Question

Rationalise the following denominator:

$\frac{3}{\sqrt{6}-\sqrt{7}}$.

• A. $-3(\sqrt{16}+\sqrt{7})$
• B. $-3(\sqrt{3}+\sqrt{7})$
• C. $-3(\sqrt{6}+\sqrt{7})$
• D. $-3(\sqrt{6}+\sqrt{17})$

Answer 1

The correct option is C $-3(\sqrt{6}+\sqrt{7})$

We need to find value of $\frac{3}{\sqrt{6}-\sqrt{7}}$

On rationalizing denominator, we get

$\frac{3(\sqrt{6}+\sqrt{7})}{(\sqrt{6}-\sqrt{7})(\sqrt{6}+\sqrt{7})}$

$=\frac{3(\sqrt{6}+\sqrt{7})}{({\left(\sqrt{6}\right)}^2-{\left(\sqrt{7}\right)}^2))}$

$=\frac{3(\sqrt{6}+\sqrt{7})}{6-7}$

$=-3(\sqrt{6}+\sqrt{7})$