elementary and intermediate algebra quadratic...
Post on 12-Jul-2020
45 Views
Preview:
TRANSCRIPT
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.
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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
Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra
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.
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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
Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra
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( )
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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
x = ± 25 = ±5 x = ± 1 = ±1The solutions are 5, –5, 1, and –1.
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
x = ± 9 = ±3 x = ± 4 = ±2The solutions are 3, –3, 2, and –2.
Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra
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
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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
The solutions are 43
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.
Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra
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
The solutions are 393
, -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 .
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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 .
Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra
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
.
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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
¯ ˜ .
Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra
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).
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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
The solutions are 15
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
Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra
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
The solutions are 15
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
The solutions are 54
and 3.
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions
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{ }
top related