Question
Rationalize the denominator $\frac{8}{3 \sqrt{2}+\sqrt{5}}$
✓
Answer
The conjugate pair of $3 \sqrt{2}+\sqrt{5}$ is $3 \sqrt{2}-\sqrt{5}$
$
\begin{aligned}
\frac{8}{3 \sqrt{2}+\sqrt{5}} & =\frac{8}{3 \sqrt{2}+\sqrt{5}} \times \frac{3 \sqrt{2}-\sqrt{5}}{3 \sqrt{2}-\sqrt{5}} \\
& =\frac{8(3 \sqrt{2}-\sqrt{5})}{(3 \sqrt{2})^2-(\sqrt{5})^2} \\
& =\frac{8 \times 3 \sqrt{2}-8 \sqrt{5}}{9 \times 2-5}=\frac{24 \sqrt{2}-8 \sqrt{5}}{18-5}=\frac{24 \sqrt{2}-8 \sqrt{5}}{13}
\end{aligned}
$
$
\begin{aligned}
\frac{8}{3 \sqrt{2}+\sqrt{5}} & =\frac{8}{3 \sqrt{2}+\sqrt{5}} \times \frac{3 \sqrt{2}-\sqrt{5}}{3 \sqrt{2}-\sqrt{5}} \\
& =\frac{8(3 \sqrt{2}-\sqrt{5})}{(3 \sqrt{2})^2-(\sqrt{5})^2} \\
& =\frac{8 \times 3 \sqrt{2}-8 \sqrt{5}}{9 \times 2-5}=\frac{24 \sqrt{2}-8 \sqrt{5}}{18-5}=\frac{24 \sqrt{2}-8 \sqrt{5}}{13}
\end{aligned}
$
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