Question

If $x=8-\sqrt{60}$, then $\frac{1}{2}[\sqrt{x}+\frac{2}{\sqrt{x}}]=$.....

• A. $\sqrt{5}$
• B. $\sqrt{3}$
• C. $2\sqrt{5}$
• D. $2\sqrt{3}$

Answer 1

The correct option is A $\sqrt{5}$

$x=8-\sqrt{60}=\left[\right(\sqrt{5}{)}^2=2\sqrt{3}\sqrt{5}+\left(\sqrt{3}{)}^2\right]$

$\Rightarrow x=(\sqrt{5}-\sqrt{3}{)}^2$

Hence $\frac{1}{2}[\sqrt{x}+\frac{2}{\sqrt{x}}]=\frac{1}{2}[\sqrt{5}-\sqrt{3}+\frac{2}{\sqrt{5}-\sqrt{3}}]$

$\Rightarrow \frac{1}{2}[\sqrt{5}-\sqrt{3}+\frac{2(\sqrt{5}+\sqrt{3})}{2}]$

$\Rightarrow \sqrt{5}$