Question

If $\alpha ,\beta$ are the roots of $x^2+pa+q=0$ and $\gamma ,\delta$ are the roots of $x^2+rx+s=o$; 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

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

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

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

$({\alpha }^2+r\alpha +s)({\beta }^2+r\beta +s)$..........(1)

Since $\alpha$ is a root of $x^2+px+q=0$

$\therefore$ ${\alpha }^2+p\alpha +q=0$

or ${\alpha }^2=-p\alpha -q$ and similarly ${\beta }^2=-p\beta -q$.

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

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

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

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

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

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

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

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

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