Question

Rationalising $\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}$ will give __________.

• A. $5-2\sqrt{6}$
• B. $5+2\sqrt{6}$
• C. $7+2\sqrt{6}$
• D. $7-2\sqrt{6}$

Answer 1

The correct option is B

$5+2\sqrt{6}$

We rationalise the denominator by dividing and multiplying the number by $\sqrt{2}+\sqrt{3}$.

So, we have

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

[$\text{Using the identity}(a+b{)}^2=a^2+b^2+2ab$]

$=5+2\sqrt{6}$.