Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals:

My approach:

I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long calculation):


which is totally not in accordance with the answer, $\sqrt{2}+\sqrt{3}-\sqrt{5}$.

Can someone please explain this/give hints to me.

share|cite|improve this question
What happens if you now multiply by $\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{12}-\sqrt{5}}$? – Henry Mar 21 '14 at 7:41
@Henry ok,trying, but please tell how did you derive this number? I mean, how would I know by which fraction to multiply to rationalize a fraction with a trinomial denominator as in this case. – Gaurang Tandon Mar 21 '14 at 7:43
You say you have a denominator of $\sqrt{12}+\sqrt{5}$, so it seemed the obvious thing to do as $(\sqrt{12}+\sqrt{5})( \sqrt{12}-\sqrt{5})= \sqrt{12}^2-\sqrt{5}^2=12-5=7$ – Henry Mar 21 '14 at 7:46
No no I mean the original case of $\sqrt{2} + \sqrt{3} + \sqrt{5}$ – Gaurang Tandon Mar 21 '14 at 7:47
In general you do what you have one: remove one of the square roots in the denominator and then remove the others – Henry Mar 21 '14 at 7:48

4 Answers 4

up vote 7 down vote accepted

What I would do is multiply by the first term plus the conjugate of the last two terms. I have coloured the important parts of the following expression to make it easier to understand. $$\frac{2\sqrt{6}}{\color{green}{\sqrt{2}+\sqrt{3}+}\color{red}{\sqrt{5}}}\cdot\frac{\color{green}{\sqrt{2}+\sqrt{3}+}\color{red}{\sqrt{5}}}{\color{green}{\sqrt{2}+\sqrt{3}+}\color{red}{\sqrt{5}}}$$ $$\frac{2\sqrt{6}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})}$$ Why do I do this, you ask? Remember the difference of squares formula: $$a^2-b^2=(a+b)(a-b)$$ I am actually letting $a=\color{green}{\sqrt{2}+\sqrt{3}}$ and $b=\color{red}{\sqrt{5}}$. Therefore, our fraction can be rewritten as: $$\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{(\color{green}{\sqrt{2}+\sqrt{3}})^2-(\color{red}{\sqrt{5}})^2}$$ $$=\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2+2\sqrt{6}+3-5}$$ Oh. How nice. The integers in the denominator cancel out! $$\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2\sqrt{6}}$$ Multiply by $\dfrac{2\sqrt{6}}{2\sqrt{6}}$ $$\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2\sqrt{6}}\cdot\frac{2\sqrt{6}}{2\sqrt{6}}$$ $$=\frac{8\sqrt{3}\sqrt{6}+12\sqrt{2}\sqrt{6}-4\sqrt{30}\sqrt{6}}{(2\sqrt{6})^2}$$ $$=\frac{\color{red}{24}\sqrt{2}+\color{red}{24}\sqrt{3}-\color{red}{24}\sqrt{5}}{\color{red}{24}}$$ $$=\frac{\color{red}{24}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\color{red}{24}}$$ Cancel $24$ out in the numerator and denominator and you get: $$\sqrt{2}+\sqrt{3}-\sqrt{5}$$ $$\displaystyle \color{green}{\boxed{\therefore \dfrac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\sqrt{2}+\sqrt{3}-\sqrt{5}}}$$

There is actually a much shorter way. Let's go back to the fraction $$\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}$$ $$=\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{2+2\sqrt{2}\sqrt{3}+3-5}$$ $$=\frac{\color{red}{2\sqrt{6}}\sqrt{2}+\color{red}{2\sqrt{6}}\sqrt{3}-\color{red}{2\sqrt{6}}\sqrt{5}}{\color{red}{2\sqrt{6}}}$$ Do you see that we can factor out $2\sqrt{6}$ in the numerator? $$\frac{\color{red}{2\sqrt{6}}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\color{red}{2\sqrt{6}}}$$ Cancel $2\sqrt{6}$ in the numerator and the denominator out, and you get: $$\sqrt{2}+\sqrt{3}-\sqrt{5}$$ $$\displaystyle \color{green}{\boxed{\therefore \dfrac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\sqrt{2}+\sqrt{3}-\sqrt{5}}}$$
Hope I helped!

share|cite|improve this answer
Exactly the guidance I needed. Thanks ! – Gaurang Tandon Mar 21 '14 at 8:08

Your "long calculation" was obviously wrong, since the denominator should be

$$(\sqrt 2 + \sqrt 3 + \sqrt 5)(\sqrt 2 - (\sqrt 3 + \sqrt 5))=\\=\sqrt2^2 - (\sqrt 3 + \sqrt 5)^2 = 2 - (3 + 5 + 2\sqrt{15}) = -6-2\sqrt{15}$$

share|cite|improve this answer




Cancelling those $2\sqrt{6}$ you would end up with required result...

Can you see it at least now?

share|cite|improve this answer
That last equality is not really well justified, don't you think? I mean, in the denominator, you should get $2+3+2\sqrt 6 - 5$, you did not explain how you get rid of $\sqrt 6$ then... – 5xum Mar 21 '14 at 7:23
Also, how would I know by which fraction to multiply to rationalize a fraction with a trinomial denominator as in this case. – Gaurang Tandon Mar 21 '14 at 7:24
@5xum : That is more than a hint... That is OP's responsibility to fill the gap which can be easily done.... – Praphulla Koushik Mar 21 '14 at 7:26
@GaurangTandon : As i Have $2\sqrt{6}$ in the numerator, First thing you should make sure is to get $2\sqrt{6}$ in the denominator if possible and cancel it out.. to get such thing you should not disturb $\sqrt{2}+\sqrt{3}$ as that would give on squaring a $2\sqrt{6}$ – Praphulla Koushik Mar 21 '14 at 7:28
@downvoter : Why down vote? – Praphulla Koushik Mar 21 '14 at 7:29

Hint $\ (\sqrt a\!+\!\sqrt b\, + \sqrt c)(\sqrt a\! +\!\sqrt b\, - \sqrt c)\, =\, a\!+\!b\!-\!c+2\sqrt{ab}\ \ (=\, 2\sqrt{ab}\ \ {\rm if}\ \ a+b=c)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.