Question 3: \((v)\) \(\quad\) If \(x=3+\sqrt{8}\) then find \(x^4+\large{\frac{1}{x^4}} \)
Solution:
\[\begin{align*}
x&=3+\sqrt{8}\\
\frac{1}{x} &=\frac{1}{3+\sqrt{8}}\\
&=\frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}}\\
&=\frac{3-\sqrt{8}}{3^{2} -\left(\sqrt{8}\right)^{2}}\\
&=\frac{3-\sqrt{8}}{9-8}\\
&=3-\sqrt{8}\\
x^{2} +\frac{1}{x^{2}} &=\small{\left( 3+\sqrt{8}\right)^{2} +\left( 3-\sqrt{8}\right)^{2}}\\
&=3^{2} +\left(\sqrt{8}\right)^{2} +2\times 3\times \\&\hspace{1.3cm}\sqrt{8} +3^{2} +\left(\sqrt{8}\right)^{2} -\\&\hspace{3cm}2\times 3\times \sqrt{8}\\
&=\small{9+8+6\sqrt{8}+9+8-6\sqrt{8}}\\
&=9+8+9+8\\
&=34\\ \end{align*}\]
Taking Square on both sides
\[\begin{align*}\left( x^{2} +\frac{1}{x^{2}}\right)^{2} &=( 34)^{2}\\
x^{4} +\frac{1}{x^{4}} +2&=1156\\
x^{4} +\frac{1}{x^{4}} &=1156-2\\
x^{4} +\frac{1}{x^{4}} &=1154\\
\end{align*}\]