Question

Evaluate $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$, it being given that $\sqrt{5}=2.236\mathrm{and}\sqrt{10}=3.162$.
Hint
$\begin{array}{rcl}\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}& =& \frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}\\ & =& \frac{15}{3\sqrt{10}-3\sqrt{5}}=\frac{5}{\sqrt{10}{\displaystyle -}{\displaystyle \sqrt{5}}}\end{array}$

Answer 1

$\begin{array}{rcl}\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}& =& \frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}\\ & =& \frac{15}{3\sqrt{10}-3\sqrt{5}}\\ & =& \frac{5}{\sqrt{10}{\displaystyle -}{\displaystyle \sqrt{5}}}\\ & =& \frac{5}{\sqrt{10}{\displaystyle -}{\displaystyle \sqrt{5}}}\times \frac{\sqrt{10}+\sqrt{5}}{\sqrt{10}+\sqrt{5}}\\ & =& \frac{5\left(\sqrt{10}+\sqrt{5}\right)}{{\left(\sqrt{10}\right)}^2{\displaystyle -}{\displaystyle {\left(\sqrt{5}\right)}^2}}\\ & =& \frac{5\left(\sqrt{10}+\sqrt{5}\right)}{10-5}\\ & =& \frac{5\left(\sqrt{10}+\sqrt{5}\right)}{5}\\ & =& \sqrt{10}+\sqrt{5}\\ & =& 3.162+2.236\left(\mathrm{given}\right)\\ & =& 5.398\end{array}$

Hence, $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$ = 5.398 .