What is the value of x in the equation $$\sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} = \frac{1}{{\sqrt 6 }}?$$

Solution (By Examveda Team)

$$\eqalign{ & \sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} = \frac{1}{{\sqrt 6 }} \cr & {\text{On squaring both sides,}} \cr & \Rightarrow {\left( {\sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} } \right)^2} = \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} - 2 = \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} = 2 + \frac{1}{6} \cr & \Rightarrow \frac{{1 + x}}{x} + \frac{x}{{1 + x}} = \frac{{13}}{6} \cr & \Rightarrow \frac{{{{\left( {1 + x} \right)}^2} + {x^2}}}{{x\left( {1 + x} \right)}} = \frac{{13}}{6} \cr & \Rightarrow \frac{{1 + {x^2} + 2x + {x^2}}}{{x + {x^2}}} = \frac{{13}}{6} \cr & \Rightarrow \left( {2{x^2} + 2x + 1} \right)6 = 13\left( {x + {x^2}} \right) \cr & \Rightarrow 12{x^2} + 12x + 6 = 13x + 13{x^2} \cr & \Rightarrow {x^2} + x - 6 = 0 \cr & \Rightarrow {x^2} + 3x - 2x - 6 = 0 \cr & \Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0 \cr & \Rightarrow \left( {x + 3} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2{\text{ as }}x \ne - 3 \cr & {\text{Or, of the given options, when }}x = 2 \cr & {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \sqrt {\frac{{1 + x}}{x}} - \sqrt {\frac{x}{{1 + x}}} \cr & = \sqrt {\frac{{1 + 2}}{2}} - \sqrt {\frac{2}{{1 + 2}}} \cr & = \frac{{\sqrt 3 }}{{\sqrt 2 }} - \frac{{\sqrt 2 }}{{\sqrt 3 }} \cr & = \frac{{3 - 2}}{{\sqrt 6 }} \cr & = \frac{1}{{\sqrt 6 }} \cr} $$