What is the value of x in the equation sqrt{f | vedclass

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(C) Let $y = \sqrt{\frac{1+x}{x}}$. Then the equation becomes $y - \frac{1}{y} = \frac{1}{\sqrt{6}}$.Multiplying by $y$,we get $y^2 - \frac{1}{\sqrt{6

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(C) Let $y = \sqrt{\frac{1+x}{x}}$. Then the equation becomes $y - \frac{1}{y} = \frac{1}{\sqrt{6}}$.Multiplying by $y$,we get $y^2 - \frac{1}{\sqrt{6}}y - 1 = 0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-\frac{1}{\sqrt{6}}, c=-1$:$y = \frac{\frac{1}{\sqrt{6}} \pm \sqrt{\frac{1}{6} + 4}}{2} = \frac{\frac{1}{\sqrt{6}} \pm \sqrt{\frac{25}{6}}}{2} = \frac{\frac{1}{\sqrt{6}} \pm \frac{5}{\sqrt{6}}}{2}$.Taking the positive root,$y = \frac{6/\sqrt{6}}{2} = \frac{\sqrt{6}}{2} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}}$.Since $y = \sqrt{\frac{1+x}{x}}$,we have $\sqrt{\frac{1+x}{x}} = \sqrt{\frac{3}{2}}$.Squaring both sides,$\frac{1+x}{x} = \frac{3}{2}$.$2(1+x) = 3x \implies 2 + 2x = 3x \implies x = 2$.

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

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