Question

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

• A. 0
• B. 1
• C. -1
• D. p

Answer 1

Answer: (A)

Since, $\alpha$ and $\beta$ are the roots of $x^2-px+1=0$.
$\therefore \alpha +\beta =p$ and $\alpha \beta =1$
Also, $\gamma$ is the root of $x^2+px+1=0$
$\therefore {\gamma }^2+p\gamma +1=0$
$\Rightarrow {\gamma }^2=-p\gamma -1$
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 $-\beta$
$\therefore (\alpha +\gamma )(\beta +\gamma )=0$