4ac − 4ac b2 = 4k 1 2 (−b − + − 4ac 2c b2 b2 (−b ) ) + 2c 4ac = = 2a k 2 1 = =± 1 k −b − 4ac b2 x2 4 + and x = 1 * k (c ) −b + − 4 b or = − 1 k (a ) −b + − 4ac b 2 2 = = 2c 2c 2 96. In to have one repeated real zero, we need −b + − 4ac −b − − 4ac b b 2 2 = ⋅ 2 2 the discriminant to be 0. − 4ac b2 2c −b − − 4ac = 0 b2 (−k )2 (1) (4) − 4 (b2 ) ( ) = 0 ( ) − − 4ac 4ac b2 − 16 = k 2c −b − 4ac 2c −b − 4ac b2 b2 2 0 − − (k (k − 4) + 4) = 0 = = 2a 1 = 4 or = −4 k k −b − 4ac b2 x1 −
b2 − 4ac 4ac 4k 2 = 1 2c (−b + b2 − 4ac ) 2c (−b + b2 − 4ac ) k 2 = = 2a = 1 k =± 1 4 and −b + b2 − 4ac x2 k = 1 or k = − 1 −b + b − 4 (c ) (a ) x * −b + b − 4ac 2 2 96. In to have one repeated real zero, we need 2 2 = = 2 2c 2c −b + b 2 − 4ac −b − b 2 − 4ac = ⋅ the discriminant to be 0. 2c −b − b2 − 4ac b2 − 4ac = 0 (−k )2 − 4 (1) (4) = 0 b2 − (b2 − 4ac ) 4ac ( ) ( ) k 2 − 16 = 0 2c −b − b2 − 4ac 2c −b − b2 − 4ac (k − 4) (k + 4) = 0 = 2a = 1 k = 4 or k = −4 −b − b2 − 4ac x1
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