Question
I. 2b2 + 31b + 99 = 0 II. 4a2
+ 8a - 45 = 0Solution
I. 2b2 + 31b + 99 = 0 2b2 + 9b + 22b + 99 = 0 b (2b + 9) + 11(2b + 9) = 0 (b + 11) (2b + 9) = 0 b = -11, -9/2 II. 4a2 + 8a - 45 = 0 4a2 - 10a + 18a - 45 = 0 2a (2a - 5) + 9 (2a - 5) = 0 (2a + 9) (2a - 5) = 0 a = -9/2, 5/2 Hence, a ≥ bÂ
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