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For my homework, I have been asked to rationalise and simplify this surd;

$$\frac{11}{3\sqrt{3}+7}$$

Each time I do this I get the wrong answer. The method I am using is;

$$ \frac{11}{3\sqrt3-7} \times \frac{3\sqrt3-7}{3\sqrt3-7} $$

I ended up with $\frac{33+11\sqrt3-77}{9+3+21+7\sqrt3-21-7\sqrt3}$

This ends up no where near the right answer, even once it is simplified, can someone tell me where I'm going wrong?

Many thanks!

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I have edited part of your question to use the math formula support that is there is on this site. Since I cannot understand the rest of your question, I have left it to you to do the edits. –  Aryabhata Sep 22 '10 at 17:32
    
Do you want to rationalize $11/(3\sqrt3-7)$ or $11/(3\sqrt3+7)$? –  KennyTM Sep 22 '10 at 17:37
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5 Answers 5

up vote 4 down vote accepted

You're mistakenly multiplying $\rm\; a * b\sqrt{3} \ =\ ab + a\sqrt{3}\:\,\;$ but $\rm\; ab\:\sqrt{3}\;$ is correct.

In other words $\rm\; b\:\sqrt{3}\;$ means $\rm b * \sqrt{3}\:,\;$ not $\rm\; b + \sqrt{3}\:.$

Also, to rationalize the denominator use $\rm\; (a+b\sqrt 3)\:(a-b\sqrt 3)\ =\ a^2 - 3 b^2$

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After multiplying the numerator and denominator by $3\sqrt3-7$ the new denominator is $$(3\sqrt 3+7)(3\sqrt3-7)=(3\sqrt3)^2-7^2=27-49=-22$$ a nice integer to divide by.

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HINT: The general trick is $$ \frac{1}{\sqrt{a}+b}=\frac{\sqrt{a}-b}{(\sqrt{a}+b)(\sqrt{a}-b)}=\frac{\sqrt{a}-b}{a-b^2}. $$

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Do you mean rationalise

$$ \frac{11}{3\sqrt{3}-7} \qquad \text{?} $$

And are you sure you're trying

$$ \frac{11}{3\sqrt{3}-7} \cdot \frac{3\sqrt{3} + 7}{3\sqrt{3} + 7}\qquad \text{?} $$

In general, Wikipedia is almost always of great help too.

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the answer is

$$-\frac{33\sqrt3-77}{22}$$

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Thank you for your answer; please edit it to use MathJax (I'd edit it myself but I'm not sure what you're trying to say). Also, please elaborate on your solution. –  vadim123 May 5 '13 at 4:35
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