Question

Let $p,qandr$ be real numbers $(p\ne q,r\ne 0)$, such that the roots of the equation $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

• A. $\frac{p^2+q^2}{2}$
• B. $2(p^2+q^2)$
• C. $p^2+q^2$
• D. $p^2+q^2+r^2$

Answer 1

The correct option is C $p^2+q^2$

Converting the given equation in general form
we get
$\left[\right(x+q)+(x+p\left)\right]r=(x+p)(x+q)$

$x^2+(p+q–2r)x+pq–pr–qr=0$

$p+q=2r...\left(1\right)$

$\alpha =-\beta$

$(\alpha +\beta {)}^2-2\alpha \beta ={\alpha }^2+{\beta }^2$

$[\because (\alpha +\beta )=0]$

${\alpha }^2+{\beta }^2=-2\alpha \beta$
$=-2(pq-pr-qr)=-2pq+2r(p+q)\text{from}\left(1\right){\alpha }^2+{\beta }^2=p^2+q^2$