( ( + 2) − 1) x x ( ) = a. f x ≤ 0 0 2 = = The zeros are and 1 . x x –2 − − + 1 = 0 x x + −1 = 0 x2 x 2 ( ) ( ) ()( ) = x − 1 ± 1 − 4 1 −1 (1) 2 = −1 ± + = −1 ± 1 4 5 2 2 { The solution set is x ⎧⎪ Solution set: ⎨ − 2 ≤ ≤ 1} or, x . −1 − 5 , −1 + 5 ⎫⎪⎬ 2 2 [−2, ⎪⎩ ⎭⎪ interval notation, 1] .
( x + 2) ( x − 1) ≤ 0 a. f ( x ) = 0 2 The zeros are x = –2 and x = 1 . − x − x + 1 = 0 x2 + x −1 = 0 2 ( ) ( ) ()( ) x = − 1 ± 1 − 4 1 −1 2 (1) = −1 ± 1 + 4 = −1 ± 5 2 2 The solution set is { x − 2 ≤ x ≤ 1} or, ⎧⎪ Solution set: ⎨ −1 − 5 , −1 + 5 ⎫⎪⎬ . 2 2 interval notation, [−2, 1] . ⎪⎩ ⎭⎪
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