Question

If the equation
$(p^2+q^2)x^2-2(pr+qs)x+r^2+s^2=0$ has equal roots then

• A. $pq=rs$
• B. $ps=rq$
• C. $ps=\sqrt{rq}$
• D. $ps=\sqrt{rs}$

Answer 1

The correct option is B $ps=rq$

$\propto =\beta$
$\frac{-b-\sqrt{b^2-4ac}}{2\propto }=\frac{-2+\sqrt{b^2-4ac}}{2\propto }$
$b^2-4ac=0$
${[-2(pq+qs)]}^2-4(p^2+q^2)(r^2+s^2)=0$
$4p^2{r}^2+4q^2{s}^2+8pqrs-4p^2{r}^2-4q^2{r}^2-4q^2{s}^2$
$p^2{s}^2+q^2{r}^2-2pqrs=0$
${(ps-qr)}^2=0$
$ps=qr$