Question

If $x=\left(5+2\sqrt{6}\right)$, show that $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)=\pm 2\sqrt{3}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ x=5+2\sqrt{6} \\ \Rightarrow \frac{1}{x}=\frac{1}{5+2\sqrt{6}}\times \frac{5-2\sqrt{6}}{5-2\sqrt{6}}=\frac{5-2\sqrt{6}}{5^2-{\left(2\sqrt{6}\right)}^2}=5-2\sqrt{6} \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ x+\frac{1}{x}=5+2\sqrt{6}+5-2\sqrt{6}=10 \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ {\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2=x+\frac{1}{x}+2 \\ \Rightarrow {\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2=10+2 \\ \Rightarrow {\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^2=12 \end{gathered}$$
$$\begin{gathered} \mathrm{Taking}\mathrm{the}\mathrm{square}\mathrm{root}\mathrm{of}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \Rightarrow \sqrt{x}+\frac{1}{\sqrt{x}}=\pm 2\sqrt{3} \end{gathered}$$