Question

If $\frac{3+\sqrt{8}}{3-\sqrt{8}}+\frac{3-\sqrt{8}}{3+\sqrt{8}}=a+b\sqrt{2}$, then find $a$ and $b$.

Answer 1

$\begin{array}{c}\frac{3+\sqrt{8}}{3-\sqrt{8}}+\frac{3-\sqrt{8}}{3+\sqrt{8}}=a+b\sqrt{2}\\ \frac{{\left(3+\sqrt{8}\right)}^2+{\left(3-\sqrt{8}\right)}^2}{9-8}=a+b\sqrt{2}\\ 9+8+6\sqrt{8}+9+8-6\sqrt{8}=a+b\sqrt{2}\\ 34=a+b\sqrt{2}\\ a=34\\ b=0\end{array}$