Question

If $x=5+2\sqrt{6}$, then the value of ${(\sqrt{x}-\frac{1}{\sqrt{x}})}^2$ is _____

• A. $4\sqrt{6}$
• B. $8$
• C. $16$
• D. $12$
• E. None of these

Answer 1

Answer: (B)

$8$

$(\sqrt{x}-\frac{1}{\sqrt{x}}{)}^2$

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

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

Given that $x=5+2\sqrt{6}$

$\Rightarrow \frac{1}{x}=\frac{1}{5+2\sqrt{6}}$

Rationalize the denominator

$\Rightarrow \frac{1}{x}=\frac{1}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}}$

$\Rightarrow \frac{1}{x}=\frac{5-2\sqrt{6}}{5^2-(2\sqrt{6}{)}^2}$

$\Rightarrow \frac{1}{x}=\frac{5-2\sqrt{6}}{25-24}=5-2\sqrt{6}$

$\Rightarrow \frac{1}{x}=5-2\sqrt{6}$

Therefore, $\Rightarrow x+\frac{1}{x}-2=(5+2\sqrt{6})+(5-2\sqrt{6})-2=8$

$\Rightarrow (\sqrt{x}-\frac{1}{\sqrt{x}}{)}^2=8$