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If one root is square of the other root of the equation $x^{2}+p x+q=0$, then the relations between $p$ and $q$ is
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Verified Answer
The correct answer is:
$p^{3}-(3 p-1) q+q^{2}=0$
Given equation $x^{2}+p x+q=0$ has roots $\alpha$ and $\alpha^{2}$
$$
\begin{array}{l}
\Rightarrow \text { Sum }=\alpha+\alpha^{2}=-\mathrm{p} \text { and Product }=\alpha^{3}=\mathrm{q} \\
\Rightarrow \alpha(\alpha+1)=-\mathrm{p} \\
\Rightarrow \alpha^{3}\left[\alpha^{3}+1+3 \alpha(\alpha+1)\right]=-\mathrm{p}^{3} \\
\Rightarrow \mathrm{q}(\mathrm{q}+1-3 \mathrm{p})=-\mathrm{p}^{3} \\
\Rightarrow \mathrm{p}^{3}-(3 \mathrm{p}-1) \mathrm{q}+\mathrm{q}^{2}=0
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow \text { Sum }=\alpha+\alpha^{2}=-\mathrm{p} \text { and Product }=\alpha^{3}=\mathrm{q} \\
\Rightarrow \alpha(\alpha+1)=-\mathrm{p} \\
\Rightarrow \alpha^{3}\left[\alpha^{3}+1+3 \alpha(\alpha+1)\right]=-\mathrm{p}^{3} \\
\Rightarrow \mathrm{q}(\mathrm{q}+1-3 \mathrm{p})=-\mathrm{p}^{3} \\
\Rightarrow \mathrm{p}^{3}-(3 \mathrm{p}-1) \mathrm{q}+\mathrm{q}^{2}=0
\end{array}
$$
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