I am asked to simplify $\frac{\sqrt{12x}}{2+2\sqrt{3}}$ and the solution is provided as $\frac{3\sqrt{x}-\sqrt{3x}}{2}$. I arrived at $-12x + 6x\sqrt{3}$ and I'm not sure how to arrive at the text book solution.
My working:
$$\frac{\sqrt{12x}}{2+2\sqrt{3}} = \frac{\sqrt{12x}}{2+2\sqrt{3}}\frac{2 – \sqrt{3}}{2 – \sqrt{3}} = \frac{12x(2-\sqrt{3})}{(2+2\sqrt{3})(2-\sqrt{3})} = \frac{24x-12x\sqrt{3}}{4+(2\cdot(-3))}=\frac{24x-12x\sqrt{3}}{-2}$$
Then, multiplying out the denominator I get:
$-12x+6x\sqrt{3}$
Is my thought process sound up to a point? Where did I go wrong and how can I arrive at $\frac{3\sqrt{x}-\sqrt{3x}}{2}$?
Best Answer
First error: $$ \sqrt{12x}(2-\sqrt{3})\ne 12x(2-\sqrt{3}) $$ Second error: $$ (2+2\sqrt{3})(2-\sqrt{3})=4+4\sqrt{3}-2\sqrt{3}-2\cdot3=2\sqrt{3}-2\ne-2 $$ Third error: in order to rationalize the denominator, you have to multiply by $2-2\sqrt{3}$.
On the other hand, you can proceed more simply: $$ \frac{\sqrt{12x}}{2+2\sqrt{3}}=\frac{2\sqrt{3x}}{2(\sqrt{3}+1)} =\frac{\sqrt{3x}}{\sqrt{3}+1}\frac{\sqrt{3}-1}{\sqrt{3}-1}= \frac{\sqrt{9x}-\sqrt{3x}}{3-1}=\frac{3\sqrt{x}-\sqrt{3x}}{2} $$