Question

Simplify $\frac{6-4\sqrt{3}}{6+4\sqrt{3}}$ by rationalizing the denominator.

Answer 1

Given, $\frac{6-4\sqrt{3}}{6+4\sqrt{3}}$

$=\frac{6-4\sqrt{3}}{6+4\sqrt{3}}\times \frac{6-4\sqrt{3}}{6-4\sqrt{3}}$ [ On Rationalizing the denominator ]

$=\frac{{(6-4\sqrt{3})}^2}{(6+4\sqrt{3})(6-4\sqrt{3})}$

$=\frac{36-48\sqrt{3}+48}{36-48}$

$=\frac{84-48\sqrt{3}}{-12}$

$=-(7-4\sqrt{3})$

$=4\sqrt{3}-7$