# Simplify a surd expression

I won't lie, I have the following questions in my maths homework but I have no idea how to solve it and I wondered if you could help me here! My teacher has taught us, what seems to be the basics of surds now, but it no seems quite apparent that he hasn't taught us everything - still, he wants to test us seeing as though we are doing a maths GCSE is November; so I get the impression that with the knowledge we already have, it would be possible to work the question out eventually. But anyway, the question is as follows:

Given that $135 = 3^3 \cdot 5$, simply the expression - Give your answer in surd form

$$\frac{\sqrt{135}}{\sqrt{7}-\sqrt{12}}$$

Any help would be greatly appreciated, although I straight answer isn't really that helpful to anyone as I really need to know how you got the answer - so any explainations would be brilliant!

Thanks in advance

Everyone,

Thank you very much for all the help you have provided (or attempted to provide :) )! I am so reluctant and somewhat afraid to announce it, but it turns out that there was an error in the question. When I went back to maths, my maths teacher admitted that even he struggled with the question because it was not the question it was supposed to be, and he said "the board that produce these example exam questions must have made a typo...", as it is supposed to be ${\sqrt{75}}$ not ${\sqrt{7}}$! Suddenly the question is considerably more easy to solve....

So, once again, much appreciation for everyone who has posted on this question, and apologies for wasting people's time!

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Just a comment: I am a professional research mathematician, and I don't know what "Give your answer in surd form" means. (I am vaguely aware that "quadratic surd" is old-fashioned terminology for something like $\sqrt{x}$...) Probably someone here will know what this means, but some clarification certainly couldn't hurt. –  Pete L. Clark Sep 25 '11 at 18:54
Just to be certain: is it $\dfrac{\sqrt{135}}{\sqrt7}-\sqrt{12}$, as you wrote it, or is it $\dfrac{\sqrt{135}}{\sqrt7-\sqrt{12}}$? –  Brian M. Scott Sep 25 '11 at 19:09
Probably what is meant by "surd form" is some linear combination of radicals with rational coefficients. So you need only rationalize the denominator. Further, based on the given factorization, you may be expected to pull square factors out of radicands. You may wish to double-check the statement of the problem for errors, since it appears too simple as stated. –  Bill Dubuque Sep 25 '11 at 19:11
Andy, please avoid using [homework] as the only tag for the question. You should add at least one more suitable tag hinting on the content of the question itself (e.g. [algebra-precalculus] might be fitting here). –  Asaf Karagila Sep 25 '11 at 19:12
The new version of the question makes it much more clear what is intended: very likely you should multiply numerator and denominator by $\sqrt{7} + \sqrt{12}$ and simplify. –  Pete L. Clark Sep 25 '11 at 19:47

## 1 Answer

Here's a similar problem worked out (assuming that "in surd form" means something like "as a sum of roots of square-free integers multiplied by rational numbers"):

\begin{align} \frac{\sqrt{875}}{\sqrt{11}-\sqrt{45}} &=\frac{\sqrt{875}(\sqrt{11}+\sqrt{45})}{(\sqrt{11}-\sqrt{45})(\sqrt{11}+\sqrt{45})}\\ &=\frac{\sqrt{875}(\sqrt{11}+\sqrt{45})}{\sqrt{11}^2-\sqrt{45}^2}\\ &=\frac{\sqrt{875}(\sqrt{11}+\sqrt{45})}{11-45}\\ &=\frac{\sqrt{5^3\cdot7}(\sqrt{11}+\sqrt{3^2\cdot5})}{-34}\\ &=\frac{5\sqrt{5\cdot7}(\sqrt{11}+3\sqrt{5})}{-34}\\ &=\frac{5\sqrt{5\cdot7\cdot11}+15\sqrt{5\cdot5\cdot7}}{-34}\\ &=\frac{5\sqrt{5\cdot7\cdot11}+75\sqrt{7}}{-34}\\ &=-\frac5{34}\sqrt{385}-\frac{75}{34}\sqrt{7}\\ \end{align}

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Could the downvoter please explain the downvote? –  joriki Sep 25 '11 at 21:37
$@$joriki: I upvoted your answer. My only guess is that the downvoter felt somehow that you were doing the OP's homework. Of course you didn't answer the problem asked but a similar problem, so in order for the OP to adapt your answer to his question it seems necessary and sufficient for him to learn the techniques behind this type of algebra problem. In my opinion, what you are doing to the OP could best be described as "teaching", so I find the downvote very puzzling. –  Pete L. Clark Sep 25 '11 at 22:28
@Pete: Interesting -- I was thinking that the downvote might be for the opposite reason, that I solved a slightly different problem and didn't answer the question as posed :-) –  joriki Sep 25 '11 at 22:29
$@$ joriki: Well then, I think we have just proved that it is, in theory, impossible to please everyone all the time. We should write this up for publication...oh, but maybe this is an already known result? –  Pete L. Clark Sep 25 '11 at 22:31
It's also worth noticing that $\sqrt{135}=\sqrt{9}\sqrt{15}$, and $\sqrt{9}$ can be simplified. –  Michael Hardy Sep 25 '11 at 22:39