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II. 8b2 + 10b + 3 = 01). a < b2). a > b3). a ≤ b4). a ≥ b |
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Answer» I. a2 - 8A + 12 = 0 ⇒ a2 - 6a - 2A + 12 = 0 ⇒ a(a - 6) - 2(a - 6) = 0 ⇒ (a - 2) (a - 6) = 0 Then, a = 2 or a = 6 II. 8b2 + 10b + 3 = 0 ⇒ 8b2 + 6b + 4B + 3 = 0 ⇒ 2B(4b + 3) + 1(4b + 3) = 0 ⇒ (2b + 1) (4b + 3) = 0 Then, b = (-1/2) or b = (-3/4) So, when a = 2, a > b for b = (-1/2) and a > b for b = (-3/4) And when a = 6, a > b for b = (-1/2) and a > b for b = (-3/4) ∴ we can observe that a > b. |
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