Question

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

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

Answer 1

Answer: (A)

$\sqrt{5}$

(A) Given, $x=8-\sqrt{60}$
$\Rightarrow \sqrt{x}=\sqrt{8-\sqrt{60}}$
$=\sqrt{8-2\sqrt{15}}$
$=\sqrt{5}-\sqrt{3}$
$\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}+\sqrt{3}}{2}$
$=\frac{2}{\sqrt{x}}=\sqrt{5}+\sqrt{3}$
$\therefore [\frac{1}{2}\sqrt{x}+\frac{2}{\sqrt{x}}]=\frac{1}{2}[\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}]$
$=\sqrt{5}$