If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r},\ (x
eq - p,\ x
eq - q,\ r
eq 0)$ are equal in magnitude but opposite in sign, then $p + q$ is equal to

Question

If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r},\ (x 
eq - p,\ x 
eq - q,\ r 
eq 0)$ are equal in magnitude but opposite in sign, then $p + q$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given, $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$
$\Rightarrow r(2x + p + q) = \left\lbrack x^{2} + (p + q)x + pq ightbrack$
$\Rightarrow x^{2} + (p + q - 2r)x + pq - r(p + q) = 0$
As we know, if roots are equal in magnitude but opposite in sign, then coefficient of $x$ will be zero
$	herefore\ \ \ p + q - 2r = 0\ \  \Rightarrow p + q = 2r$

Q. If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r},\ (x \neq - p,\ x \neq - q,\ r \neq 0)$ are equal in magnitude but opposite in sign, then $p + q$ is equal to

Explanation

Reviewed by Dr. Abubacker

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