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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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Verified Answer
The correct answer is:
$-r$
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$$\ldots$ (iii)
and $\quad \alpha \beta \gamma=-r$ $\ldots$ (iv)
Now, $\quad p q=(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)$
$=(0+\gamma)(\alpha \beta+\gamma(\alpha+\beta))$
$[\therefore \alpha+\beta=0$ given $]$
$=\gamma(\alpha \beta+0)=\alpha \beta \gamma=-r$
$x^3+p x^2-q x+r=0$ $\ldots$ (i)
Let $\alpha, \beta, \gamma$ are roots of this equation
$\alpha+\beta+\gamma=-p$ $\ldots$ (ii)
$\alpha \beta+\beta \gamma+\gamma \alpha=-q$$\ldots$ (iii)
and $\quad \alpha \beta \gamma=-r$ $\ldots$ (iv)
Now, $\quad p q=(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)$
$=(0+\gamma)(\alpha \beta+\gamma(\alpha+\beta))$
$[\therefore \alpha+\beta=0$ given $]$
$=\gamma(\alpha \beta+0)=\alpha \beta \gamma=-r$
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