Question 3: \((iii)\) If \(x=3+\sqrt{8}\) then find \(x^2+\large{\frac{1}{x^2}} \)
Solution:
\[\begin{align*}
x&=3+\sqrt{8}\\
\frac{1}{x} &=\frac{1}{3+\sqrt{8}}\\
&=\frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}}\\
&=\frac{3-\sqrt{8}}{3^{2} -\left(\sqrt{8}\right)^{2}}\\
&=\frac{3-\sqrt{8}}{9-8}\\
&=3-\sqrt{8}\\
x^{2} +\frac{1}{x^{2}} &=\left( 3+\sqrt{8}\right)^{2} +\left( 3-\sqrt{8}\right)^{2}\\
&=3^{2} +\left(\sqrt{8}\right)^{2} +2\times 3\times \\ & \hspace{1.7cm} \sqrt{8} +3^{2} +\left(\sqrt{8}\right)^{2} – \\& \hspace{2.5cm}2\times 3\times \sqrt{8}\\
&=9+8+9+8\\
&=34\\
\end{align*}\]