If the sum of two roots of $x^3+p x^2+q x-5=0$ is equal to its third root, then $p\left(p^2-4 q\right)=$

Question

Aicharya · Accepted Answer

$x^3+p x^2+q x-5=0 \quad \alpha+\beta+\gamma=-p$ $\Rightarrow 2 \alpha=-p \Rightarrow \alpha=\frac{-p}{2} \quad[\because \beta+\gamma=\alpha]$ $	herefore \frac{-p^3}{8}+\frac{p^3}{4}-\frac{p q}{2}-5=0 \Rightarrow \frac{p^3}{8}+\frac{p q}{2}-5=0$ $\Rightarrow p\left(p^2-4 p qight)=40$

If the sum of two roots of $x^3+p x^2+q x-5=0$ is equal to its third root, then $p\left(p^2-4 q\right)=$

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