Question

If $x$ $=$ $5+2$ $\sqrt{6}$, find the value of $\sqrt{x}+\frac{1}{\sqrt{x}}$

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

Answer 1

The correct option is A $2\sqrt{3}$

$\sqrt{x}+\frac{1}{\sqrt{x}}=?$

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

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

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

$=7+2\sqrt{6}+\frac{5-2\sqrt{6}}{5^2-(2\sqrt{6}{)}^2}=7+2\sqrt{6}+\frac{5-2\sqrt{6}}{25-24}$

$=7+2\sqrt{6}+5-2\sqrt{6}=12$

$\therefore \sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{12}=2\sqrt{3}$