5

x

7 6 5 4 3 2 1 1

x

2

3

x

4

52. y  1x  22 2  2 y

y

1 2 3 4 5 6 7

1

2 3

51. y  1x  32 2  3

2 1 1 2

5 4 3 2 1 1 2 3

6 5 4 3 2 1

4 5

y

y

2 3

1

3 2 1 1

50. y  1x  22 2  4

2 1

y

y

2

49. y  1x  42 2  2

48. y  1x  32 2  1

5 4 3 2 1

8 7 6 5 4 3 2 1

2

2 1 1

47. y  1x  12 2  3

5 4 3 2 1

5 4 3 2 1 7 6 5 4 3 2 1 1 2 3 4 5

1 2

3

x

6 54 3 2 1 1 2 3 4 5

1 2 3 4

x

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53. y  1x  12 2  1

54. y  1x  42 2  4 y

y

2 1 1 1 2

3 4

5

x

1

2

x

4 3 2 1 1 2

58. y  4x2  3

1 2 3

4 3 2 1 1 2 1

2 3 4

4

5

x

3 4

5 4 3 2 1 1

1 61. y   1x  12 2  2 4

1 2 3

4

5

x

2 3 4

2 3 4

5

6

x

y

7 6 5 4 3 2 1 1 2 3 1

2

3

x

1 64. y   1x  22 2  1 3 y 5 4 3 2 1

2

2

1

3 4

3 4

5

5

6

3

7

2 5

2

5 6

y

2 3 4

1

4

x

5 4 3 2 1

1

x

3 2 1

4 3 2 1 1

4 3 2 1 1

6

60. y  21x  32 2  1

1 63. y  1x  22 2  1 3

x

5

4 5

9 8 7 6 5 4 3

65. Compare the graphs of the following equations. a. y  x2  3 b. y  1x  32 2

7 6 5 4 3 2 1 1 2

2 3 4

3

3

y

y

x

1

y

1 62. y  1x  12 2  2 4

6 5 4 3 2 1

6

7 6 5 4 3 2 1

2 1

5

59. y  21x  32 2  1

9 8 7 6 5 4 3

2

c. y  3x2

5 4 3 2 1

y

y

1

y

8 7 6 5 4 3 2 1

2 3

6 5 4 3 2 1

4 3 2 1 1

5 6 7 8

6

57. y  4x2  3

5 4 3 2 1 1

3 4

4 5

2 3

56. y  31x  12 2

y

4 3 2 1

7 6 5 4 3 2 1 5 4 3 2 1 1

55. y  31x  12 2

1

2 3 4

5

6

x

4 3 2 1 1

66. Compare the graphs of the following equations. a. y  1x  22 2 b. y  2x2 c. y  x2  2

1

2 3 4

5

6

x

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Graphs of Quadratic Functions

For Exercises 67–78, write the coordinates of the vertex and determine if the vertex is a maximum point or a minimum point. Then write the maximum or minimum value. 67. f 1x2  41x  62 2  9

68. g1x2  31x  42 2  7

3 70. h1x2   1x  52 2  10 7

71. k1x2 

73. n1x2  6x2 

21 4

76. B1x2  51x  32 2 

1 1x  82 2 2

74. q1x2  4x2  1 4

1 6

77. F1x2  7x 2

2 69. p1x2   1x  22 2  5 5 72. m1x2 

2 1x  112 2 9

75. A1x2  21x  72 2 

3 2

78. G1x2  7x2

79. True or false: The function defined by g1x2  5x2 has a maximum value but no minimum value. 80. True or false: The function defined by f 1x2  21x  52 2 has a maximum value but no minimum value. 81. True or false: If the vertex 12, 82 represents a minimum point, then the minimum value is 2. 82. True or false: If the vertex 12, 82 represents a maximum point, then the maximum value is 8.

Expanding Your Skills 83. A suspension bridge is 120 ft long. Its supporting cable hangs in a shape that resembles a parabola. The function defined by H1x2  901 1x  602 2  30 1where 0 x 1202 approximates the height of the supporting cable a distance of x ft from the end of the bridge (see figure).

H(x) 120 ft

a. What is the location of the vertex of the parabolic cable?

0

b. What is the minimum height of the cable?

x 0

c. How high are the towers at either end of the supporting cable? 84. A 50-m bridge over a crevasse is supported by a parabolic arch. The function defined by f 1x2  0.161x  252 2  100 1where 0 x 502 approximates the height of the supporting arch x meters from the end of the bridge (see figure). a. What is the location of the vertex of the arch? b. What is the maximum height of the arch (relative to the origin)?

f(x) 50 m

0

0

x

85. The staging platform for a fireworks display is 6 ft above ground, and the mortars leave the platform at 96 ft/sec. The height of the mortars, h(t), can be modeled by h(t)  16(t  3)2  150, where t is the time in seconds after launch. a. If the fuses are set for 3 sec after launch, at what height will the fireworks explode? b. Will the fireworks explode at their maximum height? Explain.

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Chapter 7 Quadratic Equations and Functions

Section 7.5 Concepts 1. Writing a Quadratic Function in the Form f (x ) ⫽ a(x ⫺ h)2 ⫹ k 2. Vertex Formula 3. Determining the Vertex and Intercepts of a Quadratic Function 4. Applications and Modeling of Quadratic Functions

Vertex of a Parabola: Applications and Modeling 1. Writing a Quadratic Function in the Form f(x) ⫽ a(x ⫺ h)2 ⫹ k The graph of a quadratic function is a parabola, and if the function is written in the form f 1x2  a1x  h2 2  k 1a  02 , then the vertex is at (h, k). A quadratic function can be written in the form f 1x2  a1x  h2 2  k 1a  02 by completing the square. The process is similar to the steps outlined in Section 7.1 except that all algebraic manipulations are performed on the right-hand side of the equation. Example 1

Writing a Quadratic Function in the Form f (x ) ⫽ a (x ⫺ h)2 ⫹ k (a ⫽ 0)

Given: f 1x2  x2  8x  13

a. Write the function in the form f 1x2  a1x  h2 2  k. b. Identify the vertex, axis of symmetry, and minimum function value.

Solution:

a. f 1x2  x2  8x  13  11x2  8x2  13

Avoiding Mistakes

 11x2  8x

2  13

Do not factor out the leading coefficient from the constant term.

Rather than dividing by the leading coefficient on both sides, we will factor out the leading coefficient from the variable terms on the right-hand side. Next, complete the square on the expression within the parentheses: 3 12 182 4 2  16.

 11x2  8x  16  162  13

Rather than add 16 to both sides of the function, we add and subtract 16 within the parentheses on the right-hand side. This has the effect of adding 0 to the right-hand side.

 11x2  8x  162  16  13

Use the associative property of addition to regroup terms and isolate the perfect square trinomial within the parentheses.

 1x  42 2  3

Factor and simplify.

b. f 1x2  1x  42 2  3 The vertex is 14, 32 . The axis of symmetry is x  4. Because a 7 0, the parabola opens upward. The minimum value is 3 (Figure 7-15). Answers

1. a. f 1x2  1x  42  17 b. Vertex: 14, 172; axis of symmetry: x  4; minimum value: 17 2

Skill Practice

y 10 f(x)  (x  4)2  3 8 6 4 2 x 10 8 6 4 2 2 4 6 8 10 2

(4, 3)

x  4

4 6 8 10

Figure 7-15

1. Given: f 1x2  x2  8x  1 a. Write the equation in the form f 1x2  a1x  h2 2  k. b. Identify the vertex, axis of symmetry, and minimum value of the function.

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Section 7.5

Example 2

Vertex of a Parabola: Applications and Modeling

Analyzing a Quadratic Function

Given: f 1x2  2x2  12x  16

a. Write the function in the form f 1x2  a1x  h2 2  k. b. Find the vertex, axis of symmetry, and maximum function value. c. Find the x- and y-intercepts. d. Sketch the graph of the function.

Solution:

a. f 1x2  2x2  12x  16 2  16

 21x2  6x

 21x2  6x  9  92  16  21x2  6x  92  122192  16

 21x  32 2  18  16

To find the vertex, write the function in the form f 1x2  a1x  h2 2  k. If the leading coefficient is not 1, factor the coefficient from the variable terms. Add and subtract the quantity 3 12 162 4 2  9 within the parentheses. To remove the term 9 from the parentheses, we must first apply the distributive property. When 9 is removed from the parentheses, it carries with it a factor of 2. Factor and simplify.

 21x  32 2  2 b. f 1x2  21x  32 2  2 The vertex is (3, 2). The axis of symmetry is x  3. Because a 6 0, the parabola opens downward and the maximum value is 2. c. The y-intercept is given by f 102  2102 2  12102  16  16. The y-intercept is 10, 162 .

To find the x-intercept(s), find the real solutions to the equation f 1x2  0. f 1x2  2x2  12x  16 0  2x2  12x  16

Substitute f 1x2  0.

0  21x2  6x  82

Factor.

0  21x  421x  22 x4

or

x2

y 4 2 4 3 2 1

f(x)  2x2  12x 16

The x-intercepts are (4, 0) and (2, 0). d. Using the information from parts (a)–(c), sketch the graph (Figure 7-16).

-2 -4

(2, 0) 1 2

(3, 2) (4, 0) 3

4 5

-6

8 10 12 14 16

x3

(0, 16)

Figure 7-16

6

x

619

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Skill Practice

2. Given: g 1x2  x2  6x  5 a. Write the equation in the form g 1x2  a 1x  h2 2  k. b. Identify the vertex, axis of symmetry, and maximum value of the function. c. Determine the x- and y-intercepts. d. Graph the function.

2. Vertex Formula

Completing the square and writing a quadratic function in the form f 1x2  a1x  h2 2  k 1a  02 is one method to find the vertex of a parabola. Another method is to use the vertex formula. The vertex formula can be derived by completing the square on f 1x2  ax2  bx  c 1a  02. f 1x2  ax2  bx  c  a ax2 

b x a

 a ax2 

b b2 b2 x  2  2b  c a 4a 4a

 a ax2 

b b2 b2 x  2 b  1a2a 2 b  c a 4a 4a b 2 b2 b  c 2a 4a

 a ax 

b 2 b2 b c 2a 4a

 a ax 

b 2 4ac b2 b   2a 4a 4a

 a ax 

b 2 4ac  b2 b  2a 4a

f 1x2  a1x

2. a. g 1x2  1x  32  4 b. Vertex: (3, 4); axis of symmetry: x  3; maximum value: 4 c. x -intercepts: (5, 0) and (1, 0); y -intercept: (0, 5) y d.

Add and subtract 3 12 1b a2 4 2  b2 14a2 2 within the parentheses. Apply the distributive property and remove the term b2 14a2 2 from the parentheses. Factor the trinomial and simplify. Apply the commutative property of addition to reverse the last two terms. Obtain a common denominator.

b 2 4ac  b2 bd  2a 4a

 h2 2



h Therefore, the vertex is a

10 8 6 4 2

8 10

Factor a from the variable terms.

k

The function is in the form f 1x2  a1x  h2 2  k, where

2

108 6 4 2 2 4 6

bc

 a ax 

 a c x  a

Answers

1a  02

b 2a

and

k

4ac  b2 4a

b 4ac  b2 , b 2a 4a

4ac  b2 , it is usually easier 4a to determine the x-coordinate of the vertex first and then find y by evaluating the function at x  b  12a2 . Although the y-coordinate of the vertex is given by

2

4

6 8 10

x

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Section 7.5

Vertex of a Parabola: Applications and Modeling

FORMULA The Vertex Formula For f 1x2 ⫽ ax2 ⫹ bx ⫹ c a

1a ⫽ 02 , the vertex is given by

⫺b 4ac ⫺ b2 , b 2a 4a

or

a

⫺b ⫺b , fa bb 2a 2a

3. Determining the Vertex and Intercepts of a Quadratic Function Example 3

Determining the Vertex and Intercepts of a Quadratic Function

Given: h1x2 ⫽ x2 ⫺ 2x ⫹ 5 a. Use the vertex formula to find the vertex. b. Find the x- and y-intercepts. c. Sketch the function.

Solution: a. h1x2 ⫽ x2 ⫺ 2x ⫹ 5 a⫽1

b ⫽ ⫺2

c⫽5

The x-coordinate of the vertex is

Identify a, b, and c. ⫺1⫺22 ⫺b ⫽ ⫽ 1. 2a 2112

The y-coordinate of the vertex is h112 ⫽ 112 2 ⫺ 2112 ⫹ 5 ⫽ 4. The vertex is (1, 4).

b. The y-intercept is given by h102 ⫽ 102 2 ⫺ 2102 ⫹ 5 ⫽ 5. The y-intercept is (0, 5). To find the x-intercept(s), find the real solutions to the equation h1x2 ⫽ 0. h1x2 ⫽ x2 ⫺ 2x ⫹ 5 0 ⫽ x2 ⫺ 2x ⫹ 5

x⫽

This quadratic equation is not factorable. Apply the quadratic formula: a ⫽ 1, b ⫽ ⫺2, c ⫽ 5.

⫺1⫺22 ⫾ 21⫺22 2 ⫺ 4112152 2112

⫽

2 ⫾ 14 ⫺ 20 2112

⫽

2 ⫾ 1⫺16 2

⫽

2 ⫾ 4i 2

⫽ 1 ⫾ 2i The solutions to the equation h1x2 ⫽ 0 are not real numbers. Therefore, there are no x-intercepts.

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c.

y 8 7 (0, 5) h(x)  x2  2x  5 6 5 4 (1, 4) 3

Because a  0, the parabola opens up.

2 1 5 4 3 2 1 1

1

2 3

4

5

x

2

TIP: The location of the vertex and the direction that the parabola opens can be used to determine whether the function has any x-intercepts. Given h(x)  x 2  2x  5, the vertex (1, 4) is above the x-axis. Furthermore, because a 7 0, the parabola opens upward. Therefore, it is not possible for the graph to cross the x-axis (Figure 7-17).

Figure 7-17

Skill Practice

3. Given: f 1x2  x 2  4x  6 a. Use the vertex formula to find the vertex of the parabola. b. Determine the x- and y-intercepts. c. Sketch the graph.

4. Applications and Modeling of Quadratic Functions Example 4

Applying a Quadratic Function

The crew from Extravaganza Entertainment launches fireworks at an angle of 60° from the horizontal. The height of one particular type of display can be approximated by: h1t2  16t 2  12813t where h(t) is measured in feet and t is measured in seconds. a. How long will it take the fireworks to reach their maximum height? Round to the nearest second. b. Find the maximum height. Round to the nearest foot.

Solution: h1t2  16t2  128 13t

This parabola opens downward; therefore, the maximum height of the fireworks will occur at the vertex of the parabola.

Answers

3. a. Vertex: 12, 22

b. x-intercepts: none; y-intercepts: (0, 6) c. y

a  16

10 8 6 4 2 108 6 4 2 2 4 6 8 10

b  12813

c0

Identify a, b, and c, and apply the vertex formula.

The x-coordinate of the vertex is

2

4

6 8 10

x

128 13 12813 b   ⬇ 6.9 2a 21162 32

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Section 7.5

Vertex of a Parabola: Applications and Modeling

The y-coordinate of the vertex is approximately h16.92  1616.92 2  1281316.92  768 The vertex is (6.9, 768). a. The fireworks will reach their maximum height in 6.9 sec. b. The maximum height is 768 ft. Skill Practice 4. An object is launched into the air with an initial velocity of 48 ft/sec from the top of a building 288 ft high. The height h(t) of the object after t seconds is given by h 1t 2 16t 2  48t  288

a. Find the time it takes for the object to reach its maximum height. b. Find the maximum height.

In Section 2.3 we wrote a linear model given two points. Now we will learn how to write a quadratic model of a parabola given three points by using a system of equations and the standard form of a parabola: y  ax2  bx  c. This process involves substituting the x- and y-coordinates of the three given points into the quadratic model. Then we solve the resulting system of three equations (Section 3.6).

Example 5

Writing a Quadratic Model

Write an equation of a parabola that passes through the points (1, 1), (1, 5), and (2, 4).

Solution: Substitute (1, 1) into the equation y  ax2  bx  c

112  a112 2  b112  c 1  a  b  c a  b  c  1

Substitute (1, 5) into the equation y  ax2  bx  c

152  a112 2  b112  c 5  a  b  c a  b  c  5

Substitute (2, 4) into the equation y  ax2  bx  c

Solve the system:

142  a122 2  b122  c 4  4a  2b  c 4a  2b  c  4

A

a  b  c  1

B

a  b  c  5

C

4a  2b  c 

4

Answers 4. a. 1.5 sec

b. 324 ft

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Notice that the c-variables all have a coefficient of 1. Therefore, we choose to eliminate the c-variable. A B

B C

a  b  c  1 a  b  c  5

Multiply by 1.

a  b  c  5 4a  2b  c  4

a  b  c  1 a  b  c  5

Multiply by 1.

Substitute b with 2 in equation E :

2b

3a  3b

5 4 3 2 1 1 2 3

(1, 5)

1

2

3 4

D

 9

E

3a  3122  9 3a  6  9 3a  3 a1

Substitute a and b in equation A to solve for c:

(2, 4)

4 2

a  b  c  5 4a  2b  c  4

y 5 4 y  x2  2x 4 3 2 1

 b

112  122  c  1 3  c  1 c  4

Substitute a  1, b  2, and c  4 in the standard form of the parabola for the final answer. 5

x

(1, 1)

y  ax2  bx  c

y  112x2  122x  142

4 5

y  x2  2x  4

A graph y  x2  2x  4 is shown in Figure 7-18. Notice that the graph of the function passes through the points (1, 1), (1, 5), and (2, 4).

Figure 7-18

Skill Practice 5. Write an equation of the parabola that passes through the points (1, 1), (2, 1), and (3, 9).

Answer 5. a  1, b  1, c  3; y  x 2  x  3

Section 7.5

Practice Exercises

Boost your GRADE at ALEKS.com!

• Practice Problems • Self-Tests • NetTutor

• e-Professors • Videos

Study Skills Exercise 1. Write the vertex formula.

Review Exercises 2. How does the graph of f 1x2  2x2 compare with the graph of y  x2? 3. How does the graph of p1x2  14x2 compare with the graph of y  x2? 4. How does the graph of Q1x2  x2  83 compare with the graph of y  x2? 5. How does the graph of r1x2  x2  7 compare with the graph of y  x2?

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Section 7.5

Vertex of a Parabola: Applications and Modeling

625

6. How does the graph of s1x2 ⫽ 1x ⫺ 42 2 compare with the graph of y ⫽ x2? 7. How does the graph of t1x2 ⫽ 1x ⫹ 102 2 compare with the graph of y ⫽ x2? 8. Find the coordinates of the vertex of the parabola defined by g1x2 ⫽ 21x ⫹ 32 2 ⫺ 4. For Exercises 9–16, find the value of n to complete the square. Factor the resulting trinomial. 9. x2 ⫺ 8x ⫹ n 2 13. b2 ⫹ b ⫹ n 9

10. x2 ⫹ 4x ⫹ n

11. y2 ⫹ 7y ⫹ n

2 14. m2 ⫺ m ⫹ n 7

15. t 2 ⫺

12. a2 ⫺ a ⫹ n

1 t⫹n 3

1 16. p2 ⫹ p ⫹ n 4

Concept 1: Writing a Quadratic Function in the Form f (x ) ⴝ a (x ⴚ h)2 ⴙ k

For Exercises 17–28, write the function in the form f 1x2 ⫽ a1x ⫺ h2 2 ⫹ k by completing the square. Then identify the vertex. (See Examples 1–2.) 17. g1x2 ⫽ x2 ⫺ 8x ⫹ 5

18. h1x2 ⫽ x2 ⫹ 4x ⫹ 5

19. n1x2 ⫽ 2x2 ⫹ 12x ⫹ 13

20. f 1x2 ⫽ 4x2 ⫹ 16x ⫹ 19

21. p1x2 ⫽ ⫺3x2 ⫹ 6x ⫺ 5

22. q1x2 ⫽ ⫺2x2 ⫹ 12x ⫺ 11

23. k1x2 ⫽ x2 ⫹ 7x ⫺ 10

24. m1x2 ⫽ x2 ⫺ x ⫺ 8

25. F1x2 ⫽ 5x2 ⫹ 10x ⫹ 1

26. G1x2 ⫽ 4x2 ⫹ 4x ⫺ 7

27. P1x2 ⫽ ⫺2x2 ⫹ x

28. Q1x2 ⫽ ⫺3x2 ⫹ 12x

Concept 2: Vertex Formula For Exercises 29–40, find the vertex by using the vertex formula. (See Example 3.) 29. Q1x2 ⫽ x2 ⫺ 4x ⫹ 7

30. T1x2 ⫽ x2 ⫺ 8x ⫹ 17

31. r1x2 ⫽ ⫺3x2 ⫺ 6x ⫺ 5

32. s1x2 ⫽ ⫺2x2 ⫺ 12x ⫺ 19

33. N1x2 ⫽ x2 ⫹ 8x ⫹ 1

34. M1x2 ⫽ x2 ⫹ 6x ⫺ 5

1 5 35. m1x2 ⫽ x2 ⫹ x ⫹ 2 2

1 36. n1x2 ⫽ x2 ⫹ 2x ⫹ 3 2

37. k1x2 ⫽ ⫺x2 ⫹ 2x ⫹ 2

38. h1x2 ⫽ ⫺x2 ⫹ 4x ⫺ 3

1 39. A1x2 ⫽ ⫺ x2 ⫹ x 3

2 40. B1x2 ⫽ ⫺ x2 ⫺ 2x 3

For Exercises 41–44, find the vertex two ways:

a. by completing the square and writing in the form f 1x2 ⫽ a1x ⫺ h2 2 ⫹ k. b. by using the vertex formula.

41. p1x2 ⫽ x2 ⫹ 8x ⫹ 1

42. F1x2 ⫽ x2 ⫹ 4x ⫺ 2

43. f 1x2 ⫽ 2x2 ⫹ 4x ⫹ 6

Concept 3: Determining the Vertex and Intercepts of a Quadratic Function For Exercises 45–52 a. Find the vertex. b. Find the y-intercept. c. Find the x-intercept(s), if they exist. d. Use this information to graph the function. (See Examples 2–3.)

44. g1x2 ⫽ 3x2 ⫹ 12x ⫹ 9

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Chapter 7 Quadratic Equations and Functions

45. y  x2  2x  3

46. y  x2  4x  3

47. y  2x2  2x  4

y

y

y 5 4 3 2 1

2

2

8 7 6 5 4 3 2 1

3 4

3 4

5 4 3 2 1 1

5

5

2

5 4 3 2 1 1

5 4 3 2 1 1 2 3

4

5

49. y  x2  3x 

x

9 4

2 3 4 5

1 2 3

4

5

x

3 9 50. y  x2  x  2 16

4

5

x

1 2

3 4

5

x

3 2 1 3

x

2 3 4 5

4 5

1

2

3 4

5

1 2

3 4

5 6

7

x

6

52. y  x2  4x

y

y

5 4 3 2 1 1 2 3

5 4 3 2 1 1

9 6 3

5 4 3 2 1

5 4 3 2 1

5 4 3 2 1 1 2 3

y 24 21 18 15 12

51. y  x2  2x  3

y

y 5 4 3 2 1 5 4 3 2 1 1

5 4 3 2 1 1

48. y  2x2  12x  19

1

2

3 4

5

x

5 4 3 2 1 1 2 3

1

2

3 4

5

x

4 5

Concept 4: Applications and Modeling of Quadratic Functions 53. Mia sells used MP3 players. The average cost to package MP3 players is given by the equation C1x2  2x2  40x  2200, where x is the number of MP3 players she packages in a week. How many players must she package to minimize her average cost? 54. Ben sells used iPods. The average cost to package iPods is given by the equation C1x2  3x2  120x  1300, where x is the number of iPods packaged per month. Determine the number of iPods that Ben needs to package to minimize the average cost. 55. A set of fireworks mortar shells is launched from the staging platform at 100 ft/sec from an initial height of 8 ft above the ground. The height of the fireworks, h(t), can be modeled by h1t2  16t 2  100t  8, where t is the time in seconds after launch. (See Example 4.) a. What is the maximum height that the shells can reach before exploding? b. For how many seconds after launch would the fuses need to be set so that the mortar shells would in fact explode at the maximum height 56. A baseball player throws a ball, and the height of the ball (in feet) can be approximated by y1x2  0.011x2  0.577x  5 y

where x is the horizontal position of the ball measured in feet from the origin. a. For what value of x will the ball reach its highest point? Round to the nearest foot. b. What is the maximum height of the ball? Round to the nearest tenth of a foot.

0 0

x

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Section 7.5

57. Gas mileage depends in part on the speed of the car. The gas mileage of a subcompact car is given by m1x2  0.04x2  3.6x  49, where 20 x 70 represents the speed in miles per hour and m(x) is given in miles per gallon.

627

Vertex of a Parabola: Applications and Modeling

58. Gas mileage depends in part on the speed of the car. The gas mileage of a luxury car is given by L1x2  0.015x2  1.44x  21, where 25 x 70 represents the speed in miles per hour and L(x) is given in miles per gallon.

a. At what speed will the car get the maximum gas mileage?

a. At what speed will the car get the maximum gas mileage?

b. What is the maximum gas mileage?

b. What is the maximum gas mileage? Round to the nearest whole unit.

59. Tetanus bacillus bacterium is cultured to produce tetanus toxin used in an inactive form for the tetanus vaccine. The amount of toxin produced per batch increases with time and then becomes unstable. The amount of toxin (in grams) as a function of time t (in hours) can be approximated by: b1t2  

1 2 1 t  t 1152 12

60. The bacterium Pseudomonas aeruginosa is cultured with an initial population of 104 active organisms. The population of active bacteria increases up to a point, and then due to a limited food supply and an increase of waste products, the population of living organisms decreases. Over the first 48 hr, the population can be approximated by: P1t2  1718.75t 2  82,500t  10,000

a. How many hours will it take to produce the maximum yield? b. What is the maximum yield?

where 0 t 48 a. Find the time required for the population to reach its maximum value. b. What is the maximum population?

For Exercises 61–66, use the standard form of a parabola given by y  ax2  bx  c to write an equation of a parabola that passes through the given points. (See Example 5.) 61. 10, 42, 11, 02, and 11, 102

62. 10, 32, 13, 02, 11, 82

63. 12, 12, 12, 52, and 11, 42

64. 11, 22, 11, 62, and 12, 32

65. 12, 42, 11, 12, and 11, 72

66. 11, 42, 11, 62, and 12, 182

Expanding Your Skills 67. A farmer wants to fence a rectangular corral adjacent to the side of a barn; however, she has only 200 ft of fencing and wants to enclose the largest possible area. See the figure. a. If x represents the length of the corral and y represents the width, explain why the dimensions of the corral are subject to the constraint 2x  y  200. b. The area of the corral is given by A  xy. Use the constraint equation from part (a) to express A as a function of x, where 0 6 x 6 100.

x

corral

x

c. Use the function from part (b) to find the dimensions of the corral that will yield the maximum area. [Hint: Find the vertex of the function from part (b).]

y

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Chapter 7 Quadratic Equations and Functions

68. A veterinarian wants to construct two equal-sized pens of maximum area out of 240 ft of fencing. See the figure.

y

a. If x represents the length of each pen and y represents the width of each pen, explain why the dimensions of the pens are subject to the constraint 3x ⫹ 4y ⫽ 240.

x

y x

b. The area of each individual pen is given by A ⫽ xy. Use the constraint equation from part (a) to express A as a function of x, where 0 6 x 6 80. x

c. Use the function from part (b) to find the dimensions of an individual pen that will yield the maximum area. [Hint: Find the vertex of the function from part (b).]

y

y

Graphing Calculator Exercises For Exercises 69–74, graph the functions in Exercises 45–50 on a graphing calculator. Use the Max or Min feature to approximate the vertex. 69. Y1 ⫽ x2 ⫹ 2x ⫺ 3

(Exercise 45)

71. Y1 ⫽ 2x2 ⫺ 2x ⫹ 4 73. Y1 ⫽ ⫺x2 ⫹ 3x ⫺

9 4

(Exercise 47) (Exercise 49)

Section 7.6

70. Y1 ⫽ x2 ⫹ 4x ⫹ 3

(Exercise 46)

72. Y1 ⫽ 2x2 ⫺ 12x ⫹ 19

(Exercise 48)

3 9 74. Y1 ⫽ ⫺x2 ⫺ x ⫺ 2 16

(Exercise 50)

Nonlinear Inequalities

Concepts

1. Solving Quadratic and Polynomial Inequalities

1. Solving Quadratic and Polynomial Inequalities 2. Solving Rational Inequalities 3. Inequalities with “Special Case” Solution Sets

In this section, we will expand our study of solving inequalities. Quadratic inequalities are inequalities that can be written in one of the following forms: ax2 ⫹ bx ⫹ c ⱖ 0

ax2 ⫹ bx ⫹ c ⱕ 0

ax2 ⫹ bx ⫹ c 7 0

ax2 ⫹ bx ⫹ c 6 0

where a ⫽ 0

Recall from Section 7.4 that the graph of a quadratic function defined by f 1x2 ⫽ ax2 ⫹ bx ⫹ c is a parabola that opens upward or downward. • The inequality ax2 ⫹ bx ⫹ c 7 0 is asking “For what values of x is the function positive (above the x-axis)?” • The inequality ax2 ⫹ bx ⫹ c 6 0 is asking “For what values of x is the function negative (below the x-axis)?” The graph of a quadratic function can be used to answer these questions.

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Section 7.6

Example 1

Nonlinear Inequalities

Using a Graph to Solve a Quadratic Inequality

Use the graph of f 1x2  x2  6x  5 in Figure 7-19 to solve the inequalities. a. x2  6x  5 6 0

b. x2  6x  5 7 0

Solution:

From Figure 7-19, we see that the graph of f 1x2  x2  6x  5 is a parabola opening upward. The function factors as f 1x2  1x  121x  52 . The x-intercepts are (1, 0) and (5, 0), and the y-intercept is (0, 5). y

x2  6x  5  0

5 4 3 2 x 1 1

x5 1 2

3 2 1 1

x2  6x  5 0

f(x)  x2  6x  5

3 4

5 6 7

x

1 x 5

2 3 4 5

Figure 7-19

a. The solution to x2  6x  5 6 0 is the set of real numbers, x, for which f 1x2 6 0. Graphically, this is the set of all x values corresponding to the points where the parabola is below the x-axis (shown in red). x2  6x  5 6 0

for 5x 0 1 6 x 6 56 or equivalently, (1, 5)

b. The solution to x  6x  5 7 0 is the set of real numbers, x, for which f 1x2 7 0. This is the set of x values where the parabola is above the x-axis (shown in blue). 2

x2  6x  5 7 0

for 5x 0 x 6 1

or

x 7 56

or

1, 12 ´ 15, 2

Skill Practice Refer to the graph of f 1x2  x 2  3x  4 to solve the inequalities. 1. x2  3x  4 7 0 2. x2  3x  4 6 0

y 3 2 1 −5 −4 −3 −2 −1

−1 −2 −3 −4 −5

1 2 3

4

5

x

f(x) = x2 + 3x − 4

−6 −7

TIP: The inequalities in Example 1 are strict inequalities. Therefore, x  1 and

x  5 (where f1x2  0) are not included in the solution set. However, the corresponding inequalities using the symbols  and  do include the values where f1x2  0. The solution to x2  6x  5  0 is 5x 0 1  x  56 or equivalently, 3 1, 54.

The solution to x2  6x  5  0 is 5x 0 x  1 or x  56 or 1, 14 ´ 3 5,  2.

Answers

1. 5x ƒ x 6 4 or x 7 16; 1, 42 ´ 11,  2 2. 5x ƒ 4 6 x 6 16; 14, 12

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Chapter 7 Quadratic Equations and Functions

Notice that x  1 and x  5 are the boundaries of the solution set to the inequalities in Example 1. These values are the solutions to the related equation x2  6x  5  0.

DEFINITION Boundary Points The boundary points of an inequality consist of the real solutions to the related equation and the points where the inequality is undefined. Recall that testing points in intervals bounded by these points is the basis of the test point method to solve inequalities.

TIP: In Section 1.7, we used the test point method to solve absolute value inequalities.

PROCEDURE Solving Inequalities by Using the Test Point Method Step 1 Find the boundary points of the inequality. Step 2 Plot the boundary points on the number line. This divides the number line into intervals. Step 3 Select a test point from each interval and substitute it into the original inequality. • If a test point makes the original inequality true, then that interval is part of the solution set. Step 4 Test the boundary points in the original inequality. • If the original inequality is strict (< or >), do not include the boundary points in the solution set. • If the original inequality is defined using  or , then include the boundary points that are well defined within the inequality. Note: Any boundary point that makes an expression within the inequality undefined must always be excluded from the solution set.

Example 2

Solving a Quadratic Inequality by Using the Test Point Method

Solve the inequality by using the test point method. 2x2  5x 6 12

Solution: 2x2  5x 6 12

2x2  5x  12

Step 1: Find the boundary points. Because polynomials are defined for all values of x, the only boundary points are the real solutions to the related equation. Solve the related equation.

2x  5x  12  0 2

12x  321x  42  0 x

3 2

x  4

The boundary points are 32 and 4.

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Section 7.6

I

II

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

Step 2: Plot the boundary points.

III 0

1

2

3

Test x ⴝ ⴚ5 2x2  5x 6 12

4

5

Step 3: Select a test point from each interval.

6

Test x ⴝ 0 2x2  5x 6 12 ?

?

2102 2  5102 6 12

2152 2  5152 6 12 ?

?

(

?

2122 2  5122 6 12 ?

0 6 12

8  10 6 12

?

25 6 12 False True

6 5 4 3 2 1

Test x ⴝ 2 2x2  5x 6 12

?

50  25 6 12

False

0

1

(

?

0 6 12 True False

2

3

4

5

18 6 12 False

Step 4: Test the boundary points. The strict inequality excludes values of x for which 2x2  5x  12. Therefore, the boundary points are not included in the solution set.

6

3 2

The solution is 5x 0 4 6 x 6 32 6 or equivalently in interval notation 14 , 32 2 .

Calculator Connections Graph Y1  2x2  5x and Y2  12. Notice that Y1 6 Y2 for 5x 0 4 6 x 6 32 6. y1  2x2  5x y2  12 3  12 2

(4, 12) (

Skill Practice Solve the inequality.

4

3. x2  x 7 6

Example 3

631

Nonlinear Inequalities

(

3 2

Solving a Polynomial Inequality by Using the Test Point Method

Solve the inequality by using the test point method.

x1x  221x  42 2 1x  42 7 0

Solution:

x1x  221x  42 2 1x  42 7 0 x1x  221x  42 2 1x  42  0 x0 x2 x  4

I

II

III

6 5 4 3 2 1

0

1

Step 1: Find the boundary points. x4

IV 2

3

V 4

5

Step 2: Plot the boundary points. 6

Step 3: Select a test point from each interval. Test x ⴝ ⴚ5: 515  2215  42 2 15  42 7 0

315 7 0

Test x ⴝ ⴚ1:

111  2211  42 2 11  42 7 0

135 7 0

Test x ⴝ 1:

111  2211  42 2 11  42 7 0

75 7 0

Test x ⴝ 3:

313  2213  42 2 13  42 7 0

147 7 0

Test x ⴝ 5:

515  2215  42 2 15  42 7 0

1215 7 0

?

?

?

?

?

?

?

?

?

?

False False True False Answer

True

3. 1, 32 ´ 12, 2

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Chapter 7 Quadratic Equations and Functions

False

False

6 5 4 3 2 1

( 0

True 1

) 2

False 3

( 4

True 5

6

Step 4: The inequality symbol, , does not include equality. Therefore, exclude the boundary points in the solution set.

The solution is 5x 0 0 6 x 6 2 or x 7 46 or, equivalently in interval notation, 10, 22 ´ 14, 2 . Skill Practice Solve the inequality. 4. t 1t  521t  22 2  0

TIP: In Example 3, one side of the inequality is factored, and the other side is zero. For inequalities written in this form, we can use a sign chart to determine the sign of each factor. Then the sign of the product (bottom row) is easily determined. Sign of x

Sign of 1x  22

Sign of 1x  42

2

Sign of 1x  42

Sign of x1x  221x  42 1x  42 2

  

    

 

 

  

  

   

 

 

  

  

   

 

 

  

  

    

 

 

  

               4 0 2 4

The solution to the inequality x1x  221x  42 2 1x  42 7 0 includes the intervals for which the product is positive (shown in blue). The solution is 10, 22 ´ 14, 2.

Example 4

Solving a Polynomial Inequality by Using the Test Point Method

Solve the inequality by using the test point method.

x2  x  4  0

Solution: x2  x  4  0 x x40 2

Answer

4. 1, 0 4 ´ 35, 2

Step 1: Find the boundary points in the related equation. Since this equation is not factorable, use the quadratic formula to find the solutions.

x

b ; 2b2  4ac 2a

x

1 ; 2112 2  4112142 2112

x

1 ; 117 2

x

1  117 1  117 ⬇ 2.56 and x  ⬇ 1.56 2 2

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Section 7.6

I

II

6 5 4 3 2 1

Step 2: Plot the boundary points.

III 0

1

2

Test x  3

3

4

5

Step 3: Select a test point from each interval.

6

Test x  0

x2  x  4  0

132 2  132  4  0

102 2  102  4  0

?

0

122 2  122  4  0

?

True 1

2

?

4  0 False

False

6 5 4 3 2 1

x2  x  4  0

?

2  0 True True

Test x  2

x2  x  4  0

?

3

4

Nonlinear Inequalities

5

6

?

2  0 True

Step 4: Test the boundary points. Both boundary points make the inequality true. Therefore, both boundary points are included in the solution set.

1  117 1  117 or x  f or equivalently 2 2 1  117 1  117 in interval notation: a, d´c , b. 2 2 The solution is e x冟 x 

Skill Practice Solve the inequality. 5. x 2  3x  1  0

2. Solving Rational Inequalities The test point method can be used to solve rational inequalities. A rational inequality is an inequality in which one or more terms is a rational expression. The solution set to a rational inequality must exclude all values of the variable that make the inequality undefined. That is, exclude all values that make the denominator equal to zero for any rational expression in the inequality.

Example 5

Solving a Rational Inequality by Using the Test Point Method

Solve the inequality.

3 7 0 x1

Solution: 3 7 0 x1 3 0 x1 1x  12  a

3 b  1x  12  0 x1 30

The only boundary point is x  1.

Step 1: Find the boundary points. Note that the inequality is undefined for x  1, so x  1 is a boundary point. To find any other boundary points, solve the related equation. Clear fractions. There is no solution to the related equation.

Answer 5. c

3  213 3  213 , d 2 2

633

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Chapter 7 Quadratic Equations and Functions

I

II

5 4 3 2 1

Step 2: Plot boundary points. 0

1

2

3

4

5

Test x ⴝ 0

Test x ⴝ 2

? 3 70 102  1

? 3 70 122  1

3 ? 7 0 False 1

3 7 0 True 1

False 5 4 3 2 1

0

(

1

Step 3: Select test points.

Step 4: The boundary point x  1 cannot be included in the solution set because it is undefined in the original inequality.

True 2

3

4

5

The solution is 5x 0 x 7 16 or equivalently in interval notation, 11, 2. Skill Practice Solve the inequality. 6.

5 6 0 y2

TIP: Using a sign chart we see that the quotient of the factors 3 and 1x  12 is positive on the interval 11,  2.

Therefore, the solution to the inequality

Sign of 3

Sign of 1x  12 3 Sign of x1

3 7 0 is 11,  2. x1

  

  

  

  

  

  

1 (undefined) y

f(x) 

5 4 3 2 1 5 4 3 2 1 1

(

1

2 3

2 3 4 5

3 0 x1

4 5

x1

3 7 0 can be confirmed from the graph of the x1 3 related rational function, f 1x2  (see Figure 7-20). x1 The solution to the inequality

x

• •

The graph is above the x-axis where f 1x2 

3 7 0 for x 7 1 (shaded red). x1 Also note that x  1 cannot be included in the solution set because 1 is not in the domain of f.

Figure 7-20

Example 6

Solving a Rational Inequality by Using the Test Point Method

Solve the inequality by using the test point method. Answer

6. 12,  2

x2 3 x4

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Section 7.6

Nonlinear Inequalities

Solution: x⫹2 ⱕ3 x⫺4

Step 1: Find the boundary points. Note that the inequality is undefined for x ⫽ 4. Therefore, x ⫽ 4 is automatically a boundary point. To find any other boundary points, solve the related equation.

x⫹2 ⫽3 x⫺4 1x ⫺ 42a

x⫹2 b ⫽ 1x ⫺ 42 132 x⫺4

Clear fractions.

x ⫹ 2 ⫽ 31x ⫺ 42

Solve for x.

x ⫹ 2 ⫽ 3x ⫺ 12 ⫺2x ⫽ ⫺14 x⫽7 The solution to the related equation is x ⫽ 7, and the inequality is undefined for x ⫽ 4. Therefore, the boundary points are x ⫽ 4 and x ⫽ 7. I ⫺3 ⫺2 ⫺1

II

0

1

2

3

4

5

Step 2: Plot boundary points.

III 6

7

Step 3: Select test points.

8

Test x ⴝ 0 x⫹2 ⱕ3 x⫺4

Test x ⴝ 5 x⫹2 ⱕ3 x⫺4

Test x ⴝ 8 x⫹2 ⱕ3 x⫺4

0⫹2 ? ⱕ3 0⫺4

5⫹2 ? ⱕ3 5⫺4

8⫹2 ? ⱕ3 8⫺4

1 ? ⫺ ⱕ3 2

7 ? ⱕ3 1

True

10 ? ⱕ3 4

False

5 ? ⱕ3 2 Test x ⴝ 4

Test x ⴝ 7

x⫹2 ⱕ3 x⫺4

x⫹2 ⱕ3 x⫺4

4⫹2 ? ⱕ3 4⫺4

7⫹2 ? ⱕ3 7⫺4

6 ? ⱕ 3 Undefined 0 ⫺3 ⫺2 ⫺1

0

1

2

9 ? ⱕ3 3 3

(

4

5

6

7

8

True

Step 4: Test the boundary points.

True

The boundary point x ⫽ 4 cannot be included in the solution set, because it is undefined in the inequality. The boundary point x ⫽ 7 makes the original inequality true and must be included in the solution set.

The solution is 5x 0 x 6 4 or x ⱖ 76, or equivalently in interval notation, 1⫺⬁, 42 ´ 37, ⬁2. Skill Practice Solve the inequality. 7.

x⫺5 ⱕ ⫺1 x⫹4

Answer 1 7. a⫺4, d 2

635

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Chapter 7 Quadratic Equations and Functions

3. Inequalities with “Special Case” Solution Sets The solution to an inequality is often one or more intervals on the real number line. Sometimes, however, the solution to an inequality may be a single point on the number line, the empty set, or the set of all real numbers. Example 7

Solving Inequalities

Solve the inequalities. a. x2  6x  9  0

b. x2  6x  9 7 0

c. x2  6x  9  0

d. x2  6x  9 6 0

Solution: a. x2  6x  9  0

1x  32 2  0

Notice that x2  6x  9 is a perfect square trinomial.

Factor x2  6x  9  1x  32 2.

The quantity 1x  32 2 is a perfect square and is greater than or equal to zero for all real numbers, x. The solution is all real numbers, 1,  2. True True

True

5 4 3 2 1

f 1x2  x 2  6x  9, or equivalently f 1x2  1x  32 2, is equal to zero at x  3 and positive (above the x-axis) for all other values of x in its domain.

1x  32 2 7 0

10 8 f(x)  x2  6x  9 6 4 2

6 8 10

3

4

5

This is the same inequality as in part (a) with the exception that the inequality is strict. The solution set does not include the point where x2  6x  9  0. Therefore, the boundary point x  3 is not included in the solution set. True

4 6

2

False

f(x)

2

1

b. x2  6x  9 7 0

TIP: The graph of

108 6 4 2 2 4

0

8 10

x

((

True

5 4 3 2 1

0

1

2

3

4

5

The solution set is 5x 0 x 6 3 or x 7 36 or equivalently 1, 32 ´ 13, 2. c. x2  6x  9  0

1x  32 2  0

A perfect square cannot be less than zero. However, 1x  32 2 is equal to zero at x  3. Therefore, the solution set is 536. False

False

6 5 4 3 2 1 True

0

1

2

3

4

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Section 7.6

Nonlinear Inequalities

d. x2  6x  9 6 0

1x  32 2 6 0

A perfect square cannot be negative; therefore, there are no real numbers x such that 1x  32 2 6 0. There is no solution, 5 6. Skill Practice Solve the inequalities.

Answers

8. x  4x  4  0

9. x  4x  4 7 0

10. x 2  4x  4  0

11. x 2  4x  4 6 0

2

8. 9. 10. 11.

2

Section 7.6

All real numbers; 1, 2 1, 22 ´ 12,  2 526 5 6

Practice Exercises • Practice Problems • Self-Tests • NetTutor

Boost your GRADE at ALEKS.com!

• e-Professors • Videos

Study Skills Exercise 1. Define the key terms. a. Quadratic inequality

b. Boundary points

c. Test point method

d. Rational inequality

Review Exercises For Exercises 2–8, solve the equations. 2. x2  10x  9

3. 2t 2  3t  4  0

5. 2m1m  421m  62  0

6. n13  n21n  72  0

7.

14 7 x3

8.

4. 3w2  w  5  0

15 5 x2

Concept 1: Solving Quadratic and Polynomial Inequalities For Exercises 9–12, estimate from the graph the intervals for which the inequality is true. (See Example 1.) 9.

10.

y

y  p(x)

y

4

9

3 2

8

1

6 5 4

5 4 3 21 1

7 1

2 3

4

5

x

2 3

y  g(x)

3 2

4

1

5 6

5 4 3 2 1 1

a. p 1x2 7 0

b. p 1x2 6 0

c. p1x2  0

d. p1x2  0

a. g 1x2 7 0 c. g 1x2  0

1

2 3

4

5

x

b. g 1x2 6 0 d. g 1x2  0

637

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Chapter 7 Quadratic Equations and Functions

11.

y

12.

y

5 4 3 y  h(x) 2 1

5 4 3 y  f(x) 2 1 5 4 3 2 1 1 2 3

1

2

3 4

4 5

a. f 1x2 7 0 c. f 1x2  0

x

5

5 4 3 2 1 1 2 3

1

2

3 4

5

x

4 5

b. f 1x2 6 0

d. f 1x2  0

a. h1x2 7 0

b. h1x2 6 0

c. h1x2  0

d. h1x2  0

For Exercises 13–18, solve the equation and related inequalities. (See Examples 2–4.) 13. a. 314  x212x  12  0

14. a. 51y  6213  5y2  0

b. 314  x212x  12 6 0

b. 51y  6213  5y2 6 0

c. 314  x212x  12 7 0

c. 51y  6213  5y2 7 0

15. a. x2  7x  30

16. a. q2  4q  5

b. x2  7x 6 30

b. q2  4q  5

c. x2  7x 7 30

c. q2  4q  5

17. a. 2p1 p  22  p  3

18. a. 3w1w  42  10  w

b. 2p1 p  22  p  3

b. 3w1w  42 6 10  w

c. 2p1 p  22  p  3

c. 3w1w  42 7 10  w

For Exercises 19–38, solve the polynomial inequality. Write the answer in interval notation. (See Examples 2–4.) 19. 1t  721t  12 6 0

20. 1 p  421 p  22 7 0

21. 614  2x215  x2 7 0

22. 812t  5216  t2 6 0

23. m1m  12 2 1m  52  0

24. w2 13  w21w  22  0

25. a2  12a  32

26. w2  20w  64

27. 5x2  2x  1 7 0

28. 3x2  2x  4 6 0

29. x2  3x  6

30. x2  8  3x

31. b2  121 6 0

32. c2  25 6 0

33. 3p1 p  22  3  2p

34. 2t1t  32  t  12

35. x3  x2  12x

36. x3  36 7 4x2  9x

37. w3  w2 7 4w  4

38. 2p3  5p2  3p

Concept 2: Solving Rational Inequalities For Exercises 39–42, solve the equation and related inequalities. (See Examples 5–6.) 39. a.

10 5 x5

40. a.

8 4 a1

41. a.

z2  3 z6

42. a.

w8 2 w6

b.

10 6 5 x5

b.

8 7 4 a1

b.

z2  3 z6

b.

w8 2 w6

c.

10 7 5 x5

c.

8 6 4 a1

c.

z2  3 z6

c.

w8 2 w6

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Nonlinear Inequalities

639

For Exercises 43–54, solve the rational inequality. Write the answer in interval notation. (See Examples 5–6.) 43.

2 ⱖ0 x⫺1

44.

⫺3 ⱕ0 x⫹2

45.

b⫹4 7 0 b⫺4

46.

a⫹1 6 0 a⫺3

47.

3 6 ⫺1 2x ⫺ 7

48.

8 7 1 4x ⫹ 9

49.

x⫹1 ⱖ4 x⫺5

50.

x⫺2 ⱕ5 x⫹6

51.

1 ⱕ2 x

52.

1 ⱖ3 x

53.

1x ⫹ 22 2 7 0 x

54.

1x ⫺ 32 2 6 0 x

Concept 3: Inequalities with “Special Case” Solution Sets For Exercises 55–70, solve the inequalities. (See Example 7.) 55. x2 ⫹ 10x ⫹ 25 ⱖ 0

56. x2 ⫹ 6x ⫹ 9 6 0

57. x2 ⫹ 2x ⫹ 1 6 0

58. x2 ⫹ 8x ⫹ 16 ⱖ 0

59. x4 ⫹ 3x2 ⱕ 0

60. x4 ⫹ 2x2 ⱕ 0

61. x2 ⫹ 12x ⫹ 36 6 0

62. x2 ⫹ 12x ⫹ 36 ⱖ 0

63. x2 ⫹ 3x ⫹ 5 6 0

64. 2x2 ⫹ 3x ⫹ 3 7 0

65. ⫺5x2 ⫹ x 6 1

66. ⫺3x2 ⫺ x 7 6

67. x2 ⫹ 22x ⫹ 121 7 0

68. y2 ⫺ 24y ⫹ 144 7 0

69. 4t 2 ⫺ 12t ⱕ ⫺9

70. 4t 2 ⫺ 12t 7 ⫺9

Mixed Exercises For Exercises 71–94, identify the inequality as one of the following types: linear, quadratic, rational, or polynomial 1degree 7 22. Then solve the inequality and write the answer in interval notation. 71. 2y2 ⫺ 8 ⱕ 24

72. 8p2 ⫺ 18 7 0

73. 15x ⫹ 22 2 7 ⫺4

74. 13 ⫺ 7x2 2 6 ⫺1

75. 41x ⫺ 22 6 6x ⫺ 3

76. ⫺713 ⫺ y2 7 4 ⫹ 2y

77.

2x ⫹ 3 ⱕ2 x⫹1

78.

5x ⫺ 1 ⱖ5 x⫹3

79. 4x3 ⫺ 40x2 ⫹ 100x 7 0

80. 2y3 ⫺ 12y2 ⫹ 18y 6 0

81. 2p3 7 4p2

82. w3 ⱕ 5w2

83. ⫺31x ⫹ 42 2 1x ⫺ 52 ⱖ 0

84. 5x1x ⫺ 221x ⫺ 62 2 ⱖ 0

85. x2 ⫺ 2 6 0

86. y2 ⫺ 3 7 0

87. x2 ⫹ 5x ⫺ 2 ⱖ 0

88. t2 ⫹ 7t ⫹ 3 ⱕ 0

89.

a⫹2 ⱖ0 a⫺5

90.

92. ⫺5p ⫹ 8 6 p

t⫹1 ⱕ0 t⫺2

93. 4x2 ⫹ 2 ⱖ ⫺x

91. 2 ⱖ t ⫺ 3 94. 5x ⫺ 4 ⱖ 2x2

Expanding Your Skills 95. A firework display is planned for the 4th of July. The fire department wants to make sure that all fireworks explode at least 120 ft above the ground. The staging platform is 8 ft high, and all the mortar shells will be launched at 108 ft/sec. The height of the mortars, h(t), can be modeled by h1t2 ⫽ ⫺16t 2 ⫹ 108t ⫹ 8

where t is the time in seconds after launch.

For what interval of time should the fuses be set? Round to the nearest tenth of a second.

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Chapter 7 Quadratic Equations and Functions

Graphing Calculator Exercises x 7 0 x2 enter Y1 as x 1x  22 and determine where the graph is above the x-axis. Write the solution in interval notation.

x 6 0 x2 enter Y1 as x  1x  22 and determine where the graph is below the x-axis. Write the solution in interval notation.

96. To solve the inequality

97. To solve the inequality

98. To solve the inequality x2  1 6 0, enter Y1 as x2  1 and determine where the graph is below the x-axis. Write the solution in interval notation.

99. To solve the inequality x2  1 7 0, enter Y1 as x2  1 and determine where the graph is above the x-axis. Write the solution in interval notation.

For Exercises 100–103, determine the solution by graphing the inequalities. 100. x2  10x  25  0 102.

101. x2  10x  25  0

8 6 0 x 2

103.

2

6 7 0 x 3 2

Problem Recognition Exercises Recognizing Equations and Inequalities At this point, you have learned how to solve a variety of equations and inequalities. Being able to distinguish the type of problem being posed is the first step in successfully solving it. For Exercises 1–20, a. Identify the problem type. Choose from • • • • • •

linear equation quadratic equation rational equation absolute value equation radical equation equation quadratic in form

• • • • • •

polynomial equation linear inequality polynomial inequality rational inequality absolute value inequality compound inequality

b. Solve the equation or inequality. Write the solution to each inequality in interval notation if possible. 1. 1z2  42 2  1z2  42  12  0

2. 3  0 4t  1 0 6 6

3. 2y1y  42  5  y

3 4. 2 11x  3  4  6

5. 5   0w  4 0

6.

7. m3  5m2  4m  20  0

8. x  4 7 5 and 2x  3  23

9. 5  233  1x  42 4  3x  14

5 3  1 x2 x2

10. 02x  6 0  0x  3 0

11.

13. 2t  8  6  t

14. 14x  32 2  10

15. 4  x 7 2 or 8 6 2x

17. x2  10x  25

18.

16.

1 3 5 x2  x 3 4 6

19. x  13 2x  36  0

3 1 x2

20. x4  13x2  36  0

12. 9 6 0x  4 0

10 6 0 x 1 2

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Group Activity

Group Activity Creating a Quadratic Model of the Form y  a1x  h2 2  k Estimated time: 20 minutes Group Size: 3 In Chapter 3, we modeled the path of a softball that was thrown from right field to third base. The data points are given in the table. The values of t represent the time in seconds after the ball was released, and y represents the height of the ball in feet. Time (sec) t

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

2.0

Height (ft) y

5

11

16

19

21

22

21

19

16

12

6

1. Graph the points defined by (t, y) in the table.

2. Select a point that you think best represents the vertex of the parabola. Label this point (h, k). (h, k) 

Height (ft)

25 20 15 10 5 0

3. Substitute the values of h and k into the formula y  a1t  h2 2  k, and write the equation here. y  a(t 

Height of Softball vs. Time

y

0

0.4

0.8

1.2 1.6 Time (sec)

2

t 2.4

)2 

4. Choose a different point (t, y) from the graph. Substitute these values into the equation in step 3 and then solve for a. 5. Substitute the values of h, k, and a that you found in steps 2 and 4 into the equation y  a1t  h2 2  k.

6. Use the function found in step 5 to approximate the height of the ball after 0.7 sec.

7. Use the function found in step 5 to approximate the height of the ball after 1.8 sec. How well does this value match the observed value of 12 ft?

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642

Chapter 7 Quadratic Equations and Functions

Chapter 7

Summary

Section 7.1

Square Root Property and Completing the Square

Key Concepts

Examples

The square root property states that

Example 1

If

x k 2

then x  1k

1x  52 2  13 x  5  113

(square root property)

x  5  i113

The solution set is 55  i1136. Follow these steps to solve a quadratic equation in the form ax2  bx  c  0 1a  02 by completing the square and applying the square root property: 1. Divide both sides by a to make the leading coefficient 1. 2. Isolate the variable terms on one side of the equation. 3. Complete the square: Add the square of one-half the linear term coefficient to both sides of the equation. Then factor the resulting perfect square trinomial. 4. Apply the square root property and solve for x.

Example 2 2x2  12x  16  0 2x2 12x 16 0    2 2 2 2 x2  6x  8  0 x2  6x  8 x2  6x  9  8  9 1x  32 2  17

x  3  117 x  3  117

The solution set is 53  1176.

Note: 3 12 162 4 2  9

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Summary

Section 7.2

Quadratic Formula

Key Concepts

Example

The solutions to a quadratic equation ax2  bx  c  0 1a  02 are given by the quadratic formula

Example 1

b  2b2  4ac x 2a

The discriminant of a quadratic equation ax  bx  c  0 is b2  4ac . If a, b, and c are rational numbers, then 2

1. If b  4ac 7 0 , then there will be two real solutions. Moreover, a. If b2  4ac is a perfect square, the solutions will be rational numbers. b. If b2  4ac is not a perfect square, the solutions will be irrational numbers. 2. If b2  4ac 6 0 , then there will be two imaginary solutions. 3. If b2  4ac  0 , then there will be one rational solution. 2

Three methods to solve a quadratic equation are 1. Factoring and applying the zero product rule. 2. Completing the square and applying the square root property. 3. Using the quadratic formula.

Section 7.3

643

3x2  2x  4  0 a3 x

b  2

c4

122  2122 2  4132142 2132



2  14  48 6



2  144 6



2  2i111 6

The discriminant is 44. Therefore, there will be two imaginary solutions.

2 11  i 1112 1



6 3



1  i111 3



1 111  i 3 3

The solution set is e

1 111  if. 3 3

Equations in Quadratic Form

Key Concepts

Example

Substitution can be used to solve equations that are in quadratic form.

Example 1 x2 3  x1 3  12  0

Let u  x13.

u2  u  12  0

1u  421u  32  0 u4

or

u  3

x1  3  4

or

x1 3  3

x  64

or

x  27 Cube both sides.

The solution set is 564, 276.

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644

Chapter 7 Quadratic Equations and Functions

Section 7.4

Graphs of Quadratic Functions

Key Concepts

A quadratic function of the form f 1x2  x  k shifts the graph of y  x2 up k units if k 7 0 and down 0k 0 units if k 6 0. 2

Examples Example 1 y

f(x)  x2  3

6 5 4 3 2 1

y  x2

6 54 3 2 1 1 2

1 2 3

4 5

6

x

3 4 5 6

A quadratic function of the form f 1x2  1x  h2 2 shifts the graph of y  x2 right h units if h 7 0 and left 0h 0 units if h 6 0.

Example 2 y 6 5 f(x)  (x  2)2 4 3 2 1

y  x2

6 54 3 2 1 1 2

1

2 3

4 5

6

x

3 4 5 6

The graph of a quadratic function of the form f 1x2  ax2 is a parabola that opens upward when a 7 0 and opens downward when a 6 0. If 0a 0 7 1, the graph of y  x2 is stretched vertically by a factor of 0a 0 . If 0 6 0a 0 6 1, the graph of y  x2 is shrunk vertically by a factor of 0a 0 .

Example 3 y 6 5 4 3 2 1 6 5 4 3 2 1 1 2 3

y  x2 1

2

3 4

5 6

x

1 f(x)  3 x2

4 5 6

A quadratic function of the form f 1x2  a1x  h2 2  k has vertex (h, k). If a 7 0, the vertex represents the minimum point. If a 6 0, the vertex represents the maximum point.

Example 4 y 9 8 7 6

f(x)  (x  2)2  3

5 4 3

minimum point (2, 3) 2 1 7 6 5 4 3 2 1 1 2 3

1

2 3 4

5

x

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Summary

Section 7.5

645

Vertex of a Parabola: Applications and Modeling

Key Concepts

Examples

Completing the square is a technique used to write a quadratic function f 1x2  ax2  bx  c 1a  02 in the form f 1x2  a1x  h2 2  k for the purpose of identifying the vertex (h, k).

Example 1 f 1x2  3x2  6x  11  31x2  2x

2  11

 31x  2x  1  12  11 2

 31x2  2x  12  3  11  31x  12 2  8

 33x  112 4 2  8

The vertex is 11, 82. Because a  3 7 0, the parabola opens upward and the vertex 11, 82 is a minimum point. The vertex formula finds the vertex of a quadratic function f 1x2  ax2  bx  c 1a  02.

Example 2

The vertex is

a3

b 4ac  b2 a , b 2a 4a

or

b b a ,fa bb 2a 2a

f 1x2  3x2  6x  11

x

b6

c  11

6  1 2132

f 112  3112 2  6112  11  8 The vertex is (1, 8).

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Chapter 7 Quadratic Equations and Functions

Section 7.6

Nonlinear Inequalities

Key Concepts

Example

The Test Point Method to Solve Polynomial and

Example 1

Rational Inequalities 1. Find the boundary points of the inequality. (Boundary points are the real solutions to the related equation and points where the inequality is undefined.) 2. Plot the boundary points on the number line. This divides the number line into intervals. 3. Select a test point from each interval and substitute it into the original inequality. • If a test point makes the original inequality true, then that interval is part of the solution set. 4. Test the boundary points in the original inequality. • If the original inequality is strict (< or >), do not include the boundary points in the solution set. • If the original inequality is defined using ⱕ or ⱖ, then include the boundary points that are well defined within the inequality.

28 ⱕ4 2x ⫺ 3

28 ⫽4 2x ⫺ 3 12x ⫺ 32 ⴢ

The inequality is undefined for x ⫽ 32. Find other possible boundary points by solving the related equation. Related equation

28 ⫽ 12x ⫺ 32 ⴢ 4 2x ⫺ 3 28 ⫽ 8x ⫺ 12 40 ⫽ 8x x⫽5

The boundaries are x ⫽ I

Note: Any boundary point that makes an expression within the inequality undefined must always be excluded from the solution set.

3 and x ⫽ 5. 2 II

III 5

3 2

Test x ⫽ 1:

28 ? ⱕ 4 True 2 112 ⫺ 3

Interval II: Test x ⫽ 2:

28 ? ⱕ 4 False 2 122 ⫺ 3

Interval III: Test x ⫽ 6:

28 ? ⱕ 4 True 2 162 ⫺ 3

Interval I:

The boundary point x ⫽ 32 is not included because 28 is undefined there. The boundary x ⫽ 5 does 2x ⫺ 3 check in the original inequality. (

3 2

The solution is 1⫺⬁, 32 2 ´ 35, ⬁2.

5

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Review Exercises

Chapter 7

Section 7.2

Section 7.1 For Exercises 1–8, solve the equations by using the square root property. 1. x2 ⫽ 5

2. 2y2 ⫽ ⫺8

3. a2 ⫽ 81

4. 3b2 ⫽ ⫺19

5. 1x ⫺ 22 2 ⫽ 72

6. 12x ⫺ 52 2 ⫽ ⫺9

7. 13y ⫺ 12 2 ⫽ 3

8. 31m ⫺ 42 2 ⫽ 15

9. The length of each side of an equilateral triangle is 10 in. Find the height of the triangle. Round the answer to the nearest tenth of an inch.

10 in.

h

24. Explain how the discriminant can determine the type and number of solutions to a quadratic equation with rational coefficients. For Exercises 25–30, determine the type (rational, irrational, or imaginary) and number of solutions for the equations by using the discriminant.

10 in.

10 in.

10. Use the square root property to find the length of the sides of a square whose area is 81 in2. 11. Use the square root property to find the length of the sides of a square whose area is 150 in2. Round the answer to the nearest tenth of an inch. For Exercises 12–15, find the value of n so that the expression is a perfect square trinomial. Then factor the trinomial.

25. x2 ⫺ 5x ⫽ ⫺6

26. 2y2 ⫽ ⫺3y

27. z2 ⫹ 23 ⫽ 17z

28. a2 ⫹ a ⫹ 1 ⫽ 0

29. 10b ⫹ 1 ⫽ ⫺25b2

30. 3x2 ⫹ 15 ⫽ 0

For Exercises 31–38, solve the equations by using the quadratic formula. 31. y2 ⫺ 4y ⫹ 1 ⫽ 0

32. m2 ⫺ 5m ⫹ 25 ⫽ 0

33. 6a1a ⫺ 12 ⫽ 10 ⫹ a

34. 3x1x ⫺ 32 ⫽ x ⫺ 8

35. b2 ⫺

4 3 ⫽ b 25 5

37. ⫺32 ⫹ 4x ⫺ x2 ⫽ 0

36. k2 ⫹ 0.4k ⫽ 0.05 38. 8y ⫺ y2 ⫽ 10

For Exercises 39–42, solve using any method. 3 w 2 ⫺ ⫽ w 8 4

12. x2 ⫹ 16x ⫹ n

13. x2 ⫺ 9x ⫹ n

39. 3x2 ⫺ 4x ⫽ 6

40.

1 14. y2 ⫹ y ⫹ n 2

2 15. z2 ⫺ z ⫹ n 5

41. y2 ⫹ 14y ⫽ ⫺46

42. 1a ⫹ 12 2 ⫽ 11

For Exercises 16–21, solve the equation by completing the square and applying the square root property. 16. w2 ⫹ 4w ⫹ 13 ⫽ 0

17. 4y2 ⫺ 8y ⫺ 20 ⫽ 0

18. 2x2 ⫽ 12x ⫹ 6

19. ⫺t2 ⫹ 8t ⫺ 25 ⫽ 0

20. 3x2 ⫹ 2x ⫽ 1

7 21. b2 ⫹ b ⫽ 2 2

22. Solve for r.

V ⫽ pr2h

23. Solve for s.

A ⫽ 6s2

1r 7 02 1s 7 02

43. The landing distance that a certain plane will travel on a runway is determined by the initial landing speed at the instant the plane touches down. The function D relates landing distance in feet to initial landing speed s: D1s2 ⫽

1 2 s ⫺ 3s ⫹ 22 for s ⱖ 50 10

where s is in feet per second.

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a. Find the landing distance for a plane traveling 150 ft/sec at touchdown. b. If the landing speed is too fast, the pilot may run out of runway. If the speed is too slow, the plane may stall. Find the maximum initial landing speed of a plane for a runway that is 1000 ft long. Round to one decimal place. 44. The recent population of Kenya (in thousands) can be approximated by P1t2  4.62t2  564.6t  13,128, where t is the number of years since 1974.

Section 7.3 For Exercises 47–56, solve the equations by using substitution, if necessary. 47. x  41x  21  0 48. n  61n  8  0 49. y4  11y2  18  0 50. 2m4  m2  3  0 51. t25  t15  6  0

a. If this trend continues, approximate the number of people in Kenya for the year 2025.

52. p25  3p15  2  0

b. In what year after 1974 will the population of Kenya reach 50 million? (Hint: 50 million equals 50,000 thousand.)

53.

2t 3  1 t1 t2

54.

1 m  2 m2 m3

45. A custom-built kitchen island is in the shape of a rectangle. The length is 1 ft more than twice the width. If the area is 22.32 ft2, determine the dimensions of the island.

55. 1x2  52 2  21x2  52  8  0 56. 1x2  32 2  51x2  32  4  0

Section 7.4 For Exercises 57–64, graph the functions.

58. f 1x2  x2  3

57. g1x2  x2  5

y

y 9

3 2 1

4 5

8 7 6 5 4 3 2 1

6 7

5 4 3 2 1 1

5 4 3 2 1 1

1 2

3 4

5

x

2 3

46. Lincoln, Nebraska, Kansas City, Missouri, and Omaha, Nebraska, form the vertices of a right triangle. The distance between Lincoln and Kansas City is 10 mi more than 3 times the distance between Lincoln and Omaha. If the distance from Omaha and Kansas City is 167 mi, find the distance between Lincoln and Omaha. Round to the nearest mile. Omaha

Lincoln

Kansas City

2

3 4

5

x

60. k1x2  1x  32 2

59. h1x2  1x  52 2

y

y

7 6 5 4 3 2 1

8 7 6 5 4 3 2 1 2 1 1 2

1

1 2

3 4

5 6 7 8

x

7 6 5 4 3 2 1 1 2 3

1

2

3

x

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Review Exercises

62. n1x2  4x2

61. m1x2  2x2

y

y

1 2

3 4

5

x

5 4 3 2 1 1

2 3

74. g1x2  4x2  8x  3 1 2

3 4

5

x

75. m1x2  3x2  3x  11

2 3

4 5

76. n1x2  3x2  2x  7

4 5

6 7

6

63. p1x2  21x  52 2  5

64. q1x2  41x  32 2  3 5 4 3 2 1

1 1

2

3 4

5 6 7 8

x

4 5

7 6 5 4 3 2 1 1 2 3

6 7 9

y 5 4 3

b. Find the x- and y-intercepts. 1

2

3

x

4 5

8

3 77. For the quadratic equation y  x2  3x, 4 a. Write the coordinates of the vertex.

y

y

2 1 1 2 3

73. f 1x2  2x2  4x  17

4 3 2 1

3 2 1 5 4 3 2 1 1

For Exercises 73–76, find the coordinates of the vertex of each parabola by using the vertex formula.

c. Use this information to sketch a graph of the parabola.

2 1 5 4 3 2 1 1 2

1

2 3

4

5

x

3 4 5

78. For the quadratic equation y  1x  22 2  4, For Exercises 65–66, write the coordinates of the vertex of the parabola and determine if the vertex is a maximum point or a minimum point. Then write the maximum or the minimum value. 1 5 65. t1x2  1x  42 2  3 3

For Exercises 67–68, write the equation of the axis of symmetry of the parabola. 3 2 2 4 67. a1x2   ax  b  2 11 13 4 3 2 2 68. w1x2   ax  b  3 16 9

Section 7.5 For Exercises 69–72, write the function in the form f 1x2  a1x  h2 2  k by completing the square. Then write the coordinates of the vertex. 69. z1x2  x2  6x  7

5

b. Find the x- and y-intercepts.

2 1

c. Use this information to sketch a graph of the parabola.

5 1 66. s1x2   1x  12 2  7 7

70. b1x2  x2  4x  44

71. p1x2  5x2  10x  13 72. q1x2  3x2  24x  54

y

a. Write the coordinates of the vertex.

4 3

5 4 3 2 1 1 2

1

2 3

4

5

3 4 5

79. The height h1t2 (in feet) of a projectile fired vertically into the air from the ground is given by the equation h1t2  16t 2  96t, where t represents the number of seconds that the projectile has been in the air. a. How long will it take the projectile to reach its maximum height? b. What is the maximum height? 80. The weekly profit, P(x) (in dollars), for a catering service is given by P1x2  0.053x2  15.9x  7.5. In this context, x is the number of meals prepared. a. Find the number of meals that should be prepared to obtain the maximum profit. b. What is the maximum profit?

x

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81. Write an equation of a parabola that passes through the points (3, 4), (2, 5), and (1, 4) 82. Write an equation of a parabola that passes through the points (4, 18), (2, 12), and (1, 8)

4x 0 x2

d. Solve the inequality.

4x 0 x2

For Exercises 85–96, solve the inequalities. Write the answers in interval notation.

Section 7.6

85. w2  4w  12 6 0

83. Solve the equation and inequalities. How do your answers to parts (a), (b), and (c) relate to the graph of g 1x2  x2  4? a. x2  4  0

87.

b. x  4 6 0 c. x2  4 7 0

6 4 2 10 8 6 4 2 2 4

g(x)  x 2  4

2

4 6

8 10

6 8 10

84. a. For what values of x is b. Solve the equation.

12 6 x2

86. t 2  6t  9  0 88.

8  4 p1

89. 3y1y  521y  22 7 0

y 10 8

2

Chapter 7

c. Solve the inequality.

4x undefined? x2

x

90. 3c1c  2212c  52 6 0 91. x2  4x 1 93.

w1 7 1 w3

95. t 2  10t  25 0

92. y2  4y 7 5 94.

2a 2 a3

96. x2  4x 6 4

4x 0 x2

Test

For Exercises 1–3, solve the equation by using the square root property. 1. 1x  32 2  25

2. 1p  22 2  12

3. 1m  12 2  1

For Exercises 5–6, solve the equation by completing the square and applying the square root property. 6. 2x2  3x  7

For Exercises 7–8, a. Write the equation in standard form ax2  bx  c  0. b. Identify a, b, and c.

d. Determine the number and type (rational, irrational, or imaginary) of solutions. 7. x2  3x  12

4. Find the value of n so that the expression is a perfect square trinomial. Then factor the trinomial d 2  11d  n.

5. 2x2  12x  36  0

c. Find the discriminant.

8. y1y  22  1

For Exercises 9–10, solve the equation by using the quadratic formula. 9. 3x2  4x  1  0 10. x1x  62  11  x 11. The base of a triangle is 3 ft less than twice the height. The area of the triangle is 14 ft2. Find the base and the height. Round the answers to the nearest tenth of a foot. 12. A circular garden has an area of approximately 450 ft2. Find the radius. Round the answer to the nearest tenth of a foot.

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Test

For Exercises 13–17, solve the equation. 13. x  1x  6  0

2/3

14. y

 2y

1/3

8

15. 13y  82 2  1313y  82  30  0 y 1 17. 3   2 y1

16. p4  15p2  54

18. Find the vertex of y  x2  6x  8 two ways, a. By completing the square.

For Exercises 19–21, graph the functions.

b. Approximate the year in which the population of India reached 1 billion (1000 million). (Round to the nearest year.)

5 4 3

26. Explain the relationship between the graphs of y  4x2 and y  4x2.

2 1 5 4 3 2 1 1 2

1

2 3

4

5

x

27. Given the function defined by

f 1x2  1x  42 2  2

3 4 5

20. f 1x2  1x  42 2

a. Identify the vertex of the parabola. b. Does this parabola open upward or downward?

y 14 12 10 8 6 4 2 3 2 1 2

c. Does the vertex represent the maximum or minimum point of the function? d. What is the maximum or minimum value of the function f ? 1 2

3 4

5 6 7

e. Write an equation for the axis of symmetry.

x

4 6

28. For the function defined by g(x)  2x2  20x  51, find the vertex by using two methods.

y

1 21. g1x2  1x  22 2  3 2

5 4 3

a. Complete the square to write g(x) in the form g(x)  a(x  h)2  k. Identify the vertex.

2 1 5 4 3 2 1 1 2

1

2 3

4

5

b. Use the vertex formula to find the vertex.

x

29. Given: f1x2  x2  4x  12

3 4 5

a. Write the equation for the function in the form f 1x2  a1x  h2 2  k.

h(x)

b. Determine the vertex. c. Find the x- and y-intercepts.

h(x) ft

22. A child launches a toy rocket from the ground. The height of the rocket can be determined by its horizontal distance from the launch pad x by x2 x 256

a. If this trend continues, approximate the number of people in India in the year 2014.

25. Explain the relationship between the graphs of y  x2 and y  (x  3)2.

y

2

h1x2  

23. The recent population of India (in millions) can be approximated by P1t2  0.135t 2  12.6t  600, where t  0 corresponds to the year 1974.

24. Explain the relationship between the graphs of y  x2 and y  x2  2.

b. By using the vertex formula.

19. h1x2  x  4

651

d. Determine the maximum or minimum value. e. Write an equation for the axis of symmetry. x feet

x

where x and h(x) are in feet. How many feet from the launch pad will the rocket hit the ground?

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Chapter 7 Quadratic Equations and Functions

a. Determine the value of x that maximizes the area of the field.

30. A farmer has 400 ft of fencing with which to enclose a rectangular field. The field is situated such that one of its sides is adjacent to a river and requires no fencing. The area of the field (in square feet) can be modeled by A1x2  

x2  200x 2

where x is the length of the side parallel to the river (measured in feet).

b. Determine the maximum area that can be enclosed. For Exercises 31–36, solve the inequalities. 31.

2x  1 0 x6

32. 50  2a2 7 0

33. y3  3y2  4y  12 6 0

34.

3 7 2 w3

35. 5x2  2x  2 6 0 36. t 2  22t  121 0 x

Chapters 1–7

Cumulative Review Exercises

1. Given: A  52, 4, 6, 8, 106 and B  52, 8, 12, 166 a. Find A ´ B.

b. Find A 傽 B.

2. Perform the indicated operations and simplify. 12x2  52  1x  3215x  22

3. Simplify completely.

1 3 4  a b  811 2 2 0

4. Perform the indicated operations. Write the answer in scientific notation: 13.0 1012 216.0 103 2

5. a. Factor completely.

x  2x  9x  18 3

2

b. Divide by using long division. Identify the quotient and remainder. 1x3  2x2  9x  182  1x  32

8. Jacques invests a total of $10,000 in two mutual funds. After 1 yr, one fund produced 12% growth, and the other lost 3%. Find the amount invested in each fund if the total investment grew by $900. 9. Solve the system of equations. 1 1 13 x y 9 3 9 1 9 x y 2 2 10. An object is fired straight up into the air from an initial height of 384 ft with an initial velocity of 160 ft/sec. The height in feet is given by h1t2  16t2  160t  384 where t is the time in seconds after launch.

1 1x  1221 1x  122

a. Find the height of the object after 3 sec.

6. Multiply. 7. Simplify.

4 12x

c. Find the time required for the object to hit the ground.

b. Find the height of the object after 7 sec.

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Cumulative Review Exercises

11. Solve.

1x  32 2  16  0

12. Solve.

2x2  5x  1  0

19. The quantity y varies directly as x and inversely as z. If y  15 when x  50 and z  10, find y when x  65 and z  5.

13. Find the value of n so that the expression is a perfect square trinomial. Then factor the trinomial x2  10x  n. 2x3  250

14. Factor completely.

b. Find the y-intercept and interpret its meaning in the context of this problem.

y 5 4 3 2 1 5 4 3 2 1 1

20. The total number of flights (including passenger flights and cargo flights) at a large airport can be approximated by F1x2  300,000  0.008x, where x is the number of passengers. a. Is this function linear, quadratic, constant, or other?

3x  5y  10

15. Graph the equation.

653

c. What is the slope of the function and what does the slope mean in the context of this problem? 1

2

3 4

5

x

21. Let m1x2  1x  4 and n1x2  x2  2. Find

2 3

a. The domain of m

4 5

16. a. Find the x-intercepts of the function defined by g1x2  2x2  9x  10.

22. Consider the function y  f 1x2 graphed here. Find y 5

b. What is the y-intercept?

1

4

2

5

3

6

2 1 5 4 3 2 1 1 2

a. The domain c. f 112

4

5

x

e. For what value(s) of x is f 1x2  0? 1 . x

23. Solve.

28x  5  22x  2

24. Solve for f.

x

b. The range d. f(0)

y

4 5

2 3

4 5

5 4 3 2 1 2 3

1

3

18. Graph the function defined by f 1x2 

1

y  f(x)

4 3

17. Explain why this relation is not a function.

5 4 3 2 1 1 2 3 4 5

b. The domain of n

25. Solve.

1 1 1   p q f

15 1 2   t4 t2 t  2t  8 2

4 y3 y4

y 26. Simplify.

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Chapter 7 Quadratic Equations and Functions

27. Given the function defined by f(x)  2(x  3)2  1

29. a. Solve the inequality.

2x2  x  10  0

a. Write the coordinates of the vertex. b. Does the graph of the function open upward or downward?

b. How does the answer in part (a) relate to the graph of the function f 1x2  2x2  x  10? y

c. Write the coordinates of the y-intercept.

10 8 6 4 2

d. Find the x-intercepts, if possible. e. Sketch the function.

5 4 3 21 2 4 6 8 10

y 20 16 12 8 4 10 8 6 4 2 4 8 10

2 4 6 8 10

1 2 3 4 5

x

12

28. Find the vertex of the parabola.

f 1x2  x2  16x  2

30. Solve the inequality.

8 4 x3

x

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Exponential and Logarithmic Functions and Applications

8

CHAPTER OUTLINE 8.1 Algebra of Functions and Composition 656 8.2 Inverse Functions 663 8.3 Exponential Functions 672 8.4 Logarithmic Functions 682 Problem Recognition Exercises: Identifying Graphs of Functions

695

8.5 Properties of Logarithms 696 8.6 The Irrational Number e and Change of Base 704 Problem Recognition Exercises: Logarithmic and Exponential Forms

717

8.7 Logarithmic and Exponential Equations and Applications 718 Group Activity: Creating a Population Model

730

Chapter 8 This chapter is devoted to the study of exponential and logarithmic functions. These functions are used to study many naturally occurring phenomena such as population growth, exponential decay of radioactive matter, and growth of investments. Are You Prepared? To prepare yourself for this chapter, practice evaluating the expressions with exponents. Use the clues to fill in the boxes labeled A–H. Then fill in the remaining part of the grid so that every row, every column, and every 2 ⫻ 3 box contains the digits 1 through 6. A. B. C. D. E. F. G. H.

Value of x in the equation 5x ⫽ 25. Value of x in the equation 3x ⫽ 81. Value of x in the equation 81/x ⫽ 2. Real value of x in the equation x 4 ⫽ 16 (assume that x 7 0). 1 Absolute value of x in the equation 2 x ⫽ . 16 Value of x in the equation 2 x ⫽ 64. Value of x in the equation 50 ⫽ x. Value of x in the equation e0 ⫽ x.

B

F

C

4

2 D

G

4

A

4

3 E

2 H

3 2

655

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Chapter 8 Exponential and Logarithmic Functions and Applications

Section 8.1

Algebra of Functions and Composition

Concepts

1. Algebra of Functions

1. Algebra of Functions 2. Composition of Functions 3. Multiple Operations on Functions

Addition, subtraction, multiplication, and division can be used to create a new function from two or more functions. The domain of the new function will be the intersection of the domains of the original functions.

DEFINITION Sum, Difference, Product, and Quotient of Functions f

Given two functions f and g, the functions f  g, f  g, f ⴢ g, and g are defined as 1f  g21x2  f 1x2  g1x2

1f  g21x2  f 1x2  g1x2 1f ⴢ g2 1x2  f 1x2 ⴢ g1x2 f 1x2 f a b1x2  g g1x2

provided g1x2  0

For example, suppose f 1x2  ƒ x ƒ and g 1x2  3. Taking the sum of the functions produces a new function denoted by f  g. In this case, 1 f  g21x2  ƒ x ƒ  3. Graphically, the y values of the function f  g are given by the sum of the corresponding y values of f and g. This is depicted in Figure 8-1. The function f  g appears in red. In particular, notice that 1 f  g2122  f 122  g122  2  3  5. ( f  g)(x)  |x|  3 f (x)  |x|

y 7 6 5 4

f (2)  g(2)

g(x)  3 2 1 5 4 3 2 1 1 2 3

1 2 3 4 5

x

Figure 8-1

Example 1 Given: g1x2  4x

Adding, Subtracting, Multiplying, and Dividing Functions

a. Find 1g  h21x2 .

h1x2  x2  3x

c. Find 1g ⴢ k21x2 .

k1x2  x  2

b. Find 1h  g21x2 .

k d. Find a b1x2 . h

Solution:

a. 1g  h21x2  g1x2  h1x2

 14x2  1x 2  3x2  x2  x

Simplify.

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Section 8.1

657

Algebra of Functions and Composition

b. 1h  g21x2  h1x2  g1x2

 1x2  3x2  14x2  x 2  7x

Simplify.

c. 1g ⴢ k21x2  g1x2 ⴢ k1x2

 14x21x  22

 4x2  8x

Simplify.

k1x2 k d. a b1x2  h h1x2 

x2 x2  3x

From the denominator we have x2  3x  0 or, equivalently, x1x  32  0. Hence, x  3 and x  0.

Skill Practice Given f 1x 2  x  1, g1x2  5x2  x, and h1x2  x2, find 1. 1f  g21x2

2. 1g  f 21x2

3. 1g ⴢ h21x2

f 4. a b1x2 h

2. Composition of Functions DEFINITION

Composition of Functions

The composition of f and g, denoted f ⴰ g, is defined by the rule 1 f ⴰ g21x2  f 1g1x22

provided that g1x2 is in the domain of f.

The composition of g and f, denoted g ⴰ f, is defined by the rule 1g ⴰ f 21x2  g1 f 1x22

provided that f 1x2 is in the domain of g.

Note: f ⴰ g is also read as “f compose g,” and g ⴰ f is also read as “g compose f.”

For example, given f 1x2  2x  3 and g1x2  x  5, we have 1f ⴰ g21x2  f 1g1x22

 f 1x  52

Substitute g1x2  x  5 into the function f.

 21x  52  3

Avoiding Mistakes Be careful to note that the notation f 1g 1x22 represents function composition not multiplication.

 2x  10  3  2x  7 In this composition, the function g is the innermost operation and acts on x first. Then the output value of function g becomes the domain element of the function f, as shown in the figure. g x

f g(x)

f(g(x))

Answers 1. 5x 2  2x  1

2. 5x 2  1

3. 5x 4  x 3

4.

x1 x2

,x0

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Example 2

Composing Functions

Given: f 1x2 ⫽ x ⫺ 5, g1x2 ⫽ x2, and n1x2 ⫽ 1x ⫹ 2, find a. 1 f ⴰ g21x2

b. 1g ⴰ f 21x2

c. 1n ⴰ f 21x2

Solution:

a. 1 f ⴰ g21x2 ⫽ f 1g1x22 ⫽ f 1x 2 2

Evaluate the function f at x2.

⫽ 1x 2 2 ⫺ 5 ⫽ x2 ⫺ 5

TIP: Examples 2(a) and 2(b) illustrate that the order in which two functions are composed may result in different functions. That is, f ⴰ g does not necessarily equal g ⴰ f.

Replace x by x2 in the function f.

b. 1g ⴰ f 21x2 ⫽ g1 f 1x22 ⫽ g1x ⫺ 52 ⫽ 1x ⫺ 52

2

⫽ x2 ⫺ 10x ⫹ 25

Evaluate the function g at 1x ⫺ 52.

Replace x by 1x ⫺ 52 in function g. Simplify.

c. 1n ⴰ f 21x2 ⫽ n1 f 1x22 ⫽ n1x ⫺ 52 ⫽ 11x ⫺ 52 ⫹ 2

Evaluate the function n at x ⫺ 5.

Replace x by the quantity 1x ⫺ 52 in function n.

⫽ 1x ⫺ 3 Skill Practice Given f 1x 2 ⫽ 2x 2, g 1x2 ⫽ x ⫹ 3, and h 1x2 ⫽ 1x ⫺ 1, find 5. 1 f ⴰ g21x2

6. 1g ⴰ f 21x2

7. 1h ⴰ g21x2

3. Multiple Operations on Functions Example 3

Combining Functions

Given the functions defined by f 1x2 ⫽ x ⫺ 7 and h1x2 ⫽ 2x3, find the function values, if possible. a. 1 f ⴢ h2132

h b. a b 172 f

Solution:

a. 1 f ⴢ h2132 ⫽ f 132 ⴢ h132

⫽ 13 ⫺ 72 ⴢ 2132 3

⫽ 1⫺42 ⴢ 21272 ⫽ ⫺216

Answers 5. 2x 2 ⫹ 12x ⫹ 18 7. 1x ⫹ 2

6. 2x 2 ⫹ 3

c. 1h ⴰ f 2122 1f ⴢ h2 (3) is a product (not a composition).

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Section 8.1

h172 h b. a b 172  f f 172 2172 3



(undefined)

77

659

Algebra of Functions and Composition

The value x  7 makes the denominator zero.

h Therefore, a b 172 is undefined. f c. 1h ⴰ f 2122  h1 f 1222

Evaluate f(2) first. f 122  2  7  5.

 h152  2152

Substitute the result into function h.

3

 211252  250 Skill Practice Given h 1x2  x  4 and k 1x2  x2  3, find 8. 1h ⴢ k2 122

k 9. a b 142 h

10. 1k ⴰ h2112

Finding Function Values from a Graph

Example 4

For the functions f and g pictured, find the function values if possible. y

a. g(2)

5 y  g(x) 4 3 2

b. 1 f  g2132 g c. a b 152 f

5 4 3 21 1 2 y  f(x) 3 4 5

d. 1 f ⴰ g 2122

1 2 3 4 5

x

Solution: a. g122  1

b. 1 f  g2132  f 132  g132  2  132

The value g(2) represents the y-value of y  g(x) (the red graph) when x  2. Because the point (2, 1) lies on the graph, g122  1. Evaluate the difference of f 132 and g132 .

TIP: For your reference, determining function values from a graph was first covered in Section 2.6.

Estimate function values from the graph.

 5 g152 g c. a b 152  f f 152

Evaluate the quotient of g(5) and f(5).

1 (undefined) 0 g The function is undefined at x  5 because the denominator is zero. f 

Answers 8. 2

9. undefined

10. 22

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d. 1 f ⴰ g 2122  f 1g1222

From the red graph, find g122 . We see that g122  3.

 f 132

From the blue graph, find the value of f at x  3.

1 y

Skill Practice Find the values from the graph. 11. g 132 f 13. a b122 g

12. 1f  g2142

5 4 3 2 1

14. 1g ⴰ f 2122

5 4 3 2 1 1 2 3 4 5

Answers 11. 2

12. 0

13. 4

14. 2

Section 8.1

Practice Exercises

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Study Skills Exercise 1. Define the key term composition of functions.

Concept 1: Algebra of Functions

2. Given f 1x2  x2 and g1x2  2x  3, find a. f 122

b. g122

c. 1 f  g2122

For Exercises 3–14, refer to the functions defined below. f 1x2  x  4

g1x2  2x2  4x

Find the indicated functions. (See Example 1.)

h1x2  x2  1

k1x2 

1 x

3. 1 f  g21x2

4. 1 f  g2 1x2

5. 1g  f 21x2

6. 1 f  h21x2

7. 1 f ⴢ h21x2

8. 1h ⴢ k21x2

9. 1g ⴢ f 21x2

10. 1 f ⴢ k21x2

h 11. a b 1x2 f

f 13. a b1x2 g

f 14. a b 1x2 h

g 12. a b1x2 f

y  f(x) y  g(x) 1 2 3 4 5

x

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Algebra of Functions and Composition

Concept 2: Composition of Functions For Exercises 15–22, find the indicated functions. Use f, g, h, and k as defined in Exercises 3–14. (See Example 2.) 15. 1f ⴰ g 21x2

16. 1f ⴰ k21x2

17. 1g ⴰ f 21x2

18. 1k ⴰ f 21x2

19. 1k ⴰ h 21x2

20. 1h ⴰ k 21x2

21. 1k ⴰ g 21x2

22. 1g ⴰ k 21x2

23. Based on your answers to Exercises 15 and 17, is it true in general that 1 f ⴰ g 21x2 ⫽ 1g ⴰ f 21x2? 24. Based on your answers to Exercises 16 and 18, is it true in general that 1 f ⴰ k21x2 ⫽ 1k ⴰ f 21x2 ? For Exercises 25–28, find 1 f ⴰ g21x2 and 1g ⴰ f 21x2. 25. f 1x2 ⫽ x2 ⫺ 3x ⫹ 1, g1x2 ⫽ 5x 28. f 1x2 ⫽

1 , g1x2 ⫽ 冟x ⫹ 2冟 x⫹2

26. f 1x2 ⫽ 3x2 ⫹ 8, g1x2 ⫽ 2x ⫺ 4

27. f 1x2 ⫽ 冟x冟, g1x2 ⫽ x3 ⫺ 1

29. For h1x2 ⫽ 5x ⫺ 4, find 1h ⴰ h21x2.

30. For k1x2 ⫽ ⫺x2 ⫹ 1, find 1k ⴰ k21x2.

Concept 3: Multiple Operations on Functions For Exercises 31–46, refer to the functions defined below. m 1x2 ⫽ x3

n1x2 ⫽ x ⫺ 3

Find each function value if possible. (See Example 3.)

r1x2 ⫽ 1x ⫹ 4

p1x2 ⫽

1 x⫹2

31. 1m ⴢ r2102

32. 1n ⴢ p2102

33. 1m ⫹ r21⫺42

34. 1n ⫺ m2142

35. 1r ⴰ n 2132

36. 1n ⴰ r 2152

37. 1p ⴰ m 21⫺12

38. 1m ⴰ n 2152

39. 1m ⴰ p 2122

40. 1r ⴰ m 2122

41. 1r ⫹ p21⫺32

42. 1n ⫹ p21⫺22

43. 1m ⴰ p 21⫺22

44. 1r ⴰ m 21⫺22

r 45. a b1122 n

46. a

n b122 m

For Exercises 47–64, approximate each function value from the graph, if possible. (See Example 4.) 47. f 1⫺42

48. f(1)

49. g1⫺22

50. g(3)

51. 1 f ⫹ g2122

52. 1g ⫺ f 2132

53. 1 f ⴢ g21⫺12

54. 1g ⴢ f 21⫺42

g 55. a b 102 f

f 56. a b 1⫺22 g

f 57. a b 102 g

g 58. a b 1⫺22 f

59. 1g ⴰ f 21⫺12

60. 1 f ⴰ g 2102

61. 1 f ⴰ g 21⫺42

62. 1g ⴰ f 21⫺42

63. 1g ⴰ g 2122

64. 1 f ⴰ f 21⫺22

y 5 4 y ⫽ g(x) 3 2 1

⫺5⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

y ⫽ f(x)

1 2 3 4 5

x

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For Exercises 65–80, approximate each function value from the graph, if possible. (See Example 4.) 65. a132

66. a(1)

67. b112

68. b(3)

69. 1a  b2112

70. 1a  b2102

71. 1b ⴢ a2112

72. 1a ⴢ b2122

73. 1b ⴰ a2102

74. 1a ⴰ b2122

75. 1a ⴰ b2142

76. 1b ⴰ a2132

b 77. a b132 a

a 78. a b142 b

79. 1a ⴰ a2122

y 5 4 3 2 1

5 4 3 2 1 1 2 3

y = a(x)

x

1 2 3 4 5

y = b(x)

4 5

80. 1b ⴰ b2112

81. The cost in dollars of producing x toy cars is C1x2  2.2x  1. The revenue received is R1x2  5.98x. To calculate profit, subtract the cost from the revenue. a. Write and simplify a function P that represents profit in terms of x. b. Find the profit of producing 50 toy cars. 82. The cost in dollars of producing x lawn chairs is C1x2  2.5x  10.1. The revenue for selling x lawn chairs is R1x2  6.99x. To calculate profit, subtract the cost from the revenue. a. Write and simplify a function P that represents profit in terms of x.

83. The functions defined by D1t2  0.925t  26.958 and R1t2  0.725t  20.558 approximate the amount of child support (in billions of dollars) that was due and the amount of child support actually received in the United States since 2000. In each case, t  0 corresponds to the year 2000. a. Find the function F defined by F1t2  D1t2  R1t2. What does F represent in the context of this problem? b. Find F(4). What does this function value represent in the context of this problem?

Amount ($ billions)

b. Find the profit in producing 100 lawn chairs.

40 35 30 25 20 15 10 5 0

Difference Between Child Support Due and Child Support Paid, United States

D(t)  0.925t  26.958 R(t)  0.725t  20.558 F(t)  0.2t  6.4 0

2 4 6 8 Year (t  0 corresponds to 2000)

10

Source: U.S. Bureau of the Census

84. The rural and urban populations in the South (in the United States) between the years 1900 and 1970 can be modeled by the following: u1t2  0.0566t 3  0.952t 2  177.8t  4593 The variable t represents the number of years since 1900, r (t) represents the rural population in thousands, and u(t) represents the urban population in thousands. a. Find the function T defined by T(t)  r(t)  u(t). What does the function T represent in the context of this problem?

Population (thousands)

r1t2  3.497t 2  266.2t  20,220

Rural and Urban Populations in the South, United States, 1900–1970

45,000 40,000 35,000 2 30,000 r(t)  3.497t  266.2t  20,220 25,000 20,000 15,000 10,000 u(t)  0.0566t 3  0.952t2  177.8t  4593 5000 t 0 0 10 20 30 40 50 60 70 Year (t  0 corresponds to 1900)

Source: Historical Abstract of the United States

b. Use the function T to approximate the total population in the South for the year 1940.

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663

85. Joe rides a bicycle and his wheels revolve at 80 revolutions per minute (rpm). Therefore, the total number of revolutions, r, is given by r1t2  80t, where t is the time in minutes. For each revolution of the wheels of the bike, he travels approximately 7 ft. Therefore, the total distance he travels, D, depends on the total number of revolutions, r, according to the function D1r2  7r. a. Find 1D ⴰ r 21t2 and interpret its meaning in the context of this problem. b. Find Joe’s total distance in feet after 10 min. 86. The area of a square is given by the function a1x2  x2, where x is the length of the sides of the square. If carpeting costs $9.95 per square yard, then the cost, C, to carpet a square room is given by C1a2  9.95a, where a is the area of the room in square yards. a. Find 1C ⴰ a 21x2 and interpret its meaning in the context of this problem.

b. Find the cost to carpet a square room if its floor dimensions are 15 yd by 15 yd.

Inverse Functions

Section 8.2

1. Introduction to Inverse Functions

Concepts

In Section 2.6, we defined a function as a set of ordered pairs (x, y) such that for every element x in the domain, there corresponds exactly one element y in the range. For example, the function f relates the weight of a package of deli meat, x, to its cost, y.

1. Introduction to Inverse Functions 2. Definition of a One-to-One Function 3. Finding an Equation of the Inverse of a Function 4. Definition of the Inverse of a Function

f  {(1, 4), (2.5, 10), (4, 16)} That is, 1 lb of meat sells for $4, 2.5 lb sells for $10, and 4 lb sells for $16. Now suppose we create a new function in which the values of x and y are interchanged. The new function, called the inverse of f, denoted f 1, relates the price of meat, x, to its weight, y.

Avoiding Mistakes

f 1  {(4, 1), (10, 2.5), (16, 4)} Notice that interchanging the x- and y-values has the following outcome. The domain of f is the same as the range of f 1, and the range of f is the domain of f 1.

2. Definition of a One-to-One Function A necessary condition for a function f to have an inverse function is that no two ordered pairs in f have different x-coordinates and the same y-coordinate. A function that satisfies this condition is called a one-to-one function. The function relating the weight of a package of meat to its price is a one-to-one function. However, consider the function g defined by g  {(1, 4), (2, 3), (2, 4)} same y different x

This function is not one-to-one because the range element 4 has two different x-coordinates, 1 and 2. Interchanging the x- and y-values produces a relation that violates the definition of a function. {(4, 1), (3, 2), (4, 2)} same x different y

This relation is not a function because for x  4 there are two different y-values, y  1 and y  2.

f 1 denotes the inverse of a function. The 1 does not represent an exponent.

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In Section 2.6, you learned the vertical line test to determine visually if a graph represents a function. Similarly, we use a horizontal line test to determine whether a function is one-to-one.

PROPERTY Using the Horizontal Line Test Consider a function defined by a set of points (x, y) in a rectangular coordinate system. Then y is a one-to-one function of x if no horizontal line intersects the graph in more than one point. To understand the horizontal line test, consider the functions f and g. f  {(1, 4), (2.5, 10), (4, 16)}

g  {(1, 4), (2, 3), (2, 4)} y

y 5 4 3 2 1

16 14 12 10 8 6 4 2 5 4 3 2 1 2 4

1 2 3 4 5

5 4 3 2 1 1 2 3 4 5

x

This function is one-to-one. No horizontal line intersects more than once. Example 1

1 2 3 4 5

x

This function is not one-to-one. A horizontal line intersects more than once.

Identifying a One-to-One Function

For each function, determine if the function is one-to-one. a.

b.

y

y

x

x

Solution: a. Function is not one-to-one. A horizontal line intersects in more than one point.

b. Function is one-to-one. No horizontal line intersects more than once. y

y

x

x

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Section 8.2

Inverse Functions

Skill Practice For each function determine if the function is one-to-one. 1.

2.

y 5 4 3 2

y 5 4 3 2 1

1 5 4 3 21 1 2 3 4 5

1 2 3 4 5

x

5 4 3 21 1 2 3 4 5

1 2 3 4 5

x

3. Finding an Equation of the Inverse of a Function

For a one-to-one function defined by y  f 1x2 , the inverse is a function y  f 1 1x2 that performs the inverse operations in the reverse order. For example, the function defined by f(x)  2x  1 multiplies x by 2 and then adds 1. Therefore, the inverse function must subtract 1 from x and divide by 2. We have f 1 1x2 

x1 2

The expression f 1 1x2 is read as “f inverse of x.”

To facilitate the process of finding an equation of the inverse of a one-to-one function, we offer the following steps.

PROCEDURE Finding an Equation of an Inverse of a Function For a one-to-one function defined by y  f (x), the equation of the inverse can be found as follows: Step 1 Step 2 Step 3 Step 4

Replace f (x) by y. Interchange x and y. Solve for y. Replace y by f 1(x).

Example 2

Finding an Equation of the Inverse of a Function

Find the inverse.

f (x)  2x  1

Solution: Foremost, we know the graph of f is a nonvertical line. Therefore, f(x)  2x  1 defines a one-to-one function. To find the inverse we have y  2x  1

Step 1:

Replace f (x) by y.

x  2y  1

Step 2:

Interchange x and y.

Step 3:

Solve for y. Subtract 1 from both sides.

x  1  2y x1 y 2 f 1 1x2 

x1 2

Divide both sides by 2. Step 4:

Replace y by f 1(x).

Skill Practice Find the inverse. 3. f 1x2  4x  6

Answers 1. Not one-to-one 2. One-to-one x6 3. f 1 1x 2  4

665

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y 5 4 3 2 1 5 4 3 2 1 1 2 (5, 3) 3 4 5

f(x)  2x  1 yx (1, 3) (3, 1)

x

1 2 3 4 5

f 1(x) 

The key step in determining the equation of the inverse of a function is to interchange x and y. By so doing, a point (a, b) on f corresponds to a point (b, a) on f 1. For this reason, the graphs of f and f 1 are symmetric with respect to the line y  x (Figure 8-2). Notice that the point (3, 5) of the function f corresponds to the point (5, 3) of f 1. Likewise, (1, 3) of f corresponds to (3, 1) of f 1.

x1 2

Example 3

(3, 5)

Finding an Equation of the Inverse of a Function

Find the inverse of the one-to-one function. Figure 8-2

3 g1x2  1 5x  4

Solution: 3 y 1 5x  4 3

x  1 5y  4 3

x  4  1 5y

1x  42 3  1 1 5y2 3

Step 1:

Replace g(x) by y.

Step 2:

Interchange x and y.

Step 3:

Solve for y. Add 4 to both sides.

3

To eliminate the cube root, cube both sides.

1x  42  5y 3

Simplify the right side.

1x  42 3 y 5 g1 1x2  g1(x) 

(x  4)3 y 5 10 8 6 4 2

108 6 4 2 2 4 6 8 10

yx

Divide both sides by 5.

1x  42 3 5

Step 4:

Replace y by g1(x).

Skill Practice Find the inverse. 3 4. h1x2  1 2x  1

3

g(x)  5x  4 2 4 6 8 10

Figure 8-3

x

The graphs of g and g 1 from Example 3 are shown in Figure 8-3. Once again we see that the graphs of a function and its inverse are symmetric with respect to the line y  x. For a function that is not one-to-one, sometimes we can restrict its domain to create a new function that is one-to-one. This is demonstrated in Example 4.

Example 4

Finding the Equation of an Inverse of a Function with a Restricted Domain

Given the function defined by m(x)  x2  4 for x  0, find an equation defining m1.

Solution: From Section 7.4, we know that y  x2  4 is a parabola with vertex at (0, 4) (Figure 8-4). The graph represents a function that is not one-to-one. However, with the restriction on the domain x  0, the graph of m(x)  x2  4, x  0, consists of only the “right” branch of the parabola (Figure 8-5). This is a one-to-one function.

Answer 4. h 1 1x2 

x3  1 2

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Section 8.2

y 10 8 6 4 2 108 6 4 2 2 4 6 8 10

y

y  x2  4

2 4 6 8 10

10 8 6 4 2 x

108 6 4 2 2 4 6 8 10

Figure 8-4

667

Inverse Functions

m(x)  x2  4; x  0

2 4 6 8 10

x

Figure 8-5

To find the inverse, we have y  x2  4

x0

Step 1: Replace m(x) by y.

xy 4

y0

Step 2: Interchange x and y. Notice that the restriction x  0 becomes y  0.

2

x  4  y2 1x  4  y

y0

Step 3: Solve for y. Subtract 4 from both sides.

y0

Apply the square root property. Notice that we obtain the positive square root of x  4 because of the restriction y  0.

m1 1x2  1x  4

Step 4: Replace y by m1(x). Notice that the domain of m1 has the same values as the range of m.

Figure 8-6 shows the graphs of m and m1.

108 6 4 2 2 4 6 8 10

Skill Practice Find the inverse. 5. g1x2  x2  2

y m(x)  x2  4; x  0

x0

yx

10 8 6 4 2

m1(x)  x  4 2 4 6 8 10

Figure 8-6

4. Definition of the Inverse of a Function DEFINITION Inverse Function If f is a one-to-one function represented by ordered pairs of the form (x, y), then the inverse function, denoted f 1, is the set of ordered pairs denoted by ordered pairs of the form (y, x).

An important relationship between a function and its inverse is shown in Figure 8-7. Domain of f

Range of f f y

x f −1

Range of f −1

Domain of f −1

Figure 8-7

Answer

5. g 1 1x 2  1x  2

x

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Recall that the domain of f is the range of f 1 and the range of f is the domain of f 1. The operations performed by f are reversed by f 1. This leads to the inverse function property.

PROPERTY Inverse Function Property If f is a one-to-one function, then g is the inverse of f if and only if ( f ⴰ g)(x)  x

for all x in the domain of g

(g ⴰ f )(x)  x

for all x in the domain of f

and

Example 5

Composing a Function with Its Inverse

Show that the functions are inverses. f 1x2  2x  1

g1x2 

x1 2

Solution:

To show that the functions f and g are inverses, confirm that 1f ⴰ g21x2  x and 1g ⴰ f 21x2  x. 1f ⴰ g21x2  f 1g1x22

1g ⴰ f 21x2  g1f 1x22

fa

x1 b 2

 g 12x  12

 2a

x1 b1 2



12x  12  1 2

2x 2

x11



x ✔

x ✔

The functions f and g are inverses because 1f ⴰ g21x2  x and 1g ⴰ f21x2  x. This also confirms our solution to Example 2. Skill Practice Show that the functions are inverses. 6. f 1x2  3x  2

Answer

6. 1f ⴰ g2 1x2  f 1g 1x22 x2 b2x  3a 3 1g ⴰ f 2 1x2  g 1f 1x22 3x  2  2  x 3

and

g 1x2 

x2 3

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Section 8.2

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Practice Exercises

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Study Skills Exercise 1. Define the key terms. a. Inverse function

b. One-to-one function

c. Horizontal line test

Review Exercises 2. Write the domain and range of the relation {(3, 4), (5, 2), (6, 1), (3, 0)}. For Exercises 3–8, determine if the relation is a function by using the vertical line test. (See Section 2.6.) 3.

y

4.

y

x

6.

y

5.

x

x

y

7.

y

8.

y

x

x

x

Concept 1: Introduction to Inverse Functions For Exercises 9–12, write the inverse function for each function. 9. g  {(3, 5), (8, 1), (3, 9), (0, 2)} 11. r  {(a, 3), (b, 6), (c, 9)}

10. f  {(6, 2), (9, 0), (2, 1), (3, 4)} 12. s  {(1, x), (2, y), (3, z)}

Concept 2: Definition of a One-to-One Function 13. The table relates a state, x, to the number of representatives in the House of Representatives, y, for a recent year. Does this relation define a one-to-one function? If so, write a function defining the inverse as a set of ordered pairs.

State x

Number of Representatives y

Colorado

7

California

53

Texas

32

Louisiana Pennsylvania

7 19

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Chapter 8 Exponential and Logarithmic Functions and Applications

14. The table relates a city, x, to its average January temperature, y. Does this relation define a one-to-one function? If so, write a function defining the inverse as a set of ordered pairs. City x

Temperature y (°C)

Gainesville, Florida

13.6

Keene, New Hampshire

6.0

Wooster, Ohio

4.0

Rock Springs, Wyoming

6.0

Lafayette, Louisiana

10.9

For Exercises 15–20, determine if the function is one-to-one by using the horizontal line test. (See Example 1.) 15.

16.

y

17.

y

x

x

18.

19.

y

y

x

20.

y

x

y

x

x

Concept 3: Finding an Equation of the Inverse of a Function For Exercises 21–32, write an equation of the inverse for each one-to-one function as defined. (See Examples 2–4.) 21. h(x)  x  4

22. k(x)  x  3

25. p(x)  x  10

26. q1x2  x 

3 29. g1x2  2 2x  1

30. f 1x2  x3  4

2 3

1 23. m1x2  x  2 3

24. n(x)  4x  2

27. f(x)  x3  1

3 28. g1x2  1 x  2

31. g1x2  x2  9

x0

32. m1x2  x2  1

x0

33. The function defined by f (x)  0.3048x converts a length of x feet into f (x) meters. a. Find the equivalent length in meters for a 4-ft board and a 50-ft wire. b. Find an equation defining y  f 1(x). c. Use the inverse function from part (b) to find the equivalent length in feet for a 1500-m race in track and field. Round to the nearest tenth of a foot.

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671

Inverse Functions

34. The function defined by s(x)  1.47x converts a speed of x mph to s(x) ft/sec. a. Find the equivalent speed in feet per second for a car traveling 60 mph. b. Find an equation defining y  s1(x). c. Use the inverse function from part (b) to find the equivalent speed in miles per hour for a train traveling 132 ft/sec. Round to the nearest tenth. For Exercises 35–41, answer true or false. 35. The function defined by y  2 has an inverse function defined by x  2. 36. The domain of any one-to-one function is the same as the domain of its inverse. 37. All linear functions with a nonzero slope have an inverse function. 38. The function defined by g1x2  0 x 0 is one-to-one. 39. The function defined by k(x)  x2 is one-to-one. 40. The function defined by h1x2  0 x 0 for x  0 is one-to-one. 41. The function defined by L(x)  x2 for x  0 is one-to-one. 42. Explain how the domain and range of a one-to-one function and its inverse are related. 43. If (0, b) is the y-intercept of a one-to-one function, what is the x-intercept of its inverse? 44. If (a, 0) is the x-intercept of a one-to-one function, what is the y-intercept of its inverse? 45. a. Find the domain and range of the function defined by f 1x2  1x  1.

b. Find the domain and range of the function defined by f 1(x)  x2  1, x  0.

46. a. Find the domain and range of the function defined by g(x)  x2  4, x  0. b. Find the domain and range of the function defined by g1 1x2   1x  4.

For Exercises 47–50, the graph of y  f(x) is given. a. State the domain of f.

b. State the range of f.

c. State the domain of f 1.

d. State the range of f 1.

e. Graph the function defined by y  f 1(x). The line y  x is provided for your reference. y

47.

y  f(x)

5 4 3 2 1

5 4 3 2 1 1 2

y

48.

yx 1 2 3 4 5

y  f(x) x

5 4 3 2 1

5 4 3 2 1 1 2

y

49. 5 4 3 2 1

yx 1 2 3 4 5

x

5 4 3 2 1 1 2

y

50. 5 4 3 2 1

y  f(x) yx

1 2 3 4 5

x

5 4 3 2 1 1 2

3

3

3

3

4 5

4 5

4 5

4 5

yx y  f(x) 1 2 3 4 5

x

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Chapter 8 Exponential and Logarithmic Functions and Applications

Concept 4: Definition of the Inverse of a Function For Exercises 51–56, verify that f and g are inverse functions by showing that a. 1 f ⴰ g21x2  x

b. 1g ⴰ f 21x2  x (See Example 5.)

51. f 1x2  6x  1 and g1x2  53. f 1x2 

x1 6

52. f 1x2  5x  2 and g1x2 

3 2 x and g1x2  8x3 2

54. f 1x2 

55. f 1x2  x2  1, x  0, and g1x2  1x  1, x  1

x2 5

3 1 x and g1x2  27x3 3

56. f 1x2  x2  3, x  0, and g1x2  1x  3, x   3

Expanding Your Skills For Exercises 57–66, write an equation of the inverse of the one-to-one function. 57. q1x2  1x  4 61. f 1x2 

58. v1x2  1x  16

x1 x1

65. n1x2  x2  9

62. p1x2  x0

3x x3

59. z1x2   1x  4 63. t1x2 

60. u1x2   1x  16

2 x1

66. g1x2  x2  1

64. w1x2 

4 x2

x0

Graphing Calculator Exercises For Exercises 67–70, use a graphing calculator to graph each function on the standard viewing window. Use the graph of the function to determine if the function is one-to-one on the interval 10  x  10. If the function is one-to-one, find its inverse and graph both functions on the standard viewing window. 3 67. f 1x2  1 x5

68. k(x)  x3  4

69. g(x)  0.5x3  2

70. m(x)  3x  4

Section 8.3

Exponential Functions

Concepts

1. Definition of an Exponential Function

1. Definition of an Exponential Function 2. Approximating Exponential Expressions with a Calculator 3. Graphs of Exponential Functions 4. Applications of Exponential Functions

The concept of a function was first introduced in Section 2.6. Since then we have learned to recognize several categories of functions, including constant, linear, rational, and quadratic functions. In this section and the next, we will define two new types of functions called exponential and logarithmic functions. To introduce the concept of an exponential function, consider the following salary plans for a new job. Plan A pays $1 million for a month’s work. Plan B starts with 2¢ on the first day, and every day thereafter the salary is doubled. At first glance, the million-dollar plan appears to be more favorable. Look, however, at Table 8-1, which shows the daily payments for 30 days under plan B.

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Exponential Functions

Table 8-1 Day

Payment

Day

Payment

Day

Payment

1

2¢

11

$20.48

21

$20,971.52

2

4¢

12

$40.96

22

$41,943.04

3

8¢

13

$81.92

23

$83,886.08

4

16¢

14

$163.84

24

$167,772.16

5

32¢

15

$327.68

25

$335,544.32

6

16

$655.36

26

$671,088.64

7

$1.28

64¢

17

$1310.72

27

$1,342,177.28

8

$2.56

18

$2621.44

28

$2,684,354.56

9

$5.12

19

$5242.88

29

$5,368,709.12

10

$10.24

20

$10,485.76

30

$10,737,418.24

Notice that the salary on the 30th day for plan B is over $10 million. Taking the sum of the payments, we see the total salary for the 30-day period is $21,474,836.46. The daily salary for plan B can be represented by the function y ⫽ 2x, where x is the number of days on the job and y is the salary (in cents) for that day. An interesting feature of this function is that for every positive 1-unit change in x, the y-value doubles. The function y ⫽ 2x is called an exponential function.

DEFINITION Exponential Function Let b be any real number such that b 7 0 and b Z 1. Then for any real number x, a function of the form y ⫽ bx is called an exponential function.

An exponential function is recognized as a function with a constant base and exponent, x. Notice that the base of an exponential function must be a positive real number not equal to 1.

2. Approximating Exponential Expressions with a Calculator Up to this point, we have evaluated exponential expressions with integer exponents and rational exponents, for example, 43 ⫽ 64 and 41Ⲑ 2 ⫽ 14 ⫽ 2. However, how do we evaluate an exponential expression with an irrational exponent such as 4p? In such a case, the exponent is a nonterminating and nonrepeating decimal. The value of 4p can be thought of as the limiting value of a sequence of approximations using rational exponents: 43.14 ⬇ 77.7084726 43.141 ⬇ 77.81627412 43.1415 ⬇ 77.87023095 o 4 ⬇ 77.88023365 p

An exponential expression can be evaluated at all rational numbers and at all irrational numbers. Therefore, the domain of an exponential function is all real numbers.

673

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Chapter 8 Exponential and Logarithmic Functions and Applications

Example 1

Approximating Exponential Expressions with a Calculator

Approximate the expressions. Round the answers to four decimal places. a. 813

b. 5117

c. 11012

Solution:

Calculator Connection On a calculator, use the , , or key to approximate an expression with an irrational exponent. a. 813 ⬇ 36.6604

b. 5117 ⬇ 0.0013

c. 11012 ⬇ 5.0946 Skill Practice Approximate the value of the expressions. Round the answers to 4 decimal places. 2. 1515

1. 9p

3. 1713

3. Graphs of Exponential Functions

The functions defined by f 1x2  2x, g1x2  112 2x, h1x2  3x, and k1x2  5x are all examples of exponential functions. Example 2 illustrates the two general shapes of exponential functions. Example 2

Graphing an Exponential Function

Graph the functions f and g. a. f 1x2  2x

b. g1x2  1 12 2 x

Solution: Table 8-2 shows several function values for f(x) and g(x) for both positive and negative values of x. The graphs are shown in Figure 8-8. g(x)  ( 12 )

Table 8-2 x 4

16

1 0

1

1

1

2

2

4

3

8

4

16

1 2 1 4 1 8 1 16

2

1. 995.0416 3. 5.3936

2. 426.4028

g(x) ⴝ (12)x

1 16 1 8 1 4 1 2

3

Answers

f(x) ⴝ 2x

8 4 2

x

y

16 14

f(x)  2 x

12 10 8 6 4 2 5 4 3 2 1 2 4

1

2 3 4

Figure 8-8

5

x

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Exponential Functions

Skill Practice Graph the functions f and g on the same grid. 4. f 1x2  3x

1 x 5. g 1x 2  a b 3

The graphs in Figure 8-8 illustrate several important features of exponential functions.

PROPERTY Graphs of f ( x ) ⴝ b x

The graph of an exponential function defined by f 1x2  bx 1b 7 0 and b Z 12 has the following properties. y

0 y  5

1 2

3 4

5

x

2x  8  3y

5 4 3 2 1 1 2

3 4

5

x

5 4 3 2 1 1 2 3

4 5

4 5

y

5 4 3 2 1 1 2 3 4 5

5 4 3 2 1 1 2 x  12 y 3 4 5

x

y

35. 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

x2

1

2

3 4

5

x

5 4 3 2 1 1 2 3 4 5 y 36.

x 25 y

1 2 3 4 5

x

5 4 3 2 1 1 2 3 4 5

1

2

x

3 4 5 6 7

39. a. x  0, y  0 b. x  y  100 c. x  4y y d. 100 90 80 70 60 50 40 30 20 10 x

{(2, 1, 5)} 41. Inconsistent system Dependent system 43. {(1, 2, 2)} The sides are 5 ft, 12 ft, and 13 ft. The pumps can drain 250, 300, and 400 gal/hr. The angles measure 113°, 38°, and 29°. 33 48. 3  2 49. 1  4 50. 3  1 1 1 1 4 1 1 3 51. c 52. £ 2 1 ` d 3 † 8§ 1 1 1 2 2 1 9 53. x  9, y  3 54. x  5, y  2, z  8 55. a. 4 1 2 0 3 1 3 1 b. c 56. a. £ 0 9 1 † 12 § ` d 0  13 2 3 2 2 5 1 2 0 3 b. £ 0 9 1 57. {(1, 2)} 58. {(3, 6)} 12 § 0 8 2 4 59. {(1, 3, 2)} 60. {(6, 1, 1)}

1. Yes 2. b 5. {(1, 4)}

3. c

4. a 6. {(2, 1)} y

y 5 4 3

3 2 1 x

3 2 1 1 2 3 4

Chapter 3 Test, pp. 309–310

5 4

1 2 3 4 5

x

1 2 3 4 5

5 4 3 2 1 1 2 3 4 5

6 5 4 3 2 1

40. 42. 44. 45. 46. 47.

y

34.

5 4 3 2 1

y

38.

5 4 3 2 1

10 20 30 40 50 60 70 80 90 100

2 3

33.

x

y

32.

5 4 3 2 1

5 4 3 2 1 1

5

6 7

y

x  3

3 4

4 5

2 3

31.

1 2

5 4 3 2 1 1 2 3

y

37.

1 2 3 4 5

x

5 4 3

(1, 4)

2 1 5 4 3 2 1 1 4x  2y  4 2

(2, 1) 1

2

3 4

5

5 4 3 2 1 1 2 3

3 4 5

x

3x  y  7

f(x)  x  3

2 1

4 5

1

2

3 4 3

5

x

g(x)   2 x  2

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Student Answer Appendix

7. 5 14, 52 6

1 1 8. e a , b f 9. 5 10, 326 2 4 10. Infinitely many solutions; 5 1x, y2 0 3x  5y  76; dependent system 11. 5 12, 62 6 12. No solution; { }; inconsistent system 13. {(5, 0)} y y 14. 15. 5 4 3 2 1

5 4 3 2 1

5 4 3 2 1 1 2 3 4 5 y

x

1 2 3 4 5

2x  5y  10

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

x

y

12.

5 4 3 2 1

13. Slope 14.

5 4 3 2 1 1

1 2

3 4

5

17. a. x  0, y  0 b. 300x  400y  1200 y c. 5 4 3 2 1

(3, 0); y-intercept 10, 152 2 y 15. 5 4 3

x  2

1

2

3 4

x

5

2 1

5 4 3 2 1 1 2

y  13 x  4

4 5

4 5

x

5

2 1

3

2 3

5 4 3 2 1 1 2 3 4 5

4 5 5 2 ; x-intercept y

5 4 3 2 1 1 2

x

3 4

5 4 3

16.

5 4 3 2 1

1 2

5 4 3 2 1 1 2 3

2

3 4

5

x

3 4 5

16. 0 17. y  4x  4 18. No solution; { }; inconsistent system 19. {(1, 2)} 20. There are 4 nickels, 6 dimes, and 9 quarters. 21. a. f 1x2  2.50x  25 b. g1x2  3x  10 c. 30 DVDs 22. {(2, 0, 1)} 2 3 1 7 23. 2  3 24. For example: c d 1 4 2 6 25. {(1, 1)}

Chapter 4 1 2 3 4 5

Chapter Opener Puzzle

x

1

18. {(16, 37, 9)} 19. {(2, 2, 5)} 20. Mix 80 L of 20% solution with 120 L of 60% solution. 21. The angles are 80° and 10°. 22. Joanne can process 142 orders, Kent can process 162 orders, and Geoff can process 200 orders. 2 1 23. For example: £ 0 4 § 2.6 7 1 2 1 3 1 2 1 3 24. a. £ 0 8 3 † 10 § b. £ 0 8 3 † 10 § 5 6 3 0 0 4 8 15 25. 5 16, 12 6 26. 512, 4, 326

5 6

s

i

2

x

e

t

e

d

i

s

t

r

8

g

i

u

b

t

1

5

19 f 12 5. 54, 16 6. 12,  2 7. 1, 2 4 ´ 14, 2 8. 5 6 9. 33, 134 10. 16, 142 11. 1, 42 ´ 10, 2 2. 56w  97

3. 5 6

4. e 

4

t

a

r

t

u

i

v

e

v

+ 8

n e

8 7

3

+

9

o

n

e f o

Chapters 1–3 Cumulative Review Exercises, pp. 311– 312 1. 14

1

u 10

7

2

r

Section 4.1 Practice Exercises, pp. 320–323 3. ab3  a ⴢ b ⴢ b ⴢ b 1ab2 3  1ab2 ⴢ 1ab2 ⴢ 1ab2  a ⴢ a ⴢ a ⴢ b ⴢ b ⴢ b  a3b3 x5 5. For example: 1xy2 3  x3y3 7. For example: 2  x3 x 1 9. For example: x0  1 1x  02 11. 3 13. 25

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SA-24 1 25 125 23.  8 15. 

Student Answer Appendix

17.

31. 132 or 169

1 25

25. 1

19. 64 27. 10a

33. y8

21.

16 81

29. y8

35. 34x8 or 81x8

37.

1 p3

1 1 1 1 or 41. 2 43. 7 45. r 2 343 73 w a 1 65 26 47. 4 49. a3b2 51. 1 53. 55. 4 25 z q2 5 3b7 x16 57. 3 59. 61. 3 63.  3 65. 2 p 2a 81y20z8 11 4 8 2 4q 4m 5x 16a 67.  2 69. 4 71. 2 73. n p y b6 27 18 27y 4x 75.  24 77. 10 79. 27x3y9 81. a. $ 8.0  109 8x y b. 3.0  106 DVDs c. 1.4  1013 eV d. 1.602  1019 J 83. a. 200,000,000,000 b. 0.000004 m c. 108,200,000,000 m 85. 3.5  105 87. Proper 89. Proper 91. 3.38  104 93. 1.608  104 95. 3.4  1013 97. 2.5  103 99. 1.204  1024 hydrogen atoms and 101. There are 2  104 or 6.02  1023 oxygen atoms 20,000 people per square mile. 103. $5.25  1010 105. a. 540 months b. $10,800 c. $55,395.45 107. y2a2 109. x2a4 111. xa6y6 39.

Section 4.2 Practice Exercises, pp. 329–333 3. 10c2 5. 1.7  104 7. 6a3  a2  a; leading coefficient 6; degree 3 9. 3x4  6x2  x  1; leading coefficient 3; degree 4 11. t2  100; leading coefficient 1; degree 2 13. For example: 3x5 15. For example: x2  2x  1 17. For example: 6x4  x2 19. m2  10m 1 21. 3x4  2x3  8x2  2x 23. 2w3  w2  0.9w 9 25. 17x2y  4xy  14 27. 4a2  11a  7 29. 9x3  5x2  2x  8 31. 30y3 3 33. 4p  2p  12 35. 11ab2  a2b 37. 6z5  6z2 39. 2x3  4x2  5 41. 2xy3  3x2y  xy  5 1 9 3 43. t3  13t2  9t  13 45. a 2  ab  b 2  8 2 10 5 47. x2  6x  16 49. 3x5  x4  4x3  11 3 2 51. 4y  5y 53. 7r4  11r 55. 9x2  5xy  11y 57. 18ab  42b2 59. 3p  9 61. 2m2  6 63. 7x3  4x  5 65. 12a2b  ab  5ab2 3 2 67. 12x  6x  1 69. 8a2b  5ab2  4ab 3 2 71. 3x  16x  10x  1 73. 2.2p5  15.5p4  8.5p3  5p2  7.9p 75. 12x3  2x 77. Yes; degree 2 79. No; the term 3   3x1 and 1 is not a whole number. x 81. Yes; degree 0 83. No; the term 0x 0 is not of the form 85. a. 17 b. 8 c. 5 d. 4 axn. 3 1 9 7 87. a. b. c.  d. 89. P1x2  4x  6 4 4 4 4 91. a. P1x2  3.78x  1 b. $188 93. a. D(0)  1636 means that in the year 1990, the annual dormitory charge was $1636. D(18)  4048 means that in the year 2008, the annual dormitory charge was $4048. b. $5041.60 95. a. W(0)  6580; W(5)  7295; W(10)  8010

b. W(10)  8010 means that in the year 2010, 8010 thousand women (8,010,000) will be due child support. 97. a. (0, 0); at t  0 sec, the position of the rocket is at the origin. b. (25, 27.3); after 1 sec, the position of the rocket is (25, 27.3). c. (50, 22.6)

Section 4.3 Practice Exercises, pp. 340–343 3. 6x2  x  6

c. 3 2 3 7. 42x y 9. 11a b c 11. a  5 5 13. 2m5n5  6m4n4  8m3n3 15. 3x2y2  4x2y3 17. x2  xy  2y2 19. 12x2  28x  5 4 2 21. 2y  21y  36 23. 25s2  5st  6t2 3 2 25. 5n  3n  50n  30 27. 3.25a2  0.9ab  28b2 3 2 2 3 29. 6x  7x y  4xy  y 31. x3  343 4 3 2 2 3 33. 4a  17a b  8a b  5ab  b4 1 1 35. a2  ab  ac  12b2  8bc  c2 2 2 1 2 3 2 37. 3x  11x  7x  5 39. y  8y  150 10 1 41. a2  64 43. 9p2  1 45. x2  47. 9h2  k2 9 49. 9h2  6hk  k2 51. t2  14t  49 1 1 53. u2  6uv  9v2 55. h2  hk  k2 3 36 57. 4z4  w6 59. 25x4  30x2y  9y2 2 2 61. a. A  B b. x2  2xy  y2  B2; Both are examples of multiplying conjugates to get a difference of squares. 63. w2  2wv  v2  4 65. 4  x2  2xy  y2 2 2 67. 9a  24a  16  b 69. Write 1x  y2 3 as 2 1x  y2 1x  y2. Square the binomial and then use the distributive property to multiply the resulting trinomial by the remaining factor of x  y. 71. 8x3  12x2y  6xy2  y3 3 2 2 3 73. 64a  48a b  12ab  b 75. Multiply and simplify the first two binomials. Then multiply the resulting trinomial by the third binomial, using the distributive property. 77. 6a4  32a3  10a2 79. x3  5x2  9x  45 81. 16x2  2x  22 83. 3y2  10y  8 85. 1r  t2 2 87. x2  y3 89. The sum of p cubed and q squared 91. The product of x and the square of y 93. A1x2  4x2  70x  300 95. a. V1x2  4x3  32x2  64x b. 36 in.3 97. x2  4x  4 99. x2  4 101. x2  9 3 2 103. 9x  30x 105. Multiply 1x  22 2 1x  22 2 by squaring the binomials. Then multiply the resulting trinomials, using the distributive property. 107. 15x  62 109. 12y  12 5 6

5. a. 3

b. 9

7 8 10

Section 4.4 Practice Exercises, pp. 350–353 a. 4a  11b b. 5a2  49ab  10b2 a. 7x2  2 b. 6x4  5x3  x2  2x For example: 15y  12 2  25y2  10y  1 11. 12  8y  2y2 4t3  t  5 2 2 15. 2y2  3y  8 x  3xy  y 1 1 1 5 17.  p3  p2  p  19. a2  5a  1  a 2 3 6 5 4 21. 3s2 t  4s  2 23. 8p2q6  9p3q5  11p  2 t pq 25. a. Divisor 1x  22; quotient 12x2  3x  12; remainder 132 b. Multiply the quotient and the divisor, then add the remainder. The result should equal the dividend. 3. 5. 7. 9. 13.

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27. x  7 

9 x4

31. 4a  11

29. 3y2  2y  2 

33. 6y  5

9 y3

22 3x  2 41. x2  2x  5

35. 6x2  4x  5 

37. 4a2  2a  1 39. x2  2x  2 8 43. x2  1  2 45. n3  2n2  4n  8 x 2 47. The divisor must be of the form x  r. 49. No. The divisor is not of the form x  r. 51. a. x  5 b. x2  3x  11 c. 58 11 53. x  6 55. t  4 57. 5y  10  y1 3 59. 3y2  2y  2  61. x2  x  2 y3 63. a4  2a3  4a2  8a  16 65. x2  6x  36 54 67. 4w3  2w2  6 69. x2  4x  13  x4 4y  5 3 71. 2x  1  73. 4y  3  2 x 3y  2y  5 5 2 3 75. 2x  3x  1 77. 2k  4k2  1  4 k 9 2 84 79. x   81. a. 84 b. 4x2  6x  16  x 5 x4 c. The values are the same. 83. P(r) equals the remainder of P1x2 1x  r2. 85. a. 8x  5 b. Yes

Chapter 4 Problem Recognition Exercises, pp. 353–354 1. a. 9x2  6x  1 b. 9x2  1 c. 2 2. a. 10 b. 81m2  25 c. 81m2  90m  25 5 5 3. a. 2x  4  b. 2x  5  x 2x  1 2 4 46 c. 4x  12  4. a. y  5  b. y  7  x1 3y 3y  6 202 c. 3y  33  5. a. 30 b. 10p  50 c. 0 y6 6. a. 8x  32 b. 20 c. 0 7. 2t2  t  1 8. 15x4  5x3  10x2 9. 36z2  25 10. 9y3  2y2  3y  1 11. 6b2  11b  4 3 2 12. 10a  19a  11a  2 13. t3  6t2  8t  3 2 2 14. 2b  4b  5 15. k  4k  25 16. 3x3  x2  5 17. 7t3  12t2  2t x 11 3 1 2 1 18. xy2  2y  19. p  p  p5 7y 12 2 5 6 5 4 20. 30.6w  15.6w  7.2w 21. 36a4  48a2b  16b2 1 1 22. z4  23. m2  8m  7 24. x2  3x  14 4 9 25. 2m4  8m3  13m2  46m  21 26. x2  4x  16 2 2 27. 25  10a  10b  a  2ab  b 28. a2  x2  2xy  y2 29. 4xy 1 1 1 30. a3  12a2  48a  64 31.  x2  x  8 3 6 1 6 4 5 3 32.  x y z w 2

Section 4.5 Practice Exercises, pp. 360–362 3. 9t4  5t3  6t2  6t 5. 5y4  5y3  7y2  3y  6 7. 3v2  6v  1 9. 31x  42 11. 2z13z  22

SA-25

13. 4p1 p5  12 15. 12x2 1x2  32 17. 9t1st  32 2 3 2 19. 9a b 1a  3ab  2b2 2 21. 5xy12x  3y  12 23. b113b  11a2  12a2 25. 11x2  10x  72 27. 3xy14x2  2x  12 29. t12t2  11t  32 31. 13z  2b2 12a  52 33. 12x  32 12x2  12 2 35. 12x  12 1y  32 37. 31x  22 2 1y  22 3 39. For example: 3x  6x2  12x4 41. For example: 6(c  d)  y(c  d) 43. a. (2x  y)(a  3b) b. (2w  1)(5w  3b) c. In part (b), 3b was factored out so that the signs in the last two terms were changed. The resulting binomial factor matches the binomial factor from the first two terms. 45. 1y  42 1y2  32 47. 1p  72 16  q2 49. 1m  n2 12x  3y2 51. 12x  3y2 15a  4b2 53. 1x2  321x  12 55. 6p 1p  32 1q  52 57. 100 1x  32 1x2  22 59. 13a  b2 12x  y2 61. 14  b21a  32 63. Cannot be factored 65. It is not possible to get a common binomial factor U regardless of the order of terms. 67. A  v  cw bx bx 69. y  71. Length  2w  1 or y  ca ac 4 73. 1a  32 16a  192 75. 18 13x  52 2 14x  52 77. 1t  42 1t  32 79. 5w2 12w  12 2 17w  32

Section 4.6 Practice Exercises, pp. 373–375

3. 6c2d4e7 16d3e4  2cde8  12 5. 13a  b212x  12 7. 1z  221wz  33a2 9. 1b  821b  42 11. 1y  1221y  22 13. 1x  1021x  32 15. 1c  821c  22 17. 12x  321x  52 19. 16a  521a  12 21. 1s  3t21s  2t2 23. 31x  1821x  22 25. 21c  421c  32 27. 21x  y21x  5y2 29. Prime 31. 13x  5y21x  3y2 33. 5uv1u  3v2 2 35. x1x  721x  22 37. 12z  5215z  12 39. Prime 41. 21t  421t  102 43. 17a  4212a  32 45. 2b13a  221a  32 47. a. x2  10x  25 b. 1x  52 2 49. a. 9x2  12xy  4y2 b. 13x  2y2 2 51. 30x 53. 16z2t 55. 1y  42 2 57. 18m  52 2 59. Not a perfect square trinomial 61. 13a  5b2 2 2 2 63. 414t  20tv  5v 2 ; Not a perfect square trinomial 65. 51b2  22 2 67. a. 1u  52 2 b. 1x2  52 2 c. 1a  42 2 69. a. 1u  1321u  22 b. 1w3  1321w3  22 c. 1y  921y  62 71. 13x  4213x  12 73. 12x  921x  12 75. 13y  1121y  62 77. 13y3  221y3  32 79. 14p2  121p2  12 81. 1x2  1221x2  32 83. The factorization 12y  1212y  42 is not factored completely because the factor 2y  4 has a GCF of 2. 85. 1w2  62 2 87. 19w  52 2 89. 31a  b21x  22 91. 2abc2 16a  2b  3c2 93. 2x15x  6212x  52 95. Prime 97. 12w2  1521w2  22 99. 11  d211  3d2 or 13d  121d  12 101. 1a  2b21x  5a2 103. 81z  4w21z  7w2 105. 1y  x21a  5c2 107. g 1x2  13x  221x  42 109. n1t2  1t  102 2 111. Q1x2  x2 1x  421x  22 2 113. k 1a2  1a  421a  22

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Section 4.7 Practice Exercises, pp. 383–385

3. 12x  52 5. 15  3y2 12x  12 7. 418p  121p  12 9. Look for a binomial of the form a2  b2. This factors as 1a  b2 1a  b2. 11. 1x  32 1x  32 13. 14  7w214  7w2 15. 212a  9b212a  9b2 17. Prime 19. 21a2  42 1a  22 1a  22 21. 17  k3 217  k3 2 23. 1x  42 1x  42 1x  12 25. 12x  12 12x  121x  32 27. 19y  72 1y  22 1y  22 29. 17x  2  y217x  2  y2 31. 1w  3n  121w  3n  12 33. 1p2  5  t 2 21p2  5  t 2 2 35. 13u2  2v2  52 13u2  2v2  52 37. Look for a binomial of the form a3  b3. This factors as 39. 12x  1214x2  2x  12 1a  b21a2  ab  b2 2. 2 41. 15c  32 125c  15c  92 43. 1x  102 1x2  10x  1002 45. 14t 2  12116t 4  4t 2  12 47. 2110y2  x2 1100y4  10y2x  x2 2 49. 2z12z  32 14z2  6z  92 1 1 51. 1p4  521p8  5p4  252 53. a6y  b a6y  b 5 5 55. 2 13d6  42 13d6  42 57. 21121v2  162 59. 41x  22 1x  22 61. 15  7q215  7q2 63. 1t  2s  62 1t  2s  62 65. 13  t2 19  3t  t2 2 1 1 3 67. a3a  b a9a2  a  b 2 2 4 69. 21m  22 1m2  2m  42 71. 1x  y21x  y21x2  y2 2 73. 1a  b21a2  ab  b2 21a6  a3b3  b6 2 1 1 1 1 1 75. a p  b a p2  p  b 77. Prime 2 5 4 10 25 1 1 1 1 79. a x  yb a x  yb 5 2 5 2 81. 1a  b21a2  ab  b2 21a  b21a2  ab  b2 2 83. 12  y2 14  2y  y2 2 12  y2 14  2y  y2 2 85. 1h2  k2 21h4  h2k2  k4 2 87. 12x2  52 14x4  10x2  252 89. 4x2  9 3 6 3 91. 8a  27 93. 64x  y 95. a. x2  y2 b. 1x  y2 1x  y2 c. 20 in.2 97. 1x  y2 1x  y  12 99. 1x  y2 1x 2  xy  y2  12 2

Chapter 4 Problem Recognition Exercises, pp. 385–387 1. An expression whose only factors are 1 and itself 2. Factor out the GCF. 3. Difference of squares a2  b2, difference of cubes 3 a  b3, or sum of cubes a3  b3 4. Look for a perfect square trinomial, a2  2ab  b2 or a2  2ab  b2. 5. Try factoring by grouping 2 terms by 2 terms or by grouping 3 terms by 1 term. 6. Let u  14x2  12. The polynomial becomes 3u2  20u  12. Factor this simpler expression and then back substitute. 7. a. Trinomial b. 312x  32 1x  52 8. a. Trinomial b. m14m  12 12m  32 9. a. Difference of squares b. 212a  52 12a  52 10. a. Grouping b. 1b  y2 1a  b2 11. a. Trinomial b. 12u  v2 17u  2v2 12. a. Perfect square trinomial b. 13p  2q2 2 13. a. Difference of cubes b. 212x  12 14x2  2x  12 14. a. Sum of squares b. Prime

a. Sum of cubes b. 13y  5219y2  15y  252 a. None of these b. Prime a. Sum of cubes b. 214p2  3q2116p4  12p2q  9q2 2 a. Perfect square trinomial b. 51b  32 2 a. Difference of squares b. 12a  1212a  1214a2  12 a. Perfect square trinomial b. 19u  5v2 2 a. Grouping b. 1 p  6  c21 p  6  c2 a. Sum of squares b. 41x2  42 a. Grouping b. 212x  y213a  b2 a. Difference of cubes b. 15y  22125y2  10y  42 a. Trinomial b. 15y  121y  32 a. Difference of squares b. 21m2  821m2  82 a. Difference of squares b. 1t  1021t  102 a. Difference of squares b. 12m  7n212m  7n2 a. Sum of cubes b. 1 y  321 y2  3y  92 a. Sum of cubes b. 1x  121x2  x  12 a. Trinomial b. 1d  421d  72 a. Trinomial b. 1c  821c  32 a. Perfect square trinomial b. 1x  62 2 a. Perfect square trinomial b. 1p  82 2 a. Grouping b. 1ax  b212x  52 a. Grouping b. 14x  a212x  b2 a. Trinomial b. 12y  1215y  42 a. Trinomial b. 13z  2214z  12 a. Difference of squares b. 101p  821p  82 a. Difference of squares b. 215a  6215a  62 a. Difference of cubes b. z1z  421z2  4z  162 a. Difference of cubes b. t1t  221t2  2t  42 a. Trinomial b. b1b  521b  92 a. Trinomial b. y1y  421y  102 a. Perfect square trinomial b. 13w  4x2 2 a. Perfect square trinomial b. 12k  5p2 2 a. Grouping b. 1012x  a213x  12 a. Grouping b. 1015x  c21x  42 a. Difference of squares b. 1w2  421w  221w  22 a. Difference of squares b. 1k2  921k  321k  32 a. Difference of cubes b. 1t2  221t4  2t2  42 a. Sum of cubes b. 1 p2  321 p4  3p2  92 a. Trinomial b. 14p  1212p  52 a. Trinomial b. 13m  4213m  52 a. Perfect square trinomial b. 16y  12 2 a. Perfect square trinomial b. 13a  72 2 a. Sum of squares b. 21x2  252 a. Sum of squares b. 41y2  162 a. Trinomial b. s2 13r  2214r  52 a. Trinomial b. w2 17z  421z  22 a. Trinomial b. 1x  3y21x  11y2 a. Trinomial b. 1s  3t21s  12t2 a. Sum of cubes b. 1m2  n21m4  m2n  n2 2 a. Difference of cubes b. 1a  b2 21a2  ab2  b4 2 a. None of these b. x1x  42 a. None of these b. y1y  92 67. 1x  y21x  y2 2 2 4 69. 1a  32 16a  192 1u  v2 1u  v2 71. 1813x  52 2 14x  52 14  b2 3 12  b2 1 1 2 72. 512y  32 2 16y  112 73. a x  b 10 7 1 1 2 2 2 74. a a  b 75. 5x 15x  62 5 6 76. x3 1x3  22 77. 14p2  q2 212p  q212p  q2

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 68. 70.

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78. 1s2t 2  92 1st  32 1st  32 1 1 1 79. ay  b ay2  y  b 4 4 16 1 1 1 80. az  b az2  z  b 5 5 25 81. 1a  b2 1a  b216a  b2 82. 12p  q212p  q21p  3q2 1 1 2 1 83. a t  b 84. a y  3 4 5 85. 1x  6  a2 1x  6  a2 87. 89. 91. 93. 95. 96. 97. 98. 99. 100. 101.

91. f 1x2  1x  521x  22 ; x  5 and x  2 represent x-intercepts. 93. f 1x2  1x  12 2; x  1 represents the x-intercept. 95. The radius is 6 ft. 97. The length is 8 ft and the width is 6 ft. 99. 1x  221x  22  0 or x2  4  0 101. 1x  021x  32  0 or x2  3x  0 1 2 b 2 86. 1a  5  b21a  5  b2

1p  q  92 1p  q  92 88. 1m  n  321m  n  32 1b  x  22 1b  x  22 90. 1p  y  321p  y  32 12  u  v2 12  u  v2 92. 15  a  b215  a  b2 13a  b2 12 x  y2 94. 1q  3215p  42 1u  22 1u  221u2  2u  421u2  2u  42 11  v2 11  v2 11  v  v2 2 11  v  v2 2 1x4  12 1x2  12 1x  12 1x  12 1y4  162 1y2  42 1y  22 1y  22 1a  b2 1a  b  12 15c  3d2 15c  3d  12 1x  y2 1x2  xy  y2 2 15w  2z2

Section 4.8 Graphing Calculator Exercises, p. 401 102. 12, 0211, 02

103. 15, 0214, 02

Section 4.8 Practice Exercises, pp. 397–401

3. x110x  32 5. 1 p  52 12p  12 7. 1t  121t2  t  12 9. The equation must be set equal to 0, and the polynomial must be factored. 11. Correct form 13. Incorrect form. Polynomial is not factored. 15. Incorrect form. Equation is not set equal to 0. 3 9 1 17. 53, 56 19. e  , f 21. e 0, 4, f 2 5 10 23. 50.4, 2.16 3 29. e 0, f 2 1 37. e 1, , 3 f 2 43. 512, 56

25. 59, 36 31. e

23 f 3

45. e 

1 f 11

1 27. e 3, f 2

33. 536

39. 55, 46

1 35. e  , 2 f 3

5 41. e , 1 f 2 47. 50, 6, –26

5 51. e 3, 3,  f 53. 5, 5 2 55. 4, 3 57. 7, 6 or 6, 7 59. 9, 7 or 7, 9 61. The length is 7 ft, and the width is 5 ft. 63. The length is 20 yd, and the width is 15 yd. 65. a. The base is 5 in., and the height is 6 in. b. The area is 15 in.2. 67. The base is 10 ft, and the height is 5 ft. 69. The integers are 4 and 5. 71. a. 14 mi b. The alternative route using superhighways 73. The lengths are 6 m, 8 m, and 10 m. 75. The radius is 2 units. 77. a. 0, 3 b. f 102  0 79. a. 7, 1 b. f 102  7 81. x-intercepts: (2, 0), (1, 0), (0, 0); y-intercept: (0, 0) 83. x-intercept: (1, 0); y-intercept: (0, 1) 85. 13, 02 13, 02; d 87. 11, 02; a 89. a. The function is in the form s1t2  at 2  bt  c. b. (0, 0) and (100, 0) c. At 0 sec and 100 sec, the rocket is at ground level 1height  02 . d. At 1 sec and 99 sec 49. 50, 4, – 46

105. 12, 02

104. (3, 0)

Chapter 4 Review Exercises, pp. 408–411 1. 35x5 or 243x5

2. 

18 x12

3. 3xy2

4.

3x3 2y2

54y24 x10y5 a4 b15 6. 7. 4 20 8. 6 6 16b 55 8a 4x 9 6 9. a. 3.6866  10 b. 1.0  10 10. a. 1.0  103 b. 5.1557  109 11. a. 0.001 b. 0.000000001 12. a. 5,230,000,000 ft2 b. $1,091,000,000,000 13. 6.25  109 14. 2.0  108 7 8 15. 3.24  10 16. 3.64  10 17. Trinomial; degree 4 18. Monomial; degree 0 19. a. 7 b. 23 c. 5 20. a. 12 b. 10 c. 2 21. a. A152 ⬇ 25 means that in the year 2005, Americans on average consumed approximately 25 gal of bottled water each. b. A1152 ⬇ 49 means that in the year 2015, Americans will consume approximately 49 gal of bottled water each if this trend continues. 22. 2x2  3x  xy  1 23. 20xy  18xz  3yz 24. 3a3  5a2  6a 3 25. 2a2  6a 26. x4  x2  1 27. x4  x2 4 28. 6x  6y 29. 5x  9y 30. 11x  1 31. 4x2  11x 32. 3x  11 33. 3x 5. 

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34. 2x3  14x2  8x

35. 18x3  15x2  12x 1 36. x2  x  42 37. x2  11x  18 38. x2  2x  5 4 1 1 39. 2y2  y  40. 27x3  125 41. x3  y3 5 25 1 42. 4x2  20x  25 43. x2  4x  16 4 4 44. 9y2  121 45. 36w2  1 46. t2  16 9 1 47. z2  48. x2  4x  4  b2 16 49. c2  w2  6w  9 50. 8x3  12x2  6x  1 51. y6  9y4  27y2  27 52. a. 4x2  12x  9 2 2 b. 16x  24x  9 c. 12x  12x 53. a. P1x2  8x  4 b. A1x2  3x2  2x 54. a. False; cannot add unlike terms True b. True False; when subtracting like terms, keep the variable the same. 55. 2x2  4xy  3y2 56. 2x2  3x  4 3 2 57. a. 3y  2y  6y  4 b. Quotient 3y3  2y2  6y  4; no remainder c. Multiply the quotient and the divisor. 32 58. x  2 59. x  4  x4 72x  4 60. 2x3  2x2  8x  24  2 x  3x 4x  4 3 2 61. 2x  2x  5x  4  2 x x 62. Synthetic division can be used with a divisor in the form 63. a. x  3 b. 2x3  11x2  31x  99 c. 298 x  r. 4 64. t2  t  6 65. x  2  x5 4 19 66. x  4  67. w2  3w  9  x4 w3 68. p3  2p2  4p  8 69. x1x2  4x  112 or 2 70. 713w3  w  22 x1x  4x  112 71. 1x  72 15x  22 72. 1t  42 13t  52 73. 1m  82 1m2  12 74. 1212x  32 1x2  32 75. x12a  b212x  32 76. 1y  621 y2  12 77. The trinomial must be of the form a2  2ab  b2 or 78. 13x  2y2 16x  5y2 a2  2ab  b2. 79. 13m  5t21m  2t2 80. 5a2 13  4a2 14  a2 or 5a2 14a  32 1a  42 81. k2 13k  2212k  12 82. 17x  62 2 83. 215z  42 2 84. 19w  7219w  12 85. 14x  32 2 86. 313a2  12 12a2  52 87. 1w2  12 13w2  52 1 1 1 88. 15  y2 15  y2 89. ax  b ax2  x  b 3 3 9 90. Prime 91. h1h2  92 92. 1a  421a2  4a  162 2 93. 1k  421k  221k  22 94. y13y  2213y  22 95. 1x  4y  32 1x  4y  32 96. 1a  6  b21a  6  b2 97. 1t  8  5c2 1t  8  5c2 98. 1y  3  4x21y  3  4x2 99. It can be written in the form ax2  bx  c  0 1a  02. 100. It is a parabola. 101. Quadratic 102. Quadratic 103. Linear 104. Quadratic 105. 53, 56 3 4 9 106. e , 7 f 107. e  , 1 f 108. e 1, 5, f 8 3 2 109. 11, 02 11, 02 10, 42; b

110. 11, 0211, 0210, 22; d 1 111. 12, 02 12, 02 10, 402; c 112. 12, 02 12, 02a0, b ; a 2 113. Length 15 ft; width 8 ft; height 10 ft

114. a.

Time t (sec)

Height h(t) (ft)

0

1280

1

624

3

592

10

3840

20

5760

30

4480

42

1280

b. The position of the missile is below sea level. c. The missile will be at sea level after 2 sec and again after 40 sec.

Chapter 4 Test, pp. 412–413 8y12 9x12 4. 7 5. 5.68 14 25y x 6. 2.3  109 7. F112  1 F122  40 F(0)  8 8. 8x2  8x  8 9. 2a3  13a2  2a  45 23 10. 2x2  x  6 11. 25x2  16y4 12. The 3 2 expression 25x  49 does not include the middle term 70x. 5 1 13. 49x2  56x  16 14. x2y3  xy  3y2  2 2 17 15. 5p2  p  1 16. y3  2y2  4y  6  y2 17. See page 385. 18. See page 389. 19. 31a  6b21a  3b2 20. 1c  121c  121c2  12 21. 1y  721x  32 22. Prime 23. 101u  221u  12 24. 312t  5212t  52 25. 51y  52 2 26. 7q13q  22 27. 12x  121x  221x  22 28. 1y  521y2  5y  252 29. 1x  4  y21x  4  y2 30. r2 1r2  1621r  421r  42 2 2 31. 1x  321x  22 32. 12  c216a  b2 3 33. e , 5 f 34. 50, 76 35. 58, 26 2 1 1 1 1 36. e , 1 f 37. e 0, ,  f 38. e  f 5 4 4 4 39. 256 ft 40. a. P142  127.292. In the year 2004, the population of Japan was approximately 127.292 million. b. 126.968 million c. 124.52 million 41. (4, 0)(2, 0)(0, 8); c 42. 14, 0213, 0213, 0210, 362; b 43. 13, 0211, 0210,62; d 44. 14, 0213, 0210, 02 ; a 1. 5a13

2. x11

3.

Chapters 1–4 Cumulative Review Exercises, pp. 413–414 1.

[5, 12] 5

12

2. x2  5x  10 y 3. a. 5 4 3 2 1 54 3 2 1 1 2 3 4 5

y  x2 1 2 3 4 5

y

b.

x

5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

y  |x| 1

2

3 4

5

x

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71 5. 1.97  108 people 8 6. Penn State scored 26 points and Florida State scored 23 points. 7. 37°, 74°, 69° 8. x2  4x  16 4 9. m  ; y-intercept (0, 3) 3 4.

y

1

2

3 4

5

x

65. 73. 81.

a6 10. y  5 11. 12. 5 10, 3, 226 8b30 13. a. Function b. Not a function 16. 586

5 3 17. y  x  2 2 18. a. The fourth test would have to be 107; therefore, it is not possible. b. Student can score between 67 and 100, inclusive. 19. 15 L of 40% solution; 10 L of 15% solution 20. 8b3  6b2  4b  3 21. 3a3  5a2  2a  5 5 1 22. 3w   w 23. 57, 26 24. 50, 4, 56 2 25. 1 , 12 ´ 14, 2 15. 7

47.

61.

4 5

14. m  7

39.

53.

5 4 3 4x  3y  9 2 1 5 4 3 2 1 1 2 3

31.

Chapter 5 Chapter Opener Puzzle To add or subtract rational expressions, we need 2 4 8 3 1 6 common denominator. a 5 7

Section 5.1 Practice Exercises, pp. 422–425 3 1 3. k102   , k112  1, k122   , k142 is undefined 4 2 1 5. n112  2, n102  1, na b  0, n112  1 3 7. a. 5x 0x is a real number, x  06 b. 1, 02 ´ 10,  2 9. a. 5v 0v is a real number, v  76 b. 1, 72 ´ 17,  2 11. a. 5x 0x is a real number, x  52 6 b. 1, 52 2 ´ 1 52,  2 13. a. 5q 0q is a real number, q  9, q  36 b. 1, 92 ´ 19, 32 ´ 13,  2 15. a. 5c 0c is a real number} b. 1, 2 17. a. 5x 0x is a real number, x  5, x  56 b. 1, 52 ´ 15, 52 ´ 15, 2 19. a. 5x 0x is a real number} b. 1, 2 21. 1, 42 ´ 14,  2 ; b 23. 1, 42 ´ 14,  2 ; d 2x 25. a. b. Cannot be simplified y 1x  42 1x  22 x2 27. a. b. x  4, x  1 c. 1x  42 1x  12 x1 1x  92 2 x9 29. a. b. x  9, x  9 c. 1x  92 1x  92 x9

87.

SA-29

25x2 w8z3 1 2 33. 35.  2 3 37. 3 2 3 9y 4m n 12p  12 2 1 1 t4 41.  43.  45. x5 3c  5 t3 p1 21z2  4z  162 3z 2x  5 49. 51.  2z  5 z4 2 1 55. 1 57. 1 59. 2 c4 1 63.  ; cannot be simplified c4 121x  y2 xy t1 4 2a 67. 69.  71.  t6 p3 b 8 1a  52 2 2x 1 x  17 75. 77. 79.  2b  3 2x  3 a2 5 x5 t 2  2t  4 1 83. 85. For example: x2 3t  5 x2 1 For example: f(x)  x5

Section 5.2 Practice Exercises, pp. 428–430 t1 1 1 bc3 5.  7. 9. t1 p1 x3 2 x12x  32 213w  72 2 3 11. 3 13. 12r 15. 2 17. 5w  4 z y 1x  12 11 r3 2 19. 15 21. y 1y  22 23. 25. 4r 2 5a4b3 1 4 2s  5t 27. 29. 31. 1 p  2216p  72 1b  221b  32 s  5t 24 2a  b2 4x 9 33. 2 35.  37. 39. 2 31x  22 a  b2 4y2 y 21x  32 2a 41. 43. 45. m  n 5 x2 3y  x x  3y a3 47.  49. 1 51. or x  3y x  3y a4 4012x  12 1x  6213x  22 53. 55. 3 3x  2 2k 5x 2 2 57. 3 cm 59. ft 4 h 3.

Section 5.3 Practice Exercises, pp. 438–441 3. 11. 19. 23. 27. 33. 39. 43. 49.

31b  12

2 5a  1 1 5. 7. 9. x 21b  12 2 x1 2x  5 13. 2 15. 40x 17. 30m4n7 12x  921x  62 21. x 2 1x  121x  72 2 1x  421x  221x  62 25. a  4 or 4  a 1x  621x  22 29. 2x3  4x2 31. y2  y 15xy 8p  15 t  s 1 35. 37. st 3 6p2 2b  20 2x 2x 41. or b1b  52 x6 6x 2 3y  1 6b  5b  4 10x 45. 47. 1b  421b  12 12x  121x  22 y4 3 11x  6 51.  53. 1 w 1x  621x  621x  32

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55. 57. 61. 67. 71. 73. 79.

Student Answer Appendix

10  x x  10 or 31x  52 1x  52 315  x215  x2 x2  7x  6 m5 59. 1m  521m  32 2x 2 2 2 3 w 3 x  13x  3 63. 65.  w2 t3 x1x  12 2 2z2  15z x2  x  6 69. 21x  12 1z  32 1z  42 2y 2y or 1x  y2 1x  y2 1y  x21y  x2 2p2  2p  3 5y  18 5 75. 77. 31p  42 y1y  62 1z  121z  12 4x2  2x  50 3x2  5x  18 cm 81. m 2 1x  321x  52 3x

Section 5.4 Practice Exercises, pp. 447–448 10  3x 2a  8 5x2 5b2 3. 5. 7. 9. 2 3a 1a  52 1a  12 27 2x 10 1 11. 13. 15. 4 17. 28y 19. 8 x 3 3p t 2a  3 21. 23. 25.  p1 4  9a t1 41w  12 1 x2 27.  29. 31. w2 y4 x1 y1 t2 a  2 33. 35. 37.  a  3 y5 1t  12 2 2b2  3a 2 1 39.  41. 43. x1x  h2 b1b  a2 x1x2  32 1 6 45. 47. 414  h2 x1x  h2 y2  y1 4 63 49. m  51. 53. x2  x1 10 3 1 1 55. 1x 1  y 1 2 1  1  11/x2  11/y2 x  y 1 xy 57. x1x  12 simplifies to yx

Chapter 5 Problem Recognition Exercises, p. 449 1. 4. 7. 10. 12. 15. 19. 22.

4y2  8y  9 x2  x  13 4x  1 2. 3. 2y12y  32 x4 4x  3 2y  5 a5 4 5. 6. 3a 1y  12 1y  12 w4 1t  321t  22 a4 a 8. 9. 2 t 2a  1 x2  4xy  y2 2xy2 x2  4xy  y2 11. or 5 1x  y21x  y2 1y  x2 1y  x2 x2  4x  14 x2  14x  99 13. 14. 31x  42 61x  22 101x  72 5t2  6t  17 1 1 16. 17. y  1 18.  3 9 1t  221t  32 3 31x  42 9x2  6x  15 3z 20. 21. 2x  3 413x  12 2x2y2 2 y  11y  1 2x  3 23. 24. a  10 1  5x 1y  32 1y  22

Section 5.5 Practice Exercises, pp. 455–457 3.

y5

1y  121y  12 2

35. 39. 41. 47. 49. 51. 53. 55. 59. 61. 69. 73.

1 2t  1

7. xy

9. 5146

15 15. 566 17. 556 f 22 566 21. 5256 23. 53, 16 25. 54, 46 29. 526 (The value 2 does not check.) 58, 26 5606 33. 5 6 (The value 5 does not check.) 31 37. 556 (The value 2 does not check.) e f 5 5 6 (The value 4 does not check.) IR FK 11 43. m  45. E  e f a 4 K E  Ir E R or R   r I I 2A 2A  hb or B  b B h h b b or t  t xa ax y y or x  x 1  yz yz  1 R1 R2 2A 57. R  h ab R2  R1 s2  s1  vt1 s2  s1 or t2   t1 t2  v v 8 11 11 63. 56, 26 65. e f 67. e f e f 5 2 2 71. 5d|d is a real number6 55, 56 2 5 75. 3 77. e f 2 5

11. 5246 19. 27. 31.

5.

13. e 

Chapter 5 Problem Recognition Exercises, pp. 457–458 2w  30 b. 5156 c. The problem in 1w  521w  52 part (a) is an expression, and the problem in part (b) is an 3x  8 8 equation. 2. a. b. e  f c. The problem 61x  22 3 in part (a) is an expression, and the problem in part (b) is an 1 1 equation. 3. 4. 5. 516 a1 c2 2 23 x  12 6. 556 7. 8. 9. 516 x1x  12 51t  42 8p  11 3 12x  121x  12 10. e , 1 f 11. 12. 2 412p  32 21x  22 x3 15a  2 4 13. 14. 15. e  f 16. 5106 15 6x2 12a2 2y  13 c7 17. 18. 2 1c  32 1c  12 1y  52 2 1y  22 19. 5 6 (The value 4 does not check.) 20. 5 6 (The value 2 does not check.) 1. a.

Section 5.6 Practice Exercises, pp. 465–468 3. 56, 16 9.

121p  12 p8

5.

9 215t  12 11. 586

6 7. e  , 3 f 7 13. 566

15. 5156

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1 20 3 3 17. e f 19. e f 21. 55, ⫺16 23. e , ⫺ f 3 9 7 7 25. 5 8, ⫺5 6 27. There must be 6 adults. 29. A 14-oz box contains 84 g of fat. 31. 1000 swordfish were caught. 33. Pam needs 11.5 gal of gas. 35. There are approximately 4000 bison in the park. 37. There are 31 men. 39. 595 are men and 500 are women. 41. a ⫽ 8 ft, b ⫽ 8.4 ft 43. x ⫽ 12 in., 48 83 45. a. x ⫹ 7 b. c. y ⫽ 13 in., z ⫽ 4.2 in. x x⫹7 47. The motorist drives 40 mph in the rain and 60 mph in sunny weather. 49. The Broadmoor truck travels 70.4 mph and the Wescott truck travels 76.8 mph. 51. The cyclist rode 10 mph against the wind. 53. Celeste walks 5 ft/sec on the ground and 7 ft/sec on the moving walkway. 55. Joe runs 6 mph and Beatrice runs 8 mph. 24 20 57. It will take them hr. 59. It will take them hr. 7 3 61. a. The new pump will take 20 hr. b. The technician should return at noon on Friday. 63. Gus would take 6 hr; Sid would take 12 hr. 65. y ⫽ 5 67. x ⫽ 5

20. 24. 27. 29. 31. 33. 35. 36. 38. 41.

Section 5.7 Practice Exercises, pp. 474–477 3.

26 5

13. b ⫽

5. k c

1 4

15. Q ⫽

19. L ⫽ kw 1v 25. k ⫽ 512

7. 7

9. Directly kx y

11. T ⫽ kq

17. c ⫽ kst

ky2 9 23. k ⫽ z 2 27. k ⫽ 1.75 29. x ⫽ 70 21. x ⫽

31. b ⫽ 6 15 33. Z ⫽ 56 35. Q ⫽ 9 37. L ⫽ 9 39. B ⫽ 2 41. a. 3.6 g b. 4.5 g c. 2.4 g 43. a. $0.40 b. $0.30 c. $1.00 45. 355,000 tons 47. 18.75 lm 49. 18.5 A 51. $3500 53. 42.6 ft 55. 150 ft2 57. 300 W 59. a. A ⫽ kl 2 b. The area is 4 times the original. c. The area is 9 times the original.

Chapter 5 Review Exercises, pp. 484–487 2 1. a. k 122 ⫽ , k 102 ⫽ 0, k 112 undefined, k 1⫺12 3 1 2 undefined, k a b ⫽ ⫺ b. 1⫺⬁, ⫺12 ´ 1⫺1, 12 ´ 11, ⬁ 2 2 3 1 1 3 2. a. h 112 ⫽ , h 102 ⫽ 0, h 1⫺12 ⫽ ⫺ h 1⫺32 ⫽ ⫺ , 2 2 10 1 2 xz2 ha b ⫽ b. 1⫺⬁, ⬁ 2 3. 2a 4. 5. x ⫺ 1 2 5 5 k⫹5 x2 ⫹ 3x ⫹ 9 6. 7. ⫺ 8. ⫺1a2 ⫹ 921a ⫹ 32 k⫺3 3⫹x y 1y ⫹ 22 2t ⫹ 5 9. ⫺ 10. 11. c; 1⫺⬁, 32 ´ 13, ⬁ 2 t⫹7 y⫺3 12. a; 1⫺⬁, ⫺22 ´ 1⫺2, ⬁ 2 13. b; 1⫺⬁, 02 ´ 10, 32 ´ 13, ⬁ 2 3 a 14. d; 1⫺⬁, ⬁2 15. 16. ⫺ 2 2 7 1k ⫺ 42 x⫺y y⫺x or 17. ⫺ 18. 1x ⫹ 82 2 19. 5x 5x 2 1k ⫺ 22

44. 48. 50.

SA-31

a⫹1 x⫺5 1 8 21. 22. ⫺ 23. a⫺1 x⫺4 b 9w2 51y2 ⫹ 12 5 25. 26. 13k ⫹ 521k ⫹ 52 2 71y2 ⫺ 2y ⫹ 42 2 13x ⫹ 42 x2 ⫹ x ⫺ 1 28. 3 1x ⫹ 221x ⫺ 22 x y⫺3 3⫺y a⫺4 or 30. 2y ⫺ 1 1 ⫺ 2y 2 1a ⫹ 32 4x2 ⫹ 17x ⫹ 11 4k2 ⫺ k ⫹ 3 32. 2 x⫹4 1k ⫹ 12 1k ⫺ 12 2 1a2 ⫺ 52 13x ⫹ 19 34. 1a ⫺ 521a ⫹ 32 1x ⫹ 121x ⫹ 321x ⫹ 22 6 17 ⫺ 4x2 ⫺6 14x ⫺ 72 or 3x ⫺ 5 3x ⫺ 5 2 116k ⫺ 92 9a2 ⫹ a ⫹ 4 37. 14k ⫹ 321k ⫺ 1214k ⫺ 32 13a ⫺ 121a ⫺ 22 1k ⫺ 221k ⫹ 22 ⫺5y2 ⫹ 17y ⫹ 9 1 39. 40. 1y ⫺ 321y ⫹ 22 x⫹1 15 x 12y ⫹ 12 1⫹a a⫹1 or ⫺ 42. ⫺y 43. 4y 1⫺a a⫺1 a 15b ⫹ 12 y 1 2x 45. 46. 47. m ⫽ x⫺y 3b 18 64 m⫽ 49. 536 1The value 1 does not check.2 23 5 6 (The values 3 and ⫺3 do not check.) 51. 50, 176

52.

53. 55, 16

54. 516

A ⫺b b or x ⫽ 56. P ⫽ c⫺a a⫺c rt ⫹ 1 15 216 7 3 f 57. e f 58. e 59. e ⫺ f 60. e ⫺ f 2 7 11 4 61. Manning would gain 231 yd. 62. Erik can buy $253.80 Canadian. 63. Tony averaged 20 mph the first day and 15 mph the second day. 64. His speed driving was 60 mph. 65. It would take 409 hr. 66. The larger pipe would take 9 hr and the smaller would take 18 hr. 67. a. F ⫽ kd b. k ⫽ 3 c. 12.6 lb 68. y ⫽ 16 69. y ⫽ 12 70. 48 km 55. x ⫽

Chapter 5 Test, pp. 487–488 2 1 1. a. h 102 ⫽ , h 152 ⫽ , h 172 is undefined, h 1⫺72 is 7 6 undefined b. 1⫺⬁, ⫺72 ´ 1⫺7, 72 ´ 17, ⬁2 2. 1⫺⬁, ⬁ 2 3. a. {x 0 x is a real number and x ⫽ 4, 91x ⫹ 12 2 2m2 4. 5. x ⫽ ⫺36 b. f 1x2 ⫽ x⫺4 3n 3x ⫹ 5 23 6. m ⫽ ⫺ 7. ⫺1x ⫺ 32 or ⫺x ⫹ 3 8. x ⫺ 4 9 2 2 2 uv x ⫹ 5x ⫹ 2 3 9. 10. 11. 21v2 ⫺ uv ⫹ u2 2 x⫹1 4 1 12. a 13. 14. 536 1x ⫹ 521x ⫹ 32 15. 50, 46 16. 5 6 (The value 4 does not check.) 1 ⫺1 Fr 2 17. T ⫽ or T ⫽ 18. m1 ⫽ p⫺v v⫺p Gm2

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1 or 2 20. a  14 m, b  15 m 6 21. The cities are 1960 mi apart. 22. Lance rides 16 mph against the wind and 20 mph with the wind. 23. It would take 207 hr working together. ky 24. x  2 25. 3.3 sec t

Chapters 1–5 Cumulative Review Exercises, pp. 488–489 1. Set

Number

ⴚ22 ✔

Real numbers Irrational numbers

P

6

ⴚ12

✔

✔

✔

✔

✔

Rational numbers

✔

✔

Integers

✔

✔

Whole numbers

✔

Natural numbers

✔

8 17. 0.9 19. 0.4 21. a. 8 b. 4 3 c. 8 d. 4 e. Not a real number f. 4 1 5 23. 3 25. 27. 2 29.  31. Not a real 2 4 number 33. 10 35. 0.2 37. 0.5 39. 0a 0 x 41. a 43. 0 a 0 45. 0 x  1 0 47. 0x 0 y2 49.  y 2 51. 53. 92 55. 2 57. 923 59. xy2 0x 0 5 hk2 t a3 61. 63.  65. 3xy2z 67. 69.  q b 4 3 71. 2y2 73. 2p2q3 75. 9 cm 77. 13 ft 79. They were 5 mi apart. 81. They are 25 mi apart. 83. a. Not a real number b. Not a real number c. 0 d. 1 e. 2 Domain: 冤2, 2 85. a. 2 b. 1 5 c. 0 d. 1 Domain: 1, 2 87. a, d 2 89. 1,  2 91. 35,  2 93. 1, 2 95. 1, 1 4 ; b 97. 310,  2 ; d 99. q  p2 6 101. 3 103. 8 in. 1x 15.

19.

1 2. e a2, b f 3. x2  x  13 2 2A  h b2 2A or b1   b2 b. 10 cm 4. a. b1  h h 5. The dimensions are 50 m by 30 m. 6. 5(1, 2, 4)6 1 7. y   x  4 3 y 8. 9. 75 mph

Section 6.1 Graphing Calculator Exercises, p. 502 105. 8.3066 108. 0.5566 111. 0.1235 113.

106. 76.1446 109. 15.6525 112. 1.0622

107. 3.7100 110. 6.2403 114.

5 4 3

2x  y  3

2 1

5 4 3 2 1 1 2

1

2

3 4

5

x

3

115.

4 5

3 2 10. 10, 02, a , 0b, a , 0b 4 3

14. 17. 19. m

11. 8 12y  z2 2 14y2  2yz2  z4 2

xa 4 1x  72 c7 x3 15. 16. 51, 1, 56 c 1x  102 1x  22 53, 16 18. The distance is 278 mi. a. Vertical line; slope is undefined b. Horizontal line; 0 20. $3500

12. 15x  22 12x  12

116.

13.

Chapter 6 Chapter Opener Puzzle I _M I ___ N _A ___ __ _A __ _G __ ___ __ _R __ _Y __ 4 8 7 1 4 3 7 6 9

B ___ E _R _N __ _U __ _M __ ___ __ 3 5 8 10 2 6

Section 6.1 Practice Exercises, pp. 499–502 3. a. 8, 8 b. 8 c. There are two square roots for every positive number. 164 identifies the positive square root. 5. a. 9 b. 9 7. There is no real number b such that b2  36. 9. 7 11. 7 13. Not a real number

Section 6.2 Practice Exercises, pp. 507–510 11. 12 1 1 13. Not a real number 15. 4 17. 19.  5 7 n m 21. am/n  1 a ; The numerator of the exponent represents the power of the base. The denominator of the exponent represents the index of the radical. 23. a. 8 b. 8 1 1 c. Not a real number d. e.  f. Not a real number 8 8 1 25. a. 125 b. 125 c. Not a real number d. 125 1 1 e.  f. Not a real number 27. 29. 27 125 512 3. a. 3

b. 27

5. 5

7. 3

9. 12

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31. ⫺

1 81

39. ⫺2

33.

27 1000

41. 6

3 2 51. 2 q

35. Not a real number 43. 10

4 3 53. 6 2 y

45.

3 4

3 2 55. 2 xy

47. 1 57.

37. ⫺2 49.

9 2

1 5

1 qr 59. x1/3 61. 10b1Ⲑ 2 63. y2Ⲑ3 65. 1a2b3 2 1Ⲑ4 1 4 67. 69. p 71. y2 73. 62Ⲑ5 75. 5/3 x t y9 25a4d 2z3 77. a7 79. 81. 8 83. 85. 5xy2z3 c w x x3y2 87. x13z4Ⲑ3 89. 5 91. a. 10.9% b. 8.8% z c. The account in part (a). 93. 2.7 in. 95. 1y 4 3 2 97. 2 99. 2 101. y 1x 103. 4x4y3 z x 15 3 6 105. 2xy2z2 2 107. 1 109. 1 w 111. 3 yz2 x 113. 0.3761 115. 2.9240 117. 31.6228

Section 6.3 Practice Exercises, pp. 514–516 p

7 4 3 5. 2 x 7. y9Ⲑ 2 9. x2 1x 11. q2 1 q q5 4 13. a2b2 1a 15. ⫺x2y3 1y 17. 217 19. 215 3 3 21. 15 12 23. 3 22 25. 5b 1ab 27. 2x2 2 5x 4 2 3 3 29. ⫺2x z 12y 31. 2wz 25z 33. x 35. p2 3 3 1 512 519 37. 5 39. 41. 43. 45. 415 2 3 6 3 47. ⫺30 13 49. 5x2y 1y 51. 3yz2x2z 53. 2w2 1 2 55. 57. 59. 2a7b4c15d11 12c b 10y6 61. 3a3b 12b 63. ⫺10a2b2 13ac 65. x2 17xy 3 15 67. 3a2b 16 69. 213 71. 4 1 1 73. 3 75. 2k3 simplifies to k 2k simplifies to 2 w 2w6 77. 2 141 ft 79. 615 m 81. The distance is 9012 ft or approximately 127.3 ft. 83. The path from A to B and B to C is faster.

3.

Section 6.4 Practice Exercises, pp. 520–522 b2 7. 1x3y2 1Ⲑ4 9. y11Ⲑ12 2 11. a. Not like radicals b. Like radicals c. Not like radicals 13. a. Both expressions can be simplified by using the distributive property. b. Neither expression can be simplified because the terms do not 3 contain like radicals or like terms. 15. 915 17. 21 tw 4 4 19. 5 110 21. 8 13 ⫺ 214 23. 21x ⫹ 21y 25. Cannot be simplified further 29 3 27. Cannot be simplified further 29. z16 18 31. 0.7x 1y 33. Simplify each radical: 312 ⫹ 3512. Then add like radicals: 38 12 35. 15 37. 813 3 39. 3 17 41. ⫺12 43. 1 45. ⫺512a 3 3 2 3 47. 8s2t2 2 49. 6x 1 51. 14p2 15 s x 3 2 53. ⫺ 2a b 55. 15x ⫹ 62 1x 57. 15x ⫺ 62 12 3 3 3 59. 33d 2 61. 2a2b 16a 63. 5x1 2c 2 ⫺ 61 x 4 3 3. ⫺xy2 x

5.

SA-33

65. False; 19 ⫹ 116 ⫽ 19 ⫹ 16; 7 ⫽ 5 67. True 69. False; 1y ⫹ 1y ⫽ 21y ⫽ 12y 71. 248 ⫹ 212 simplifies to 623 3 6 73. 52 75. The difference of x ⫺ x2 simplifies to 4x2 the principal square root of 18 and the square of 5 77. The sum of the principal fourth root of x and the cube of y 79. 916 cm ⬇ 22.0 cm 81. x ⫽ 212 ft 83. a. 1015 yd b. 22.36 yd c. $105.95

Section 6.5 Practice Exercises, pp. 527–529 3 1. ⫺2xy2z2 2 2x2z

q2

3 7. 22 7 p1/8 3 4 3 9. 2 11. 225 13. 42 15. 82 21 4 20 17. ⫺24ab1a 19. 6110 21. 6x12 3 23. 10a3b2 1b 25. ⫺24x2y2 12y 27. 6ab2 2a2b2 7 29. 12 ⫺ 6 13 31. 213 ⫺ 16 33. ⫺2x ⫺ 1x 3 35. ⫺8 ⫹ 7 130 37. x ⫺ 51x ⫺ 36 3 2 3 39. 2 41. 9a ⫺ 281ab ⫹ 3b y ⫺1 y⫺6 43. 81p ⫹ 3p ⫹ 51pq ⫹ 161q ⫺ 2q 45. 15 47. 3y 49. 6 51. 709 53. a. x2 ⫺ y2 b. x2 ⫺ 25 55. 29 ⫹ 8113 57. p ⫺ 217p ⫹ 7 59. 2a ⫺ 6 12ab ⫹ 9b 61. 3 ⫺ x2 63. 4 4 1 65. x ⫺ y 67. a. 3 ⫺ x b. 3 ⫹ 213x ⫹ x 9 4 c. 3 ⫺ 2 13x ⫹ x 69. True 71. False; 73. False; 5 is multiplied 1x ⫺ 252 2 ⫽ x2 ⫺ 2x25 ⫹ 5 by 3 only. 75. True 77. 6x 79. 3x ⫹ 1 81. x ⫹ 19 ⫺ 8 1x ⫹ 3 83. 2t ⫹ 1012t ⫺ 3 ⫹ 22 4 3 85. 1225 ft2 87. 18215 in.2 89. 2 x 15 6 6 8 2 91. 2 12z2 93. p 2p 95. u2u 12 6 2 4 3 4 97. 2 99. 2 101. 22453x7y7 xy 2 ⴢ 32 or 2 72 6 3 103. 2m23 2mn5 105. a ⫹ b

3. 2yz2 12z

5.

Chapter 6 Problem Recognition Exercises, p. 530 3 3 1. a. 226 b. 22 3 2. a. 326 b. 32 2 3 3 3. a. 10y3 22 b. 2y2 2 4. a. 4z7 22z b. 2z5 2 25 4 3 4 3 5. a. 425 b. 2210 c. 225 6. a. 423 b. 22 6 4 3 2 4 c. 22 7. a. x2y3 2x b. xy2 2 3 x c. xy2 xy2 3 4 2 3 8. a. a5b6 b. a3b4 2 9. a. 2st2 2 a c. a2b3 2 a 4s2 4 5 2 2 6 3 2 b. 2st 22st c. 2st 2t 10. a. 2v w 212vw 4 5 b. 2vw5 2 11. a. 225 b. 5 6v3 c. 2vw4 2 3v2 12. a. 2210 b. 10 13. a. ⫺3 26 b. 60 14. a. ⫺727 b. 210 15. a. 322 b. 4 16. a. 323 b. 6 17. a. 722 b. 240 3 3 18. a. ⫺22 b. 60 19. a. 142 3 b. 482 9 3 3 20. a. 22 b. 3024

Section 6.6 Practice Exercises, pp. 537–540 3. 12y15 9. ⫺5 17. 3b2 25.

3 2 2 x 3 2 2 x

5. ⫺18y ⫹ 31y ⫹ 3 7. 64 ⫺ 161t ⫹ t 3 ⫺2 j1 2 7x2 2a12 11. 3 13. 15. 3 k y x z13y 1 15 19. 21. 23. 2b2 w2 15 4 13z 2 2a2 13 27. 29. 4 31. 2 3 13z 22a

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33. 41. 49. 57. 63. 67. 73. 77. 83. 85. 87. 93. 99.

Student Answer Appendix

3 12y 3 12 a12a 1x 35. 37. 39. x y 2 2 3 4 3 3 3 32 71 2 4 ⫺6 2 x 2 4w 43. 45. 47. x w 2 2 4 3 1 2x 2 16 1x 2127 51. 53. 55. 4 x 3 21 x 12x 59. 12 ⫹ 16 61. 1x ⫺ 23 2x3 4 12 ⫺ 12 ⫺412 ⫹ 12 or 65. 4 16 ⫹ 8 ⫺7 7 ⫺1p ⫹ 1q ⫺121 ⫹ 2 17 69. 71. 1x ⫺ 15 p⫺q 4 1xy ⫺ 3x ⫺ y w ⫹ 111w ⫹ 18 75. y⫺x 81 ⫺ w 5 ⫹ 121 5 ⫺ 110 79. 81. ⫺6 15 ⫺ 13 4 16 3 simplifies to 81 2 3 1 4 4 4x ⫹ 412 simplifies to x2 ⫺ 2 x ⫺ 12 p 12 2 16 17 115 sec ⬇ 1.11 sec 89. 91. 4 3 15 3 ⫺33 a⫺b 8125 95. 97. 5 2 13 ⫺ 12 a ⫹ 2 1ab ⫹ b 3 5 101. 15 ⫹ 3h ⫹ 15 14 ⫹ 5h ⫹ 2

Section 6.7 Practice Exercises, pp. 547–549 3w 1w 3 5. 3c 1 2c 7. 4x ⫺ 6 9. 9p ⫹ 7 4 11. 546 13. 536 15. 5426 17. 5296 19. 51406 21. 5 6 (The value 25 does not check.) 23. 576 (The value 25. 5 6 (The value 3 does not check.) ⫺1 does not check.) 4pr 3 r 2 ⫺ p2 r 2 r2 27. V ⫽ 29. h2 ⫽ or h2 ⫽ 2 ⫺ r 2 2 3 p p 31. a2 ⫹ 10a ⫹ 25 33. 5a ⫺ 6 15a ⫹ 9 35. r ⫹ 22 ⫹ 10 1r ⫺ 3 37. 5⫺36 39. 5 6 (The value 12 does not check.) 41. 526 (The value 18 does not check.) 43. 596 9 45. 5⫺46 47. e f 49. 526 5 51. 5 6 (The value 9 does not check.) 11 53. e ⫺ f 55. 5 6 (The value 83 does not check.) 4 1 57. 5⫺16 (The value 3 does not check.) 59. e , ⫺1 f 3 61. 5 6 (The values 3 and 23 do not check.) 63. 566 65. 526 (The value 18 does not check.) 67. a. 30.25 ft b. 34.5 m 69. a. $2 million b. $1.2 million c. 50,000 passengers 71. a. 12 lb b. t 1182 ⫽ 5.1. An 18-lb turkey will take about 5.1 hr to cook. 73. b ⫽ 225 ⫺ h2 75. a ⫽ 2k2 ⫺ 196 3.

Section 6.7 Graphing Calculator Exercises, p. 549 76. The x-coordinate of the 77. The x-coordinate of the point of intersection is the point of intersection is solution to the equation. the solution to the equation.

78.

79.

Section 6.8 Practice Exercises, pp. 557–559 1 f 9. 1⫺1 ⫽ i and 25 11. 12i 13. i13 15. 2i 15 ⫺11 ⫽ ⫺1 17. ⫺60 19. 13i17 21. ⫺7 23. ⫺12 25. ⫺3110 27. i12 29. 3 31. ⫺i 33. 1 35. i 37. 1 39. ⫺i 41. ⫺1 43. a ⫺ bi 45. Real: ⫺5; imaginary: 12 47. Real: 0; imaginary: ⫺6 3 49. Real: 35; imaginary: 0 51. Real: ; imaginary: 1 5 3 3 53. 7 ⫹ 6i 55. 57. 5 ⫹ 0i ⫹ i 10 2 59. ⫺1 ⫹ 10i 61. ⫺24 ⫹ 0i 63. 18 ⫹ 6i 65. 26 ⫺ 26i 67. ⫺29 ⫹ 0i 69. ⫺9 ⫹ 40i 13 71. 35 ⫹ 20i 73. ⫺20 ⫹ 48i 75. ⫹ 0i 16 21 1 3 3 4 20 77. ⫺ i 79. 81. ⫺ i ⫹ i 5 5 25 25 29 29 17 1 1 5 1 1 83. ⫺ ⫺ i 85. ⫺ ⫺ i 87. ⫺ i 10 10 2 2 2 3 1 1 89. 1 ⫹ 10i 91. ⫹ i 93. ⫺1 ⫹ i12 4 2 1 12 95. ⫺2 ⫺ i13 97. ⫺ ⫹ 99. Yes 101. Yes i 2 2 3. 9 ⫺ x

5. 5⫺86

7. e

Chapter 6 Review Exercises, pp. 567–569 3 1. a. False; 10 ⫽ 0 is not positive. b. False; 1 ⫺8 ⫽ ⫺2 2 2. 21⫺32 ⫽ 19 ⫽ 3 3. a. False b. True 5 4. 5. 5 6. 6 7. a. 3 b. 0 c. 17 4 d. 31, ⬁ 2 8. a. 0 b. 1 c. 3 d. 3⫺5, ⬁2 3 12x 9. 4 10. a. 0 x 0 b. x c. 0 x 0 d. x ⫹ 1 ⫹4 1 2x 11. a. 2 0 y 0 b. 3y c. 0 y 0 d. y 12. 8 cm 13. Yes, provided the expressions are defined. For example: 14. n represents the root. x 5 ⴢ x 3 ⫽ x 8 and x1Ⲑ5 ⴢ x2Ⲑ5 ⫽ x3Ⲑ5

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15. Take the reciprocal of the base and change the 1 exponent to positive. 16. 5 17. 18. 4 2 16y12 a3c 19. b10 20. 21. 2 22. x 34 9 xz b 23. 12y2 2 1 3 24. 2.1544 25. 6.8173 26. 54.1819 4 27. See page 511. 28. 6 13 29. xz1 xy 2a 3 30. 10ab3 1 31.  2b b 32. a. The principal square root of the quotient of 2 and x b. The cube of the sum of x and 1 33. 31 ft 34. they are like radicals. 35. Cannot be combined; the indices are different. 36. Cannot be combined; one term has a radical, but the other does not. 37. Can be combined: 4 38. Can be added after simplifying: 1912 31 3xy 3 3 39. 5 17 40. 9 1 41. 10 12 42. 13x2 28 2x2 43. False; 5 and 3 1x are not like radicals. 44. False; 1y  1y  2 1y (add the coefficients). 4 45. 6 46. 2 1 47. 2121  6133 2 48. 6 115  15 49. 4x  9 50. y  16 51. 7y  2121xy  3x 52. 12w  2013w  25 53. 2z  9 16z  42 54. 3a  5 15a  10 y2 13y 6 5 4 55. u2 2 56. 2 57. u 4w3 5x3 2 2 3 114y 2x y 1 2x t4 2 58. 59. 60. 61. 9w 3 4 2y z 3 3 42 3p 115w 2 4x2 62. 63. 64.  x 3w 3p 65. 110  115 66. 3 15  3 17 67. 1t  13 68. 1w  17 69. The quotient of the principal square root of 2 and the square of x 49 70. e f 71. 5316 72. 5126 73. 576 2 74. 596 75. 5 6 (The value 2 does not check.) 1 76. e , 4 f 77. 526 (The value 2 does not check.) 2 78. 615 m ⬇ 13.4 m 79. a. v(20) ⬇ 25.3 ft/sec. When the water depth is 20 ft, a wave travels about 25.3 ft/sec. b. 8 ft 80. a  bi, where a and b are real numbers and 81. a  bi, where b  0 82. In each case i  1 1 we simplify the expression by multiplying the numerator and denominator by the conjugate of the denominator. 83. 4i 84. i15 85. 15 86. 2i 87. 1 88. i 89. i 90. 0 91. 5  5i 92. 9  17i 93. 25  0i 94. 24  10i 17 17 95.   i; real part:  ; imaginary part: 1 4 4 96. 2  i; real part: 2; imaginary part: 1 4 7 5 3 97. 98. 3  4i 99.   i  i 13 13 2 2 2 4 16 210 100.   101.   102. 2  4i i i 17 17 3 6

SA-35

Chapter 6 Test, pp. 570–571 1. a. 6 d. Real

b. 6

2. a. Real

3. a. y

7. a2bc2 1bc

b. Not real c. Real 4 3 4. 3 5. 6. 21 4 3

b. ƒ y ƒ

8. 3x2yz2 12xy

9. 4w2 1w

10.

x2 5y

322 12. a. f 182  213; f 162  212; 2 13. 0.3080 f 142  2; f 122  0 b. 1, 2 4 20x2y3 12 6 3 14. 3 15. 16. 27y 17. 110 z 18. 315 19. 312x  315x 20. 40  1015x  3x 3 2 2 2 x x  6  5 1x 21. 22. 23. a. 2i12 b. 8i x 9x 1 22 c.  24. 1  11i 25. 30  16i 26. 28 i 2 2 17 6 27. 33  56i 28. 104 29.  i 25 25 15 9 30.   31. r1102 ⬇ 1.34; the radius of a sphere i 17 17 of volume 10 cubic units is approximately 1.34 units. 17 32. 21 ft 33. 5166 34. e f 5 35. 526 (The value 42 does not check.) 11.

Chapters 1–6 Cumulative Review Exercises, pp. 571–572

1. 54 2. 15x  5y  5 3. 536 4. 1, 52 5. 39, 32 6. 5 6 7. y  2x  5 8. x  5 1 1 1 9. e a , b f 10. a2, 2, b is not a solution. 2 3 2 11. Together it will take them 178 hr. 1 12. a. f 122  10; f 102  2; f 142  14; f a b  0 2 1 b. 12, 102 , 10, 22 , 14, 142 , a , 0b 2 y c. 5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

f(x)  4x  2

1 2

3 4

5

x

13. a6b12c4 14. a. 1.4  104 b. 3.14  109 2 15. 2x  x  15; 2nd degree 16. 115  312  3 3 1 3c 1 2 17. x  4 18.  19. 20. 32b 15b 6 d x  15 21. 2  3i 22. 5106 23. x1x  521x  52 1 24. 25. 14x2  26x  17 12a  121a  22 6 10 7 3 26. 213  215 27. 28. e ,  f  i 17 17 6 2 29. 1x  3  y21x  3  y2 30. 1x2  221x4  2x2  42

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Chapter 7 Chapter Opener Puzzle Q ___ U ___ A ___ D ___ R ___ A ___ T ___ I ___ C ___ 2 1 6 5 8 6 4 7 3

Section 7.1 Practice Exercises, pp. 580–583

3. 526 5. 5176 7. 55i6 9. 51, 56 3  17 133 11. e f 13. 55  3i126 15. e  f 2 3 4 13 17. e  19. 1. Factoring and applying the zero if 5 5 product rule. 2. Applying the square root property. 566 21. n  9; 1x  32 2 23. n  16; 1t  42 2 1 1 2 25 5 2 25. n  ; ac  b 27. n  ; ay  b 4 2 4 2 1 1 2 1 2 1 29. n  ; ab  b 31. n  ; ap  b 25 5 9 3 33. See page 577. 35. 53, 56 37. 53  i276 16 if 39. 52  i 126 41. 55, 26 43. e 1  2 2 1 13 165 3 f f 45. e 2  i f 47. e  49. e   3 5 5 4 4 2d 51. 52  1116 53. 51, 76 55. a. t  b. 8 sec 4 A 1Ap or r  p 57. r  59. a  2d 2  b2  c2 Bp 3V 13Vph or r  63. 4.2 ft 65. 7.1 in. B ph ph 67. a. 4.5 thousand textbooks or 35.5 thousand textbooks b. Profit increases to a point as more books are produced. Beyond that point, the market is “flooded,” and profit decreases. There are two points at which the profit is $20,000. Producing 4.5 thousand books makes the same profit using fewer resources as producing 35.5 thousand books. 61. r 

47. The times are

3  15 sec ⬇ 2.62 sec or 2

3  15 sec ⬇ 0.38 sec. 49. a. x2  2x  1  0 2 b. 0 c. 1 rational solution 51. a. 19m2  8m  0  0 b. 64 c. 2 rational solutions 53. a. 5p2  0p  21  0 b. 420 c. 2 irrational solutions 55. a. n2  3n  4  0 b. 7 c. 2 imaginary solutions 57. Discriminant: 16; two x-intercepts 59. Discriminant: 0; one x-intercept 61. Discriminant: 39; no x-intercepts 5  113 5  113 , 0b, a , 0b; 63. x-intercepts: a 2 2 y-intercept: (0, 3) 65. x-intercepts: none; y-intercept: 10, 12 5  141 5  141 , 0b, a , 0b; 67. x-intercepts: a 4 4 3 17 if y-intercept: (0, 2) 69. e   2 2 1 1 5 1 71. e  , 6 f 73. e  , f 75. e   2i f 3 2 2 2 3  17 f 77. e 79. 51  3i126 81. 52  126 2 1 3 83. e  , f 85. 53, 36 87. a. 53  1146 8 4 b. 53  1146 c. Answers will vary.

Section 7.2 Graphing Calculator Exercises, p. 597 89.

90.

91.

92.

Section 7.2 Practice Exercises, pp. 594–597 3. 4  215 5. 2  i13 7. The form ax2  bx  c  0 9. 512, 16 1 2 111 114 1 11. e  13. e  15. 57, 56 if if 9 9 6 6 3  141 2 17. e 19. e f 21. 53  i156 f 2 5 1  129 1 113 23. e 25. e f f 2 4 5  113 2 2 12 27. e   29. e if f 3 3 2 7  1109 31. 52, 46 33. e f 6 3 13 3 if 35. a. 1x  321x2  3x  92 b. e 3,   2 2 2 37. a. 3x1x  2x  22 b. 50, 1  i6 39. The length of each side is 3 ft. 41. The legs are approximately 1.6 in. and 3.6 in. 43. The legs are approximately 8.2 m and 6.1 m. 45. a. Approximately 4.5 fatalities per 100 million miles driven b. Approximately 1 fatality per 100 million miles driven c. Approximately 4.2 fatalities per 100 million miles driven d. For drivers 26 yr old and 71 yr old

Section 7.3 Practice Exercises, pp. 601–603 1. An equation is in quadratic form if it can be written as au 2  bu  c  0 where u represents an algebraic expression. 3 17 f 3. e  5. 546 7. 52  1266 2 2 9. a. 54, 66 b. 54, 1, 2, 36 11. 51  i12, 1  126 13. 55, 1, 26 15. 527, 86 1 17. e , 243 f 19. 546 (The value 64 does not check.) 32 1 21. e f (The value 4 does not check.) 4 23. 576 25. 5166; yes 27. 53i, 156 5  113 13 29. e 31. e , i 15 f f 2 2

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9  173 f 4 7 3 39. e  ,  f 4 2

33. e

43. 47. 53. 57.

35. 52  136

37. 5 2,  2i6

1 1 41. e ,  , 12, 12 f 2 2 1 13 e 2, 1, 1  i 13,   45. 55, 46 if 2 2 17 50, 26 49. e 1, f 51. 564, 1256 2 55. 5  4i, 1 6 5 12, 2i6 8 59. e , 2 f 54, i 156 3

Section 7.3 Graphing Calculator Exercises, p. 603

61. a. 5 i 126 b. Two imaginary solutions; no real solutions c. No x-intercepts d.

6. a. Quadratic b. 54, 56 7. a. Quadratic form b. 53, 1, 1, 36 8. a. Quadratic form b. 52, 2, i, i6 9. a. Quadratic form (or radical) b. 5166 (The value 1 9  189 does not check.) 10. a. Quadratic b. e f 2 11. a. Linear b. 546 12. a. Linear b. 516 13. a. Quadratic b. 5  i126 14. a. Quadratic b. 5  i166 15. a. Rational b. 52, 16 16. a. Rational b. 52, 46 17. a. Quadratic b. 510  11016 18. a. Quadratic b. 59  1776 1  117 19. a. Quadratic b. e f 2 1  113 20. a. Quadratic b. e f 2 21. a. Radical b. 536 (The value 1 does not check.) 22. a. Radical b. 566 (The value 1 does not check.) 23. a. Quadratic form (or radical) b. 5125, 276 24. a. Quadratic form (or radical) b. 564, 16

Section 7.4 Practice Exercises, pp. 613–617 5  110 f 7. 527, 86 5 9. The value of k shifts the graph of y  x2 vertically. y y 11. 13. 3. 53  2i6

62. a. 51, 16 b. Two real solutions; no imaginary solutions c. Two x-intercepts d.

5. e

5 4 3 2 1

5 4 3 2 1

1 2 3 4 5 5 4 3 2 1 1 2 f(x)  x2  2 3

63. a. 50, 3, 26 b. Three real solutions; no imaginary solutions c. Three x-intercepts d.

15.

17.

y

Chapter 7 Problem Recognition Exercises, pp. 603–604 1. 55  1226 4. e 

3 171  if 8 8

2. 58  1596

3. e 

1 147  if 6 6

5. a. Quadratic b. 52, 76

x

x

5 4 3 2 1 1

n(x)  x2  13 1 2

3 4

5

x

2 3 4 5

21.

y

54 3 2 1 1 2 3

5

y 5 4 3 2 1

4 5

7 6 5 4 3 2 1

3 4

4 q(x)  x2  4 5

5 4 3 2 1

19.

1 2

5 4 3 2 1 1 2 3

4 5

1 2 3 4 5 5 4 3 2 1 1 2 3 S(x)  x2  2 3

64. a. 5 1, 36 b. Four real solutions; no imaginary solutions c. Four x-intercepts d.

x

r(x)  (x  1)2 1 2 3 4 5

y 7 6 5 4 3 2 1

x

3 2 1 1 2 3

k(x)  (x  3)2

1 2

3 4

5 6 7

x

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23.

25.

y 7 6 5 4 3 2 1

A(x)  (x  34 )2 1 2

5 4 3 2 1 1

3 4

5

53.

y 7 6 5 4 3 2 1

x

1 2

5 4 3 2 1 1

5

x

W(x)  (x  1.25)2

2 3

2 3

3 4

27. The value of a vertically stretches or shrinks the graph of y  x2. y y 29. 31. 8 7 6

5 4 3 2 1

5 4 3

f(x)  4x2

2 1 5 4 3 2 1 1 2

33.

1 2 3

4

5

x

5 4 3 2 1 1 1 h(x)  x2 2 4 3

35.

y

4

5

x

5 4 3 2 1 1

39. g

41. a

43. e 47.

8 7 6 5 4 3 2 1

(3, 2) 1 2

3 4

5 6 7

x

5 4 3 2 1 1 2 3

(1, 3)

51.

y

2 3 4 5 6 7 8

x4 1 2

3 4

5

x

3 4

5 6 7 8

x

(4, 2)

6 5 4 3 2 1

(0, 3)

5 4 3 2 1 1

1 2 3

4

5

1 2

3 4

5

x

4 5

5 4 3 2 1

4 5

2 3 4

x1

63.

1

2 3 4

5

6

2

3

y

1

4 3 2 1 1

(2, 1) 1

2 3 4

5

1 2

3

x

6

x

2 3 4 5

x1

x2

65. a. y  x  3 is y  x shifted up 3 units. b. y  (x  3)2 is y  x2 shifted left 3 units. c. y  3x2 is y  x2 with a vertical stretch. 67. Vertex 16, 92 ; minimum point; minimum value: 9 69. Vertex (2, 5); maximum point; maximum value: 5 71. Vertex 18, 02 ; minimum point; minimum value: 0 21 21 73. Vertex a0, b; maximum point; maximum value: 4 4 3 3 75. Vertex a7,  b; minimum point; minimum value:  2 2 77. Vertex (0, 0); minimum point; minimum value: 0 79. True 81. False 83. a. (60, 30) b. 30 ft c. 70 ft 85. a. The fireworks will explode at a height of 150 ft. b. Yes, because the ordered pair (3, 150) is the vertex. 2

Section 7.5 Practice Exercises, pp. 624–628

x

x

y 5 4 3 2 1

x

6

3

x3

(1, 2)

5

7 6 5 4 3 2 1 7 6 5 4 3 2 1 1 2 (3, 1)

y

4 3 2 1 1

(1, 0) 1

x

2

2

7 6 5 4 3 2 1 1 2 3

x  3

59.

y

3 4

y

(3, 3)

57.

2 1

x

2

y

2

2 1 1

1 2

5 x  1 4 3 2 1

x3

5

4 3 2 1 1 2

6 5 4 3 2 1

4 5

y

3 4

2 x  1 3

61.

2 3

c(x)  x2

1 2

5 4 3 2 1 1

x0

2 1 v(x)   x2 1 5 1 2 3

(1, 1)

y 8 7 6 5 4 3

3 4

y

4 5

2 1

5

4 3

3

49.

3 4

5

5 4 3 2 1 1 2

3 2 1 1

2

4 5

5 4 3 2 1

37. d 45.

1

x

55.

y 7 6 5 4 3 2 1

1. For f 1x2  ax2  bx  c, the vertex 1h, k2 is given by b 4ac  b2 b b a , b or a ,fa bb . 3. The graph of 2a 4a 2a 2a p is the graph of y  x2 shrunk vertically by a factor of 14. 5. The graph of r is the graph of y  x2 shifted up 7 units. 7. The graph of t is the graph of y  x2 shifted to the left 49 7 2 10 units. 9. 16; 1x  42 2 11. ; ay  b 4 2 1 1 2 1 2 1 13. 15. ; ab  b ; at  b 81 9 36 6 17. g1x2  1x  42 2  11; 14, 112 19. n1x2  21x  32 2  5; 13, 52

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21. p1x2  31x  12 2  2; 11, 22 7 89 7 2 89 ; a ,  b 23. k1x2  ax  b  2 4 2 4 25. F 1x2  51x  12 2  4; 11, 42 1 2 1 1 1 27. P1x2  2 ax  b  ; a , b 29. (2, 3) 4 8 4 8 31. 11, 22 33. 14, 152 35. 11, 22 37. (1, 3) 3 3 39. a , b 41. 14, 152 43. 11, 42 2 4 45. a. 11, 42 b. 10, 32 47. a. 1 12, 72 2 b. (0, 4) c. (1, 0), (3, 0) c. No x-intercepts y y d. d.

(3, 0)

5 4 3 2 1

5 4 3 2 1 1 2

(1, 4)

49. a. c. 1 32, 02 d.

1 32,

3 4 5

02

4

5

x

(0, 3)

5 4 3 2 1 1

5 4 3 2 1 1

1 2 3

(3, 0) 4

5

1 2

3 4

5

x

5 4 (0, 3) 3 2 1

(1, 0)

5 4 3 2 1 1

2

1

2

3 4

5

2 3

3 4

4 5

5

53. Mia must package 10 MP3 players. 55. a. 164.25 ft b. 3.125 sec 57. a. 45 mph b. 32 mpg 59. a. 48 hr b. 2 g 61. a  9, b  5, c  4; y  9x2  5x  4 63. a  2, b  1, c  5; y  2x2  x  5 65. a  3, b  4, c  0; y  3x2  4x 67. a. The sum of the sides must equal the total amount of fencing. b. A  x 1200  2x2 c. 50 ft by 100 ft

Section 7.5 Graphing Calculator Exercises, p. 628 69.

x

51. a. (1, 4) b. (0, 3) c. (1, 0), (3, 0) y d. (1, 4)

( 32 , 0)

( 12 , 72 )

2

b. 10, 94 2

y 5 4 3 2 1

(0,  94)

(1, 0) 1 2 3

8 7 6 5 (0, 4) 4 3 2 1

70.

x

73.

74.

Section 7.6 Practice Exercises, pp. 637–640 3 223  if 5. 50, 4, 66 7. 556 4 4 9. a. 1, 22 ´ 13, 2 b. 12, 32 c. 3 2, 34 d. 1, 2 4 ´ 33, 2 11. a. 12, 02 ´ 13, 2 b. 1, 22 ´ 10, 32 c. 1, 2 4 ´ 30, 34 d. 32, 04 ´ 3 3,  2 13. a. 54, 12 6 b. 1, 12 2 ´ 14, 2 1 c. 12, 42 15. a. 510, 36 b. 110, 32 1 1 c. 1, 102 ´ 13,  2 17. a. e  , 3 f b. c  , 3 d 2 2 1 c. a,  d ´ 3 3,  2 19. 11, 72 2 21. 1, 22 ´ 15,  2 23. 35, 0 4 25. 3 4, 84 1  16 1  16 27. a, b´a , b 5 5 3  133 3  133 , d 29. c 31. 111, 112 2 2 1 33. a,  d ´ 33,  2 35. 1, 3 4 ´ 3 0, 44 3 37. 12, 12 ´ 12, 2 39. a. 576 b. 1, 52 ´ 17, 2 c. 15, 72 41. a. 546 b. 3 4, 62 c. 1, 4 4 ´ 16, 2 43. 11,  2 45. 1, 42 ´ 14, 2 7 1 47. a2, b 49. 15, 74 51. 1, 02 ´ c , b 2 2 53. 10, 2 55. 1,  2 57. 5 6 59. 506 61. 5 6 63. 5 6 65. 1, 2 3 67. 1, 112 ´ 111,  2 69. e f 71. Quadratic; 2 73. Quadratic; 1,  2 75. Linear; 152, 2 34, 4 4 77. Rational; 1, 12 79. Polynomial 1degree 7 22; 81. Polynomial 1degree 7 22; 12, 2 10, 52 ´ 15,  2 83. Polynomial 1degree 7 22; 1, 5 4 85. Quadratic; 122, 222 5  233 5  233 87. Quadratic; a, d´c , b 2 2 89. Rational; 1, 2冥 ´ 15,  2 91. Linear; 1, 5 4 93. Quadratic; 1, 2 95. The fuses should be set for between 1.3 sec and 5.4 sec after launch. 3. e

Section 7.6 Graphing Calculator Exercises, p. 640 71.

72.

SA-39

96. 1, 02 ´ 12,  2

97. 10, 22

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98. 11, 12

99. 1, 12 ´ 11,  2

9 2 81 ; ax  b 4 2 1 2 1 15. n  ; az  b 25 5 13. n 

18. 53  2 136 100. 556

101. 556

102. 5 6

103. 5 6

Chapter 7 Problem Recognition Exercises, p. 640 1. a. Equation quadratic in form and polynomial equation b. 52 22, 16 2. a. Absolute value inequality b. 112, 12 3. a. Polynomial inequality b. 3 12, 5 4 4. a. Radical equation b. 516 5. a. Absolute value equation b. 59, 16 6. a. Rational equation b. 54  2266 7. a. Polynomial inequality b. 35, 24 ´ 32, 2 8. a. Compound inequality b. 1, 12 9. a. Linear inequality b. 3 23, 2 10. a. Absolute value equation b. 59, 16 11. a. Rational inequality b. 1, 22 ´ 35,  2 12. a. Absolute value inequality b. 1, 132 ´ 15, 2 13. a. radical equation b. 546 (The value 7 does not 210 3 check.) 14. a. Quadratic equation b. e  if 4 4 15. a. Compound inequality b. 1, 62 ´ 14,  2 16. a. Linear equation and rational equation

b. e 

Chapter 7 Review Exercises, pp. 647–650 157 if 3 5 3 1  13 f 5. 52  6 126 6. e  i f 7. e 2 2 3 9. 5 23 in. 艐 8.7 in. 10. 9 in. 8. 54  156 11. 5 26 in. 艐 12.2 in. 12. n  64; 1x  82 2 1. 5 156

2. 52i6

3. 596

4. e 

1 2 1 ; ay  b 16 4

16. 52  3i6

19. 54  3i6

17. 51  166

1 20. e , 1 f 3 1Vph or r  ph

V 1 21. e , 4 f 22. r  B ph 2 16A A 23. s  or s  24. See page 588. 6 B6 25. Two rational solutions 26. Two rational solutions 27. Two irrational solutions 28. Two imaginary solutions 29. One rational solution 30. Two imaginary solutions 5 513 5 31. 52  136 32. e  33. e 2,  f if 2 2 6 1 4 4 1 1 34. e 2, f 35. e ,  f 36. e ,  f 3 5 5 10 2 2  122 37. 52  2i 176 38. 54166 39. e f 3 40. 58, 26 41. 57  136 42. 51  1116 43. a. 1822 ft b. 115 ft/s 44. a. ⬇ 53,939 thousand b. 2021 45. The dimensions are approximately 3.1 ft by 7.2 ft. 46. The distance between Lincoln and Omaha is approximately 50 mi. 47. 5496 1The value 9 does not check.2 16 48. 54, 166 49. 53, 126 50. e  , i f 2 51. 5243, 326 52. 532, 16 53. 53  1106 1  1181 54. e 56. 5 27, 26 f 55. 53i, i 236 6 57. 58. y y 3 2 1

9

1 2

5 4 3 2 1 1

3 4

5

8 7 6 5 4 3 f(x)  x2  3 2 1

x

2 3 4 5

g(x)  x2  5

6 7

59.

2 1 1

5 4 3 2 1 1

60.

y

k(x)  (x  3)2

h(x)  (x  5)2

8 7 6 5 4 3 2 1

11 f 2

17. a. Polynomial inequality b. 556 18. a. Rational inequality b. 5 6 19. a. Radical equation and equation quadratic in form b. 516, 816 20. a. Polynomial equation and equation quadratic in form b. 52, 36

14. n 

1 2

3 4

5 6 7 8

x

62.

y

2 3 4 5 6 7

1 2

3 4

5

m(x)  2x2

3 4

5

1

2

3

3 4

5

x

y 7 6 5 4 3 2 1 x

2 3

3 2 1 5 4 3 2 1 1

2

7 6 5 4 3 2 1 1

2

61.

1

x

y 4 3 2 1

5 4 3 2 1 1 2 3 4 5 6

1 2

n(x)  4x2

x

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63.

y

64.

1 2 1 1 2 3 4 5

1

2

3 4

x

5 6 7 8

y

Chapter 7 Test, pp. 650–652

5 4 3 2 1

p(x)  2(x  5)2  5

7 6 5 4 3 2 1 1 2 3

6 7

1

2

3

x

4 5

8 9

q(x)  4(x  3)2  3

5 5 65. a4, b is the minimum point. The minimum value is . 3 3 1 66. a1,  b is the maximum point. The maximum value is 7 1 2 3  . 67. x   68. x  7 11 16 69. z1x2  1x  32 2  2; 13, 22 70. b1x2  1x  22 2  48; 12, 482 71. p1x2  51x  12 2  8; 11, 82 72. q1x2  31x  42 2  6; 14, 62 73. 11, 152 1 22 1 41 74. 11, 72 75. a , b 76. a ,  b 2 4 3 3 77. a. (2, 3) b. (0, 0), (4, 0) c. y 5 4 3

1. 52, 86 2. 52  2 136 3. 51  i6 11 2 121 4. n  5. 53  3 136 ; ad  b 4 2 3 147 6. e  7. a. x2  3x  12  0 if 4 4 b. a  1, b  3, c  12 c. 39 d. Two imaginary solutions 8. a. y2  2y  1  0 b. a  1, b  2, c  1 c. 0 d. One rational solution 7  15 1 9. e 1, f 10. e 11. The height is f 3 2 approximately 4.6 ft and the base is 6.2 ft. 12. The radius is approximately 12.0 ft. 13. 596 (The value 4 does not 11 check.) 14. 58, 646 15. e , 6 f 3 5  157 16. 5  16, 36 17. e f 2 18. The vertex is (3, 17). y y 19. 20. 5 4 3 2 1 5 4 3 2 1 1 2

h(x) 

2 1 5 4 3 2 1 1 y  34 x2  3x2

1

2 3

4

5

x

x2 

4

4

5

x

3

3 2 1 2

4 5

4 6

1 2

3 4

5 6 7

x

y

21. 5

2 1

4 5

78. a. (2, 4) y c.

b. (0, 0) (4, 0)

5 4 3 2 1 1 2

5

3

4 3

4 5

5 4 3 2 1 1 2

2 3

f(x)  (x  4)2

4 3

3

2 1

1

14 12 10 8 6 4 2

y  (x  2)2  4

1

2 3

4

5

x

3 4 5

79. a. 3 sec b. 144 ft 80. a. 150 meals b. $1200 81. a  1, b  4, c  1; y  x2  4x  1 82. a  1, b  1, c  6; y  x2  x  6 83. a. 52, 26; The x-intercepts are 12, 02 and 12, 02 . b. On the interval 12, 22 the graph is below the x-axis. c. On the intervals 1, 22 and 12, 2 the graph is above the x-axis. 84. a. x  2 b. 506 c. 1, 0 4 ´ 12,  2 d. 3 0, 22 85. 12, 62 86. 1, 2 87. 1, 22 ´ 3 0,  2 88. 1, 1 4 ´ 11,  2 89. 12, 02 ´ 15,  2 5 90. 12, 02 ´ a , b 2 91. 1, 2  134 ´ 32  13,  2 92. 1, 52 ´ 11, 2 93. 13,  2 94. 13,  2 95. 556 96. 1, 22 ´ 12,  2

1

2 3

4

5

x

g(x)  12 ( x  2)2  3

22. The rocket will hit the ground 256 ft away. 23. a. 1320 million b. 1999 24. The graph of y  x2  2 2 is the graph of y  x shifted down 2 units. 25. The graph of y  1x  32 2 is the graph of y  x2 shifted 3 units to the left. 26. The graph of y  4x2 is the graph of y  4x2 opening downward instead of upward. 27. a. (4, 2) b. Downward c. Maximum point d. The maximum value is 2. e. x  4 28. a. g1x2  21x  52 2  1; (5, 1) b. (5, 1) 29. a. f 1x2  1x  22 2  16 b. Vertex: 12, 162 c. x-intercepts: (6, 0) and (2, 0); y-intercept: (0, 12) d. The minimum value is 16. e. x  2 1 30. a. 200 ft b. 20,000 ft2 31. c , 6b 32. 15, 52 2 3 33. 1, 32 ´ 12, 22 34. a3,  b 2 35. 5 6 36. 5116

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Chapters 1–7 Cumulative Review Exercises, pp. 652–654 1. a. {2, 4, 6, 8, 10, 12, 16} b. {2, 8} 2. 3x2  13x  1 3. 16 4. 1.8  1010 5. a. 1x  22 1x  321x  32 b. Quotient: x2  5x  6; remainder: 0 6. x  2 2 12x 7. 8. $8000 in 12% account; $2000 in 3% account x 9. 5 18, 72 6 10. a. 720 ft b. 720 ft c. 12 sec 5  133 f 11. 53  4i6 12. e 13. n  25; 1x  52 2 4 14. 21x  521x2  5x  252 y 15.

Chapter 8 Chapter Opener Puzzle 5

1

2

3 4

5

5 16. a. a , 0b, 12, 02 b. (0, 10) 17. The domain element 3 2 has more than one corresponding range element. y 18. 5 4 3 2 1

5 4 3 2 1 1 2 3 4 5

1

2 3

4 5

3

6

2

4

5

2

4

2

1

1

5

3

6

2

4

6

5

1

3

6

2

5

4

3

1

4

3

2

5

6

D

H

E

1

Section 8.1 Practice Exercises, pp. 660–663

4 5

1

C

x

2 3x  5y  10 3

f (x)  x

4

A

5 4 3 2 1 5 4 3 2 1 1

F

B

6

3 G

1

1 f  g21x2  2x 2  5x  4 1g  f 21x2  2x2  3x  4 1 f ⴢ h21x2  x3  4x2  x  4 1g ⴢ f 21x2  2x3  12x2  16x x2  1 h 11. a b 1x2  , x  4 f x4 f x4 13. a b1x2  2 , x  0, x  2 g 2x  4x 2 15. 1 f ⴰ g 21x2  2x  4x  4 3. 5. 7. 9.

17. 1g ⴰ f 21x2  2x2  20x  48

x

19. 1k ⴰ h21x2 

1 x2  1

21. 1k ⴰ g21x2 

19. y  39 20. a. Linear b. (0, 300,000). If there are no passengers, the airport runs 300,000 flights per year. 8 c. m  0.008 or m  1000 . There are eight additional flights per 1000 passengers. 21. a. 3 4, 2 b. 1,  2 22. a. 1, 2 4 b. 1, 4 4 c. 4 d. 3 e. x  1 pq 1 1 23. e , f 24. f  25. 576 18 2 pq y1 26. 27. a. (3, 1) b. Upward c. (0, 19) y3 d. No x-intercept e. y 20 16 12 8 4 10 8 6 4 2 4 8 10

f(x)  2(x  3)2  1

2 4 6 8 10

x

Section 8.2 Practice Exercises, pp. 669–672

12

5 29. a. a,  d ´ 3 2, 2 2 these intervals, the graph is on or above the x-axis 30. 1, 32 ´ 35,  2 28. Vertex: (8, 62)

1 23. No , x  0, x  2 2x2  4x 2 25. 1f ⴰ g21x2  25x  15x  1; 1g ⴰ f 21x2  5x2  15x  5 27. 1f ⴰ g21x2  冟x3  1冟; 1g ⴰ f 21x2  冟x冟3  1 29. 1h ⴰ h21x2  25x  24 31. 0 33. 64 35. 2 1 37. 1 39. 41. 0 43. Undefined 64 4 45. 47. 2 49. 2 51. 0 53. 1 9 55. 0 57. Undefined 59. 1 61. 2 63. 2 65. 1 67. 2 69. 3 71. 6 73. 1 75. 4 77. 0 79. 2 81. a. P1x2  3.78x  1 b. $188 83. a. F 1t2  0.2t  6.4; F represents the amount of child support (in billion dollars) not paid. b. F 142  7.2 means that in 2004, $7.2 billion of child support was not paid. 85. a. 1D ⴰ r 21t2  560t; This function represents the total distance Joe travels as a function of time that he rides. b. 5600 ft

b. On

3. Yes 5. No 7. Yes 9. g1  {(5, 3), (1, 8), (9, 3), (2, 0)} 11. r1  {(3, a), (6, b), (9, c)} 13. The function is not one-to-one. 15. Yes 17. No 19. Yes

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21. h1(x)  x  4 23. m1(x)  3(x  2) 3 25. p1(x)  x  10 27. f 1 1x2  1 x1 3 x  1 29. g1 1x2  31. g1 1x2  1x  9 2 x 33. a. 1.2192 m, 15.24 m b. f 1 1x2  c. 4921.3 ft 0.3048 35. False 37. True 39. False 41. True 43. (b, 0) 45. a. Domain [1, ), range [0, ) b. Domain [0, ), range [1, ) 47. a. [4, 0] b. [0, 2] c. [0, 2] d. [4, 0]

3 69. g 1 1x2  1 2x  4

70. m 1 1x2 

x4 3

y 5 4 3 2 1

y  f(x)

Section 8.3 Practice Exercises, pp. 678–681 3. 2x2  2x  3

yx

5 4 3 2 1 1 2

1 2 3 4 5

x

4 5

49. a. [0, 2]

b. [0, 4]

c. [0, 4]

d. [0, 2]

y 5 4 3 2 1 5 4 3 2 1 1 2

y  f(x) yx y  f 1(x) 1 2 3 4 5

x

3 4 5

x1 b1x 6 16x  12  1 b. 1g ⴰ f 21x2  x 6 3 3 3 28x 2x 3 53. a. 1 f ⴰ g2 1x2   x b. 1g ⴰ f 21x2  8 a b x 2 2 55. a. 1 f ⴰ g21x2  1 1x  12 2  1  x b. 1g ⴰ f 21x2  21x2  12  1  x 57. q1 1x2  x2  4, x 0 59. z1 1x2  x2  4, x 0 x1 x2 61. f 1 1x2  63. t1 1x2  x 1x 65. n1 1x2  1x  9 51. a. 1 f ⴰ g2 1x2  6 a

Section 8.2 Graphing Calculator Exercises, p. 672 67. f 1 1x2  x3  5

3 68. k 1 1x2  1 x4

3x  1 2x2  x  2

7. 6x2  3x  5

1 13. 6 15. 8 17. 5.8731 1000 19. 1385.4557 21. 0.0063 23. 0.8950 25. a. x  2 b. x  3 c. Between 2 and 3 27. a. x  4 b. x  5 c. Between 4 and 5 1 1 29. f 102  1, f 112  , f 122  , f 112  5, f 122  25 5 25 1 31. h102  1, h112  3, h112  , h1 122 ⬇ 4.73, 3 33. If b 7 1, the graph is increasing. If h1p2 ⬇ 31.54 0 6 b 6 1, the graph is decreasing. y y 35. 37. 9. 25

y  f 1(x)

3

5.

f(x)  4 x (1, 14 )

11.

8 7 6 5 4 (1, 4) 3 2 1 (0, 1)

5 4 3 2 1 1

1 2 3 4

9

(1, 8)

m(x)  ( 18 )x

5

x

5 4 3 2 1 1

2

39.

y

41.

8 7 6 5 (1, 4) h(x)  2x1 4 3 2 (0, 2) (1, 1) 1 5 4 3 2 1 1

8 7 6 5 4 3 2 (0, 1) 1 1 (1, 8 )

1 2 3 4

5

x

1

2

3 4

5

x

y

8 7 6 (1, 5) 5 4 3 x g(x)  5 2 (0, 1) 1 (1, 1 ) 5

5 4 3 2 1 1

2

1 2 3 4

5

x

2

43. a. 0.25 g b. ⬇ 0.16 g 45. a. 758,000 b. 379,000 c. 144,000 47. a. P 1t2  153,000,00011.01252 t b. P1412 ⬇ 255,000,000 49. a. $1640.67 b. $2691.80 c. A102  1000. The initial amount of the investment is $1000. A172  2000. The amount of the investment doubles in 7 yr.

Section 8.3 Graphing Calculator Exercises, p. 681 51.

52.

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53.

54.

73. a. f a

1 1 1 b  3, f a b  2, f a b  1, f 112  0, 64 16 4 f 142  1, f 1162  2, f 1642  3 y b. 5 4 3 2 1

55.

56.

f(x)  log4 x

2 1 1

1

2

3 4

x

5 6 7 8

2 3 4 5 y

75. 3  x y

57.

58.

5 4 3 2 1

x

y

1 9 1 3

1 3

2 1 0 1

9

2

y  log3 x

2 1 1

1

2

3 4

x

5 6 7 8

2 3 4 5

1 y 77. a b  x 2 y 5

Section 8.4 Practice Exercises, pp. 690–694

3

3. i 5. a. s 122  b.

25 5 2 4 , s 112  , s 102  1, s 112  , s 122  4 2 5 25

y

y  log1/2 x

2 1 1

1 2 3 4

x

5 6 7 8

y

4 2 1

2 1 0 1

1 2 1 4

4 5

79. 15,  2

3 2 1

5 4 3 2 1 1

2 1

x

2 3

8 7 6 5 4

s(x)  ( 25 )x

4

1 2 3 4

5

x

2

1 1 7. g122  , g 112  , g 102  1, g 112  3, g 122  9 9 3 9. by  x 11. 54  625 13. 104  0.0001 15. 62  36 17. bx  15 19. 3x  5 10 1 21. a b  x 23. log3 81  x 25. log5 25  2 4 1 27. log7 a b  1 29. logb y  x 31. loge y  x 7 1 33. log1/3 9  2 35. 2 37. 1 39. 41. 0 2 1 43. 5 45. 1 47. 3 49. 51. 1 53. 3 2 55. 6 57. 2 59. 0.7782 61. 0.4971 63. 1.5051 65. 2.2676 67. 5.5315 69. 7.4202 71. a. Slightly less than 2 b. Slightly more than 1 c. log 93 ⬇ 1.9685, log 12 ⬇ 1.0792

2

81. 1, 22

1 83. a ,  b 85. 11.2, 2 3 89. 1, 02 ´ 10,  2 91. ⬇7.35

87. 1, 22 93. a. t (months)

0

1

2

6

12

24

S1(t)

91 82.0 76.7 65.6 57.6 49.1

S2(t)

88 83.5 80.8 75.3 71.3 67.0

b. Group 1: 91; Group 2: 88

c. Method II

Section 8.4 Graphing Calculator Exercises, p. 694 95. Domain: 16,  2; asymptote: x  6

96. Domain: 12, 2; asymptote: x  2

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97. Domain: 12,  2; asymptote: x  2

99. Domain: 1, 22; asymptote: x  2

98. Domain: 18,  2; asymptote: x  8

100. Domain: 1, 32; asymptote: x  3

Section 8.4 Problem Recognition Exercises, p. 695 1. e 7. k

2. l 8. b

3. j 9. i

4. g 10. h

5. c 11. f

6. a 12. d

Section 8.5 Practice Exercises, pp. 701–704 3. 13. 25. 37. 43. 45. 51. 53. 55. 57. 61. 67. 71. 75. 83. 89.

1 4 5. 7. 0.9031 9. 5.0475 11. a 6 c 15. a, b, c 17. 1 19. 4 21. 11 23. 3 0 27. 9 29. 0 31. 3 33. 1 35. 2x 4 39. 0 41. Expressions a and c are equivalent. Expressions a and c are equivalent. 47. log 2  log x 49. 4 log5 x log3 x  log3 5 log4 a  log4 b  log4 c 1 logb x  logb y  3 logb z  logb w 2 1 log2 1x  12  2 log2 y  log2 z 2 1 2 1 59. 5 log w log a  log b  log c 3 3 3 1 x2z 63. 3 65. log3 a 3 b logb a  3  log b c 2 y a2c 2 69. logb x log3 a 4 b 2b 3 1x  62 2 2 y 73. log c log8 18a5 2 or log8 a5  1 d 5 z 1 77. 1.792 79. 2.485 81. 4.396 logb a b x1 0.916 85. 13.812 87. 16.09 a. B  10 log I  10 log I0 b. 10 log I  160

Section 8.5 Graphing Calculator Exercises, p. 704 91. a. Domain: 1, 02 ´ 10,  2

b. Domain: 10, 2

c. They are equivalent for all x in the intersection of their domains, 10,  2 . 92. a. Domain: 1, 12 ´ 11,  2 b. Domain: 11, 2

c. They are equivalent for all x in the intersection of their domains, 11,  2 .

Section 8.6 Practice Exercises, pp. 712–717 3. log3 a 7.

a 1d b c5

x 4

y 0.05

3

0.14

2

0.37

1

1

0

2.72

1

7.39

5.

1 1 3 log6 x  log6 y  log6 z 4 2 4 y 8 7 6 5 4 3 2 1 5 4 3 2 1 1

y  e x1

1

2 3

4

5

x

2

9.

Domain: 1, 2 x 2 1

y 2.14 2.37

0

3

1

4.72

2

9.39

3 22.09

y

8 7 6 5 4 3 2 1 5 4 3 2 1 1 2

y  ex  2

1 2 3 4 5

x

Domain: 1, 2 11. a. $12,209.97 b. $13,488.50 c. $14,898.46 d. $16,050.09 An investment grows more rapidly at higher interest rates. 13. a. $12,423.76 b. $12,515.01 c. $12,535.94 d. $12,546.15 e. $12,546.50 More money is earned at a greater number of compound periods per year. 15. a. $6920.15 b. $9577.70 c. $13,255.84 d. $18,346.48 e. $35,143.44 More money is earned over a longer period of time.

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y

17.

x

y

2.25 1.39 2.50 0.69 2.75 0.29 3

0

4

0.69

5

1.10

6

1.39

19.

x

2 1 1 2 3

2

3 4

5 6 7 8

x

y  ln (x  2)

c. They appear to be the same. 67. 68.

Domain: 12,  2 y

5 4 3 2 1

0.25 2.39 1.69

0.75 1.29 1

1.00

2

0.31

3

0.10

4

0.39

y  ln x  1

2 1 1 2 3 4 5

1 2 3 4 5 6 7

x

69.

Domain: 10, 2

y

21. a.

1

4 5

y

0.5

66. a–b

5 4 3 2 1

10 8 6 4 2 10 8 6 4 2 2 4

f(x)  10 x g(x)  log x 2

4 6 8 10

x

Section 8.6 Problem Recognition Exercises, p. 717

6 8 10

b. Domain: 1, 2; range: 10,  2 c. Domain: 10, 2; range: 1,  2 27. 6 35. ln a

29. 2x  3 2

a

3

b1c

b

41. 2 ln b  1

37. ln a

23. 1

31. ln 1p 1q2 6 3

4

x b y3z

43. 4 ln a 

25. 0 x 33. ln B y3

39. 2 ln a  2 ln b 1 ln b  ln c 2

1 2 1 47. a. 2.9570 b. 2.9570 ln a  ln b  ln c 5 5 5 c. They are the same. 49. 2.8074 51. 1.5283 53. 2.1269 55. 0 57. 3.3219 59. 3.8124 61. a. 15.4 yr b. 6.9 yr c. 13.8 yr 63. a. 19.8 yr b. 13.9 yr c. 27.8 yr 45.

Section 8.6 Graphing Calculator Exercises, pp. 716–717 65. a–b

Exponential Form

Logarithmic Form

1.

25  32

log2 32  5

2.

3  81

log3 81  4

3.

z x

logz x  y

4.

bc  a

logb a  c

y

5.

103  1000

6.

10  10

7.

e b

ln b  a

8.

eq  p

ln p  q

9. 10. 11.

1

a

1 12 2 2



1 4

1 13 2 2  9 10

2

 0.01

log 1000  3 log 10  1

log1/2 1 14 2  2

log1/3 9  2

log 0.01  2

12.

10x  4

log 4  x

13.

e 1

ln 1  0

14.

e e

ln e  1

15.

1/2

5

log25 5  12

1/4

2

log16 2  14

17.

et  s

ln s  t

18.

e w

ln w  r

19.

2



1

p

16.

20. c. They appear to be the same.

4

0

1

25 16

r

15 3

1 225

1 log15 1 225 2  2

log3 p  1

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Section 8.7 Practice Exercises, pp. 726–730 x2 3. logb 3x12x  32 4 5. logb a 7. 596 b 3x  5 42 0.08 9.2 9. 510 6 11. 5e 6 13. 510 15. 556  406 17. 5106 19. 5256 21. 5596 23. 516 25. 506 37 27. e  f 29. 536 (The value 3 does not check.) 9 31. 546 33. 536 35. 526 37. 5 6 (The value 3 does not check.) 39. 546 41. 566 11 1 43. e f 45. 526 47. e f 49. 516 2 12 4 ln 21 51. e f 53. 52, 16 55. e f 19 ln 8 ln 15 57. 5ln 8.12546 59. 5log 0.01386 61. e f 0.07 ln 3 ln 3 2 ln 2 63. e 65. e 67. e f f f 1.2 ln 5  ln 3 ln 6  ln 2 ln 4 125 ln 20 69. e 71. e 3 ln a 73. e f bf  2f 0.04 6 ln 5 75. a. ⬇1285 million (or 1,285,000,000) people b. ⬇1466.5 million (or 1,466,500,000) people c. The year 2049 (t ⬇ 50.8) 77. a. 500 bacteria b. ⬇ 660 bacteria c. ⬇25 min 79. It will take 9.9 yr for the investment to double. 81. a. 7.8 g b. 18.5 days 83. The intensity of sound of heavy traffic is 103.07 W/m2. 85. It will take 38.4 yr. 87. a. 1.42 kg b. No 89. 510 5, 103 6 1 1 91. e , f 27 9

y 5 4 3 q1(x)  4 (x  2) 2 3 1 5 4 3 2 1 1 2 3 4 5

8 7 6 5 4 3 2 1

x

q(x)  3 x  2 4

8 7 6 5 4 3 g(x)  ( 14 )x 2 1

f(x)  3x

5 4 3 2 1 1 2

1 2 3 4 5

x

y

41.

1. 1 f  g21x2  2x3  9x  7 2. 1 f  g21x2  2x3  7x  7 x7 ,x2 3. 1 f ⴢ n21x2  4. 1 f ⴢ m21x2  x3  7x2 x2 f x7 5. a b1x2  ,x0 g 2x3  8x g 2x3  8x 6. a b1x2  ,x7 f x7 2 7. 1m ⴰ f 2 1x2  1x  72 or x2  14x  49 1 1 8. 1n ⴰ f 2 1x2  11. 167 , x  9 9. 100 10. x9 8 12. 10 13. a. 12x  12 2 or 4x2  4x  1 b. 2x2  1 1 14. 15. 3 c. No, f ⴰ g  g ⴰ f 16. 0 4 18. 4 19. 1 20. No 21. Yes 17. 1

1 2 3 4 5

29. a. Domain: 31,  2; range: 30, 2 b. Domain: 3 0, 2 range: 31,  2 30. p1(x)  (x  2)2, x 2 1 ⬇ 0.028 31. 1024 32. 33. 2 34. 10 36 35. 8.825 36. 16.242 37. 1.627 38. 0.681 y y 39. 40.

94.

Chapter 8 Review Exercises, pp. 736–740

4 1x  22 3

3 25. f 1 1x2  1 x1 24. g1(x)  (x  3)5 2x  4 1 26. n 1x2  x 2 1 27. 1f ⴰ g21x2  5 a x  b  2  x  2  2  x 5 5 1 2 2 2 1g ⴰ f21x2  15x  22   x    x 5 5 5 5 28. The graphs are symmetric about the line y  x.

Section 8.7 Graphing Calculator Exercises, p. 730 93.

23. q1 1x2 

22. {(5, 3), (9, 2), (1, 0), (1, 4)}

1 2 3 4 5

x

y

42.

8 7 6 5 4 k(x)  (25)x 3 2 1

8 7 6 5 x 4 h(x)  5 3 2 1 5 4 3 2 1 1 2

5 4 3 2 1 1 2

1 2

3 4 5

x

5 4 3 2 1 1 2

1 2 3 4 5

x

43. a. Horizontal b. y  0 44. a. 15,000 mrem b. 3750 mrem c. Yes 45. 3 46. 0 47. 1 48. 8 51. 5 52. 1 49. 4 50. 4 53. y 54. y 5 4 3

5 4

q(x)  log3 x

3

2

2

1

1

2 1 1

1 2 3 4

5 6 7 8

x

2 1 1

2 3

2 3

4 5

4 5

55. a. Vertical asymptote b. x  0

r(x)  log1/2 x 1 2 3 4

5 6 7 8

56. a. 2.5

x

b. 9.5

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58. 6

59. 0

60. 7 x 61. a. logb x  logb y b. logb a b c. p logb x y 4 3 2 xy 1ab 62. logb a b 63. log3 a 2 4 b 64. log 5 z cd 65. 0 68. 148.4132 69. 14.0940 66. b 67. a 70. 32.2570 71. 57.2795 72. 1.7918 73. 2.1972 74. 1.3424 75. 1.3029 76. 3.3219 77. 1.9943 78. 0.8370 79. 3.6668 80. a. $33,361.92 b. $33,693.90 c. $33,770.48 d. $33,809.18 81. a. S102  95; the student’s score is 95 at the end of the course. b. S162 ⬇ 23.7; the student’s score is 23.7 after 6 months. c. S1122 ⬇ 20.2; the student’s score is 20.2 after 1 yr. 82. 1, 2 83. 1,  2 84. 1,  2 4 85. 10, 2 86. 15, 2 87. 17,  2 88. a ,  b 3 1 89. 1, 52 90. 51256 91. e f 92. 52166 49 1001 93. 53112 6 94. e 95. 596 f 2 190 96. 556 (The value 2 does not check.) 97. e f 321 4 ln 21 ln 18 98. 516 99. e f 100. e 101. e f f 7 ln 4 ln 5 log 1512 ln 0.06 102. 5ln 0.16 103. e  104. e f f 2 2 log 821 3 ln 2 105. e 106. e f f 3 ln 7  ln 2 5 ln 14 107. e 108. a. 1.09 mg b. 0.15 mg f ln 14  ln 6 c. 16.08 days 109. a. 150 bacteria b. ⬇ 185 bacteria c. ⬇ 99 min 110. a. V102  15,000; the initial value of the car is $15,000. b. V1102  3347; the value of the car after 10 years is $3347. c. 7.3 yr

Chapter 8 Test, pp. 740–742 x4 x2  2 f 1. a g b1x2  2 2. 1h ⴢ g21x2  ,x0 x x 2 3. 1g ° f 2 1x2  x2  8x  18 1 ,x4 4. 1h ⴰ f 2 1x2  x4 3 1 5. 48 6.  7. 8. 18 2 18 g x2  2 9. a b1x2  10. A function is one-to-one ,x4 f x4 11. b if it passes the horizontal line test. 12. f 1 1x2  4x  12 13. g1 1x2  1x  1 y 14. y  p(x)

5 4 3 2 1

5 4 3 2 1 1 2 3 4 5

1 2 3 4 5

x

y  p1(x)

15. a. 4.6416 b. 32.2693 c. 687.2913

y

16.

8 7 6 5 4 3 2 1

f(x)  4x1

5 4 3 2 1 1 2

17. a. log16 8  18.

3 4

1 2 3 4 5

x

b. x5  31

y 5 4 3 2 1 2 1 1 2 3 4 5

g(x)  log3 x

1 2 3 4 5 6 7 8

x

loga n 20. a. 1.3222 b. 1.8502 c. 2.5850 loga b 21. a. 1  log3 x b. 5 22. a. logb 1 2x y3 2 1 b. log a 3 b or log a3 23. a. 1.6487 b. 0.0498 a c. 1.0986 d. 1 24. a. y  ln x b. y  e x 25. a. p142 ⬇ 59.8; 59.8% of the material is retained after 4 months. b. p1122 ⬇ 40.7; 40.7% of the material is retained after 1 yr. c. p102  92; 92% of the material is retained at the end of the course. 26. a. 9762 thousand (or 9,762,000) people b. The year 2020 1t ⬇ 20.42 27. a. P 102  300; there are 300 bacteria initially. b. 35,588 bacteria c. 1,120,537 bacteria d. 1,495,831 bacteria 28. 5256 (The value 4 does not check.) ln 50 29. 5326 30. 5e2.4  76 31. 576 32. e f ln 4 ln 250 ln 58 7 ln 4 33. e 34. e 35. e f  3f f 2.4 ln 2 ln 5  ln 4 36. a. P125002  560.2; at 2500 m the atmospheric pressure is 560.2 mm Hg. b. 760 mm Hg c. 1498.8 m 37. a. $2909.98 b. 9.24 yr to double 19.

Chapters 1–8 Cumulative Review Exercises, pp. 742–743 p 3p3 5 2. 1   3. Quotient: 4 2 4 3 2 t  2t  9t  18; remainder: 0 4. 0 x  3 0 3 21 25 4d110 5. 6. 2 27 in. 7. c 5 26 7 8. 1 115  16  130  2132 m2 9.   i 29 29 3 10. e 6, f 11. 4.8 L 12. 24 min 13. 517, 126 2 12 sg 23 cd 14. e  f 15. x  16. t  g 11 ab 1. 

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4  234 18. (7, 0), (3, 0) f 3 2 19. a. 30t b. 10t c. 2t 2  5t 20. 506 1 21. a. x  2 b. y  6 c. y  x  5 2 22. 40°, 80°, 60° 23. Infinitely many solutions; 5 1x, y2 02x  y  46 ; dependent system 1 1 2 24. c , b 25. f 1 1x2  x  26. 40 m3 2 5 15 x2  5x  25 27. 28. 1  x 29. a. Yes; 12x  12 1x  22 x  4, x  2 b. 586 c. 1, 22 ´ 14, 8 4 3 30. The numbers are  , 4. 31. 546 (The value 9 4 32. 1, 2 does not check.) 33. a. P162  2,000,000, P1122  1,000,000, P1182  500,000, P1242  250,000, P1302  125,000 b. 48 hr 34. a. 2 b. 3 c. 6 d. 3 35. a. 217.0723 b. 23.1407 c. 0.1768 d. 3.7293 e. 0.4005 f. 2.6047 2 36. e f 37. 5ln 11002 6 38. 536 (The value 9 3 2 1z 1 does not check.) 39. log a 2 3 b 40. ln x  ln y 3 3 xy 17. e

S ___ E ___ C ___ T ___ I ___ O ___ N ___ S ___ 1 7 5 2 3 4 6 1

Section 9.1 Practice Exercises, pp. 751–756 173 9. 4 11. 8 8 13. 4 12 15. 142 17. Subtract 5 and 7. This becomes 5  (7)  12. 19. y  13, y  1 21. x  0, x  8 23. Yes 25. No 27. Center (4, 2); r  3 29. Center (1, 1); r  1 5. 134

4 5

4 5

3 4

5

y

3 4

5 6 7

8

4

6

8 10

x

2 3

54 3 2 1 1 2 3 4 5

4 5 6 7

31. Center (0, 2); r  2

5 4 3 2 1 1 2

8 10

3 4

4 3

43. Center (0, 3); r 

x

3 2 1 1 2 3 4 5

1

2

3 4

5

x

y 5 4 3

4 3 2 1 5 4 3 2 1 1 2 3

1 2

3 4 5

x

2 1 5 4 3 2 1 1 2

1

2 3

4

5

x

3 4 5

y 5 4 3 2 1

1 2 3 4 5

x

1 2 3 4 5 6 7

5 4 3 2 1 1 2

1

2 3

4

5

x

3 4 5

x 2  y2  4 51. x2  1y  22 2  4 55. x2  y2  49 1x  22 2  1y  22 2  9 1x  32 2  1y  42 2  36 61. 11, 02 63. 11, 62 65. 10, 32 11, 22 1 3 67. a , b 69. 10.4, 12 71. (1, 3) 2 2 73. 140, 712 2 ; they should meet 40 mi east, 712 mi north of the warehouse. 75. x2  1y  32 2  4 49. 53. 57. 59.

y

1 2 3 4 5

x

47. Center (0, 0); r  1

5 4 3 2 1

5 4 3 2 1

5

45. Center: (1, 2); r  3

6

33. Center (3, 0); r  2 12

y

3 4

6 5 4 3 2 1

4 6

4 5

5 4 3 2 1

x

2

y

2

10 8 6 4 2 2

1

41. Center: (0, 3); r  2

10 8 6 4 2

y

y

54 3 2 1 1 2 3 4 5

2 3

2

39. Center (1, 3); r  6

7.

3 2 1 2

x

5 4 3 2 1 1 2 3

1

y

C ___ O ___ N ___ I ___ C ___ 5 4 6 3 5

1

5 4 3 2 1

5 4 3 2 1 1

Chapter Opener Puzzle

2 1 1

y

y 5 4 3 2 1

Chapter 9

3. 3 15

4 8 37. Center a , 0b; r  5 5

35. Center (0, 0); r  16

x

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77. 1x  42 2  1y  42 2  16

y

9.

8 7 6 5 4 3 2 1 1 1 2

2 1

(2, 1)

2

3 4

5 6 7 8

9

x2

x

2 3

4

5

x

5 4 3 2 1 1 2 3

4 5

4 5 y

15.

9

2

81.

1

y

y  2(x  2)2  8

Section 9.1 Graphing Calculator Exercises, p. 756

1

86.

2 3

4

5

x

x 2

x

y

21.

2 1

2 1 5 4 3 2 1 1 2 3 4 5

7. a

11 , 1b 2

1

2 3

4

5

x

x

4 5 y 7 6 5 3

1

2 3

4

5

2 1

3 4 5

2 3

y

1

5 4 3

y 2

2 3

y 1 4

5

x

1

2

2 1

(1, 1)

x 5 4 3 2 1 1 2 3 4 5 1 2 x  (y  1)2  1 3 4 5

27. 15, 12

7 3 1 29. a2, b 31. a ,  b 4 2 4 33. 110, 12 35. The maximum height of the water is 22 ft. 37. A parabola whose equation is in the form y  a1x  h2 2  k has a vertical axis of symmetry. A parabola whose equation is in the form x  a1y  k2 2  h has a horizontal axis of symmetry. 39. Vertical axis of symmetry; opens upward 41. Vertical axis of symmetry; opens downward 43. Horizontal axis of symmetry; opens right 45. Horizontal axis of symmetry; opens left 47. Vertical axis of symmetry; opens downward 49. Horizontal axis of symmetry; opens right

Section 9.3 Practice Exercises, pp. 771–775 3. Center: 18, 62; radius: 10 5. Vertex: 13, 12; x  3 1 2 5 2 1 7. ax  b  ay  b  2 2 4

x

y

23.

4 5

y

5

32

x

x  (y  2) 3

5 4 3

4

( 3, 3) 4 y  3 y0

5 4 3

25. 12, 12

2 3

8 7 6 5 4 3 2 1 1

5 4 3 2 1 1 2 2

3. 5 5. Center: (0, 1); radius: 4

1

x  (y  3)2  3

2 (0, 2) 1

Section 9.2 Practice Exercises, pp. 763–766

x

x 2

19.

5 4 3

5 4 3 2 1 1 (3, 0) 2

85.

5

3

5 4 3 2 1 1

x  y2  3

4

(2, 1)

5 4 3 2 1 1 2

2 1

84.

2 3

2 1

5 4 3

17.

1

5 3 17 4 y  (x  2 )2  4 3

(23 , 174 )

(2, 8)

8 7 6

82.

83.

2 1

3

13.

79. 1x  12  1y  12  29 2

5 4 3

y  (x  2)2  1

5 4 3 2 1 1 2

1

y

11.

5 4 2 3 y  (x  2)  1

y

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y

9. x2 y2  1 4 9

2 1

(2, 0)

x2 y2  1 16 9

(2, 0)

5 4 3 2 1 1 2

1

3

y

11.

5 4 (0, 3) 3

2 3

4

5

(4, 0)

x

4 5

7 6 y2 x2  16  1 25 5 4 3

(4, 0)

5 4 3 2 1 1 2

1

3

(0, 3)

y

33.

5 4 (0, 3) 3 2 1

2 3

4

2 1

x

5

(5, 0)

7 6 5 4 3 2 1 1 2

(0, 3)

4 5

5 4 3

y2

2 1

(1, 0)

2 3

4

5 6

(0, 2) y2 x2  1 4 4

4

5 6

x x2 

19. Center: (1, 2)

y

(1, 5) 6

(4, 8)

4 3 2

(6, 5)

(6, 2)

(1, 2)

3

(4, 2) 2

3 4

5 6 7

8

9

x

21. Center: (2, 3)

6 5 4 3 2 1 1 (1, 1) 2 3 4

1

2 3

(2, 2)

4

5

x

(2, 3)

(1, 3) 3 4 5

25. Vertical 31. Vertical

2

3 4

x

5 4 3 y2 x2  1 16 4

(6, 1)

2 (0, 1)

10 8 6 4 2 2 4 6

(2, 4)

(5, 3)

27. Horizontal

2 3

4

5 6

7

x

7 6

2

4

(0, 4)

8 10

29. Horizontal

3

(6, 1) 6 8 10

4 5 6 7

x

41. 45. 49. 53.

(0, 4)

2 1

7 6 5 4 3 2 1 1 2

10 8 (0, 6) 6 4

2 1 5 4 3 2 1 1 2

1

y

5 4 3

(1, 0) 1

7 y

39.

23. Center: (0, 1)

y

2 1

4 5 6

(4, 2)

1

4

1

7

5 4 3

3

5

7

1 1

5 6

7 6

7 6 5 4 3 2 1 1 2

y

(4, 5)

y2 1 36

(1, 0)

17. Center: (4, 5)

1

4

x

y

37.

(5, 0)

4 5 6

2

2 3

4 (0, 2) 5 6 7

2 (0, 1) 1

(2, 5)

1

3

5 4 3

3

5

7

2 1

7 6 5 4 3 2 1 1 2

6 5 4 3 2 1 1 2 3 1 (0, 1) 2

6

5 6

5 4 3 (0, 2)

x

y

8

4

x

6

(1, 0) 1

6

9

7

7

4 5 6

(5, 0)

5 6

y

35.

(0, 2)

3

x2  y2  1 25

4

7

6 5 4 3 2 1 1 2

15.

2 3

4 5 6

6

x2  4  1

(5, 0) 1

3

y

13.

SA-51

1

2 3

x

(0, 4)

Hyperbola 43. Ellipse Ellipse 47. Hyperbola Ellipse 51. Hyperbola The height 10 ft from the center is approximately 49 ft.

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y

55.

y

57.

5 4 3 2 1

(⫺3, 3)

⫺4 ⫺3 ⫺2 ⫺1 ⫺1 (⫺2, ⫺2) ⫺2 ⫺3 ⫺4 ⫺5

1

2

3

4

(1, ⫺2)

5

6

x

(⫺3, 1)

5 4 3 2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

(4, ⫺2)

35. 5 12, 02, 1⫺2, 026 5 11, 02, 1⫺1, 026 5 13, 22, 1⫺3, 22, 13, ⫺22, 1⫺3, ⫺226 41. 5 10, ⫺22, 10, 226 5 18, 62, 1⫺8, ⫺626 45. 5 14, 52, 1⫺4, ⫺526 5 12, 02, 10, 126 22 22 47. e a⫺22, 49. 3 and 4 b, a 22, ⫺ bf 2 2 51. 5 and 17, ⫺5 and 17, 5 and ⫺ 17, or ⫺5 and ⫺17 33. 37. 39. 43.

6

1

2

3 4

5

x

⫺2

Section 9.4 Graphing Calculator Exercises, p. 783

(⫺3, ⫺1) ⫺3 ⫺4

53.

Section 9.3 Problem Recognition Exercises, pp. 775–776 1. Standard equation of a circle 2. Ellipse centered at the origin 3. Distance between two points 4. Midpoint between two points 5. Parabola with vertical axis of symmetry 6. Hyperbola with horizontal transverse axis 7. Hyperbola with vertical transverse axis 8. Parabola with horizontal axis of symmetry 9. Parabola 10. Parabola 11. Circle 12. Circle 13. Hyperbola 14. Hyperbola 15. Circle 16. Circle 17. Parabola 18. Parabola 19. Ellipse 20. Ellipse 21. Circle 22. Hyperbola 23. Parabola 24. Ellipse 25. Hyperbola 26. Circle 27. Ellipse 28. Parabola

54.

Section 9.4 Practice Exercises, pp. 780–783 1 7 7. a⫺ , ⫺ b 2 2 9. Zero, one, or two 11. Zero, one, or two 13. Zero, one, two, three, or four 15. Zero, one, two, three, or four 17. 5 1⫺3, 02, 12, 52 6 19. 5 10, 12, 1⫺1, 026 5. 1x ⫹ 52 2 ⫹ 1 y ⫺ 32 2 ⫽ 64

3. 7 12

y

y

10 9 8 7 6

(⫺3, 0)

5 4 3 2 1

(⫺1, 0)

(2, 5)

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3

1 2

3 4

5

x

5 4 3 2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

y

5 4 3

(√2, 2)

2 1

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

x

⫺2 ⫺3 ⫺4 ⫺5

23. 5 14, 22 6

25. 510, 026

29. 5 10, 02, 1⫺2, ⫺82 6

56. (0, 1) 1 2 3

4

5

x

57. 5 6

21. 5 1 12, 22, 1⫺ 12, 22 6

(⫺√2, 2)

55.

3 9 27. e a , b f 2 4 31. 5 11, 426

58. 5 6

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Section 9.5 Practice Exercises, pp. 786–791 1. k 3. e 5. i 13. True 15. False y 17. a.

7. j

9. a

11. d

4 2

5 4 3 2 1 1

2 5 4 3 2 1 1

1

2

3

4

5

x

5 4 3 2 1 1

5 4 3 2 1

5 4 3 2 1 1

1

2

3

4

5

37.

y

y ln x

5 4 3 2 1 1 2 3 4 5

39.

x

1

2

3

4 5

x

4 5

b. The parabola y  x2  1 would be drawn as a dashed curve. 21. 1x  32 2  1y  42 2 625 y y 23. 25. 5 4 3

x y2

2x  y  1 2

1 3

4

x

5

5 4 3 2 1

5 4 3 2 1 1 2 3 4

2 3 4

5

5 y 5 4

1 2 3 4 5

41.

2 1

5 4 3 2 1 1

1

2

3

4

5

x

2 3 4

1

y 5 4 3 2 1 1

2

3

4

x  y2  4

5

x

3

4 5

x

3

1

2

3 4 5

1

2

3

4

5

1

2

3

4

5

4 5

x

y

5 4 3 2 1 1 2 3 4 5

x

y

45.

y

9 8 7 6 5 4 3

3 2 1 5 4 3 2 1 1 2 3 4 5

1

2

3

4 5

x

2 1 5 4 3 2 1 1 y

51.

5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

5

2

5 4

47. 5 6 49.

3

2

5 4 3 2 1

5 4 3 2 1 1 2 3 4 5

x

1

x

2 3 4 5 y

43.

3 4 5

5 4 3 2 1

5 4 3 2 1 1

4 3 2 1

3

2

y

y 5x

5

2

1

2 3 4 5

5 4 3 2 1

b. The set of points on and “outside” the circle x2  y2  9 c. The set of points on the circle x2  y2  9 y 19. a.

5 4 3 2 1 1 2 3 4 5

5

5

35.

5

29.

4

4

4

(x  1)2  (y  2)2 9

3

3

x

2 3

27.

1 2

2

1

2

3 2 x  16y2 16 2 1

9x2  y2 9

1

3

1

y 5 4

3

5

5 4 3 2 1 1

33.

5

x2  y2 9 4

y  x2  1

y

31.

1

2

3

4

5

x

x

y 8 7 6 5 4 3 2 1

5 4 3 2 1 1 2

x

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y

53.

y

55.

5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

6

1 2 3 4 5

x

x2 9

5 4 3 2 1 1 2

1

2

3

4 5

x

5

1. 2 110 2. 189 3. x  5 or x  1 4. x  3 5. Center (12, 3); r  4 6. Center 17, 52; r  9 7. Center 13, 82; r  2 15 8. Center 11, 62; r  412 9. a. x2  y2  64 b. 1x  82 2  1y  82 2  64 10. 1x  62 2  1 y  52 2  10 11. 1x  22 2  1y  82 2  8 1 2 12. ax  b  1 y  22 2  4 2 1 2 49 13. 1x  32 2  ay  b  9 14. x2  y2  3 4 15. x2  1 y  22 2  9 16. 14, 22 17. 11, 82 18. Vertical axis of symmetry; parabola opens downward 19. Horizontal axis of symmetry; parabola opens right 20. Horizontal axis of symmetry; parabola opens left 21. Vertical axis of symmetry; parabola opens upward y y 22. 23. 5 4 3 2 1

5 4 3 2 1 1 2 3 4 5 1 2 3 x  (y  1)2 4 5

x

y

24. 1 2 y x 4

5 4 3 2 1

5 4 3 2 1 (2, 0) 1 2 3 4 x  2 5

2

x  2y 1 (0, 0) 3

4

5

x

2

3

1

4

5

31.

(1, 0)

5 4 3 2 1 1 2 3 4 5

26. y  1x  32  4; vertex: (3, 4); axis of symmetry: x  3 27. x  1y  22 2  2; vertex: (2, 2); axis of symmetry: y  2 1 1 2 28. x  4 ay  b  1; vertex: a1, b ; 2 2 1 axis of symmetry: y  2 1 2 1 1 1 29. y  2 ax  b  ; vertex: a , b ; 2 2 2 2 1 axis of symmetry: x   2

5 4 (0, 3) 3 2 1

x2  4y2  36

(6, 0)

y0 3

4

5

6

x

4

5

6

x

32. Center: (5, 3) y 2 1

(5, 1)

x 2 1 1 2 3 4 5 6 7 8 1 2 3 (3, 3) (7, 3) (5, 3) 4 5 6 7 8

(5, 7)

33. Center: (0, 2) y

(5, 2)

2

5

(6, 0)

6 5 4 3 2 1 1 2 3 1 2 3 (0, 3) 4 5 6

5 4 3

1

4

y

x

y  (x  2)2

5 4 3 2 1

(3, 0)

6

y

25.

5 4 3 2 1 1 2 1 2 3 4 x0 5 2

1



(0, 5)

5 4 3 2 1

(3, 0)

3 4

5 4 3 2 (0, 1) 1

y2 25

6 5 4 3 2 1 1 2 3 1 2 3 4 5 (0, 5) 6

Chapter 9 Review Exercises, pp. 798–801

y1

y

30.

5 4 3 2 1

x

(0, 5)

2 (0, 2) 1

(5, 2)

5 4 3 2 1 1 2 3 1 (0, 1) 2

4

5

x

3 4 5

34. Horizontal 35. Vertical 36. Vertical 37. Horizontal y 38. 6 2

x 4

 y2  1

(2, 0)

5 4 3 2 1

6 5 4 3 2 1 1 2 3 4 5 6

(2, 0) 1

2

3

4

5

6

x

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y

39.

47. a. Circle and line y b.

6 5 4 (0, 4) 3 2 1

y2  x2  16

6 5 4 3 2 1 1 2 3 1 2 3 ( 0, 4) 4 5 6

4

5

5 4 3 2 1 2

1 2 3 4

48.

5

8

10 8 6 4 2 2

5 5 c. e a , b, 14, 112 f 2 4 45. a. Line and parabola y b.

(3, 1) 1 2

3 4

x

12

5 4

1 3 2 1

2

4 6

8 10

x

5 4 3 2 1 1 2

1

2

3

4 5

1

2 3

4 5

x

3 4 5

2

y

57.

5 4 (0, 3) 3 2 1

12 9 ,  bf 5 5

y2 x2 25  4

2 3 4 5

c. 5 15, 152, 13, 12 6 46. a. Circle and line y b.

c. e 10, 32, a

1

2 2 3 (x  3)  (y  1)  9 2 1 x 5 4 3 2 1 1 2 3 4 5 6 1

9

4 5

(165 , 125 )

5 4

6

2 3

x

y

56.

6 3

1

5

4 6 8 10

18 15 12 9

5 4 3 2 1 1

3 4

y2 x2 16  81 

8 6 4 2

6

6 5 4 3 2 1 3

2

16 12 ,  bf 5 5 7 13 16 14 e 10, 22, a , b f 49. e a , b, 15, 12 f 9 9 5 5 5 18, 42, 12, 226 51. 5 12, 42, 12, 42, 1 12, 22, 112, 226 5 13, 12, 13, 12, 13, 12, 13, 126 5 16, 52, 16, 52, 16, 52, 16, 526 y y 55. 10

x

4

(5, 15)

1

c. e 10, 42, a

50. 52. 53. 54.

(52 , 54 )

2

5 43 2 1 1

4 (0, 4) 5

12 10 8 6 4

x

2 3

40. Hyperbola 41. Ellipse 42. Ellipse 43. Hyperbola 44. a. Line and parabola y b. (4, 11)

6

5 4 3 2 1

3 4

5

x

5 4 3 (x  2)2  (y  1)2 4 2 1 x 5 4 3 2 1 1 2 3 4 5 1

2 3

(125,  95)

4 5

58.

y2 x2  4 ⱕ

1 5 4 3 2 1 1 2 3 4 5 1 2 y (x  1)2 3 4 5

y

59.

y 5 4 3 2

1

5 4 3 2 1

x

5 4 3 2 1 1 2 3 4 5

x

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60.

y

61.

y

5 4 3 2 1 1 2 3 4 5

1

2

3

4

5

x

5 4 3 2 1 1 2 3 4 5

y

9. 7 6

5 4 3 2 1

5 4 3 2 1

(4, 3) 1

2 3

4

5

x

(4, 2)

5 1 5 2. Center: a ,  b; r  6 3 7 3. Center: 10, 22; r  3 4. a. 15 b. x2  1y  42 2  5 5. The center is the midpoint (3.8, 1.95). y 6.

11109 8 7 6 5 4 3 2 1 1

y2

(3, 2)

6 5 4 3 2 1 1 2 3 4 5 6 1 2 3 4 x  (y  2)2  3 5 6

5 4 3 2 1

6 5 4 3 2 1 1 2 x2 3 4 5 6

y

10. 2

7

2

y x  1 1 4

6 5 4 3 2 1

1

x y  1 16 49

(4, 0)

2

3

4

7

2

3

4

5

6

x

1

2

3

4

5

x

2 3 4

y  (x  2)2 1

5 7

17.

(0, 7)

6

7 6 5 4 3 2 1 1 2 3 4 5 6 7

(4, 0) 1

2

(0, 7)

3

4

5

6

7

x

5 4 3 2 1 1 2 3 4 5

5 4 3 2 1 1 2 1 y  3 x  13 4 5

18.

y

1 2

3

4 5

1

3

4

x

y 5 4 3 2 1

5 4 3 2 1

5 4 3 2 1

x

5 4 3 2 1

2 3 x y 1 2 1

5 4 3 2 1 1

y 2

1

5 4

8. 2

6

5 6 7

7

6

(2, 1)

5

x

11. 5 13, 02, 10, 426; graph b 12. 5 6; graph a 13. The addition method can be used if the equations have corresponding like terms. 14. 5 12, 02, 12, 026 y y 15. 16.

y

7.

2 3

3 4

7 6 5 4 3 2 1 1 2 3 4 5 6

x

1

2

6 5 4 3 2 1

(1, 3)

1

Chapter 9 Test, pp. 801–802 1. 2 110

5 4 3 2

(4, 4)

(9, 3)

1

2

3

4

5

x

5 4 3 2 1 1 2 3 4 5

2

Chapters 1–9 Cumulative Review Exercises, pp. 803–804 1. 5x | x is a real number6 10 ) 2. a ,  b 10 3 3

5

x

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4. a. 15, 02, 10, 32

3. 10, 15

b. m 

3 5

39. y2  x2  1

y

c.

5 4 3 2 1

5y  3x  15  0

7 6 5 4 3 2 1 1 2 3

1

2

3

40.

y

5 4 3 2 1 1 2 3 4 5

x

y 5 4 3 2 1

5 4 3 2 1 1 2

3

4

5

x

5 4 3 2 1 1 2 3 4 5

1

2

3

4

x

5

4 5

3 3 5. slope: ; y-intercept: a0,  b 6. 12 dimes, 5 quarters 4 2 5 3 7. 5 12, 3, 126 8. e a , b f 9. 5 16, 326 2 2 10. f 102  12; f 112  16; f 122  10; f 142  16 11. 14, 22 12. z  32 13. 1g ⴰ f 21x2  x  7; 14. a. 5 b. 1x  12 1x2  12; 5 x  1 c. They are the same. 15. 1x  y21x  y  62 a1 3 3 2 16. x  2x  7x  4 17. e 6, f 18. 2 a2 x2  x  4 19. 20. 50, 16 21. a. 5 6 1x  32 1x  22 12 15 b. 5116 22. 30  24i 23.  i 41 41 24. 2 17 m 25. a. 17.6 ft; 39.6 ft; 70.4 ft b. 8 sec 1 1 13 26. e  , 27. 52  2116  if 5 10 10 28. 15, 362 29. a. 13, 02, 11, 02 b. 10, 32 c 11, 42 y

5 4 (0, 3) 3 2 1

(1, 4)

(3, 0)

(1, 0)

6 5 4 3 2 1 1 2 3

1

2

3 4

x

31. a,

1 9 d ´ c , b 2 2

5 4 3 2 1

1 37. a , 0b 2

1. 4. 7. 9. 11.

2. 26, 34, 42 3. 48, 96, 192 5. 25, 36, 49 6. 324, 972, 2916 x y , x y , xy6 8. a5b4, a4b5, a3b6 2 2 a  2ab  b 10. a3  3a2b  3ab2  b3 n1 13, 16, 19

1 1 1 9 , 27 , 81 3 4 2 5

Section 10.1 Practice Exercises, pp. 811–812 3. a3  3a2b  3ab2  b3 5. 1  4g  6g2  4g3  g4 7 6 2 5 4 7. p  7p q  21p q  35p4q6  35p3q8  21p2q10  7pq12  q14 9. 625  500u3  150u6  20u9  u12 11. 120 13. 1 15. False 17. True 19. 6!  6  15  4  3  2  12  6  5! 21. 1680 23. 6 25. 56 27. 1 29. m11  11m10n  55m9n2 31. u24  12u22v  66u20v2 33. 9 terms 35. s6  6s5t  15s4t 2  20s3t 3  15s2t 4  6st 5  t 6 37. b3  9b2  27b  27 39. 16x4  32x3y  2 2 3 4 14 41. c  7c12d  21c10d 2  24x y  8xy  y 8 3 6 4 4 5 35c d  35c d  21c d  7c 2d 6  d 7 1 5 5 5 5 5 43. a  a4b  a3b2  a2b3  ab4  b5 32 16 4 2 2 45. 462m6n5 47. 495u16v4 49. g9 5. 16x4  32x3z  24x2z2  8xz3  z4 2 3 4 5 7. 4, 7, 10, 13 9. 13, 2, 15, 16 11.  , ,  , 3 4 5 6 13. 0, 3, 8, 15 15. 0, 2, 6, 12 17. 3, 9, 27, 81 19. When n is odd, the term is negative. When n is even, the term is positive. 21. an  2n 23. an  2n  1 1 n1 25. an  2 27. an  112 29. an  112n 2n n 3 31. an  n 33. $60, $58.80, $57.62, $56.47 5 35. 25,000; 50,000; 100,000; 200,000; 400,000; 800,000; 1,600,000 37. A sequence is an ordered list of terms. A series is the sum of the terms of a sequence. 31 73 39. 90 41. 43. 30 45. 10 47. 49. 38 16 12 3. 28

2 3 33. e f 34. h1 1x2  1 x1 3 35. x2  1y  52 2  16 y 36.

5 4 3 2 1 1 2 3 4 5

Chapter Opener Puzzle

Section 10.2 Practice Exercises, pp. 818–820

4 5

30. 11, 192

Chapter 10

1 2

3

4 5

x

32. log8 32 

5 3

51. 1

6

53. 55

4

38. 5 10, 426

61. a 112 k1

k1

1 3k

5

55. a n

57. a 4

n1 5

i1

n

63. a x n1

5

59. a 4j j1

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65. 3, 2, 7, 12, 17 67. 5, 21, 85, 341, 1365 69. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55

Chapter 10 Review Exercises, pp. 839–840

Section 10.3 Practice Exercises, pp. 824–826 3. 1, 1, 1, 1 11. 17. 21. 27. 33. 41. 49. 59.

5. 55

9. 3 7 5 2 13. 3, 8, 13, 18, 23 15. 2, , 3, , 4 2 2 2, 2, 6, 10, 14 19. an  5  5n 1 3 an  2n 23. an   n 25. an  25  4n 2 2 31. a9  47 29. a6  17 an  14  6n a7  30 35. a11  48 37. 19 39. 22 23 43. 11 45. a1  2, a2  5 47. 670 290 51. 15 53. 95 55. 924 57. 300 210 61. 5050 63. 980 seats; $14,700 7. 2

Section 10.4 Practice Exercises, pp. 831–834 3. an  4n  8

5. 25

7. 2

9. 

1 4

11. 2

3 3 , 4 8 an  3142 n1 1 21. an  5132 n1 23. an  12 142 n1 25. 64 1 243 27.  29. 48 31. 9 33. 35. 4 8 8 37. A geometric sequence is an ordered list of numbers in which the ratio between each term and its predecessor is constant. A geometric series is the sum of the terms of such a sequence. 1562 11 3124 665 39. 41.  43. 45. 125 8 27 243 47. 172 49. a. $1050.00, $1102.50, $1157.63, $1215.51 b. a10  $1628.89; a20  $2653.30; a40  $7039.99 1 9 1 6 51. r  ; 53. r   ;  6 5 3 4 3 55. r   ; sum does not exist 2 57. The sum is $800 million. 59. The total vertical distance traveled is 28 ft. 7 1 7 61. a. b. c. 63. a. $1,429,348 10 10 9 b. $1,505,828 c. $76,480 3 15. 6, 3, , 2 17. 1, 6, 36, 216, 1296 19.

13. 3, 6, 12, 24, 48

Section 10.4 Problem Recognition Exercises, p. 834 1. Geometric, r  

2. Geometric, r  

3 2

2 1 4. Arithmetic, d  2 3 Geometric, r  2 6. Geometric, r  3 Arithmetic, d  2 8. Arithmetic, d  4 Neither 10. Neither 11. Neither Neither 13. Geometric, r  1 Geometric, r  1

3. Arithmetic, d  5. 7. 9. 12. 14.

1 2

1. 5. 6. 7. 9.

40,320 2. 120 3. 66 4. 84 x10  20x8  160x6  640x4  1280x2  1024 c4  12c3d  54c2d 2  108cd 3  81d 4 8. 1512x10y3 a11  22a10b  220a9b2 10. 160a3b3 11. 1, 2, 5, 8, 11 15,000x3y21 2 4 1 1 3 2 8 16 12. 2, 16, 54 13. ,  , ,  14.  , ,  , 3 2 5 3 3 9 27 81 n 3 15. an  16. an  112 n  n 17. k 18. 5 n1 2 7 3i n 19. 5 20. 22 21. a 22. an  n2  a bh i 2 i1 4 5 2 23. 12, 10.5, 9, 7.5, 6 24. , 1, , , 2 3 3 3 9 25. an  10n  9 26. an  14n  20 27. a17  2 28. a25  155 29. 24 terms 30. 19 terms 31. d  10 32. an  5n  19 33. 620 34. 92.5 35. 294 36. 5989 37. r  3 1 1 1 38. r  1.2 39. 1, ,  , 40. 10, 20, 40, 80 4 16 64 1 n1 128 41. an  4122 n1 42. an  6 a b 43. a6  3 243 44. a4  270 45. a1  4 46. a1  1 47. 1275 25 182 48.  49. 18 50. 3 2 51. a. $10,700, $11,449, $12,250.43 b. The account is worth $12,250.43 after 3 yr. c. $19,671.51

Chapter 10 Test, pp. 840–841 1. 1 2. 210 3. a4  4a3b  6a2b2  4ab3  b4 4 3 2 4. 81y  216y x  216y2x4  96yx6  16x8 3 3 1 5. 56a3c15 6. 1,  ,  ,  7. 65 4 5 2 8. a. 8, 9.5, 11, 12.5, 14 b. an  1.5n  6.5

4

9. a x3n n1

1 1 10. d  11. r  12. a. 15, 18, 21, 24 4 3 b. an  3n  12 c. a40  132 13. a. 4, 8, 16, 32 3 1 n1 n1 b. an  4122 c. 2048 14. an   a b 5 2 15. an  2n  22 16. 31 terms 17. 7 terms 1023 1 18. 10,900 19. 20. 8 21. a1  64 27 22. a18  202 23. a. $1095 b. $1160.70 c. $5933.13 d. $86,568.71

Chapters 1–10 Cumulative Review Exercises, pp. 842–843 1. x2  x  13 2c  5 21c  42 a1 9.  a 5.

12. x  9y

20a6 b10 xy 6.  xy 2.

10.

3.

1 16

7. 4

4.

110 2

3 8. 3xy4 2 2x2z2

2x  5 1x  521x  221x  102

13. 2x3  3x2  6x  17 

11. 7x2 22x 27 x2

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c. 13, 42 39. y  5x  18 2 d. 1, 42 e. 1, 3 4 43. 511, 2, 326 2A  hb2 44. a. 2213 b. 16, 22 45. b1  h 46. t192  4.2; This means that it will take the rat 4.2 min if the rat has practiced 9 times. 47. The car is moving 75 mph. 48. The angles are 57º and 123º. 49. She invested $8000 at 6.5% and $5000 at 5%. 50. The plane travels 500 mph in still air. The wind speed is 50 mph.

14. 12a  72 13a  22 15. 3c15c  2215c  22 16. 13x  4y2 2 19x2  12xy2  16y4 2 17. 1w  921w  221w  22 18. 546 19. 506 5 23 2 if 20. e f 21. e  22. 54, 4, 1, 16 4 5 5 7 3 23. e ,  f 24. 516 25. 526 (The value 5 2 4 x1z does not check.) 26. ln a 5 b 27. 3.4929 y 28. 3 2, 52 29. 3 5, 0 4 ´ 31,  2 30. 556 31. 1, 42 ´ 31,  2 32. 1, 12 ´ 13,  2 y y 33. 34. 5 4 (x  2)2  y2  9 3 2

1

1 1

2

3 4

5

x

3

4 5

4 5

36.

y

x

5 4 3 2 1 1 2

3

35. y2

37. 38. 40. 41.

5 4 3x  2y  6 3 2 5 4 3 2 1 1 2

2

3 4

5

1

2

3 4

5

Section A.1 Practice Exercises, pp. A9–A11 x

y

1

1 5 4 3 2 1 1 2

1

1

2

3 4

5

x

5 4 3 2 1 1 2

3

3

4 5

4 5

a. 11, 02, 15, 02 b. 10, 52 slope 4; y-intercept 10, 62 a. 2 b. undefined c. x  42. 513, 126 1, 5 4

Additional Topics Appendix

5 4 3 2

5 4 3 2

SA-59

x

3. 16 5. 1 7. 4 9. 6 11. 26 13. 6 15. 46 17. a. 30 b. 30 19. Choosing the row or column with the most zero elements simplifies the arithmetic when evaluating a determinant. 21. 12 23. 15 25. 0 27. 8a  2b 29. 4x  3y  6z 31. 0 33. D  16; Dx  63; Dy  66 35. {(2, 1)} 23 9 37. {(4, 1)} 39. e a , b f 41. x  1 13 13 16 1 43. z  45. y  2 41 47. Cramer’s rule does not apply when the determinant D  0. 49. { }; inconsistent system 51. {(0, 0)} 53. Infinitely many solutions; 5 1x, y2 0 x  5y  36; dependent system 55. {(3, 2, 1)} 57. { }; inconsistent system 59. x  19 61. w  1 63. 36 65. a. 2 b. 2 c. x  1

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Application Index Biology/Health/ Life Sciences Adenosine deaminase levels in humans, 126 Age range for average height of boys, 84 Amount of medicine prescribed by weight of patient, 475 Average weight for boys/girls based on age, 155 Bacteria population as function of time, 627, 681, 728, 740–741, 743, 819 Bison population in Wyoming, 466 Blood pressure, normal range of, 16 Calcium intake program, 310 Calories in candy, 465 Cholesterol level ranges, 124, 459 Epidemic deaths as function of time, 716 Fat amount in foods, 465 Feasible cage configurations at animal shelter, 272 Femur length vs. height in women, 182, 186 Hemoglobin range in humans, 96 Life expectancy vs. year of birth, 229 Longevity of animals, by species, 183 Maximum heart rate vs. age, 124, 189 Normal level of TSH for adults, 92 Number of manatees in Florida, 466 pH level in ammonia, 688 pH level in antacid tablet, 738 pH level in blood, 693 pH level in shampoo, 689 pH ranges in food substances, 16, 738 Platelets range in humans, 96 Pollution in atmosphere, 476 Radioactive decay, 710–711 Ratio of cats to dogs in shelter, 459

Sodium content of foods, from total sodium intakes, 283 Time for rat to run maze, 843 Weight vs. age for children, 155 White blood cell range in humans, 96

Business and Economics Airline flights as function of passenger volume, 653 Amount of each nut in nut mixture, from relative amounts, 287 Cell phone subscription trends, 155 Commission, 66 Cost of airline operation, 548 Cost of cab ride per mile, 143 Cost of running business, 224 Cost per item as function of number produced, 400, 476 Cost to package product, 626 Currency exchange rates, 486 Depreciation, 135–136, 740 Drinks, price of related to number sold, 180 Earnings needed to meet average salary level, 84 Hot dogs, price of related to number sold, 179 Interest compound, 582, 713, 728, 739, 742 simple, 60, 61–62, 66–67, 116, 263, 474, 476, 489 Markups/markdowns, 60–61, 66 Minimum wage trends, 171–172 Mixing foods for resale, 67, 287 Mixing money among investments, 61–62 Monthly salary, 81 Number of fish, from weight of total catch, 465 Number of registered passenger cars in U.S., 81–82 Orders processed per day by each worker, 310 Profit as function of number of items produced, 84, 277, 332, 582, 662

Profit as function of number of items sold, 84, 476, 649, 662 Restaurant chain opening trend analysis, 176 Salary plan, 672–673 Salary with commissions, vs. sales, 66, 143, 175 Salary with tips, vs. number of tables served, 226 Sales as function of advertising expenditures, 694 Sales by each salesperson, from relative totals, 283 Tickets sold, by type, 262, 286, 306 Time required for each person working alone to finish job, 468 Time required for two workers to finish job, 278, 468 Total economic impact of tourist spending, 833 Total revenue from sold-out theater, 826 Total salary over period, 834

Construction and Design Bridge support design, 617 Bulge height in heat-expansion of bridge, 568 Cell tower location, 788 Cost of carpeting room, 663 Dimensions of fenced area, 627–628 Elevator passenger capacity, 126 Fencing needed to enclose area, 73, 277, 516, 627 Length of roof, from structure dimensions, 571 Length of screw, 125 Length of tower guy wires, 569 Margin of error in measurement, 112, 560 Maximum enclosable area with given amount of fence, 73 Outdoor fountain water projection, 766

I-1

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Application Index

Pump performance, from relative performance, 308, 468, 483 Slope of hill, 153 Slope of ladder, 146, 152, 399 Slope of leaning telephone pole, 414 Slope of roof, 146, 152 Slope of steps or ramp, 146, 223 Slope of treadmill, 153 Strength of beam as function of dimensions, 477 Time required for two workers to finish job, 278, 463–464, 486 Time required to fill tank, from pipe size, 486 Window film needed for project, 26

Consumer Applications Car rental cost vs. mileage, 265 Cellular phone charges, 172 Cellular phone subscriptions in U.S., 155 Coins by type, from relative number, 262, 265, 311 Cook time vs. weight for turkey Cost of bottle of laundry detergent by size, 465 Cost of cab ride, 143 Cost of car rental, 175 Cost of items, from money spent, 262, 287, 466 Dog run dimensions, 73, 277 Gas mileage, 177, 466 Gas mileage as function of speed, 627 Home cost trends, 167, 177, 833 Income after taxes, 122 Income tax owed, 122 Lengths of rope needed for rope trick, 122 Loan amount borrowed, from interest paid, 67, 116 Loan payment amounts, 819 Measurement error in contents of food packaging, 112 Mixing products, 67, 287 Number of books sold inversely varies by price, 476 Number of people served by turkey, by weight, 476 Property tax trends, 176 Rabbit pen dimensions, 74 Sales tax, 66

Satellite dish depth, 766 Spending on snacks at movie, 231, 257 Telephone company charges compared, 265, 306 Television audience viewing American Idol, 125 Textbook cost vs. size, 174 Theater seating, 826 Video rental costs compared, 311 Wedding spending per year, 66

Distance/Rate/Time Airplane landing distance vs. speed, 647–648 Altitude gain over horizontal distance, 153 Average speed of airplane, 68 Average speed of boats, 68 Average speed of bus and train, 462–463 Average speed of car and train, 463 Average speed of car overtaking bus, 468 Average speed of racecar, 75 Average speed of walker, 467 Car braking distance vs. speed, 546, 596 Distance as function of time, 804 Distance as function of wheel revolutions, 663 Distance between cities, from map, 466, 488, 489, 586, 648 Distance between vehicles/walkers traveling apart, 498, 501, 586 Distance between vertices in arch, 774 Distance from earth to celestial objects, 319–320, 322 Distance of lightning strike, calculating, 176 Distance to location, 125, 399, 755 Distance traveled, from time and speeds, 201 Distance traveled by bungee jumper, 833 Gas remaining vs. time, 136 Height vs. growth rate, 840 Height vs. horizontal distance, 333, 412, 516, 626, 651, 766, 774 Height vs. rebounds of ball, 816, 830–831, 833

Height vs. time in free fall, 201, 397, 549 Height vs. velocity, 548, 587, 596, 617, 639, 652 Maximum height of launched object, 617, 626, 649 Speed of airplane discounting wind speed, 260, 263, 306, 843 Speed of airplane with tailwind, 163, 260–261 Speed of bicyclist against wind, 285, 467, 486, 488 Speed of bicyclist discounting wind speed, 285, 486, 488 Speed of boat as function of current, 263, 264, 467 Speed of boat discounting water speed, 263, 264, 467 Speed of car as function of time, 488, 843 Speed of cars/travelers, from distance of separation, 68, 122, 467 Speed of driving and walking, 486 Speed of hiker, 64, 68 Speed of motorist in rainstorm, 467 Speed of moving sidewalk, 263, 467 Speed of runner and biker, 461–462 Speed of travelers, from relative speeds, 516 Speed of triathlon bicyclist, 263 Speed of triathlon runner, 467 Speed of walking and cycling, 467 Speed of walking on nonmoving ground, 263, 467 Speed vs. time for vehicle, 467 Stopping distance vs. speed, 476 Time of car races, 75 Time of height after launch, 396, 400, 411, 622–623 Time required for object to fall, 397, 539, 546, 580, 583 Time required for pendulum swing, 477, 488, 539, 548 Time required for rat to run a maze, 843 Time required for travelers to meet, 122 Time required for trip, 125 Time required for two workers to finish job, 278, 468 Time required to overtake traveler, 742 Time required to reach investment goal, 75, 84 Wind speed, 260–261, 285, 306, 843

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Application Index

Education and School Associate degrees conferred, trends in, 179 College attendance by males, trends in, 122 College enrollment, from relative enrollments, 286 College enrollment trends, 223 Dormitory fee trends, 332 Grade average relating to final exam, 96–97, 213 Grades needed to make desired average, 80–81, 84, 124, 414 Medical degrees awarded to women, 66 Number of men enrolled in math, 466 Number of women graduated from law school, 172–173 Ratio of children to adults, 465 Ratio of passing to failing students, 460 Ratio of single men to single women in college, 466 Retention of material as function of time, 689–690, 739, 741 Student-teacher ratios Study time availability analysis, 270–272 Study time vs. grade, analysis of, 180–181, 202 Teaching methods of vocabulary, 693–694 Textbook cost vs. size, 174

Environment/Earth Science/Geography Annual loss due to earthquakes in California, 323 Atmospheric pressure as function of altitude, 742 Average low/high temperature, 29, 97 Average temperature by city, 670 Conversion of degrees Celsius to/from Fahrenheit, 29 Hurricane intensity, 12 pH level of rainwater, 688 Population relating to pollution, 476 Precipitation relating to month of year, 189 Richter scale for earthquakes, 704 Slope of altitude increase of airplane, 153

Slope of hill, 153 Snowfall rate and totals, 170 Visibility vs. altitude, 487 Weak earthquakes, 788 Wind chill vs. wind speed, 178 Yearly rainfall for two cities, 231

Gardening and Landscaping Acreage for planting fruit trees, 307 Area of walkway and garden as function of walkway width, 342, 409 Dimensions of area enclosable with available fencing/edging, 73, 277 Dimensions of garden, from perimeter, 73, 117 Edging around tree, 74 Fertilizer mixtures, 67 Height of tree, from shadow cast, 460–461 Perimeter of garden as function of side length, 332 Radius of garden, from area, 650 Time required to dig ditch for piping, 571

Geometry Angle measure of supplementary angles, 264 Angle measures of complementary angles, 264, 306, 310 Angle measures of triangle, 264, 282–283, 286, 308, 565 Area as function of side length, 26, 38, 340, 342, 409, 477, 509, 652 Area of region, 343, 385, 408, 476 Dimensions of region, from area, 399, 582, 648, 650 Dimensions of region, from perimeter, 68–69, 401 Dimensions of region, from volume, 582, 595 Height of triangle, 399, 647 Length of triangle side, 73, 125, 308, 394, 399, 466, 516, 582, 595 Perimeter as function of side length, 265, 286, 328, 332, 409, 488 Pythagorean Theorem, 393, 522

I-3

Radius of circle, from area, 70–71, 399, 401, 650 Radius of sphere, from volume, 509, 582 Volume as function of side length, 339–340, 342–343, 411, 545 Volume of figure, 30

Investment Amount in account as function of time, 681, 729, 833 Amount invested, from return, 122, 125, 259, 263, 265, 301, 306, 362, 652 Annual rate of return, 507, 509, 840 Bond yield vs. price, 841 Interest compound, 582, 713, 728, 739, 742, 819, 841 simple, 60, 61–62, 67, 263, 474, 476, 489, 843 Savings deposits per month, 323 Savings totals with monthly deposits, 171, 478–479, 841 Time required to reach investment goal, 75, 84, 716

Politics and Public Policy Child support due vs. child support paid, 333 Economic bailout, 319 Margin of error in polls, 112 Minimum wage trends, 171–172 Number of representatives in House of Representatives, 669

Science and Technology Acceleration of mass influenced by force, 75 Atoms in substance, number of, 322–323 Conversion of degrees Celsius to/from Fahrenheit, 29 Depth of water, from wave velocity, 569 Distance from earth to celestial objects, 319–320, 322 Distance of lightning strike, calculating Electrical current vs. voltage and resistance, 476, 477

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Application Index

Electrical resistance vs. wire size, 476 Force on spring vs. distance stretched, 176, 209, 477, 487 Force required to stretch spring, 176, 209 Frequency vs. length in vibrating string, 476 Height vs. time in free fall, 201, 397 High-energy particle accelerator, 322 Intensity of light as function of distance, 476 Kinetic energy, 473–474 Mass of object, 126 Mixing solutions, 62–63, 67, 122, 125, 231, 258, 262–264, 306, 310, 466 Pendulum period vs. length, 477, 488, 539, 548 pH measurement, 688–689, 693, 738 Radioactive decay, 676–677, 680, 725–726, 728–730, 738–740, 819 Sequence of heights in bouncing ball, A–21, A–33 Sound intensity in decibels, 472–473, 704, 729 Thickness of metal measured by machine, 110 Volume of gas as function of temperature and pressure, 743 Weight of ball as function of radius, 477

Sports/Exercise/ Entertainment

Statistics and Demographics

Daytona 500 car race speed, 75 Dimensions of court, 73, 393 Distance from home plate to second base, 516 Dominoes array, 826 Golf holes by par type, 287 Height of object vs. time, 297–298 Indianapolis 500 car race speed, 75 Length of track for track and field racing, 670 Men’s swimming times, trends in, 178 Points scored, by type of shot, 264 Points scored, from relative scores, 413 Projected yardage gain from number of passes, 486 Spending on snacks at amusement park, 257 Tickets sold by type, 264, 286, 306 Women’s track and field records, trends in, 190 Yards rushed vs. margin of victory, 224–225

Alcohol-related deaths, trends in, 122 Bottled water consumption, 227, 408 Child support due vs. child support paid, 333, 662 Fatality rate for drivers, 596 Males employed full-time in U.S., 151 Margin of error in polls, 112 Miles driven per year, trends in, 177 Number of representatives in House of Representatives, 183 Numbers, from relative size, 125 Per capita yearly income in U.S., 219 Percentage of males who smoke, 329 Population density, 151, 322–323 Population density trends, 189, 413, 662 Population growth trends, 82, 329, 333, 408, 413, 648, 681, 724–725, 728, 741 Prisoners in federal or state facilities, 179 Ratio of male to female police officers, 459 Soft drink consumption per capita, trends in, 229 Weekly earnings for women, 213

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Subject Index A Absolute value evaluation of, 17, 97 explanation of, 16–17, 39 solving equations containing two, 101–102 translating to/from words, 109–110 valuation of, 17 Absolute value bars, 496 Absolute value equations explanation of, 97, 120 method to solve, 97–102, 120 Absolute value functions, 203, 205 Absolute value inequalities explanation of, 103, 120 methods to solve, 103–107, 121 test point method to solve, 107–109 AC-method, to factor trinomials, 362–364, 405 Addition associative property of, 32 commutative property of, 32, 35 of complex numbers, 554–555 distributive property of multiplication over, 32, 33, 40–41 of functions, 656–657 identity property of, 32 inverse property of, 32 of like terms, 33–35, 780 of polynomials, 324–325, 403 of radicals, 517–519, 563 of rational expressions, 431–432, 435–438, 481 of real numbers, 17–18, 39 Addition method to solve nonlinear systems of equations, 779–780 to solve systems of linear equations, 249–252, 300 Addition property of equality, 45 Addition property of inequality, 77 Algebraic expressions evaluation of, 26 order of operations for, 24–25

simplification of, 30–35, 40–41, 314–318 Alternating sequences, 814 Angles application involving, 69–70 complementary, 69, 117 supplementary, 69, 117 Applications. See also Applications Index of binomial theorem, 809–810 of complex fractions, 446 of consecutive integers, 58–59 of determinants, A-1 of distance, rate, and time, 63–64, 260–261, 461–463 of exponential functions, 676–678, 705–706, 724–726 of geometric series, 830–831 of geometry, 68–70, 117, 261 of linear equations in two variables, 219 of linear inequalities, 80–82 of literal equations, 70–71, 117 of midpoint formula, 751 of mixtures, 62–63, 258 of natural logarithmic functions, 710–711 of percents and rates, 59–61 of polynomial functions, 328, 329 of principal and interest, 61–62, 259–260 of product of polynomials, 339–340 of proportions, 459–461 of quadratic equations, 392–394 of quadratic formula, 586–587 of radical equations, 545 of radical functions, 546 of rational equations, 458–464, 483 of rational exponents, 507 of relations, 186 of sequences, 816 of similar triangles, 460–461 of slope, 146, 151 of systems of linear equations, 256–261, 301 of systems of linear equations in three variables, 282–283

of variation, 471–474 of work, 463–464 Approximately equal to, 9 Area formulas for, 117 of trapezoid, 71 Argument, of logarithmic expressions, 682 Arithmetic sequences explanation of, 820 finding number of terms in, 822 finding specified term of, 821–822 nth term of, 821 Arithmetic series explanation of, 822–823 sum of, 823–824 Associative property of addition, 32 Associative property of multiplication, 32 Asymptotes, 204 Augmented matrices, 289, 304 Axis horizontal transverse, 769, 795 transverse, 769, 770 vertical transverse, 769, 795 Axis of symmetry, of parabolas, 604, 757–760, 794

B b0, 314 Base of exponential expressions, 22 of logarithmic expressions, 682 Binomial expansions binomial theorem for, 808–810 explanation of, 806, 836 factorial notation to find coefficients of, 807–808 finding specific term in, 810–811 Pascal’s triangle and, 807 Binomial factors, 356, 359 Binomials. See also Polynomials difference of squares to factor, 376–379, 382 explanation of, 324, 403 square of, 806 sum and difference of cubes to factor, 379–380, 382 I-5

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Subject Index

Binomials—Cont. summary of factoring methods for, 380–382, 406–407 Binomial theorem applications of, 809–810 explanation of, 808–809 bn, 314, 315 Boundary points, 107–109, 121, 630, 635 Braces, 24 Brackets, 24 Branches, 769

C Calculators. See Graphing calculators Center, of circle, 747 Change-of-base formula explanation of, 709–710, 735 on graphing calculators, 710 Circles

explanation of, 747, 757 on graphing calculators, 748 graphs of, 747–749 radius of, 747 standard equation of, 747, 793 writing equation of, 750 Clearing decimals, 51 Clearing fractions, 49–51, 450 Clearing parentheses, 34–35 Coded messages, 215 Coefficients explanation of, 31 of polynomial, 323, 324, 403 Column matrices, 288 Commission, 59, 116 Common difference, of arithmetic sequence, 820, 821 Common logarithmic functions, 684–685, 733 Common ratio, of geometric sequence, 826–828, 838 Commutative property of addition, 32, 35 Commutative property of multiplication, 32 Complementary angles, 69, 117 Completing the square explanation of, 576–577, 642 to find vertex formula, 620 to graph parabolas, 758–759, 761–762 to solve quadratic equations, 577–579 Complex fractions in applications, 446 explanation of, 441, 482 methods to simplify, 441–446, 482

Complex numbers addition of, 554–555 division of, 556 explanation of, 553, 577 multiplication of, 555 real and imaginary parts of, 554 simplification of, 556, 557 subtraction of, 554–555 Composition, of functions, 657–658, 731 Compound inequalities applications of, 92–93 explanation of, 89, 118–119 joined by word and, 87–89 joined by word or, 91–92 linear in two variables, 268–270 method to solve, 89–90 Compound interest, 705–706 Concentration rate, in mixture problem, 62 Conic sections, 757. See also Circles; Ellipses; Hyperbolas; Parabolas Conjugates, 336, 526, 536, 566 Consecutive integers, 58–59 Consistent systems of linear equations, 235, 298 Constant functions, 202–203, 221 Constant terms, 31, 40, 170 Continuously compounded interest, 706 Contradictions explanation of, 52, 115 identifying, 52–53 Cost problems, 256–257 Cramer’s Rule explanation of, A-5–A-8 dependent systems of equations and, A-8–A-9 Cubes difference of, 379–382 perfect, 379, 495, 496 sum of, 379–381 Cubic functions, 203, 205

D Decay functions, exponential, 675, 676–677 Decimals, clearing, 51 Degree, of polynomials, 323, 324, 403 Denominators least common, 432–433 rational expressions with like, 431–432 rational expressions with unlike, 435–438 rationalizing, 532–537, 564

Dependent systems of linear equations Cramer’s rule and, A-8–A-9 explanation of, 235, 298 Gauss-Jordan method to solve, 293 in three variables, 284 in two variables, 247 Determinants applications of, A-1 explanation of, A-1 ongraphing calculators, A-4 of 2  2 matrix, A-1–A-2 of 3  3 matrix, A-2–A-4 Difference of cubes, 379–382 Difference of squares explanation of, 336, 337 to factor binomials, 376–379, 381, 382 Directrix, of parabola, 757 Direct variation applications involving, 471, 472 explanation of, 469 Discriminant, of quadratic equation, 587–590, 643 Distance with absolute value, 109 between two points, 746–747 Distance, rate, and time problems, 63–64, 260–261, 461–463 Distance formula, 746–747, 793 Distributive property of multiplication over addition application of, 33, 72 explanation of, 32, 40–41 Division of complex numbers, 556 of functions, 656–657 of inequalities, 77 long, 344–347 of polynomials, 343–347, 404–405 of rational expressions, 427–428, 480 of real numbers, 21–22, 40 synthetic, 348–350 Division property of equality, 45 Division property of inequality, 78 Division property of radicals, 531–532, 564 Domain of functions, 196–197, 656 of inverse functions, 666–667 of logarithmic functions, 685, 687–688 of rational functions, 416, 418 of relations, 182–185, 220 Dot Mode (graphing calculator), 417

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E e (irrational number), 704–705 Elimination method. See Addition method Ellipses explanation of, 757, 766, 795 foci of, 766 graphs of, 767–768 standard form of equation of, 766–767, 795 Empty sets explanation of, 52 symbol for, 86 Equality addition property of, 45 division property of, 45 multiplication property of, 45 subtraction property of, 45 Equal to, 9 Equations. See also Linear equations in one variable; Linear equations in two variables; Literal equations; Quadratic equations; Rational equations; Systems of nonlinear equations; specific types of equations absolute value, 97–102, 120 of circles, 747, 793 conditional, 52–53 contradictions as, 52–53, 115 of ellipse, 766–768, 795 equivalent, 45, 117 explanation of, 44 exponential, 721–724, 736 of hyperbola, 769–770, 795 identities as, 52–53, 115 of inverse of functions, 665–667, 732 of lines, 156–164, 218 literal, 70–72, 580 logarithmic, 718–721, 735 of parabola, 758, 760, 794 quadratic, 388–397, 407, 574–579, 600–601 in quadratic form, 598–600 radical, 540–545, 565 rational, 449–455, 483 solutions to, 44 Equivalent equations, 45 Equivalent rational expressions, 434 Expanding minors, A-3–A-5 Exponential equations applications of, 724–726 explanation of, 721, 736 method to solve, 721–724, 736

Exponential expressions equivalence of, 721 evaluation of, 23 explanation of, 22 on graphing calculators, 23, 673–674 Exponential functions applications of, 676–678, 705–706, 724–726 base 10, 704 decay, 675–677 explanation of, 672–673, 686, 733 graphs of, 674–676 growth, 675, 678 Exponents explanation of, 22, 40 integer, 314–315, 403 properties of, 314–316 rational, 503–507, 562 simplifying expressions with, 314–318 Expressions. See Algebraic expressions; Radical expressions; Rational expressions Extraneous solutions, 452, 540

F Factorial notation, 807–808 Factors/factoring binomials, 356, 359, 376–382, 406–407 difference of squares, 376–379, 381, 382 explanation of, 72 greatest common, 354–355, 357, 359 by grouping, 357–359, 378–379 negative, 355–356 to solve quadratic equations, 389–392 sum and difference of cubes, 379–382 within terms, 31 trinomials, 362–373 using substitution, 272–273 Feasible regions, 270–272 Finite sequences, 813, 835 First-degree polynomial equations. See Linear equations in one variable Focus of ellipse, 766 of hyperbola, 769 of parabola, 757

I-7

Form fitting, 71 Formulas. See also specific formulas as literal equations, 454–455 Fractions clearing, 49–51, 450 complex, 441–446, 482 Function notation explanation of, 193–195, 203, 221 finding intercepts using, 207–208 Functions. See also Exponential functions; Inverse functions; Logarithmic functions; Polynomial functions; Quadratic functions; Radical functions; Rational functions absolute value, 203, 205 algebra of, 656–657, 731 combining, 658–660 composition of, 657–658, 731 constant, 202–203, 221 cubic, 203, 205 domain of, 196–197, 656 evaluation of, 195 explanation of, 191, 220 exponential, 672–678, 733 graphs of, 196, 203–209, 221, 659–660 identity, 203, 205 inverse, 665–668, 732 linear, 202–203, 221 logarithmic, 682–690, 733 multiple operations on, 658–660 natural logarithmic, 706–709 nonlinear, 205 one-to-one, 663–665 polynomial, 327–329 quadratic, 203, 206–207, 221, 394–397, 407, 604–612 radical, 498–499, 546 range of, 196, 220 rational, 416–418, 479 reciprocal, 203, 205 relations vs., 191–192 square root, 203, 205 vertical line test for, 192–193, 220 Function values, 195 Fundamental principle of rational expressions, 419, 421 Future value, 478–479

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Subject Index

G Gauss-Jordan method explanation of, 288–289 to solve system of linear equations, 290–294 GCF. See Greatest common factor (GCF) Geometric sequences explanation of, 826, 838 finding specified term of, 827–828 nth term of, 827 Geometric series applications of, 830–831 explanation of, 828, 838 infinite, 830 sum of, 828–830, 838 Geometry applications, 68–70, 117, 261, 282–283 Graph (graphing calculator feature), 138 Graphing calculator features det function, A-4 Dot Mode, 417 Graph, 138 Intersect, 238–239, 722, 778 matrix editor, A-4 Minimum and Maximum, 612 p (pi) key on, 7 Seq, 814, 815 standard viewing window, 138 Sum, 815 Table, 138, 195, 417 Trace feature, 239 Value, 160 zero, 396 Graphing calculators change-of-base formula on, 709, 710 circles on, 748 determinants of matrix on, A-4 exponential expressions on, 23, 673–674 factorial notation on, 807, 808 finite series on, 816 functions on, 195, 748 logarithms on, 685, 688 matrices on, 294 minimum and maximum values on, 612 nonlinear functions on, 205 nonlinear systems of equations on, 778 nth roots on, 495 nth term of sequence on, 814, 816 order of operations on, 25 parabolas on, 606, 761

radical expressions on, 512 radical functions on, 499 square roots on, 493 systems of equations on, 238–239 x- and y-intercepts of functions on, 208 Graphs/graphing. See also Rectangular coordinate system of circles, 747–749 of ellipses, 767–768 of exponential functions, 674–676 of feasible region, 270–272 of functions, 196, 203–209, 221, 659–660 of function values, 196 of f(x)=ex, 705 of horizontal lines, 137 of hyperbolas, 769, 770 of irrational numbers, 705 of linear equations in two variables, 130–133 of linear inequalities in two variables, 268–270, 302 of logarithmic functions, 685–688, 707 of nonlinear inequalities in two variables, 784–785, 797 of nonlinear systems of equations, 776–778 of parabolas, 395, 604–612 of quadratic functions, 395, 604–612, 644 of quadratic inequalities, 629 to solve systems of linear equations, 235–238 of systems of nonlinear inequalities in two variables, 785–786 of vertical line, 136–137 Greater than, 9 Greater than or equal to, 9 Greatest common factor (GCF) explanation of, 354–355, 405 factoring out, 355, 357, 359, 364, 381 Grouping, factoring by, 357–359, 378–379, 405 Grouping symbols, 24 Growth functions, exponential, 675, 678

H Horizontal asymptotes, 204 Horizontal lines, 136, 137, 164, 216, 218

Horizontal line test, 664, 732 Horizontal transverse axis, 769, 795 Hyperbolas explanation of, 757, 769, 795 focus of, 769 graphs of, 769, 770 standard form of equation of, 769–770, 795 vertex of, 769, 794

I Identities explanation of, 52, 115 identifying, 52–53 Identity functions, 203, 205 Identity property of addition, 32 Identity property of multiplication, 32, 421 Imaginary numbers (i) explanation of, 550–551 powers of, 552–553 Inconsistent systems of linear equations explanation of, 235, 298 Gauss-Jordan method to solve, 293 in three variables, 285 in two variables, 246, 253 Independent systems of linear equations, 235, 298 Index of summation, 816, 817, 835 Inequalities. See also Systems of nonlinear inequalities; specific types of inequalities absolute value, 103–109, 120–121 addition property of, 77 boundary points of, 630, 635 compound, 85–93, 118–119, 268–270 division property of, 78 explanation of, 9–10 linear in one variable, 76–82, 118 linear in two variables, 265–268, 302 multiplication property of, 78 nonlinear, 628–633, 783–785 polynomial, 631–633, 646 properties of, 77, 118 quadratic, 628–631 rational, 633–635, 646 with “special case” solution sets, 636–637 subtraction property of, 77 symbols for, 9 translating to/from words in, 12 Infinite geometric series, 830, 838

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Infinite sequences, 813, 835 Integer exponents, 403 Integers consecutive, 58–59 explanation of, 6, 8, 39 Interest applications involving, 61–62 compound, 705–706 compounded continuously, 706 simple, 59, 116 Intersect (graphing calculator feature), 238–239, 722 Intersection, of sets, 85–97, 118–119 Interval notation domain in, 197 explanation of, 10 expressing sets by using, 11–12, 53 union and intersection in, 87 Inverse functions explanation of, 667–668, 732 finding equation of, 665–667 graphs of, 664–667 one-to-one, 663–665, 667 property of, 668, 732 Inverse property of addition, 32 Inverse property of multiplication, 32 Inverse variation applications involving, 471–472 explanation of, 469 Irrational numbers e, 704–705 explanation of, 7, 8, 39, 493, 735

J Joint variation applications involving, 473–474 explanation of, 470, 484

L LCD. See Least common denominator (LCD) Leading coefficient, of polynomial, 324, 369–370, 403 Leading term, of polynomial, 324, 369, 403 Least common denominator (LCD) explanation of, 49 of rational expressions, 432–433, 435 Least common multiple, 251 Less than, 9 Less than or equal to, 9 like radicals, 517, 563

Like terms addition and subtraction of, 33–35 explanation of, 31, 40, 324 Linear equations in one variable applications of, 58–64, 116 clearing fractions and decimals to solve, 49–51 conditional, 52–53 with consecutive integers, 58–59 contradictions as, 52–53 explanation of, 44, 115 identities as, 52–53 methods to solve, 45–49, 115 problem-solving steps for, 57–58, 116 Linear equations in two variables. See also Systems of linear equations applications of, 219 explanation of, 129, 216 graphs of, 130–133 horizontal and vertical lines and, 136–137 solutions to, 130 x- and y-intercepts of, 133–136 Linear functions, 202–203, 221 Linear inequalities in one variable applications of, 80–82 explanation of, 76 methods to solve, 76–80 properties of, 77, 118 Linear inequalities in two variables compound, 268–270 explanation of, 265, 302 graphs of, 265–270, 302 Linear models explanation of, 170, 219 interpretation of, 171–172 method to write, 170–171 from observed data points, 172–174 Lines equations of, 156–164, 218 horizontal, 136, 137, 164, 216, 218 slope of, 145–151 vertical, 136–137, 164, 216, 218 x- and y-intercepts of, 134–136 Literal equations. See also Quadratic formula application of, 70–71, 117 explanation of, 70, 117 involving rational expressions, 454–455 methods to solve, 71–72, 580

I-9

Logarithmic equations explanation of, 718 method to solve, 718–721, 735 Logarithmic expressions equivalence of, 620 evaluation of, 683–684 in expanded form, 698–699, 709 as single logarithm, 700–701 Logarithmic functions applications of, 688–689 common, 684–685, 688–690, 733 conversion to exponential form, 682–683 domain of, 685, 687–688 explanation of, 682, 733 graphs of, 685–688 natural, 706–709 Logarithms change-of-base formula and, 709–710, 735 on graphing calculators, 685, 688 power property of, 697, 698, 701, 707, 734 product property of, 696–698, 707, 734 quotient property of, 697, 698, 707, 734 sum or difference of logarithms as single, 700–701, 708 Long division of polynomials, 344–346 synthetic division vs., 50

M Mathematical models, 170 Matrices augmented, 289, 304 coefficient, 288 column, 288 determinants of, A-1–A-5 explanation of, 288, 304 minor of element of, A-2–A-4 on graphing calculators, 294 order of, 288, 304 reduced row echelon form of, 290–294, 304 row, 288 to solve systems of linear equations, 288, 290–293 square, 288 2  2, A-1–A-2 2  3, 294 3  3, A-2–A-4 Maximum (graphing calculator feature), 612

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Subject Index

Maximum value, 611 Mean, 835 Measurement error, expressing with absolute value, 110 Midpoint formula, 750–751, 793 Minimum (graphing calculator feature), 612 Minimum value, 611 Minors of element of matrix, A-2–A-4 expanding, A-3–A-5 Mixture problems, 62–63, 258 Models. See Linear models Monomials. See also Polynomials division by, 343–344 explanation of, 324, 403 multiplication of, 334 Multiplication associative property of, 32 commutative property of, 32 of complex numbers, 555 distributive property of, over addition, 32, 33, 40–41 of functions, 656–657 identity property of, 32, 421 of inequalities, 77 inverse property of, 32 of polynomials, 334–336, 404 of radical expressions, 526, 527, 563 of rational expressions, 426–427, 480 of real numbers, 20, 40 Multiplication property of equality, 45, 50 Multiplication property of inequality, 78 Multiplication property of radicals, 510–513, 522–524, 563

N Natural logarithmic functions applications of, 710–711 explanation of, 706 graphs of, 707 properties of, 707–709 Natural numbers, 6, 8 Negative factors, 355–356 Negative number, square root of, 24 Negative slope, 147 Nonlinear functions, on graphing calculators, 205 Nonlinear inequalities. See also Systems of nonlinear inequalities graphs of, 784–785, 797 polynomial, 631–633

quadratic, 628–631 rational, 633–635 with “special case” solution sets, 636–637 systems of, 270–272, 785–786 test point method for, 631–635, 646, 797 in two variables, 783–785 Notation/symbols absolute value bars, 496 approximately equal to, 9 determinant, A-1 empty set, 86 equal to, 9 factorial, 807–808 function, 193–195, 203, 207–208, 221 function composition, 657 greater than, 9 greater than or equal to, 9 grouping, 24 inequalities, 9 interval, 10–12, 53 less than, 9 less than or equal to, 9 not equal to, 9 p (pi), 7 radical, 23, 492, 505 scientific, 318–320, 403 set-builder, 6 sigma, 816, 817 Not equal to, 9 nth roots explanation of, 494–495, 561 on graphing calculators, 495 identification of, 495 of real numbers, 495 simplification of, 497 of variable expressions, 495–496 nth term of sequence explanation of, 813, 821, 826–827, 837 formula for, 814 on graphing calculators, 814, 816 method to find, 815 Number line. See Real number line Numbers. See also Complex numbers; Integers; Real numbers imaginary, 550–551 irrational, 7, 8, 39, 493, 704–705, 735 natural, 6, 8, 39 rational, 6–8, 39 whole, 6, 8, 39

O Observed data, linear models from, 171–174 One-to-one functions, 663–665 Opposites, of real numbers, 16, 39 Ordered pairs in rectangular coordinate system, 129, 131 relations as set of, 182–184 as solution to system of linear equations, 234 Ordered triples, 278, 303 Order of matrix, 288, 304 Order of operations application of, 25 explanation of, 24, 40 on graphing calculators, 25 simplifying radicals by using, 513–514 Origin, of rectangular coordinate system, 128

P Parabolas. See also Quadratic equations axis of symmetry of, 604, 757–760 completing the square to find vertex of, 620, 758–759, 761–762 directrix of, 757 explanation of, 206, 221, 394, 604, 618, 757, 794 focus of, 757 on graphing calculators, 660, 762 graphs of, 395, 757–760 standard form of equation of, 758, 760, 794 vertex of, 604, 618–624, 645, 757, 762–763, 794 Parallel lines in slope-intercept form, 158–159 slope of, 149–151, 162–163 Parentheses clearing, 34–35 explanation of, 11, 24 Pascal, Blaise, 807 Pascal’s triangle, 402, 807 Percents applications, 60–61 explanation of, 59 Perfect cubes, 379, 495, 496 Perfect fifth powers, 495 Perfect fourth powers, 495 Perfect squares, 493, 495, 496

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Perfect square trinomials explanation of, 337, 406, 576 factoring of, 370–371 Perimeter application involving, 68–69 formula for, 69, 70 Perpendicular lines in slope-intercept form, 158–159 slope of, 149–151, 163 p (pi), 7, 71 Plotting points, 129 Point-slope formula explanation of, 160–161, 164, 218 use of, 161–163 Polynomial functions in applications, 328, 329 evaluation of, 327–328 explanation of, 327 Polynomial inequalities, 631–633, 646 Polynomials. See also Binomials; Factors/factoring; Monomials; Trinomials addition of, 324–325, 403 applications of product of, 339–340 degree of, 323, 324, 403 division of, 343–347, 404–405 explanation of, 323–324, 403 higher-degree, 391, 392 leading coefficient of, 324, 369–370 leading term of, 324, 369, 403 multiplication of, 334–336, 404 prime, 363, 366 subtraction of, 325–327, 403 translations involving, 338–339 Population growth, 677–678, 724–725 Population models, 730–731 Positive slope, 147 Power, 22 Power property of logarithms, 697, 698, 701, 707 Prime polynomials, 363, 366 Principal and interest problems, 61–62, 259–260 Principal square roots, 23, 40, 492 Problem solving, 57–58, 116. See also Translating to/from English form; Word problems Product property of logarithms, 696–698, 707 Proportions. See also Variation in applications, 459–460

explanation of, 458, 483 methods to solve, 458–459 with similar triangles, 460–461 Pythagorean theorem applications of, 394, 497–498, 746 explanation of, 393, 497, 561

Q Quadrants, of rectangular coordinate system, 128 Quadratic equations. See also Parabolas applications of, 392–394 completing the square to solve, 576–579 discriminant of, 587–590, 643 equations reducible to, 600–601 explanation of, 388, 407 factoring to solve, 389–392 methods to solve, 591–594 quadratic formula to solve, 584–585, 591 solutions to, 579 square root property to solve, 574–575, 577–579, 591 zero product rule to solve, 388–389, 407, 591 Quadratic form, 598–600, 643 Quadratic formula applications using, 586–587 derivation of, 583–584 explanation of, 583, 643 to solve quadratic equations, 584–585, 591 Quadratic functions applications of, 396–397, 622–624 determining vertex and intercepts of, 621–622 explanation of, 203, 206–207, 221, 394–395, 407, 604 in form f(x)  a(x  h)2  k, 610–612, 618–620, 644 graphs of, 395, 604–612, 644 (See also Parabolas) x- and y-intercepts of, 591 Quadratic inequalities explanation of, 628–629 graphs to solve, 629 test point method to solve, 630–633 Quadratic in form, equations, 598–600

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Quotient property of logarithms, 697, 698, 707 Quotients, of functions, 656

R Radical equations applications of, 545 explanation of, 540 method to solve, 540–543, 565 Radical expressions addition of, 517–519, 563 conjugate, 526, 536 division property of, 531–532 like, 517, 563 multiplication of, 526, 527, 563 multiplication property of, 510–513, 522–524, 563 order of operations to simplify, 513–514 rationalizing the denominator of, 532–537 simplification of, 506, 511–514, 525, 562 simplified form of, 530–531 squaring two-term, 525–526 subtraction of, 517–519, 563 Radical functions applications, 546 domain of, 498–499 explanation of, 498 on graphing calculators, 499 Radical notation, 23, 492, 505 Radical sign, 23 Radicand, 534 Radius, of circle, 747 Range of functions, 196, 220 of relations, 182–185 Rate of change, 170, 217 Rate problems, 63–64, 260–261 Rational equations applications of, 458–464, 483 explanation of, 449–450 formulas involving, 454–455 methods to solve, 450–454, 483 Rational exponents applications of, 507 converting between radical notation and, 505 explanation of, 503–504, 562 properties of, 505–506 Rational expressions addition of, 431–432, 435–438, 481 division of, 427–428, 480 equivalent, 434

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Rational expressions—Cont. explanation of, 416, 479 fundamental principle of, 419, 421 least common denominator of, 432–433, 435–438 multiplication of, 426–427, 480 simplification of, 418–422 solving literal equations involving, 454–455 subtraction of, 431–432, 437, 481 Rational functions domain of, 416, 418 evaluation of, 416–417 explanation of, 416, 479 on graphing calculators, 417 Rational inequalities explanation of, 633 test point method to solve, 633–635, 646 Rationalizing the denominator explanation of, 532–533, 564 one term, 533–535 two terms, 535–537 Rational numbers explanation of, 7, 39 identification of, 7 set of, 6, 7 Ratios of -1, 421–422 common, 826–828, 838 proportions and, 458, 483 Real number line absolute value on, 97, 103, 104 explanation of, 6 inequalities and, 9, 76, 77 Real numbers absolute value of, 16–17 addition of, 17–18, 39 division of, 21–22, 40 explanation of, 6 multiplication of, 20, 40 opposite of, 16 ordering, 9–10 properties of, 31–34 set of, 5–10, 16–17, 39 subtraction of, 19, 40 Reciprocal functions, 203, 205 Reciprocals, 20, 39 Rectangles area of, 117 perimeter of, 69, 70 Rectangular coordinate system. See also Graphs/graphing explanation of, 128–129 Reduced row echelon form, 290–294, 304

Relations applications involving, 186 domain and range of, 182–185, 220 explanation of, 182, 220 functions vs., 191–192 Right triangles, 393 Roots nth, 494–497, 561 square, 7, 23–24, 40, 492–494 of variable expressions, 495–497 Row matrices, 288

S Sales tax, 59, 116 Sample mean, 835 Sample standard deviation, 835 Scientific notation application of, 319–320 explanation of, 303, 318 on graphing calculators, 319 writing numbers in, 318–319 Seq (graphing calculator feature), 814, 815 Sequences alternating, 814 in applications, 816 arithmetic, 820–822, 837 explanation of, 812–813, 837 finite, 813, 837 geometric, 826–828 infinite, 813 nth term of, 815 terms of, 813–815 Series arithmetic, 822–824, 838 explanation of, 816–817, 837 finite, 816 geometric, 828–831, 838 Set-builder notation, 6, 11 Sets classifying numbers by, 8 empty, 52, 86 explanation of, 6, 39 intersection of, 85, 118–119 in interval notation, 10–12 of ordered pairs, 182–184 of real numbers, 5–10, 39 solution, 44 union of, 85 Sigma notation, 816, 817 Similar triangles, 460–461 Simple interest, 59, 116

Simplification of algebraic expressions, 30–35, 40–41, 314–318 of complex fractions, 441–446, 482 of complex numbers, 556, 557 of radical expressions, 506, 511–514, 525, 562 of rational expressions, 418–422 of ratios of -1, 421–422 Slope explanation of, 145–146, 217 formula for, 147–149 interpretation of, 151 methods to find, 146, 446 negative, 147 of parallel lines, 149–151 of perpendicular lines, 149–151 positive, 147 undefined, 148 zero, 148 Slope-intercept form equations in, 236 explanation of, 156–157, 164, 218 to find equation of line, 159–160 graph of, 158–160 Solutions to equations, 44, 130 extraneous, 452, 540 to systems of linear equations, 234–235, 279, 298 Solution sets explanation of, 44 inequalities with “special case,” 636–637 Special case products, 336–339, 525–526 Special product formulas, 337 Square matrices, 288, A-1 Square root functions, 203, 205 Square root property explanation of, 574, 642 to solve quadratic equations, 574–575, 577–579, 642 Square roots evaluation of, 24 explanation of, 23–24, 40, 492 on graphing calculators, 493 identification of, 492 as irrational numbers, 7 method to simplify, 493–494 of negative number, 24 positive and negative, 492 principal, 23, 40, 492

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Squares. See also Completing the square area of, 117 of binomials, 337 difference of, 336, 337, 376–379, 382 perfect, 493, 495, 496 sum of, 377 Standard deviation, 835 Standard equation of circle, 747. See also Circles Standard form of ellipse equation, 767 of equations of line, 156, 164, 218 of hyperbola equation, 769 of parabola equation, 760 Study skills, 1–3 Subscripts, 26 Subsets, 6 Substitution method to factor polynomials, 372–373 to solve dependent systems, 247 to solve equations in quadratic form, 598–600 to solve inconsistent systems, 246 to solve nonlinear systems of equations, 776–779 to solve systems of linear equations, 243–245, 299 Subtraction of complex numbers, 554–555 of functions, 656–657 of like terms, 33–35 of polynomials, 325–327, 403 of radicals, 517–519, 563 of rational expressions, 431–432, 437, 481 of real numbers, 19, 40 Subtraction property of equality, 45 Subtraction property of inequality, 77 Sum of arithmetic series, 823, 824 of functions, 656 of geometric series, 828–830 Sum (graphing calculator feature), 815 Summation, index of, 816, 817, 835 Sum of cubes, 379–381 Sum of squares, 377 Superscripts, 6 Supplementary angles, 69, 117 Symbols. See Notation/symbols

Synthetic division to divide polynomials, 349–350 explanation of, 348 long division vs., 350 Systems of linear equations addition method to solve, 249–252, 300 applications of, 256–261, 301 consistent, 235, 298 dependent, 235, 246, 253, 298 explanation of, 234 Gauss-Jordan method to solve, 290–293 on graphing calculators, 238–239 graphs to solve, 235–238, 298 inconsistent, 235, 246, 253, 298 independent, 235, 298 matrices to solve, 288–294 solutions to, 234–235, 298 substitution method to solve, 243–245, 299 Systems of linear equations in three variables applications of, 282–283 Cramer’s rule to solve, A-5–A-9 dependent, 284 explanation of, 278, 303 inconsistent, 285 methods to solve, 280–282 solutions to, 279 Systems of nonlinear equations addition method to solve, 779–780 explanation of, 776–777, 796 on graphing calculators, 778 graphs of, 776–778 substitution method to solve, 776–779 Systems of nonlinear inequalities feasible region and, 270–272 in two variables, 785–786, 797

T Table (graphing calculator feature), 138, 195, 417 Terms coefficient of, 31 constant, 31 explanation of, 30, 40 factors within, 31 like, 31, 33–35 variable, 31 Test point method for absolute value inequalities, 107–109

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for linear inequalities in one variable, 80 for nonlinear inequalities, 631–635, 646, 797 for polynomial inequalities, 631–633 for quadratic inequalities, 630–631 for rational inequalities, 633–635 Time problems, 63–64, 260–261 Trace (graphing calculator feature), 239 Translating to/from English form. See also Problem solving; Word problems for absolute value, 109–110 basic guidelines for, 57 for compound inequalities, 92–93 for inequalities, 12 for polynomials, 338–339 for variation, 469–470 Transverse axis, of hyperbola, 769, 770 Trapezoids, area of, 71, 117 Trial-and-error method, 365–370, 406 Triangles area of, 117 Pascal’s, 402, 807 right, 393 similar, 460–461 sum of measures of angles with, 69 Trinomials. See also Polynomials AC-method to factor, 362–364, 405 explanation of, 324, 403 perfect square, 337, 370–371, 406, 576 steps to factor, 373 substitution to factor, 372–373 trial-and-error method to factor, 365–370, 406 2  3 matrix, 294

U Undefined slope, 148 Union, of sets, 85–87, 118 Unlike terms, 31

V Value (graphing calculator feature), 160 Variable expressions, roots of, 495–497 Variables, 26 Variable terms, 31, 40, 170

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Variation. See also Proportions applications of, 471–474 direct, 469, 471, 472, 484 inverse, 469, 472–473, 484 joint, 470, 473–474, 484 translations involving, 469–470 Vertex of hyperbola, 769, 794 of parabola, 604, 618–624, 645, 794 of quadratic function, 618–622 Vertex formula, 620–621, 645, 762–763 Vertical asymptotes, 204 Vertical lines, 136–137, 164, 216, 218 Vertical line test, 192–193 Vertical transverse axis, 769, 795

W

Y

Whole numbers, 6, 8 Word problems, 57–58, 116. See also Translating to/from English form Work problems, 463–464

y-axis, 128 y-coordinates explanation of, 129, 136 in one-to-one functions, 663 y-intercepts explanation of, 133, 216 of functions, 207–209 of linear equation, 133–135 of quadratic functions, 395, 621–622 slope-intercept form to find, 156–158

X x-axis, 128 x-coordinates explanation of, 129, 136 in one-to-one functions, 663 x-intercepts explanation of, 133, 216 of functions, 207–209 of linear equation, 133–135 of quadratic functions, 395, 621–622

Z Zero (graphing calculator feature), 160, 396 Zero product rule, 388–389, 407 Zero slope, 148