Question:
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to
(a) $\frac{1}{2}(3-2 \sqrt{2})$
(b) $\frac{1}{3+2 \sqrt{2}}$
(c) $3-2 \sqrt{2}$
(d) $3+2 \sqrt{2}$
Solution:
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2 \sqrt{2}}=\frac{1}{3-2 \sqrt{2}} \cdot \frac{3+2 \sqrt{2}}{3+2 \sqrt{2}}$ $[\because \sqrt{8}=\sqrt{2 \times 2 \times 2}=2 \sqrt{2}]$
[multiplying numerator and denominator by $3+2 \sqrt{2}]$
$=\frac{3+2 \sqrt{2}}{9-(2 \sqrt{2})^{2}}$ [using identity $(a-b)(a+b)=a^{2}-b^{2}$ ]
$=\frac{3+2 \sqrt{2}}{9-8}=3+2 \sqrt{2}$