Question

$\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$

• A. $\sqrt{10}+\sqrt{2}$
• B. $\frac{\sqrt{10}+\sqrt{5}}{\sqrt{3}}$
• C. $\sqrt{10}+\sqrt{5}$
• D. $\sqrt{10}+\sqrt{11}$

Answer 1

The correct option is C

$\sqrt{10}+\sqrt{5}$

Simplification

$$\begin{gathered} \frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}} \\ =\frac{15}{\sqrt{10}+\sqrt{2^2\times 5}+\sqrt{2^2\times 10}-\sqrt{5}-\sqrt{4^2\times 5}} \\ =\frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}} \\ =\frac{15}{3\sqrt{10}-3\sqrt{5}} \\ =\frac{15}{3\left(\sqrt{10}-\sqrt{5}\right)} \\ =\frac{5}{\sqrt{10}-\sqrt{5}} \end{gathered}$$ $\left[\sqrt{a^2}=a\right]$

$=\frac{5\left(\sqrt{10}+\sqrt{5}\right)}{\left(\sqrt{10}-\sqrt{5}\right)\left(\sqrt{10}+\sqrt{5}\right)}$ [Rationalization]

$$\begin{gathered} =\frac{5\left(\sqrt{10}+\sqrt{5}\right)}{{\left(\sqrt{10}\right)}^2-{\left(\sqrt{5}\right)}^2} \\ =\frac{5\left(\sqrt{10}+\sqrt{5}\right)}{10-5} \\ =\frac{5\left(\sqrt{10}+\sqrt{5}\right)}{5} \\ =\sqrt{10}+\sqrt{5} \end{gathered}$$ $\left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right]$

Therefore, $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$ is equal to $\sqrt{10}+\sqrt{5}$

Hence, option (C) is correct.