If α,β are the roots of the equation x2+Px+P3=0, P≠0 such that α=β2 then the roots of the given equation are

If α,β are roots x2+px+p3=0 thenα+β=−p,αβ=p3and Given α=β2⇒ β3=p3 ⇒ β=p∴ α+β=−p⇒ β2+β=−p⇒ p2+p=−p⇒ p2+2p=0⇒ p(p+2)=0p≠0⇒ p=−2∴ α=4, β=−2

If α,β are the roots of the equation x2+Px+P3=0, P≠0 such that α=β2 then the roots of the given equation are

If α,β are roots x2+px+p3=0 thenα+β=−p,αβ=p3and Given α=β2⇒ β3=p3 ⇒ β=p∴ α+β=−p⇒ β2+β=−p⇒ p2+p=−p⇒ p2+2p=0⇒ p(p+2)=0p≠0⇒ p=−2∴ α=4, β=−2

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$I f α, β a r e t h e r o o t s o f t h e e q u a t i o n x^{2} + P x + P^{3} = 0, P \neq 0 s u c h t h a t α = β^{2} t h e n t h e r o o t s o f t h e g i v e n e q u a t i o n a r e$

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$4, - 2$

$1, - 1$

$1, 1$

$4, 2$

answer is A.

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Detailed Solution

$\begin{array}{l} I f α, β a r e r o o t s x^{2} + p x + p^{3} = 0 t h e n \\ α + β = - p, α β = p^{3} \\ a n d G i v e n α = β^{2} \\ \Rightarrow β^{3} = p^{3} \Rightarrow β = p \\ ∴ α + β = - p \\ \Rightarrow β^{2} + β = - p \\ \Rightarrow p^{2} + p = - p \\ \Rightarrow p^{2} + 2 p = 0 \\ \Rightarrow p (p + 2) = 0 \\ p \neq 0 \Rightarrow p = - 2 \\ ∴ α = 4, β = - 2 \end{array}$