Question

If $x=7+4\sqrt{3}$, find the value of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$.

Answer 1

$$\begin{gathered} \text{Given}, \\ x=7+4\sqrt{3} \\ \Rightarrow x=4+4\sqrt{3}+3 \\ \Rightarrow x={\left(2\right)}^2+2\times 2\times \sqrt{3}+{\left(\sqrt{3}\right)}^2 \\ \Rightarrow x={\left(2+\sqrt{3}\right)}^2 \\ \Rightarrow \sqrt{x}=\pm \left(2+\sqrt{3}\right) \\ \mathrm{When}\sqrt{x}=2+\sqrt{3}, \\ \frac{1}{\sqrt{x}}=\frac{1}{2+\sqrt{3}} \\ \Rightarrow \frac{1}{\sqrt{x}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ \Rightarrow \frac{1}{\sqrt{x}}=\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \\ \Rightarrow \frac{1}{\sqrt{x}}=\frac{2-\sqrt{3}}{4-3} \\ \Rightarrow \frac{1}{\sqrt{x}}=2-\sqrt{3}\mathrm{Therefore},\mathrm{when}\sqrt{x}=-\left(2+\sqrt{3}\right),\frac{1}{\sqrt{x}}=-\left(2-\sqrt{3}\right)=-2+\sqrt{3}\mathrm{Therefore},\sqrt{x}+\frac{1}{\sqrt{x}}=2+\sqrt{3}+2-\sqrt{3}\mathrm{or}-\left(2+\sqrt{3}\right)+\left(-2+\sqrt{3}\right)=4\mathrm{or}-4 \end{gathered}$$