Let a, β be the roots of the equation x2-px+r=0 and α2,2β be the roots of the equation x2−qx+r =0. Then the value of r is :

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29(p−q)(2q−p)

29(q−p)(2p−q)

29(q−2p)(2q−p)

29(2p−q)(2q−p)

answer is D.

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Detailed Solution

α+β=p,αβ=rand α2+2β=qBut α+β=p and α+4β=2q ⇒ β=13(2q−p) and α=23(2p−q)Thus, aβ=r⇒ 29(2p−q)(2q−p)=r

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