Question

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

• A. $\frac{6\sqrt{2}-19}{-17}$
• B. $\frac{6\sqrt{2}-19}{17}$
• C. $\frac{17\sqrt{2}-6}{17}$

Answer 1

The correct option is B $\frac{6\sqrt{2}-19}{17}$

For rationalization, apply $a^2-b^2=(a+b)(a-b).$

$\frac{\sqrt{3}-3\sqrt{6}}{\sqrt{3}+3\sqrt{6}}=\frac{\sqrt{3}(1-3\sqrt{2})}{\sqrt{3}(1+3\sqrt{2})}$
(cancelling common terms from numberator and denominator)
$=\frac{1-3\sqrt{2}}{1+3\sqrt{2}}=\frac{(1-3\sqrt{2})\times (1-3\sqrt{2})}{(1+3\sqrt{2})\times (1-3\sqrt{2})}=\frac{(1-3\sqrt{2}{)}^2}{(1{)}^2-(3\sqrt{2}{)}^2}=\frac{(1{)}^2-2\times 1\times 3\sqrt{2}+(3\sqrt{2}{)}^2}{1-9\times 2}=\frac{1-6\sqrt{2}+9\times 2}{-17}=\frac{19-6\sqrt{2}}{-17}=\frac{6\sqrt{2}-19}{17}$