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Q.
If $\alpha, \beta$ are the roots of $x^2 - px + 1 = 0$ and $\gamma$ is a root of $x^2 + px + 1 = 0$, then $\left(\alpha+\gamma\right)\left(\beta+\gamma\right)$ is
WBJEEWBJEE 2015Complex Numbers and Quadratic Equations
Solution:
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 $x^{2}+p x+1=0$
$\therefore \gamma^{2}+p \gamma+1=0$
$\Rightarrow \gamma^{2}=-p \gamma-1$
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 $-\beta$
$\therefore (\alpha+\gamma)(\beta+\gamma)=0$