Question: If the roots of equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude ...

Question

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

Answer 1

Given equation can be written as

$x^{2} + (p + q - 2r)x + \lbrack pq - (p + q)r\rbrack = 0$

Since the roots are equal and of opposite sign,

∴ Sum of roots = 0

⇒ $- (p + q - 2r) = 0$ ⇒ $p + q = 2r$

Question: If the roots of equation \(\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}\) are equal in magnitude ...