Question

Find the values of $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$.

• A. $\sqrt{5}$$(\sqrt{3}+2)$
• B. $\sqrt{5}$$(\sqrt{3}+\sqrt{2})$
• C. $\sqrt{5}$$(\sqrt{2}+2)$
• D. $\sqrt{5}$$(\sqrt{2}+1)$

Answer 1

The correct option is D $\sqrt{5}$$(\sqrt{2}+1)$

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

= $\frac{15}{\sqrt{10}+2\sqrt{5}+2\sqrt{10}-\sqrt{5}-4\sqrt{5}}$

= $\frac{15}{3\sqrt{1}0-3\sqrt{5}}$

= $\frac{15}{3(\sqrt{1}0-\sqrt{5})}$

= $\frac{5}{(\sqrt{1}0-\sqrt{5})}$

Now, rationalising the denominator, we get,
= $\frac{5}{(\sqrt{10}-\sqrt{5})}\times \frac{(\sqrt{10}+\sqrt{5})}{(\sqrt{10}+\sqrt{5})}$

= $\frac{5(\sqrt{10}+\sqrt{5})}{10-5}$

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

= $\sqrt{2}\sqrt{5}+\sqrt{5}$

= $\sqrt{5}(\sqrt{2}+1)$