Question

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

Answer 1

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

$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$

$\frac{x+p+x+q}{(x+p)(x+q)}=\frac{1}{r}$

$(2x+p+q)r=x^2+px+qx+pq$

$x^2+(p+q-2r)x+pq-pr-qr=0$

Let $\alpha$ and $\beta$ be the roots.

$⟹\alpha +\beta =-(p+q-2r)$ ...[1]

$⟹\alpha \beta =pq-pr-qr$ ...[2]

Roots are equal in magnitude and opposite in sign

$⟹\alpha +\beta =0$ .

$⟹-(p+q-2r)=0$ ...[3]

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

$=(-(p+q-2r){)}^2-2(pq-pr-qr)$ ...(from [1] and [2])

$=p^2+q^2+4r^2+2pq-4pr-4qr-2pq+2pr+2qr$

$=p^2+q^2+4r^2-2pr-2qr$

$=p^2+q^2+2r(2r-p-q)$ ...(from [3])

$=p^2+q^2+0$

$=p^2+q^2$

Hence, answer is option (B)