Question

If the roots of $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ are equal in magnitude but opposite in sign and the product of roots is $k(p^2+q^2)$, then $k=$

• A. $-1$
• B. $-\frac{1}{2}$
• C. $-\frac{3}{2}$
• D. $-2$

Answer 1

The correct option is B $-\frac{1}{2}$

$r(x+p)(x+q)=2x+p+q$
$x^2+(p+q)x+pq-2rx-pr-qr=0$
$x^2+x(p+q-2r)-pr-qr+pq=0$
Sum of the roots will be $0$.
Therefore,
$p+q-2r=0$
$(p+q)=2r$
Product of the roots
$=pq-r(p+q)$
$=pq-\frac{p^2+q^2+2pq}{2}$
$=\frac{-1}{2}(p^2+q^2)=k(p^2+q^2)$