Question

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

Answer 1

$\alpha +\beta =-p,\gamma +\delta =-p,\alpha \beta =-q,\gamma \delta =r$
$\therefore \alpha +\beta =\gamma +\delta$
Hence $(\alpha -\gamma )(\alpha -\delta )={\alpha }^2-\alpha (\gamma +\delta )+\gamma \delta$
$={\alpha }^2-\alpha (\alpha +\beta )+\gamma \delta =-\alpha \beta +\gamma \delta =q+r$
Similarly $(\beta -\gamma )(\beta -\delta )=q+r$