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#### bergausstein

##### Active member

- Jul 30, 2013

- 191

what is the rule on how to group this kind of denominator to get the conjugate.

$\displaystyle \frac{2+\sqrt{3}+\sqrt{5}}{2+\sqrt{3}-\sqrt{5}}$

thanks!

- Thread starter bergausstein
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- Thread starter
- #1

- Jul 30, 2013

- 191

what is the rule on how to group this kind of denominator to get the conjugate.

$\displaystyle \frac{2+\sqrt{3}+\sqrt{5}}{2+\sqrt{3}-\sqrt{5}}$

thanks!

- Mar 22, 2013

- 573

- Thread starter
- #3

- Jul 30, 2013

- 191

what do you mean by "with resperct to $\sqrt{5}$ and $\sqrt{3}$?

- Mar 22, 2013

- 573

- Thread starter
- #5

- Jul 30, 2013

- 191

my answer is

$\displaystyle \sqrt{3}+\frac{2\sqrt{5}}{11}+\frac{3\sqrt{15}}{22}$

is this correct?

$\displaystyle \sqrt{3}+\frac{2\sqrt{5}}{11}+\frac{3\sqrt{15}}{22}$

is this correct?

- Mar 22, 2013

- 573

No. Can you check your calculations?

Last edited:

- Mar 1, 2012

- 249

\(\displaystyle \dfrac{2+\sqrt3+\sqrt5}{2+\sqrt3-\sqrt5}\)

\(\displaystyle \dfrac{2+\sqrt3+\sqrt5}{(2+\sqrt3)-\sqrt5} \times \dfrac{2+\sqrt3+\sqrt5}{(2+\sqrt3)+\sqrt5}\)

\(\displaystyle =\dfrac{(2+\sqrt3+\sqrt5)^2}{(2+\sqrt3)^2-5}\)

I would now expand \(\displaystyle (2+\sqrt3)^2\) to get \(\displaystyle 7 + 4\sqrt(3)\)

If you put that in the fraction and collect like terms (ie: 7-5=2):

\(\displaystyle =\dfrac{(2+\sqrt3+\sqrt5)^2}{2+4\sqrt3}\)

Are you aware of the conjugate in this last expression? Generally when you rationalise the denominator nobody cares about the numerator - unless explicitly told otherwise leave it as is

- Feb 15, 2012

- 1,967

$\dfrac{2+\sqrt{3}+\sqrt{5}}{2+\sqrt{3}-\sqrt{5}}$

$= \dfrac{2+\sqrt{3}+\sqrt{5}}{2+\sqrt{3}-\sqrt{5}}\cdot\dfrac{2-(\sqrt{3}-\sqrt{5})}{2-(\sqrt{3}-\sqrt{5})}$

$=\dfrac{6+4\sqrt{5}}{4 - (\sqrt{3} - \sqrt{5})^2}$

$=\dfrac{6+4\sqrt{5}}{4 - (8 - 2\sqrt{15})}$

$= \dfrac{6 + 4\sqrt{5}}{2\sqrt{15} - 4}$

At this point, you want to multiply by $2\sqrt{15} + 4$ top and bottom:

$= \dfrac{6 + 4\sqrt{5}}{2\sqrt{15} - 4}\cdot\dfrac{2\sqrt{15} + 4}{2\sqrt{15} + 4}$

whereupon you will arrive at an answer equivalent to SuperSonic4's (if you were to follow his next step).