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II. 8b2 + 10b + 3 = 01). a < b2). a > b3). a ≤ b4). a ≥ b

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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