( ) 83. = 0 x 87. ( ) = 4 + − 20 = x x V x 0 ( g − 2)2 = 4 x Let = u − = ± → = 2 4 x x u x 2 + − = 20 u u 2 0 − 2 = ±2 x (u (u = 2 ± 2 + 5) − 4) x = 0 + 5 = 0 u − 4 = = 4 or = 0 or u x x 0 = 0 since that is not a feasible length x −5 = u for the original sheet. Therefore, the original = 4 u = −5 x sheet should measure 4 feet on each side. = 4 x = not real x ( ) = = = V x 42 16 x (16) Check: = 16 88. g 4 − − 16 20 = 16 + 4 20 + 2 = 0 ( ) The only zero of = g x x = ( 16 − 20 is 16. The x − x + 2) only of the graph of is 16. x-intercept g − 2 = ± 16 x − 2 = ±4 x ) = 2 ± 4 84. x = 0 ( f x + − 2 = x x 0 = 6 or = −2 x x = Let u → = Discard = −2 since width cannot be negative. x u x x 2 + − 2 = u u 2 Therefore, the original sheet should measure 6 (u (u 0 − 1) + 2) feet on each side. = 0 − 1 = 0 u 89. a. When the ball strikes the ground, the + 2 = 0 u or distance from the ground will be 0. = −2 u = 1 u Therefore, we solve = = −2 0 s x = 1 x = 96 + 80t − 16t = 0 not real x = 12 = 1 2 x Check: − = + − 1 2 1 1 2 −16t + 80t + 96 = 0 (1) 2 = 1 f = 0 + The only zero of − 2 is 1. The x ( ) = + f x x
83. g ( x ) = 0 x + x − 20 = 0 87. V ( x ) = 4 ( x − 2)2 = 4 Let u = x → u 2 = x x − 2 = ± 4 u 2 + u − 20 = 0 (u + 5) (u − 4) = 0 x − 2 = ±2 x = 2 ± 2 u + 5 = 0 or u − 4 = 0 x = 4 or x = 0 u = −5 x = −5 x = not real Check: g (16) = 16 + u = 4 x = 4 x = 42 = 16 16 − 20 = 16 + 4 − 20 = 0 88. x = 0 since that is not a feasible length for the original sheet. Therefore, the original sheet should measure 4 feet on each side. V ( x ) = 4 2 The only zero of g ( x ) = x + x − 20 is 16. The ( x − 2) = 16 only x-intercept of the graph of g is 16. x − 2 = ± 16 x − 2 = ±4 84. f ( x ) = 0 x = 2 ± 4 x + x − 2 = 0 x = 6 or x = −2 Let u = x → u 2 = x Discard x = −2 since width cannot be negative. u 2 + u − 2 = 0 (u − 1) (u + 2) = 0 u − 1 = 0 or u = 1 x = 1 x = 12 = 1 u + 2 = 0 u = −2 x = −2 x = not real Therefore, the original sheet should measure 6 feet on each side. 89. a. When the ball strikes the ground, the distance from the ground will be 0. Therefore, we solve s = 0 96 + 80t − 16t 2 = 0 Check: f (1) = 1 + 1 − 2 = 1 + 1 − 2 = 0 −16t 2 + 80t + 96 = 0 The only zero of f ( x ) = x + x − 2 is 1. The