Question

If $x=\left(2+\sqrt{3}\right)$, find the value of $\left(x^2+\frac{1}{x^2}\right)$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ x=2+\sqrt{3} \\ \mathrm{Now}, \\ \frac{1}{x}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-{\left(\sqrt{3}\right)}^2}=2-\sqrt{3} \\ \Rightarrow x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4 \\ \Rightarrow x+\frac{1}{x}=4 \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ {\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2=4^2 \\ \Rightarrow {\mathrm{x}}^2+{\left(\frac{1}{\mathrm{x}}\right)}^2+2\times x\times \frac{1}{x}=16 \\ \Rightarrow x^2+\frac{1}{x^2}=14 \end{gathered}$$