Question

lf $\alpha ,\beta$ are the roots of $x^2+px-q=0$ and $\gamma ,\delta$ that of $x^2+px+r=0$, then $(\alpha -\gamma )(\beta -\gamma )(\alpha -\delta )(\beta -\delta )=$

• A. $(q-r{)}^2$
• B. $(q+r{)}^2$
• C. $-(q+r{)}^2$
• D. $-(q-r{)}^2$

Answer 1

The correct option is B $(q+r{)}^2$

Since, $\alpha ,\beta$ are the roots of $x^2+px-q=0$
$\alpha +\beta =-p$ and $\alpha \beta =-q$
Also, $\gamma ,\delta$ are roots of $x^2+px+r=0$,
$\gamma +\delta =-p$ and $\gamma \delta =r$
Consider $(\alpha -\gamma )(\beta -\gamma )(\alpha -\delta )(\beta -\delta )$
$=({\gamma }^2-(\alpha +\beta )\gamma +\alpha \beta )({\delta }^2-(\alpha +\beta )\delta +\alpha \beta )$
$=({\gamma }^2+p\gamma -q)({\delta }^2+p\delta -q)$
$=(-r-q)(-r-q)(\because \gamma ,\delta$ are roots of $x^2+px+r=0)$
$=(r+q{)}^2$