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Question

Let α,β be the roots of ax2+bx+c=0, a0. If 1,α+β,αβ are in A.P. and 1α,12,1β are also in A.P., then the value of α2+β22α2β2α2+β2 is

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Solution

α,β be the roots of ax2+bx+c=0
Then using sum and product of roots, we get
α+β=baαβ=ca
Now,
1,α+β,αβ are in A.P., so
1+αβ=2(α+β)a+2b+c=0(1)
Also,
1α,12,1β are in A.P., so
1α+1β=1α+βαβ=1b=c(2)
From equation (1) and (2), we get
a=b=c
Then the equation becomes,
a(x2x+1)=0x2x+1=0 (a0)
Therefore,
α2+β22α2β2α2+β2=12α2+β2=12(α+β)22αβ=3

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