Question

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

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

Answer 1

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

$\sqrt{A}=\sqrt{5+2\sqrt{6}}=\sqrt{3+2+2\sqrt{3\times 2}}=\sqrt{(\sqrt{3}{)}^2+\sqrt{2}{)}^2+2\sqrt{3}\sqrt{2}}=\sqrt{(\sqrt{3}+\sqrt{2}{)}^2}=\sqrt{3}+\sqrt{2}$

$\therefore \sqrt{A}+\frac{1}{\sqrt{A}}=(\sqrt{3}+\sqrt{2})+\frac{1}{(\sqrt{3}+\sqrt{2})}$

$=(\sqrt{3}+\sqrt{2})+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$
$=(\sqrt{3}+\sqrt{2})+\frac{\sqrt{3}-\sqrt{2}}{3-2}$

$=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}$

$=2\sqrt{3}$