Question

Solve
$\frac{1}{(a+b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} ,[x \neq0,x \neq - (a + b)]$
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Answer 1

Answer

$\dashrightarrow \sf \: \dfrac{1}{(a+b + x)} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{x}$

$\sf \dashrightarrow \dfrac{1}{(a + b + x)} - \dfrac{1}{x} = \dfrac{1}{a} + \dfrac{1}{b}$

$\sf \dashrightarrow \dfrac{x - (a + b + x)}{x \: (a + b + x)} = \dfrac{b + a}{ab}$

$\sf \dashrightarrow \dfrac{- (a + b)}{x \: (a + b + x)} = \dfrac{(a + b)}{ab}$

On dividing both sides by (a + b),

${\sf \dashrightarrow \dfrac{-1}{x (a + b + x)} = \dfrac{1}{ab}}$

By cross multiplication,

$\sf \dashrightarrow x (a + b + x) = - ab$

$\sf \dashrightarrow {x} ^{2} + ax + bx + ab = 0$

$\sf \dashrightarrow x (x + a) + b (x + a) = 0$

$\sf \dashrightarrow (x + a) (x + b) = 0$

$\sf \dashrightarrow x + a = 0 \: \: or \: \: x + b =0$

$\sf \dashrightarrow x = - a \: \: or \: \: x = - b$

Thus, - a and - b are the roots of the given equation.