Question

Simplify of the following by rationalizing the denominator:
$\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$

Answer 1

For a given expression $\frac{a-\sqrt{b}}{a+\sqrt{b}}$, we multiply and divide the expression by its rationalizing factor which in this case is $a-\sqrt{b}$, to free the denominator from the root.

Given:$\frac{6-4\sqrt{2}}{6+4\sqrt{2}}$

Thus, multiplying by $(6-4\sqrt{2})$ numerator and denominator we get,

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

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

$=\frac{36-48\sqrt{2}-32}{36-32}$

$=\frac{4(9-12\sqrt{2}-8)}{4}$

$=1-12\sqrt{2}$