Question

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

• A. 5-2
• B. 5+2
• C. 7+2
• D. None of these

Answer 1

The correct option is B

5+2

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

So, we have

$\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{(\sqrt{3}+\sqrt{2}{)}^2}{{\sqrt{3}}^2-{\sqrt{2}}^2}$
$={\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+2\sqrt{6}$
[Using the identity $(a+b{)}^2=a^2+b^2+2ab$]
$=5+2\sqrt{6}$