420

Chapter 12

Exercise Set 12.1

1. The two square roots of 36 are ± 36 = ±6 .

3. The square root property is: If x2 = a , where ais a real number, then x = ± a .

5. A trinomial, x2 + bx + c , is a perfect square

trinomial if b2

Ê Ë Á ˆ

¯ ˜

2= c .

7. a. Yes, x = 9 is the solution to the equation. Itis the only real number that satisfies theequation.

b. No, x = 3 is not the solution. Both –3 and 3satisfy the equation.

9. Multiply the equation by 1/2.

11. You should add the square of half the coefficient

of the first degree term:

†

-62( )

2= -3( )2 = 9 .

13. x2 = 49x = ± 49 = ±7

15.

†

x 2 + 49 = 0x 2 = -49

x = ± -49 = ±7i

17.

†

y 2 + 48 = 0y 2 = -48y = ± -48 = ±4i 3

19.

†

y 2 +11= -50y2 = -61y = ± -61 = ±i 61

21. p - 4( )2 = 16p - 4 = ± 16p - 4 = ±4

p = 4 ± 4p = 4 + 4 or P = 4 - 4P = 8 P = 0

23.

†

x + 3( )2 + 49 = 0

x + 3( )2 = -49

x + 3= ± -49x + 3= ± -49

x = -3± 7i

25.

†

a- 4( )2 + 45= 0

a - 4( )2 = -45

a- 4 = ± -45a- 4 = ±3i 5

a = 4± 3i 5

27.

†

b+ 13( )

2= 4

9

b+ 13

= ± 49

b+ 13 = ±23

b = - 13

± 23

†

b = - 13

+ 23

or b = - 13

- 23

b = 13 b = -33

b = -1

29.

†

b- 13( )2+ 49 = 0

b- 13( )2

= - 49

b- 13 = ± -49

b- 13 = ±23 i

b= 13±23 i or b=

1± 2i3

31. x + 1.8( )2 = 0.81x + 1.8 = ± 0.81x + 1.8 = ±0.9

x = -1.8± 0.9x = -1.8 + 0.9 or x = -1.8 - 0.9x = -0.9 x = -2.7

SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

421

33. 2a - 5( )2 = 122a - 5 = ± 122a - 5 = ±2 3

2a = 5 ± 2 3

a =5 ± 2 3

2

35.2y + 1

2Ê Ë Á ˆ

¯ ˜

2=

425

2y +12

= ±425

2y + 12

= ±25

2y + 12

=25

or 2y + 12

= -25

2y = - 12

+25

2y = - 12

-25

2y = -1

102y = -

910

y = - 120

y = - 920

37.

†

x 2 + 3x - 4 = 0x2 + 3x = 4

x 2 + 3x + 94 = 4+94

x + 32( )2

= 254

x + 32 = ±254

x + 32 = ±52

x = - 32 ±52

†

x = - 32

+ 52

or x = - 32

- 52

x = 22 x =-82

x =1 x = -4

39. x2 + 2x - 15 = 0x 2 + 2x = 15

x2 + 2x + 1= 15+ 1x + 1( )2 = 16

x + 1= ±4x = ±4 - 1

x = 4 -1 or x = -4 - 1x = 3 x = -5

41. x2 - 6x + 8 = 0x 2 - 6x = -8

x2 - 6x + 9 = -8 + 9x2 - 6x + 9 = 1

x - 3( )2 = 1x - 3 = ±1

x = ±1 + 3x = 1 + 3 or x = -1+ 3x = 4 x = 2

43.

†

x2 - 6x + 5= 0x 2 - 6x = -5

x 2 - 6x + 9= -5+ 9

x 2 - 6x + 9= 4

x - 3( )2 = 4x - 3= ±2

x = ±2+ 3

†

x = 2+ 3 or x = -2+ 3x = 5 x =1

45.

†

2x2 + x -1= 012 2x

2 + x -1( ) = 12 0( )x 2 + 12 x -

12 = 0

x 2 + 12 x =12

x 2 + 12 x +1

16 =12 +

116

x + 14( )2

= 916

x + 14 = ±9

16x + 14 = ±

34

x = ± 34 -14

†

x = 34

- 14

or x = - 34

- 14

x = 24 x = -44

x = 12 x = -1

Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra

422

47.

†

2z 2 - 7z - 4 = 012 2z

2 - 7z- 4( ) = 12 0( )z 2 - 72 z- 2 = 0

z 2 - 72 z = 2

z 2 - 72 z+4916 = 2+

4916

z- 74( )2

= 8116

z - 74 = ±8116

z - 74 = ±94

z = ± 94 +74

†

z = 94

+ 74

or z = - 94

+ 74

z = 164 z = -24

z = 4 z = - 12

49.

†

x2 -11x + 28= 0x2 -11x = -28

x 2 -11x + 1214 = -28+1214

x - 112( )2

= 94

x - 112 = ±94

x - 112 = ±32

x = ± 32 +112

†

x = 32

+ 112

or x = - 32

+ 112

x = 142 x =82

x = 7 x = 4

51.

†

-x 2 + 3x + 4 = 0 ‹ multiply by -1x 2 - 3x - 4 = 0

x2 - 3x = 4

x 2 - 3x + 94

= 4 + 94

x - 32( )

2= 25

4

x - 32

= ± 254

x - 32 = ±52

x = 32

± 52

†

x = 32

+ 52

or x = 32

- 52

x = 82 x =-22

x = 4 x = -1

53.

†

-z 2 + 9z- 20 = 0 ‹ multiply by -1

z 2 - 9z+ 20 = 0z 2 - 9z = -20

z 2 - 9z+ 814 = -20+814

z - 92( )2

= 14z- 92 = ±

12

z = ± 12 +92

†

z = 12

+ 92

or z = - 12

+ 92

z = 102 z =82

z = 5 z = 4

55. b2 = 3b + 28b2 - 3b = 28

b2 - 3b +94

=112

4+

94

b - 32

Ê Ë Á ˆ

¯ ˜

2=

1214

b - 32

= ±112

b = ±112

+32

b = -112

+32

or b =112

+32

b = -82

b =142

b = -4 b = 7

SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

423

b = -112

+32

or b =112

+32

b = -82

b =142

b = -4 b = 7

57.

†

x 2 + 9x = 10

x 2 + 9x + 814 = 10+814

x + 92( )2

= 404 +814

x + 92( )2

= 1214x + 92 = ±

112

x = ± 112 -92

†

x = 112

- 92

or x = -112

- 92

x = 22 x = -202

x =1 x = -10

59.

†

x 2 - 4x -10 = 0x 2 - 4x =10

x2 - 4x + 4 =10+ 4

x - 2( )2 =14x - 2 = ± 14

x = 2± 14

61.

†

r 2 +8r + 5= 0r 2 + 8r = -5

r 2 + 8r+16 = -5+16

r + 4( )2 =11r + 4 = ± 11

r = -4 ± 11

63.

†

c 2 - c - 3= 0c2 -c = 3

c 2 - c + 14 = 3+14

c - 12( )2

= 134

c - 12 = ±134

c = 12 ±132

c = 1± 132

65. x2 + 3x + 6 = 0x2 + 3x = -6

x2 + 3x + 94

= -244

+94

x + 32

Ê Ë Á ˆ

¯ ˜

2=

-154

x + 32

= ±-154

x + 32

= ±i 15

2

x = - 32

±i 15

2

x =-3 ± i 15

2

67.

†

2x 2 - 2x = 0x2 - x = 0

x 2 - x + 14 =14

x - 12( )2

= 14x - 1

2= ± 1

2x = ± 12 +

12

†

x = 12

+ 12

or x = - 12

+ 12

x = 22 x = 0x =1 x = 0

Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra

424

69.

†

- 14

b2 - 12

b= 0

-4 - 14 b2 - 12b = 0( )

†

b2 + 2b = 0b2 + 2b+1= 0+1

b+1( )2 =1b+1= ±1

†

b+1= 1 or b+1= -1b= 0 b= -2

71.

†

18z 2 - 6z = 0

z 2 - 13 z = 0

z 2 - 13

z+ 136

= 0+ 136

z - 16( )

2= 1

36z- 16 = ±

16

z = ± 16

+ 16

†

z = 16

+ 16

or z = - 16

+ 16

z = 26 z = 0

z = 13 z = 0

73.

†

- 12 p2 - p+ 32 = 0

p2 + 2p- 3= 0p2 + 2 p= 3

p2 + 2p+1= 3+1

p+1( )2 = 4p+1= ±2

p= ±2-1

†

p= 2-1 or p = -2-1p= 1 p = -3

75. 2x2 = 8x + 90x2 = 4x + 45

x2 - 4x = 45x2 - 4x + 4 = 45 + 4

x - 2( )2 = 49x - 2 = ±7

x = ±7 + 2x = 7 + 2 or x = -7 + 2x = 9 or x = -5

x = 7 + 2 or x = -7 + 2x = 9 or x = -5

77. 3x 2 + 33x + 72 = 0x2 + 11x + 24 = 0

x 2 +11x = -24

x2 +11x + 1214

= -24 + 1214

x + 112

Ê Ë Á ˆ

¯ ˜

2= -

964

+1214

x -112

Ê Ë Á ˆ

¯ ˜

2=

254

x +112

= ±52

x = ± 52

-112

x = -52

-112

or x =52

-112

x = -162

or x = -62

x = -8 or x = -3

79. 3w2 + 2w - 1 = 03w 2 + 2w = 1

w2 +23

w =13

w 2 +23

w +19

=13

+19

w + 13

Ê Ë Á ˆ

¯ ˜

2=

49

w + 13

= ±23

w = ± 23

-13

w = - 23

-13

or w = 23

-13

w = -33

w =13

w = -1

SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

425

81. 2x2 - x = -5

x2 -12

x = -52

x2 - 12

x + 116

= -4016

+116

x -14

Ê Ë Á ˆ

¯ ˜

2= -

3916

x - 14

= ±i 39

4

x = 14

±i 39

4

x = 1 ± i 394

83. 52

x 2 +32

x -54

= 0

25

52

x2 + 32

x - 54

= 0È Î Í

˘ ˚ ˙

x2 + 35

x - 12

= 0

x2 +35

x =12

x2 + 35

x + 9100

=12

+9

100

x +3

10Ê Ë Á ˆ

¯ ˜

2=

59100

x + 310

= ±59

10

x = - 310

±59

10

x =-3 ± 59

10

85. a.

†

21= x + 2( ) x - 2( )

b.

†

21= x + 2( ) x - 2( )21= x2 - 2x + 2x - 4

0 = x 2 - 250 = x + 5( ) x - 5( )x + 5 = 0 or x - 5 = 0x = -5 x = 5

Disregard the negative answer since xrepresents a distance. x = 5.

87. a.

†

18= x + 4( ) x + 2( )

b.

†

18 = x + 4( ) x + 2( )18 = x 2 + 2x + 4x + 80 = x 2 + 6x - 10Using the quadratic formula:

x =- 6( ) ± 62 - 4 1( ) -10( )

2 1( )

x = -6± 762

x = -6± 2 192

x = -3± 19Disregard the negative answer since xrepresents a distance. x =

†

-3+ 19 .

89.

†

d = 16 x2

24 = 16 x2

6 ⋅24 = x 2

144 = x2

x =12

The car’s speed was about 12 mph.

91. Let x be the first integer. Then x + 2 is the nextconsecutive odd integer.

x x + 2( ) = 63x2 + 2x = 63

x2 + 2x + 1 = 63 + 1x + 1( )2 = 64

x + 1 = ±8x = -1 ± 8

x = -1+ 8 or x = -1 - 8x = 7 x = -9Since it was given that the integers are positive,one integer is 7 and the other is 7 + 2 = 9.

Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra

426

93. Let x be the width of the rectangle. Then 2x + 2is the length. Use length · width = area.

(2x + 2)x = 602x2 + 2x = 60

x2 + x = 30

x2 + x +14

= 30 +14

x +12

Ê Ë Á ˆ

¯ ˜

2=

1204

+14

x + 12

Ê Ë Á ˆ

¯ ˜

2=

1214

x +12

= ±112

x = - 12

±112

x = -12

+112

or x = -12

-112

x =102

= 5 x = -122

= 6

Since the width cannot be negative,the width is 5 ft.Length = 2 5( ) + 2 = 10 + 2 = 12 ft.The rectangle is 5 ft by 12 ft.

95. Let s be the length of the side. Then s + 6 is thelength of the diagonal (d). Use s2 + s2 = d2 .

2s2 = s + 6( )2

2s2 = s2 + 12s + 36s2 = 12s + 36

s2 - 12s = 36s2 - 12s + 36 = 36 + 36

s - 6( )2 = 72s - 6 = ±6 2

s = 6 ± 6 2Length is never negative. Thus,s = 6 + 6 2 ª 14.49 .The room is about 14.49 ft by 14.49 ft.

97. Since the radius is 10 inches, the diameter (d) is20 inches. Use the formula s2 + s2 = d2 to findthe length (s) of the other two sides.s2 + s2 = d2

2s2 = 202

2s2 = 400s2 = 200s = ± 200 = ±10 2

Length is never negative.Thus s = 10 2 ª 14.14 inches.

Thus s = 10 2 ª 14.14 inches.

99. A = pr224p = pr2

24 = r2

± 24 = r±2 6 = rLength is never negative.Thus r = 2 6 ª 4.90 feet .

101.A = P 1 +

rn

Ê Ë Á ˆ

¯ ˜

nt

†

540.80 = 500 1+ r1( )1 2( )

540.80 = 500 1+ r( )2

1.0816 = 1+ r( )2

±1.04 = 1+ r-1±1.04 = rAn interest rate is never negative. Thus

†

r = -1+1.04 = 0.04 = 4% .

103.A = P 1 +

rn

Ê Ë Á ˆ

¯ ˜

nt

1432.86 = 1200 1+ r2

Ê Ë Á ˆ

¯ ˜

2 3( )

1432.86 = 1200 1+r2

Ê Ë Á ˆ

¯ ˜

6

1.19405 = 1 + r2

Ê Ë Á ˆ

¯ ˜

6

±1.03 ª 1+r2

-1± 1.03 ª r2

-2 ± 2.06 ª rAn interest rate is never negative.Thus Steve Rodi’s annual interest rate is about–2 + 2.06 = 0.06 = 6%.

105. a. To find the surface area, we must firstdetermine the radius. Use V = pr2h withV = 160 and h = 10 to get160 = pr2 10( )16 = pr216p

= r2

4p

= r

Since the radius equals 4p

, use the formula

S = 2pr2 + 2prh to calculate the surface

area.

S = 2p4p

Ê

Ë Á

ˆ

¯ ˜

2+ 2p

4p

Ê

Ë Á

ˆ

¯ ˜ (10)

= 2p16p

Ê Ë Á ˆ

¯ ˜ +

80pp

= 32 + 80 pª 173.80

SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

427

160 = pr2 10( )16 = pr216p

= r2

4p

= r

Since the radius equals 4p

, use the formula

S = 2pr2 + 2prh to calculate the surface

area.

S = 2p4p

Ê

Ë Á

ˆ

¯ ˜

2+ 2p

4p

Ê

Ë Á

ˆ

¯ ˜ (10)

= 2p16p

Ê Ë Á ˆ

¯ ˜ +

80pp

= 32 + 80 pª 173.80

The surface area is about173.80 square inches.

b. Use V = pr2h with V = 160 and h = 10 toobtain 160 = pr2 10( ) . In part (a) this wassolved for r to get

r =4p

=4p

⋅pp

=4 p

pª 2.26

The radius is about 2.26 inches.

c. Use S = 2pr2 + 2prh with S = 160 and h =10.

160 = 2pr2 + 2pr 10( )160 = 2pr2 + 20pr1602p

=2pr2

2p+

20pr2p

80p

= r2 +10 r80p

+ 25 = r2 +10r + 2580 + 25p

p= r + 5( )2

±80 + 25p

p= r + 5

±80 + 25p

p- 5 = r

The radius is never negative.Thus r ≈ 2.1 inches.

107.

†

-4 2z- 6( ) = -3 z- 4( ) + z-8z+ 24 = -3z +12+ z-8z+ 24 = -2z+12

-6z = -12z = 2

108. Let x = the amount invested at 7%. Then theamount invested at

†

6 14 % will be 10,000 – x. Theinterest earned at 7% will be 0.07x. and theamount of interest earned at 6.25% will be.0625(10,000 – x). The total interest earned is$656.50.

†

0.07x + 0.0625 10,000 - x( ) = 656.500.07x + 625- 0.0625x = 656.500.0075x = 31.5x = 4200

$4200 was invested at 7% and $10,000"–"$4200= $5800 was invested at

†

6 14 %

109.

†

m = y2 - y1x2 - x1m = 4 - 4

-1- -3( )= 0

-1+ 3 =02 = 0

110.

†

4x 2 + 9x - 2x - 2

-8x2 -18x + 44x3 + 9x2 - 2x

4x 3 + x2 - 20x + 4

111.

†

x + 3 = 2x - 7x + 3= 2x - 7 or x + 3= - 2x - 7( )

- x = -10 x + 3= -2x + 7x = 10 3x = 4

x = 43