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in Mathematical reasoning by (15 points)
If alpha, beta are roots of the equation ax2 + 3x + 2 = 0 (a< 0), then

Alpha square by beta +beta square by alpha greater than

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1 Answer

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by (15 points)
Since a << 0, therefore discriminant D=9−8a>0D=9-8a>0.
So, αandβαandβ are real.
We have, α+β=−3aandαβ=2aα+β=-3aandαβ=2a
∴ α2β+β2α=α3+β3αβ=(α+β)3−3α(α+β)αβ∴ α2β+β2α=α3+β3αβ=(α+β)3-3α(α+β)αβ
⇒ α2β+β2α=(α+β)3αβ−3(α+β)=−272a2+9a<0 [∵a<0]

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