Question

If the sum of the roots of the quadratic equation $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ is zero, then the product of the roots is $-\left[\frac{p^2+q^2}{2}\right]$
Say yes or no.

• A. Yes
• B. No
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is A Yes

$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$
$r(x+q+x+p)=(x+p)(x+q)$
$2rx+rp+rq=x^2+px+qx+pq$
$x^2+x(p+q-2r)+pq-rp-rq=0$
Sum of the roots $=-[p+q-2r]=0$ or $r=\frac{p+q}{2}$
Product of roots $=pq-r(p+q)$
= $pq-(p+q)\left(\frac{(p+q)}{2}\right)$
= $pq-\frac{(p+q{)}^2}{2}$
= $\frac{2pq-p^2-q^2-2pq}{2}$
= $\frac{-p^2-q^2}{2}$