Question

If $\alpha ,\beta$ are the roots of $x^2+px+1=0$ and $\gamma ,\delta$ are the roots of $x^2+qx+1=0$, then $(\alpha -\gamma )(\beta -\gamma )(\alpha -\delta )(\beta +\delta )=$

• A. $2q^2$
• B. $2p^2$
• C. $p^2-q^2$
• D. $q^2-p^2$

Answer 1

The correct option is D $q^2-p^2$

$\alpha ,\beta$ are the roots of $x^2+px+1=0$
$\Rightarrow \alpha +\beta =-p,\alpha \beta =1$

$\gamma ,\delta$ are the roots of $x^2+qx+1=0$
$\Rightarrow \gamma \delta =1,{\gamma }^2+q\gamma +1=0$,
${\delta }^2+q\delta +1=0$

$(\alpha -\gamma )(\beta -\gamma )(\alpha +\delta )(\beta +\delta )$
$=[\alpha \beta -\gamma (\alpha +\beta )+{\gamma }^2][\alpha \beta +\delta (\alpha +\beta )+{\delta }^2]$
$=(1+p\gamma +{\gamma }^2)(1-p\delta +{\delta }^2)$
$=(p\gamma -q\gamma )(-p\delta -q\delta )$
$=-\gamma \delta (p-q)(p+q)$
$=-(p^2-q^2)=q^2-p^2$

Hence, option D.