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Question

If α, β are the roots of the equation x2px+q=0 and α1,β1 be the roots of the equation x2qx+p=0, then form the quadratic equation whose roots are 1α1β+1αβ1 and 1αα1+1ββ1

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Solution

Given, α and β roots of x2px+q=0
α+β=p
αβ=q
And α, and β are roots of x2qx+p=0
α1+β1=q
α1.β1=p
We have to find quadratic equation whose whose roots are
1α1β+1αβ1 and 1αα1+1ββ1
Now,
1α1β+1αβ1+1αα1+1ββ1=1α1[1α+1β]+1β1[1α+1β]
=α+βαβ×α1+β1α1β1=pq×qp=1
and
(1αβ+1αβ1)×(1α1α+1β1β)=1α21αβ+1α1β1β2+1β1α1α2+1β21αβ
=1αβ[1α21+1β21]+1α1β1[1β2+1α2]
=1q[β21+α21α21.β21]+1p[α2+β2α2β2]
=1q.p2[(α1+β1)22α1β1]+1pq2[(α+β)22αβ]
=1q.p2[q22p]+1pq2[p22q]
=q3+p3pqp2q2
Therefore Equation is
x2x+q3+β24pqp2q2

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