Question

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

Answer 1

Answer: 3

$ta{n}^{-1}\frac{1}{1+2x}+ta{n}^{-1}\frac{1}{1+4x}=ta{n}^{-1}\frac{2}{x^2}$
or $ta{n}^{-1}\left[\frac{\frac{1}{1+2x}+\frac{1}{1+4x}}{1-\frac{1}{1+2x}\frac{1}{1+4x}}\right]=ta{n}^{-1}\frac{2}{x^2}$
or $\frac{2+6x}{6x+8x^2}=\frac{2}{x^2}$
or $6x^3-14x^2-12x=0$
or $x\left(x-3\right)\left(3x+2\right)=0$
$orx=3$ 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 $x=3$.