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

Expression : $$\sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}}=\frac{1}{\sqrt{6}}$$

Squaring both sides, we get :

=> $$(\sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}})^2=(\frac{1}{\sqrt{6}})^2$$

=> $$(\frac{1+x}{x})-(\frac{x}{1+x})-2(\sqrt{\frac{1+x}{x}})(\sqrt{\frac{x}{1+x}})=\frac{1}{6}$$

=> $$\frac{(1+x)^2+x^2}{x(1+x)}=\frac{1}{6}+2$$

=> $$\frac{x^2+2x+1+x^2}{x^2+x}=\frac{13}{6}$$

=> $$\frac{2x^2+2x+1}{x^2+x}=\frac{13}{6}$$

=> $$13x^2+13x=12x^2+12x+6$$

=> $$x^2+x-6=0$$

=> $$x^2+3x-2x-6=0$$

=> $$x(x+3)-2(x+3)=0$$

=> $$(x+3)(x-2)=0$$

=> $$x=2,-3$$

=> Ans - (C)