Question

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

Answer 1

Given $x=5-2\sqrt{6}=3+2-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)=(\sqrt{3}-\sqrt{2}{)}^2$

$⟹\sqrt{x}=\pm (\sqrt{3}-\sqrt{2})$

$\sqrt{x}+\frac{1}{\sqrt{x}}=\pm (\sqrt{3}-\sqrt{2}+\frac{1}{\sqrt{3}-\sqrt{2}})=\pm (\sqrt{3}-\sqrt{2}+\frac{\sqrt{3}+\sqrt{2}}{3-2})=\pm (\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2})=\pm 2\sqrt{3}$