Question

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

Answer 1

$$\begin{gathered} x=2+\sqrt{3} \\ \mathrm{Now}, \\ \frac{1}{\mathrm{x}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ =\frac{2-\sqrt{3}}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =2-\sqrt{3} \\ \mathrm{So}, \\ x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3} \\ =4 \\ \therefore x+\frac{1}{x}=4 \\ \mathrm{Squaring}\mathrm{both}\mathrm{side}s,\mathrm{we}\mathrm{get}: \\ {x}^2+\frac{1}{x^2}+2=16 \\ \Rightarrow x^2+\frac{1}{x^2}=14 \\ \mathrm{Then},\mathrm{again}\mathrm{squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ {\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)}^2=196 \end{gathered}$$