For Thought. 3.1 Exercises 142 CHAPTER 3 POLYNOMIAL AND RATIONAL FUNCTIONS. 1. False, the range of y = x 2 is [0, ).

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1 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS For Thought. False, the range of = is [0, ).. False, the verte is the point (, ) True. True 5. True, since b a = 6 =. 6. True, the -intercept of = ( + ) is the verte ( /, 0) and the -intercept is (0, ). 7. True 8. True, since ( ) is alwas nonnegative. 9. True, since if and p are the length and the width, respectivel, of a rectangle with perimeter p, then the area is = p. This is a parabola opening down with verte ( ) p, p. Thus, the maimum area is p False. Eercises. upward. downward. verte. verte - 0. Completing the square, we get. = - -9 = = ( ( ) ) ( ) = ( ) 9. ( 6 ) ( + 9 ) 9 ( = ) 9 5. minimum 6. maimum 7. ais of smmetr 8. -intercept. = -9/.5 ( ) 5 ( = + 5 ) 5 9. Completing the square, we get = = ( ( ) + + ) ( ) + + = ( + ). ( ) Copright 0 Pearson Education, Inc.

2 . QUADRATIC FUNCTIONS AND INEQUALITIES. Completing the square, we get ( ( ) 6 = 6 + ) = ( ) +. ( ) Completing the square, we find ( ( ) ( ) = ) = ( + ) Completing the square, we get ( ( ) = + ) = ( ). ( ) + 7. = ( ) ( = + ) + 8. = ( + ) ( + = ) Completing the square, we find ( ( ) = + ) = ( ). ( ) ( 9. = + 9 ) = ( ) Copright 0 Pearson Education, Inc.

3 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS ( 0. = + + ) ( 9 + = + ) +. Since it opens up with verte (, ), the range is [, ), minimum value is, decreasing on (, ), and increasing on (, ). -/ /. Since b a = = and f() = + = 6, the verte is (, ).. Since b a = 8 = and f( ) = = 7, the verte is (, 7).. Verte: (, ). Verte: ( 6, /) 5. Since b a = / = and f( /) = = ( 8, the verte is, ) Since b a = / = and f( ) = / / / = 5/, the verte is (, 5/). 7. Up, verte (, ), ais of smmetr =, range [, ), minimum value, decreasing on (, ), inreasing on (, ). 8. Down, verte (, ), ais of smmetr =, range (, ], maimum value, increasing on (, ), decreasing on (, ). 9. Since it opens down with verte (0, ), the range is (, ], maimum value is, decreasing on (0, ), and increasing on (, 0). 0. Since it opens down with verte (0, 5), range is (, 5], maimum value is 5, decreasing on (0, ), and increasing on (, 0).. Since it opens up with verte (, ), the range is [, ), minimum value is, decreasing on (, ), and increasing on (, ).. Since it opens up with verte (, 8), range is [ 8, ), minimum value is 8, decreasing on (, ), and increasing on (, ).. Since it opens up with verte (, ), range is [, ), minimum value is, decreasing on (, ), and increasing on (, ). 5. Since it opens up with verte is (, ), the range is [, ), minimum value is, decreasing on (, ), and increasing on (, ). 6. Since it opens down with verte is ( 6, 7), range is (, 7], maimum value is 7, decreasing on ( 6, ), and increasing on (, 6). 7. Since it opens down with verte (/, 7/), the range is (, 7/], maimum value is 7/, decreasing on (/, ), and increasing on (, /). 8. Since it opens down with verte is ( /, /), the range is (, /], maimum value is /, decreasing on ( /, ), and increasing on (, /). 9. Since it opens down with verte is (/, 9), the range is (, 9], maimum value is 9, decreasing on (/, ), and increasing on (, /). 0. Since it opens up with verte is (/, 6), range is [ 6, ), minimum value is 6, decreasing on (, /), and increasing on (/, ).. Verte (0, ), ais = 0, -intercept (0, ), -intercepts (±, 0), opening up Copright 0 Pearson Education, Inc.

4 . QUADRATIC FUNCTIONS AND INEQUALITIES 5-6. Verte (, 9), ais =, -intercept (0, 0), -intercepts (0, 0), (6, 0), opening up Verte (0, 8), ais = 0, -intercept (0, 8), -intercepts (±, 0), opening down Verte (, ), ais =, -intercept (0, 5), -intercepts (, 0), (5, 0), opening up 5 -. Verte (/, /), ais = /, -intercept (0, 0), -intercepts (0, 0), (, 0), opening up Verte (, 9), ais =, -intercept (0, 8), -intercepts (, 0), (, 0), opening up Verte (, ), ais =, -intercept (0, 0), -intercepts (0, 0), (, 0), opening down 9. Verte (, ), ais =, -intercept (0, 0), -intercepts (0, 0), (, 0), opening down 5 5. Verte (, 0), ais =, -intercept (0, 9), -intercept (, 0), opening up Verte (, 8), ais =, -intercept (0, 0), -intercepts ( 5, 0), (, 0), opening down Copright 0 Pearson Education, Inc.

5 6 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 5. Verte (, ), ( ais =, ) -intercept (0, ), 6 -intercepts ±, 0, opening down 5. Verte (, 5), ais =, -intercept (0, 6), no -intercepts, opening down The -intercepts are =,. Since the parabola opens upward, the solution set for 0 is (, ) (, ). 5. The -intercepts are =,. Since the parabola opens upward, the solution set for + 0 is (, ] [, ). 55. Since + = ( ), the -intercepts are = ±. Since the parabola opens upward, the solution set for + 0 is (, + ). 56. Since = ( ), the -intercepts are = ±. Since the parabola opens upward, the solution set for 0 is [, + ]. 57. Since 6 = ( + )( ), the - intercepts are = /, /. Then the solution set for + 6 or 6 0 is (, /) (/, ). 58. Since 5 6 = ( + )(5 6), the - intercepts are =, 6/5. Then the solution set for or is (, 6/5). 59. (, ] [, ) 60. (, ) 6. (, ) 6. (, ] [, ) 6. [, ] 6. [, ] 65. The roots of + = 0 are = and = +. If = 5, then ( 5) ( 5) + 0. If =, then () () + 0. If = 5, then (5) (5) The solution set of + 0 is (, + ) and its graph follows. ( + ) 66. The roots of + = 0 are = and = +. If = 5, then ( 5) ( 5) + 0. If =, then () () + 0. If = 5, then (5) (5) The solution set of + 0 is (, + ) and its graph is [ 67. The roots of 0 = 0 are = 0 and = 0. If =, then ( ) 0 0. If = 0, then (0) 0 0. If =, then () The solution set of 9 is (, 0] [ 0, ) and its 0 0 graph is ] [ + ] Copright 0 Pearson Education, Inc.

6 . QUADRATIC FUNCTIONS AND INEQUALITIES The roots of 7 = 0 are = 7 and = 7. If = 5, then ( 5) 7 0. If = 0, then (0) 7 0. If = 5, then (5) The solution set of 7 0 is (, 7) ( 7, ) and its 7 7 graph is ) ( 69. The roots of = 0 are = 5 7 and = If =, then () 0() If = 5, then (5) 0(5) If = 8, then (8) 0(8) The solution set of is (, 5 7) (5 + 7, ) and its graph is 5 7 ) ( 70. The roots of 6 + = 0 are = 6 and = + 6. If = 0, then (0) 6(0) + 0. If =, then () 6() + 0. If = 9, then (9) 6(9) The solution set of is (, 6] [ + 6, ) and its graph is 6 ] + 6 [ 7. Note, p + 9 = 0 has no real roots. If p = 0, then (0) The signs of p + 9 are shown below. + 0 The solution set of p is (, ) and its graph follows. 7. Note, 5 s = 0 has no real roots. If p = 0, then 5 (0) 0. The signs of 5 s are shown below. 0 The solution set of 5 s 0 is (, ) and its graph follows. 7. Note, a 8a + 0 = 0 has no real roots. If a = 0, then (0) 8(0) The signs of a 8a + 0 are shown below. + 0 The solution set of a 8a is. 7. Note, t 6t + 5 = 0 has no real roots. If t = 0, then (0) 6(0) The signs of t 6t + 5 are shown below. + 0 The solution set of t 6t is (, ) and its graph follows. 75. Note, w 5w + 6 = 0 has no real roots. If w = 0, then (0) 5(0) The signs of w 5w + 6 are shown below. + 0 The solution set of w 5w+6 0 is (, ) and its graph follows. Copright 0 Pearson Education, Inc.

7 8 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 76. Note, z + z + 5 = 0 has no real roots. If z = 0, then (0) + (0) The signs of z + z + 5 are shown below. + 0 The solution set of z + z is the empt set. 77. The zeros of f() = ( )( + ) are = and = /. If =, then f( ) 0. If = 0, then f(0) 0. If =, then f() / The solution set is (, /) 78. The zeros of f() = ( + )( ) are = / and =. If =, then f( ) 0. If = 0, then f(0) 0. If =, then f() The solution set is [ /, ] 79. The zeros of f() = 5 = (+)( 5) are = and = 5. If =, then f( ) 0. If = 0, then f(0) 0. If = 6, then f(6) The solution set is (, ) (5, ). 80. The zeros of f() = 5+ = ( )( ) are = and =. If = 0, then f(0) 0. If =, then f() 0. If = 5, then f(5) The solution set is (, ) (, ). 8. The zeros of f(w) = w w = (w + )(w 6) are w = and w = 6. If w =, then f( ) 0. If w = 0, then f(0) 0. If w = 7, then f(7) The solution set is (, ] [6, ). 8. The zeros of f() = = ( + 5)( + ) are = 5 and =. If = 6, then f( 6) 0. If =, then f( ) 0. If = 0, then f(0) The solution set is [ 5, ]. 8. The zeros of f(t) = t 6 = (t + )(t ) are t = and t =. If t = 5, then f( 5) 0. If t = 0, then f(0) 0. If t = 5, then f(5) The solution set is [, ]. 8. The zeros of f(h) = h 6 = (h + 6)(h 6) are h = 6 and h = 6. If h = 7, then f( 7) 0. If h = 0, then f(0) 0. If h = 7, then f(7) Copright 0 Pearson Education, Inc.

8 . QUADRATIC FUNCTIONS AND INEQUALITIES 9 The solution set is (, 6] [6, ). 85. The zero of f(a) = (a + ) is a =. If a =, then f( ) 0. If a = 0, then f(0) The solution set is { }. 86. The zero of f(c) = c c + = (c ) is c =. If c = 0, then f(0) 0. If c =, then f() The solution set is {}. 87. The zero of f(z) = (z ) is z = /. If z = 0, then f(0) 0. If z =, then f() / The solution set is (, /) (/, ). c) If =, then ( ) ( ) 0 0. If = 0, then (0) (0) 0 0. If = 6, then (6) (6) 0 0. The signs of 0 are shown below The solution set of 0 0. is (, ) (5, ) d) Using the sign graph of 0 given in part c), the solution set of 0 0 is [, 5]. e) B using the method of completing the square, one obtains 0 = 0 = ( ) 0 9 ( ) 9. The graph of f is obtained from the graph of = b shifting to the right b unit, and down b 9. [ f) Domain is (, ), range is 9 ),, minimum -value is The zero of f(s) = (s + ) is s = /. If s =, then f( ) 0. If s = 0, then f(0) / 0 The solution set is (, ). 89. a) Since 0 = ( 5)( + ) = 0, the solution set is {, 5}. b) Since 0 = 0, we get = 0 or ( ) = 0. The solution set is {0, } g) The solution to f() 0 ma be obtained b considering the part of the parabola that is above the -ais, i.e., when lies in (, ) (5, ). While, the solution to f() 0 ma be obtained b considering the part of the parabola on or below the -ais. i.e., when is in [, 5]. Copright 0 Pearson Education, Inc.

9 50 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS h) -intercepts are (5, 0) and (, 0), the -intercept is (0, 0), ais of smmetr = ( ), verte, 9, opens up, ( ) increasing on,, decreasing on (, ) 90. a) B using the method of completing the square we get ( ) + = ( ) + + = ( ) +. If ( ) + = 0, then = ( ) and b the square root propert we have ± =. The solution set is { ± }. b) If + + = 0, we get = 0. B using the method of completing the square we get ( ) = 0 or ( ) =. Then the solution set is { ± } or { ± }. c) Let = + and =. If =, then ( ) + ( ) + 0. If = 0, then (0) + (0) + 0. If =, then () + () + 0. The signs of ++ are shown below The solution set of is (, + ). d) Using the sign graph of + + given in part c), the solution set of is (, ] [ +, ). e) From the solution in part a), we obtained f() = ( ) +. The graph of f is obtained from the graph of = b shifting to the right b unit, reflecting about the -ais, and shifting up b units. f) Domain is (, ), range is (, ], maimum -value is - g) The solution to f() 0 ma be obtained b considering the part of the parabola above the -ais, i.e., when lies in (, + ). While, the solution to f() 0 ma be obtained b considering the part of the parabola on or below the -ais, i.e., when is in (, ] [ +, ). h) -intercepts are (±, 0), the -intercept is (0, ), ais of smmetr =, verte (, ), opens down, increasing on (, ), decreasing on (, ) 9. Since b a = height is h() = 6 ft. 8 =, the maimum ( 6) 9. Since the verte of h(t) = 6t + 6t + 6 is (, 70), the maimum height is 70 ft. 9. a) Finding the verte involves the number b a = 60 = 5. Thus, the maimum height is h(5) = 6(5) + 60(5) + 8 = 08 ft. b) When the arrow reaches the ground, one has h(t) = 0. 6t + 60t + 8 = 0 t 0t = (t 5) = t = 5 ± Copright 0 Pearson Education, Inc.

10 . QUADRATIC FUNCTIONS AND INEQUALITIES 5 Since t 0, the arrow reaches the ground in = sec. 9. a) In finding the verte, one uses the number b a = 56 = 8. Thus, the maimum height is h(8) = 6(8) + 56(8) = 0 ft. b) When the RPG reaches the ground, one has h(t) = 0. 6t + 56t = 0 6t(t 6) = 0 t = 0, 6 Thus, the RPG reaches the ground after 6 sec. 95. a) About 00 mph b) The value of A that would maimize M is A = b a = mph c) Lindbergh fling at 97 mph would use lbs. of fuel or lbs. gallons per hour. 6. lbs per gal 96. (a) Since b a =.06, the average size (0.7) reached a minimum in 998 (= 99+). (b) = = 0 Appling the quadratic formula to solve for, we obtain =.06 ± (0.7) 6, Thus, and the average size will be 70 acres in the ear 008 (= 99 + ). (c) From the figure, we find that the average size was increasing during 99-98, and decreasing in (d) = = 0 Appling the quadratic formula to solve for, we obtain =.06 ± (0.7).7, 5.80 Since the parabola is opening upward, the average size is less than 5 acres when , i.e., during the ears Let and be the length and width, respectivel. Since + = 00, we find = 00. The area as a function of is f() = (00 ) = 00. The graph of f is a parabola and its verte is (50, 500). Thus, the maimum area is 500 d. Using = 50 from the verte, we get = 00 = = 50. The dimensions are 50 d b 50 d. 98. Let and be the length and width, respectivel of the rectangular mirror. Since there are 0 ft of frame molding, we have + = 0 or = 5. Then the area of the rectangle is A = = (5 ) = + 5. The graph of A as a function of is a parabola opening down. The -coordinate of the verte is = b a = 5 =.5. Note, = 5 = 5.5 =.5. Since =.5 and =.5 are the length and width, the dimensions of the mirror that maimize the area are.5 ft-b-.5 ft. Copright 0 Pearson Education, Inc.

11 5 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 99. Let the length of the sides be,,,,. 0 Then + = 0 and =. The area of rectangular enclosure is ( ) 0 A() = = ( = (0 ). This is a parabola opening down. Since b 0 (0) = 0 and = = 0, the a optimal dimensions are 0 ft b 0 ft. 00. Let and be the overall length and overall width, respectivel of the large rectangle. Note, there are four separate pens and consequentl there are three sides with length and three sides with length. Since there are 00 ft of fencing, we obtain + = 00 or = 00. Then the area of the large rectangle is ( ) 00 A = = The graph of A as a function of is a parabola opening down. The -coordinate of the verte is = b a = 00/ = 00. Note, = 00 = = 00. Since and are the length and width of the large rectangle, the overall dimensions that maimize the total area are 00/ ft-b-00/ ft. 0. Let the length of the sides be,, and 0. The area of rectangular enclosure is A() = (0 ) = 0. We have a parabola opening down. Since b = 7.5 and = 0 (7.5) = 5, the a optimal dimensions are 5 ft b 7.5 ft. 0. Let and be the number of eight-foot railroad ties to be used for the length and width, respectivel. Since + =, we get = 7. The area as a function of is f() = 6(7 ). The permissible values of are,,,, 5, 6. Note, the maimum of f() for these permissible values is 768 and this is obtained when = or =. Hence, the dimensions that maimizes area is ft. b ft. 0. Let be the length of a folded side. The area of the cross-section is A = (0 ). This is a parabola opening down with b/a =.5. The dimensions of the cross-section are.5 in. high and 5 in. wide. 0. Let and l be the height and length of the bo in ft, respectivel. Since the board is ft long, + l = and so l = ( ). The volume of the cage in ft is = l = ( ) = 6. This is a parabola opening down where = b a = 6 =. Since l = ( ) =, the dimensions are ft b ft b ft high. 05. Let n and p be the number of persons and the price of a tour per person, respectivel. a) The function epressing p as a function of n is p = 50 n. b) The revenue is R = (50 n)n or R = 50n n. c) Since the graph of R is a parabola opening down with verte (5, 65), we find that 5 persons will give her the maimum revenue of $ Let n and p be the number of tickets and price per ticket, respectivel. a) Simplifing n = (p 0), we find n = p. b) The total revenue R as a function of p is the quadratic function R = ( p) p or R = 500p +, 000p. Copright 0 Pearson Education, Inc.

12 . QUADRATIC FUNCTIONS AND INEQUALITIES 5 c) Since b =, the ticket price of $ will a maimize revenue. 07. Since v = 50p 50p is a parabola opening down, to maimize v choose p = b a = = /. 08. Let be the number of people who have been infected b the flu virus and let be the dail rate of infection. Then = k(5000 ) = 5000k k is a parabola opening down where k is the constant of proportion. The rate of infection is at a maimum when = b a = 5000k k = (a) A graph of atmospheric pressure versus altitude is given below. -. a,70 (b) From the graph, one finds the atmospheric pressure is decreasing as the went to h = 9, 09 feet. (c) Since b a = ( , 70 ) feet, the function is decreasing on the interval [0,, 70] and increasing on [, 70, ). (d) No, it does not make sense to speak of atmospheric pressure at heights that lie in (, 70, ) since it is higher than the summit of 9,09 ft. (e) It is valid for altitudes less than 0, 000 feet which is less than the height of the summit. h 0. a) Since R is a parabola opening down with -intercepts at (0, 0) and (000, 0), R() 0 for in (0, 000). b) Since b a = 000 = 500, R() is ( ) increasing on [0, 500] and decreasing on [500, ). c) From part b), we conclude the number of items that maimizes revenue is 500 items. d) From part b), since b = 500 it follows a that the maimum revenue is R(500) = $500, 000. e) The marginal revenue is given b MR() = (000( + ) ( + ) ) (000 ) = 998. f) Note, 998 = MR() 0 if lies in (0, 99.5) and MR() 0 if lies in (99.5, ).. (a) Using a graphing calculator, we find that the equation of the regression line is = , The equation of the quadratic regression curve is = , 65.6 (b) Judging from the graph, it is too close to tell which function seems more reasonable. [The green graph is the line, and the pink graph is the parabola.] The quadratic regression curve seems more reasonable since it is closer to the value of a 9-ear-old car Copright 0 Pearson Education, Inc.

13 5 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS (c) We substitute = into the regression equations in part (a). Using the linear function, the price of an eleven ear old car is $995. Using the quadratic curve, we obtain $,5.. (a) Let be the number of ears since 99. Using a graphing calculator, we obtain the equation of the quadratic regression is = (b) Yes, the quadratic function appears to be a good fit as shown below No, since a parabola opening upward does not satisf the Horizontal Line Test. 5. Since f(9) = (9) 9 = 9 and g(9) = 9 =, we obtain (f + g)(9) = 9 + =. 6. = ( + ) Domain {,,, }, range {, 5} 8. Since 0, the domain is [, ) and the range of is (, ]. Thinking Outside the Bo XXXI D (c) Using the quadratic equation in part (a), we obtain b a 0.75 (0.007) 9 In the ear 00 (= ), the number of homicides was a minimum. (d) Substitute = 0 into the quadratic equation in part (a): 0 = Using the quadratic formula, we find 0 or.5. Since 0, there will be 0 homicides per 00,000 in 0 (= ).. Let = k/. If = and =, we find k = 6. Thus, if = 0 then = k = 6 0 = 9 0. The vertices of the quadrilateral where the triangle and square overlap are A(8/0, 8/0), B(8/0, 9/0), C(/0, 9/0), and D(96/0, /0). Subdividing the quadrilateral into two triangles and a rectangle, we find that the area of the quadrilateral is /.. Pop Quiz A B. Completing the square, we get ( ( ) 8 = ). (, 8) = ( + ). C ( ) 8. Since the verte is (, ) and the parabola opens downward, the range is (, ].. Since the verte is (.5,.5) and the parabola opens upward, the minimum -value is Since = ( )( + ), the -intercepts are (, 0) and (, 0). Note, the line = is equidistant from the -intercepts. Thus, = is the ais of smmetr. Copright 0 Pearson Education, Inc.

14 . QUADRATIC FUNCTIONS AND INEQUALITIES The zeros of f() = ( + ) are = and = 0. If = 5, then f( 5) 0. If =, then f( ) 0. If =, then f() The solution set is (, 0).. Linking Concepts a) Let C represent the rectangular center section. Since the length around the track is mile, we get + π =. The area of C is ( ) = which can be rewritten as π f() = π ( ). The graph of f is a parabola opening down whose verte has -coordinate =. Then = π = (/) π = π. The dimensions that will maimize the area of C are = mile and = π mile. b) Rounded to the nearest tenth of a foot, we get = (580) = 0 ft and = (580) 80. ft. π A sketch of the track is given. c) The total area enclosed b the track is ( ) + π. Since = as shown π in part a), the total area is ( ) ( ) T () = + π. π π This simplifies to T () = π ( ). Note, the graph of T is a parabola and the -coordinate of the verte is = 0. Then = π = (0) π = π. Thus, the dimensions that maimize the total area are = 0 and = π mile. ( d) Note, + π + 60 ) =. Solving for, 580 we obtain = π/. π The area of the rectangular center section is A c () = = (88 + π ). π The graph of A c is a parabola opening down whose -coordinate of the verte is = π mile 76 = π (580) ft 76 = 0 0π ft 5.8 ft. Substituting into = we obtain π/, π = π 88 mile = ft π 780. ft. Copright 0 Pearson Education, Inc.

15 56 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS Net, the total area of the track is Since = as ( T () = + π + 60 ). 580 π/, we can rewrite T () π T () = + π. π Note, the graph of T is a parabola opening down with verte in the second quadrant. Since 0, the maimum value of T occurs when = 0. π/ Substituting = 0 into =, π we obtain = π π mile = π (580) ft π = ft π ft. A sketch of the track when the mile is measured on the outside edge is shown below. For Thought. False. True, b the Remainder Theorem.. True, b the Factor Theorem.. True, since 5 =. 5. False, since P () = has no zero. 6. False, rather c c + c 5 = b. 7. False, since P () = True 9. True, since is a root. 0. False, since is not a root.. Eercises. zero. remainder. zero, factor. zero, comple numbers 5. Quotient, remainder ) Quotient +, remainder + ) Quotient + 6, remainder ) Quotient w 7w, remainder 70 w 7w w ) w + 5w + 0w 7 w + w 7w + 0w 7w + w w 7 w Copright 0 Pearson Education, Inc.

16 . ZEROS OF POLYNOMIAL FUNCTIONS Quotient s +, remainder 6 s + s 5 )s s + 6 s 5s s + 6 s Quotient h + h +, remainder 0h + h + h + h )h + h + 0h + h 5 h + 0h h h + h + h h + 0h 9h h + 0h 5 h + 0h 9 0h +. Quotient + 6, remainder 6. Quotient + 7, remainder Quotient + 6, remainder Quotient + 8, remainder Quotient +, remainder 0 / Quotient , remainder 7 / Quotient a a + 6, remainder 0 -/ Quotient b, remainder 0 -/ Quotient + + +, remainder Quotient , remainder Quotient +, remainder Copright 0 Pearson Education, Inc.

17 58 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS. Quotient , remainder f() = f( ) = f( ) = f() = g() = g( ) = g( /) = 55/8 -/ / 9/ -9/8-9/ 9/ 55/8 0. g(/) = 57/8 / / -7/ -7/8-7/ -7/ 57/8. h( ) = h() = h() = h( ) = Yes, ( + )( + ) = ( + )( + )( ) Copright 0 Pearson Education, Inc.

18 . ZEROS OF POLYNOMIAL FUNCTIONS Yes, (+5)( + ) = (+5)(+)( ) Yes, ( )( +8+5) = ( )(+5)(+) Yes, ( )( 0+) = ( )( 6)( ) Yes, since the remainder below is zero No, since the remainder below is not zero No, since the remainder below is not zero Yes, since the remainder below is zero Yes, since the remainder below is zero No, since the remainder below is not zero No, since the remainder below is not zero / -5 7 / 7/ -/8 7/ -/ /8 6. Yes, since the remainder below is zero -/ ±{,,,, 6, 8,, } 8. ±{,,, 6} 9. ±{,, 5, 5} 50. ±{, } { 5. ±,, 5, 5,,, 8,,, 8, 5, 5, 5 8, 5, 5, 5 } 8 { 5. ±,,,, 6, 9, 8,, } 9 { 5. ±,,,, 6, 9, 8,, } 9 { 5. ±,,, 5, 5, 5 } 55. Zeros are,, since Copright 0 Pearson Education, Inc. and 7 + = ( )( )

19 60 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 56. Zeros are,, since = ( )( + ) 57. Zeros are, ± i since = ( ) + = 0 or = ±i 58. Zeros are, ± i since and + + = 0 implies = ± 59. Zeros are /, /, 5/ since / and 8a a + 0 = (a 6a + 5) = (a )(a 5) 60. Zeros are /, /, / since -/ and 8b 8b + = (9b 9b + ) = (b )(b ) 6. Zeros are, ± i since / and the zeros of 8t t + are (b the quadratic formula) ± i 6. Zeros are, ± i since -/ and the zeros of t t + 0 are (b the quadratic formula) ± i 6. Zeros are 6, ± i since and the zeros of are (b the quadratic formula) ± i 6. Zeros are, ± i since and the zeros of + are (b the quadratic formula) ± i 65. Zeros are,, ±i since Copright 0 Pearson Education, Inc.

20 . ZEROS OF POLYNOMIAL FUNCTIONS and the zeros of w + are ±i 66. Zeros are,, ±i since / and the zeros of v + 6 = (v + ) are ±i 67. Zeros are, ± since and the zeros of are ± 68. Zeros are, ± since and the zeros of are ± 69. Zeros are /, /, / since / and ( 7 + ) = ( )( ) 70. Zeros are /, /5, / since -/ and ( ) = (5 )( ) 7. Rational zero is /6 since / and b the quadratic formula has imaginar zeros ± i. 7. Rational zero is 7/5 since 7/ and b the quadratic formula has imaginar zeros ± i. 7. Rational zeros are 7/, 6/7 since 7/ / and + = 0 has imaginar zeros ±i. Copright 0 Pearson Education, Inc.

21 6 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 7. Rational zeros are /7, 8/7 since -/ / and has imaginar zeros ±i. 75. Dividing b, we find Moreover, = ( + 5)( + ). Note, the zeros of + 9 are ±i. Thus, the zeros of f() are = 5,,, ±i. 76. Dividing b, we find B using the method of completing the square, we find + = ( ). Then the zeros of ( ) are ±. Thus, all the zeros are =,, 5, ±. 78. Dividing b, we find For the quotient, we find 0 = ( 5)( + ). Then the zeros of are =,, 5. B using the method of completing the square, we find + = ( ) + 9. Then the zeros of ( ) + 9 are ± i. Thus, all the zeros are =,, 5, ± i. + = + 5 since Moreover, 8 = ( 6)( + ). Note, the zeros of 5 are ± 5. Then the zeros of f() are =,, 6, ± Dividing b, we find For the quotient, we find = ( 5)( ). Then the zeros of are =,, = + + since Copright 0 Pearson Education, Inc.

22 . ZEROS OF POLYNOMIAL FUNCTIONS a a + 5 a - 5 = a + 5 a since b b + b c c c = b b + since = + c c c )c c c + 0c c h + h h = + h h h )h + h h + 0h h t 5 t + = + 7 t + t + )t 5 t = + )6 6 since since since since 87. a) Note, P (t) t = t + t 58t The drug will be eliminated in t = 6 hr. b) About 0 ppm c) About hours d) Between and 5 hours approimatel, the concentration is above 80 ppm. Thus, the concentration is above 80 ppm for about hours. 88. The difference between the intended and the actual volume is or equivalentl 8 (7 )(6 ) Using snthetic division, one finds / Solving + 0 = 0, we find 0 ± 0 ( )( ) 8 5 ± =., 0.7. There are two possible values of, namel, =.5 in. or 0.7 in. for which there is no difference between the intended and actual volume. Note,. in. had to be ecluded since it is too long. 89. If w is the width, then w(w + )(w + 9) = 60. This can be re-written as w + w + 6w 60 = 0. Copright 0 Pearson Education, Inc.

23 6 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS Using snthetic division, we find Since w + 8w + 6 = 0 has non-real roots, the width of the HP bo is w = 5 in. The dimensions are 5 in. b 9 in. b in. 90. Let h, h +, and h + 5 be the height, width, and depth, respectivel. One can rewrite as h(h + )(h + 5) = 6 h + 8h + 0h 6 = 0. B appling snthetic division, one derives Since h + h + 7 = 0 has onl non-real zeros, h = 6. The dimensions of the bo are 6 in. b 6 in. b 7 in. 9. The quotient and remainder when + + is divided b are the same as the quotient and remainder, respectivel, when + + is divided b. Then one can use sthetic division for the latter case since the divisor is. 9. If c is a factor of P () then P () = Q() ( c) for some polnomial Q(). Thus, P (c) = Q(c) 0 = 0 and c is a zero of P (). 9. Using the method of completing the square, we find f() = ( ) + 9. Factoring, we have ( = ) ( = ) 8. + = ( )( ) Then the solution set is {/, }. 95. Since an absolute value is alwas nonnegative, the solution set of is (, ). 96. Solving for : 9 = 5 8 = 6 The solution set is {/7}. 97. a) 6a(a + a 0) = 6a(a 5)(a + ) b) ( 6) = ( + )( ) = ( + )( )( + ) 98. We find a contradiction as follows: The solution set is. Thinking Outside the Bo XXXII When a point on a circle with radius r is rotated through an angle of π/, the distance the point rotates is s = r π. The sum of the distances traveled b point A is π 5 + π π = ( 5 + 9)π. Copright 0 Pearson Education, Inc.

24 . THE THEORY OF EQUATIONS 65. Pop Quiz. Linking Concepts. The quotient is + and the remainder is 7 since (a) {[( + 5) + ] + } + = { } [ ] + + = { } = The quotient is + and the remainder is 9 since f() = 0 since the remainder is 0 as shown below 7 0. The factors of the constant in f() = +8 are p = ±, ±, ±, ±8. The factors of the leadig coefficient in f() are q = ±, ±. Thus, the list of possible rational zeros are ± {,,, 8, /}. 5. Divide f() = b. / The quotient is + 8 or ( + 9). The zeros of the quotient are = ±i. Thus, the zeros of f() are =, ±i. (b) (c) { } = { } [ ] 8 + = { } [(6 + 9) + 6] 8 + = {[((6 ) + 9) + 6] 8} + Using Horner s method we obtain P () = {[((6 ) + 9) + 6] 8} +. = 6 B substitution, we get P () = 6() 5 () + 9() + 6() 8() + = 6. (d) Fewer arithmetic operations were done in part (c) in finding P () b using Horner s method. (e) A polnomial can be re-written without using powers of higher than and this can be done b successivel factoring an out of all the non-constant terms. (f) The new form shows how to evaluate a polnomial b using onl the operations of addition and multiplication. These two operations are used in the same wa in snthetic division. For Thought. False, the multiplicit is.. True. True. False, it factors as ( 5) ( + ). Copright 0 Pearson Education, Inc. 5. False, rather +5i is also a solution. 6. True

25 66 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 7. False, since the are solutions to a polnomial with real coefficients of degree at least. 8. False, is not a solution. 9. True, since 5 6 = 0 has no sign changes. 0. True. Eercises. multiplicit. n. a bi. upper bound 5. Degree ; 5 with multiplicit since ( 5) = 0 6. Degree ; 9 with multiplicit since ( 9) = 0 7. Degree 5; 0 with multiplicit, and ± since ( )( + ) = 0 8. Degree 6; 0 with multiplicit, and ±i since ( i)( + i) = 0 9. Degree ; 0 and each have multiplicit since ( + ) = ( ) 0. Degree 5; 0 with multiplicit, with multiplicit since ( 6 + 9) = ( ). Degree ; / and / each with multiplicit. Degree 8; 0 and / each with multiplicit, and / with multiplicit since ( + ) ( ) = 0. Degree ; the roots are 0, ± 0 since ( 6) = (( ) 0) = 0. Degree ; the roots are 0, ± since ( 8 + ) = (( ) ) = [ ( ) ] [ ( ) + ] = ( ) = [ 8. ( ) + ] [ 5 ( ) ] 5 = ( ) 5 = [( ) i] [( ) + i] = ( ) + = [( ) + i] [( ) i] = ( ) + = ( ) [( ) i] [( ) + i] = ( ) [ ( ) + 6 ] = ( + ) ( ( i)) ( ( + i)) = ( + )( ( i) ( + i)+ ( i)( + i)) = ( ) ( + ) + ( i)( + i) = ( + )( + ) = +. ( + )( 5) = 0 or 5 = 0. ( 6)( + ) = 0 or 5 6 = 0 5. ( + i)( i) = 0 or + 6 = 0 6. ( + 9i)( 9i) = 0 or + 8 = 0 7. ( ( i))( (+i)) = 0 or 6+0 = 0 8. ( (+i))( ( i)) = 0 or 8+7 = 0 9. (+)( i)(+i) = 0 or + ++ = 0 0. ( )( i)(+i) = 0 or + = 0. ( i )( + i ) = 0 or + = 0. ( + )( i )( + i ) = 0 or = 0. ( ) [ ( i)] [ ( + i)] = 0 or = 0. ( 5) [ ( i)] [ ( + i)] = 0 or = 0 Copright 0 Pearson Education, Inc.

26 . THE THEORY OF EQUATIONS ( )( )( ) = 0 or = 0 6. ( + )( )( + ) = 0 or = 0 7. ( ) [ ( i)] [ ( + i)] = 0 or = 0 8. ( + ) [ ( i)] [ ( + i)] = 0 or = 0 9. ( )( )( ) = 0 or = 0 0. ( + )( + )( ) = 0 or 6 = 0. ( i)( + i) [ ( + i)] [ ( i)] = 0 or + + = 0. ( i)( + i) [ ( i)] [ ( + i)] = 0 or = 0. P () = has no sign change and P ( ) = has sign changes. There are (a) negative roots, or (b) negative root & imaginar roots.. P () = has sign changes and P ( ) = 5 6 has no sign changes. There are (a) positive roots, or (b) positive root and imaginar roots. 5. P () = +7+6 has sign change and P ( ) = has sign changes. There are (a) positive root and negative roots, or (b) positive root and imaginar roots. 6. P () = has sign change and P ( ) = has sign change. There are positive root, negative root, and imaginar roots. 8. P () = has sign change and P ( ) = has sign change. There are positive root, negative root, and imaginar roots. 9. P (t) = t t +t 5t+7 has sign changes and P ( t) = t + t + t + 5t + 7 has no sign change. There are (a) positive roots, or (b) positive roots and imaginar roots, or (c) imaginar roots. 50. P (r) = 5r +r +7r 6 has sign changes and P ( r) = 5r r 7r 6 has no sign change. There are (a) positive roots & imaginar roots, or (b) imaginar roots. 5. P () = and P ( ) = 5 5 have no sign changes; imaginar roots and P () = + = P ( ) has sign changes. There are (a) positive roots and negative roots, or (b) positive roots and imaginar roots, or (c) negative roots and imaginar roots, or (d) imaginar roots. 5. Best integral bounds are. One checks that, are not upper bounds and that is a bound is a lower bound since P () = = P ( ) has no sign change. There are imaginar roots. Copright 0 Pearson Education, Inc.

27 68 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 5. Best integral bounds are. One checks that is not an upper bound and that is a bound is not a lower bound but is a bound since Best integral bounds are. One checks that is not an upper bound and that is a bound , are not lower bounds but is a bound since Best integral bounds are. One checks that, are not upper bounds and that is a bound , are not lower bounds but is a bound since Best integral bounds are 5. One checks that,,, are not upper bounds and that 5 is a bound is a lower bound since Best integral bounds are. One checks that is not an upper bound and that is a bound is not a lower bound but is a bound since Best integral bounds are. Multipl equation b, this makes the leading coefficient positive. One checks that, are not upper bounds and that is a bound is a lower bound since Copright 0 Pearson Education, Inc.

28 . THE THEORY OF EQUATIONS Best integral bounds are. Multipl equation b to to make the leading coefficient positive. One checks that, are not upper bounds and that is a bound ,, are not lower bounds but is a lower bound since Roots are, 5, since and 0 = ( 5)( + ) 6. Roots are,, since and = ( + )( + ). 6. Roots are, ± since and b the quadratic formula the roots of = 0 are ±. 6. Roots are, ± i since and b the quadratic formula the roots of + + = 0 are ± i. 65. Roots are,, ±i since and the root of + = 0 are ±i. 66. Roots are,, ±i since and the roots of + = 0 are ±i. 67. Roots are /, /, 5 since / and = ( )( + 5). Copright 0 Pearson Education, Inc.

29 70 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 68. Roots are /, /, since and = ( + )( + ). 69., each have multiplicit since and + = ( + )( ). 70. Roots are 0, ±i, ±i since ( + + ) = ( + )( + ). 7. Use snthetic division on the cubic factor in ( 6 + 8) = Since + = ( ), the roots are (with multiplicit ) and Use snthetic division on the cubic factor in ( ) = Since = ( + ), the roots are (with multiplicit ) and Use snthetic division on the 5th degree factor in ( ) = Note, the roots of + = 0 are ±i. Thus, the roots are = 0,, ±, ±i. 7. Use snthetic division on the 5th degree factor in ( ) = Factoring, we obtain + 7 = ( )( + ) = 0. The roots are 0 (with multiplicit ),, ± and ±i. 75. The roots are ±,, /, / since Copright 0 Pearson Education, Inc

30 . THE THEORY OF EQUATIONS and the last quotient is 8 + = ( )( ). 76. The roots are, ±, /, / since and the last quotient is = ( )( ). 77. B the Theorem of Bounds and the graph below the intervals are 5 6, B the Theorem of Bounds and the graph below the intervals are 6 7, B the Theorem of Bounds and the graph below the intervals are 6 6, Copright 0 Pearson Education, Inc.

31 7 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 80. B the Theorem of Bounds and the graph below B the Theorem of Bounds and the graph below the intervals are 5,. 8. B the Theorem of Bounds and the graph below the intervals are,. the intervals are, Multipling b 00, t 8t + t + 0 = Since t t 5 = (t 5)(t + ) = 0, we obtain t = hr and t = 5 hr. 8. Solve = B the quadratic formula, the zeros of are = 7 ± 9.,.68 Then = or.68 since Let be the radius of the cone. The volume of the cone is π and the volume of the clinder with height is π (). So π = π + π (). Copright 0 Pearson Education, Inc.

32 . THE THEORY OF EQUATIONS 7 Divide b π, multipl b, and simplif to obtain + = The radius is = in. since +8+ = 0 has no real roots. 86. Let r and r + be the radius of the hemispherical roof and height of the circular wall, respectivel, in feet. Since the volume of a hemisphere is π r, we obtain πr (r + ) + π r = 68π r + r + r = 68 5r + 6r = (68) 5r + 6r 950 = 0. From the application of snthetic division as shown below, and since 5r +66r+79 = 0 has no real roots, the radius is r = feet. 87. From the following and (a + bi) + (c + di) = (a + c) + i(b + d) = (a + c) i(b + d) a + bi + c + di = (a bi) + (c di) = (a + c) i(b + d) we obtain that the conjugate of the sum of two comple numbers is equal to the sum of their conjugates. 88. The calculations and (a + bi)(c + di) = ac bd + i(ad + bc) = ac bd i(ad + bc) a + bi c + di = (a bi)(c di) = ac bd i(ad + bc) show that the conjugate of the product of two comple numbers is equal to the product of their conjugates. 89. If a is real then a = a + 0i = a 0i = a. 90. Using the results in Eercises 8 85, we get a n (a + bi) n a (a + bi) + a 0 = a n (a + bi) n a (a + bi) + a 0 = a n (a bi) n a (a bi) + a 0. The st epression is equal to 0 since a + bi is a root. Thus, the last line is 0 and a bi is a root. 9. Let f() = a( )( )( ). Since f(0) = 6a =, we find a =. Then substitute a = and multipl out f() to obtain f() = There is no such polnomial. For if there was one, then let it be f() = a( )( )( ). Since f(0) = 6a = 6, a =. Subsitute this back into f. So f( ) =, a contradiction. 9. The divisors of the constant term 6 are p = ±, ±, ±, ±6. The divisors of the leading coefficient are q = ±, ±. Then the possible rational zeros are p/q = ±, ±, ±, ±6, ±/, ±/. 9. The -intercepts are =,. Since the parabola opens upward, the solution set to + 0 is (, ] [, ). Copright 0 Pearson Education, Inc.

33 7 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS 95. b is a function of a since there is a unique number of primes less than a. a is not a function of b since the ordered pairs (b, a) = (, 6) and (a, b) = (, 7) have the same first coordinates and different second coordinates. 96. If s is the average speed to Atlantic Cit, then s/ is the average speed to Newark. Since time = distance/(average speed). Then the driving time to Newark increased b d s/ d s 00% = d s 00% = 50% where d is the distance between Atlantic Cit and Newark. 97. (f g)() = ( ) 9 = The vertical line =. Thinking Outside the Bo XXXIII c) The distance between a verte of the cubic bo and the center of the ball that is closest to the verte is. This is obtained b using the Pthagorean theorem. Then the diagonal of the cube holding the five balls has a length of +. If is the length of an edge of the cube then + + = ( + ). Solving for, we find = +. Then the volume of the cube is ( ) + Volume = =. Simplifing, we obtain Volume = The bo with the least volume is b) a) Consider the top laer of four balls and the ball that sits above it. Connecting the centers of the five balls gives a pramid with a -b- square base and a slanted height of. The diagonal of the square base is. B looking at a slanted height and half of the base, the height of the pramid is. Then the height of the bo is +. Multipling b the area of the -b- base of the bo which is 6, we obtain the volume of the bo. That is, the volume of the bo is 6( + ) = b) Using the same pramid in part a), we obtain that the height of the bo is +. Multipling b the area of the -b- base of the bo which is 6, we find that the volume of the bo is 6( + ) = Pop Quiz. 0 with multiplicit, and ± since ( ) = ( )( + ) = 0. Let the roots be ±i, 5. A polnomial is ( i)( + i)( 5) = ( + 6)( 5) and which simplifies to Since there are two sign changes in the coefficients of , there are either two or no positive roots.. Replacing b, we find that there is eactl one sign change in the coefficients of ( ) +5( ) ( )+9 = Thus, there is eactl one negative root. Copright 0 Pearson Education, Inc.

34 . MISCELLANEOUS EQUATIONS The roots are,, since and the last quotient factors into 8 = ( )( + ).. Linking Concepts (a) (b) (c) Let r, l, t, and V be the radius of the sphere, length of the side of the base, and the thickness of the base, and total volume, respectivel. Then we have V = πr + tl. If V = 96π, l = r, and t = π, then 96π = πr + π(r). Divide b π, multipl b, and simplif to obtain r + r 888 = 0. From the following and since r + 8r + = 0 has non-real roots, the radius is r = 9 in. If r = l, t =, and V = 5000, then l + π ( ) l = Solving for l, we find l The radius is r inches. For Thought. False, since ( + ) = ( ) + ( ) +.. False, since is a solution of the first and not of the second equation.. False, since 7 is a solution of the first equation but not of the second.. False, rather let u = / and u = /. 5. True, since = ± /. ( 6. False, ) /5 = ( ) /5 = ( ) =. 7. False, = is not a solution. 8. True 9. True 0. False, since ( ) = 6.. Eercises. Factor: ( + ) ( + ) = 0 ( )( + ) = ( )( + )( + ) = 0 The solution set is {±, }.. Factor: ( ) 5( ) = 0 ( 5)( ) = ( 5)( + 5)( ) = 0 The solution set is {± 5, }.. Factor: ( + 500) ( + 500) = 0 ( )( + 500) = { 0 } The solution set is ±, Factor the left-hand side. ( 00) ( 00) = 0 ( )( 00) = 0 = or = 00 { } 6 The solution set is ±, Set the right-hand side to 0 and factor. a(a 5a + 5) = 0 a = 5 ± 5 ()(5) a = 5 ± 05 or a = 0 or a = 0 Copright 0 Pearson Education, Inc.

35 76 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS { } 5 ± 05 The solution set is, Factor: b(b 9b + 0) = b(b 5)(b ) = 0 The solution set is {0,, 5}. 7. Factor: ( ) = ( )( + ) = 0 The solution set is {0, ±}. 8. Factor: 5m (m m + ) = 5m (m ) = 0 The solution set is {0, }. 9. Factor: (a )(a + ) = (a )(a + )(a i)(a + i) = 0. The solution set is {±, ±i}. 0. Factoring, we obtain w(w + ) = w(w + )(w w + ) = 0. Then w = 0, or w = ± ()() w = 0, or w = ± w = 0, or w = ± i and the solution set is. Squaring each side, we get + = { 0,, ± i }. 0 = + = ( 8)( ). Checking =, we get. Then = is an etraneous root. The solution set is {8}.. Squaring each side, = = = ( 5)( 0). Checking = 5, we get. Then = 5 is an etraneous root. The solution set is {0}.. Isolate the radical and then square each side. = = + 00 = ( 5)( 6) Checking = 6, we get 6. Then = 6 is an etraneous root. The solution set is {5}.. Isolate the radical and then square each side. = + 0 = 5 + = ( )( ) Checking =, we get. Then = is an etraneous root. The solution set is {}. 5. Isolate the radical and then square each side. w = w w = w w + w = (w )(w + ) = 0 w =, Checking w =, we get. Then w = is an etraneous root. { The solution set is. } 6. Isolate the radical and then square each side. t = t 9t = t 9t + t = (t )(t + ) = 0 t =, Checking t =, we get. Then t = is an etraneous root. { Solution set is. } 7. Multipl both sides b z z + and square each side. z + = z z + = 9z B the quadratic formula, 0 = 9z z z = ± 6 (9)( ) 8 z = ± 5 8 = ± 8 = ± 9 Copright 0 Pearson Education, Inc.

36 . MISCELLANEOUS EQUATIONS 77 Since z = 0 and the right-hand 9 side of the original equation is nonnegative, z = is an etraneous root. 9 { } + The solution set is Multipl both sides b p 9p +. 9p + p = 0 9p + = p 9p + = p p 9p = 0 B the quadratic formula, p = 9 ± 8 ()( ) 8 = 9 ± 97 8 If p = 9 97 then the left-hand side of the 8 original equation becomes negative and so p = 9 97 is an etraneous root. 8 { } The solution set is Squaring each side, one obtains 5 = 9 = ( 6)( + ) = 0. The solution set is {, 6}. 0. Squaring each side, one obtains + 5 = + 5 = ( )( + ) = 0 =,. Checking =, notice that the left-hand side and the right hand-side of the original equation have opposite signs. Then = { is an etraneous root. The solution set is. }. Isolate a radical and square each side. + 0 = = = 8 = 9 = The solution set is {9}.. Isolate a radical and square each side. 6 = 6 = + 0 = 0 = 00 = Checking = 00 we get 8 =, a contradiction. There is no solution and the solution set is the empt set.. Isolate a radical and square each side. n + = 5 n n + = 5 0 n + (n ) 0 = 0 n = n = n The solution set is {5}.. Isolate a radical and square each side. + 0 = = + + ( ) 8 = = = The solution set is {6}. 5. Isolate a radical and square each side. + 5 = = ( + 6) 8 = = ( + 6) = 0 ( 0)( 78) = 0 Copright 0 Pearson Education, Inc.

37 78 CHAPTER POLYNOMIAL AND RATIONAL FUNCTIONS Checking = 78 we get 5 9 and = 78 is an etraneous root. The solution set is {0}. 6. Isolate a radical and square each side. = + = + + ( ) = = + = ( ) 8 + = 0 ( 6)( ) = 0 = 6, The solution set is {, 6}. 7. Raise each side to the power /. Then = ± / = ±8 / = ± {. The solution set is ± }. 8. Raise each side to the power /. ( ) / ( ) / Then = ± = ± = 8 ± { } = ±. The solution set is ±. 9. Raise each side to the power /. Thus, w = ± (6) / = ± () = ± 8. { The solution set is ± } Raise each side to the power /. Thus, w = (7) / = () = { 9. The solution set is. 9}. Raise each side to the power. So, t = (7) = 9. The solution set is { 9. Raise each side to the power. ( ) So, t = =. The solution set is {}.. Raise each side to the power. Then s = () = and s = +. { 5 The solution set is. } }.. Raise each side to the power. ( ) Then s = = 9 and the solution set is {}. 5. Since { ( 9)( ) = 0, the solution set is ±, ± }. 6. Since = 0, ( 5)( ) = 0. { The solution set is ± 5, ± }. 7. Since { ( + 7)( ) = 0, the solution set is ±, ±i } Since { ( + )( ) = 0, the solution set is ±, ±i }. 9. Since ( + 9)( 9) = 0, the solution set is {±, ±i}. 0. Since ( + 5)( 5) = 0, the solution set is {±5, ±5i}.. Let u = c ( ) c and u =. Then 5 5 u + u 8 = 0 (u + )(u ) = 0 u =, c c = or = 5 5 c = 0 or c = 0 c = 7 or c =. { The solution set is 7, }.. Let u = b 5 6 ( ) b 5 and u =. Then 6 u u 6 = 0 (u )(u + ) = 0 u =, b 5 b 5 = or = 6 6 b 5 = 8 or b 5 = b = or b = 7. The solution set is {, 7}. Copright 0 Pearson Education, Inc.

38 . MISCELLANEOUS EQUATIONS 79. Let u = ( ) 5 and u =. Then 5 u + u = (u + )(u ) = 0 u =, 5 = or 5 = = 0 + or = 5 = or 0 5 =. { The solution set is 0, }. 5. Let u = ( ) and u =. Then u + u = (u + 6)(u ) = 0 u = 6, = 6 or = = or = = 7 or 6 =. { 7 The solution set is 6, }. 5. Let u = v v and u = ( v v ). Then u 7u + 60 = (u 5)(u ) = 0 u = 5, v v = 5 or v v = v v 5 = 0 or v v = 0 (v 5)(v + ) = 0 or (v 6)(v + ) = 0. The solution set is {,, 5, 6}. 6. Let t = u + u and t = ( u + u ). Then t t = (t )(t + ) = 0 t =, u + u = or u + u = u + u = 0 or u + u + = 0 (u )(u + ) = 0 or (u + ) = 0. The solution set is {±, }. 7. Factor the left-hand side. ( ) ( ) = 0 =, The solution set is {, 9}. = 9, 8. Factor the left-hand side as ( 5) ( + ) = 0. Since is nonnegative then onl the first factor can be zero and so = 5. The solution set is { 5 9. Factor the left-hand side as ( q ) ( q ) = 0. Then q =, and the solution set is {9, 6}. 50. Set the right-hand side to 0 and factor. Then h h / + = 0 ( h ) = 0 The solution set is {}. h = h = 5. Set the right-hand side to 0 and factor. / 7 / + 0 = 0 ( ) ( ) / 5 / = 0 The solution set is {8, 5}. / = 5, 5. Factor the left-hand side. ( ) ( ) Then / / = 0. Since / =, the solution set is {, 6}. 5. An equivalent statement is w = or w = w = 7 or w =. { The solution set is ± } 7, ±. 5. An equivalent statement is a = or a = a = or a = 0. { The solution set is ± }, 0. }. Copright 0 Pearson Education, Inc.