Elementary and Intermediate Algebra Quadratic …wps.prenhall.com/wps/media/objects/794/813466/ssm/ch12_3.pdfSSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions

437

7. E = i2r , for iEr

= i 2

Er

= i

9.

†

d = 16t 2, for td16 = t 2

d16 = t fi t = d

4

11. E = mc2 , for cEm

= c2

Em

= c

13. V =13

pr2h , for r

3V = pr2h3Vph

= r2

3Vph

= r

15.

†

d = L2 +W 2 , for Wd2 = L2 +W 2

d2 - L2 =W 2

d 2 - L2 =W

17. a2 + b2 = c2 , for bb2 = c2 - a2

b = c2 - a2

19.

†

d = L2 +W 2 + H 2 , for Hd2 = L2 +W 2 + H 2

d2 - L2 -W 2 = H 2

d 2 - L2 -W 2 = H

21.

†

h = -16t2 + s0 , for t16t 2 = s0 - h

t 2 =s0 - h

16

t =s0 - h

16

t =s0 - h4

23. E =12

mv 2 , for v

2E = mv2

2Em

= v2

2Em

= v

25.a =

v22 - v1

2

2d, for v1

2ad = v22 - v1

2

2ad + v12 = v2

2

v12 = v2

2 - 2ad

v1 = v22 - 2ad

27. ¢ v = c2 - v2 , for c¢ v ( )2 = c2 - v2

¢ v ( )2 + v2 = c2

c = ¢ v ( )2+ v2

29. a.

†

P n( ) = 2.4 n2 + 9n - 3

P 6( ) = 2.4 6( )2+ 9 6( )- 3

=137.4 fi $13,740The profit would be $13,740.

b.

†

P n( ) = 2.4n2 + 9n- 3

200= 2.4n2 + 9n- 32.4n2 + 9n- 203= 0

x =-9± 92 - 4 2.4( ) -203( )

2 2.4( )

x = -9± 2029.84.8

x ª 8 or x ª -11Eight tractors must be sold.

Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra

438

Eight tractors must be sold.

31. T = 6.2t2 +12t + 32

a. When the car is turned on, t = 0.T = 6.2 0( )2 +12 0( ) + 32 = 32The temperature is 32°F.

b. T = 6.2 1( )2 + 12 1( ) + 32 = 50.2The temperature after 1 minute is 50.2°F.

c. 120 = 6.2t2 +12t + 320 = 6.2t2 +12t - 88

t =-12 ± 122 - 4 6.2( ) -88( )

2 6.2( )

t ª-12 ± 48.23

12.4t ª 2.92 or t ª -4.86The radiator temperature will reach 120°Fabout 2.92 min. after the engine is started.

33. a. C = -0.011500( )2 + 80 1500( ) + 20,000= 117, 500

The cost is about $117,500.

b. 150, 000 = -0.01s2 + 80s + 20, 0000 = -0.01s2 + 80s - 130, 000

s =-80 ± 802 - 4 -0.01( ) -130,000( )

2 -0.01( )

s ª-80 ± 34.64

-0.02s ª 2268 or s ª 5732Notice that s must fall between 1200 and4000. Therefore, Mr. Boyle can purchasea 2268 sq ft house for $150,000.

35. a. 0 = -3. 3t2 - 2.3t + 62

t =-(-2.3)± (-2.3)2 - 4(-3. 3)(62)

2(-3.3)t = 2.3 ± 28. 7

-6.6t ≈ –4.7 or t = 4Since time must be positive, the car takes4 seconds for the drop.

b. s"= 6.74(4) + 2.3 = 29.26The speed is 29.26 feet per second.

37. a. m t( ) = 0.05t 2 - 0.32t + 3.15In 2003, t"= 21.

†

m 21( ) = 0.05 212( ) - 0.32 21( ) + 3.15

= 22.05- 6.72 + 3.15ª 18.5

Veterinary bills for dogs in 2003 amountedto about $18.5 billion.

b. m t( ) = 0.05t 2 - 0.32t + 3.15 , m t( ) = 12

†

25= 0.05t 2 - 0.32 t+ 3.15

0 = 0.05t 2 - 0.32 t- 21.85

†

t =0.32 ± 0.32( )2

- 4 0.05( ) -21.85( )2 0.05( )

= 0.32 ± 0.1024 + 4.370.1

= 0.32± 4.47240.1

ª 0.32 ± 2.11480500.1

†

t ª 0.32+ 2.11480500.1 = 2.4348050

0.1 ª 24.3or

†

t ª 0.32- 2.11480500.1 = -1.794805

0.1 ª -17.9Since

†

1£ t £ 28 , the only solution is t ≈24.3. Thus $25 billion will be spent onveterinary bills for dogs approximately24.3 years after 1982 or in 2006.

39. Let x be the width of the playground. Thenthe length is given by x + 5.

†

Area = length ¥ width500= x x + 5( )x 2 + 5x - 500 = 0x - 20( ) x + 25( ) = 0

x = 20 or x = -25Disregard the negative value. The width ofthe playground is 20 meters and the length is20 + 5 = 25 meters.


439

41. Let r be the rate at which the present equipmentdrills.

d r t =dr

presentequipment 64 r 64

rnewequipment 64 r + 1 64

r +1They would have hit water in 3.2 hours less timewith the new equipment.

64r +1

=64r

- 3.2

r(r + 1) 64r + 1

Ê Ë Á ˆ

¯ ˜ = r(r + 1) 64

rÊ Ë Á ˆ

¯ ˜ - r(r + 1)(3.2)

64r = 64(r + 1) - 3.2r( r + 1)64r = 64r + 64 - 3.2r2 - 3.2r

0 = 64 - 3.2r2 - 3.2r3.2r2 + 3.2r - 64 = 0

r2 + r - 20 = 0r + 5( ) r - 4( ) = 0

r + 5 = 0 or r - 4 = 0r = -5 r = 4

Use the positive value. The present equipmentdrills at a rate of 4 ft/hr.

43. Let x"be Latoya’s rate going uphill so x + 2 isher rating going downhill. Using

dr

= t givestuphill + tdownhill = 1.75

6x

+6

x + 2= 1.75

x(x + 2) 6x

Ê Ë Á ˆ

¯ ˜ + x(x + 2) 6

x + 2Ê Ë Á ˆ

¯ ˜ = x(x + 2)(1.75)

6(x + 2) + 6x = 1.75x(x + 2)6x + 12 + 6x = 1.75x 2 + 3.5x

0 = 1.75x 2 - 8.5x -12

x =- -8.5( ) ± -8.5( )2 - 4 1.75( ) -12( )

2 1.75( )

=

8 ± 156.253.5

=8.5 ±12.5

3.5x = 6 or x ª -1.14Since the time must be positive, Latoya’s uphillrate is 6 mph and her downhill rate isx + 2 = 8 mph.

45. Let x be the time of the experienced mechanicthen x"+ 1 is the time of the inexperienced

mechanic.6x

+6

x +1= 1

x(x + 1) 6x

Ê Ë Á ˆ

¯ ˜ + x(x + 1) 6

x + 1Ê Ë Á ˆ

¯ ˜ = x(x + 1)(1)

6(x +1) + 6x = x2 + x6x + 6 + 6x = x2 + x

0 = x2 - 11x - 6

x =-(-11) ± (-11)2 - 4(1)(-6)

2(1)

=11 ± 121 + 24

2

=11 ± 145

2

x =11 + 145

2or x =

11- 1452

ª 11.52 ª -0.52Since the time must be positive, it takes Bonitaabout 11.52 hours and Pamela about 12.52 hoursto rebuild the engine.

47. Let r be the speed of the plane in still air.

d r t =dr

With wind 80 r"+ 30 80r + 30

Against wind 80 r – 30 80r - 30

The total time is 1.3 hours

†

80r+ 30

+ 80r - 30

= 1.3

r+ 30( ) r - 30( ) 80r + 30 + 80

r- 30 =1.3( )80 r - 3( ) + 80 r + 30( ) = 1.3 r2 - 900( )

80r - 240 + 80r + 240 = 1.3r2 -1170160r = 1.3r2 -1170

0 = 1.3r2 -160 r - 1170

r =- -160( ) ± -160( )2 - 4 1.3( ) -1170( )

2 1.3( )

=160 ± 25, 600 + 6084

2.6

=160 ± 31,684

2.6=

160 ±1782.6


440

r =160 +178

2.6or r =

160 - 1782.6

=3382.6

=-182.6

= 130 ª -6.92Since speed must be positive, the speed of theplane in still air is 130 mph.

49. Let t be the number of hours for Chris to cleanalone. Then t + 0.5 is the number of hours forJohn to clean alone.

Rate ofwork

Timeworked

Part of Taskcompleted

Chris1t

66t

John1

t + 0.56

6t + 0.5

6t

+6

t + 0.5= 1

t(t + 0.5) 6t

Ê Ë Á ˆ

¯ ˜ + t(t + 0.5) 6

t + 0.5Ê Ë Á ˆ

¯ ˜ = t(t + 0.5)(1)

6(t + 0.5) + 6t = t(t + 0.5)6t + 3 + 6t = t 2 + 0.5t

12t + 3 = t 2 + 0.5t0 = t 2 -11.5t - 3

t =11.5 ± -11.5( )2 - 4 1( ) -3( )

2 1( )

=11.5 ± 132.25 + 12

2

=11.5 ± 144.25

2

t =11.5 + 144.25

2or t =

11.5 - 144.252

ª 11.76 ª -0.26Since the time must be positive, it takesChris about 11.76 hours and John about11.76 + 0.5 = 12.26 hours to clean alone.

51. Let x be the speed of the trip from Lubbock toPlainview. Then x"– 10 is the speed fromPlainview to Amarillo.

d r t

first part 60 x 60x

second part 100 x – 10 100x -10

Including the 2.5 hours she spent in Plainview,the entire trip took Lisa 5.5 hours.

†

60x + 2.5+ 100

x -10 = 5.5

60x + 100

x -10 = 3

x(x -10) 60x + 100

x -10 = 3( )60(x -10) +100x = 3(x 2 -10x)60x - 600+100x = 3x 2 - 30x

-600+160x = 3x 2 - 30x0= 3x 2 -190x + 6000 = (x - 60)(3x -10)

x - 60 = 0 or 3x - 10 = 0x = 60 x =

103

103

miles per hour is too slow for a car, so the

speed of the trip from Lubbock to Plainview was60 mph.

53. Answers will vary.

55. Let l = original length and w = originalwidth. A system of equations that describesthis situation is

†

l ⋅ w = 18l+ 2( ) w + 3( ) = 48

If you solve for l in the first equation youget

†

l = 18w . Substitute

†

18w into the l in the

second equation. The result is an equationin only one variable which can be solved.

†

l+ 2( ) w + 3( ) = 4818w + 2( ) w + 3( ) = 48

18+ 54w

+ 2w + 6 = 48

2w - 24+ 54w = 0

w 2w - 24 + 54w

= 0( )2w2 - 24w + 54 = 02 w - 3( ) w - 9( ) = 0w = 3 or w = 9

If w = 3, then

†

l = 183 = 6 . One possible set

of dimensions for the original rectangle is 6m by 3 m.


441

If w = 9, then

†

l = 189 = 2 . Another possible

set of dimensions for the original rectangleis 2 m by 9 m.

57.

†

- 4 5- 3( )3[ ] + 23

- 4 2( )3[ ] + 23

- 4 8( )[ ]+ 8-32+ 8

-24

58.

†

IR+ Ir = E, for RIR = E - Ir

R = E - IrI

59.

†

2rr- 4 - 2r

r+ 4 + 64r 2 -16

= 2rr- 4 ⋅ r + 4

r + 4 - 2rr + 4 ⋅ r- 4

r- 4 + 64r+ 4( ) r- 4( )

= 2r 2 +8rr+ 4( ) r- 4( )

- 2r 2 - 8rr + 4( ) r - 4( )

+ 64r+ 4( ) r - 4( )

=2r 2 + 8r - 2r 2 -8r( )+ 64

r+ 4( ) r- 4( )

= 16r + 64r+ 4( ) r- 4( )

=16 r+ 4( )

r+ 4( ) r- 4( ) or 16

r- 4

60.

†

x3/4 y-2

x1/2y 2Ê

Ë Á

ˆ

¯ ˜

4

= x(3/4 )-(1/2)y-2-2( )4

= x1/4 y-4( )4

= x1y-16

= xy16

61.

†

x2 + 3x + 9 = xx 2 + 3x + 9 = x2

3x + 9= 03x = -9 fi x = -3Upon checking, this value does not satisfythe equation. There is no real solution.

Exercise Set 12.4

1. A given equation can be expressed as an equation in quadratic form if the equation can be written in the formau2 + bu + c = 0 .

3. Let

†

u = x2 . Then

†

3x4 - 5x2 +1= 0 fi 3 x2( )2- 5x 2 +1= 0 fi 3u2 - 5u+1= 0

5. Let

†

u = z -1 . Then

†

z-2 - z -1 = 56 fi z -1( )2- z -1 = 56 fi u2 - u = 56


442

7.

†

x 4 + x 2 - 6

Let u = x2

x 2( )2+ x 2 - 6

u2 + u - 6u + 3( ) u - 2( )

Back substitute x 2 for ux 2 + 3( ) x 2 - 2( )

9.

†

x 4 + 5x2 + 6

Let u = x2

x 2( )2+ 5x2 + 6

u2 + 5u + 6u + 2( ) u + 3( )

Back substitute x 2 for ux 2 + 2( ) x2 + 3( )

11.

†

6a4 + 5a2 - 25

Let u = a2

6 a2( )2+ 5a2 - 25

6u2 + 5u - 252u + 5( ) 3u - 5( )

Back substitute a2 for u2a2 + 5( ) 3a2 - 5( )

13.

†

4 x + 1( )2 + 8 x + 1( ) + 3Let u = x + 1

4u2 + 8u + 32u + 3( ) 2u + 1( )

Back substitute x +1 for u2(x + 1) + 3( ) 2(x + 1) + 1( )2x + 2 + 3( ) 2x + 2 + 1( )2x + 5( ) 2x + 3( )

15.

†

6 a + 2( )2 - 7 a + 2( ) - 5Let u = a + 26u2 - 7u - 52u + 1( ) 3u - 5( )

Back substitute a + 2 for u2 a + 2( ) + 1( ) 3 a + 2( ) - 5( )2a + 4 + 1( ) 3a + 6 - 5( )2a + 5( ) 3a + 1( )

17.

†

a2b2 + 8ab + 15Let u = ab

ab( )2 + 8ab + 15

u2 + 8u + 15u + 3( ) u + 5( )

Back substitute ab for uab + 3( ) ab + 5( )

19.

†

3x2 y2 - 2xy - 5Let u = xy

3 xy( )2- 2xy - 5

3u2 - 2u - 53u - 5( ) u + 1( )

Back substitute xy for u3xy - 5( ) xy + 1( )

21.

†

2a2 5- a( ) - 7a 5 - a( ) + 5 5 - a( )Factor out 5- a( )

5- a( ) 2a2 - 7a + 5( )5- a( ) 2a - 5( ) a - 1( )

23.

†

2x 2 x - 3( ) + 7x x - 3( ) + 6 x - 3( )Factor out x – 3

x - 3( ) 2x2 + 7x + 6( )x - 3( ) 2x + 3( ) x + 2( )


443

25.

†

y 4 + 13y2 + 30

Let u = y2

y 2( )2+ 13y2 + 30

u2 + 13u + 30u + 3( ) u + 10( )

Back substitute y 2 for uy 2 + 3( ) y 2 + 10( )

27.

†

x 2 x + 3( ) + 3x x + 3( ) + 2 x + 3( )Factor out x + 3

x + 3( ) x 2 + 3x + 2( )x + 3( ) x + 1( ) x + 2( )

29.

†

5a5b2 - 8a4b3 + 3a3b4

Factor out a3b2

a3b2 5a2 - 8ab + 3b2( )a3b2 5a - 3b( ) a - b( )

31. x4 - 10x2 + 9 = 0x2( )2 - 10x2 + 9 = 0

u2 - 10u + 9 = 0 ¨ Replace x2 with uu - 9( ) u -1( ) = 0

u - 9 = 0 or u - 1 = 0u = 9 u = 1

x 2 = 9 x2 = 1 ¨ Replace u with x2

x = ± 9 = ±3 x = ± 1 = ±1The solutions are 3, –3, 1, and –1.

33.

†

x 4 - 26x 2 + 25= 0

†

x 2( )2- 26x 2 + 25= 0

u2 - 26u+ 25= 0 ¨ Replace x2 with uu- 25( ) u-1( ) = 0

†

u- 25= 0 u-1= 0u = 25 u = 1

x2 = 25 x2 = 1 ¨ Replace u with x2


35. x4 - 13x 2 + 36 = 0x2( )2 - 13x 2 + 36 = 0

u2 - 13u + 36 = 0 ¨ Replace x2 with uu - 9( ) u - 4( ) = 0

u - 9 = 0 u - 4 = 0u = 9 u = 4

x 2 = 9 x2 = 4 ¨ Replace u with x 2



444

37.

†

a4 - 7a2 +12 = 0

†

a2( )2- 7a2 +12 = 0

u2 - 7u+12 = 0 ¨ Replace a2 with uu - 4( ) u - 3( ) = 0

†

u- 4 = 0 or u - 3= 0u = 4 u = 3

a2 = 4 a2 = 3 ¨ Replace u with a2

a = ± 4 = ±2 a = ± 3

The solutions are 2, –2, 3 , and - 3 .

39.

†

4x4 - 5x2 +1= 0

†

4 x 2( )2- 5x2 +1= 0

4u2 - 5u +1= 0¨ Replace x2 with u4u -1( ) u-1( ) = 0

†

4u -1= 0 or u-1= 0u = 1

4u = 1

x 2 = 14

x2 = 1¨ Replace u with x2

x = ± 14 = ± 1

2 x = ± 1 = ±1

The solutions are 12

, -12

, 1, and –1.

41. r4 - 8r2 = -15r4 - 8r2 + 15 = 0

r2( )2- 8r2 + 15 = 0

u2 - 8u + 15 = 0 ¨ Replace r2 with uu - 3( ) u - 5( ) = 0

u - 3 = 0 u - 5 = 0u = 3 u = 5

r2 = 3 r2 = 5 ¨ Replace u with r2

r = ± 3 r = ± 5

The solutions are 3, - 3, 5, and - 5 .

43.

†

z 4 - 7z 2 =18

†

z 4 - 7z 2 -18 = 0

z 2( )2

- 7z 2 -18 = 0

u2 - 7u-18 = 0 ¨ Replace z2 with uu - 9( ) u+ 2( ) = 0

†

u- 9 = 0 u+ 2 = 0u = 9 u = -2

z 2 = 9 z 2 = -2 ¨ Replace u with z 2

z = ±3 z = ± -2 = ±i 2The solutions are

†

3, - 3, i 2, and - i 2 .

45.

†

-c 4 = 4c 2 - 5

†

c 4 + 4c2 - 5= 0

c 2( )2

+ 4c2 - 5= 0

u2 + 4u - 5= 0 ¨ Replace c 2 with uu -1( ) u + 5( ) = 0

†

u-1= 0 u + 5= 0u = 1 u = -5

c2 = 1 c 2 = -5 ¨ Replace u with c 2

c = ±1 c = ± -5 = ±i 5The solutions are

†

1, -1, i 5, and -i 5 .

47.

†

x = 2x - 6

†

2x - x - 6 = 0

2 x1/2( )2

- x1/2 - 6 = 0

2u2 - u- 6 = 0 ¨ Replace x1/2 with u2u+ 3( ) u- 2( ) = 0

†

2u+ 3= 0 or u- 2 = 0u = - 3

2 or u = 2

x1/2 = - 32 or x1/2 = 2 ¨ Replace u with x1/2

x = 22 = 4

†

x1/2 = - 32 has no solution since there is no value

of x for which

†

x1/2 = - 32 .

The solution is 4.

49.

†

x + x = 6

†

x + x - 6 = 0

x1/2( )2

+ x1/2 - 6 = 0

u2 + u- 6 = 0 ¨ Replace x1/2 with uu + 3( ) u - 2( ) = 0

†

u+ 3= 0 or u- 2 = 0u = -3 or u = 2

x1/2 = -3 or x1/2 = 2 ¨ Replace u with x1/2


445

x = 22 = 4

†

x1/2 = -3 has no solution since there is no valueof x for which

†

x1/2 = -3.The solution is 4.

51. 9x + 3 x = 29x + 3 x - 2 = 0

9 x1 /2( )2+ 3x1/2 - 2 = 0

9u2 + 3u - 2 = 0 ¨ Replace x1/2 with u3u -1( ) 3u + 2( ) = 0

3u -1 = 0 or 3u + 2 = 0u =

13

u = -23

x1/ 2 =13

x1 /2 = -23

¨ Replace u with x1/2

x =19

x1 /2 = -23

has no solution since there is no value of x for which x1 /2 = -23

.

The solution is 19

.

53. (x + 3)2 + 2(x + 3) = 24(x + 3)2 + 2(x + 3) - 24 = 0

u2 + 2u - 24 = 0 ¨ Replace x + 3 with u(u - 4)(u + 6) = 0

u - 4 = 0 or u + 6 = 0u = 4 u = -6

x + 3 = 4 x + 3 = -6 ¨ Replace u with x + 3x = 1 x = -9

The solutions are 1 and –9.

55.

†

6 a - 2( )2 = -19 a - 2( ) - 10

6 a - 2( )2 + 19 a - 2( ) + 10 = 06u2 + 19u + 10 = 0 ¨ Replace a - 2 with u3u + 2( ) 2u + 5( ) = 0

†

3u + 2= 0 or 2u + 5= 0u = - 2

3 u = - 52

a - 2= - 23

a- 2 = - 52

¨ Replace u with a - 2

a = 43

a = - 12


and -12

.

57. (x2 - 1)2 - (x2 - 1) - 6 = 0u2 - u - 6 = 0 ¨ Replace x 2 - 1 with u

u + 2( ) u - 3( ) = 0u + 2 = 0 or u - 3 = 0

u = -2 u = 3x2 - 1 = -2 x2 -1 = 3 ¨ Replace u with x2 - 1

x2 = -1 x2 = 4x = -1 = ±i x = ± 4 = ±2

The solutions are i, –i, 2, and –2.


446

u + 2 = 0 or u - 3 = 0u = -2 u = 3

x2 - 1 = -2 x2 -1 = 3 ¨ Replace u with x2 - 1x2 = -1 x2 = 4x = -1 = ±i x = ± 4 = ±2

The solutions are i, –i, 2, and –2.

59.

†

2(b+ 2)2 + 5(b+ 2) - 3= 02u2 + 5u - 3= 0 ¨ Replace b + 2 with u

(u + 3)(2u -1) = 0

†

u+ 3= 0 or 2u -1= 0u = -3 u = 1

2b+ 2= -3 b+ 2 = 1

2¨ Replace u with b + 2

b= -5 b = - 32

The solutions are –5 and -32

.

61. 18(x2 - 5)2 + 27(x2 - 5) +10 = 018u2 + 27u +10 = 0 ¨ Replace x2 - 5 with u

3u + 2( ) 6u + 5( ) = 03u + 2 = 0 or 6u + 5 = 0

u = -23

u = -56

x2 - 5 = -23

x 2 - 5 = -56

¨ Replace u with x 2 - 5

x2 =133

x2 =256

x = ±133

x = ±256

= ±133

⋅33

= ±56

⋅66

= ±393

= ±5 6

6


, -393

,5 6

6, and -

5 66

.

63.

†

x-2 +10x-1 + 25= 0

x-1( )2+10 x-1( ) + 25= 0

u2 +10u + 25= 0 ¨ Replace x-1 with uu+ 5( ) u+ 5( ) = 0

†

u+ 5= 0u = -5

x-1 = -5 ¨ Replace u with x-1

x = - 15

The solution is

†

- 15 .


447

65.

†

6b-2 - 5b-1 +1= 0

6 b-1( )2

- 5 b-1( )+1= 0

6u2 - 5u+1= 0 ¨ Replace b-1 with u2u-1( ) 3u-1( ) = 0

†

2u-1= 0 or 3u-1= 0u = 1

2 u = 13

b-1 = 12 b-1 = 1

3 ¨ Replace u with b-1

b= 2 b= 3

The solutions are 2 and 3.

67.

†

2b-2 = 7b-1 - 32b-2 - 7b-1 + 3= 0

2 b-1( )2- 7 b-1( ) + 3= 0

2u2 - 7u + 3= 0 ¨ Replace b-1 with u2u -1( ) u - 3( ) = 0

†

2u-1= 0 or u- 3= 0u = 1

2 u = 3

b-1 = 12 b-1 = 3 ¨ Replace u with b-1

b= 2 b= 13

The solutions are 2 and

†

13 .

69.

†

x-2 + 9x-1 = 10x-2 + 9x-1 -10= 0

x-1( )2+ 9 x-1( ) -10= 0

u2 + 9u -10= 0 ¨ Replace x-1 with uu+10( ) u -1( ) = 0

†

u+10 = 0 or u -1= 0u = -10 u =1

x-1 = -10 x-1 =1 ¨ Replace u with x-1

x = - 110

x =1

The solutions are -1

10 and 1.

71.

†

x-2 = 4x-1 +12x-2 - 4x-1 -12 = 0

x-1( )2- 4 x-1( )-12 = 0

u2 - 4u-12 = 0 ¨ Replace x-1 with uu+ 2( ) u - 6( ) = 0

†

u+ 2 = 0 or u - 6= 0u = -2 u = 6

x-1 = -2 x-1 = 6 ¨ Replace u with x-1

x = - 12

x = 16

The solutions are

†

- 12 and 1

6 .


448

†

u+ 2 = 0 or u - 6= 0u = -2 u = 6

x-1 = -2 x-1 = 6 ¨ Replace u with x-1

x = - 12

x = 16

The solutions are

†

- 12 and 1

6 .

73.

†

x 2/3 - 3x1/3 = -2

x1/3( )2

- 3x1/3 + 2= 0

u2 - 3u + 2= 0 ¨ Replace x1/3 with uu -1( ) u - 2( ) = 0

†

u-1= 0 or u - 2= 0u = 1 u = 2

x1/3 = 1 x1/3 = 2 ¨ Replace u with x1/3

x = 13 x = 23

= 1 = 8

The solutions are 1 and 8.

75.

†

b2/3 +11b1/3 + 28= 0

†

b1/3( )2+11b1/3 + 28= 0

u2 +11u + 28= 0 ¨ Replace b1/3 with uu + 7( ) u+ 4( ) = 0

†

u+ 7 = 0 or u+ 4 = 0u = -7 or u = -4

b1/3 = -7 or b1/3 = -4 ¨ Replace u with b1/3

b = -7( )3 or b= -4( )3

= -343 = -64The solutions are –343 and –64.

77. -2a - 5a1/2 + 3 = 0-2 a1/ 2( )2 - 5a1/2 + 3 = 0

-2u2 - 5u + 3 = 0 ¨ Replace a1/2 with u2u2 + 5u - 3 = 0 2u - 1( ) u + 3( ) = 0

†

2u-1= 0 or u+ 3= 0u = 1

2 u = -3a1/2 = 2 a1/2 = -3 ¨ Replace u with a1/2

a = 12( )

2= 1

4

a1/ 2 = -3 has no solution since there is no value of a for which a12 = -3 . The solution is

14

.


449

79.

†

c 2/5 - 3c1/5 + 2= 0

c1/5( )2

- 3c1/5 + 2= 0

u2 - 3u + 2= 0 ¨ Replace c1/5 with uu - 2( ) u-1( ) = 0

†

u- 2 = 0 u -1= 0u = 2 u =1

c1/5 = 2 c1/5 =1 ¨ Replace u with c1/5

c = 25 c =15

= 32 =1The solutions are 32 and 1.

81. f x( ) = x - 5 x + 4 , f x( ) = 0

0 = x1 /2( )2 - 5x1/2 + 4

0 = u2 - 5u + 4 ¨ Replace x1 /2 with u0 = u - 1( ) u - 4( )u - 1 = 0 or u - 4 = 0

u = 1 u = 4x1/2 = 1 x1/2 = 4 ¨ Replace u with x1/2

x = 1 x = 16The x-intercepts are (1, 0) and (16, 0).

83.

†

h x( ) = x +13 x + 36, h x( ) = 0

0 = x1/2( )2+13x1/2 + 36

0 = u2 +13u+ 36 ¨ Replace x1/2 with u0 = u+ 9( ) u+ 4( )

†

u+ 9 = 0 or u+ 4 = 0u = -9 u = -4

x1/2 = -9 x1/2 = -4 ¨ Replace u with x1/2

There are no values of x for which

†

x1/2 = -9 or x1/2 = -4 . There are no x-intercepts.

85.

†

p x( ) = 4x-2 -19x-1 - 5 ,

†

p x( ) = 0

0 = 4 x -1( )2 - 19x-1 - 5

0 = 4u2 - 19u - 5 ¨ Replace x -1 with u0 = 4u +1( ) u - 5( )4u +1 = 0 or u - 5 = 0

u = -14

u = 5

x -1 = -14

x -1 = 5 ¨ Replace u with x-1

x = -4 x =15

The x-intercepts are (–4, 0) and 15

, 0Ê Ë Á ˆ

¯ ˜ .


450

4u +1 = 0 or u - 5 = 0u = -

14

u = 5

x -1 = -14

x -1 = 5 ¨ Replace u with x-1

x = -4 x =15

The x-intercepts are (–4, 0) and 15

, 0Ê Ë Á ˆ

¯ ˜ .

87. f x( ) = x 2/3 + x1/3 - 6 , f x( ) = 0

0 = x1 /3( )2+ x1/3 - 6

0 = u2 + u - 6 ¨ Replace x1 /3 with u0 = u + 3( ) u - 2( )u + 3 = 0 or u - 2 = 0

u = -3 u = 2x1/ 3 = -3 x1 /3 = 2 ¨ Replace u with x1/3

x = -27 x = 8The x-intercepts are (–27, 0) and (8, 0).

89. g x( ) = x2 - 3x( )2+ 2 x 2 - 3x( ) - 24 , g x( ) = 0

0 = u2 + 2u - 24 ¨ Replace x2 - 3x with u0 = (u + 6)(u - 4)

u + 6 = 0 or u - 4 = 0x2 - 3x + 6 = 0 x2 - 3x - 4 = 0 ¨ Replace u with x 2 - 3x

( x - 4)( x +1) = 0 x - 4 = 0 or x + 1 = 0

x = 4 x = -1There are no x-intercepts for x2 - 3x + 6 = 0 since b2 - 4ac = -3( )2 - 4 1( ) 6( ) = 9 - 24 = -15 .The x-intercepts are (4, 0) or (–1, 0).

91.

†

f x( ) = x4 - 20x + 64, f x( ) = 0

0 = x 2( )2- 20x 2 + 64

0 = u2 - 20u + 64 ¨ Replace x2 with u0 = u-16( ) u- 4( )

†

u-16 = 0 or u- 4 = 0u =16 u = 4

x 2 =16 x2 = 4 ¨ Replace u with x 2

x = ± 16 = ±4 x = ± 4 = ±2The x-intercepts are ( 4, 0), (–4, 0), ( 2, 0), and (–2, 0).


451

The x-intercepts are ( 4, 0), (–4, 0), ( 2, 0), and (–2, 0).

93. When solving an equation of the form ax 4 + bx2 + c = 0 , let u = x 2 .

95. When solving an equation of the form

†

ax-2 +bx-1 + c = 0, let u = x-1.

97. If the solutions are ±2 and ±4, the factors must be x - 2( ), x + 2( ) , x - 4( ) and x + 4( ) .0 = x - 2( ) x + 2( ) x - 4( ) x + 4( )

0 = x2 - 4( ) x 2 -16( )0 = x4 - 20x 2 + 64

99. If the solutions are

†

± 2 and ± 3 , the factors must be

†

x + 2( ), x - 2( ), x + 3( ), x - 3( ) .

†

0= x + 2( ) x - 2( ) x + 3( ) x - 3( )0= x 2 - 2( ) x2 - 3( )0= x 4 - 5x2 + 6

101. No. An equation of the form ax 4 + bx2 + c = 0 can have no imaginary solutions, two imaginary solutions, orfour imaginary solutions.

103. a. 3x2 -

3x

= 60 The LCD is x2

x2 3x 2

Ê Ë Á

ˆ ¯ ˜ - x2 3

xÊ Ë Á ˆ

¯ ˜ = x2 60( )

3 - 3x = 60x2

0 = 60x2 + 3x - 30 = 3 20x2 + x - 1( )0 = 3 5x - 1( ) 4x +1( )

5x -1 = 0 or 4x + 1 = 0

x =15

x = -14


and -14

b. 3x2 -

3x

= 60

3x-2 - 3x-1 = 60

3 x -1( )2 - 3x-1 - 60 = 0

3u2 - 3u - 60 = 0 ¨ Replace x-1 with u3 u2 - u - 20( ) = 03 u - 5( ) u + 4( ) = 0


452

u - 5 = 0 or u + 4 = 0u = 5 u = -4

x -1 = 5 x-1 = -4 ¨ Replace u with x -1

x =15

x = -14


and -14

.

112. 15 r + 2( ) + 22 = -8

r + 215 r + 2( ) r + 2( ) + 22 r + 2( ) = -

8r + 2

r + 2( )

15 r + 2( )2 + 22 r + 2( ) = -815 r + 2( )2 + 22 r + 2( ) + 8 = 0

15u2 + 22u + 8 = 0 ¨ Replace r + 2 with u5u + 4( ) 3u + 2( ) = 0

5u + 4 = 0 or 3u + 2 = 0u = -

45

u = -23

r + 2 = -45

r + 2 = -23

¨ Replace u with r + 2

r = -145

r = -83

The solutions are -145

and -83

.

107. 4 - x - 2( )-1 = 3 x - 2( )-2

4 - u-1 = 3u-2 ¨ Replace x - 2 by u

4 -1u

=3

u2

u2 4 -1u

Ê Ë Á ˆ

¯ ˜ = u2 3

u2Ê Ë Á

ˆ ¯ ˜

4u2 - u = 34u2 - u - 3 = 0

4u + 3( ) u -1( ) = 04u + 3 = 0 or u - 1 = 0

u = -34

u = 1

x - 2 = -34

x - 2 = 1 ¨ Replace u by x - 2

x =54

x = 3


and 3.


453

109. x6 - 9x 3 + 8 = 0x3( )2

- 9x3 + 8 = 0

u2 - 9u + 8 = 0 ¨ Replace x3 with uu - 8( ) u - 1( ) = 0

u - 8 = 0 or u -1 = 0u = 8 u = 1

x 3 = 8 x3 = 1 ¨ Replace u with x3

x = 2 x = 1The solutions are 2 and 1.

111. x2 + 2x - 2( )2 - 7 x2 + 2x - 2( ) + 6 = 0

u2 - 7u + 6 = 0 ¨ Replace x2 + 2x - 2 with uu - 6( ) u -1( ) = 0

u - 6 = 0 or u - 1= 0u = 6 u = 1

x 2 + 2x - 2 = 6 x 2 + 2x - 2 = 1 ¨ Replace u with x2 + 2x - 2x2 + 2x - 8 = 0 x2 + 2x - 3 = 0

x + 4( ) x - 2( ) = 0 x + 3( ) x -1( ) = 0

x + 4 = 0 or x - 2 = 0 x + 3 = 0 or x - 1= 0

x = -4 x = 2 x = -3 x = 1The solutions are –4, 2, –3, and 1.

113. 2n4 - 6n2 - 3 = 02 n2( )2

- 6n2 - 3 = 0

2u2 - 6u - 3 = 0 ¨ Replace n2 with u

u =6 ± -6( )2 - 4 2( ) -3( )

2 2( )

=6 ± 60

4

=6 ± 2 15

4

=3 ± 15

2

n2 =3 ± 15

2 ¨ Replace u with n2

n = ±3 ± 15

2

115.

†

45 - 3

4 - 23( ) = 4

5 - 912 - 8

12( )= 4

5- 1

12( )= 48

60 - 560

= 4360

116.

†

6 x + 4( )- 4 3x + 3( ) = 66x + 24 -12x -12 = 6

-6x +12 = 6-6x = -6

x =1

117.

†

D : all real numbersR : y y ≥ 0{ }

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Elementary and Intermediate Algebra Quadratic …wps.prenhall.com/wps/media/objects/794/813466/ssm/ch12_3.pdfSSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions