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The question is $$\frac{4-3\sqrt3}{4\sqrt6}$$ The first attempt of solving it, I got -6 but when I checked my answer using Mathway.com, it came out as: $$\frac{4\sqrt{6}-9\sqrt{2}}{24}$$ So is this correct and how would I get this answer if it is? ​​

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    What did you do? – 2017-01-07
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    Can you please show us your steps? – 2017-01-07
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    You could use a calculator and easily see which answer, if any, is correct. So we'll leave that question to you. You need to show us more of your work before we show you our methods. We want to prevent this from becoming a homework-answering site. – 2017-01-07

2 Answers 2

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Hint: Try multiplying both numerator and denominator by $\sqrt{6}$:

$$ \frac{(4 - 3\sqrt{3})\sqrt{6}}{4\sqrt{6}\times \sqrt{6}}$$

$\sqrt{a} \times \sqrt{a} = a$, $\sqrt{a}\times \sqrt{b} = \sqrt{ab}$, $ \ \forall \ a \ge 0$

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    this is only right for $$a\geq 0$$ – 2017-01-07
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Well, by rationalization, you just want the denominator to be a rational number, so you multiply top and bottom by $\sqrt 6$, to eliminate the radical in the denominator: $$ \frac{4-3\sqrt 3}{4 \sqrt 6} = \frac{(4-3\sqrt 3)\sqrt 6}{4 \sqrt 6 \times \sqrt 6} = \frac{4\sqrt 6 -3\sqrt 18}{4 \times 6} = \frac{4\sqrt 6 -9\sqrt 2}{24} $$ Which is the answer you have been given by mathway.com. $(3 \sqrt 18 = 3 \sqrt 9 \sqrt 2 = 3 \times 3 \sqrt 2 = 9 \sqrt 2)$

Now, if you have got the answer $-6$, and my answer is not informative, then write down your attempt in your question, so that we can go over what's wrong with that. If you wanted the answer, then this is it.

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    I multiplied everything by the square root of 6 and got (4sqrt(6)-3sqrt(18))/(24). Then I took 4 out of four to get 0 and took 4 out of 24 to get six So (\sqrt(6)-3\sqrt(18))/(6). Then took 3 out of the three and six. Subtracted the leftover roots on top and divided the rest of it to get -6. – 2017-01-07
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    If you take out a factor you need to factor it out of the every term in the enumerator, i.e., not just of $4\sqrt{6}$ but of $3\sqrt{18}$. Also how do you get 5 when you take 4 out of 24? – 2017-01-07
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    6, sorry. I'll fix that. – 2017-01-07
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    I see! Thank you for explaining yourself, it only helps me clear your doubt. So this is what you did: you have $\frac {4 \sqrt 6 - 3\sqrt{18}}{24}$, and you decided you were going to remove four from both the top and bottom. You did this by removing $4$ from the bottom, to get six. On the top, you removed four from ONE of the terms, leaving the other term $(3 \sqrt{18})$ intact. "This is incorrect". When removing four from top and bottom, you must remove four from ALL TERMS involved (terms are separated by addition). Hence, you must divide $3 \sqrt{18}$ by four as well! – 2017-01-07
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    Oh and when you take 4 out of $4\sqrt{6}$ the remaining part is $\sqrt{6}$, not 0. – 2017-01-07
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    Oh, okay. I see now. Thanks so much everyone! – 2017-01-07
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    To help you reason this out, suppose we did this with numbers, say $\frac{8+12}{4}$ is in consideration. If you remove $4$ from only $4$ and $8$, you would get $\frac{2+12}{1}$, which is fourteen. On the other hand, adding $8$ and $12$, gives $20$, and $\frac{20}{4} = 5$, which is the correct answer, and not $12$. I hope you see why EVERY term, needs to be divided by the denominator, rather than just one, for cancellation. Great, I'm happy you have understood, I know many who still have not, so that's an achievement. – 2017-01-07