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LV超3A名牌購物網
(a) (3x + 4)(x - 1) = 0
x = - 4/3 or 1
Case I
When x < - 4/3
(3x + 4) < 0
(x - 1) < 0
so (3x + 4)(x - 1) > 0 acceptable.
Case II
- 4/3 < x < 1
(3x + 4) > 0
(x - 1) < 0
so (3x + 4)(x - 1) < 0 not acceptable.
Case III
x > 1
(3x + 4) > 0
(x - 1) > 0
so (3x + 4)(x - 1) > 0 acceptable.
Answer : x < - 4/3 or x > 1.
(b) 5x^2 + 8x - 4 < 0
(5x - 2)(x + 2) < 0
Using the same method as in (a)
Answer : - 2 < x < 2/5.
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