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II. 28b2 + 15b + 2 = 01). a < b2). a > b3). a ≤ b4). a ≥ b |
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Answer» I. a2 + 10a + 21 = 0 ⇒ a2 + 7a + 3A + 21 = 0 ⇒ a(a + 7) + 3(a + 7) = 0 ⇒ (a + 3) (a + 7) = 0 Then, a = (-3) or a = (-7) II. 28b2 + 15B + 2 = 0 ⇒ 28b2 + 7b + 8B + 2 = 0 ⇒ 7b(4b + 1) + 2(4b + 1) = 0 ⇒ (7b + 2) (4b + 1) = 0 Then, b = (-2/7) or b = (-1/4) So, when a = (-3), a < b for b = (-2/7) and a < b for b = (-1/4) And when a = (-7), a < b for b = (-2/7) and a < b for b = (-1/4) ∴ we can OBSERVE that a < b. |
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