Question
Write the correct answer in the following:
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to
  1. $\frac{1}{2}(3-2\sqrt{2})$
  2. $\frac{1}{3+2\sqrt{2}}$
  3. $3-2\sqrt{2}$
  4. $3+2\sqrt{2}$

Answer

  1. $3+2\sqrt{2}$
Solution:
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}=\frac{1}{3-2\sqrt{2}}\cdot\frac{3+2\sqrt{2}}{3+2\sqrt{2}}$ $[\because\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}]$
$[$multiplying numerator and denominator by $3+2\sqrt{2}]$
$\frac{3+2\sqrt{2}}{9-(2-\sqrt{2})^2}$$[\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$
$=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$

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