Question

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

• A. $5+2\sqrt{5}$
• B. $5-2\sqrt{6}$
• C. $5-2\sqrt{5}$
• D. $5+2\sqrt{6}$

Answer 1

The correct option is D $5+2\sqrt{6}$

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

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

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

$=5+2\sqrt{6}$