Question

When denominator is rationalised, then the number $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$ becomes?

• A. $3\sqrt{2}+2\sqrt{3}$
• B. $3\sqrt{2}-2\sqrt{3}$
• C. $3\sqrt{2}+2\sqrt{2}$
• D. $3\sqrt{2}-2\sqrt{2}$

Answer 1

The correct option is B $3\sqrt{2}-2\sqrt{3}$

Given, $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

Rationalizing the denominator,

$=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}$

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

$=\frac{\sqrt{12}-\sqrt{18}}{2-3}$

$=\frac{\sqrt{2\times 2\times 3}-\sqrt{2\times 3\times 3}}{-1}$

$=\frac{2\sqrt{3}-3\sqrt{2}}{-1}$

$=3\sqrt{2}-2\sqrt{3}$