Question 3: \((iv)\) If \(x=3+\sqrt{8}\) then find \(x^2-\large{\frac{1}{x^2}} \)
Solution:
\[\begin{align*}
x&=3+\sqrt{8}\\
\frac{1}{x} &=\frac{1}{3+\sqrt{8}}\\
&=\frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}}\\
&=\frac{3-\sqrt{8}}{3^{2} -\left(\sqrt{8}\right)^{2}}\\
&=\frac{3-\sqrt{8}}{9-8}\\
&=3-\sqrt{8}\\
x^{2} -\frac{1}{x^{2}} &=\left( 3+\sqrt{8}\right)^{2} -\left( 3-\sqrt{8}\right)^{2}\\&=3^{2} +\left(\sqrt{8}\right)^{2} +2\times 3\times \\&\hspace{1.3cm}\sqrt{8} -(3^{2} +(\sqrt{8})^{2} -2\\&\hspace{4cm}\times 3\times \sqrt{8})\\
&=9+8+6\sqrt{8} -9-8+\\&\hspace{5cm}6\sqrt{8}\\
&=12\sqrt{8}\\
\end{align*}\]