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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!

  • 0
    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.2010-09-22
  • 0
    Do you want to rationalize $11/(3\sqrt3-7)$ or $11/(3\sqrt3+7)$?2010-09-22

6 Answers 6

5

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 $\rm\; (a+b\sqrt 3)\:(a-b\sqrt 3)\ =\ a^2 - 3 b^2$

3

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.

3

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

2

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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Apparently you made a minor operational mistake when interpreting $ 3\sqrt{3} + 7 $ as its conjugate, $ 3\sqrt{3} - 7 $. The proper way to rationalize the denominator containing a radical is to remove that radical by multiplying the numerator and denominator by the original radical's denominator's conjugate, $ 3\sqrt{3} - 7 $:

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

Multiplying the above expression yields:

$$ \frac{33\sqrt{3} - 77}{9\cdot3 - 21\sqrt{3} + 21\sqrt{3} - 49} $$

Canceling the $ 21\sqrt{3} $ in the denominator and simplifying,

$$ \frac{33\sqrt{3} - 77}{9\cdot3 - 21\sqrt{3} + 21\sqrt{3} - 49} = \frac{33\sqrt{3} - 77}{27 - 49} = \frac{33\sqrt{3} - 77}{-22} = -\frac{33\sqrt{3} - 77}{22}$$

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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.2013-05-05