Chapter 12: Quadratic Functions SSM: Elementary...

Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra

428

Exercise Set 12.2

1. The quadratic formula is x =-b ± b2 - 4ac

2awhich gives the values of x!whereax 2 + bx + c = 0 .

3. a = –3, b = 6, and c = 5

5. Yes, multiply both sides of the equation-6x2 +

12

x - 5 = 0 by –1 to obtain

6x2 -12

x + 5 = 0 . The equations are

equivalent so they will have the samesolutions.

7. a. For a quadratic equation in standard form,ax 2 + bx + c = 0 , the discriminant is theexpression under the square root symbol inthe quadratic formula, b2 - 4ac .

b. 3x 2 - 6x + 20 = 0 , a = 3, b = –6, and c = 20.b2 - 4ac = -6( )2 - 4 3( ) 20( )

= 36 - 240= -204

c. If b2 - 4ac > 0 , then the quadratic equationwill have two distinct real solutions. Sincethere is a positive number under the radicalsign in the quadratic formula, the value of theradical will be real and there will be two realsolutions. If b2 - 4ac = 0 , then the equationhas the single real solution

-b2a

. If

b2 - 4ac < 0 , the expression under the radicalsign in the quadratic formula is negative. Thusthe equation has no real solution.

9.

†

x 2 + 3x + 2 = 0

†

b2 - 4ac = 3( )2- 4 1( ) 2( )

= 9-8= 1

Since 1 > 0, there are two real solutions.

11.

†

3z 2 + 4z+ 5= 0

†

b2 - 4ac = 42 - 4 3( ) 5( )= 16- 60= -44

Since –44 < 0, there is no real solution.

13. 5p2 + 3p - 7 = 0b2 - 4ac = 32 - 4 5( ) -7( )

= 9 + 140= 149

Since 149 > 0 there are two real solutions.

15. -5x 2 + 5x - 6 = 0b2 - 4ac = 52 - 4 -5( ) -6( )

= 25 - 120= -95

Since –95 < 0, there is no real solution.

17. x2 + 10.2x + 26.01 = 0b2 - 4ac = 10.2( )2 - 4 1( ) 26.01( )

= 104.04 - 104.04= 0

There is one real solution.

19. b2 = -3b -94

b2 + 3b +94

= 0

b2 - 4ac = 32 - 4 1( )94

Ê Ë Á ˆ

¯ ˜

= 9 - 9= 0

There is one real solution.

21.

†

x 2 - 9x +18= 0

†

x =9± 92 - 4 1( ) 18( )

2 1( )

= 9± 81- 722

= 9± 92

= 9± 32

†

x = 9+ 32

or x = 9- 32

= 122 = 6

2= 6 = 3

The solutions are 6 and 3.

SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

429

23.

†

a2 - 6a +8 = 0

†

a = -b± b2 - 4ac2a

=6± -6( )2

- 4 1( ) 8( )2 1( )

= 6± 36- 322

= 6± 42

= 6± 22

†

a = 6- 22

or a = 6+ 22

= 42 = 8

2= 2 = 4


25.

†

x 2 = -2x + 3

x 2 + 2x - 3= 0

x =-2 ± 22 - 4 1( ) -3( )

2 1( )

=-2 ± 4 + 12

2

=-2 ± 16

2=

-2 ± 42

x =-2 + 4

2or x =

-2 - 42

=22

=-62

= 1 = -3The solutions are 1 and –3.

27.

†

-b2 = 4b- 20

b2 + 4b- 20 = 0

†

b =-4± (4)2 - 4(1)(-20)

2(1)

= -4± 16+ 802

= -4± 962

= -4± 4 62

= -2± 6The solutions are

†

-2+ 2 6 and

†

-2- 2 6 .

29.

†

b2 - 49 = 0

†

b =0± 02 - 4 1( ) -49( )

2 1( )

= ± 1962

= ±142

†

b = 142 or b = -14

2= 7 = -7

The solutions are 7 and –7.

31. 3w 2 - 4w + 5 = 0

w =-(-4) ± (-4)2 - 4(3)(5)

2(3)

=4 ± 16 - 60

6

=4 ± -44

6

=4 ± 2i 11

6

=2 2 ± i 11( )

6

=2 ± i 11

3

The solutions are 2 - i 113 and 2 + i 11

3 .

33.

†

c 2 - 3c = 0

†

c =3± -3( )2

- 4 1( ) 0( )2 1( )

= 3± 92

= 3± 32

†

c = 3+ 32

or c = 3- 32

= 62 = 0

2= 3 = 0



430

35. 4s2 - 8s + 6 = 012 (4s2 - 8s + 6 = 0)

2s2 - 4s + 3 = 0

s =-(-4) ± (-4)2 - 4(2)(3)

2(2)

=4 ± 16 - 24

4

=4 ± -8

4

=4 ± 2i 2

4

=2 ± i 2

2

The solutions are 2 - i 22 and 2 + i 2

2 .

37.

†

a2 + 2a +1= 0

†

a =- 2( ) ± 2( )2

- 4 1( ) 1( )2 1( )

= -2± 4- 42

= -2± 02

= -2± 02

= -22

= -1

The solution is –1.

39.

†

x 2 -10x + 25= 0

†

x =- -10( ) ± -10( )2

- 4 1( ) 25( )2 1( )

= 10± 100-1002

= 10± 02

= 10± 02

= 102

= 5

The solution is 5.

41. x2 - 2x - 1= 0

x =- -2( ) ± -2( )2 - 4 1( ) -1( )

2 1( )

=2 ± 4 + 4

2

=2 ± 8

2

=2 ± 2 2

2= 1 ± 2

The solutions are 1 - 2 and 1 + 2 .

43.

†

2- 3r 2 = -4r

3r 2 - 4r- 2 = 0

†

r =-(-4)± (-4)2 - 4(3)(-2)

2(3)

= 4 ± 16+ 246

= 4 ± 406

= 4 ± 2 106

= 2± 103

The solutions are

†

2+ 103 and 2- 10

3 .

45.

†

2x 2 + 5x - 3= 0

†

x =-(5)± (5)2 - 4(2)(-3)

2(2)

= -5± 25+ 244

= -5± 494

= -5±74

†

x = -5+74 or x = -5- 7

4= 2

4= 1

2

= -124

= -3

The solutions are

†

12 and –3.


431

47. (2a + 3)(3a – 1) = 26a2 + 7a - 3 = 26a2 + 7a - 5 = 0

a =-(7) ± (7)2 - 4(6)(-5)

2(6)

=-7 ± 49 + 120

12

=-7 ± 169

12=

-7 ±1312

a =-7 -13

12or a =

-7 + 1312

=-2012

=6

12= -

53

=12

The solutions are

†

12 and -

53

.

49.

†

12 t 2 + t -12= 0

2 12 t 2 + t-12 = 0( )

t 2 + 2t - 24 = 0

†

t =-(2) ± (2)2 - 4(1)(-24)

2(1)

= -2± 4+ 962

= -2± 1002

= -2±102

†

t = -2+102

or t = -2-102

= 82 = -12

2= 4 = -6

The solutions are 4 and –6.

51.

†

9r 2 - 9r+ 2 = 0

†

r =-(-9) ± (-9)2 - 4(9)(2)

2(9)

= 9± 81- 7218

= 9± 918

= 9± 318

†

r = 9+ 318

or r = 9- 318

= 1218 = 6

18= 2

3 = 13

†

r =-(-9) ± (-9)2 - 4(9)(2)

2(9)

= 9± 81- 7218

= 9± 918

= 9± 318

†

r = 9+ 318

or r = 9- 318

= 1218 = 6

18= 2

3 = 13

The solutions are

†

23 and 1

3 ..

53. 12

x 2 + 2x +23

= 0

6 12

x2 + 2x +23

= 0Ê Ë Á ˆ

¯ ˜

3x 2 + 12x + 4 = 0

x =-12 ± (12)2 - 4(3)(4)

2(3)

=-12 ± 144 - 48

6

=-12 ± 96

6

=-12 ± 4 6

6

=2(-6 ± 2 6)

2(3)

=-6 ± 2 6

3

The solutions are -6 + 2 63 and -6 - 2 6

3 .

55.

†

a2 - a5 - 1

3 = 0

†

15 a2 - a5 - 1

3 = 0( )15a2 - 3a- 5= 0

†

a =-(-3)± (-3)2 - 4(15)(-5)

2(15)

= 3± 9+ 30030

= 3± 30930

The solutions are 3 - 30930 and 3+ 309

30 .


432

57.

†

c = 6-cc - 4

c c - 4( ) = -c+ 6

c 2 - 4c = -c+ 6c 2 - 3c - 6 = 0

†

c =-(-3)± (-3)2 - 4(1)(-6)

2(1)

= 3± 9+ 242

= 3± 332

The solutions are 3 + 332 and 3 - 33

2 .

59.

†

2x 2 - 4 x + 3= 0

†

x =- -4( )± -4( )2

- 4 2( ) 3( )2 2( )

= 4 ± 16- 244

= 4 ± -84

= 4 ± 2i 24

= 2± i 22

The solutions are

†

2+i 22 and 2-i 2

2 .

61.

†

2y2 + y = -32y 2 + y + 3= 0

†

y =-1± (1)2 - 4(2)(3)

2(2)

= -1± 1- 244

= -1± -234

= -1± i 234

The solutions are -1+ i 234 and -1- i 23

4 .

63. 0.1x 2 + 0.6x - 1.2 = 010(0.1x 2 + 0.6x -1.2 = 0)

x2 + 6x - 12 = 0

x =-6 ± 62 - 4(1)(-12)

2(1)

=-6 ± 36 + 48

2

=-6 ± 84

2

=-6 ± 2 21

2= -3 ± 21

The solutions are -3 + 21 or - 3 - 21 .

65.

†

f x( ) = x2 - 2x + 4 ,

†

f x( ) = 4

†

x 2 - 2x + 4 = 4

x2 - 2x = 0

†

x =2± -2( )2

- 4 1( ) 0( )2 1( )

= 2± 42

= 2± 22

†

x = 2+ 22

or x = 2- 22

= 42 = 0

2= 2 = 0

The values of x are 2 and 0.

67.

†

k x( ) = x 2 - x -10, k x( ) = 20

x 2 - x -10= 20x 2 - x - 30= 0

†

x =1± -1( )2

- 4 1( ) -30( )2 1( )

= 1± 1+1202

= 1± 1212

= 1±112

x = 1+112 or x = 1-11

2= 12

2 = -102

= 6 = -5The values of x are 6 and –5.


433

The values of x are 6 and –5.

69. h t( ) = 2t2 - 7t + 1 , h t( ) = -32t 2 - 7t + 1= -32t 2 - 7t + 4 = 0

t =7 ± -7( )2 - 4 2( ) 4( )

2 2( )

=7 ± 49 - 32

4

=7 ± 17

4

The values of t are 7 + 17

4 and

7 - 174

.

71. g a( ) = 2a2 - 3a + 16 , g a( ) = 142a2 - 3a + 16 = 142a2 - 3a + 2 = 0

a =3 ± -3( )2 - 4 2( ) 2( )

2 2( )

=3 ± 9 - 16

4

=3 ± -7

4There are no real values of a for which g(a) =14.

73. If 2 and 5 are solutions, the factors must be

†

x - 2( ) and x - 5( ) .

†

f x( ) = x - 2( ) x - 5( )f (x) = x2 - 5x - 2x +10f x( ) = x2 - 7x +10

75. If 4 and –6 are solutions, the factors must bex - 4( ) and

†

x + 6( ) .

†

f x( ) = x - 4( ) x + 6( )f x( ) = x2 + 6x - 4x - 24

f x( ) = x2 + 2x - 24

77. If -35

and 23

are solutions, the factors must

be 5x + 3( ) and 3x - 2( ) .

†

f (x) = (5x + 3)(3x - 2)

f (x) = 15x 2 - 10x + 9x - 6f (x) = 15x 2 - x - 6

79. If 3 and - 3 are solutions, the factors mustbe x - 3( ) and x + 3( ) .

f x( ) = x - 3( ) x + 3( )f x( ) = x2 - 3

81. 2i and –2i are solutions, the factors must bex - 2i( ) and x + 2i( ).f x( ) = x - 2i( ) x + 2i( )f x( ) = x 2 - 4i 2

f x( ) = x 2 + 4

83. If 3 + 2 and 3 - 2 are solutions, thefactors must be x - 3+ 2( )( ) and

x - 3- 2( )( ) .

f x( ) = x - 3 + 2( )( ) x - 3- 2( )( )f x( ) = x - 3 - 2( ) x - 3+ 2( )f x( ) = x - 3( )2 - 2( )2

f x( ) = x2 - 6x + 9 - 2f x( ) = x2 - 6x + 7

85. If 2 + 3i and 2 - 3i are solutions, the factorsmust be x - 2 + 3i( )( ) and x - 2 - 3i( )( ) .f x( ) = x - 2 + 3i( )( ) x - 2 - 3i( )( )f x( ) = x - 2 - 3i( ) x - 2 + 3i( )f x( ) = x - 2( )2 - 9i2

f x( ) = x2 - 4x + 4 + 9f x( ) = x2 - 4x +13


434

87. a.

†

n 10- 0.02n( ) = 450

b.

†

n 10- 0.02n( ) = 450

10n- 0.02n 2 = 4500.02n2 -10n + 450= 0

n =10± -10( )2 - 4 0.02( ) 450( )

2 0.02( )

n = 10± 100- 360.04

n = 10± 640.04

n = 10± 80.04 fi n = 450 or n = 50

Since n ≤ 65, the number of lamps thatmust be sold is 50.

89. a.

†

n 50- 0.2n( ) =1680

b.

†

n 50- 0.2n( ) =1680

50n- 0.2n 2 = 16800.2n2 - 50n +1680= 0

n =50± -50( )2 - 4 0.2( ) 1680( )

2 0.2( )

n = 50± 2500-13440.4

n = 50± 11560.4

n = 50± 340.4 fi n = 210 or n = 40

Since n ≤ 50, the number of chairs thatmust be sold is 40.

91. Any quadratic equation for which thediscriminant is a non-negative perfect squarecan be solved by factoring. Any quadraticequation for which the discriminant is apositive number but not a perfect square canbe solved by the quadratic formula but not byfactoring over the set of integers.

93. Yes. If the discriminant is a perfect square, thesimplified expression will not contain aradical and the quadratic equation can besolved by factoring.

95. Let x be the number.2x2 + 3x = 14

2x2 + 3x - 14 = 0

†

x =-3± 32 - 4 2( ) -14( )

2 2( )

=-3± 9 + 112

4

=-3± 121

4

=-3± 11

4

x =-3+ 11

4 since x is positive

x =84

x = 2

97. Let x be the width. Then 3x – 2 is the length.Use A = (length)(width).

†

21 = (3x - 2)( x)21 = 3x2 - 2x

3x2 - 2x - 21 = 0

x =-(-2) ± (-2)2 - 4(3)(-21)

2(3)

=2 ± 4 + 252

6

=2 ± 256

6=

2 ± 166

Since width is positive, usex =

2 + 166

=186

= 3

3x!–!2 = 3(3) –!2 = 9 – 2 = 7The width is 3 feet and the length is 7 feet.

99. Let x be the amount by which each side is tobe reduced.Then 6 – x is the new widthand 8 – x is the new lengthnew area =

12

(old area)

=12

6 ⋅8( )

=12

48( )

= 24new area = (new width)(new length)24 = (6 – x)(8 – x)0 = 48 - 14x + x2 - 240 = x2 - 14x + 24

x =-(-14) ± (-14)2 - 4(1)(24)

2(1)

=14 ± 196 - 96

2

=14 ± 100

2=

14 ±102


435

=12

6 ⋅8( )

=12

48( )

= 24new area = (new width)(new length)24 = (6 – x)(8 – x)0 = 48 - 14x + x2 - 240 = x2 - 14x + 24

x =-(-14) ± (-14)2 - 4(1)(24)

2(1)

=14 ± 196 - 96

2

=14 ± 100

2=

14 ±102

x =14 +10

2or x =

14 - 102

=242

=42

= 12 = 2We reject x = 12, since this would givenegative values for width and length.The only meaningful value is x = 2 inchessince this gives positive values for the newwidth and length.

101. a. h = 12 at2 + v0t + h0

20 = 12 (-32)t2 + 60t + 80

20 = -16t2 + 60 t + 800 = -16t2 + 60 t + 600 = 16t2 - 60t - 60

t =-(-60) ± (-60)2 - 4(16)(-60)

2(16)

t = 60 ± 744032

Since time must be positive, use

t =60 + 7440

32ª 4.57

The horseshoe is 20 feet from the ground afterabout 4.57 seconds.

b. 0 = 12 (-32)t2 + 60t + 80

0 = -16t2 + 60 t + 80

t =-60 ± 602 - 4(-16)(80)

2(-16)

t = -60 ± 8720-32

t =60 ± 8720

32Since time must be positive, use

t =60 + 8720

32ª 4.79

The horseshoes strike the ground afterabout 4.79 seconds.

103. x2 - 5x - 10 = 0, a = 1, b = - 5, c = -10

x =- - 5( ) ± - 5( )2

- 4(1)(-10)

2(1)

=5 ± 5 + 40

2

=5 ± 45

2

=5 ± 3 5

2

x =5 + 3 5

2or x =

5 - 3 52

=4 5

2=

-2 52

= 2 5 = - 5The solutions are 2 5 and - 5 .

105. Let s be the length of the side of the originalcube. Then s + 0.2 is the length of the side ofthe expanded cube. (s + 0.2)3 = s3 + 6s3 + 0.6s2 + 0.12s + 0.008 = s3 + 6

0.6s2 + 0.12s + 0.008 = 60.6s2 + 0.12s - 5.992 = 0

s =-0.12 ± (0.12)2 - 4(0.6)(-5.992)

2(0.6)

=-0.12 ± 0.0144 + 14.3803

1.2

=-0.12 ± 14.3952

1.2

Use the positive value since a length cannot benegative.

s =-0.12 + 14.3952

1.2ª

-0.12 + 3.79411.2

ª 3.0618The original side was about 3.0618millimeters long.


436

107. a. h = 12 at2 + v0t + h0

†

0 =12

(-32)t 2 + 0 t + 60

0 = -16t 2 + 0 t + 60

t =- 0( ) ± 02 - 4(-16)(60)

2(-16)

t =0 ± 3840

-32t ª -1.94 or t ª 1.94Since time must be positive, use 1.94 sec.

b. h = 12 at2 + v0t + h0

†

0 =12

(-32)t 2 + 0 t +120

0 = -16t 2 + 0 t +120

t =- 0( ) ± 02 - 4(-16)(120)

2(-16)

t =0 ± 7680

-32t ª -2.74 or t ª 2.74Since time must be positive, use 2.74 sec.

c. Courtney’s rock will strike the groundfirst.

d. The height of Travis’ rock is given by

†

h t( ) = -16t 2 +100t + 60 . The height ofCourtney’s rock is given by

†

h t( ) = -16t 2 + 60t +120 . We want toknow when these will be equal.

†

-16t 2 +100 t+ 60 = -16t2 + 60t +120100t+ 60 = 60 t+12040t = 60

t = 6040 fi t = 1.5

The rocks will be the same distance abovethe ground after 1.5 seconds.

108.

†

3.33¥103

1.11¥101 = 3.331.11 ¥ 103

101 = 3¥102 or 300

109.

†

f x( ) = x2 + 2x - 5

f 3( ) = (3)2 + 2(3) - 5= 9+ 6- 5=10

110.

†

3x + 4y = 22x = -5y -1Rewrite the system in standard form.

†

3x + 4y = 22x + 5y = -1To eliminate the x variable, multiply the firstequation by 2 and the second equation by –3,and then add.

†

6x + 8y = 4-6x -15y = 3

- 7y = 7 fi y = -1Substitute –1 for y in the first equation tofind x.

†

3x + 4 -1( ) = 23x - 4 = 23x = 6 fi x = 2The solution is (2, –1).

111.

†

x + yx - y

⋅x + yx + y

=x 2 + x y + x y + yx2 + x y - x y - y

=x 2 + 2x y + y

x 2 - y

112.

†

x2 + 6x - 4 = x

x2 + 6x - 4 = x2

6x - 4 = 06x = 4

x =46

or 23

Exercise Set 12.3

1. Answers will vary.

3. A = s2 , for s.A = s


437

5.

†

A = S2 - s 2, for S

A + s2 = S2

A+ s 2 = S fi S = A + s2

Chapter 12: Quadratic Functions SSM: Elementary...

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