First slide

Theory of equations

Question

If the roots of the equation $\frac{1}{x + a} + \frac{1}{x + b} = \frac{1}{c}$
are equal in magnitude but opposite in sign, then their product is

Moderate

A

$\frac{1}{2} (a^{2} + b^{2})$
B

$- \frac{1}{2} (a^{2} + b^{2})$
C

$\frac{1}{2} a b$
D

$- \frac{1}{2} a b$

Solution

The equation (1) can be written as
$c (x + b + x + a) = (x + a) (x + b)$
or $x^{2} + (a + b - 2 c) x + a b - a c - b c = 0$ (2)
Let $α$ and $- α$ be the roots of (2) then
$0 = α + (- α) = a + b - 2 c \Rightarrow c = \frac{1}{2} (a + b)$
Also, $2 α (- α) = 2 a b - 2 (a + b) c = 2 a b - (a + b)^{2}$
$\begin{array}{l} = - (a^{2} + b^{2}) \\ \Rightarrow α (- α) = - \frac{1}{2} (a^{2} + b^{2}) \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App