If one root of the equation x2+px+q=0 is the square of the other, then

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p3+q2-q(3p+1)=0

p3+q2+q(1+3p)=0

p3+q2+q(3p-1)=0

p3+q2+q(1-3p)=0

answer is D.

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Detailed Solution

Let root of the given equation x2+px+q=0 are α and α2Now, αα2=α3=q, α+α2=-pCubing both sides, α3+(α2)3+3α.α2(α+α2)=-p3q+q2+3q(-p)=-p3p3+q2+q(1-3p)=0

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