Question

Question 13
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to

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

Answer 1

The answer is D.
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}=\frac{1}{3-2\sqrt{2}}.\frac{3+2\sqrt{2}}{3+2\sqrt{2}}[\because \sqrt{8}=\sqrt{2\times 2\times 2}=2\sqrt{2}]$
[Multiplying numerator and denominator by $3+2\sqrt{2}]$
[Using identity $(a-b)(a+b)=a^2-b^2]$
$=\frac{3+2\sqrt{2}}{9-(2\sqrt{2}{)}^2}=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$