Question

Find the equivalent form of $\frac{4\sqrt{2}+3\sqrt{6}}{5\sqrt{6}+6\sqrt{2}}$.

• A. $\frac{2+21\sqrt{3}}{39}$
• B. $\frac{21+2\sqrt{3}}{39}$
• C. $\frac{21-2\sqrt{3}}{39}$
• D. $\frac{2-21\sqrt{3}}{39}$

Answer 1

The correct option is B $\frac{21+2\sqrt{3}}{39}$

Rationalize the denominator of $\frac{4\sqrt{2}+3\sqrt{6}}{5\sqrt{6}+6\sqrt{2}}$, to find its equivalent form.

In order to rationalize the denominator, multiply and divide the given fraction by $5\sqrt{6}-6\sqrt{2}$.

$\begin{array}{cc}\frac{4\sqrt{2}+3\sqrt{6}}{5\sqrt{6}+6\sqrt{2}}& =\frac{4\sqrt{2}+3\sqrt{6}}{5\sqrt{6}+6\sqrt{2}}\times \frac{5\sqrt{6}-6\sqrt{2}}{5\sqrt{6}-6\sqrt{2}}\\ & =\frac{4\sqrt{2}(5\sqrt{6}-6\sqrt{2})+3\sqrt{6}(5\sqrt{6}-6\sqrt{2})}{(5\sqrt{6}+6\sqrt{2})(5\sqrt{6}-6\sqrt{2})}\\ & =\frac{20\sqrt{12}-24\left(2\right)+15\left(6\right)-18\sqrt{12}}{(5\sqrt{6}{)}^2-(6\sqrt{2}{)}^2}\\ & =\frac{(20\times 2)\sqrt{3}-48+90-(18\times 2)\sqrt{3}}{150-72}\\ & =\frac{40\sqrt{3}+42-36\sqrt{3}}{78}\\ & =\frac{42+4\sqrt{3}}{78}\\ & =\frac{2(21+2\sqrt{3})}{78}\\ & =\frac{21+2\sqrt{3}}{39}\end{array}$