Question : What is the value of $x$ in the equation $\sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}}=\frac{1}{\sqrt6}$?
Option 1: –2
Option 2: 3
Option 3: 2
Option 4: None of these
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Correct Answer: 2
Solution :
Given: $\sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}}=\frac{1}{\sqrt6}$
Squaring both sides
⇒ $\left(\sqrt{\frac{1+x}{x}}-\sqrt{\frac{x}{1+x}}\right)^{2}=\left(\frac{1}{\sqrt6}\right)^{2}$
⇒ $\frac{1+x}{x} +\frac{x}{1+x}-2=\frac{1}{6}$
⇒ $\frac{1+x}{x} +\frac{x}{1+x}=\frac{13}{6}$
⇒ $\frac{1+x^{2}+2x+x^{2}}{x+x^{2}}=\frac{13}{6}$
⇒ $6+6x^{2}+12x+6x^{2}=13x+13x^{2}$
⇒ $6+12x^{2}+12x=13x+13x^{2}$
⇒ $x^{2}+x-6=0$
⇒ $x^{2}+3x-2x-6=0$
⇒ $x(x+3)-2(x+3)=0$
⇒ $(x+3)(x-2)=0$
⇒ $x=-3, 2$
From the given options we can say that the value of $x$ is 2.
Hence, the correct answer is 2.
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