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α(x-β)2+β(x-α)2=0 ---(1)
Given α and β are real and of opposite sign
Therefore αβ= -ve
From equation 1
α(x2+β2-2xβ) + β(x2+α2-2αx)=0
(α+β)x2-4αβx+αβ(α+β)=0
product of the roots = αβ(α+β)/(α+β)= αβ = -ve
discriminant Δ=16 α2β2-4 αβ( α+β)2
as 16 α2β2 and ( α+β)2 are positive and αβ is negative, Δ is positive
so the roots of the equation are real and opposite in sign.
Ans: c
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