Question

Rationalise the denominator of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$.

• A. $-\left(\frac{\sqrt{2}+\sqrt{6}+2}{4}\right)$
• B. $\left(\frac{\sqrt{2}+\sqrt{6}+2}{4}\right)$
• C. $-\left(\frac{\sqrt{2}-\sqrt{6}+2}{4}\right)$
• D. $\left(\frac{\sqrt{2}-\sqrt{6}+2}{4}\right)$

Answer 1

The correct option is A $-\left(\frac{\sqrt{2}+\sqrt{6}+2}{4}\right)$

$\frac{1}{\sqrt{3}-\sqrt{2}-1}=\frac{1}{\sqrt{3}-\sqrt{2}-1}\times \frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{3}-\sqrt{2}+1}$
$=\frac{\sqrt{3}-\sqrt{2}+1}{(\sqrt{3}-\sqrt{2}-1)(\sqrt{3}-\sqrt{2}+1)}$
$=\frac{\sqrt{3}-\sqrt{2}+1}{{(\sqrt{3}-\sqrt{2})}^2-1^2}$
$=\frac{\sqrt{3}-\sqrt{2}+1}{3+2-2\sqrt{3}\times \sqrt{2}-1}=\frac{\sqrt{3}-\sqrt{2}+1}{4-2\sqrt{6}}$
$=\frac{\sqrt{3}-\sqrt{2}+1}{2(2-\sqrt{6})}=\frac{\sqrt{3}-\sqrt{2}+1}{2(2-\sqrt{6})}\times \frac{2+\sqrt{6}}{2+\sqrt{6}}$
$=\frac{(\sqrt{3}-\sqrt{2}+1)(2+\sqrt{6})}{2(2^2-{\left(\sqrt{6}\right)}^2)}$
$=\frac{2\sqrt{3}-2\sqrt{2}+2+\sqrt{3}\times \sqrt{6}-\sqrt{2}\times \sqrt{6}+\sqrt{6}}{2(4-6)}$
$=\frac{2\sqrt{3}-2\sqrt{2}+2+\sqrt{3\times 6}-\sqrt{2\times 6}+\sqrt{6}}{-4}$
$=\frac{2\sqrt{3}-2\sqrt{2}+2+\sqrt{3^2\times 2}-\sqrt{2^2\times 3}+\sqrt{6}}{-4}$
$=\frac{2\sqrt{3}-2\sqrt{2}+2+3\sqrt{2}-2\sqrt{3}+\sqrt{6}}{-4}$
$=\frac{\sqrt{2}+\sqrt{6}+2}{-4}=-\left(\frac{\sqrt{2}+\sqrt{6}+2}{4}\right)$