Question

The number obtained on rationalising the denominator of $\frac{1}{\sqrt{9}-\sqrt{8}}$ is ?

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

Answer 1

Answer: (C)

We use the identity $(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$.

So, $\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{\sqrt{9}-\sqrt{8}}\times \left(\frac{\sqrt{9}+\sqrt{8}}{\sqrt{9}+\sqrt{8}}\right)$

$=\frac{\sqrt{9}+\sqrt{8}}{9-8}$

$=\frac{\sqrt{9}+\sqrt{8}}{1}$

$=\sqrt{3^2}+\sqrt{4\times 2}$

$=3+2\sqrt{2}$