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 $\frac{3\sqrt{x}-\sqrt{3x}}{2}$?

  • 2
    $\begingroup$ Why do you get $\sqrt {12x}(2-\sqrt{3}) = 12x(2-\sqrt 3)$? You arbitrarily just removed the radical sign over the $\sqrt{12x}$. That was your error. Do it again without removing the radical sign. $\endgroup$ – fleablood Jan 6 '19 at 19:02
  • 2
    $\begingroup$ Your second error is multiplying numerator and denominator by $2-\sqrt 3$ instead of $2-2\sqrt 3$. $\endgroup$ – TonyK Jan 6 '19 at 19:07
  • $\begingroup$ And your third and fourth errors occur when you try to calculate $(2+2\sqrt 3)(2-\sqrt 3)$. $\endgroup$ – TonyK Jan 6 '19 at 19:08
  • $\begingroup$ @TonyK good catch. $\endgroup$ – fleablood Jan 6 '19 at 19:08

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: $$ \frac{\sqrt{12x}}{2+2\sqrt{3}}=\frac{2\sqrt{3x}}{2(\sqrt{3}+1)} =\frac{\sqrt{3x}}{\sqrt{3}+1}\frac{\sqrt{3}-1}{\sqrt{3}-1}= \frac{\sqrt{9x}-\sqrt{3x}}{3-1}=\frac{3\sqrt{x}-\sqrt{3x}}{2} $$


$\frac{\sqrt{12x}}{2+2\sqrt{3}}=\frac{\color{green}{\sqrt{12x}}}{2+2\sqrt{3}}\frac{2 - \sqrt{3}}{2 - \sqrt{3}}=\frac{\color{red}{12x}(2-\sqrt{3})}{\color{purple}{(2+2\sqrt{3})(2-\sqrt{3})}} = \frac{24x-12x\sqrt{3}}{\color{orange}{4+(2\cdot(-3))}}=\frac{24x-12x\sqrt{3}}{-2}$

You magically turned $\color{green}{\sqrt{12x}}$ into $\color{red}{12x}$ for no reason whatsoever.

And you incorrectly calculated $\color{purple}{(2+2\sqrt{3})(2-\sqrt{3})} = \color{orange}{4+(2\cdot(-3))}$.

You should have done:

$\frac{\sqrt{12x}}{2+2\sqrt{3}}=\frac{\sqrt{12x}}{2+2\sqrt{3}}\color{green}{\frac{2 - 2\sqrt{3}}{2 - 2\sqrt{3}}}=\frac{\sqrt{12x}(2-2\sqrt{3})}{\color{purple}{(2+2\sqrt{3})(2-2\sqrt{3})}} = \frac{...}{\color{purple}{4-4*3}}=...$

  • $\begingroup$ Thanks for the feedback and explanation. My first error re 12x was an outright error. I was not sure how to multiply out the denominator so thanks for clarifying how I do that here. $\endgroup$ – Doug Fir Jan 6 '19 at 19:19

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