Question

The sum of roots of the equation $tan^{-1} \frac{1}{1+2x} + tan^{-1} \frac{1}{1+4x} = tan^{-1} \frac{2}{x^{2}} $ is_______.

Answer 1

$tan^{-1} \frac{1}{1+2x} + tan^{-1} \frac{1}{1+4x} = tan^{-1} \frac{2}{x^{2}}$
or $tan^{-1}\left[\frac{\frac{1}{1+2x}+\frac{1}{1+4x}}{1-\frac{1}{1+2x} \frac{1}{1+4x}}\right] = tan^{-1} \frac{2}{x^{2}} $
or $ \frac{2+6x}{6x+8x^{2}} = \frac{2}{x^{2}} $
or $6x^{3} - 14x^{2} - 12x = 0 $
or $x\left(x-3\right)\left(3x+2\right) = 0$
$or x = 3$ or $x = -2/3$ (as $x \ne 0)$
But for $x = -2/3$, L.H.S. $< 0$ and R.H.S. $> 0$
Hence, the only solution is $x = 3$.