If the sum of two of the roots of $x^3+p x^2-q x+r=0$ is zero, then $p q$ is equal to

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Aicharya · Accepted Answer

Given that, $x^3+p x^2-q x+r=0$ $\ldots$ (i) Let $\alpha, \beta, \gamma$ are roots of this equation $\alpha+\beta+\gamma=-p$ $\ldots$ (ii) $\alpha \beta+\beta \gamma+\gamma \alpha=-q$\ldots$ (iii) and $\quad \alpha \beta \gamma=-r$ $\ldots$ (iv) Now, $\quad p q=(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)$ $=(0+\gamma)(\alpha \beta+\gamma(\alpha+\beta))$ $[	herefore \alpha+\beta=0$ given $]$ $=\gamma(\alpha \beta+0)=\alpha \beta \gamma=-r$

If the sum of two of the roots of $x^3+p x^2-q x+r=0$ is zero, then $p q$ is equal to

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