Question

If the roots of the equation $\frac{1}{x+p}$ + $\frac{1}{x+q}$ = $\frac{1}{r}$ are equal in magnitude and opposite in sign, then

• A. $p+q=r$
• B. $p+q=2r$
• C. Product of roots $=$ $-\frac{1}{2}(p^2+q^2)$
• D. Sum of roots $=1$

Answer 1

Answer: (B), (C)

The correct options are
B $p+q=2r$
C Product of roots $=$ $-\frac{1}{2}(p^2+q^2)$

Given equation is : $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$

$\Rightarrow$ $r(2x+p+q)=(x+p)(x+q)$

$\Rightarrow$ $x^2+(p+q-2r)x+pq-pr-qr=0\dots \left(1\right)$

From the equation we get, sum of roots $=\frac{-(p+q-2r)}{1}=$$2r-p-q$

Product of roots $=\frac{pq-pr-qr}{1}=pq-pr-qr$

As the roots are equal and opposite,we get sum of roots $=0\Rightarrow 2r-p-q=0$

$\Rightarrow p+q=2r\dots \left(2\right)$

Substituting $r$ from equation $\left(2\right)$ in equation $\left(1\right)$, we get Product of roots

$=pq-p\left(\frac{p+q}{2}\right)-q\left(\frac{p+q}{2}\right)$

$=pq-\frac{p^2}{2}-\frac{pq}{2}-\frac{qp}{2}-\frac{q^2}{2}$

$=-\frac{p^2}{2}-\frac{q^2}{2}$

$=-\frac{1}{2}\left(p^2{+q}^2\right)\dots \left(3\right)$

Hence, options (B) and (C) are the correct choices.