chapter 8bhhs.bhusd.org/ourpages/auto/2006/8/11/1155323019648/aat soluti… · the form y 5 ax. 2....

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469Algebra 2

Worked-Out Solution Key

Prerequisite Skills (p. 548)

1. The asymptote of the graph is y 5 0.

2. Two variables x and y show direct variation provided y 5 ax where a is a nonzero constant.

3. An extraneous solution of a transformed equation is not an actual solution of the original equation.

4. y 5 ax y 5 4x

8 5 a(2) 5 4(22)

4 5 a 5 28

y 5 4x

5. y 5 ax y 5 24x

4 5 a(21) 5 24(22)

24 5 a 5 8

y 5 24x

6. y 5 ax y 5 1 } 6 x

2 5 a(12) 5 1 } 6 (22)

2 }

12 5 a 5 2

2 } 6

1 }

6 5 a 5 2

1 } 3

y 5 1 } 6 x

7. x2 2 11x 2 26 5 (x 2 13)(x 1 2)

8. 2x3 2 4x2 1 2x 5 2x(x2 2 2x 1 1) 5 2x(x 2 1)(x 2 1)

5 2x(x 2 1)2

9. 6x4 2 4x3 2 24x 1 16 5 2(3x4 2 2x3 2 12x 1 8) 5 2[x3(3x 2 2) 2 4(3x 2 2)] 5 2(3x 2 2)(x3 2 4) 10. (3x2 2 6) 1 (7x2 2 x) 5 3x2 1 7x2 2 x 2 6

5 10x2 2 x 2 6

11. (22x2 1 6) 2 (x2 2 x) 5 22x2 1 6 2 x2 1 x

5 23x2 1 x 1 6

12. (x 1 2)(x 2 9)2 5 (x 1 2)[x2 2 2(x)(9) 1 92] 5 (x 1 2)(x2 2 18x 1 81) 5 (x 1 2)x2 1 (x 1 2)(218x) 1

(x 1 2)(81)

5 x3 1 2x2 2 18x2 2 36x 1 81x 1 162

5 x3 2 16x2 1 45x 1 162

Lesson 8.1

Investigating Algebra Activity 8.1 (p. 550)

1. No, because the ratios of apparent height to the distance are not approximately equal.

2. Answers will vary. The products are approximately equal.

3. Answers will vary. The equation should be of the form xy 5 a, where a is the approximate value of each product in Exercise 2.

4. Answers will vary.

8.1 Guided Practice (pp. 552–554)

1. y 5 3x

x and y show direct variation because the equation is of the form y 5 ax.

2. xy 5 0.75

y 5 0.75

} x

x and y show inverse variation because the equation is of

the form y 5 a } x .

3. y 5 x 2 5

x and y show neither direct variation nor inverse variation

because the equation is not of the form y 5 ax or y 5 a } x .

4. y 5 ax y 5 12

} x 5. y 5 ax y 5 2 8 } x

3 5 a } 4 5

12 } 2 21 5

a } 8 5 2

8 } 2

12 5 a 5 6 28 5 a 5 24

y 5 12

} x y 5 2 8 } x

6. y 5 a } x y 5

6 } x

12 5 a

} 1 }

2 5

6 } 2

6 5 a 5 3

y 5 6 } x

7. n 5 a } s

3000 5 a } 5

15,000 5 a

A model is n 5 15,000

} s .

8. c 5 26,000

} A 5 26,000

} 79 ø 329

The number of chips per wafer for a chip with an area of 79 square millimeters is about 329.

9. z 5 axy z 5 7 } 2 xy

7 5 a(1)(2) 5 7 } 2 (22)(5)

7 5 2a 5 235

7 }

2 5 a

z 5 7 } 2 xy

Chapter 8

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470Algebra 2Worked-Out Solution Key

10. z 5 axy z 5 22xy

24 5 a(4)(23) 5 22(22)(5)

24 5 212a 5 20

22 5 a

z 5 22xy

11. z 5 axy z 5 2 3 } 2 xy

18 5 a(22)(6) 5 2 3 } 2 (22)(5)

18 5 212a 5 15

2 3 } 2 5 a

z 5 2 3 } 2 xy

12. z 5 axy z 5 7 } 3 xy

56 5 a(26)(24) 5 7 } 3 (22)(5)

56 5 24a 5 2 70

} 3

7 }

3 5 a

z 5 7 } 3 xy

13. x 5 aw

} y 14. p 5 aqr

} s

8.1 Exercises (pp. 555–557)

Skill Practice

1. If z varies directly with the product of x and y, then z is said to vary jointly with x and y.

2. Find the product x p y for each data pair (x, y). If the products are constant or approximately constant, then the set of data pairs shows inverse variation.

3. xy 5 1 } 5

y 5 1 } 5x



4. y 5 x 1 4


because the equation is not of the form y 5 ax or y 5 a } x .

5. y }

x 5 8

y 5 8x


6. 4x 5 y


7. y 5 2 } x



8. x 1 y 5 6

y 5 2x 1 6

x and y show neither direct nor inverse variation because

the equation is not of the form y 5 ax or y 5 a } x .

9. 8y 5 x

y 5 x } 8


10. xy 5 12

y 5 12

} x



11. C; xy 5 5

y 5 5 } x

12. y 5 a } x y 5 2

20 } x 13. y 5

a } x y 5

9 } x

24 5 a } 5 5 2

20 } 3 9 5

a } 1 5

9 } 3

220 5 a 9 5 a 5 3

y 5 2 20

} x y 5 9 } x

14. y 5 a } x y 5 2

24 } x 15. y 5

a } x y 5

14 } x

8 5 a }

23 5 2 24

} 3 2 5 a } 7 5

14 } 3

224 5 a 5 28 14 5 a

y 5 2 24

} x y 5 14

} x

16. y 5 a } x y 5

21 } x 17. y 5

a } x y 5

5 } x

28 5 a

} 3 }

4 5

21 } 3 2

5 } 4 5

a }

24 5 5 } 3

28 1 3 } 4 2 5 a 5 7 2

5 } 4 (24) 5 a

21 5 a 5 5 a

y 5 21

} x y 5 5 } x

18. y 5 a } x y 5

2 } x

2 1 } 6 5

a }

212 5 2 } 3

2 1 } 6 (212) 5 a

2 5 a

y 5 2 } x

Chapter 8, continued

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471Algebra 2



19. y 5 a } x y 5 2

35 } 3x

27 5 a

} 5 }

3 5 2

35 }

3(3)

27 1 5 } 3 2 5 a 5 2

35 } 9

2 35

} 3 5 a

y 5 2

35 } 3 }

x 5 2

35 } 3x

20. x p y: 1.5(40) 5 60, 2.5(24) 5 60, 4(15) 5 60, 7.5(8) 5 60, 10(6) 5 60

y }

x :

40 }

1.5 5

80 } 3 ø 26.7,

24 }

2.5 5

48 } 5 5 9.6,

15 }

4 5 3.75,

8 } 7.5 5

16 } 15 ø 1.1,

6 }

10 5

3 } 5 5 0.6

x and y show inverse variation because the products x p y are equal.

21. x p y: 12(132) 5 1584, 18(198) 5 3564, 23(253) 5 5819, 29(319) 5 9251, 34(374) 5 12,716

y }

x :

132 }

12 5 11,

198 }

18 5 11,

253 }

23 5 11,

319 }

29 5 11,

374 }

34 5 11

x and y show direct variation because the ratios y }

x

are equal.

22. x p y: 4(16) 5 64, 5(11) 5 55, 6.2(10) 5 62, 7(9) 5 63, 11(6) 5 66

y }

x :

16 }

4 5 4,

11 } 5 5 2.2,

10 }

6.2 ø 1.6,

9 } 7 ø 1.3,

6 }

11 ø 0.5


because neither the products x p y nor the ratios y }

x

are equal.

23. x p y: 4(21) 5 84, 6(14) 5 84, 8(10.5) 5 84, 8.4(10) 5 84, 12(7) 5 84

y }

x :

21 }

4 5 5.25,

14 }

6 5

7 } 3 ø 2.3,

10.5 }

8 5

21 } 16 ø 1.3,

10

} 8.4

5 25

} 21 ø 1.2, 7 }

12 ø 0.6

x and y show inverse variation because the products x p y are equal.

24. z 5 axy z 5 22xy

24 5 a(2)(26) 5 22(24)(5)

24 5 212a 5 40

22 5 a

z 5 22xy

25. z 5 axy z 5 1 } 4 xy

12 5 a(8)(6) 5 1 } 4 (24)(5)

12 5 48a 5 215

1 }

4 5 a

z 5 1 } 4 xy

26. z 5 axy z 5 20xy

15 5 a 1 2 1 } 4 2 (23) 5 20(24)(5)

15 5 3 } 4 a 5 2400

20 5 a

z 5 20xy

27. z 5 axy z 5 1 } 14 xy

23 5 a(16)(27) 5 1 } 14 (24)(5)

23 5 242a 5 2 20

} 14

1 }

14 5 a 5 2

10 } 7

z 5 1 } 14 xy

28. z 5 axy z 5 2 1 } 3 xy

6 5 a(9)(22) 5 2 1 } 3 (24)(5)

6 5 218a 5 20

} 3

2 1 } 3 5 a

z 5 2 1 } 3 xy

29. z 5 axy z 5 25xy

75 5 a(5)(23) 5 25(24)(5)

75 5 215a 5 100

25 5 a

z 5 25xy

30. A; z 5 axy

236 5 a(23)(24)

236 5 12a

23 5 a

31. x 5 ay

} z 32. y 5 axz2 33. w 5 axz

} y

34. The variables x and y should be in the numerator and the variable w should be in the denominator because z varies directly with x and y and inversely with w. The correct

equation is z 5 axy3

} Ï

}

w .

35. Sample answer: f (x) 5 2x, g(x) 5 2 } x , h(x) 5 2x 1

2 } x ;

h(2) 5 2(2) 1 2 } 2 5 4 1 1 5 5

36. x varies directly with z.

x 5 a } y , y 5

b } z

x 5 a

} b } z 5

a }

b z 5 cz

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Problem Solving

37. n 5 a }

s n 5

103.68 }

s

54 5 a } 1.92 5

103.68 } 3.87

103.68 5 a ø 26.8

A model is n 5 103.68

} s . You can store 26 photos on your

camera when the average photo size is 3.87 megapixels.

38. Sample answer: 7.4(1200) 5 8880, 8.9(1000) 5 8900, 12.1(750) 5 9075, 17.9(500) 5 8950

I p R 5 9000 or R 5 9000

} I

R 5 9000

} I 5 9000

} 34 ø 265

When I 5 34 milliamps, the resistance is about 265 ohms.

39. P 5 a } A P 5

172 } A

0.43 5 a } 400 5

172 } 60

172 5 a ø 2.87

P 5 172

} A

The pressure if you wear the boots is about 2.87 pounds per square inch.

40. a. f 5 a Ï

} T }

Ld

262 5 a Ï

}

670 }

62(0.1025)

1665.01 5 a Ï}

670

64.3 ø a

f 5 64.3 Ï

} T }

Ld

b. f 5 64.3 Ï

} T }

Ld 5

64.3 Ï}

1629 }

201.6(0.49) ø 26.3

The frequency of the note is about 26.3 hertz.

41. a. F 5 Gm1m2

} d2

b. F 5 Gm1m2

} d 2

3.53 3 1022 5 G(5.98 3 1024)(1.99 3 1030)

}}} (1.50 3 1011)2

3.53 3 1022 5 G(5.98 3 1.99)(1024 3 1030)

}}} (1.50)2(1011)2

3.53 3 1022 5 G(11.9002 3 1054)

}} 2.25 3 1022

3.53 3 1022 5 G(1.19002 3 1055)

}} 2.25 3 1022

(3.53 3 1022)(2.25 3 1022) 5 G(1.19002 3 1055) 7.9425 3 1044 5 G(1.19002 3 1055)

7.9425 3 1044

}} 1.19002 3 1055 5 G

1 7.9425 }

1.19002 2 1 1044

} 1055 2 5 G

6.7 3 10211 ø G

c. As the masses of the two objects increase and the distance between them is held constant, the gravitational force decreases. As the masses of the two objects are held constant and the distance between them increases, the gravitational force increases.

42. a. P 5 aWD2

} L

a(2W )D2

} 2L

5 aWD2

} L 5 P

P stays the same when the width and length of the beam are doubled.

b. P 5 aWD2

} L

a(2W)(2D)2

} L 5 8aWD2

} L 5 8P

P is multiplied by 8 when the width and depth of the beam are doubled.

c. P 5 aWD2

} L

a(2W )(2D)2

} 2L

5 8aWD2

} 2L 5 4aWD2

} L 5 4P

P is quadrupled when all three dimensions are doubled.

d. Sample answer: If the safe load of a beam is increased by a factor of 4, you can double the length, width, and depth, or you can quadruple the width, or you can double the depth.

Mixed Review

43. y 5 x2 1 8x 2 20 44. y 5 2x2 1 4x 1 3

4

x

y

22

3

x

y

21


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473Algebra 2


45. f (x) 5 x3 1 1 46. g(x) 5 2 Ï}

x

2

x

y

22

1

x

y

21

47. y 5 Ï}

x 1 4 2 1 48. y 5 1 } 2 Ï}

x 2 1 1 1

2

x

y

21

1

x

y

21

49. y 5 23x 50. h(x) 5 3x 1 1 1 2

1

x

y

22

1

x

y

21

51. y 5 1 3 } 4 2 x 2 1

1

x

y

21

52. Ï}

x 5 14 Check: Ï}

196 0 14

( Ï}

x )2 5 142 14 5 14 ✓

x 5 196

53. 2 Ï}

x 2 5 5 45 Check: 2 Ï}

625 2 5 0 45

2 Ï}

x 5 50 2(25) 2 5 0 45

Ï}

x 5 25 50 2 5 0 45

( Ï}

x )2 5 252 45 5 45 ✓

x 5 625

54. x2/3 2 9 5 0 Check: 272/3 2 9 0 0

x2/3 5 9 (271/3)2 2 9 0 0

(x2/3)3/2 5 93/2 32 2 9 0 0

x 5 (91/2)3 9 2 9 0 0

5 33 5 27 0 5 0 ✓

55. x 2 6 5 Ï}

3x

(x 2 6)2 5 ( Ï}

3x )2

x2 2 12x 1 36 5 3x

x2 2 15x 1 36 5 0

(x 2 12)(x 2 3) 5 0

x 2 12 5 0 or x 2 3 5 0

x 5 12 or x 5 3

Check x 5 12: Check x 5 3:

12 2 6 0 Ï}

3(12) 3 2 6 0 Ï}

3(3)

6 0 Ï}

36 23 0 Ï}

9

6 5 6 ✓ 23 Þ 3

The only solution is 12.

56. Ï}

2x 5 x 2 4

( Ï}

2x )2 5 (x 2 4)2

2x 5 x2 2 8x 1 16

0 5 x2 2 10x 1 16

0 5 (x 2 8)(x 2 2)

x 2 8 5 0 or x 2 2 5 0

x 5 8 or x 5 2


Ï}

2(8) 0 8 2 4 Ï}

2(2) 0 2 2 4

Ï}

16 0 4 Ï}

4 0 22

4 5 4 ✓ 2 Þ 22

The only solution is 8.

57. Ï}

4x 2 3 5 Ï}

2x 1 13 Check:

( Ï}

4x 2 3 )2 5 ( Ï}

2x 1 13 )2 Ï}

4(8) 2 3 0 Ï}

2(8) 1 13

4x 2 3 5 2x 1 13 Ï}

32 2 3 0 Ï}

16 1 13

2x 2 3 5 13 Ï}

29 5 Ï}

29 ✓

2x 5 16

x 5 8

Lesson 8.2


1. f (x) 5 2 4 } x The domain is all real numbers

1x

y

21

except 0, and the range is all real numbers except 0.


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2. y 5 8 } x 2 5 The domain is all real numbers

2y

22 x


3. y 5 1 } x 2 3 1 2 The domain is all real numbers

1

y

21 x


4. y 5 x 2 1

} x 1 3 The domain is all real numbers

2

y

21 x


5. y 5 2x 1 1

} 4x 2 2 The domain is all real numbers

1

x

y

1

except 1 }

2 , and the range is all

real numbers except 1 }

2 .

6. f (x) 5 23x 1 2


1

y

22 x


7. If the cost of the 3-D printer is $21,000, then the function

is c 5 300m 1 21,000

}} m . The graph of c 5 300m 1 21,000

}} m

lies closer to the axes than the graph of

c 5 300m 1 24,000

}} m . Both graphs have the same

asymptotes, domain, and range.


Skill Practice

1. The function y 5 7 } x 1 4 1 3 has a range of all real

numbers except 3 and a domain of all real numbers except 24.

2. The function f (x) 5 23x 1 5

} 2x 1 1 is not a rational function

because the expression 2x 1 1 is not a polynomial.

3. y 5 3 } x The graph lies farther from the

1

x

y

21

axes than the graph of y 5 1 } x .

Both graphs lie in the fi rst and third quadrants and have the same asymptotes, domain, and range.

4. y 5 10

} x The graph lies farther from the

2

x

y

22

axes than the graph of y 5 1 } x .


5. y 5 25


1

x

y

21

axes than the graph of y 5 1 } x and it

lies in Quadrants II and IV instead of Quadrants I and III. Both graphs have the same asymptotes, domain, and range.

6. y 5 20.5

} x The graph lies closer to the

1

x

y

21



7. y 5 0.1

} x The graph lies closer to the

0.2

20.2 x

y axes than the graph of y 5 1 } x .



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475Algebra 2


8. f (x) 5 15


4

x

f(x)

21

axes than the graph of y 5 1 } x . Both

graphs lie in the fi rst and third quadrants and have the same asymptotes, domain, and range.

9. g(x) 5 26


2x

g(x)

21



10. h(x) 5 23


1x

h(x)

21




1

x

y

1



1

x

y

21



2

x

y

21


14. f (x) 5 1 } x 1 2 The domain is all real numbers

2

y

21 x


15. y 5 25

} x 2 7 The domain is all real numbers y

24

22 x


16. y 5 26


1

y

21 x


17. y 5 23


1x

y

21


18. g(x) 5 22


1x

1

g(x) except 7, and the range is all real numbers except 0.

19. y 5 24

} x 1 4 1 3 The domain is all real numbers

1

y

21 x



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20. y 5 10 } x 1 7 2 5 The domain is all real numbers

3x

y

29


21. y 5 23

} x 2 4 2 1 The domain is all real numbers

1

x

y

1


22. h(x) 5 11 } x 2 9 1 9 The domain is all real numbers

2

x2

h(x) except 9, and the range is all real numbers except 9.

23. D; Because the rational function y 5 3 } x 1 8 2 3 is of the

form y 5 a } x 2 h 1 k, where h 5 28 and k 5 23, the

asymptotes are x 5 28 and y 5 23.

24. a. The graph of y 5 23

} x 2 3 2 2 is a refl ection in the line

y 5 22 and lies farther from the asymptotes than the

graph of y 5 1 } x 2 3 2 2.

b. The graph of y 5 1 } x 1 1 2 2 is the graph of

y 5 1 } x 2 3 2 2 translated to the left 4 units.

c. The graph of y 5 1 } x 2 3 1 2 is the graph of

y 5 1 } x 2 3 2 2 translated up 4 units.

25. The graph shown is the graph

2x

y

22

of y 5 8 } x . The graph should

be y 5 2 8 } x .

26. The vertical asymptote should 1

x

y

21

be x 5 1, not x 5 21.

27. y 5 x 1 4


2

x

y

2


28. y 5 x 2 1


2

y

26 x


29. y 5 x 1 6


1

x

y

1

except 2, and the range is all real

numbers except 1 }

4 .


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477Algebra 2


30. y 5 8x 1 3


2

y

22 x


31. y 5 25x 1 2


3

y

22

x

except 2 5 } 4 , and the range is all real

numbers except 2 5 } 4 .

32. f (x) 5 6x 2 1


1

x21

f(x) except 1 }

3 , and the range is all real

numbers except 2.

33. g(x) 5 5x } 2x 1 3 The domain is all real numbers

2

x22

g(x) except 2 3 } 2 , and the range is all real

numbers except 5 }

2 .

34. h(x) 5 5x 1 3


10x

22

h(x) except 10, and the range is all

real numbers except 25.

35. Sample answer: The function y 5 1 } x 1 8 1 3 has a

domain of all real numbers except 28 and a range of all real numbers except 3.

36. f (x) 5 a }

x 2 h 1 k

5 a 1 k(x 2 h)

} x 2 h

5 a 1 kx 2 kh

} x 2 h

5 kx 1 a 2 kh

} x 2 h

Problem Solving

37. An equation is c 5 43m 1 50

} m , where c is the average cost

per month and m is the number of months of service.

Months of service

Ave

rag

e m

on

thly

co

st

0 1 2 3 4 5 6 7 8 9 m

c

0102030405060708090

53 5 43m 1 50

} m

53m 5 43m 1 50

10m 5 50

m 5 5

After 5 months, the average cost will be $53.

38. An equation is c 5 59m 1 100

} m , where c is the average

cost per month and m is the number of months of membership.

Months of membership

Ave

rag

e m

on

thly

co

st

0 1 3 5 72 4 6 80

40

80

120

160140

100

60

20

m

c

69 5 59m 1 100

} m

69m 5 59m 1 100

10m 5 100

m 5 10



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39. a. t 5 1000 } 0.6T 1 331 5

1000 }}

0.6(25) 1 331 5

1000 } 346 ø 2.89

It takes about 2.89 seconds for sound to travel 1 kilometer when the air temperature is 258C. So, for sound to travel 5 kilometers, it takes about 5(2.89) 5 14.45 seconds.

b.

10 2 3 4 5 6 T

t

2.97

0

2.982.993.003.013.02

Air Temperature (8C)

Tim

e (s

eco

nd

s)

From the graph, you can estimate the air temperature to be 3.98C.

40. a.

Percent (as a decimal)

Co

st (

tho

usa

nd

s o

f d

olla

rs)

0 0.2 0.4 0.6 0.80

30

60

90

120

y

x

A reasonable domain for the function y 5 15x } 1.1 2 x is

all real numbers from 0 to 1. A reasonable range for the function is all real numbers from 0 to 150.

b. y 5 15x } 1.1 2 x 5

15(0.2) } 1.1 2 0.2 5

3 } 0.9 ø 3.3

It costs about $3300 to remove 20% of the pollutant.

y 5 15x } 1.1 2 x 5

15(0.4) } 1.1 2 0.4 5

6 } 0.7 ø 8.6

It costs about $8600 to remove 40% of the pollutant.

y 5 15x } 1.1 2 x 5

15(0.8) } 1.1 2 0.8 5

12 } 0.3 5 40

It costs $40,000 to remove 80% of the pollutant. Doubling the percent of the pollutant removed does not double the cost of removing the pollutant. For example, to remove 20% of the pollutant it costs $3300 and to remove 40% it costs $8600 and 2 p 3300 Þ 8600.

41. a. Approaching: fl 5 1,480,000

} 740 2 r ;

Moving away: fl 5 1,480,000

} 740 1 r

b.

Speed (mi/h)

Freq

uen

cy (

Hz)

0 12 24 36 48 r

f

0180018501900195020002050210021502200

1,480,000740 2 rfl 5

1,480,000740 1 rfl 5

c. The frequency of a sound that is approaching is greater than that of a sound moving away.

42. Average speed 5 Total distance

}} Total time

Total distance 5 d1 1 d2, where d1 is the distance traveled at 10 knots and d2 is the distance traveled at 15 knots.

d1 5 10(3) 5 30

d2 5 15t, where t is the time that it uses motor power

Total time 5 t 1 3

So, an equation for the boat’s average speed for the entire trip is:

s 5 30 1 15t

} t 1 3

Time using motor(hours)

Ave

rag

e sp

eed

(kn

ots

)

0 4 62 8 100

42

68

101214

s

t

Mixed Review

43. m2 1 18m 1 65 5 (m 1 5)(m 1 13)

44. p2 1 15p 1 56 5 ( p 1 7)( p 1 8)

45. q2 2 49 5 (q 1 7)(q 2 7)

46. r22 20r 1 100 5 (r 2 10)(r 2 10) 5 (r 2 10)2

47. x2 2 4x 2 21 5 (x 2 7)(x 1 3)

48. z 2 2 9z 1 20 5 (z 2 5)(z 2 4)

49. f (x) 5 2x4 50. f (x) 5 x3 1 5

20

x21

f(x)

10

x

f(x)

21

51. f (x) 5 x5 2 1 52. f (x) 5 3x4 1 2

5

x22

f(x)

5

x21

f(x)


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479Algebra 2


53. f (x) 5 2x6 54. f (x) 5 22x3 1 3

10

x22

f(x)

10

x21

f(x)

55. e5 p e29 5 e5 2 9 5 e24 5 1 }

e4

56. 3e6 p ex 5 3e6 1 x

57. ex p e3x 1 2 5 e x 1 3x 1 2 5 e 4x 1 2

Graphing Calculator Activity 8.2 (p. 564)

1. y 5 5 } x 1 2 2. y 5 7 2

3 } x

3. y 5 4 1 2 } x 2 5 4. y 5

6 } x 1 1 1 2

5. y 5 7 } 2x 1 8 6. y 5

9 2 2x } x 2 3

7. f (x) 5 x 2 4

} x 1 2 8. g(x) 5 5x 2 2

} 3x 1 9

9. A function that gives the average cost y per visit after x

visits is y 5 4x 1 100

} x .

As the number of visits increases, the average cost gets closer to $4. A reasonable domain for the function is all real numbers greater than 0 and a reasonable range is all real numbers greater than 0.

Lesson 8.3

8.3 Guided Practice (p. 567)

1. y 5 4 }

x2 1 2

The numerator has no zeros, so there is no x-intercept. The denominator has no real zero, so there is no vertical asymptote.

m 5 0 and n 5 2, so m < n and y 5 0 is a horizontal asymptote.

x y

22 2 }

3

21 4 }

3

0 2

1 4 }

3

2 2 }

3

1

x

y

21

2. y 5 3x2

} x2 2 1

3x2 5 0 → x 5 0, so 0 is an x-intercept.

x2 2 1 5 0 → x 5 61, so x 5 1 and x 5 21 are vertical asymptotes. m 5 2 and n 5 2, so m 5 n and

y 5 3 } 1 5 3 is a horizontal asymptote.

x y

24 16

} 5

2 3 } 2

27 } 5

2 1 } 2 21

0 0

1 }

2 21

3 }

2

27 } 5

4 16

} 5

2

x

y

22


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3. y 5 x2 2 5

} x2 1 1

x2 2 5 5 0 → x 5 6 Ï}

5 , so 6 Ï}

5 are x-intercepts.

The denominator has no real zeros, so there is no vertical

asymptote. m 5 2 and n 5 2, so m 5 n and y 5 1 } 1 5 1 is

a horizontal asymptote.

x y

2 Ï}

5 0

21 22

0 25

1 22

Ï}

5 0

2

x

f(x)

23

4. y 5 x2 2 2x 2 3

} x 2 4

x2 2 2x 2 3 5 0 → (x 2 3)(x 1 1) 5 0 →

x 5 3 or x 5 21, so 3 and 21 are x-intercepts.

x 2 4 5 0 → x 5 4, so x 5 4 is a vertical asymptote.m 5 2 and n 5 1, so m > n and the graph has no horizontal asymptote.

x y

21 0

0 3 }

4

2 3 }

2

3 0

5 12

9 12

2

x

y

2

5. V 5 πr2h S 5 2πr2 1 2πrh

544 5 πr2h 5 2πr2 1 2πr 1 544 }

πr2 2

544

} πr2 5 h 5 2πr2 1

1088 } r

S 5 2πr2 1 1088

} r

MinimiumX=4.4239125 Y=369.90438

Using the minimum feature, you get a minimum value of about 369, which occurs when r ø 4.42 and

h ø 544 }

π (4.42)2 ø 8.86. So, the soup can using the least

amount of material has a radius of about 4.42 centimeters and a height of about 8.86 centimeters.


Skill Practice

1. The graph of a rational function f has no horizontal asymptote when the degree of the function’s numerator is greater than the degree of its denominator.

2. To fi nd the x-intercepts of the graph of f (x) 5 p(x)

} q(x)

, fi nd

the real zeros of p(x). To fi nd the vertical asymptotes of f, fi nd the real zeros of q(x).

3. C; The graph of y 5 210

} x2 2 9

has no x-intercepts, vertical

asymptotes at x 5 23 and x 5 3, and a horizontal asymptote at y 5 0 as shown in graph C.

4. A; The graph of y 5 x2 2 10

} x2 1 3

has x-intercepts 2 Ï}

10 and

Ï}

10 , no vertical asymptotes, and a horizontal asymptote at y 5 1 as shown in graph A.

5. B; The graph of y 5 x3 }

x2 2 4 has x-intercept 0, vertical

asymptotes at x 5 22 and x 5 2, no horizontal asymptote, and end behavior the same as the graph of y 5 x3 2 2 5 x as shown in graph B.

6. A; The graph of y 5 3 }

x2 2 4 has no x-intercepts, vertical

asymptotes at x 5 22 and x 5 2, and a horizontal asymptote at y 5 0 as shown in the given graph.

7. y 5 5 }

x2 2 1 5

5 }}

(x 1 1)(x 2 1)

The numerator has no zeros, so there is no x-intercept. The zeros of the denominator are 61, so x 5 1 and x 5 21 are vertical asymptotes.

8. y 5 x 1 1

} x2 1 5

The zero of the numerator is 21, so 21 is an x-intercept. The denominator has no real zeros, so there is no vertical asymptote.

9. f (x) 5 x2 1 9 }

x2 2 2x 2 15 5

x2 1 9 }}

(x 2 5)(x 1 3)

The numerator has no real zeros, so there is no x-intercept. The zeros of the denominator are 5 and 23, so x 5 5 and x 5 23 are vertical asymptotes.

10. y 5 x2 2 7x 2 60

} x 1 3 5 (x 2 12)(x 1 5)

}} x 1 3

The zeros of the numerator are 12 and 25, so 12 and 25 are x-intercepts. The zero of the denominator is 23, so x 5 23 is a vertical asymptote.

11. y 5 x3 1 27

} 3x2 1 x

5 (x 1 3)(x2 2 3x 1 9)

}} x(3x 1 1)

The real zero of the numerator is 23, so 23 is an x-intercept. The zeros of the denominator are 0 and 2

1 } 3 ,

so x 5 0 and x 5 2 1 } 3 are vertical asymptotes.


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481Algebra 2


12. g(x) 5 2x2 2 3x 2 20

}} x2 1 1

5 (2x 1 5)(x 2 4)

}} x2 1 1

The zeros of the numerator are 2 5 } 2 and 4, so 2

5 } 2 and 4

are x-intercepts. The denominator has no real zeros, so there is no vertical asymptote.

13. The vertical asymptote occurs at the zeros of the denominator, not the numerator. The denominator x2 2 8x 1 7 factors as (x 2 7)(x 2 1). So, the zeros of the denominator are 7 and 1 and the vertical asymptotes are at x 5 7 and x 5 1.

14. C; The numerator and denominator have the same

degree, so the horizontal asymptote is y 5 am

} bn

5 4 } 1 5 4.

15. y 5 2x }

x2 2 1

2x 5 0 → x 5 0, so 0 is an x-intercept.

x2 2 1 5 0 → x 5 6 1, so x 5 1 and x 5 21 are vertical asymptotes. m 5 1 and n 5 2, so m < n and y 5 0 is a horizontal asymptote.

x y

23 2 3 }

4

22 2 4 }

3

2 1 } 2

4 }

3

0 0

1 }

2 2

4 } 3

2 4 }

3

3 3 }

4

3

x

y

22

16. y 5 8 }

x2 2 x 2 6

The numerator has no zeros, so there is no x-intercept.

x2 2 x 2 6 5 0 → (x 2 3)(x 1 2) 5 0 →

x 5 3 or x 5 22, so x 5 3 and x 5 22 are vertical asymptotes.


x y

25 1 }

3

23 4 }

3

21 22

0 2 4 } 3

2 22

4 4 }

3

6 1 }

3

x2

1

y

17. f (x) 5 x2 2 9

} 2x2 1 1

x2 2 9 5 0 → x 5 63, so 3 and 23 are x-intercepts.

The denominator has no real zeros, so there is no vertical asymptote.

m 5 2 and n 5 2, so m 5 n and y 5 1 } 2 is a

horizontal asymptote.

x y

23 0

21 2 8 } 3

0 29

1 2 8 } 3

3 0

21x

f(x)

1

18. y 5 x 2 4

} x2 2 3x

x 2 4 5 0 → x 5 4, so 4 is an x-intercept.

x2 2 3x 5 0 → x(x 2 3) 5 0 → x 5 0 or x 5 3, so x 5 0 and x 5 3 are vertical asymptotes.


x y

22 2 3 } 5

21 2 5 } 4

1 3 }

2

2 1

5 }

2

6 } 5

4 0

5 1 }

10

1

x

y

21

19. y 5 x2 1 11x 1 18

}} 2x 1 1

x2 1 11x 1 18 5 0 →

(x 1 9)(x 1 2) 5 0 → x 5 29 or x 5 22, so 29 and 22 are x-intercepts.

2x 1 1 5 0 → x 5 2 1 } 2 , so x 5 2

1 } 2 is a vertical

asymptote.

m 5 2 and n 5 1, so m > n and the graph has no horizontal asymptote.


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x y

29 0

25 4 }

3

22 0

21 28

0 18

1 10

8 10

x

y

2

3

20. g(x) 5 x3 2 8

} 6 2 x2

x3 2 8 5 0 → x 5 2, so 2 is an x-intercept.

6 2 x2 5 0 → 6 Ï}

6 5 x, so x 5 Ï}

6 and x 5 2 Ï}

6 are vertical asymptotes.


x y

25 7

23 35

} 3

22 28

0 2 4 } 3

2 0

3 2 19

} 3

4 2 28

} 5

2

x

g(x)

24

21. y 5 x2 1 3

} 2x3

The numerator has no real zeros, so there is no x-intercept.

2x3 5 0 → x 5 0, so x 5 0 is a vertical asymptote.


x y

23 2 2 } 9

21 22

1 2

3 2 }

9

1

x

y

1

22. y 5 x2 2 5x 2 36

} 3x

x2 2 5x 2 36 5 0 → (x 2 9)(x 1 4) 5 0 → x 5 9 or x 5 24, so 9 and 24 are x-intercepts.

3x 5 0 → x 5 0, so x 5 0 is a vertical asymptote.


x y

29 2 10

} 3

24 0

21 10

2 27

9 0

12 4 }

3

x

y

3

2

23. h(x) 5 3x2 1 10x 2 8

}} x2 1 4

3x2 1 10x 2 8 5 0 → (3x 2 2)(x 1 4) 5 0 →

x 5 2 } 3 or x 5 24, so

2 }

3 and 24 are x-intercepts.

The denominator has no real zeros, so there is no vertical asymptote.

m 5 2 and n 5 2, so m 5 n and y 5 3 } 1 5 3 is a horizontal

asymptote.

x y

26 1

24 0

22 22

21 23

0 22

1 1

2 3

4 4

x

h(x)

2

1

24. Sample answer: The graphs of y 5 3x3 2 3

} x 1 1 and

y 5 3x3 1 6

} x 2 4 have the same end behavior as the graph

of y 5 3x2 because the degree of the numerator is

greater than the degree of the denominator and

y 5 am

} bn

xm 2 n 5 3 } 1 x3 2 1 5 3x2.


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483Algebra 2


25. y 5 15 }

x2 1 2

The range of the function is all real numbers such that 0 < y ≤ 7.5.

26. y 5 3x2

} x2 2 9

The range of the function is all real numbers except 0 < y ≤ 3.

27. y 5 x2 2 2x

} 2x 1 3

The range of the function is all real numbers except 20.209 < y < 24.791.

28. f (x) 5 a }

x2 1 b

f (0) 5 a }

02 1 b 5

a }

b 5 2 → a 5 2b

f (1) 5 a }

12 1 b 5

a }

1 1 b 5 1

2b }

1 1 b 5 1 a 5 2b

2b 5 1(1 1 b) 5 2(1)

2b 5 1 1 b 5 2

b 5 1

a 5 2 and b 5 1 so f (x) 5 2 }

x2 1 1 .

29. f (x) 5 a }

x2 1 b

f (22) 5 a }

(22)2 1 b 5

a }

4 1 b 5 2

f (1) 5 a }

12 1 b 5

a }

1 1 b 5 4 → a 5 4(1 1 b)

4(1 1 b)

} 4 1 b

5 2 a 5 4(1 1 2)

4(1 1 b) 5 2(4 1 b) 5 4(3)

4 1 4b 5 8 1 2b 5 12

2b 5 4

b 5 2

a 5 12 and b 5 2, so f (x) 5 12 }

x2 1 2 .

30. f (x) 5 a }

x2 1 b

f (23) 5 a }

(23)2 1 b 5

a }

9 1 b 5 21

→ a 5 2(9 1 b)

f (2) 5 a }

22 1 b 5

a }

4 1 b 5 26

2(9 1 b)

} 4 1 b

5 26 a 5 2[9 1 (23)]

(29 1 b) 5 26(4 1 b) 5 26

29 2 b 5 224 2 6b

5b 5 215

b 5 23

a 5 26 and b 5 23, so f (x) 5 26 }

x2 2 3 .

Problem Solving

31. a. V 5 πr2l

100 5 πr2l

100

} π r2 5 l

b. S 5 2πr2 1 2πrl

5 2πr2 1 2πr 1 100 }

πr2 2 5 2πr2 1

200 } r

c. S 5 2πr2 1 200

} r

MinimiumX=2.5153956 Y=119.26542


l ø 100 }

π (2.515)2 ø 5.032. So, the bale using the least

amount of plastic has a radius of about 2.515 feet and a length of about 5.032 feet.

32. a. V 5 3 Ï

}

3 } 2 s2h

24 5 3 Ï

}

3 } 2 s2h

24 p 2 }

3 Ï}

3 5

3 Ï}

3 } 2 p 2

} 3 Ï

}

3 s2h

16

} Ï

}

3 5 s2h

16 }

Ï}

3 s2 5 h


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b. S 5 3 Ï

}

3 } 2 s2 1 6sh

5 3 Ï

}

3 } 2 s2 1 6s 1 16

} Ï

}

3 s2 2

5 3 Ï

}

3 } 2 s2 1

96 }

Ï}

3 s

S 5 3 Ï

}

3 } 2 s2 1

96 }

Ï}

3 s

MinimiumX=2.2012856 Y=37.768142

Using the minimum feature, you get a minimum value of about 38, which occurs when s ø 2.201 and

h 5 16 }

Ï}

3 (2.201)2 ø 1.907. So, the aquarium using the

least amount of material has a side length of about 2.201 feet and a height of about 1.907 feet.

33. a. Depth, d Temp., T

1000 4.7634

1050 4.5796

1100 4.4094

1150 4.2515

1200 4.1044

1250 3.9672

1300 3.8389

b. T 5 17,800d 1 20,000

}} 3d2 1 740d 1 1000

0 1000 1100 1200 1300 d

T

3.6

0

3.84.04.24.44.64.8

Depth (meters)

Tem

per

atu

re (

8C)

From the graph, you can estimate that the depth at which the mean temperature is 48C is 1238 meters.

34. a. n 5 1054t 1 6204

}} 26.62t 1 100

Years since 1993

Sh

ares

of

sto

ck s

old

(in

bill

ion

s)

0 2 31 4 5 6 7 8 9 t

n

050

100150200250300350400450

b. The number of shares of stock sold on the New York Stock Exchange increased each year from 1993 to 2002.

c. Because the model represents the total number of shares of stock sold each year, you need to look at the points on the graph for t 5 1, 2, 3, etc. From the graph, you can see that n is less than 100 at t 5 2 and n is greater than 100 at t 5 3. So you can estimate that the year when the number of shares of stock sold was fi rst greater than 100 billion was 1996.

35. a. g 5 3.99 3 1014

}}} h2 1 (1.28 3107)h 1 (4.07 31013)

Altitude above sea level (m)

Acc

eler

atio

n d

ue

to g

ravi

ty (

m/s

2 )

0 4000 8000 12000 16000 h

g

02468

1012

b. g 5 3.99 3 1014

}}} h2 1 (1.28 3 107)h 1 (4.07 3 1013)

5 3.99 3 1014

}}}} 80002 1 (1.28 3 107)(8000) 1 (4.07 3 1013)

ø 3.99 3 1014 }

4.08 3 1013 5 3.99

} 4.08 p 1014 }

1013 5 0.978 p 101 5 9.78

The acceleration due to gravity at 8000 meters is about 9.78 meters per second squared.

c. 112 km 5 1000 p 112 5 112,000 m

g 5 3.99 3 1014

}}} h2 1 (1.28 3 107)h 1 (4.07 3 1013)

5 3.99 3 1014

}}}} 112,0002 1 (1.28 3 107)(112,000) 1 (4.07 31013)

ø 3.99 3 1014

} 4.21 3 1013 5

3.99 } 4.21 p 1014

} 1013 ø 0.947 p 101 5 9.47

The acceleration due to gravity at 112 kilometers is about 9.47 meters per second squared.

d. As altitude increases, the acceleration due to gravity g decreases slowly.


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485Algebra 2


36. a. Volume of tank 5

Volume of outer tank 2

Volume of inner tank

100 5 π(r 1 1)2(h 1 1) 2 πr2h

100 5 π(r2 1 2r 1 1)(h 1 1) 2 πr2h

100 5 π(r2h 1 r2 1 2rh 1 2r 1 h 1 1) 2 πr2h

100 5 πr2h 1 πr2 1 2πrh 1 2πr 1 πh 1 π 2 πr2h

100 5 πr2 1 2πrh 1 2πr 1 πh 1 π

100 2 πr2 2 2πr 2 π 5 2πrh 1 πh

100 2 πr2 2 2πr 2 π 5 h(2πr 1 π)

100 2 πr2 2 2πr 2 π

}} 2πr 1 π

5 h

b. V 5 πr2h

V 5 πr2 1 100 2 πr2 2 2πr 2 π }}

2πr 1 π 2

c. V 5 πr2 1 100 2 πr2 2 2πr 2 π }}

2πr 1 π 2

MaximumX=2.744536 Y=64.946444

Using the maximum feature, you get a maximum value of about 65, which occurs when r ø 2.74 and

h ø 100 2 π r2 2 2π r 2 π

}} 2π r 1 π

ø 2.75. So the tank has a

maximum volume when the radius is about 2.74 feet and the height is about 2.75 feet.

Mixed Review

37. x2 2 64 5 (x 1 8)(x 2 8)

38. x2 2 8x 2 48 5 (x 2 12)(x 1 4)

39. 18x2 2 37x 2 20 5 (9x 1 4)(2x 2 5)

40. 12x2 2 15x 2 18 5 3(4x2 2 5x 2 6) 5 3(4x 1 3)(x 2 2)

41. 5x2 1 22x 2 30 cannot be factored.

42. 5x3 1 40 5 5(x3 1 8) 5 5(x3 1 23) 5 5(x 1 2)(x2 2 2x 1 4) 43. x3 2 4x2 1 8x 2 32 5 x2(x 2 4) 1 8(x 2 4)

5 (x2 1 8)(x 2 4)

44. x3 1 2x2 2 35x 5 x(x2 1 2x 2 35) 5 x(x 2 5)(x 1 7)

45. x5 2 9x3 2 36x 5 x(x4 2 9x2 2 36) 5 x(x2 2 12)(x2 1 3)

46. x5y

} x2y4 5 x5 2 2y1 2 4 Quotient of powers property

5 x3y23

5 x3

} y3 Negative exponent property

47. 48x21y4

} 6x2y3 5 8x21 2 2y4 2 3 Quotient of powers property

5 8x23y

5 8y

} x3 Negative exponent property

48. 1 x2y4

} xy5 2

2

5 1 x2y4 2 2

} 1 xy5 2 2

Power of a quotient property

5 (x2)2(y4)2

} x2(y5)2

Power of a product property

5 x4y8

} x2y10 Power of a power property

5 x4 2 2y8 2 10 Quotient of powers property

5 x2y22

5 x2

} y2 Negative exponent property

49. 1 72x3y21

} 12x21y2 2

21

5 12x21y2

} 72x3y21 Negative exponent property

5 1 } 6 x21 2 3y2 2 (21) Quotient of powers property

5 1 } 6 x24y3

5 y3

} 6x4 Negative exponent property

50. 6x22y2

} 36xy23 5

1 } 6 x22 2 1y2 2 (23) Quotient of powers property

5 1 } 6 x23y5

5 y5

} 6x3 Negative exponent property

51. 1 x5y4

} x7y8 2 22

5 1 x7y8

} x5y4 2 2 Negative exponent property

5 (x7y8)2

} (x5y4)2 Power of quotient property

5 (x7)2(y8)2

} (x5)2(y4)2


5 x14y16



5 x4y8


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52. 1 90x3y21

} 18x21y22 2

2

5 (90x3y21)2

} (18x21y22)2


5 902(x3)2(y21)2

}} 182(x21)2(y22)2


5 8100x6y22

} 324x22y24 Power of a power property

5 25x6 2 (22)y 22 2 (24) Quotient of powers property

5 25x8y2

53. 1 xy6

} x2y5 2 3 5

(xy6)3

} (x2y5)3


5 x3( y6)3

} (x2)3( y5)3


5 x3y18



5 x23y3

5 y3

} x3 Negative exponent property

Quiz 8.1–8.3 (p. 571)

1. y 5 a } x y 5

24 } x

3 5 a } 8 5

24 }

24

24 5 a 5 26

y 5 24

} x

2. y 5 a } x y 5

218 } x

29 5 a } 2 5

218 }

24

218 5 a 5 9 } 2

y 5 218

} x

3. y 5 a } x y 5 2

40 } 3x

8 }

3 5

a }

25 5 2 40 }

3(24)

8 }

3 p (25) 5 a 5

240 }

212

2 40

} 3 5 a 5 10

} 3

y 5 240

} 3 } x

5 2 40

} 3x

4. y 5 a } x y 5

8 } x

232 5 a }

21

} 4 5

8 }

24

232 1 2 1 } 4 2 5 a 5 22

8 5 a

y 5 8 } x

5. y 5 3 } 2x

1

x

y

21

6. y 5 4 } x 2 2 1 1

2

x

y

1

7. f (x) 5 22x

} 3x 2 6

3x 2 6 5 0 → 3(x 2 2) 5 0 → x 5 2, so x 5 2 is a vertical asymptote.

y 5 a } c 5 2

2 } 3 is a horizontal asymptote.

1x

f(x)

1

8. y 5 28 }

x2 2 1


x2 2 1 5 0 → x 5 61, so x 5 1 and x 5 21 are vertical asymptotes.

m 5 0 and n 5 2, m < n and y 5 0 is a horizontal asymptote.


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487Algebra 2



x y

23 21

22 2 8 } 3

2 1 } 2 32 }

3

0 8

1 } 2 32

} 3

2 2 8 } 3

3 21

1x

22

9. y 5 x2 2 6

} x2 1 2

x2 2 6 5 0 → x 5 6 Ï}

6 , so Ï}

6 and 2 Ï}

6 are x-intercepts. The denominator has no real zeros, so there is no vertical asymptote.


asymptote.

x y

24 5 } 9

2 Ï}

6 0

21 2 5 } 3

0 23

1 5 } 3

Ï}

6 0

4 5 } 9

1

x

y

21

10. g(x) 5 x3 2 8

} 2x2

x3 2 8 5 0 → x 5 2, so 2 is an x-intercept.

2x2 5 0 → x 5 0, so x 5 0 is a vertical asymptote.


x y

24 2 9 } 4

22 22

21 2 9 } 2

1 2 7 } 2

2 0

4 7 } 4

2

x

g(x)

22

11. Strike percentage 5

Total strikes 4

Total pitches

y 5 x 1 16

} x 1 38

(17, 0.60)

Consecutive strikes (after 38)

Str

ike

per

cen

t (a

s a

dec

imal

)

0 10 20 30 40 x

y

00.10.20.30.40.50.60.70.80.9

From the graph, you can estimate that the pitcher must throw 17 consecutive strikes to reach a strike percentage of 0.60.

Mixed Review of Problem Solving (p. 572)

1. a. Let c be the average cost and p be the number of photos printed.

c 5 Unit cost p Number printed 1 Cost of printer

}}}} Number printed

c 5 0.60p 1 200

} p

b.

0 4 8 12 16 20 24 32 p

c

10

0

203040506070

Number of photos

Ave

rag

e co

st (

do

llars

)

(21, 10)

From the graph you can estimate that you have to print 22 photos before the average cost drops to $10 per printed photo.

c. As the number of photos printed increases, the average cost decreases.

2. a. V 5 πr2h

1663 5 πr2h

1663

} πr2 5 h

b. S 5 2πr2 1 2πrh

5 2πr2 1 2πr 1 1663 }

πr2 2 5 2πr2 1 3326

} r

c. S 5 2πr2 1 3326

} r

MinimiumX=6.4205274 Y=777.03881

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h ø 1663

} π (6.42)2 ø 12.84.

So, the oatmeal canister using the least amount of material has a radius of about 6.42 centimeters and a height of about 12.84 centimeters.

3. Yes; 5(800) 5 4000

6(667) 5 4002

7(571) 5 3997

8(500) 5 4000

Each product is approximately equal to 4000. So, the data show inverse variation.

4. Sample answer: The function y 5 1 }

x2 2 5x 1 6 has a

domain of all real numbers except x 5 2 and x 5 3.

5. y 5 a } x

4 5 a } 5

20 5 a

y 5 20

} x 5 20

} 10 5 2

6. a. b 5 aw

} h2

20 5 a(51.2)

} 1.62

20 5 51.2a

} 2.56

20(2.56) 5 51.2a

51.2 5 51.2a

1 5 a

b 5 w

} h2

b. b 5 w

} h2

20 5 45

} h2

20h2 5 45

h2 5 2.25

h 5 1.5

b 5 w

} h2 5

41 }

1.42 5 41

} 1.96 ø 21

b 5 w

} h2

19 5 w }

1.52

19 5 w } 2.25

19(2.25) 5 w

42.75 5 w

Body mass index

Height (m)

Weight (Kg)

20 1.5 45

21 1.4 41

19 1.5 42.75

c. Let h equal the height of the shorter person and let 0.1h 1 h 5 1.1h be the height of the taller person. The body mass index of the taller person is given by

b 5 w }

(1.1h)2 5 w }

1.21h2 and the body mass index of the

shorter person is b 5 w

} h2 . So, the body mass index of

the shorter person is 1.21 times the body mass index of the taller person.

7. I 5 a }

d 2

10 5 a }

12

10 5 a } 1

10 5 a

I 5 10

} d2 5

10 }

152 ø 0.04

The intensity of the sound you hear is about 0.04 watts per square meter if you are 15 meters from the stage.

8. a. M(t) 5 3500

} t 1 500

M(5) 5 3500

} 5 1 500 5 700 1 500 5 1200

The motorcycle’s value is $1200 after 5 years.

b. The horizontal asymptote of the function is M 5 500, so the value of the motorcycle approaches $500 as time passes.

Lesson 8.4


1. 2(x 1 1)

}} (x 1 1)(x 1 3)

5 2(x 1 1)

}} (x 1 1)(x 1 3)

5 2 } x 1 3

2. 40x 1 20

} 10x 1 30

5 20(2x 1 1)

} 10(x 1 3)

5 2 p 10 p (2x 1 1)

}} 10 p (x 1 3)

5 2(2x 1 1)

} x 1 3

3. The expression 4 }

x(x 1 2) cannot be simplifi ed.

4. x 1 4

} x2 2 16

5 x 1 4 }}

(x 1 4)(x 2 4) 5

1 } x 2 4

5. x2 2 2x 2 3

} x2 2 x 2 6

5 (x 2 3)(x 1 1)

}} (x 2 3)(x 1 2)

5 x 1 1

} x 1 2

6. 2x2 1 10x

}} 3x2 1 16x 1 5

5 2x(x 1 5)

}} (3x 1 1)(x 1 5)

5 2x } 3x 1 1

7. New tin: S 5 2(2s)2 1 4(2s)(h) 5 2(4s2) 1 8sh

5 8s2 1 8sh

V 5 (2s)2(h) 5 4s2h

S }

V 5

8s2 1 8sh }

4s2h 5

4s(2s 1 2h) }

4s(sh) 5

2s 1 2h }

sh


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489Algebra 2


8. 3x5y2

} 8xy

p 6xy2

} 9x3y

5 18x6y4

} 72x4y2 5

2 p 9 p x4 p x2 p y2 p y2

}} 2 p 9 p 4 p x4 p y2 5

x2y2

} 4

9. 2x2 2 10x

} x2 2 25

p x 1 3 }

2x2 5 2x(x 2 5)

}} (x 1 5)(x 2 5)

p x 1 3 }

2x p x

5 2x(x 2 5)(x 1 3)

}} (2x)(x)(x 1 5)(x 2 5)

5 x 1 3

} x(x 1 5)

10. x 1 5

} x3 2 1

p (x2 1 x 1 1) 5 x 1 5

} x3 2 1

p x2 1 x 1 1

} 1

5 x 1 5 }}

(x 2 1)(x2 1 x 1 1) p

(x2 1 x 1 1) }

1

5 (x 1 5)(x2 1 x 1 1)

}} (x 2 1)(x2 1 x 1 1)

5 x 1 5

} x 2 1

11. 4x }

5x 2 20 4

x2 2 2x }

x2 2 6x 1 8 5

4x } 5x 2 20 p x

2 2 6x 1 8 }

x2 2 2x

5 4x }

5(x 2 4) p

(x 2 4)(x 2 2) }}

x(x 2 2)

5 4x(x 2 4)(x 2 2)

}} 5x(x 2 4)(x 2 2)

5 4 } 5

12. 2x2 1 3x 2 5

} 6x

4 (2x2 1 5x) 5 2x2 1 3x 2 5

} 6x p 1 }

2x2 1 5x

5 (2x 1 5)(x 2 1)

}} 6x p 1 }

x(2x 1 5)

5 (2x 1 5)(x 2 1)

}} (6x)(x)(2x 1 5)

5 x 2 1

} 6x2


Skill Practice

1. To divide one rational expression by another, multiply the fi rst rational expression by the reciprocal of the second rational expression.

2. A rational expression is simplifi ed when its numerator and denominator have no common factors (other than 61).

3. B; x2 2 9x 1 14

} x2 2 5x 2 14

5 (x 2 7)(x 2 2)

}} (x 2 7)(x 1 2)

5 x 2 2

} x 1 2

4. A; x2 2 4 }

x2 1 9x 1 14 5

(x 1 2)(x 2 2) }}

(x 1 7)(x 1 2) 5

x 2 2 } x 1 7

5. C; x2 1 5x 2 14

} x2 2 4x 1 4

5 (x 1 7)(x 2 2)

}} (x 2 2)(x 2 2)

5 x 1 7

} x 2 2

6. 4x2 }

20x2 2 12x 5

4x p x }

4x(5x 2 3) 5

x } 5x 2 3

7. x2 2 x 2 20

} x2 1 2x 2 15

5 (x 2 5)(x 1 4)

}} (x 1 5)(x 2 3)

The expression cannot be simplifi ed.

8. x2 1 2x 2 24

} x2 1 7x 1 6

5 (x 1 6)(x 2 4)

}} (x 1 6)(x 1 1)

5 x 2 4

} x 1 1

9. x2 2 11x 1 24

}} x2 2 3x 2 40

5 (x 2 8)(x 2 3)

}} (x 2 8)(x 1 5)

5 x 2 3

} x 1 5

10. x2 1 4x 1 4

} x2 2 5x 1 4

5 (x 1 2)(x 1 2)

}} (x 2 4)(x 2 1)


11. 2x2 1 2x 2 4

} x2 2 5x 2 14

5 2(x2 1 x 2 2)

}} x2 2 5x 2 14

5 2(x 1 2)(x 2 1)

}} (x 1 2)(x 2 7)

5 2(x 2 1)

} x 2 7

12. x 2 4

} x3 2 64

5 x 2 4 }}

(x 2 4)(x2 1 4x 1 16) 5

1 }

x2 1 4x 1 16

13. x2 2 36

}} x2 1 12x 1 36

5 (x 1 6)(x 2 6)

}} (x 1 6)(x 1 6)

5 x 2 6

} x 1 6

14. 3x3 1 6x2 1 12x

}} x3 2 8

5 3x(x2 1 2x 1 4)

}} (x 2 2)(x2 1 2x 1 4)

5 3x } x 2 2

15. 8x2 1 10x 2 3

}} 6x2 1 13x 1 6

5 (4x 2 1)(2x 1 3)

}} (3x 1 2)(2x 1 3)

5 4x 2 1

} 3x 1 2

16. 5x2 1 18x 2 8

}} 10x2 2 x 2 2

5 (5x 2 2)(x 1 4)

}} (5x 1 2)(2x 2 1)


17. x3 2 5x2 2 3x 1 15

}} x2 2 8x 1 15

5 x2(x 2 5) 2 3(x 2 5)

}} (x 2 5)(x 2 3)

5 (x2 2 3)(x 2 5)

}} (x 2 5)(x 2 3)

5 x2 2 3

} x 2 3

18. Variable terms that are not factors were factored out.

x2 1 16x 2 80

}} x2 2 16

5 (x 1 20)(x 2 4)

}} (x 1 4)(x 2 4)

5 x 1 20

} x 1 4

19. You can only divide out common factors. Since the factors that are divided out are not common factors of the entire numerator and denominator, you cannot divide them out.

x2 1 16x 1 48

}} x2 1 8x 1 16

5 (x 1 12)(x 1 4)

}} (x 1 4)(x 1 4)

5 x 1 12

} x 1 4

20. B; The numerator and denominator factor

as (x 1 4)(x 1 2)

}} (x 1 3)(x 2 1)

. The numerator and denominator

have no common factors (other than 61), so the

expression x2 1 6x 1 8

} x2 1 2x 2 3

is in simplifi ed form.

21. P 5 4(2x) 5 8x

A 5 (2x)(2x) 5 4x2

P

} A

5 8x

} 4x2 5

4 p 2 p x } 4 p x p x 5

2 } x

22. P 5 x 1 (3x 2 x) 1 x 1 x 1 x 1 x 1 3x

5 x 1 2x 1 7x

5 10x

A 5 x(3x) 1 x(x) 5 3x2 1 x2 5 4x2

P

} A

5 10x

} 4x2 5

5 p 2 p x } 2 p 2 p x p x 5

5 } 2x

23. The fi eld in Exercise 21 because the perimeter of the fi eld in Exercise 21 is smaller and the areas of both fi elds are the same.

24. 5x3y

} x2y2 p

y3

} 15x2 5

5x3y4

} 15x4y2 5

5 p x3 p y2 p y2

}} 5 p 3 p x3 p x p y2 5

y2

} 3x

25. 48x5y3

} y4 p

x2y }

6x3y2 5 48x7y4

} 6x3y6 5

8 p 6 p x3 p x4 p y4

}} 6 p x3 p y4 p y2 5

8x4

} y2

26. x(x 2 3)

} x 2 2

p (x 1 3)(x 2 2)

}} x 5 x(x 2 3)(x 1 3)(x 2 2)

}} x(x 2 2)

5 (x 2 3)(x 1 3)

27. 4(x 1 5)

} x2 p

x(x 1 1) }

2(x 1 5) 5

2(2)(x)(x 1 5)(x 1 1) }}

2(x)(x)(x 1 5) 5

2(x 1 1) } x

28. 3x 2 12

} x 1 5 p x 1 6 }

2x 2 8 5

3(x 2 4) } x 1 5 p x 1 6

} 2(x 2 4)

5 3(x 2 4)(x 1 6)

}} 2(x 1 5)(x 2 4)

5 3(x 1 6)

} 2(x 1 5)


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29. x 1 5

} 4x 2 16

p 2x2 2 32 }

x2 2 25 5

x 1 5 }

4(x 2 4) p

2(x2 2 16) }}

(x 1 5)(x 2 5)

5 2(x 1 5)(x 1 4)(x 2 4)

}} 2(2)(x 2 4)(x 1 5)(x 2 5)

5 x 1 4

} 2(x 2 5)

30. x2 1 3x 2 4

} x2 1 4x 1 4

p 2x2 1 4x }

x2 2 4x 1 3

5 (x 1 4)(x 2 1)

}} (x 1 2)(x 1 2)

p 2x(x 1 2)

}} (x 2 3)(x 2 1)

5 2x(x 1 4)(x 2 1)(x 1 2)

}}} (x 1 2)(x 1 2)(x 2 3)(x 2 1)

5 2x(x 1 4)

}} (x 1 2)(x 2 3)

31. x2 2 3x 2 10

} x2 2 2x 2 15

p (x2 1 10x 1 21)

5 x2 2 3x 2 10

} x2 2 2x 2 15

p x2 1 10x 1 21

}} 1

5 (x 2 5)(x 1 2)(x 1 7)(x 1 3)

}}} (x 2 5)(x 1 3)

5 (x 1 2)(x 1 7)

32. x2 1 5x 2 36

} x2 2 49

p (x2 2 11x 1 28)

5 x2 1 5x 2 36

} x2 2 49

p x2 2 11x 1 28

}} 1

5 (x 1 9)(x 2 4)(x 2 4)(x 2 7)

}}} (x 1 7)(x 2 7)

5 (x 1 9)(x 2 4)2

}} x 1 7

33. 4x2 1 20x

} x3 1 4x2 p (x2 1 8x 1 16) 5

4x2 1 20x }

x3 1 4x2 p x2 1 8x 1 16

} 1

5 4x(x 1 5)(x 1 4)(x 1 4)

}} x p x(x 1 4)

5 4(x 1 5)(x 1 4)

}} x

34. 5x2y3

} x7

4 30xy4

} y3 5

5x2y3

} x7 p

y3

} 30xy4

5 5x2y6

} 30x8y4 5

5 p x2 p y4 p y2

}} 5 p 6 p x2 p x6 p y4 5

y2

} 6x6

35. 8x2y2z

} xz3 4

10xy }

x4z 5

8x2y2z }

xz3 p x4z }

10xy 5

8x6y2z2

} 10x2yz3

5 2 p 4 p x2 p x4 p y p y p z2

}} 2 p 5 p x2 p y p z2 p z

5 4x4y

} 5z

36. (x 1 3)(x 2 2)

}} x(x 1 1)

4 x 1 3

} x 5 (x 1 3)(x 2 2)

}} x(x 1 1)

p x }

x 1 3

5 x(x 1 3)(x 2 2)

}} x(x 1 1)(x 1 3)

5 x 2 2

} x 1 1

37. 8x2

} x 1 4

4 x }

2(x 2 4) 5

8x2

} x 1 4 p 2(x 2 4)

} x

5 16(x)(x)(x 2 4)

}} x(x 1 4)

5 16x(x 2 4)

} x 1 4

38. x2 2 6x 2 27

} 2x2 1 2x

4 x2 2 14x 1 45

}} x2

5 x2 2 6x 2 27

} 2x2 1 2x

p x2 }}

x2 2 14x 1 45

5 (x 2 9)(x 1 3)

}} 2x(x 1 1)

p x2 }}

(x 2 9)(x 2 5)

5 x p x p (x 2 9)(x 1 3)

}} 2x(x 1 1)(x 2 9)(x 2 5)

5 x(x 1 3)

}} 2(x 1 1)(x 2 5)

39. x2 2 4x 2 5

} x 1 5 4 (x2 1 6x 1 5) 5 x2 2 4x 2 5

} x 1 5 p 1 }

x2 1 6x 1 5

5 (x 2 5)(x 1 1)

}} (x 1 5)(x 1 5)(x 1 1)

5 x 2 5

} (x 1 5)2

40. 3x2 1 13x 1 4

}} x2 2 4

4 4x 1 16

} x 1 2 5 3x2 1 13x 1 4

}} x2 2 4

p x 1 2 }

4x 1 16

5 (3x 1 1)(x 1 4)

}} (x 1 2)(x 2 2)

p x 1 2 }

4(x 1 4)

5 (3x 1 1)(x 1 4)(x 1 2)

}} 4(x 1 2)(x 2 2)(x 1 4)

5 3x 1 1

} 4(x 2 2)

41. x2 2 x 2 2

} x2 1 4x 2 5

4 x 2 2

} 5x 1 25 5 x2 2 x 2 2

} x2 1 4x 2 5

p 5x 1 25 }

x 2 2

5 (x 2 2)(x 1 1)

}} (x 1 5)(x 2 1)

p 5(x 1 5)

} x 2 2

5 5(x 2 2)(x 1 1)(x 1 5)

}} (x 1 5)(x 2 1)(x 2 2)

5 5(x 1 1)

} x 2 1

42. x2 2 8x 1 15

} x2 1 4x

4 (x2 2 x 2 20)

5 x2 2 8x 1 15

} x2 1 4x

p 1 }

x2 2 x 2 20

5 (x 2 5)(x 2 3)

}} x(x 1 4)(x 2 5)(x 1 4)

5 x 2 3

} x(x 1 4)2

43. x2 1 12x 1 32 }}

6x 1 42 4

x2 1 4x }

x2 2 49 5

x2 1 12x 1 32 }} 6x 1 42 p x

2 2 49 }

x2 1 4x

5 (x 1 8)(x 1 4)

}} 6(x 1 7)

p (x 1 7)(x 2 7)

}} x(x 1 4)

5 (x 1 8)(x 1 4)(x 1 7)(x 2 7)

}}} 6x(x 1 7)(x 1 4)

5 (x 1 8)(x 2 7)

}} 6x

44. y 5 x2 1 10x 1 21

}} x 1 3 5 (x 1 7)(x 1 3)

}} x 1 3 5 x 1 7

The graph of y 5 x2 1 10x 1 21

}} x 1 3 is the same as the graph

of y 5 x 1 7 except that there is a hole at (23, 4).

y 5 x2 1 10x 1 21

}} x 1 3

2

x

y

22

(23, 4)

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491Algebra 2


45. y 5 x2 2 36

} x 2 6 5 (x 1 6)(x 2 6)

}} x 2 6 5 x 1 6

The graph of y 5 x2 2 36

} x 2 6 is the same as the graph of

y 5 x 1 6 except that there is a hole at (6, 12).

y 5 x2 2 36

} x 2 6

2

x

y

2

(6, 12)

46. y 5 2x2 2 x 2 10

} x 1 2 5 (2x 2 5)(x 1 2)

}} x 1 2 5 2x 2 5

The graph of y 5 2x2 2 x 2 10

} x 1 2 is the same as the graph

of y 5 2x 2 5 except that there is a hole at (22, 29).

y 5 2x2 2 x 2 10

} x 1 2

2

x

y

21

(22, 29)

47. Let a and b be the unknown side lengths of the triangle.

8x

6x 15x

a b

a2 5 (6x)2 1 (8x)2 b2 5 (15x)2 1 (8x)2

a2 5 36x2 1 64x2 b2 5 225x2 1 64x2

a2 5 100x2 b2 5 289x2

a 5 10x b 5 17x

P 5 10x 1 17x 1 6x 1 15x 5 48x

A 5 1 } 2 (6x 1 15x)(8x) 5

1 } 2 (21x)(8x) 5

1 } 2 (168x2) 5 84x2

P

} A

5 48x

} 84x2 5

4 p 12 p x } 7 p 12 p x p x 5

4 } 7x

Problem Solving

48. Vp 5 1 } 3 Bh 5

1 } 3 (2r)2h 5

1 } 3 (4)r2h

Vc 5 1 } 3 πr2h

Vp

} Vc 5

1 } 3 (4)r2h

} 1 }

3 πr2h

5 4 } π

49. Average amount 5

Gross ticket sales 4

Total attendance

P 5 S 4 A

5 26420t 1 292,000

}} 6.02t2 2 125t 1 1000

4 2407t 1 7220

}} 5.92t2 2 131t 1 1000

5 26420t 1 292,000

}} 6.02t2 2 125t 1 1000

p 5.92t2 2 131t 1 1000 }}

2407t 1 7220

For 1999, t 5 7:

26420(7) 1 292,000

}} 6.02(7)2 2 125(7) 1 1000

p 5.92(7)2 2 131(7) 1 1000

}} 2407(7) 1 7220

5 247,060

} 419.98 p 373.08 }

4371 ø 50.21

The average amount a person paid per ticket in 1999 was $50.21.

50. a. ha

} hr

5 k1H3V2

} k2H2 5

k1 p H2 p H p V 2

}} k2 p H2 5

k1 p HV2

} k2

b. k1HV2

} k2

5 1

k1HV 2 5 k2

V 2 5 k2

} k1H

The shorter runner has the advantage. The larger the height the smaller the fraction representing velocity.

51. a. Vsphere 5 4 } 3 πr3

Vcylinder 5 πr2h

4 }

3 πr3 5 πr2h

4 }

3 p πr3

} πr2 5 h

4 }

3 r 5 h

b. Ssphere 5 4πr2

Scylinder 5 2πr2 1 2πrh 5 2πr2 1 2πr 1 4 } 3 r 2

5 2πr2 1 8 } 3 πr2 5

14 } 3 πr2

c. Ssphere

} Scylinder

5 4πr2

} 14

} 3 πr2

5 6 } 7

Because 6 < 7, the spherical tank uses less material.

52. a. S }

V 5

2πr2 1 2πrh }

πr2h

5 2πr(r 1 h)

} π p r p r p h 5

2(r 1 h) }

rh

b. Soup can: S }

V 5

2(r 1 h) }

rh 5

2(3.4 1 10.2) }}

(3.4)(10.2)

5 2(13.6)

} 34.68 ø 0.784


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Coffee can: S } V 5

2(r 1 h) }

rh 5

2(7.8 1 15.9) }}

(7.8)(15.9)

5 2(23.7)

} 124.02 ø 0.382

Paint can: S }

V 5

2(r 1 h) }

rh 5

2(8.4 1 19.4) }}

(8.4)(19.4)

5 2(27.8)

} 162.96 ø 0.341

c. The paint can is the most effi cient, then the coffee can, and the soup can is the least effi cient. The lower the effi ciency ratio, the more that can be put into the can, while using less material to make the can.

53. S 5 4πr2 1 2πrl

V 5 4 } 3 πr3 1 πr2l

S }

V 5

4π r2 1 2πrl }}

4 }

3 π r3 1 π r2l

5 2π r(2r 1 l )

} π r2 1 4 }

3 r 1 l 2

5 2(2r 1 l )

} r 1 4 }

3 r 1 l 2

5 3[2(2r 1 l )]

} 3r 1 4 }

3 r 1 l �2

5 6(2r 1 l )

} r (4r 1 3l )

Mixed Review

54. Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factors of 80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

The GCF is 16.

Multiples of 48: 48, 96, 144, 192, 240, …

Multiples of 80: 80, 160, 240, …

The LCM is 240.

55. Factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Factors of 155: 1, 5, 31, 155

The GCF is 5.

Multiples of 120: 120, 240, 360, …, 3720, …

Multiples of 155: 155, 310, 465, …, 3720, …

The LCM is 3720.

56. Factors of 38: 1, 2, 19, 38

Factors of 95: 1, 5, 19, 95

The GCF is 19.

Multiples of 38: 38, 76, 114, 152, 190, …

Multiples of 95: 95, 190, …

The LCM is 190.

57. Factors of 52: 1, 2, 4, 13, 26, 52

Factors of 91: 1, 7, 13, 91

The GCF is 13.

Multiples of 52: 52, 104, 156, 208, 260, 312, 364, …

Multiples of 91: 91, 182, 273, 364, …

The LCM is 364.

58. Factors of 66: 1, 2, 3, 6, 11, 22, 33, 66

Factors of 154: 1, 2, 7, 11, 14, 22, 77, 154

The GCF is 22.

Multiples of 66: 66, 132, 198, 264, 330, 396, 462, …

Multiples of 154: 154, 308, 462, …

The LCM is 462.

59. Factors of 360: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360

Factors of 450: 1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 150, 225, 450

The GCF is 90.

Multiples of 360: 360, 720, 1080, 1440, 1800, …

Multiples of 450: 450, 900, 1350, 1800, …

The LCM is 1800.

60. x(x2 1 4x 2 7) 5 x(x2) 1 x(4x) 2 x(7) 5 x3 1 4x2 2 7x

61. (x 1 9)(x 2 5) 5 (x)(x) 1 (x)(25) 1 (9)(x) 1 (9)(25)

5 x2 2 5x 1 9x 2 45 5 x2 1 4x 2 45

62. (x 1 11)(x 2 7)

5 (x)(x) 1 (x)(27) 1 (11)(x) 1 (11)(27)

5 x2 2 7x 1 11x 2 77 5 x2 1 4x 2 77

63. (x 1 2)(x2 2 6x 1 10) 5 (x 1 2)x2 2 (x 1 2)6x 1 (x 1 2)10

5 x3 1 2x2 2 6x2 2 12x 1 10x 1 20

5 x3 2 4x2 2 2x 1 20

64. (3x 2 7)(x2 2 5x) 5 (3x)(x2) 1 (3x)(25x) 1 (27)(x2) 1 (27)(25x)

5 3x3 2 15x2 2 7x2 1 35x 5 3x3 2 22x2 1 35x

65. (x 1 5)(x3 1 8x2) 5 (x)(x3) 1 (x)(8x2) 1 (5)(x3) 1 (5)(8x2) 5 x4 1 8x3 1 5x3 1 40x2 5 x4 1 13x3 1 40x2

Graphing Calculator Activity 8.4 (p. 581)

1. x2 2 5x

} x2 2 7x 1 10

5 x(x 2 5)

}} (x 2 5)(x 2 2)

5 x } x 2 2

X Y1 Y21 -1 -12 ERROR ERROR3 3 34 2 25 ERROR 1.66676 1.5 1.57 1.4 1.4X=1


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493Algebra 2


2. 3x2 1 6x

} x2 2 2x 2 8

5 3x(x 1 2)

}} (x 2 4)(x 1 2)

5 3x } x 2 4

X Y1 Y2-2 ERROR 1-1 .6 .60 0 0 1 -1 -12 -3 -33 -9 -94 ERROR ERRORX=-2

3. x2 1 5x 1 4

} x2 1 x 2 12

5 (x 1 1)(x 1 4)

}} (x 1 4)(x 2 3)

5 x 1 1

} x 2 3

X Y1 Y2-4 ERROR .42857-3 .33333 .33333-2 .2 .2-1 0 00 -.3333 -.33331 -1 -12 -3 -3X=-4

4. x 1 3

} 5x2 p x 2 1

} x 1 3

5 (x 1 3)(x 2 1)

}} 5x2(x 1 3)

5 x 2 1

} 5x2

X Y1 Y2-3 ERROR -7.2-2 -2.4 -2.4-1 -.4 -.40 0 01 0 02 .8 .83 3.6 3.6X=-3

5. 4x2 2 8x

} 5x 1 15

4 x 2 2

} x 1 3 5 4x2 2 8x

} 5x 1 15 p x 1 3 }

x 2 2

5 4x(x 2 2)(x 1 3)

}} 5(x 1 3)(x 2 2)

5 4x

} 5

X Y1 Y2-3 ERROR -2.4-2 -1.6 -1.6-1 -.8 -.80 0 01 .8 .82 ERROR 1.63 2.4 2.4X=-3

6. x2 2 3x 2 10

} x2 1 3x 1 3

p x2 1 2x 2 3

} x2 1 x 2 2

5 (x 2 5)(x 1 2)

}} x2 1 3x 1 3

p (x 1 3)(x 2 1)

}} (x 1 2)(x 2 1)

5 (x 2 5)(x 1 2)(x 1 3)(x 2 1)

}}} (x2 1 3x 1 3)(x 1 2)(x 2 1)

5 (x 2 5)(x 1 3)

}} x2 1 3x 1 3

X Y1 Y2-2 ERROR -7-1 -12 -120 -5 -51 ERROR -2.2862 -1.154 -1.1543 -.5714 -.57144 -.2258 -.2258X=-2

Lesson 8.5


1. 7 }

12x 2

5 } 12x 5

7 2 5 } 12x 5

2 } 12x 5

1 } 6x

2. 2 }

3x2 1 1 }

3x2 5 2 1 1

} 3x2 5

3 }

3x2 5 1 }

x2

3. 4x }

x 2 2 2

x } x 2 2 5

4x 2 x } x 2 2 5

3x } x 2 2

4. 2x2 }

x2 1 1 1

2 }

x2 1 1 5

2x2 1 2 }

x2 1 1 5

2(x2 1 1) }

x2 1 1 5 2

5. 5x3

10x2 2 15x 5 5x(2x 2 3)

LCM 5 5x3(2x 2 3)


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6. 8x 2 16 5 8(x 2 2) 5 23(x 2 2)

12x2 1 12x 2 72 5 12(x2 1 x 2 6) 5 (22)(3)(x 1 3)(x 2 2)

LCM 5 (23)(3)(x 1 3)(x 2 2) 5 24(x 1 3)(x 2 2)

7. 3 }

4x 2

1 } 7 5

3 } 4x p 7 } 7 2

1 } 7 p 4x

} 4x

5 21

} 28x 2 4x

} 28x 5 21 2 4x

} 28x

8. 1 }

3x2 1 x }

9x2 2 12x 5

1 }

3x2 1 x }

3x(3x 2 4)

5 1 }

3x2 p 3x 2 4

} 3x 2 4

1 x }

3x(3x 2 4) p x }

x

5 3x 2 4

} 3x2(3x 2 4)

1 x2

} 3x2(3x 2 4)

5 x2 1 3x 2 4

} 3x2(3x 2 4)

9. x }

x2 2 x 2 12 1

5 } 12x 2 48

5 x }}

(x 2 4)(x 1 3) 1

5 }

12(x 2 4)

5 x }}

(x 2 4)(x 1 3) p 12

} 12

1 5 }

12(x 2 4) p x 1 3

} x 1 3

5 12x }}

12(x 2 4)(x 1 3) 1

5x 1 15 }}

12(x 2 4)(x 1 3)

5 17x 1 15

}} 12(x 2 4)(x 1 3)

10. x 1 1 }

x2 1 4x 1 4 2

6 }

x2 2 4

5 x 1 1 }}

(x 1 2)(x 1 2) 2

6 }}

(x 1 2)(x 2 2)

5 x 1 1 }}

(x 1 2)(x 1 2) p x 2 2

} x 2 2

2 6 }}

(x 1 2)(x 2 2) p x 1 2

} x 1 2

5 x2 2 x 2 2

}} (x 1 2)2(x 2 2)

2 6x 1 12

}} (x 1 2)2(x 2 2)

5 x2 2 x 2 2 2 (6x 1 12)

}} (x 1 2)2(x 2 2)

5 x2 2 x 2 2 2 6x 2 12

}} (x 1 2)2(x 2 2)

5 x2 2 7x 2 14

}} (x 1 2)2(x 2 2)

11. x }

6 2

x } 3 }

x } 5 2

7 } 10 5

x } 6 2

x } 3 }

x } 5 2

7 } 10 p 60

} 60

5 10x 2 20x

} 12x 2 42

5 210x

} 12x 2 42

5 2(25x)

} 2(6x 2 21)

5 25x

} 6x 2 21

5 25x }

3(2x 2 7)

12. 2 }

x 2 4

} 2 }

x 1 3

5 2 } x 2 4

} 2 }

x 1 3

p x } x 5

2 2 4x } 2 1 3x 5

2(1 2 2x) } 2 1 3x

13.

3 } x 1 5 }}

2 }

x 2 3 1

1 } x 1 5

5

3 } x 1 5 }}

2 }

x 2 3 1

1 } x 1 5

p (x 2 3)(x 1 5)

}} (x 2 3)(x 1 5)

5 3(x 2 3) }}

2(x 1 5) 1 x 2 3

5 3(x 2 3) }} 2x 1 10 1 x 2 3

5 3(x 2 3)

} 3x 1 7


Skill Practice

1. A fraction that contains a fraction in its numerator or denominator is called a complex fraction.

2. To add rational expressions with unlike denominators, fi rst fi nd a common denominator. Then rewrite each rational expression using the common denominator. Finally, add the numerators.

3. 15

} 4x

1 5 } 4x 5

15 1 5 } 4x 5

20 } 4x 5

5 } x

4. x }

16x2 2 4 }

16x2 5 x 2 4

} 16x2

5. 9 }

x 1 1 2

2x } x 1 1 5

9 2 2x } x 1 1

6. 3x2

} x 2 8

1 6x } x 2 8 5

3x2 1 6x } x 2 8 5

3x(x 1 2) } x 2 8

7. 5x }

x 1 3 1

15 } x 1 3 5

5x 1 15 } x 1 3 5

5(x 1 3) } x 1 5 5 5

8. 4x2 }

2x 2 1 2

1 } 2x 2 1 5

4x2 2 1 } 2x 2 1 5

(2x 1 1)(2x 1 1) }} 2x 2 1 5 2x 1 1

9. 3x, 3(x 2 2)

LCM 5 3x(x 2 2)

10. 2x2

4x 1 12 5 4(x 1 3) 5 22(x 1 3)

LCM 5 4x2(x 1 3)

11. 2x, 2x(x 2 5)

LCM 5 2x(x 2 5)

12. 24x2 5 23(3)(x2) 8x2 2 16x 5 8x(x 2 2) 5 23(x)(x 2 2)

LCM 5 23(3)(x2)(x 2 2) 5 24x2(x 2 2)

13. x

x 2 5

x2 2 25 5 (x 1 5)(x 2 5)

LCM 5 x(x 1 5)(x 2 5)

14. 9x2 2 16 5 (3x 1 4)(3x 2 4)

3x2 2 2x 2 8 5 (3x 1 4)(x 2 2)

LCM 5 (3x 1 4)(3x 2 4)(x 2 2)


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495Algebra 2


15. D; 3x2 2 9x 5 3x(x 2 3)

6x2 5 2(3)(x2)

LCM 5 6x2(x 2 3)

16. 12

} 5x 1 7 } 6x 5

12 } 5x p 6 }

6 1

7 } 6x p 5 } 5 5

72 } 30x 1

35 } 30x 5

107 } 30x

17. 8 }

3x2 2 5 } 4x 5

8 }

3x2 p 4 }

4 2

5 } 4x p 3x

} 3x

5 32

} 12x2 2

15x }

12x2 5 32 2 15x

} 12x2

18. x 2 4

} 5x 2 12 }

5(x 2 4) 5

x 2 4 } 5x p x 2 4

} x 2 4

2 12 }

5(x 2 4) p x }

x

5 x2 2 8x 1 16

} 5x(x 2 4)

2 12x }

5x(x 2 4)

5 x2 2 20x 1 16

}} 5x(x 2 4)

19. 12 }

x2 1 5x 2 24 1

3 } x 2 3 5

12 }}

(x 1 8)(x 2 3) 1

3 } x 2 3

5 12 }}

(x 1 8)(x 2 3) 1

3 } x 2 3 p x 1 8

} x 1 8

5 12 }}

(x 1 8)(x 2 3) 1

3x 1 24 }}

(x 1 8)(x 2 3)

5 3x 1 36

}} (x 1 8)(x 2 3)

5 3(x 1 12)

}} (x 1 8)(x 2 3)

20. 3 }

x 1 4 2

1 } x 1 6 5

3 } x 1 4 p x 1 6

} x 1 6

2 1 } x 1 6 p x 1 4

} x 1 4

5 3x 1 18

}} (x 1 4)(x 1 6)

2 x 1 4 }}

(x 1 4)(x 1 6)

5 3x 1 18 2 (x 1 4)

}} (x 1 4)(x 1 6)

5 3x 1 18 2 x 2 4

}} (x 1 4)(x 1 6)

5 2x 1 14

}} (x 1 4)(x 1 6)

5 2(x 1 7)

}} (x 1 4)(x 1 6)

21. 9 }

x 2 3 1

2x } x 1 1 5

9 } x 2 3 p x 1 1

} x 1 1

1 2x } x 1 1 p x 2 3

} x 2 3

5 9x 1 9 }}

(x 2 3)(x 1 1) 1

2x2 2 6x }}

(x 2 3)(x 1 1)

5 2x2 1 3x 1 9

}} (x 2 3)(x 1 1)

22. x 1 4

} x2 2 4

2 15 } x 2 2 5

x 1 4 }}

(x 1 2)(x 2 2) 2

15 } x 2 2

5 x 1 4 }}

(x 1 2)(x 2 2) 2

15 } x 2 2 p x 1 2

} x 1 2

5 x 1 4 }}

(x 1 2)(x 2 2) 2

15x 1 30 }}

(x 1 2)(x 2 2)

5 x 1 4 2 (15x 1 30)

}} (x 1 2)(x 2 2)

5 x 1 4 2 15x 2 30

}} (x 1 2)(x 2 2)

5 214x 2 26

}} (x 1 2)(x 2 2)

5 22(7x 1 13)

}} (x 1 2)(x 2 2)

23. 215x }

x2 2 8x 1 16 1

12 } x 2 4 5

215x }}

(x 2 4)(x 2 4) 1

12 } x 2 4

5 215x }}

(x 2 4)(x 2 4) 1

12 } x 2 4 p x 2 4

} x 2 4

5 215x }}

(x 2 4)(x 2 4) 1

12x 2 48 }}

(x 2 4)(x 2 4)

5 23x 2 48

} (x 2 4)2

5 23(x 1 16)

} (x 2 4)2

24. x2 2 5 }

x2 1 5x 2 14 2

x 1 3 } x 1 7 5

x2 2 5 }}

(x 1 7)(x 2 2) 2

x 1 3 } x 1 7

5 x2 2 5 }}

(x 1 7)(x 2 2) 2

x 1 3 } x 1 7 p x 2 2

} x 2 2

5 x2 2 5 }}

(x 1 7)(x 2 2) 2

x2 1 x 2 6 }}

(x 1 7)(x 2 2)

5 x2 2 5 2 (x2 1 x 2 6)

}} (x 1 7)(x 2 2)

5 x2 2 5 2 x2 2 x 1 6

}} (x 1 7)(x 2 2)

5 2x 1 1

}} (x 1 7)(x 2 2)

25. You must have a common denominator before you can add values in the numerator.

x }

x 1 2 1

4 } x 2 5 5

x } x 1 2 p x 2 5

} x 2 5 1 4 } x 2 5 p x 1 2

} x 1 2

5 x2 2 5x

}} (x 1 2)(x 2 5)

1 4x 1 8 }}

(x 1 2)(x 2 5)

5 x2 2 x 1 8

}} (x 1 2)(x 2 5)

26. C; 2x }

x 1 4 2

x2 1 4 }

x2 2 16 5

2x } x 1 4 2

x2 1 4 }}

(x 1 4)(x 2 4)

5 2x } x 1 4 p x 2 4

} x 2 4

2 x2 1 4 }}

(x 1 4)(x 2 4)

5 2x2 2 8x

}} (x 1 4)(x 2 4)

2 x2 1 4 }}

(x 1 4)(x 2 4)

5 2x2 2 8x 2 (x2 1 4)

}} (x 1 4)(x 2 4)

5 2x2 2 8x 2 x2 2 4

}} (x 1 4)(x 2 4)

5 x2 2 8x 2 4

}} (x 1 4)(x 2 4)


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27. x }

x2 2 9 1

x 1 1 }

x2 1 6x 1 9

5 x }}

(x 1 3)(x 2 3) 1

x 1 1 }}

(x 1 3)(x 1 3)

5 x }}

(x 1 3)(x 2 3) p x 1 3

} x 1 3

1 x 1 1 }}

(x 1 3)(x 1 3) p x 2 3

} x 2 3

5 x2 1 3x

}} (x 1 3)(x 1 3)(x 2 3)

1 x2 2 2x 2 3

}} (x 1 3)(x 1 3)(x 2 3)

5 2x2 1 x 2 3

}} (x 1 3)(x 1 3)(x 2 3)

5 (2x 1 3)(x 2 1)

}} (x 1 3)(x 1 3)(x 2 3)

5 (2x 1 3)(x 2 1)

}} (x 1 3)2(x 2 3)

28. x 1 3 }

x2 2 2x 2 8 2

x 2 5 }}

x2 2 12x 1 32

5 x 1 3 }}

(x 2 4)(x 1 2) 2

x 2 5 }}

(x 2 4)(x 2 8)

5 x 1 3 }}

(x 2 4)(x 1 2) p x 2 8

} x 2 8

2 x 2 5 }}

(x 2 4)(x 2 8) p x 1 2

} x 1 2

5 x2 2 5x 2 24

}} (x 2 4)(x 1 2)(x 2 8)

2 x2 2 3x 2 10

}} (x 2 4)(x 1 2)(x 2 8)

5 x2 2 5x 2 24 2 (x2 2 3x 2 10)

}}} (x 2 4)(x 1 2)(x 2 8)

5 x2 2 5x 2 24 2 x2 1 3x 1 10

}}} (x 2 4)(x 1 2)(x 2 8)

5 22x 2 14

}} (x 2 4)(x 1 2)(x 2 8)

5 22(x 1 7)

}} (x 2 4)(x 1 2)(x 2 8)

29. x 1 2

} x 2 4

1 2 } x 1

5x } 3x 2 1

5 x 1 2

} x 2 4 p x(3x 2 1)

} x(3x 2 1)

1 2 } x p

(x 2 4)(3x 2 1) }}

(x 2 4)(3x 2 1)

1 5x } 3x 2 1 p

x(x 2 4) }

x(x 2 4)

5 x(3x2 1 5x 2 2)

}} x(x 2 4)(3x 2 1)

1 2(3x2 2 13x 1 4)

}} x(x 2 4)(3x 2 1)

1 5x2(x 2 4)

}} x(x 2 4)(3x 2 1)

5 3x3 1 5x2 2 2x

}} x(x 2 4)(3x 2 1)

1 6x2 2 26x 1 8

}} x(x 2 4)(3x 2 1)

1 5x3 2 20x2

}} x(x 2 4)(3x 2 1)

5 8x3 2 9x2 2 28x 1 8

}} x(x 2 4)(3x 2 1)

30. x 1 3

} x2 2 25

2 x 2 1

} x 2 5 1 3 } x 1 3

5 x 1 3 }}

(x 1 5)(x 2 5) 2

x 2 1 } x 2 5 1

3 } x 1 3

5 x 1 3 }}

(x 1 5)(x 2 5) p x 1 3

} x 1 3

2 x 2 1

} x 2 5 p (x 1 5)(x 1 3)

}} (x 1 5)(x 1 3)

1 3 } x 1 3 p

(x 1 5)(x 2 5) }}

(x 1 5)(x 2 5)

5 x2 1 6x 1 9

}} (x 1 5)(x 2 5)(x 1 3)

2 (x 2 1)(x2 1 8x 1 15)

}} (x 1 5)(x 2 5)(x 1 3)

1 3(x2 2 25) }}

(x 1 5)(x 2 5)(x 1 3)

5 x2 1 6x 1 9 2 (x3 1 7x2 1 7x 2 15) 1 3x2 2 75

}}}} (x 1 5)(x 2 5)(x 1 3)

5 x2 1 6x 1 9 2 x3 2 7x2 2 7x 1 15 1 3x2 2 75

}}}} (x 1 5)(x 2 5)(x 1 3)

5 2x3 2 3x2 2 x 2 51

}} (x 1 5)(x 2 5)(x 1 3)

31. x }

3 2 6

} 10 1

4 } x 5

x } 3 2 6

} 10 1

4 } x p 3x

} 3x

5 x2 2 18x

} 30x 1 12 5 x(x 2 18)

} 6(5x 1 2)

32. 15 2

2 } x }

x } 5 1 4

5 15 2

2 } x }

x } 5 1 4

p 5x } 5x 5

75x 2 10 }

x2 1 20x 5

5(15x 2 2) }

x(x 1 20)

33.

16 }

x 2 2 }

4 }

x 1 1 1

6 } x 5

16 } x 2 2 }

4 }

x 1 1 1

6 } x p

x(x 2 2)(x 1 1) }}

x(x 2 2)(x 1 1)

5 16x(x 1 1)

}}} 4x(x 2 2) 1 6(x 2 2)(x 1 1)

5 16x(x 1 1)

}} 4x2 2 8x 1 6(x2 2 x 2 2)

5 16x(x 1 1)

}} 4x2 2 8x 1 6x2 2 6x 2 12

5 16x(x 1 1)

}} 10x2 2 14x 2 12

5 16x(x 1 1)

}} 2(5x2 2 7x 2 6)

5 8x(x 1 1)

}} (5x 1 3)(x 2 2)

34.

1 }

2x 2 5 2

7 } 8x 2 20 }}

x }

2x 2 5 5

1 } 2x 2 5 2

7 }

4(2x 2 5) }}

x }

2x 2 5

5

1 } 2x 2 5 2

7 }

4(2x 2 5) }}

x }

2x 2 5 p

4(2x 2 5) }

4(2x 2 5)

5 4 2 7

} 4x

5 23

} 4x


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497Algebra 2


35.

3 }

x 2 2 2

6 }

x2 2 4 }}

3 }

x 1 2 1

1 } x 2 2

5

3 } x 2 2 2

6 }}

(x 1 2)(x 2 2) }}

3 }

x 1 2 1

1 } x 2 2

5

3 } x 2 2 2

6 }}

(x 1 2)(x 2 2) }}

3 }

x 1 2 1

1 } x 2 2

p (x 1 2)(x 2 2)

}} (x 1 2)(x 2 2)

5 3(x 1 2) 2 6

}} 3(x 2 2) 1 (x 1 2)

5 3x 1 6 2 6

}} 3x 2 6 1 x 1 2

5 3x } 4x 2 4

5 3x }

4(x 2 1)

36.

1 }

3x2 2 3 }}

5 }

x 1 1 2

x 1 4 }

x2 2 3x 2 4

5

1 }

3(x2 2 1) }}

5 }

x 1 1 2

x 1 4 }}

(x 2 4)(x 1 1)

5

1 }}

3(x 1 1)(x 2 1) }}

5 }

x 1 1 2

x 1 4 }}

(x 2 4)(x 1 1)

5

1 }}

3(x 1 1)(x 2 1) }}

5 }

x 1 1 2

x 1 4 }}

(x 2 4)(x 1 1) p

3(x 1 1)(x 2 1)(x 2 4) }}

3(x 1 1)(x 2 1)(x 2 4)

5 x 2 4 }}}

15(x 2 1)(x 2 4) 2 3(x 1 4)(x 2 1)

5 x 2 4 }}}

15(x2 2 5x 1 4) 2 3(x2 1 3x 2 4)

5 x 2 4 }}}

15x2 2 75x 1 60 2 3x2 2 9x 1 12

5 x 2 4 }}

12x2 2 84x 1 72

5 x 2 4 }}

12(x2 2 7x 1 6)

5 x 2 4 }}

12(x 2 6)(x 2 1)

37. Sample answer:

x2 2 x 2 6

} x2 1 4x

}

x 1 2

} x

5 x2 2 x 2 6

} x(x 1 4)

}

x 1 2

} x

5 x2 2 x 2 6

} x(x 1 4)

}

x 1 2

} x

p x(x 1 4)

} x(x 1 4)

5 x2 2 x 2 6

}} (x 1 2)(x 1 4)

5 (x 2 3)(x 1 2)

}} (x 1 2)(x 1 4)

5 x 2 3

} x 1 4

x2 1 3x 2 18

} 4 }

x2 1 10x 1 24

}} 4

5 x2 1 3x 2 18} 4

}

x2 1 10x 1 24

}} 4

p 4 } 4

5 x2 1 3x 2 18

}} x2 1 10x 1 24

5 (x 1 6)(x 2 3)

}} (x 1 6)(x 1 4)

5 x 2 3

} x 1 4

38. 1 }

x 2

x }

x21 1 1 }

5 }

x 5

1 } x 2

x }

1 } x 1 1

} 5 }

x

5

1 } x 2

x }

1 1 x

} x

} 5 }

x

5

1 } x 2

x }

1 1 x

} x

p x } x

} 5 }

x

5 1 } x 2

x2

} 1 1 x }

5 }

x

5 1 } x 2

x2

} 1 1 x }

5 }

x p

x(1 1 x) }

x(1 1 x)

5 1 1 x 2 x3

} 5(1 1 x)

5 2x3 1 x 1 1

} 5(x 1 1)

39. 3 2 2x

} x3 }

2 }

x2 2 1 }

x3 1 x2 5

3 2 2x

} x3 }

2 }

x2 2 1 }

x3 1 x2 p

x3(x3 1 x2) }

x3(x3 1 x2)

5 (3 2 2x)(x3 1 x2)

}} 2x(x3 1 x2) 2 x3

5 3x3 1 3x2 2 2x4 2 2x3

}} 2x4 1 2x3 2 x3

5 22x4 1 x3 1 3x2

}} 2x4 1 x3

5 x2(22x2 1 x 1 3)

}} x3(2x 1 1)

5 22x2 1 x 1 3

}} x(2x 1 1)

5 2(2x2 2 x 2 3)

}} x(2x 1 1)

5 2(2x 2 3)(x 1 1)

}} x(2x 1 1)


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40. 3x22 1 (2x 2 1)21

}}

6 }

x21 1 2 1 3x 2 1

5 3 }

x2 1 1 } 2x 2 1 }

6 }

1 }

x 1 2

1 3 } x

5 3 }

x2 1 1 } 2x 2 1 }

6 }

1 1 2x

} x 1

3 } x

5 3 }

x2 1 1 } 2x 2 1 }}

x }

x p 6

} 1 1 2x

} x

1 3 } x

5 3 }

x2 1 1 } 2x 2 1 }

6x }

1 1 2x 1

3 } x

5 3 }

x2 1 1 } 2x 2 1 }

6x }

1 1 2x 1

3 } x p

x2(2x 2 1)(1 1 2x) }}

x2(2x 2 1)(1 1 2x)

5 3(2x 2 1)(1 1 2x) 1 x2(1 1 2x)

}}} 6x3(2x 2 1) 1 3x(2x 2 1)(1 1 2x)

5 3(4x2 2 1) 1 x2 1 2x3

}} 12x4 2 6x3 1 3x(4x2 2 1)

5 12x2 2 3 1 x2 1 2x3

}} 12x4 2 6x3 1 12x3 2 3x

5 2x3 1 13x2 2 3

}} 12x4 1 6x3 2 3x

5 2x3 1 13x2 2 3

}} 3x(4x3 1 2x2 2 1)

5 (2x 1 1)(x2 1 6x 2 3)

}} 3x(2x 2 1)(2x2 1 2x 1 1)

Problem Solving

41. T 5 d } a 2 j 1

d } a 1 j

5 d } a 2 j p

a 1 j }

a 1 j 1

d } a 1 j p

a 2 j }

a 2 j

5 ad 1 dj

}} (a 2 j)(a 1 j)

1 ad 2 dj

}} (a 1 j)(a 2 j)

5 2ad }}

(a 2 j)(a 1 j)

When d 5 2468, a 5 510, and j 5 115,

T 5 2ad }}

(a 2 j)(a 1 j) 5

2(510)(2468) }}

(510 2 115)(510 1 115)

5 2,517,360

} (395)(625)

ø 10.2.

The total time needed to fl y from New York to Los Angeles and back if d 5 2468 miles, a 5 510 miles per hour, and j 5 115 miles per hour is about 10.2 hours.

42. Rt 5 1 }

1 }

R1 1

1 } R2 5

1 }

1 } R1 1

1 } R2 p

R1R2 } R1R2 5

R1R2 } R2 1 R1

When R1 5 2000 and R2 5 5600,

Rt 5 R1R2

} R2 1 R1 5

(2000)(5600) } 5600 1 2000 5

11,200,000 } 7600 ø 1473.7

The total resistance when R1 5 2000 ohms and R2 5 5600 ohms is about 1473.7 ohms.

43. a. M 5 Pi }}

1 2 1 1 }

1 1 i 2

12t 5 Pi }

1 2 1 }

(1 1 i)12t

5 Pi }

1 2 1 }

(1 1 i)12t p

(1 1 i)12t

} (1 1 i)12t 5

Pi(1 1 i)12t

} (1 1 i)12t 2 1

b. When P 5 15,500, i 5 0.06, and t 5 4,

M 5 Pi(1 1 i)12t

}} (1 1 i)12t 2 1

5 15,500(0.06)(1 1 0.06)12(4)

}} (1 1 0.06)12(4) 2 1

5 930(1.06)48

} 1.0648 2 1

ø 990.41

Your monthly payment will be $990.41 if you borrow $15,500 at an annual interest rate of 6% and repay the loan over 4 years.

44. a. A 5 391t2 1 0.112

}} 0.218t4 1 0.991t2 1 1

b. A 5 391(t 2 1)2 1 0.112

}}} 0.218(t 2 1)4 1 0.991(t 2 1)2 1 1

c.

Total amount of aspirin

5

Amount after one dose

1

Amount after two doses

5 391t2 1 0.112

}} 0.218t4 1 0.991t2 1 1

1 391(t 2 1)2 1 0.112

}}} 0.218(t 2 1)4 1 0.991(t 2 1)2 1 1

d. Using the graph from part (c) and the maximum feature, you get a maximum value of about 369, which occurs when t ø 2.209. The time after the second dose has been taken is t 2 1 ø 2.209 2 1 ø 1.209. Since 0.209 3 60 ø 13, the greatest amount of aspirin is in the bloodstream about 1 hour 13 minutes after the second dose has been taken.


n2ws-08a.indd 498 6/27/06 10:58:11 AM

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499Algebra 2


45. 4th expression: 1 1 1 }}

2 1 1 }}

2 1 1 }

2 1 1 }

2 1 1 }

2

5th expression: 1 1 1 }}

2 1 1 }}

2 1 1 }}

2 1 1 }

2 1 1 }

2 1 1 } 2

1 1 1 }

2 1 1 } 2 5 1 1

1 }

2 1 1 } 2 p 2 }

2

5 1 1 2 } 4 1 1 5 1 1

2 } 5 5 1

2 } 5 5 1.4

1 1 1 }

2 1 1 }

2 1 1 } 2 5 1 1

1 }

2 1 2 } 5

5 1 1 1 }

2 1 2 } 5 p 5 } 5 5 1 1

5 } 10 1 2

5 1 1 5 } 12 5 1

5 }

12 5 1.41

} 6

1 1 1 }}

2 1 1 }

2 1 1 }

2 1 1 } 2 5 1 1

1 }

2 1 5 } 12

5 1 1 1 }

2 1 5 } 12 p 12

} 12

5 1 1 12 } 24 1 5

5 1 1 12

} 29 5 1 12

} 29

ø 1.4137931

1 1 1 }}

2 1 1 }}

2 1 1 }

2 1 1 }

2 1 1 }

2 5 1 1

1 }

2 1 12

} 29 5 1 1

1 }

2 1 12

} 29 p 29

} 29

5 1 1 29 } 58 1 12

5 1 1 29

} 70

5 1 29

} 70

ø 1.4142857

1 1 1 }}

2 1 1 }}

2 1 1 }}

2 1 1 }

2 1 1 }

2 1 1 } 2 5 1 1

1 }

2 1 29

} 70

5 1 1 1 }

2 1 29

} 70 p 70

} 70

5 1 1 70 } 140 1 29 5 1 1

70 } 169

5 1 70

} 169

ø 1.4142012

The expressions approach Ï}

2 ø 1.4142135 . . ..

Mixed Review

46. 1 }

3 x 1 4 5 15

1 }

3 x 5 11

3 1 1 } 3 x 2 5 11(3)

x 5 33

47. 2x 5 2 5 } 8 x 2 18

21

} 8 x 5 218

8 }

21 1 21

} 8 x 2 5 218 1 8 }

21 2

x 5 2 144

} 21 5 2 48

} 7

48. 12x 1 7 5 14

} 3 x

22

} 3 x 1 7 5 0

22

} 3 x 5 27

3 }

22 1 22

} 3 x 2 5 27 1 3 }

22 2

x 5 2 21

} 22

49. x2 1 9x 2 36 5 0

(x 1 12)(x 2 3) 5 0

x 1 12 5 0 or x 2 3 5 0

x 5 212 or x 5 3

50. 3x2 1 x 2 14 5 0

(3x 1 7)(x 2 2) 5 0

3x 1 7 5 0 or x 2 2 5 0

x 5 2 7 } 3 or x 5 2

51. 4x(x 2 5) 5 4x 2 35

4x2 2 20x 5 4x 2 35

4x2 2 24x 1 35 5 0

(2x 2 7)(2x 2 5) 5 0

2x 2 7 5 0 or 2x 2 5 5 0

x 5 7 } 2 or x 5

5 } 2

52. 6x2 2 25 5 x2

5x2 2 25 5 0

5x2 5 25

x2 5 5

x 5 6 Ï}

5


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53. 4(x 2 2)2 5 144

(x 2 2)2 5 36

x 2 2 5 6 Ï}

36

x 2 2 5 66

x 5 6 1 2 5 8 or x 5 26 1 2 5 24

54. 3(x 1 5)2 2 10 5 182

3(x 1 5)2 5 192

(x 1 5)2 5 64

x 1 5 5 6 Ï}

64

x 1 5 5 68

x 5 8 2 5 5 3 or x 5 28 2 5 5 213

55. y 5 4x

2

x

y

21

56. y 5 22 p 3x

1

x

y

23

57. f (x) 5 2 } 3 p 2x

2

x

y

21

58. y 5 4 1 1 } 2 2

x

1

x

y

21

59. y 5 23 1 1 } 4 2

x

1

x

y

21

60. g(x) 5 5 1 3 } 8 2

x

1

x

g(x)

21

Lesson 8.6


1. 3 } 5x 5

2 } x 2 7

3(x 2 7) 5 2(5x)

3x 2 21 5 10x

27x 2 21 5 0

27x 5 21

x 5 23

Check:

3 }

5(23) 0

2 }

23 2 7

3 }

215 0

2 }

210

2 1 } 5 5 2

1 } 5 ✓

2. 24

} x 1 3

5 5 } x 2 3

24(x 2 3) 5 5(x 1 3)

24x 1 12 5 5x 1 15

29x 1 12 5 15

29x 5 3

x 5 3 }

29

x 5 2 1 } 3

Check:

24 }

2 1 } 3 1 3

0 5 }

2 1 } 3 2 3

24

} 8 }

3 0

5 }

2 10

} 3

2 3 } 2 5 2

3 } 2 ✓


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501Algebra 2


3. 1 }

2x 1 5 5

x } 11x 1 8

11x 1 8 5 x(2x 1 5)

11x 1 8 5 2x2 1 5x

0 5 2x2 2 6x 2 8

0 5 2(x2 2 3x 2 4) 0 5 2(x 2 4)(x 1 1)

x 2 4 5 0 or x 1 1 5 0

x 5 4 or x 5 21

Check x 5 4:

1 }

2(4) 1 5 0

4 }

11(4) 1 8

1 }

13 0

4 }

52

1 }

13 5

1 } 13 ✓

Check x 5 21:

1 }

2(21) 1 5 0

21 }

11(21) 1 8

1 }

3 0

21 }

23

1 }

3 5

1 } 3 ✓

4. 7.5

} 100

5 0.2(10)

} 10 1 x

7.5(10 1 x) 5 100(0.2)(10)

75 1 7.5x 5 200

7.5x 5 125

x 5 50

} 3 ø 16.7

You should mix about 16.7 ounces of pure silver with the jewelry silver.

5. 7 }

2 1

3 } x 5 3

2x 1 7 } 2 1

3 } x 2 5 2x(3)

7x 1 6 5 6x

x 1 6 5 0

x 5 26

Check:

7 }

2 1

3 }

26 0 3

7 }

2 2

1 } 2 0 3

6 }

2 0 3

3 5 3 ✓

6. 2 }

x 1

4 } 3 5 2 7. 3 }

7 1

8 } x 5 1

3x 1 2 } x 1

4 } 3 2 5 3x(2) 7x 1 3 } 7 1

8 } x 2 5 7x(1)

6 1 4x 5 6x 3x 1 56 5 7x

6 2 2x 5 0 24x 1 56 5 0

22x 5 26 24x 5 256

x 5 3 x 5 14

Check: Check:

2 }

3 1

4 } 3 0 2

3 } 7 1

8 } 14 0 1

6 }

3 0 2

3 } 7 1

4 } 7 0 1

2 5 2 ✓ 1 5 1 ✓

8. 3 }

2 1

4 } x 2 1 5

x 1 1 } x 2 1

2(x 2 1) 1 3 } 2 1

4 } x 2 1 2 5 2(x 2 1) 1 x 1 1

} x 2 1

2 3(x 2 1) 1 2(4) 5 2(x 1 1)

3x 2 3 1 8 5 2x 1 2

3x 1 5 5 2x 1 2

x 1 5 5 2

x 5 23

Check:

3 }

2 1

4 }

23 2 1 0 23 1 1

} 23 2 1

3 }

2 1

4 }

24 0 22

} 24

3 }

2 2 1 0

1 }

2

1 }

2 5

1 } 2 ✓

9. 3x }

x 1 1 2

5 } 2x 5

3 } 2x

2x(x 1 1) 1 3x }

x 1 1 2

5 } 2x 2 5 2x(x 1 1) 1 3 }

2x 2

2x(3x) 2 5(x 1 1) 5 3(x 1 1)

6x2 2 5x 2 5 5 3x 1 3

6x2 2 8x 2 8 5 0

2(3x2 2 4x 2 4) 5 0

2(3x 1 2)(x 2 2) 5 0

3x 1 2 5 0 or x 2 2 5 0

x 5 2 2 } 3 or x 5 2


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Check x 5 2 2 } 3 :

3 1 2

2 } 3 2 }

2 2 } 3 1 1

2 5 }

2 1 2 2 } 3 2 0

3 }

2 1 2 2 } 3 2

22

} 1 }

3 2

5 }

2 4 } 3 0

3 }

2 4 } 3

26 1 15

} 4 0 2 9 } 4

2 9 } 4 5 2

9 } 4 ✓

Check x 5 2:

3(2)

} 2 1 1

2 5 }

2(2) 0

3 }

2(2)

6 }

3 2

5 } 4 0

3 }

4

3 }

4 5

3 } 4 ✓

10. 5x }

x 2 2 5 7 1

10 } x 2 2

(x 2 2) 1 5x }

x 2 2 2 5 (x 2 2) 1 7 1

10 } x 2 2 2

5x 5 7(x 2 2) 1 10

5x 5 7x 2 14 1 10

5x 5 7x 2 4

22x 5 24

x 5 2

Check:

5(2)

} 2 2 2

0 7 1 10 } 2 2 2

10

} 0 0 7 1

10 } 0

Division by 0 is undefi ned, so the equation has no solution.

11. S(t) 5 848t2 1 3220

}} 115t2 1 1000

4.5 5 848t2 1 3220

}} 115t2 1 1000

4.5(115t2 1 1000) 5 848t2 1 3220

517.5t2 1 4500 5 848t2 1 3220

4500 5 330.5t2 1 3220

1280 5 330.5t2

3.87 ø t2

1.97 ø t

The total sales of entertainment software were about $4.5 billion about 2 years after 1995, or in 1997.


Skill Practice

1. When you write x }

3 5

x 1 2 } 5 as 5x 5 3(x 1 2), you are

cross-multiplying.

2. The solution x 5 4 is extraneous because after substituting this value into the original equation you

obtain 5 }

4 2 4 0

4 }

4 2 4 →

5 }

0 0

4 }

0 . Division by zero

is undefi ned, so 4 is an extraneous solution.

3. Graph both sides of the original equation in the same coordinate plane. If the graphs intersect at a possible solution, then it is a solution. If the graphs do not intersect at a possible solution, then it is an extraneous solution.

4. 4 }

2x 5

5 } x 1 6 5.

9 }

3x 5

4 } x 1 2

4(x 1 6) 5 5(2x) 9(x 1 2) 5 4(3x)

4x 1 24 5 10x 9x 1 18 5 12x

26x 1 24 5 0 23x 1 18 5 0

26x 5 224 23x 5 218

x 5 4 x 5 6

Check: Check:

4 }

2(4) 0

5 } 4 1 6

9 }

3(6) 0

4 }

6 1 2

4 }

8 0

5 }

10

9 }

18 0

4 }

8

1 }

2 5

1 } 2 ✓

1 }

2 5

1 } 2 ✓

6. 6 }

x 2 1 5

9 } x 1 1

6(x 1 1) 5 9(x 2 1)

6x 1 6 5 9x 2 9

23x 1 6 5 29

23x 5 215

x 5 5

Check:

6 }

5 2 1 0

9 }

5 1 1

6 }

4 0

9 }

6

3 }

2 5

3 } 2 ✓


n2ws-08a.indd 502 6/27/06 10:58:34 AM

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503Algebra 2


7. 8 }

3x 2 2 5

2 } x 2 1

8(x 2 1) 5 2(3x 2 2)

8x 2 8 5 6x 2 4

2x 2 8 5 24

2x 5 4

x 5 2

Check:

8 }

3(2) 2 2 0

2 }

2 2 1

8 }

4 0

2 }

1

2 5 2 ✓

8. x }

x 1 1 5

3 } x 1 1

x(x 1 1) 5 3(x 1 1)

x2 1 x 5 3x 1 3

x2 2 2x 2 3 5 0

(x 2 3)(x 1 1) 5 0

x 2 3 5 0 or x 1 1 5 0

x 5 3 or x 5 21


3 }

3 1 1 0

3 }

3 1 1

21 }

21 1 1 0

3 }

21 1 1

3 }

4 5

3 } 4 ✓ 2

1 } 0 0

3 }

0

Division by 0 is undefi ned, so x 5 21 is an extraneous solution and x 5 3 is the only solution.

9. x 2 3

} x 1 5 5 x } x 1 2

(x 2 3)(x 1 2) 5 x(x 1 5)

x2 2 x 2 6 5 x2 1 5x

26x 2 6 5 0

26x 5 6

x 5 21

Check:

21 2 3

} 21 1 5

0 21 }

21 1 2

24

} 4 0

21 }

1

21 5 21 ✓

10. x }

x2 2 2 5

21 } x

x(x) 5 2(x2 2 2) x2 5 2x2 1 2

2x2 5 2

x2 5 1

x 5 61


1 }

12 2 2 0

21 }

1

21 }

(21)2 2 2 0

21 }

21

1 }

21 0 21

21 }

21 0 1

21 5 21 ✓ 1 5 1 ✓

11. 4(x 2 4)

} x2 1 2x 2 8

5 4 } x 1 4

4(x 2 4)(x 1 4) 5 4(x2 1 2x 2 8) 4(x2 2 16) 5 4x2 1 8x 2 32

4x2 2 64 5 4x2 1 8x 2 32

264 5 8x 2 32

232 5 8x

24 5 x

Check:

4(24 2 4)

}} (24)2 1 2(24) 2 8

0 4 }

24 1 4

4(28)

} 0 0

4 }

0

Division by 0 is undefi ned, so x 5 24 is an extraneous solution and the equation has no solution.

12. 9 } x2 2 6x 1 9

5 3x }

x2 2 3x

9(x2 2 3x) 5 3x(x2 2 6x 1 9) 9x2 2 27x 5 3x3 2 18x2 1 27x

0 5 3x3 2 27x2 1 54x

0 5 3x(x2 2 9x 1 18) 0 5 3x(x 2 6)(x 2 3)

x 5 0 or x 2 6 5 0 or x 2 3 5 0

x 5 0 or x 5 6 or x 5 3

Check x 5 0:

9 }}

02 2 6(0) 1 9 0

3(0) }

02 2 3(0)

9 }

9 Þ

0 }

0

Division by 0 is undefi ned, so 0 is an extraneous solution.


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Check x 5 6:

9 }}

62 2 6(6) 1 9 0

3(6) }

62 2 3(6)

9 }

9 0

18 }

18

1 5 1 ✓

Check x 5 3:

9 }}

32 2 6(3) 1 9 0

3(3) }

32 2 3(3)

9 }

0 0

9 }

0

Division by 0 is undefi ned, so 3 is an extraneous solution. The only solution is x 5 6.

13. A; 3 }

x 1 2 5

6 } x 2 1

3(x 2 1) 5 6(x 1 2)

3x 2 3 5 6x 1 12

23x 2 3 5 12

23x 5 15

x 5 25

14. 4 }

x 1 x 5 5

x 1 4 } x 1 x 2 5 5x

4 1 x2 5 5x

x2 2 5x 1 4 5 0

(x 2 1)(x 2 4) 5 0

x 2 1 5 0 or x 2 4 5 0

x 5 1 or x 5 4


4 }

1 1 1 0 5

4 }

4 1 4 0 5

4 1 1 0 5 1 1 4 0 5

5 5 5 ✓ 5 5 5 ✓

15. 2 }

3x 1

1 } 6 5

4 } 3x

6x 1 2 } 3x

1 1 } 6 2 5 6x 1 4 }

3x 2

4 1 x 5 8

x 5 4

Check:

2 }

3(4) 1

1 } 6 0

4 }

3(4)

2 }

12 1

1 } 6 0

4 }

12

1 }

3 5

1 } 3 ✓

16. 5 }

x 2 2 5

2 } x 1 3

x(x 1 3) 1 5 } x 2 2 2 5 x(x 1 3) 1 2

} x 1 3

2 5(x 1 3) 2 2x(x 1 3) 5 2x

5x 1 15 2 2x2 2 6x 5 2x

22x2 2 x 1 15 5 2x

0 5 2x2 1 3x 2 15

x 5 23 6 Ï

}}

32 2 4(2)(215) }}

2(2)

5 23 6 Ï

}

129 } 4

Check x 5 23 1 Ï

}

129 } 4 :

5 }

23 1 Ï

}

129 }

4

2 2 0 2 }}

23 1 Ï

}

129 }

4 1 3

0.39297 5 0.39297 ✓

Check x 5 23 2 Ï

}

129 } 4 :

5 }

23 2 Ï

}

129 }

4

2 2 0 2 }}

23 2 Ï

}

129 }

4 1 3

23.39297 5 23.39297 ✓

17. 1 }

2x 1

3 } x 1 7 5

21 } x

2x(x 1 7) 1 1 } 2x

1 3 } x 1 7 2 5 2x(x 1 7) 1 21

} x 2

x 1 7 1 3(2x) 5 22(x 1 7)

x 1 7 1 6x 5 22x 2 14

7x 1 7 5 22x 2 14

9x 1 7 5 214

9x 5 221

x 5 2 7 } 3

Check:

1 }

2 1 2 7 } 3 2 1

3 }

2 7 } 3 1 7

0 21

} 2

7 } 3

1 }

2 14

} 3 1

3 }

14

} 3 0

3 } 7

2 3 } 14 1

9 } 14 0

3 } 7

6 }

14 0

3 } 7

3 } 7 5

3 } 7 ✓


n2ws-08a.indd 504 6/27/06 10:58:44 AM

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505Algebra 2


18. 1 }

x 2 2 1 2 5

3x } x 1 2

(x 2 2)(x 1 2) 1 1 }

x 2 2 1 2 2 5 (x 2 2)(x 1 2) 1 3x

} x 1 2

2 x 1 2 1 2(x 2 2)(x 1 2) 5 3x(x 2 2)

x 1 2 1 2(x2 2 4) 5 3x2 2 6x

x 1 2 1 2x2 2 8 5 3x2 2 6x

0 5 x2 2 7x 1 6

0 5 (x 2 6)(x 2 1)

x 2 6 5 0 or x 2 1 5 0

x 5 6 or x 5 1


1 }

6 2 2 1 2 0

3(6) }

6 1 2

1 }

1 2 2 1 2 0

3(1) }

1 1 2

1 }

4 1 2 0

18 }

8

1 }

21 1 2 0

3 }

3

9 }

4 5

9 } 4 ✓ 1 5 1 ✓

19. 5 }

x2 1 x 2 6 5 2 1

x 2 3 } x 2 2

5 }}

(x 1 3)(x 2 2) 5 2 1

x 2 3 } x 2 2

(x 1 3)(x 2 2) p 5 }}

(x 1 3)(x 2 2)

5 [(x 1 3)(x 2 2)] 1 2 1 x 2 3

} x 2 2

2 5 5 2(x 1 3)(x 2 2) 1 (x 1 3)(x 2 3)

5 5 2(x2 1 x 2 6) 1 x2 2 9

5 5 2x2 1 2x 2 12 1 x2 2 9

5 5 3x2 1 2x 2 21

0 5 3x2 1 2x 2 26

x 5 22 6 Ï

}}

22 2 4(3)(226) }}

2(3)

5 22 6 Ï

}

316 } 6

5 22 6 Ï

}

4 p 79 }} 6

5 22 6 2 Ï

}

79 } 6

5 2(21 6 Ï

}

79 ) }} 6

5 21 6 Ï

}

79 } 3

Check x 5 21 1 Ï

}

79 } 3 :

5 }}}

1 21 1 Ï}

79 }

3 2 2 1 1 21 1 Ï

}

79 }

3 2 26

0 2 1 1 21 1 Ï

}

79 }

3 2 2 3

}}

1 21 1 Ï}

79 }

3 2 2 2

1.41118 5 1.41118 ✓

Check x 5 21 2 Ï

}

79 } 3 :

5 }}}

1 21 2 Ï}

79 }

3 2 2 1 1 21 2 Ï

}

79 }

3 2 2 6

0 2 1 1 21 2 Ï

}

79 }

3 2 2 3

}}

1 21 2 Ï}

79 }

3 2 2 2

3.18882 5 3.18882 ✓

20. x 1 1

} x 1 6

1 1 } x 5

2x 1 1 } x 1 6

x(x 1 6) 1 x 1 1 }

x 1 6 1

1 } x 2 5 x(x 1 6) 1 2x 1 1

} x 1 6

2 x(x 1 1) 1 x 1 6 5 x(2x 1 1)

x2 1 x 1 x 1 6 5 2x2 1 x

x2 1 2x 1 6 5 2x2 1 x

0 5 x2 2 x 2 6

0 5 (x 2 3)(x 1 2)

x 2 3 5 0 or x 1 2 5 0

x 5 3 or x 5 22


3 1 1

} 3 1 6

1 1 } 3 0

2(3) 1 1 }

3 1 6

22 1 1 }

22 1 6 1

1 }

22 0 2(22) 1 1

} 22 1 6

4 }

9 1

1 } 3 0

7 }

9 2

1 } 4 2

1 } 2 0

23 }

4

7 }

9 5

7 } 9 ✓ 2

3 } 4 5 2

3 } 4 ✓

21. 2 }

x 2 3 1

1 } x 5

x 2 1 }

x 2 3

x(x 2 3) 1 2 }

x 2 3 1

1 } x 2 5 x(x 2 3) 1 x 2 1

} x 2 3

2 2x 1 x 2 3 5 x(x 2 1)

3x 2 3 5 x2 2 x

0 5 x2 2 4x 1 3

0 5 (x 2 3)(x 2 1)

x 2 3 5 0 or x 2 1 5 0

x 5 3 or x 5 1

Check x 5 3:

2 }

3 2 3 1

1 } 3 0

3 2 1 }

3 2 3

2 }

0 1

1 } 3 0

2 }

0

Division by 0 is undefined, so 3 is an extraneous solution.

Check x 5 1:

2 }

1 2 3 1

1 } 1 0 1 2 1

} 1 2 3

2 }

22 1 1 0

0 }

22

21 1 1 0 0

0 5 0 ✓

The only solution is x 5 1.


n2ws-08a.indd 505 6/27/06 10:58:49 AM

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22. 6x }

x 1 4 1 4 5

2x 1 2 } x 2 1

(x 1 4)(x 2 1) 1 6x }

x 1 4 1 4 2 5 (x 1 4)(x 2 1) 1 2x 1 2

} x 2 1

2 6x(x 2 1) 1 4(x 1 4)(x 2 1) 5 (x 1 4)(2x 1 2)

6x2 2 6x 1 4(x2 1 3x 2 4) 5 2x2 1 10x 1 8

6x2 2 6x 1 4x2 1 12x 2 16 5 2x2 1 10x 1 8

10x2 1 6x 2 16 5 2x2 1 10x 1 8

8x2 2 4x 2 24 5 0

4(2x2 2 x 2 6) 5 0

4(2x 1 3)(x 2 2) 5 0

2x 1 3 5 0 or x 2 2 5 0

x 5 2 3 } 2 or x 5 2

Check x 5 2 3 } 2 :

6 1 2

3 } 2 2 }

2 3 } 2 1 4

1 4 0 2 2 1 2

3 } 2 2 1 2 }}

2 3 } 2 2 1

29

} 5 }

2 1 4 0

21 }

2 5 } 2

2 18

} 5 1 4 0 2 } 5

2 } 5 5

2 } 5 ✓

Check x 5 2:

6(2)

} 2 1 4

1 4 0 2(2) 1 2

} 2 2 1

12

} 6 1 4 0

6 }

1

6 5 6 ✓

23. 10

} x 1 3 5

x 1 9 } x 2 4

x(x 2 4) 1 10 }

x 1 3 2 5 x(x 2 4) 1 x 1 9

} x 2 4

2 10(x 2 4) 1 3x(x 2 4) 5 x(x 1 9)

10x 2 40 1 3x2 2 12x 5 x2 1 9x

3x2 2 2x 2 40 5 x2 1 9x

2x2 2 11x 2 40 5 0

(2x 1 5)(x 2 8) 5 0

2x 1 5 5 0 or x 2 8 5 0

x 5 2 5 } 2 or x 5 8

Check x 5 2 5 } 2 :

10

} 2

5 } 2 1 3 0

2 5 } 2 1 9 }

2 5 } 2 2 4

24 1 3 0 13

} 2 }

2 13

} 2

21 5 21 ✓

Check x 5 8:

10

} 8 1 3 0

8 1 9 }

8 2 4

17

} 4 5

17 } 4 ✓

24. 18 }

x2 2 3x 2

6 } x 2 3 5

5 } x

18 }

x(x 2 3) 2

6 } x 2 3 5

5 } x

x(x 2 3) F 18 }

x(x 2 3) 2

6 } x 2 3 G 5 x(x 2 3) 1 5 }

x 2

18 2 6x 5 5(x 2 3)

18 2 6x 5 5x 2 15

18 2 11x 5 215

211x 5 233

x 5 3

Check:

18 }

32 2 3(3) 2

6 } 3 2 3 0

5 }

3

18

} 0 2

6 } 0 0

5 }

3

Division by 0 is undefined, so 3 is an extraneous solution and the equation has no solution.

25. x 1 3

} x 2 3

1 x } x 2 5 5

x 1 5 } x 2 5

(x 2 3)(x 2 5) 1 x 1 3 }

x 2 3 1

x } x 2 5 2 5 (x 2 3)(x 2 5) 1 x 1 5

} x 2 5 2 (x 2 5)(x 1 3) 1 x(x 2 3) 5 (x 2 3)(x 1 5)

x2 2 2x 2 15 1 x2 2 3x 5 x2 1 2x 2 15

2x2 2 5x 2 15 5 x2 1 2x 2 15

x2 2 7x 5 0

x(x 2 7) 5 0

x 5 0 or x 2 7 5 0

x 5 0 or x 5 7


0 1 3

} 0 2 3

1 0 } 0 2 5 0

0 1 5 }

0 2 5

7 1 3 }

7 2 3 1

7 } 7 2 5 0

7 1 5 } 7 2 5

3 }

23 1

0 }

25 0 5 }

25 10

} 4 1

7 } 2 0

12 }

2

21 1 0 0 21 6 5 6 ✓

21 5 21 ✓


n2ws-08a.indd 506 6/27/06 10:58:55 AM

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507Algebra 2


26. Cross-multiplying was used incorrectly. The original equation is not expressed as a proportion, so you must solve the equation by multiplying each side of the equation by the LCD, 2x2.

2x2 1 3 } 2x

1 4

} x2 2 5 2x2(1)

3x 1 8 5 2x2

27. The student simply added numerators and denominators on the left side of the equation. Both sides of the equation should have been multiplied by the LCD, 6x.

6x 1 5 } x 1

23 } 6 2 5 6x 1 45

} x 2

30 1 23x 5 270

28. C; 2 }

x 2 3 5

1 }

x2 2 2x 2 3

2 }

x 2 3 5

1 }}

(x 2 3)(x 1 1)

(x 2 3)(x 1 1) 1 2 }

x 2 3 2 5 (x 2 3)(x 1 1) F 1

}} (x 2 3)(x 1 1)

G 2(x 1 1) 5 1

2x 1 2 5 1

2x 5 21

x 5 2 1 } 2

29. Sample answer: The equation 3 }

x 5

2 } x 1 1 can be solved

using cross-multiplication because each side of the equation is a single rational expression. The equation

3 }

x 1

6 } 7 5

4 } 7x can be solved by multiplying each side of

the equation by the LCD of the fractions because the equation is not expressed as a proportion.

30. The statement is always true. By solving the equation, you obtain the following.

1 }

x 2 a 5

x } x 2 a

x 2 a 5 x(x 2 a)

0 5 x(x 2 a) 2 (x 2 a)

0 5 (x 2 1)(x 2 a)

x 2 1 5 0 or x 2 a 5 0

x 5 1 or x 5 a

Check x 5 1: Check x 5 a:

1 }

1 2 a 5

1 } 1 2 a ✓

1 }

a 2 a 0

a }

a 2 a

1 }

0 0

a }

0

Because division by 0 is undefi ned, x 5 a is an extraneous solution.

31. The statement is sometimes true. By solving the equation, you obtain the following.

3 }

x 2 a 5

x } x 2 a

3(x 2 a) 5 x(x 2 a)

0 5 x(x 2 a) 2 3(x 2 a)

0 5 (x 2 3)(x 2 a)

x 2 3 5 0 or x 2 a 5 0

x 5 3 or x 5 a

Check x 5 3: Check x 5 a:

3 }

3 2 a 5

3 } 3 2 a ✓

3 }

a 2 a 0

3 }

a 2 a

3 }

0 0

3 }

0

Because division by 0 is undefined, x 5 a is an extraneous solution. So, the only solution is x 5 3. However, when a 5 3, the equation has one possible solution, x 5 3. Substituting this into the equation

3 }

x 2 3 5

x } x 2 3 , you obtain

3 }

3 2 3 0

3 }

3 2 3

3 }

0 0

3 }

0 .

Because division by 0 is undefined, x 5 3 is an extraneous solution. So, when a 5 3, the original equation has no solution.

32. The statement is always true. By solving the equation, you obtain the following.

1 }

x 2 a 5

2 } x 1 a 1

2a }

x2 2 a2

1 }

x 2 a 5

2 } x 1 a 1

2a }}

(x 1 a)(x 2 a)

(x 1 a)(x 2 a) 1 1 } x 2 a 2

5 (x 1 a)(x 2 a) F 2 } x 1 a 1

2a }}

(x 1 a)(x 2 a) G

x 1 a 5 2(x 2 a) 1 2a

x 1 a 5 2x 2 2a 1 2a

x 1 a 5 2x

a 5 x

Check:

1 }

a 2 a 0

2 } a 1 a 1

2a }

a2 2 a2

1 }

0 0

a }

2a 1

2a } 0

Because division by 0 is undefined, x 5 a is an extraneous solution. So, the original equation has no solution.


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Problem Solving

33. 90

} 100

5 37 1 x

} 44 1 x

90(44 1 x) 5 100(37 1 x)

3960 1 90x 5 3700 1 100x

260 5 10x

26 5 x

You need to put 26 consecutive serves into play in order to raise your service percentage to 90%.

34. a. Distance for skater 1

}} Skater 1 speed

5 Distance for skater 2

}} Skater 2 speed

9 }

x 1 4.38 5

8 } x

x is the speed of skater 2.

b. 9 }

x 1 4.38 5

8 } x

9x 5 8(x 1 4.38)

9x 5 8x 1 35.04

x 5 35.04

The speed of skater 2 is 35.04 kilometers per hour and the speed of skater 1 is 35.04 1 4.38 5 39.42 kilometers per hour.

c. Skater 1 traveled 9 kilometers at a rate of 39.42 kilometers per hour. To fi nd how long the skater skated, use the distance formula d 5 rt and solve for t.

d 5 rt

9 5 39.42t

0.228 ø t

Because skater 1 traveled 9 kilometers in the same amount of time it took skater 2 to travel 8 kilometers, both skaters skated for about 0.228 hour or 60(0.228) ø 13.7 minutes.

35. n 5 635t2 2 7350t 1 27,200

}} t2 2 11.5t 1 39.4

720 5 635t2 2 7350t 1 27,200

}} t2 2 11.5t 1 39.4

720(t2 2 11.5t 1 39.4) 5 635t2 2 7350t 1 27,200

720t2 2 8280t 1 28,368 5 635t2 2 7350t 1 27,200

85t2 2 930t 1 1168 5 0

t 5 2(2930) 6 Ï

}}

(2930)2 2 4(85)(1168) }}}

2(85)

t 5 930 6 Ï

}

467,780 }} 170

t ø 9.49 or t ø 1.45

Because 9.49 is not in the domain (0 ≤ t ≤ 9), t ø 1.45 is the only solution. So, the total number of CDs shipped was about 720 million about 1 year after 1994, or in 1995.

36. a. Work rate p Time 5 Work done

You 1 room

} 8 hours

5 hours 5 }

8 room

Friend 1 room

} t hours

5 hours 5 }

t room

b. The sum 5 }

8 1

5 } t or

5t 1 40 } 8t represents the total work

done by you and your friend while working together for 5 hours.

c. Because the total work done is 1 room, set the sum from part (b) to 1 and solve for t.

5t 1 40

} 8t

5 1

5t 1 40 5 8t

40 5 3t

40

} 3 5 t

Your friend would take 40

} 3 hours or 13

1 }

3 hours to paint

the room when working alone.

37. w

} l

5 l }

l 1 w

1 } l� 5

l } l 1 1

l 1 1 5 l2

0 5 l2 2 l 2 1

l 5 2(1) 6 Ï

}}

(21)2 2 4(1)(21) }}}

2(1) 5

1 6 Ï}

5 } 2

Because the ratio must be positive,

l 5 1 1 Ï

} 5 } 2 .

l

} w

5 1 1 Ï

} 5 } 2 } 1 5

1 1 Ï}

5 } 2

38. a. Average price per minute 5 Average monthly bill

}}} Average number of minutes

f (x) 5 g(x)

} h(x)

5 20.27x3 1 1.40x2 1 1.05x 1 39.4

}}} 28.25x3 1 53.1x2 2 7.82x 1 138

b. Because 1998 is 0 years since 1998, x 5 0.

f (0) 5 20.27(0)3 1 1.40(0)2 1 1.05(0) 1 39.4

}}} 28.25(0)3 1 53.1(0)2 2 7.82(0) 1 138

5 39.4

} 138 ø 0.29

The average price per minute in 1998 was about $.29.

c. Use a graphing calculator and enter the function

f (x) 5 20.27x3 1 1.4x2 1 1.05x 1 39.4

}}} 28.25x3 1 53.1x2 2 7.82x 1 13.8

.

Then, using the table feature, you see that f (4) ø 0.11375. So, the average price per minute fell to 11 cents about 4 years after 1998, or 2002.


n2ws-08b.indd 508 6/27/06 11:00:10 AM

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509Algebra 2


Mixed Review

39. y 5 22x 1 7 40. y 5 x2 2 8x 1 21

1

x21

y

1

x21

y

41. f (x) 5 x3 2 3 42. y 5 2 Ï}

x 2 4 1 1

2

x21

f(x) 1

x1

y

43. y 5 log 4x 44. g(x) 5 2 } x 1 3 1 6

1

x1

y

2

x21

g(x)

45. Ï}

52 5 Ï}

4 p Ï}

13 5 2 Ï}

13

46. Ï}

24 5 Ï}

4 p Ï}

6 5 2 Ï}

6

47. Ï}

125 5 Ï}

25 p Ï}

5 5 5 Ï}

5

48. Ï}

252 5 Ï}

36 p Ï}

7 5 6 Ï}

7

49. Ï}

8 p Ï}

90 5 Ï}

720 5 Ï}

144 p Ï}

5 5 12 Ï}

5

50. Ï}

5 p Ï}

80 5 Ï}

400 5 Ï}

202 5 20

51. Ï}

8 }

20 5 Ï

}

2 } 5 5

Ï}

2 }

Ï}

5 5

Ï}

2 }

Ï}

5 p Ï

} 5 }

Ï}

5 5

Ï}

10 } 5

52. Ï}

60

} 9 5 Ï

}

20

} 3 5

Ï}

20 }

Ï}

3 5

Ï}

20 }

Ï}

3 p Ï

}

3 }

Ï}

3

5 Ï

}

60 } 3 5

Ï}

4 p Ï}

15 } 3 5

2 Ï}

15 } 3

Quiz 8.4–8.6 (p. 595)

1. x2 2 2x 2 24

} x2 1 3x 2 10

p 3x2 2 6x }

x3 1 4x2 5 (x 2 6)(x 1 4)

}} (x 1 5)(x 2 2)

p 3x(x 2 2)

} x2(x 1 4)

5 (x 2 6)(x 1 4)(3x)(x 2 2)

}}} (x 1 5)(x 2 2)(x)(x)(x 1 4)

5 3(x 2 6)

} x(x 1 5)

2. x2 2 10x 1 16

}} x2 2 1

p (x 2 1) 5 x2 2 10x 1 16

}} x2 2 1

p x 2 1 }

1

5 (x 2 8)(x 2 2)(x 2 1)

}} (x 1 1)(x 2 1)

5 (x 2 8)(x 2 2)

}} x 1 1

3. x2 1 9x 1 20

}} x2 2 11x 1 28

4 x2 1 8x 1 15

} x2 2 3x 2 4

5 x2 1 9x 1 20

}} x2 2 11x 1 28

p x2 2 3x 2 4

} x2 1 8x 1 15

5 (x 1 5)(x 1 4)

}} (x 2 7)(x 2 4)

p (x 2 4)(x 1 1)

}} (x 1 5)(x 1 3)

5 (x 1 5)(x 1 4)(x 2 4)(x 1 1)

}}} (x 2 7)(x 2 4)(x 1 5)(x 1 3)

5 (x 1 4)(x 1 1)

}} (x 2 7)(x 1 3)

4. x2 1 12x 1 36

}} x2 2 8x 1 12

4 (x2 2 36)

5 x2 1 12x 1 36

}} x2 2 8x 1 12

p 1 }

x2 2 36

5 (x 1 6)(x 1 6)

}} (x 2 6)(x 2 2)

p 1 }}

(x 1 6)(x 2 6)

5 (x 1 6)(x 1 6)

}}} (x 2 6)(x 2 2)(x 1 6)(x 2 6)

5 x 1 6 }}

(x 2 6)2(x 2 2)

5. 1 }

x 1 4 1

1 } x 2 4 5

1 } x 1 4 p x 2 4

} x 2 4

1 1 } x 2 4 p x 1 4

} x 1 4

5 x 2 4 }}

(x 1 4)(x 2 4) 1

x 1 4 }}

(x 2 4)(x 1 4)

5 2x }}

(x 1 4)(x 2 4)

6. 4x 1 3

} x2 2 16

1 2 } x 2 4 5

4x 1 3 }}

(x 1 4)(x 2 4) 1

2 } x 2 4

5 4x 1 3 }}

(x 1 4)(x 2 4) 1

2 } x 2 4 p x 1 4

} x 1 4

5 4x 1 3 }}

(x 1 4)(x 2 4) 1

2(x 1 4) }}

(x 1 4)(x 2 4)

5 4x 1 3 }}

(x 1 4)(x 2 4) 1

2x 1 8 }}

(x 1 4)(x 2 4)

5 6x 1 11

}} (x 1 4)(x 2 4)

7. 4 } x 1 5 2

6x 2 1 }}

x2 1 10x 1 25 5

4 } x 1 5 2

6x 2 1 }

(x 1 5)2

5 4 } x 1 5 p x 1 5

} x 1 5 2 6x 2 1

} (x 1 5)2

5 4x 1 20

} (x 1 5)2 2

6x 2 1 }

(x 1 5)2

5 4x 1 20 2 (6x 2 1)

}} (x 1 5)2

5 4x 1 20 2 6x 1 1

}} (x 1 5)2

5 22x 1 21

} (x 1 5)2


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8. x 2 4

} x 2 1

5 10 } x 1 7

(x 2 4)(x 1 7) 5 10(x 2 1)

x2 1 3x 2 28 5 10x 2 10

x2 2 7x 2 18 5 0

(x 2 9)(x 1 2) 5 0

x 2 9 5 0 or x 1 2 5 0

x 5 9 or x 5 22

Check x 5 9:

9 2 4

} 9 2 1

0 10 }

9 1 7

5 }

8 0

10 }

16

5 }

8 5

5 } 8 ✓

Check x 5 22:

22 2 4

} 22 2 1

0 10 }

22 1 7

26

} 23

0 10

} 5

2 5 2 ✓

9. x 2 4

} x 2 2

2 2x 2 1

} x 2 2 5 2

(x 2 2) 1 x 2 4 }

x 2 2 2

2x 2 1 } x 2 2 2 5 2(x 2 2)

x 2 4 2 (2x 2 1) 5 2x 2 4

x 2 4 2 2x 1 1 5 2x 2 4

2x 2 3 5 2x 2 4

23x 2 3 5 24

23x 5 21

x 5 1 } 3

Check: 1 }

3 2 4

} 1 }

3 2 2

2 2 1 1 }

3 2 2 1 }

1 }

3 2 2

0 2

2

11 } 3 }

2 5 } 3 2

2 1 } 3 }

2 5 } 3 0 2

11

} 5 2 1 } 5 0 2

2 5 2 ✓

10. 3x 1 6

} x2 2 4

5 x 1 1

} x 2 2

(3x 1 6)(x 2 2) 5 (x2 2 4)(x 1 1)

3x2 2 12 5 x3 1 x2 2 4x 2 4

0 5 x3 2 2x2 2 4x 1 8

0 5 x2(x 2 2) 2 4(x 2 2)

0 5 (x2 2 4)(x 2 2)

0 5 (x 1 2)(x 2 2)(x 2 2)

x 1 2 5 0 or x 2 2 5 0

x 5 22 or x 5 2

Check x 5 22:

3(22) 1 6

} (22)2 2 4

0 22 1 1

} 22 2 2

0 }

0 0

21 }

24

Division by 0 is undefined, so x 5 22 is an extraneous solution.

Check x 5 2:

3(2) 1 6

} 22 2 4

0 2 1 1

} 2 2 2

12

} 0 0

3 }

0

Division by 0 is undefined, so x 5 2 is an extraneous solution. Because both possible solutions are extraneous, the original equation has no solution.

11. 5 }

x 1

x 1 1 } x 1 2 5

2x 1 9 } x 1 2

x(x 1 2) 1 5 } x 1

x 1 1 } x 1 2 2 5 x(x 1 2) 1 2x 1 9

} x 1 2

2 5(x 1 2) 1 x(x 1 1) 5 x(2x 1 9)

5x 1 10 1 x2 1 x 5 2x2 1 9x

x2 1 6x 1 10 5 2x2 1 9x

0 5 x2 1 3x 2 10

0 5 (x 1 5)(x 2 2)

x 1 5 5 0 or x 2 2 5 0

x 5 25 or x 5 2

Check x 5 25:

5 }

25 1 25 1 1

} 25 1 2 0

2(25) 1 9 }

25 1 2

21 1 24

} 23 0

21 }

23

21 1 4 } 3 0

1 }

3

1 }

3 5

1 } 3 ✓

Check x 5 2:

5 }

2 1

2 1 1 } 2 1 2 0

2(2) 1 9 }

2 1 2

5 }

2 1

3 } 4 0

13 }

4

13

} 4 5

13 } 4 ✓


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511Algebra 2


12.x 2 3}x 1 2

5 x 2 1}3x 2 1

(x 2 3)(3x 2 1) 5 (x 1 2)(x 2 1)

3x2 2 10x 1 3 5 x2 1 x 2 2

2x2 2 11x 1 5 5 0

(2x 2 1)(x 2 5) 5 0

2x 2 1 5 0 or x 2 5 5 0

x 5 1}2 or x 5 5

Check x 5 1}2 :

1 }

2 2 3} 1 }

2 1 2

0 1 }

2 2 1

}

3 1 1 } 2 2 2 1

2 5 } 2

} 5 }

2

02

1 } 2

} 1 }

2

21 5 21 ✓

Check x 5 5:

5 2 3}5 1 2

0 5 2 1}3(5) 2 1

2}7 0

4}14

2}7 5

2}7 ✓

13.x 2 1}

x 1

2x 2 1}x 1 3 5

x 1 6}x 1 3

x(x 1 3) 1x 2 1}

x 1

2x 2 1}x 1 3 2 5 x(x 1 3) 1x 1 6

}x 1 3 2

(x 1 3)(x 2 1) 1 x(2x 2 1) 5 x(x 1 6)

x2 1 2x 2 3 1 2x2 2 x 5 x2 1 6x

3x2 1 x 2 3 5 x2 1 6x

2x2 2 5x 2 3 5 0

(2x 1 1)(x 2 3) 5 0

2x 1 1 5 0 or x 2 3 5 0

x 5 21}2 or x 5 3

Check x 5 21}2 :

2 1 } 2 2 1

}2

1 } 2

1 2 1 2

1 } 2 2 2 1

}2

1 } 2 1 3

02

1 } 2 1 6

}2

1 } 2 1 3

2 3 } 2

}2

1 } 2 1

22} 5 }

2

0 11

} 2

} 5 }

2

3 2 4}5 0

11}5

11}5 5

11}5 ✓

Check x 5 3:

3 2 1

} 3 1

2(3) 2 1 } 3 1 3 0

3 1 6 }

3 1 3

2 }

3 1

5 } 6 0

9 }

6

3 }

2 5

3 } 2 ✓

14. 0.360 5 12 1 x

} 60 1 x

0.360(60 1 x) 5 12 1 x

21.6 1 0.360x 5 12 1 x

21.6 5 12 1 0.640x

9.6 5 0.640x

15 5 x

You have to get 15 consecutive hits to raise your batting average to 0.360.

Problem Solving Workshop 8.6 (p. 597)

1. 80x2 1 300

} 15x2 1 200

5 4.2

X Y1 -6 4.2973 -5.9 4.2717 -5.8 4.2452 -5.7 4.2179 -5.6 4.1897 -5.5 4.1606 -5.4 4.1305 X=-5.6

X Y1 5.3 4.0995 5.4 4.1305 5.5 4.1606 5.6 4.1897 5.7 4.2179 5.8 4.2452 5.9 4.2717 X=5.6

x ø 65.6

y1 5 80x2 1 300

} 15x2 1 200

, y2 5 4.2

Using the intersection feature on a graphing calculator, you see that x ø 65.6.


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2. 5x 1 5

} x2 1 4

5 2

X Y1 0 1.25 .5 1.7647 1 2 1.5 2 2 1.875 2.5 1.7073 3 1.5385 X=1

x 5 1, x 5 1.5

y1 5 5x 1 5

} x2 1 4

, y2 5 2

Using the intersection feature on a graphing calculator, you see that x 5 1 and x 5 1.5.

3. 9x 1 2

} x 2 5 5 20.75

X Y1 5 ERROR 6 56 7 32.5 8 24.667 9 20.75 10 18.4 11 16.833 X=9

x 5 9

y1 5 9x 1 2

} x 2 5 , y2 5 20.75

Using the intersection feature on a graphing calculator, you see that x 5 9.

4. 6x2 }

2x 2 3 5 18

X Y1 0 0 1 -6 2 24 3 18 4 19.2 5 21.429 6 24 X=3

x 5 3

y1 5 6x2

} 2x 2 3 , y2 5 18

Using the intersection feature on a graphing calculator, you see that x 5 3.

5. 14x2 1 60

} 5x2 1 7

5 3.5

X Y1 3 3.5769 3.1 3.5339 3.2 3.4942 3.3 3.4574 3.4 3.4235 3.5 3.3919 3.6 3.3627 X=3.2

X Y1 -3.5 3.3919 -3.4 3.4235 -3.3 3.4574 -3.2 3.4942 -3.1 3.5339 -3 3.5769 -2.9 3.6236 X=-3.2

x ø 63.2

y1 5 14x2 1 60

} 5x2 1 7

, y2 5 3.5

Using the intersection feature on a graphing calculator, you see that x ø 63.2

6. 4.5 5 848x2 1 3220

}} 115x2 1 1000

X Y1 0 3.22 1 3.6484 2 4.5288 3 5.3327 4 5.9113 5 6.30196 6.5658 X=2

Because x 5 2 represents the number of years after 1995, total sales of entertainment software were about $4.5 billion in 1997.


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513Algebra 2


y1 5 848x2 1 3220

}} 115x2 1 1000

, y2 5 4.5

Using the intersection feature on a graphing calculator, you can see that x ø 2, which represents 1997.

7. a. 5 5 6.60

} d 1 33

X Y1 95 5.1563 96 5.1163 97 5.0769 98 5.0382 99 5 100 4.9624 101 4.9254 X=99

At a depth of 99 feet, the recommended percent of oxygen in the air that a diver breathes is 5%.

b. y1 5 660

} x 1 33 , y2 5 10

Using the intersection feature on a graphing calculator, you can see that x 5 33.

So, at a depth of 33 feet, the recommended percent of oxygen in the air that a diver breathes is 10%.

8.6 Extension (p. 600)

1. 5 }

x 2 2 < 0

X Y1 -2 -1.25 -1 -1.667 0 -2.5 1 -5 2 ERROR3 54 2.5 X=-2

The value of y is undefined when x 5 2 and appears to be negative when x < 2. The solution is x < 2.

2. x 2 5

} x 1 3

> 1

x 2 5

} x 1 3

2 1 > 0

X Y1 -7 2 -6 2.6667 -5 4 -4 8 -3 ERROR-2 -8-1 -4 X=-7

The value of y is undefined when x 5 23 and appears to be positive when x < 23. The solution is x < 23.

3. x2 2 3x 1 2

} x 2 3

< x

x2 2 3x 1 2

} x 2 3

2 x < 0

X Y1 0 -.6667 1 -1 2 -2 3 ERROR 4 25 16 .66667 X=0

The value of y is undefined when x 5 3 and appears to be negative when x < 3. The solution is x < 3.

4. 10 }

x 1 2 > 0

X Y1 -4 -5 -3 -10 -2 ERROR -1 10 0 51 3.3333 2 2.5 X=-4

The value of y is undefi ned when x 5 22 and appears to be positive when x > 22. The solution is x > 22.

5. 22x 2 3

} x 2 4

> 0

X Y1 -3 -.4286 -2.5 -.3077 -2 -.1667 -1.5 0 -1 .2-.5 .44444 0 .75 X=-3

X Y1 2.5 5.3333 3 9 3.5 20 4 ERROR 4.5 -245 -13 5.5 -9.333 X=2.5

The value of y is undefi ned when x 5 4 and appears to be positive between x 5 21.5 and x 5 4. The solution is 21.5 < x < 4.


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6. x2 2 4x 1 8

} x 2 1

< x

x2 2 4x 1 8

} x 2 1

2 x < 0

X Y1 .7 -19.67 .8 -28 .9 -53 1 ERROR 1.1 471.2 22 1.3 13.667 X=.7

X Y1 2.63 .06748 2.64 .04878 2.65 .0303 2.66 .01205 2.67 -.0062.68 -.0238 2.69 -.0414 X=2.63

The value of y is undefined when x 5 1 and appears to be negative when x < 1 and when x is greater than

approximately 2.67. Test 2 2 }

3 by entering x 5

8 } 3 into the

table. You will see that when x 5 8 } 3 then y 5 0. So, you

can conclude that the solution is x < 1 or x > 2 2 }

3 .

7. 2 4 } x 1 5 < 0

The graph lies below the x-axis when x > 25. So, the solution is x > 25.

8. 4 }

x 2 3 < 0

The graph lies below the x-axis when x < 3. So, the solution is x < 3.

9. 8 }

x2 1 1 ≥ 4

y1 5 8 }

x2 1 1

y2 5 4

Using the intersect feature, the graph of y1 lies on or above the graph of y2 when 21 ≤ x ≤ 1. The solution is 21 ≤ x ≤ 1.

10. 20 }

x2 1 1 < 2

y1 5 20 }

x2 1 1

y2 5 2

Using the intersect feature, the graph of y1 lies below the graph of y2 when x < 23 and when x > 3. The solution is x < 23 or x > 3.

11. 3x 1 2

} x 2 1

< 22

y1 5 3x 1 2

} x 2 1

y2 5 22

Using the intersect feature, the graph of y1 lies below the graph of y2 when 0 < x < 1. The solution is 0 < x < 1.

12. 3x 1 2

} x 2 1

> x

y1 5 3x 1 2

} x 2 1

y2 5 x

Using the intersect feature, the graph of y1 lies above the graph of y2 when x < 20.45 and when 1 < x < 4.45. The solution is x < 20.45 or 1 < x < 4.45.

13. 3 }

x 1 2 > 0

Critical x-value:

x 1 2 5 0

x 5 22

Test x 5 23: Test x 5 21:

3 }

23 1 2 >?0

3 }

21 1 2 >?0

23 > 0 ✗ 3 > 0 ✓

24 23 22 21 0 21

The solution is x > 22.


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515Algebra 2


14. 2 1 } x 1 5 ≤ 22

2 1 } x 1 5 1 2 ≤ 0

21 1 2(x 1 5)

}} x 1 5 ≤ 0

2x 1 9

} x 1 5 ≤ 0

Critical x-values:

2x 1 9 5 0 x 1 5 5 0

x 5 2 9 } 2 x 5 25

Test x 5 26: Test x 5 2 19

} 4 : Test x 5 24:

21 }

26 1 5 <? 22

21 }

2 19

} 4 1 5 <? 22

21 }

24 1 5 < 22

1 < 22 ✗ 24 < 22 ✓ 21 < 22 ✗

242526 23 22 21 0

922

The solution is 25 < x ≤ 2 9 } 2 .

15. 2 }

x 1 2 >

1 }

x 1 3

2 }

x 1 2 2

1 } x 1 3 > 0

2(x 1 3) 2 (x 1 2)

}} (x 1 2)(x 1 3)

> 0

2x 1 6 2 x 2 2

}} (x 1 2)(x 1 3)

> 0

x 1 4 }}

(x 1 2)(x 1 3) > 0

Critical x-values:

x 1 4 5 0 (x 1 2)(x 1 3) 5 0

x 5 24 x 1 2 5 0 or x 1 3 5 0

x 5 22 or x 5 23

Test x 5 25: Test x 5 2 7 } 2 :

2 }

25 1 2 >?

1 }

25 1 3

2 }

2 7 } 2 1 2

>? 1 }

2 7 } 2 1 3

2 2 } 3 > 2

1 } 2 ✗ 2

4 } 3 > 22 ✓

Test x 5 2 5 } 2 : Test x 5 2

3 } 2 :

2 }

2 5 } 2 1 2

>? 1 }

2 5 } 2 1 3

2 }

2 3 } 2 1 2

>? 1 }

2 3 } 2 1 3

24 > 2 ✗ 4 > 2 }

3 ✓

242526 23 22 21 0

The solution is 24 < x < 23 or x > 22.

16. 5 }

x 2 4 <

1 }

x 1 4

5 }

x 2 4 2

1 } x 1 4 < 0

5(x 1 4) 2 (x 2 4)

}} (x 2 4)(x 1 4)

< 0

5x 1 20 2 x 1 4

}} (x 2 4)(x 1 4)

< 0

4x 1 24

}} (x 2 4)(x 1 4)

< 0

Critical x-values:

4x 1 24 5 0 (x 2 4)(x 1 4) 5 0

4x 5 224 x 2 4 5 0 or x 1 4 5 0

x 5 26 x 5 4 or x 5 24


5 }

27 2 4 <?

1 }

27 1 4

5 }

25 2 4 <?

1 }

25 1 4

2 5 } 11 < 2

1 } 3 ✓ 2

5 } 9 < 21 ✗


5 }

0 2 4 <?

1 }

0 1 4

5 }

5 2 4 <?

1 }

5 1 4

2 5 } 4 <

1 }

4 ✓ 5 <

1 }

9 ✗

2325 2427 26 22 21 0 1 2 3 4 5

The solution is x < 26 or ≤ 24 < x < 4.

17. 5 }

x 1 3 ≥ 4 }

x 1 2

5 }

x 1 3 2

4 } x 1 2 ≥ 0

5(x 1 2) 2 4(x 1 3)

}} (x 1 3)(x 1 2)

≥ 0

5x 1 10 2 4x 2 12

}} (x 1 3)(x 1 2)

≥ 0

x 2 2 }}

(x 1 3)(x 1 2) ≥ 0

Critical x-values:

x 2 2 5 0 (x 1 3)(x 1 2) 5 0

x 5 2 x 1 3 5 0 or x 1 2 5 0

x 5 23 or x 5 22

Test x 5 24: Test x 5 2 5 } 2 :

5 }

24 1 3 ≥?

4 }

24 1 2

5 }

2 5 } 2 1 3

≥? 4 }

2 5 } 2 1 2

25 ≥ 22 ✗ 10 ≥ 28 ✓


5 }

0 1 3 ≥?

4 }

0 1 2

5 }

3 1 3 ≥?

4 }

3 1 2

5 }

3 ≥ 2 ✗

5 }

6 ≥ 4 }

5 ✓

23 22 21 0 1 2 3

The solution is 23 < x < 22 or x ≥ 2.


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18.2

}x 1 6

> 23}x 2 3

2}x 1 6

1 3

}x 2 3 > 0

2(x 2 3) 1 3(x 1 6)}}

(x 1 6)(x 2 3) > 0

2x 2 6 1 3x 1 18}}

(x 1 6)(x 2 3) > 0

5x 1 12}}(x 1 6)(x 2 3)

> 0

Critical x-values:

5x 1 12 5 0 (x 1 6)(x 2 3) 5 0

5x 5 212 x 1 6 5 0 or x 2 3 5 0

x 5 212}5 x 5 26 or x 5 3

5 22.4


2}27 1 6

>? 23}27 2 3

2}23 1 6

>? 23}23 2 3

22 > 3 }

10 ✗

2 }

3 >

1 }

2 ✓


2 }

0 1 6 >?

23 }

0 2 3

2 }

4 1 6 >?

23 }

4 2 3

1 }

3 > 1 ✗

1 } 5 > 23 ✓

2325 2427 26 22 21 0 1 2 3 4 5

22.4

The solution is 26 < x < 22.4 or x > 3.

19. 23680

} t 2 50

> 80

23680

} t 2 50

2 80 > 0

X Y1 4 0 5 1.7778 6 3.6364 7 5.5814 8 7.6199 9.756110 12 X=4

The value of y is 0 when x 5 4 and appears to be positive when 5 ≤ x ≤ 8 (the domain is 0 ≤ x ≤ 8). So, the number of eggs produced was greater than 80 billion from 1999 to 2002.

20. Your phone plan average <

Second phone plan average

5 1 0.05m

} m

< 0.07

5 1 0.05m

} m

2 0.07 < 0

5 1 0.05m 2 0.07m

}} m

< 0

5 2 0.02m

} m

< 0

Critical m-values:

5 2 0.02m 5 0 m 5 0

20.02m 5 25

m 5 250

Test m 5 50: Test m 5 300:

5 1 0.05(50)

} 50

<? 0.07 5 1 0.05(300)

}} 300

<? 0.07

0.15 < 0.07 ✗ 1 }

15 < 0.07 ✓

150100500 200 250 300

The solution is m > 250. So, you must talk more than 250 minutes each month so that your average cost is less than $.07 per minute.

21. 43t 1 50

} t ≤ 47

43t 1 50

} t 2 47 ≤ 0

X Y1 11 .54545 11.5 .34783 12 .16667 12.5 0 13 -.153813.5 -.296314 -.4286 X=11

The value of y is 0 when x 5 12.5 and appears to be negative when x > 12.5. So, the average monthly cost is at most $47 after 13 or more months.

y1 5 43x 1 50

} x , y2 5 47

Using the intersect feature, the graph of y1 lies below the graph of y2 when x ≥ 12.5. So, the average monthly cost is at most $47 after 13 or more months.


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517Algebra 2


22. a. Let c be the number of calendars.

Average cost per calendar

< Desired cost per calendar

710 1 4.50c

} c < 10

b. y1 5 710 1 4.50x

} x , y2 5 10

Using the intersect feature, the graph of y1 lies below the graph of y2 when x > 129.09. So, 130 or more calendars need to be printed to bring the average cost per calendar below $10.

c. y1 5 710 1 4.50x

} x , y2 5 6

Using the intersect feature, the graph of y1 lies below the graph of y2 when x > 473.33. So, 474 or more calendars need to be printed to bring the average cost per calendar below $6.

Mixed Review of Problem Solving (p. 601)

1. a. Time 5 Distance

} Speed

5 50

} s

b. Time 5 Distance

} Speed

5 50 } s 1 5

c. Total time 5 50

} s 1 50 } s 1 5

5 50

} s p s 1 5

} s 1 5 1 50 } s 1 5 p s }

s

5 50s 1 250

} s(s 1 5)

1 50s }

s(s 1 5)

5 100s 1 250

} s(s 1 5)

5 50(2s 1 5)

} s(s 1 5)

2. Total time 5 Distance

} Rate withcurrent

1 Distance

} Rate againstcurrent

1.25 5 2 } r 1 3 1

2 } r 2 3

1.25(r 1 3)(r 2 3) 5 (r 1 3)(r 2 3) 1 2 }

r 1 3 1

2 } r 2 3 2

1.25(r2 2 9) 5 2(r 2 3) 1 2(r 1 3)

1.25r2 2 11.25 5 2r 2 6 1 2r 1 6

1.25r2 2 4r 2 11.25 5 0

4(1.25r2 2 4r 2 11.25) 5 0

5r2 2 16r 2 45 5 0

(5r 1 9)(r 2 5) 5 0

5r 1 9 5 0 or r 2 5 5 0

r 5 2 9 } 5 or r 5 5

Speed cannot be negative, so reject the negative value,

2 9 } 5 . Your speed in still water is 5 miles per hour.

3. a. Rectangular container:

S 5 2[(2r)2 1 2r(h) 1 2r(h)] 5 2(4r2 1 2rh 1 2rh) 5 2(4r2 1 4rh)

V 5 (2r)(2r)(h) 5 4r2h

S }

V 5

2(4r2 1 4rh) }

4r2h 5

2(4r)(r 1 h) }

4r p r p h 5 2(r 1 h)

} rh

Cylinder:

S 5 2πr2 1 2πrh

V 5 πr2h

S }

V 5

2πr2 1 2πrh }

πr2h 5

2πr(r 1 h) } π (r)(r)(h) 5

2(r 1 h) }

rh

b. 2(r 1 h)

} rh

5 2(r 1 h)

} rh

The ratio of surface area to volume for each container is the same. So, the containers have equal effi ciencies.

4. Sample answer:

r(x) p s(x) 5 x 2 3

} x 1 4 , where r(x) 5 x 2 5 }

x2 1 5x 1 4 and

s(x) 5 x2 2 2x 2 3

} x 2 5 .

x 2 5 }

x2 1 5x 1 4 p x

2 2 2x 2 3 } x 2 5 5

x 2 5 }}

(x 1 1)(x 1 4) p

(x 2 3)(x 1 1) }} x 2 5

5 (x 2 5)(x 2 3)(x 1 1)

}} (x 1 1)(x 1 4)(x 2 5)

5 x 2 3

} x 1 4


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5. a. Amount of copper 5 Amount of zinc

}} % zinc

2 Amount of zinc

25 5 x } 0.45 2 x

b. 25 5 x } 0.45 2 x

0.45(25) 5 0.45 1 x }

0.45 2 x 2

11.25 5 x 2 0.45x

11.25 5 0.55x

20.45 ø x

You need about 20.45 ounces of zinc to make brass.

c. c 5 x } 0.45 2 x

0.45c 5 0.45 1 x }

0.45 2 x 2

0.45c 5 x 2 0.45x

0.45c 5 0.55x

0.818c 5 x

You need 0.818c ounces of zinc to make brass, where c is the amount of copper.

6. a. Let s be the speed of the truck.

Distance for car

}} Speed of car

5 Distance for truck

}} Speed of truck

120

} s 1 10

5 100

} s

b. 120

} s 1 10

5 100

} s

120s 5 100(s 1 10)

120s 5 100s 1 1000

20s 5 1000

s 5 50

The speed of the truck is 50 miles per hour and the speed of the car is 50 1 10 5 60 miles per hour.

c. To fi nd how long the car spent traveling, use the distance formula d 5 rt and solve for t.

d 5 rt

120 5 60t

21 5 t

Because the car traveled 120 miles in the same amount of time it took the truck to travel 100 miles, both vehicles spent 2 hours traveling.

7. Volume of sphere

}} Volume of cube

5 4πr3

} 3 }

s3 5 4πr3

} 3 }

s3 p 3 } 3 5 4πr3

} 3s3

The side length of the cube is 2r, so the ratio is

4π r3

} 3s3 5

4π r3

} 3(2r)3 5

4π r3

} 24r3 5

π } 6 ø 0.52.

Chapter 8 Review (pp. 603–606)

1. If two variables x and y are related by an equation of

the form y 5 a } x where a Þ 0, then x and y show inverse

variation.

2. The expression z }

xy represents the constant of variation.

3. A function of the form f (x) 5 p (x)

} q(x)

where p(x) and

q(x) are polynomials and q(x) Þ 0 is called a rational

function.

4. Sample answer: The fractions 1 }

x }

2 }

x 1 4

and

2 }

x 1 1 }}

3 }

x 2 4 1

1 } x 1 1

are

complex fractions.

5. When you rewrite the equation 3 }

x 5

2 } x 2 1 as

3(x 2 1) 5 2x, you are cross-multiplying.

6. y 5 a } x y 5

5 } x

5 5 a } 1 5

5 }

23

5 5 a 5 2 5 } 3

y 5 5 } x

7. y 5 a } x y 5

24 } x

26 5 a }

24 5 24

} 23

24 5 a 5 28

y 5 24

} x

8. y 5 a } x y 5

45 } x

18 5 a }

5 }

2 5

45 }

23

18 1 5 } 2 2 5 a 5 215

9(5) 5 a

45 5 a

y 5 45

} x

9. y 5 a } x y 5

28 } x

2 }

3 5

a }

212 5 28

} 23

212 1 2 } 3 2 5 a 5

8 } 3

24(2) 5 a

28 5 a

y 5 28

} x


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519Algebra 2


10. y 5 4 } x 2 3

1

x

y

21

The domain is all real numbers except 3, and the range is all real numbers except 0.

11. y 5 1 } x 1 5 1 2

1

x

y

21


12. f (x) 5 3x 2 2

} x 2 4

x 2 4 5 0 → x 5 4, so x 5 4 is a vertical asymptote.

y 5 a }

c 5

3 } 1 5 3 is a horizontal asymptote.

2

x

y

22


13. y 5 5 }

x2 1 1

The numerator has no real zeros, so there is no x-intercept. The denominator has no real zeros, so there is no vertical asymptote.


x y

23 1 }

2

22 1

0 5

2 1

3 1 }

2

1

x

y

21

14. y 5 4x2

} x 2 1




x y

21 22

0 0

1 }

2 22

3 }

2 18

2 16

3 18

5

x

y

22

15. h(x) 5 6x2

} x 2 2




x y

21 22

0 0

1 26

3 54

4 48

5 50

10

x

f(x)

4

16. y 5 28 }

x2 1 3



x y

23 2 2 } 3

21 22

0 2 8 } 3

1 22

3 2 2 } 3

1x

y

21


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17. y 5 x2 1 6 }

x2 2 3x 2 40


x2 2 3x 2 40 5 0

(x 2 8)(x 1 5) 5 0

x 2 8 5 0 or x 1 5 5 0

x 5 8 or x 5 25

The vertical asymptotes are x 5 8 and x 5 25.

m 5 2 and n 5 2, so m 5 n and y 5 1 } 1 5 1 is a

horizontal asymptote.

x y

28 35

} 24

26 3

22 2 1 } 3

0 2 3 }

20

1 2 1 }

6

10 53

} 15

15 33

} 20

2

6 x

y

18. g(x) 5 x2 2 1

} x 1 4

x2 2 1 5 0 → x2 5 1 → x 5 61, so 1 and 21 are x-intercepts.


m 5 2 and n 5 1, so m > n and there is no horizontal asymptote.

x y

27 216

25 224

23 8

22 3 }

2

0 2 1 } 4

x

g(x)

2

5

19. 80x4

} y3 p

xy }

5x2 5 80x5y

} 5x2y3 5

5 p 16 p x2 p x3 p y }}

5 p x2 p y p y2 5 16x3

} y2

20. x 2 3

} 2x 2 8

p 6x2 2 96 }

x2 2 9 5

x 2 3 }

2(x 2 4) p

6(x2 2 16) }}

(x 1 3)(x 2 3)

5 x 2 3

} 2(x 2 4)

p 6(x 1 4)(x 2 4)

}} (x 1 3)(x 2 3)

5 (2)(3)(x 2 3)(x 1 4)(x 2 4)

}}} (2)(x 2 4)(x 1 3)(x 2 3)

5 3(x 1 4)

} x 1 3

21. 16x2 2 8x 1 1

}} x3 2 7x2 1 12x

4 20x2 2 5x

} 15x3

5 16x2 2 8x 1 1

}} x3 2 7x2 1 12x

p 15x3 }

20x2 2 5x

5 (4x 2 1)(4x 2 1)

}} x(x2 2 7x 1 12)

p 15x3 }

5x(4x 2 1)

5 (4x 2 1)(4x 2 1)(5x)(3x)(x)

}}} x(x 2 4)(x 2 3)(5x)(4x 2 1)

5 3x(4x 2 1)

}} (x 2 4)(x 2 3)

22. x2 2 13x 1 40

}} x2 2 2x 2 15

4 (x2 2 5x 2 24)

5 x2 2 13x 1 40

}} x2 2 2x 2 15

p 1 }

x2 2 5x 2 24

5 (x 2 8)(x 2 5)

}} (x 2 5)(x 1 3)

p 1 }}

(x 2 8)(x 1 3)

5 (x 2 8)(x 2 5)

}}} (x 2 5)(x 1 3)(x 2 8)(x 1 3)

5 1 }

(x 1 3)2

23. 5 }

6(x 1 3) 1

x 1 4 } 2x 5

5 }

6(x 1 3) p x }

x 1

x 1 4 }

2x p

3(x 1 3) }

3(x 1 3)

5 5x }

6x(x 1 3) 1

3(x 1 4)(x 1 3) }}

6x(x 1 3)

5 5x }

6x(x 1 3) 1

3(x2 1 7x 1 12) }}

6x(x 1 3)

5 5x }

6x(x 1 3) 1

3x2 1 21x 1 36 }}

6x(x 1 3)

5 3x2 1 26x 1 36

}} 6x(x 1 3)

24. 5x }

x 1 8 1

4x 2 9 }

x2 1 5x 2 24 5

5x } x 1 8 1

4x 2 9 }}

(x 1 8)(x 2 3)

5 5x } x 1 8 p x 2 3

} x 2 3

1 4x 2 9 }}

(x 1 8)(x 2 3)

5 5x2 2 15x

}} (x 1 8)(x 2 3)

1 4x 2 9 }}

(x 1 8)(x 2 3)

5 5x2 2 11x 2 9

}} (x 1 8)(x 2 3)

25. x 1 2 }

x2 1 4x 1 3 2

5x }

x2 2 9

5 x 1 2 }}

(x 1 3)(x 1 1) 2

5x }}

(x 1 3)(x 2 3)

5 x 1 2 }}

(x 1 3)(x 1 1) p x 2 3

} x 2 3

2 5x }}

(x 1 3)(x 2 3) p x 1 1

} x 1 1

5 x2 2 x 2 6

}} (x 1 3)(x 2 3)(x 1 1)

2 5x2 1 5x

}} (x 1 3)(x 2 3)(x 1 1)

5 x2 2 x 2 6 2 (5x2 1 5x)

}} (x 1 3)(x 2 3)(x 1 1)

5 x2 2 x 2 6 2 5x2 2 5x

}} (x 1 3)(x 2 3)(x 1 1)

5 24x2 2 6x 2 6

}} (x 1 3)(x 2 3)(x 1 1)

5 22(2x2 1 3x 1 3)

}} (x 1 3)(x 2 3)(x 1 1)


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521Algebra 2


26. 2x

} 9 5

2 } x

2x2 5 18

x2 5 9

x 5 63


2(3)

} 9 0

2 }

3

2(23) }

9 0

2 }

23

6 }

9 0

2 }

3

26 }

9 0 2

2 } 3

2 }

3 5

2 } 3 ✓ 2

2 } 3 5 2

2 } 3 ✓

27. 5 }

x 5

7 } x 1 2 28.

x 2 1 }

4 5

3x } 9

5(x 1 2) 5 7x 9(x 2 1) 5 4(3x)

5x 1 10 5 7x 9x 2 9 5 12x

10 5 2x 29 5 3x

5 5 x 23 5 x

Check: Check:

5 } 5 0

7 }

5 1 2

23 2 1 }

4 0

3(23) }

9

1 0 7 } 7

24 }

4 0

29 }

9

1 5 1 ✓ 21 5 21 ✓

29. 2 }

x 1 2 5

6 } 2x 1 5

2(2x 1 5) 5 6(x 1 2)

4x 1 10 5 6x 1 12

22x 1 10 5 12

22x 5 2

x 5 21

Check:

2 }

21 1 2 0

6 }

2(21) 1 5

2 }

1 0

6 }

3

2 5 2 ✓

30. x 1 12

} 3 5

2x 1 3 } x 1 2

(x 1 12)(x 1 2) 5 3(2x 1 3)

x2 1 14x 1 24 5 6x 1 9

x2 1 8x 1 15 5 0

(x 1 5)(x 1 3) 5 0

x 1 5 5 0 or x 1 3 5 0

x 5 25 or x 5 23


25 1 12

} 3 0

2(25) 1 3 }

25 1 2

23 1 12 }

3 0

2(23) 1 3 }

23 1 2

7 }

3 0

27 }

23

9 }

3 0

23 }

21

7 }

3 5

7 } 3 ✓ 3 5 3

31. 2x }

x 1 4 5

23x } 4x 2 3

2x(4x 2 3) 5 (23x)(x 1 4)

8x2 2 6x 5 23x2 2 12x

11x2 1 6x 5 0

x(11x 1 6) 5 0

x 5 0 or 11x 1 6 5 0

x 5 2 6 } 11

Check x 5 0:

2(0)

} 0 1 4

0 23(0)

} 4(0) 2 3

0 }

4 0

0 }

23

0 5 0 ✓

Check x 5 2 6 } 11 :

2 1 2

6 } 11 2 }

2 6 }

11 1 4

0 23 1 2

6 } 11 2 }

4 1 2 6 } 11 2 2 3

2

12 } 11 }

38

} 11

0

18

} 11

}

2 57

} 11

2 12

} 38 0 2 18

} 57

2 6 } 19 5 2

6 } 19 ✓

32. 5 }

2 1

3 } x 5 3

2x 1 5 } 2 1

3 } x 2 5 2x(3)

5x 1 6 5 6x

6 5 x

Check:

5 }

2 1

3 } 6 0 3

5 }

2 1

1 } 2 0 3

6 }

2 0 3

3 5 3 ✓


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33. 8(x 2 1)

} x2 2 4

5 4 } x 1 2

8(x 2 1)

}} (x 1 2)(x 2 2)

5 4 } x 1 2

(x 1 2)(x 2 2) F 8(x 2 1) }}

(x 1 2)(x 2 2) G 5 (x 1 2)(x 2 2) 1 4

} x 1 2

2 8(x 2 1) 5 4(x 2 2)

8x 2 8 5 4x 2 8

4x 2 8 5 28

4x 5 0

x 5 0

Check:

8(0 2 1)

} 02 2 4

0 4 }

0 1 2

28

} 24

0 4 }

2

2 5 2 ✓

34. 3x }

x 1 1 5

12 }

x2 2 1 1 2

3x }

x 1 1 5

12 }}

(x 1 1)(x 2 1) 1 2

(x 1 1)(x 2 1) 1 3x }

x 1 1 2 5 (x 1 1)(x 2 1) F 12

}} (x 1 1)(x 2 1)

1 2 G 3x(x 2 1) 5 12 1 2(x 1 1)(x 2 1)

3x2 2 3x 5 12 1 2(x2 2 1) 3x2 2 3x 5 12 1 2x2 2 2

x2 2 3x 2 10 5 0

(x 2 5)(x 1 2) 5 0

x 2 5 5 0 or x 1 2 5 0

x 5 5 or x 5 22


3(5)

} 5 1 1

0 12 }

52 2 1 1 2

3(22) }

22 1 1 0

12 }

(22)2 2 1 1 2

15

} 6 0

12 }

24 1 2

26 }

21 0

12 }

3 1 2

5 }

2 5

5 } 2 ✓ 6 5 6 ✓

35. 2(x 1 7)

} x 1 4

2 2 5 2x 1 20

} 2x 1 8

2(x 1 7)

} x 1 4

2 2 5 2x 1 20

} 2(x 1 4)

2(x 1 4) F 2(x 1 7) }

x 1 4 2 2 G 5 2(x 1 4) F 2x 1 20

} 2(x 1 4)

G 4(x 1 7) 2 4(x 1 4) 5 2x 1 20

4x 1 28 2 4x 2 16 5 2x 1 20

12 5 2x 1 20

28 5 2x

24 5 x

Check:

2(24 1 7)

} 24 1 4

2 2 0 2(24) 1 20

} 2(24) 1 8

6 }

0 2 2 0

12 }

0

Division by 0 is undefined, so x 5 24 is an extraneous solution, and the equation has no solution.

36. a. Total free throws attempted 4

Total freethrows made 5

60 1 x } 75 1 x

b. 60 1 x

} 75 1 x 5 0.82

60 1 x 5 0.82(75 1 x)

60 1 x 5 61.5 1 0.82x

60 1 0.18x 5 61.5

0.18x 5 1.5

x 5 8. } 3

The player must make 9 consecutive free throws to raise her free-throw percentage to at least 82%.

Chapter 8 Test (p. 607)

1. y 5 a } x y 5

10 } x

2 5 a } 5 5

10 } 4

10 5 a 5 5 } 2

y 5 10

} x

2. y 5 a } x y 5

216 } x

8 5 a }

22 5 216

} 4

216 5 a 5 24

y 5 216

} x

3. y 5 a } x y 5

15 } x

10 5 a }

3 }

2 5

15 } 4

10 1 3 } 2 2 5 a 5

15 }

4

15 5 a

y 5 15

} x


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523Algebra 2


4. y 5 a } x y 5

18 } x

6 5 a } 3 5

18 } 4

18 5 a 5 9 } 2

y 5 18

} x

5. y 5 a } x y 5

214 } x

7 }

2 5

a }

24 5 214

} 4

7 }

2 (24) 5 a 5 2

7 } 2

214 5 a

y 5 214

} x

6. y 5 a } x y 5

15 } 32x

5 }

8 5

a }

3 }

4 5

15 }

32(4)

5 }

8 1 3 } 4 2 5 a 5

15 }

128

15

} 32

5 a

y 5 15

} 32 } x 5

15 } 32x

7. y 5 2 } x 1 5 2 3

2

x

y

22


8. y 5 21

} x 2 4 2 1

1

x

y

5


9. f (x) 5 6 2 x

} 2x 1 1

2x 1 1 5 0 → x 5 2 1 } 2 , so x 5 2

1 } 2 is a

vertical asymptote.

y 5 a } c 5

21 } 2 5 2

1 }

2 is a horizontal asymptote.

2

2 x

f(x)

The domain is all real numbers except 2 1 } 2 , and the range

is all real numbers except 2 1 } 2 .

10. y 5 4 }

x2 1 2



x y

22 2 }

3

21 4 }

3

0 2

1 4 }

3

2 2 }

3

1

x

y

21


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11. y 5 x2 2 4 }

x2 1 8x 1 15

x2 2 4 5 0 → x2 5 4 → x 5 62, so 2 and 22 are x-intercepts.

x2 1 8x 1 15 5 0

(x 1 5)(x 1 3) 5 0

x 1 5 5 0 or x 1 3 5 0

x 5 25 or x 5 23

The vertical asymptotes are x 5 25 and x 5 23.


asymptote.

x y

28 4

26 32

} 3

24 212

2 7 } 2 211

2 5 } 2

9 } 5

22 0

1 2 1 } 8

2 0

5

x

y

21

12. g(x) 5 x2 1 3

} 2x 2 1


2x 2 1 5 0 → x 5 1 } 2 , so x 5

1 } 2 is a vertical asymptote.

m 5 2 and n 5 1, so m > n and there is no horizontal asymptote.

x f (x)

26 23

21 2 4 } 3

0 23

1 4

2 7 }

3

7 4

1

x

g(x)

22

13. (x 2 3)(x 1 5)

x(x 1 5)

LCM 5 x(x 2 3)(x 1 5)

14. 4x2(x 2 2) 5 (22)(x2)(x 2 2)

8x(x 1 2) 5 23(x)(x 1 2)

LCM 5 23(x2)(x 2 2)(x 1 2) 5 8x2(x 2 2)(x 1 2)

15. x2 2 4x 5 x(x 2 4)

x2 2 2x 2 8 5 (x 2 4)(x 1 2)

LCM 5 x(x 2 4)(x 1 2)

16. 2x 1 6 5 2(x 1 3)

x3 1 10x2 1 21x 5 x(x2 1 10x 1 21) 5 x(x 1 3)(x 1 7)

LCM 5 2x(x 1 3)(x 1 7)

17. 3x2y

} 4x3y5 4

6y2

} 2xy3 5

3x2y }

4x3y5 p 2xy3

} 6y2

5 6x3y4

} 24x3y7

5 6x3y4

}} 6 p 4 p x3 p y4 p y3

5 1 }

4y3

18. x2 2 3x 2 4

} x2 2 3x 2 18

p x 2 6 }

x 1 1 5

(x 2 4)(x 1 1) }}

(x 2 6)(x 1 3) p x 2 6

} x 1 1

5 (x 2 4)(x 1 1)(x 2 6)

}} (x 2 6)(x 1 3)(x 1 1)

5 x 2 4

} x 1 3

19. x2 2 8x 1 15

}} x2 1 12x 1 32

p x 1 4 }

x2 2 25 5

(x 2 5)(x 2 3) }}

(x 1 4)(x 1 8) p x 1 4

}} (x 1 5)(x 2 5)

5 (x 2 5)(x 2 3)(x 1 4)

}}} (x 1 4)(x 1 8)(x 1 5)(x 2 5)

5 x 2 3 }}

(x 1 8)(x 1 5)

20. x2 2 11x 1 28

}} x2 1 5x 1 4

4 (x2 2 16)

5 x2 2 11x 1 28

}} x2 1 5x 1 4

p 1 }

x2 2 16

5 (x 2 7)(x 2 4)

}} (x 1 4)(x 1 1)

p 1 }}

(x 1 4)(x 2 4)

5 (x 2 7)(x 2 4)

}}} (x 1 4)(x 1 1)(x 1 4)(x 2 4)

5 x 2 7 }}

(x 1 4)2(x 1 1)

21. 3x } x 1 5 2

4x 1 1 } x 1 5 5

3x 2 (4x 1 1) }} x 1 5

5 3x 2 4x 2 1

} x 1 5

5 2x 2 1

} x 1 5

22. 4 }

x 2 3 1

2 } x 1 6 5

4 } x 2 3 p x 1 6

} x 1 6

1 2 } x 1 6 p x 2 3

} x 2 3

5 4x 1 24

}} (x 2 3)(x 1 6)

1 2x 2 6 }}

(x 2 3)(x 1 6)

5 6x 1 18

}} (x 2 3)(x 1 6)

5 6(x 1 3)

}} (x 2 3)(x 1 6)


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525Algebra 2



23. 3x }

x2 1 x 2 12 2

6 } x 1 4 5

3x }}

(x 1 4)(x 2 3) 2

6 } x 1 4

5 3x }}

(x 1 4)(x 2 3) 2

6 } x 1 4 p x 2 3

} x 2 3

5 3x }}

(x 1 4)(x 2 3) 2

6x 2 18 }}

(x 1 4)(x 2 3)

5 3x 2 (6x 2 18)

}} (x 1 4)(x 2 3)

5 3x 2 6x 1 18

}} (x 1 4)(x 2 3)

5 23x 1 18

}} (x 1 4)(x 2 3)

5 23(x 2 6)

}} (x 1 4)(x 2 3)

24. 4 } x 1 5 1

2x }

x2 2 25 5

4 } x 1 5 1

2x }}

(x 1 5)(x 2 5)

5 4 } x 1 5 p x 2 5

} x 2 5 1 2x }}

(x 1 5)(x 2 5)

5 4x 2 20

}} (x 1 5)(x 2 5)

1 2x }}

(x 1 5)(x 2 5)

5 6x 2 20

}} (x 1 5)(x 2 5)

5 2(3x 2 10)

}} (x 1 5)(x 2 5)

25. 3 }

x 1 2 5

x 2 3 } 2x 1 4

3(2x 1 4) 5 (x 1 2)(x 2 3)

6x 1 12 5 x2 2 x 2 6

0 5 x2 2 7x 2 18

0 5 (x 2 9)(x 1 2)

x 2 9 5 0 or x 1 2 5 0

x 5 9 or x 5 22


3 }

9 1 2 0

9 2 3 }

2(9) 1 4

3 }

22 1 2 0

22 2 3 }

2(22) 1 4

3 }

11 0

6 }

22

3 }

0 0

25 }

0

3 }

11 5

3 } 11 ✓

Division by 0 is undefined, so 22 is an extraneous solution. The only solution is x 5 9.

26. 1 }

x 1 6 1

x 1 1 } x 5

13 } x 1 6

x(x 1 6) 1 1 }

x 1 6 1

x 1 1 } x 2 5 x(x 1 6) 1 13

} x 1 6

2 x 1 (x 1 6)(x 1 1) 5 13x

x 1 x2 1 7x 1 6 5 13x

x2 1 8x 1 6 5 13x

x2 2 5x 1 6 5 0

(x 2 3)(x 2 2) 5 0

x 2 3 5 0 or x 2 2 5 0

x 5 3 or x 5 2


1 }

3 1 6 1

3 1 1 } 3 0

13 }

3 1 6

1 }

2 1 6 1

2 1 1 } 2 0

13 }

2 1 6

1 }

9 1

4 } 3 0

13 }

9

1 }

8 1

3 } 2 0

13 }

8

13

} 9 5

13 } 9 ✓

13 }

8 5

13 } 8 ✓

27. x 2 2

} x 2 1

5 x 1 2

} x 1 4

(x 2 2)(x 1 4) 5 (x 2 1)(x 1 2)

x2 1 2x 2 8 5 x2 1 x 2 2

2x 2 8 5 x 2 2

x 2 8 5 22

x 5 6

Check: 6 2 2

} 6 2 1

0 6 1 2

} 6 1 4

4 } 5 0

8 }

10

4 } 5 5

4 } 5 ✓

28. I 5 a }

r2

29. Let c be the average cost and m be the number of months.

c 5 Monthly fee p Number of months 1 Installation fee

}}}} Number of months

5 50m 1 30

} m

Ave

rage

co

st (

do

llars

)

020 4 6 8

20

60

80

40

Number of months

m

C

(5, 56)


30. Let c be the average cost and m be the number of months.

c 5 Monthly fee p Number of months 1 Setup fee


5 99m 1 50

} m

100 5 99m 1 50

} m

100m 5 99m 1 50

m 5 50

You would need to use this service for 50 months in order for your average monthly cost to fall to $100.

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Standardized Test Preparation (p. 609)

1. D;

18x

7x9x

6x12x

2x

Perimeter 5 6x 1 2x 1 12x 1 7x 1 18x 1 9x 5 54x

Area 5 6x(2x) 1 18x(7x) 5 12x2 1 126x2 5 138x2

Perimeter

} Area

5 54x

} 138x2 5

6x(9) }

6x(23x) 5

9 } 23x

2. A;

2(180) 5 360

4(90) 5 360

6(60) 5 360

8(45) 5 360

10(36) 5 360

Each product is equal to 360. So, the data show inverse variation. A model relating x and y is x p y 5 360 or

y 5 360

} x .

Standardized Test Practice (pp. 610–611)

1. B; From the graph, you can see that the graph does not pass through y 5 2, so the range is all real numbers except 2.

2. C; From the graph, you can see that the graph does pass through x 5 2, so the function is defined for x 5 2.

3. A;

212

} 2(22)

5 3

3 }

1(1) 5 3

30 }

25(22) 5 3

21.5

} 20.5(1)

5 3

Each expression is equal to 3. So, the data show joint

variation. A model relating p, q, and r is p }

qr 5 3 or

p 5 3qr.

4. C; p }

qr 5 3

20

} 24r

5 3

20 5 212r

2 5 } 3 5 r

5. D; From the graph, you can see that the graph crosses the x-axis when x 5 25 and when x 5 4.

6. D; From the graph, you can see that the graph’s end behavior approaches y 5 1.

7. B; l 5 a } w

1.72 5 a } 0.67

1.72(0.67) 5 a

1.1524 5 a

l 5 1.1524

} w

lw 5 1.1524

8. A;

Volume of prism 5 lwh

5 18.5(3.2)(5)

5 296 cm3

Volume of cylinder 5 πr2h

296 5 πr2h

296

} πr2 5 h

Surface area of cylinder 5 2πr2 1 2πrh

S 5 2πr2 1 2πr 1 296 }

πr2 2 5 2πr2 1

592 } r

S 5 2πr2 1 592

} r

MinimiumX=3.611635 Y=245.87197

Using the minimum feature, you get a minimum value of about 246, which occurs when r ø 3.61. The radius of the cylinder is about 3.61 centimeters.

9. 4 }

x 1 1 2

1 } x 5 1

x(x 1 1) 1 4 }

x 1 1 2

1 } x 2 5 1(x)(x 1 1)

4x 2 (x 1 1) 5 x2 1 x

4x 2 x 2 1 5 x2 1 x

3x 2 1 5 x2 1 x

0 5 x2 2 2x 1 1

0 5 (x 2 1)(x 2 1)

x 2 1 5 0

x 5 1

10. n 5 a } m

28 5 a }

22

16 5 a

n 5 16

} m 5 16

} 5 5 3.2


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527Algebra 2


11. q 5 ars2

} Ï

} t

21.75 5 a(22)(12)

} Ï

}

16

21.75 5 22a

} 4

27 5 22a

7 }

2 5 a

q 5 7rs2

} 2 Ï

} t 5

7(5)(23)2

} 2 Ï

}

25 5

315 } 2 p 5 5

315 } 10 5 31.5

12. f (x) 5 x 2 3 }

2x2 2 4x 1 2 5

x 2 3 }}

2(x2 2 2x 1 1) 5

x 2 3 }}

2(x 2 1)(x 2 1)

The domain of the function is all real numbers except 1.

13. 0.95 5 15 1 x

} 17 1 x

0.95(17 1 x) 5 15 1 x

16.15 1 0.95x 5 15 1 x

16.15 5 15 1 0.05x

1.15 5 0.05x

23 5 x

The student must turn in 23 consecutive assignments to raise her homework percentage up to 95%.

14. T 5 12

} s 1 12 } w 2 s

9 5 12

} s 1 12 } 6 2 s

s(6 2 s)(9) 5 s(6 2 s) 1 12 }

s 1

12 } 6 2 s 2

9s(6 2 s) 5 12(6 2 s) 1 12s

54s 2 9s2 5 72 2 12s 1 12s

54s 2 9s2 5 72

0 5 9s2 2 54s 1 72

0 5 9(s2 2 6s 1 8) 0 5 9(s 2 4)(s 2 2)

s 2 4 5 0 or s 2 2 5 0

s 5 4 or s 5 2

Check s 5 4: Check s 5 2:

9 0 12

} 4 1

12 } 6 2 4 9 0

12 }

2 1

12 } 6 2 2

9 0 3 1 12

} 2 90 6 1 12

} 4

9 5 9 ✓ 9 5 9 ✓

The speed of the escalator is either 4 feet per second or 2 feet per second.

15. To find a model for the daily per capita water consumption D as a function of the year, divide the model for the daily water consumption c by the model for the population p.

D 5 c }

p 5

2.89x2 1 61.0

}} 0.0293x2 1 1

}} 2460x 1 180,000

The model for D is a complex fraction. To simplify the fraction, divide the numerator by the denominator.

D 5 2.89x2 1 61.0

}} 0.0293x2 1 1

4 (2460x 1 180,000)

5 2.89x2 1 61.0

}} 0.0293x2 1 1

p 1 }}

2460x 1 180,000

5 2.89x2 1 61.0

}}} (0.0293x2 1 1)(2460x 1 180,000)

Since C is in billions (109) and P is in thousands (103), when you simplify these values you must multiply the answer by 109 2 3 5 106. So,

D 5 (2.89x2 1 61)106

}}} (0.0293x2 1 1)(2460x 1 180,000)

16. a. x n

0 7.25

2 7.13

4 6.96

6 6.71

8 6.35

10 5.93

12 5.84

14 6.48

16 7.22

18 7.64

20 7.83

b. n 5 0.0441x2 2 1.08x 1 7.25

}}} 0.00565x2 2 0.142x 1 1.00

Mal

es e

nro

lled

in h

igh

sc

ho

ol (

mill

ion

s)

040 8 12 16 20

2

6

8

4

Years since 1980

x

n


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c. From the graph, 7.7 million males were enrolled in high school about 18 years after 1980, or in 1998.

d. Sample answer: The graph decreases from 1980 to about 1991, then the graph increases. It has a horizontal

asymptote at n 5 0.0441

} 0.00565 ø 7.8, which means as x

increases, n gets closer and closer to 7.8. This trend will most likely not continue indefi nitely because the U.S. population increases each year, and the number of males enrolled in high school should increase each year as well.

17. a. C 5 Cost of recorder 1 Service cost p Number of months


5 99.99 1 12.95m

}} m

b. C 5 99.99 1 12.95m

}} m

(95, 14)

Ave

rage

co

st p

er m

on

th (

do

llars

)

0200 40 60 80 100

20

60

80

40

Number of months

m

C

From the graph, you can estimate that after 96 months, the average cost drops to $14 per month.

c. The horizontal asymptote is C 5 12.95

} 1 5 12.95.

As you subscribe for more and more months, the average cost per month approaches $12.95.


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chapter 8bhhs.bhusd.org/ourpages/auto/2006/8/11/1155323019648/aat soluti… · the form y 5 ax. 2....

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