Question

If $x=(3+\sqrt{8})$, find the value of $\left(x^2+\frac{1}{x^2}\right)$.

Answer 1

$$\begin{gathered} x=3+\sqrt{8} \\ \Rightarrow \frac{1}{x}=\frac{1}{3+\sqrt{8}} \\ \Rightarrow \frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}} \\ \Rightarrow \frac{1}{x}=\frac{3-\sqrt{8}}{9-8} \\ \Rightarrow \frac{1}{x}=3-\sqrt{8} \end{gathered}$$
$$\begin{gathered} \mathrm{and}x+\frac{1}{x}=3+\sqrt{8}+3-\sqrt{8}=6 \\ \mathrm{So},x+\frac{1}{x}=6 \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \Rightarrow {\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2=6^2 \\ \Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+2\times x\times \frac{1}{x}=36 \\ \Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+2=36 \\ \Rightarrow {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}=36-2 \\ \therefore {\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}=34 \end{gathered}$$