VIEWS: 36 PAGES: 17 CATEGORY: Nutrition & Healthy Eating POSTED ON: 5/27/2010
SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions 21. h = -16t 2 + s0 , for t 7. E = i2 r , for i 16t 2 = s0 - h E s -h = i2 t2 = 0 r 16 E s -h =i r t= 0 16 9. d = 16t 2 , for t s0 - h t= d = t2 4 16 23. 1 d =t ﬁ t = d E= mv 2 , for v 2 16 4 2E = mv 2 2E 11. E = mc 2 , for c = v2 E m = c2 m 2E =v E m =c m 25. 2 v 2 - v1 2 1 2 a= , for v1 13. 2d V= pr h , for r 2ad = v2 - v12 3 2 3V = pr2 h 2 2ad + v1 = v2 2 3V = r2 2 v1 = v 2 - 2ad 2 ph 2 3V v1 = v2 - 2ad =r ph 27. v ¢ = c 2 - v 2 , for c 15. 2 2 d = L +W , for W ( v ¢) 2 = c 2 - v 2 d 2 = L2 +W 2 ( v ¢) 2 + v 2 = c 2 d 2 - L2 =W 2 2 c = (v ¢ ) + v 2 d 2 - L 2 =W 29. a. P( n) = 2.4 n 2 + 9n - 3 17. 2 2 2 † a + b = c , for b 2 P( 6) = 2.4 ( 6) + 9( 6) - 3 b 2 = c2 - a 2 =137.4 ﬁ $13,740 b = c2 - a 2 The profit would be $13,740. 19. d = L 2 +W 2 + H 2 , for H b. P( n) = 2.4n 2 + 9n- 3 d 2 = L2 +W 2 + H 2 200= 2.4n 2 + 9n- 3 d 2 - L2 -W 2 = H 2 2.4n 2 + 9n- 203= 0 d 2 - L 2 -W 2 = H -9± 92 - 4( 2.4 )(-203) x= 2( 2.4 ) x = -9± 2029.8 4.8 x ª 8 or x ª -11 Eight tractors must be sold. 437 Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra Eight tractors must be sold. 37. a. m (t) = 0.05t 2 - 0.32t + 3.15 In 2003, t = 21. 31. T = 6.2t 2 +12t + 32 ( ) m( 21) = 0.05 212 - 0.32( 21) + 3.15 a. When the car is turned on, t = 0. = 22.05- 6.72 + 3.15 T = 6.2( 0) 2 +12 ( 0) + 32 = 32 ª 18.5 The temperature is 32°F. Veterinary bills for dogs in 2003 amounted to about $18.5 billion. b. T = 6.2(1) 2 + 12(1) + 32 = 50.2 The temperature after 1 minute is 50.2°F. b. m (t) = 0.05t 2 - 0.32t + 3.15, m (t) = 12 25= 0.05t 2 - 0.32t + 3.15 c. 120 = 6.2t 2 +12t + 32 0 = 0.05t 2 - 0.32t - 21.85 0 = 6.2t 2 +12t - 88 2 0.32 ± ( 0.32) - 4 (0.05)( -21.85) -12 ± 12 2 - 4(6.2)( -88) t= t= 2( 0.05) 2(6.2) tª -12 ± 48.23 = 0.32 ± 0.1024 + 4.37 0.1 12.4 t ª 2.92 or t ª -4.86 = 0.32± 4.4724 The radiator temperature will reach 120°F 0.1 about 2.92 min. after the engine is started. ª 0.32 ± 2.1148050 0.1 tª 0.32+ 2.1148050 = 2.4348050 ª 24.3 0.1 0.1 33. a. C = -0.011500 )2 + 80(1500) + 20,000 ( or = 117, 500 t ª 0.32- 2.1148050 = -1.794805 ª -17.9 0.1 0.1 The cost is about $117,500. Since 1£ t £ 28, the only solution is t ≈ 24.3. Thus $25 billion will be spent on b. 150, 000 = -0.01s 2 + 80s + 20, 000 veterinary bills for dogs approximately 0 = -0.01s2 + 80s - 130, 000 24.3 years after 1982 or in 2006. -80 ± 802 - 4( -0.01)(-130, 000) s= 39. Let x be the width of the playground. Then 2(-0.01) the length is given by x + 5. † -80 ± 34.64 Area = length ¥ width sª -0.02 500= x ( x + 5) s ª 2268 or s ª 5732 Notice that s must fall between 1200 and x 2 + 5x - 500 = 0 4000. Therefore, Mr. Boyle can purchase ( x - 20)( x + 25) = 0 a 2268 sq ft house for $150,000. x = 20 or x = -25 † 2 Disregard the negative value. The width of 35. a. 0 = -3. 3t - 2.3t + 62 the playground is 20 meters and the length is -(-2 .3)± (-2.3)2 - 4(- 3. 3)(62) 20 + 5 = 25 meters. t= 2(- 3.3) t = 2.3-6.6 7 ± 28. t ≈ –4.7 or t = 4 Since time must be positive, the car takes 4 seconds for the drop. † b. s = 6.74(4) + 2.3 = 29.26 The speed is 29.26 feet per second. † † † 438 SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions 41. Let r be the rate at which the present equipment mechanic. drills. 6 6 + =1 d x x +1 d r t= 6 6 ˆ r x(x + 1)Ê ˆ + x(x + 1)Ê Á ˜ Á ˜ = x(x + 1)(1) Ë x¯ Ë x + 1¯ present 64 64 r 6(x +1) + 6x = x 2 + x equipment r 64 6x + 6 + 6x = x 2 + x new 64 r+1 equipment r +1 0 = x 2 - 11x - 6 They would have hit water in 3.2 hours less time -(-11) ± (-11)2 - 4(1)(-6) x= with the new equipment. 2(1) 64 64 11 ± 121 + 24 = - 3.2 r +1 r = 2 64 ˆ 64 r(r + 1)ÊÁ ˜ = r(r + 1)Ê ˆ - r(r + 1)(3.2) Á ˜ 11 ± 145 Ë r + 1¯ Ë r ¯ = 2 64r = 64(r + 1) - 3.2r( r + 1) 11 + 145 11- 145 64r = 64r + 64 - 3.2r 2 - 3.2r x= or x = 2 2 0 = 64 - 3.2r 2 - 3.2r ª 11.52 ª -0.52 3.2r2 + 3.2r - 64 = 0 Since the time must be positive, it takes Bonita r2 + r - 20 = 0 about 11.52 hours and Pamela about 12.52 hours to rebuild the engine. ( r + 5)( r - 4) = 0 r + 5 = 0 or r - 4 = 0 47. Let r be the speed of the plane in still air. r = -5 r=4 d d r t= Use the positive value. The present equipment r drills at a rate of 4 ft/hr. 80 With wind 80 r + 30 43. Let x be Latoya’s rate going uphill so x + 2 is r + 30 d 80 her rating going downhill. Using = t gives Against wind 80 r – 30 r r - 30 tuphill + tdownhill = 1.75 The total time is 1.3 hours 6 6 80 + 80 = 1.3 + = 1.75 x x+2 r + 30 r - 30 6 6 ˆ x(x + 2)Ê ˆ + x(x + 2)Ê Á ˜ Á ˜ = x(x + 2)(1.75) ( 80 ( r + 30)( r - 30) r 80 + r - 30 =1.3 ) Ë x¯ Ë x + 2¯ + 30 6(x + 2) + 6x = 1.75x( x + 2) ( 80( r - 3) + 80( r + 30) = 1.3 r 2 - 900 ) 6x + 12 + 6x = 1.75x 2 + 3.5x 2 80r - 240 + 80r + 240 = 1.3r -1170 0 = 1.75x 2 - 8.5x -12 160r = 1.3r2 -1170 - ( -8.5) ± (-8.5) 2 - 4(1.75)(-12) 0 = 1.3r2 -160 r - 1170 x= 2(1.75) - (-160) ± ( -160) 2 - 4(1.3)( -1170) 8 ± 156.25 r= = 2(1.3) 3.5 160 ± 25, 600 + 6084 8.5 ±12.5 = = 2.6 3.5 x = 6 or x ª -1.14 160 ± 31,684 = Since the time must be positive, Latoya’s uphill 2.6 rate is 6 mph and her downhill rate is 160 ±178 = x + 2 = 8 mph. 2.6 45. Let x be the time of the experienced mechanic then x + 1 is the time of the inexperienced 439 Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra 160 +178 160 - 178 Including the 2.5 hours she spent in Plainview, r= or r = 2.6 2.6 the entire trip took Lisa 5.5 hours. 338 -18 60 + 2.5+ 100 = 5.5 = = x x -10 2.6 2.6 = 130 ª -6.92 60 + 100 = 3 Since speed must be positive, the speed of the x x -10 plane in still air is 130 mph. ( x(x -10) 60 + 100 = 3 x x -10 ) 49. Let t be the number of hours for Chris to clean 60(x -10) +100x = 3(x 2 -10x) alone. Then t + 0.5 is the number of hours for John to clean alone. 60x - 600+100x = 3x 2 - 30x Rate of Time Part of Task -600+160x = 3x 2 - 30x work worked completed 0= 3x 2 -190x + 600 1 6 0 = (x - 60)(3x -10) Chris 6 t t x - 60 = 0 or 3x - 10 = 0 10 1 6 x = 60 x= John 6 3 t + 0.5 t + 0.5 10 6 6 miles per hour is too slow for a car, so the + =1 3 t t + 0.5 speed of the trip from Lubbock to Plainview was 6 6 ˆ t(t + 0.5)Ê ˆ + t(t + 0.5)Ê Á ˜ Á ˜ = t(t + 0.5)(1) 60 mph. Ët¯ Ë t + 0.5¯ 6(t + 0.5) + 6t = t(t + 0.5) 53. Answers will vary. 6t + 3 + 6t = t 2 + 0.5t 55. Let l = original length and w = original 12t + 3 = t 2 + 0.5t width. A system of equations that describes 0 = t 2 -11.5t - 3 this situation is 11.5 ± ( -11.5)2 - 4(1)(-3) l ⋅ w = 18 t= 2(1) ( l + 2)( w + 3) = 48 11.5 ± 132.25 + 12 If you solve for l in the first equation you = 2 get l = 18 . Substitute 18 into the l in the w w 11.5 ± 144.25 second equation. The result is an equation = 2 in only one variable which can be solved. 11.5 + 144.25 11.5 - 144.25 t= or t = ( l + 2)( w + 3) = 48 2 2 ª 11.76 ª -0.26 Since the time must be positive, it takes ( 18 + 2)( w + 3) = 48 w Chris about 11.76 hours and John about 18+ 54 + 2w + 6 = 48 11.76 + 0.5 = 12.26 hours to clean alone. w 2w - 24+ 54 = 0 † w 51. Let x be the speed of the trip from Lubbock to Plainview. Then x – 10 is the speed from ( w 2w - 24 + 54 = 0 w ) 2 Plainview to Amarillo. 2w - 24w + 54 = 0 2( w - 3) (w - 9) = 0 d r t w = 3 or w = 9 60 first part 60 x x If w = 3, then l = 18 = 6. One possible set 3 100 of dimensions for the original rectangle is 6 second part 100 x – 10 x -10 m by 3 m. 440 † † † SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions 4 If w = 9, then l = 18 = 2. Another possible 60. Ê x 3/4 y -2 ˆ 4 9 set of dimensions for the original rectangle Á 1/2 2 ˜ = x Ë x y ¯ ( (3/4 )-(1/2) -2-2 y ) is 2 m by 9 m. 4 ( = x 1/4 y -4 ) 57. 3 [ - 4( 5- 3) + 23 ] = x1 y -16 [ ] = x 3 - 4( 2) + 23 y16 -[ 4( 8) ] + 8 61. -32+ 8 x 2 + 3x + 9 = x -24 x 2 + 3x + 9 = x 2 3x + 9= 0 58. IR+ Ir = E, for R 3x = -9 ﬁ x = -3 IR = E - Ir Upon checking, this value does not satisfy R = E - Ir the equation. There is no real solution. I 59. 2r - 2r + 64 r - 4 r + 4 r 2 -16 = 2r ⋅ r + 4 - 2r ⋅ r - 4 + 64 r - 4 r + 4 r + 4 r - 4 ( r + 4 )( r - 4) = 2r 2 +8r - 2r 2 - 8r + 64 ( r + 4 )( r - 4) ( r + 4)( r - 4) ( r + 4) (r - 4) = ( 2r 2 + 8r - 2r 2 -8r + 64 ) ( r + 4) ( r - 4) = 16r + 64 ( r + 4 )( r - 4) 16( r + 4 ) 16 = or ( r + 4 )( r - 4) r- 4 † 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 form au 2 + bu + c = 0. 2 † 3. Let u = x 2 . Then 3x 4 - 5x 2 +1= 0 ﬁ 3 x2( ) - 5x 2 +1= 0 ﬁ 3u 2 - 5u+1= 0 -1 2 5. Let u = z -1 . Then z-2 - z -1 = 56 ﬁ (z ) - z -1 = 56 ﬁ u 2 - u = 56 441 Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra 7. x4 + x2 - 6 15. 6( a + 2) 2 - 7( a + 2) - 5 Let u = x 2 Let u = a + 2 2 6u 2 - 7u - 5 ( x 2) + x2 - 6 ( 2u + 1)( 3u - 5) u2 + u - 6 Back substitute a + 2 for u ( u + 3)( u - 2) ( 2(a + 2) + 1) (3( a + 2) - 5) Back substitute x 2 for u ( 2a + 4 + 1) (3a + 6 - 5) ( x 2 + 3)( x 2 - 2) ( 2a + 5)( 3a + 1) 9. x 4 + 5x 2 + 6 17. a 2b 2 + 8ab + 15 2 Let u = x Let u = ab 2 ( x 2) + 5x 2 + 6 ( ab) 2 + 8ab + 15 u 2 + 5u + 6 u 2 + 8u + 15 ( u + 2)( u + 3) ( u + 3)( u + 5) Back substitute ab for u Back substitute x 2 for u ( ab + 3)( ab + 5) ( x 2 + 2)( x 2 + 3) 19. 3x 2 y 2 - 2xy - 5 11. 6a 4 + 5a 2 - 25 Let u = xy Let u = a 2 2 3( xy) - 2xy - 5 2 2 ( ) 6a + 5a 2 - 25 3u 2 - 2u - 5 2 6u + 5u - 25 ( 3u - 5)( u + 1) ( 2u + 5)( 3u - 5) Back substitute xy for u Back substitute a 2 for u ( 3xy - 5) ( xy + 1) ( 2a 2 )( 2 + 5 3a - 5 ) 2 21. 2a ( 5- a) - 7a (5 - a ) + 5(5 - a ) 13. 4( x + 1) 2 + 8( x + 1) + 3 Factor out ( 5- a) Let u = x + 1 ( 5- a)( 2a 2 - 7a + 5) † 4u 2 + 8u + 3 ( 5- a)( 2a - 5)( a - 1) ( 2u + 3)( 2u + 1) 23. 2x 2 ( x - 3) + 7x ( x - 3) + 6( x - 3) Back substitute x +1 for u Factor out x – 3 ( 2(x + 1) + 3)( 2(x + 1) + 1) ( 2x + 2 + 3)( 2x + 2 + 1) ( x - 3)( 2x 2 + 7x + 6) ( 2x + 5)( 2x + 3) ( x - 3)( 2x + 3)( x + 2) † 442 SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions 25. y 4 + 13y 2 + 30 5 2 4 3 3 4 29. 5a b - 8a b + 3a b Let u = y 2 Factor out a 3b 2 2 ( y 2) + 13y 2 + 30 ( a 3b 2 5a 2 - 8ab + 3b 2 ) 3 2 u 2 + 13u + 30 a b ( 5a - 3b) (a - b) ( u + 3)( u + 10) Back substitute y 2 for u ( y 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) 31. x 4 - 10x 2 + 9 = 0 2 (x2 ) - 10x 2 + 9 = 0 u2 - 10u + 9 = 0 ¨ Replace x 2 with u (u - 9)(u -1) = 0 u-9=0 or u - 1 = 0 u=9 u =1 2 x =9 x 2 = 1 ¨ Replace u with x 2 x = ± 9 = ±3 x = ± 1 = ±1 The solutions are 3, –3, 1, and –1. 33. x 4 - 26x 2 + 25= 0 2 ( ) x 2 - 26x 2 + 25= 0 † u 2 - 26u+ 25= 0 ¨ Replace x 2 with u ( u- 25)( u-1) = 0 u- 25= 0 u-1= 0 u = 25 u =1 2 x = 25 x 2 = 1 ¨ Replace u with x 2 x = ± 25 = ±5 x = ± 1 = ±1 The solutions are 5, –5, 1, and –1. 35. x 4 - 13x 2 + 36 = 0 2 (x2 ) - 13x 2 + 36 = 0 u2 - 13u + 36 = 0 ¨ Replace x 2 with u (u - 9)(u - 4) = 0 u-9=0 u-4=0 u=9 u=4 x2 = 9 x 2 = 4 ¨ Replace u with x 2 x = ± 9 = ±3 x = ± 4 = ±2 The solutions are 3, –3, 2, and –2. 443 Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra 37. a 4 - 7a 2 +12 = 0 u- 9 = 0 u+ 2 = 0 2 ( ) a 2 - 7a 2 +12 = 0 u =9 2 u = -2 z =9 z 2 = -2 ¨ Replace u with z 2 u 2 - 7u+12 = 0 ¨ Replace a 2 with u z = ±3 z = ± -2 = ±i 2 ( u - 4)( u - 3) = 0 The solutions are 3, - 3, i 2, and - i 2 . u- 4 = 0 or u - 3= 0 u=4 u =3 45. -c 4 = 4c 2 - 5 2 a =4 a 2 = 3 ¨ Replace u with a 2 c 4 + 4c 2 - 5= 0 2 a = ± 4 = ±2 a =± 3 (c ) 2 + 4c 2 - 5= 0 The solutions are 2, –2, 3 , and - 3 . u 2 + 4u - 5= 0 ¨ Replace c 2 with u ( u -1)( u + 5) = 0 39. 4x 4 - 5x 2 +1= 0 2 u-1= 0 u + 5= 0 ( ) 4 x 2 - 5x 2 +1= 0 u =1 u = -5 4u 2 - 5u +1= 0¨ Replace x 2 with u c2 = 1 c 2 = -5 ¨ Replace u with c 2 ( 4u -1)( u-1) = 0 c = ±1 c = ± -5 = ±i 5 4u -1= 0 or u-1= 0 The solutions are 1, -1, i 5, and -i 5 . u= 1 u =1 4 x2 = 1 x 2 = 1¨ Replace u with x 2 47. x = 2x - 6 4 2x - x - 6 = 0 x =± 1 =± 1 x = ± 1 = ±1 4 2 2 ( ) 2 x1/2 - x 1/2 - 6 = 0 1 1 The solutions are , - , 1, and –1. 2u 2 - u- 6 = 0 ¨ Replace x1/2 with u 2 2 (2u+ 3)( u- 2) = 0 4 2 2u+ 3= 0 or u- 2 = 0 41. r - 8r = -15 r 4 - 8r 2 + 15 = 0 u =- 3 or u =2 † 2 2 ) ( r2† - 8r 2 + 15 = 0 x1/2 = - 3 2 or x 1/2 = 2 ¨ Replace u with x1/2 u 2 - 8u + 15 = 0 ¨ Replace r2 with u † ( u - 3)(u - 5) = 0 x = 22 = 4 u-3=0 u -5 = 0 x 1/2 = - 3 has no solution since there is no value 2 u=3 u=5 2 of x for which x 1/2 = - 3 . r =3 r 2 = 5 ¨ Replace u with r 2 2 The solution is 4. r=± 3 r=± 5 † The solutions are 3, - 3, 5, and - 5 . 49. x + x = 6 x+ x -6 =0 2 † 4 43. z - 7z =18 2 (x ) 1/2 + x 1/2 - 6 = 0 †4 - 7z 2 -18 = 0 z u 2 + u- 6 = 0 ¨ Replace x 1/2 with u 2 (u + 3) (u - 2) = 0 † (z ) 2 - 7z 2 -18 = 0 u+ 3= 0 or u- 2 = 0 u 2 - 7u-18 = 0 ¨ Replace z 2 with u u = -3 or u =2 1/2 x = 2 ¨ Replace u with x 1/2 1/2 ( u - 9)( u+ 2) = 0 x = -3 or 444 † SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions x = 22 = 4 x 1/2 = -3 has no solution since there is no value of x for which x 1/2 = -3. The solution is 4. 9x + 3 x = 2 51. 9x + 3 x - 2 = 0 2 ( ) 9 x1 /2 + 3x 1/2 - 2 = 0 9u2 + 3u - 2 = 0 ¨ Replace x 1/2 with u (3u -1)(3u + 2) = 0 3u -1 = 0 or 3u + 2 = 0 1 2 u= u= - 3 3 1/ 2 = 1 1 /2 = - 2 ¨ Replace u with x 1/2 x x 3 3 1 x= 9 1 /2 2 2 x = - has no solution since there is no value of x for which x1 /2 = - . 3 3 1 The solution is . 9 53. (x + 3)2 + 2(x + 3) = 24 (x + 3) 2 + 2(x + 3) - 24 = 0 u 2 + 2u - 24 = 0 ¨ Replace x + 3 with u (u - 4)(u + 6) = 0 u - 4 = 0 or u + 6 = 0 u=4 u = -6 The solutions are 1 and –9. x +3 = 4 x + 3 = -6 ¨ Replace u with x + 3 x =1 x = -9 55. 6( a - 2) 2 = -19( a - 2) - 10 6( a - 2) 2 + 19( a - 2) + 10 = 0 6u 2 + 19u + 10 = 0 ¨ Replace a - 2 with u ( 3u + 2)( 2u + 5) = 0 3u + 2= 0 or 2u + 5= 0 u =- 2 u =-5 4 1 3 2 The solutions are and - . a - 2= - 2 a- 2 = - 5 ¨ Replace u with a - 2 3 2 3 2 a= 4 a =-1 3 2 57. (x 2 - 1) 2 - (x 2 - 1) - 6 = 0 u 2 - u - 6 = 0 ¨ Replace x 2 - 1 with u ( u + 2)( u - 3) = 0 u +2 = 0 or u - 3 = 0 u = -2 u=3 445 2 - 1 = -2 2 -1 = 3 ¨ Replace u with x 2 - 1 x x x 2 = -1 x2 = 4 x = -1 = ±i x = ± 4 = ±2 The solutions are i, –i, 2, and –2. Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra u +2 = 0 or u - 3 = 0 u = -2 u=3 2 - 1 = -2 2 -1 = 3 ¨ Replace u with x 2 - 1 x x x 2 = -1 x2 = 4 x = -1 = ±i x = ± 4 = ±2 The solutions are i, –i, 2, and –2. 59. 2(b+ 2) 2 + 5(b+ 2) - 3= 0 2u 2 + 5u - 3= 0 ¨ Replace b + 2 with u (u + 3)(2u -1) = 0 u+ 3= 0 or 2u -1= 0 u = -3 u=1 2 b+ 2= -3 b+ 2 = 1 ¨ Replace u with b + 2 2 b= -5 b =- 3 2 3 The solutions are –5 and - . 2 61. 18(x 2 - 5)2 + 27(x 2 - 5) +10 = 0 18u 2 + 27u +10 = 0 ¨ Replace x 2 - 5 with u ( 3u + 2)(6u + 5) = 0 3u + 2 = 0 or 6u + 5 = 0 2 5 u=- u=- 3 6 2 2 5 x - 5 = - ¨ Replace u with x 2 - 5 2 x -5=- 3 6 2 13 2 25 x = x = 3 6 13 25 x=± x =± 3 6 13 3 5 6 =± ⋅ =± ⋅ 3 3 6 6 39 5 6 =± =± 3 6 39 39 5 6 5 6 The solutions are ,- , , and - . 3 3 6 6 63. x -2 +10x -1 + 25= 0 2 (x ) -1 ( ) +10 x -1 + 25= 0 2 u +10u + 25= 0 ¨ Replace x -1 with u ( u+ 5)( u+ 5) = 0 u+ 5= 0 u = -5 -1 The solution is - 1 . x = -5 ¨ Replace u with x -1 5 x=-1 5 446 SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions 65. 6b-2 - 5b-1 +1= 0 2 ( ) - 5(b ) +1= 0 6 b-1 -1 6u 2 - 5u+1= 0 ¨ Replace b-1 with u ( 2u-1)( 3u-1) = 0 2u-1= 0 or 3u-1= 0 u= 1 u=1 2 3 The solutions are 2 and 3. b-1 = 1 b-1 = 1 ¨ Replace u with b-1 2 3 b= 2 b= 3 67. 2b-2 = 7b-1 - 3 2b-2 - 7b-1 + 3= 0 2 ( ) - 7( b ) + 3= 0 2 b-1 -1 2u 2 - 7u + 3= 0 ¨ Replace b-1 with u ( 2u -1)( u - 3) = 0 2u-1= 0 or u- 3= 0 u= 1 u=3 2 b-1 = 1 b-1 = 3 ¨ Replace u with b-1 The solutions are 2 and 1 . 2 3 b= 2 b= 1 3 69. x -2 + 9x -1 = 10 x -2 + 9x -1 -10= 0 -1 2 ( x ) + 9( x ) -10= 0 -1 u 2 + 9u -10= 0 ¨ Replace x -1 with u ( u+10)( u -1) = 0 u+10 = 0 or u -1= 0 u = -10 u =1 1 -1 The solutions are - and 1. x =1 ¨ Replace u with x -1 -1 x = -10 10 x =- 1 x =1 10 71. x -2 = 4x -1 +12 x -2 - 4x -1 -12 = 0 -1 2 (x ) ( ) - 4 x -1 -12 = 0 u 2 - 4u-12 = 0 ¨ Replace x -1 with u ( u+ 2)( u - 6) = 0 u+ 2 = 0 or u - 6= 0 u = -2 u=6 -1† The solutions are - 1 and 1 . x = 6 ¨ Replace u with x -1 -1 x = -2 2 6 x =-1 x= 1 2 6 447 Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra u+ 2 = 0 or u - 6= 0 u = -2 u=6 -1 The solutions are - 1 and 1 . x = 6 ¨ Replace u with x -1 -1 x = -2 2 6 x =-1 x= 1 2 6 73. x 2/3 - 3x1/3 = -2 2 (x ) 1/3 - 3x1/3 + 2= 0 u 2 - 3u + 2= 0 ¨ Replace x1/3 with u ( u -1)( u - 2) = 0 u-1= 0 or u - 2= 0 u =1 u=2 1/3 x = 2 ¨ Replace u with x 1/3 1/3 x =1 The solutions are 1 and 8. x = 13 x = 23 =1 =8 75. b 2/3 +11b1/3 + 28= 0 2 ( ) b1/3 +11b1/3 + 28= 0 u 2 +11u + 28= 0 ¨ Replace b1/3 with u ( u + 7)( u+ 4) = 0 u+ 7 = 0 or u+ 4 = 0 u = -7 or u = -4 † b1/3 = -7 or b1/3 = -4 ¨ Replace u with b1/3 3 3 b = (-7) or b= (-4) = -343 = -64 The solutions are –343 and –64. 77. -2a - 5a1/2 + 3 = 0 2 -2 ( ) a1/ 2 - 5a1/2 + 3 = 0 -2u 2 - 5u + 3 = 0 ¨ Replace a1/2 with u 2u 2 + 5u - 3 = 0 ( 2u - 1)(u + 3) = 0 2u-1= 0 or u+ 3= 0 u= 1 u = -3 2 1/2 a =2 1/2 a = -3 ¨ Replace u with a1/2 2 a= 1 =1 2 () 4 1 1 a1/ 2 = -3 has no solution since there is no value of a for which a 2 = -3. The solution is . 4 448 SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions 79. c 2/5 - 3c1/5 + 2= 0 2 (c ) 1/5 - 3c1/5 + 2= 0 u 2 - 3u + 2= 0 ¨ Replace c1/5 with u ( u - 2)( u-1) = 0 u- 2 = 0 u -1= 0 u =2 u =1 1/5 c =1 ¨ Replace u with c 1/5 1/5 c =2 c = 25 c =15 = 32 =1 The solutions are 32 and 1. 81. f (x) = x - 5 x + 4 , f (x) = 0 2 ( ) 0 = x1 /2 - 5x 1/2 + 4 0 = u2 - 5u + 4 ¨ Replace x1 /2 with u 0 = (u - 1)(u - 4) u - 1 = 0 or u - 4 = 0 u =1 u=4 x 1/2 = 1 x 1/2 = 4 ¨ Replace u with x 1/2 x =1 x = 16 The x-intercepts are (1, 0) and (16, 0). 83. h( x ) = x +13 x + 36, h ( x ) = 0 2 ( ) 0 = x 1/2 +13x1/2 + 36 0 = u 2 +13u+ 36 ¨ Replace x 1/2 with u 0 = ( u+ 9)( u+ 4 ) u+ 9 = 0 or u+ 4 = 0 u = -9 u = -4 1/2 x = -9 1/2 x = -4 ¨ Replace u with x 1/2 There are no values of x for which x 1/2 = -9 or x1/2 = -4 . There are no x-intercepts. 85. p( x) = 4x -2 -19x -1 - 5 , p( x) = 0 2 ( ) 0 = 4 x -1 - 19x -1 - 5 0 = 4u 2 - 19u - 5 ¨ Replace x -1 with u 0 = (4u +1)(u - 5) 4u +1 = 0 or u - 5 = 0 1 u=- u=5 4 1 x -1 = - x -1 = 5 ¨ Replace u with x -1 4 449 1 x = -4 x= 5 Ê1 ˆ The x-intercepts are (–4, 0) and Á , 0˜ . Ë5 ¯ Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra 4u +1 = 0 or u - 5 = 0 1 u=- u=5 4 1 x -1 = - x -1 = 5 ¨ Replace u with x -1 4 1 x = -4 x= 5 Ê1 ˆ The x-intercepts are (–4, 0) and Á , 0˜ . Ë5 ¯ 87. f ( x ) = x 2/3 + x 1/3 - 6, f ( x ) = 0 2 ( ) 0 = x1 /3 + x 1/3 - 6 0 = u2 + u - 6 ¨ Replace x1 /3 with u 0 = (u + 3)(u - 2) u+3=0 or u - 2 = 0 u = -3 u=2 x1/ 3 = -3 x1 /3 = 2 ¨ Replace u with x 1/3 x = -27 x=8 The x-intercepts are (–27, 0) and (8, 0). 89. 2 ( g( x ) = x 2 - 3x ) ( ) + 2 x 2 - 3x - 24 , g( x ) = 0 0= u2 + 2u - 24 ¨ Replace x 2 - 3x with u 0 = (u + 6)(u - 4) u + 6 = 0 or u-4 =0 2 - 3x + 6 = 0 2 - 3x - 4 = 0 ¨ Replace u with x 2 - 3x x x ( x - 4)( x +1) = 0 x - 4 = 0 or x + 1 = 0 x =4 x = -1 There are no x-intercepts for x 2 - 3x + 6 = 0 since b 2 - 4ac = ( -3)2 - 4(1)( 6) = 9 - 24 = -15. The x-intercepts are (4, 0) or (–1, 0). 91. f ( x ) = x 4 - 20x + 64, f ( x ) = 0 2 ( ) 0 = x 2 - 20x 2 + 64 0 = u 2 - 20u + 64 ¨ Replace x 2 with u 0 = ( u-16)( u- 4 ) u-16 = 0 or u- 4 = 0 u =16 u=4 2 x =16 x 2 = 4 ¨ Replace u with x 2 x = ± 16 = ±4 x = ± 4 = ±2 The x-intercepts are ( 4, 0), (–4, 0), ( 2, 0), and (–2, 0). 450 SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions The x-intercepts are ( 4, 0), (–4, 0), ( 2, 0), and (–2, 0). 93. When solving an equation of the form ax 4 + bx 2 + 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 = x 2 - 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)( x - 3) 2 2 0= x 4 - 5x 2 + 6 101. No. An equation of the form ax 4 + bx 2 + c = 0 can have no imaginary solutions, two imaginary solutions, or four imaginary solutions. 3 3 103. a. - = 60 The LCD is x 2 x2 x Ê 3 ˆ 3 x 2 Á 2 ˜ - x 2 Ê ˆ = x 2 (60) Á ˜ Ëx ¯ Ë x¯ 3 - 3x = 60x 2 0 = 60x 2 + 3x - 3 ( 0 = 3 20x 2 + x - 1 ) 0 = 3( 5x - 1)( 4x +1) 5x -1 = 0 or 4x + 1 = 0 1 1 x= x =- 5 4 1 1 The solutions are and - 5 4 3 3 b. - = 60 x2 x 3x -2 - 3x -1 = 60 2 † ( ) 3 x -1 - 3x -1 - 60 = 0 3u2 - 3u - 60 = 0 ¨ Replace x -1 with u ( ) 3 u2 - u - 20 = 0 3(u - 5)(u + 4) = 0 451 Chapter 12: Quadratic Functions SSM: Elementary and Intermediate Algebra u-5=0 or u + 4 = 0 u=5 u = -4 -1 = 5 -1 = -4 ¨ Replace u with x -1 x x 1 1 x= x =- 5 4 1 1 The solutions are and - . 5 4 8 112. 15( r + 2 ) + 22 = - r +2 8 15( r + 2 )(r + 2) + 22( r + 2) = - (r + 2 ) r+2 15( r + 2)2 + 22( r + 2) = -8 15(r + 2 )2 + 22(r + 2 ) + 8 = 0 15u 2 + 22u + 8 = 0 ¨ Replace r + 2 with u (5u + 4)(3u + 2) = 0 5u + 4 = 0 or 3u + 2 = 0 4 2 u= - u= - 5 3 4 2 r+2= - r + 2 = - ¨ Replace u with r + 2 5 3 14 8 r= - r=- 5 3 14 8 The solutions are - and - . 5 3 107. 4 - ( x - 2 )-1 = 3( x - 2) -2 4 - u -1 = 3u -2 ¨ Replace x - 2 by u 1 3 4- = 2 u u 2Ê 1ˆ Ê 3ˆ u Á4 - ˜ = u2Á 2 ˜ Ë u¯ Ëu ¯ 2 -u=3 4u 2 - u -3 = 0 4u ( 4u + 3)(u -1) = 0 4u + 3 = 0 or u - 1 = 0 3 u=- u =1 4 3 x -2 = - x - 2 = 1 ¨ Replace u by x - 2 4 5 x= x =3 4 5 The solutions are and 3. 4 452 SSM: Elementary and Intermediate Algebra Chapter 12: Quadratic Functions 109. x 6 - 9x 3 + 8 = 0 2 (x3 ) - 9x 3 + 8 = 0 u2 - 9u + 8 = 0 ¨ Replace x 3 with u (u - 8)( u - 1) = 0 u - 8 = 0 or u -1 = 0 u=8 u =1 x3 = 8 x 3 = 1 ¨ Replace u with x 3 x =2 x =1 The solutions are 2 and 1. 2 ( 111. x 2 + 2x - 2 ) ( ) - 7 x 2 + 2x - 2 + 6 = 0 - 7u + 6 = 0 ¨ Replace x 2 + 2x - 2 with u u2 (u - 6)(u -1) = 0 u-6=0 or u - 1= 0 u=6 u=1 2 + 2x - 2 = 6 2 + 2x - 2 = 1 ¨ Replace u with x 2 + 2x - 2 x x 2 + 2x - 8 = 0 x x 2 + 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=1 The solutions are –4, 2, –3, and 1. 113. 2n4 - 6n2 - 3 = 0 115. ( ) 4- 3-2 =4 - 9 - 8 ( ) 2 5 4 3 5 12 12 2( )n2 - 6n2 - 3 = 0 2u 2 - 6u - 3 = 0 ¨ Replace n2 with u = 4- 1 5 12 ( ) 48 - 5 6± (-6)2 - 4( 2)( -3) = 60 60 u= 2( 2) = 43 6 ± 60 60 = 4 116. 6( x + 4) - 4 (3x + 3) = 6 6 ± 2 15 = 6x + 24 -12x -12 = 6 4 -6x +12 = 6 3 ± 15 = -6x = -6 2 3 ± 15 x =1 n2 = ¨ Replace u with n2 2 D : all real numbers 117. 3 ± 15 n= ± R: {y y ≥ 0} 2 453