Question 4: Find the rational numbers \(p\) and \(q\) such that \(\frac{8-3\sqrt{2}}{4+3\sqrt{2}}=p+q\sqrt{2}\)
Solution:
\[p+q\sqrt{2}=\frac{8-3\sqrt{2}}{4+3\sqrt{2}} \hspace{4cm}\]
\[\begin{align*}
&=\frac{8-3\sqrt{2}}{4+3\sqrt{2}}\times \frac{4-3\sqrt{2}}{4-3\sqrt{2}}\\
&=\frac{\left(8-3\sqrt{2}\right)\left(4-3\sqrt{2}\right)}{4^2-\left(3\sqrt{2}\right)^2}\\
&=\frac{8\left( 4-3\sqrt{2}\right) -3\sqrt{2}\left( 4-3\sqrt{2}\right)}{16-9\times 2}\\
&=\frac{32-24\sqrt{2} -12\sqrt{2} +9\left(\sqrt{2}\right)^{2}}{16-18}\\
&=\frac{32-36\sqrt{2} +9\times 2}{-2}\\
&=\frac{32-36\sqrt{2} +18}{-2}\\
&=\frac{50-36\sqrt{2}}{-2}\\
&=\frac{-2\left( -25+18\sqrt{2}\right)}{-2}\\
&=-25+18\sqrt{2}\\
\end{align*}\]
Comparing both sides, we have
\[p=-25\hspace{1cm},\hspace{1cm}q=18\]