Question

If $\alpha ,\beta$ are the roots of $x^2+px+q=0$ and $\gamma ,\delta$ are the roots of $x^2+px+q=0;$ evaluate $(\alpha -\gamma )(\alpha -\delta )(\beta -\gamma )(\beta -\delta )$ in terms of p, q, r and s. Deduce the condition that the equations have a common root.

Answer 1

$\alpha +\beta =-q,\gamma +\delta =-r,\gamma \delta =s$

$(\alpha -\gamma )(\alpha -\delta )(\beta -\gamma )(\beta -\delta )$

$=[{\alpha }^2-\alpha (\gamma +\delta )+\gamma \delta ][{\beta }^2-\beta (\gamma +\delta )+\gamma \delta ]\to$ $({\alpha }^2+r\alpha +s)({\beta }^2+r\beta +s)\cdots \left(1\right)$

Since $\alpha$ is a root of $x^2+px+q=0\Rightarrow {\alpha }^2+p\alpha +q=0\Rightarrow {\alpha }^2=-p\alpha -q$
and similarly ${\beta }^2=-p\beta -q.$

Hence from (1) we have to evaluate the value of

$(-p\alpha -q+ra+s)(-p\beta -q+r\beta +s)$

$=[(r-p)\alpha -(q-s)][(r-p)-\beta (q-s)]$

$={(r-p)}^2\alpha \beta -(r-p)(q-s)(\alpha +\beta )+{(q-s)}^2$

$={(r-p)}^{2}q-(r-p)(q-s)+{(q-s)}^2$

$=(r-p)\left[(qr-pq)\right]+(pq-ps)+{(q-s)}^2$ $={(q-s)}^2-(p-r)(qr-ps)\cdots \left(2\right)$

In case the equations have a common root then $\alpha =\gamma$ or $\alpha =\delta$
and in either case $\alpha -\gamma =0$ or $\alpha -\delta =0$
and hence the given expression is zero

Therefore from (2) the required condition is

${(q-s)}^2=(p-r)(qr-ps).$