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If $\alpha$ and $\beta$ are the roots of $x^{2}-p x+1=0$ and $\gamma$ is a root of $x^{2}+p x+1=0,$ then $(\alpha+\gamma)(\beta+\gamma)$ is
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Since, $\alpha$ and $\beta$ are the roots of $x^{2}-p x+1=0$.
$\therefore$
$\alpha+\beta=p$ and $\alpha \beta=1$
Also, $\gamma$ is the root of $x^{2}+p x+1=0$ $\therefore \quad \gamma^{2}+p{\gamma}+1=0 \Rightarrow \gamma^{2}=-p \gamma-1$
Now, $(\alpha+\gamma)(\beta+\gamma)=\alpha \beta+\alpha \gamma+\beta \gamma+\gamma^{2}$
$=1+\gamma(\alpha+\beta)-p \gamma-1=\gamma(\alpha+\beta-p)$
$=\gamma \times 0=0$
$$
[\because \alpha+\beta=p]
$$
Alternate Method
Since, $\gamma=-\alpha$ or $-\beta$
$\therefore \quad(\alpha+\gamma)(\beta+\gamma)=0$
$\therefore$
$\alpha+\beta=p$ and $\alpha \beta=1$
Also, $\gamma$ is the root of $x^{2}+p x+1=0$ $\therefore \quad \gamma^{2}+p{\gamma}+1=0 \Rightarrow \gamma^{2}=-p \gamma-1$
Now, $(\alpha+\gamma)(\beta+\gamma)=\alpha \beta+\alpha \gamma+\beta \gamma+\gamma^{2}$
$=1+\gamma(\alpha+\beta)-p \gamma-1=\gamma(\alpha+\beta-p)$
$=\gamma \times 0=0$
$$
[\because \alpha+\beta=p]
$$
Alternate Method
Since, $\gamma=-\alpha$ or $-\beta$
$\therefore \quad(\alpha+\gamma)(\beta+\gamma)=0$
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