So I got this question on my math homework.

$$ \frac{2\sqrt{2}-2\sqrt{3}}{4\sqrt{3}+4\sqrt{2}} $$

The instructions are to simplify and rationalize the denominator. I've been trying to get the correct answer for a while now, but the answers I get never look right. Any pointers in the right direction would be greatly appreciated. Thank you in advance!

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    $\begingroup$ What methods have you tried so far? What are your thoughts? $\endgroup$ – Mnifldz Jan 12 '17 at 23:07
  • $\begingroup$ I've tried multiplying the bottom terms by the top and foiling them. $\endgroup$ – Grimestock Jan 12 '17 at 23:08
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    $\begingroup$ What can you multiply the bottom by to get a difference of squares? $\endgroup$ – Mnifldz Jan 12 '17 at 23:09
  • $\begingroup$ I suggest you put the terms in the same order top and bottom. i.e. $\frac {2\sqrt 2 - 2\sqrt 3}{4\sqrt 2 + 4\sqrt 3}$ it doesn't change anything, but it makes it easier to keep organized as you simplify. $\endgroup$ – Doug M Jan 12 '17 at 23:12

$$\frac{2\sqrt{2}-2\sqrt{3}}{4\sqrt{3}+4\sqrt{2}}=\frac{2(\sqrt{2}-\sqrt{3})}{4(\sqrt{3}+\sqrt{2})}$$ $$=\frac{\sqrt{2}-\sqrt{3}}{2(\sqrt{3}+\sqrt{2})}$$ $$=\frac{(\sqrt{2}-\sqrt{3})\cdot (\sqrt{3}-\sqrt{2})}{2(\sqrt{3}+\sqrt{2}) \cdot (\sqrt{3}-\sqrt{2})}$$ $$=\frac{2\sqrt{6}-5}{2}$$



  • Divide numerator and denominator by $2$;

  • When one has a fraction with $\sqrt{a} \pm \sqrt{b}$ in the denominator, it helps if you multiply both numerator and denominator of the fraction by the conjugate of the denominator, $\sqrt{a} \mp \sqrt{b}$. That is, if the denominator is $\sqrt{a}+\sqrt{b}$ multiply by $\sqrt{a}-\sqrt{b}$ and if the denominator is $\sqrt{a}-\sqrt{b}$ then multiply by $\sqrt{a}+\sqrt{b}$

  • $\begingroup$ $(a- b)(a+b)= a^2- b^2$ $\endgroup$ – user247327 Jan 12 '17 at 23:14
  • $\begingroup$ @user247327 Indeed, that is the motivation; $\endgroup$ – RGS Jan 12 '17 at 23:17
  • $\begingroup$ +1 for full answer that is well responded to homework questions. $\endgroup$ – Simply Beautiful Art Jan 12 '17 at 23:19
  • $\begingroup$ @SimpleArt thanks, but what do you mean? $\endgroup$ – RGS Jan 12 '17 at 23:22
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    $\begingroup$ I like good answers that provide how to solve the problem without actually solving it in the right places. $\endgroup$ – Simply Beautiful Art Jan 12 '17 at 23:23

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