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Question:

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

$$\frac{\sqrt{24}+\sqrt{40}-\sqrt{16}}{\sqrt{12}+\sqrt{5}}$$

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

Can someone please explain this/give hints to me.

Community
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Gaurang Tandon
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4 Answers4

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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: $$\sqrt{2}+\sqrt{3}-\sqrt{5}$$ $$\displaystyle \color{green}{\boxed{\therefore \dfrac{2\sqrt{6}}{\sqrt{2}+\sqrt3+\sqrt5}=\sqrt{2}+\sqrt{3}-\sqrt{5}}}$$
Hope I helped!
dbanet
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Anonymous Computer
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Your "long calculation" was obviously wrong, since the denominator should be

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

5xum
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$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$

$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-5}=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{5+2\sqrt{6}-5}$$

$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{2\sqrt{6}}$$

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

Can you see it at least now?

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    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 07:23
  • 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 07:24
  • @5xum : That is more than a hint... That is OP's responsibility to fill the gap which can be easily done.... –  Mar 21 '14 at 07:26
  • @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 $2\sqrt{6}$ –  Mar 21 '14 at 07:28
  • @downvoter : Why down vote? –  Mar 21 '14 at 07:29
  • Sorry, but then you are just confuzing the OP. The way you wrote the equation made it look as though all of the equalities are simple to check. With no indication that one of them requires at least 2 or 3 more lines to check, this is just very confuzing. I am the downvoter as I do not consider this answer helpful. – 5xum Mar 21 '14 at 07:29
  • @5xum : you do not have to check anything... It is already well written except the last equality for which only thing you need is the formula $(a+b)^2=a^2+b^2+2ab$ –  Mar 21 '14 at 07:31
  • I was confused when you jumped so quickly at the final step. It took me a while to catch what you actually meant. Maybe making it more explicit would greatly benefit the OP. – Yiyuan Lee Mar 21 '14 at 07:34
  • @PraphullaKoushik The last equality you wrote is nontrivial. It requires work to check. I am not talking about just squaring, but removing the $\sqrt 6$ from the denominator. At no point in your answer do you mention that a $\sqrt 6$ will appear, so the last equality is pure confusion. – 5xum Mar 21 '14 at 07:36
  • @5xum : Common, you have $\sqrt{6}$ in denominator and $\sqrt{6}$ in the numerator... Can't you cancel both of them?? –  Mar 21 '14 at 07:36
  • it may seem intuitive to us, but not necessarily to the rest... might as while edit and make more explicit your answer. – Yiyuan Lee Mar 21 '14 at 07:38
  • @YiyuanLee : My intention was not to give an answer.. As he said "after long calculation" I thought to check if it would be really long.... As i do not have pen and paper i have just checked it here (thanks to latex) and it worked well and i have submitted it... That is it... It is fixed now!! –  Mar 21 '14 at 07:48
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Hint $\ (\sqrt a\!+\!\sqrt b\, + \sqrt c)(\sqrt a\! +\!\sqrt b\, - \sqrt c)\, =\, a\!+\!b\!-\!c+2\sqrt{ab}\ \ (=\, 2\sqrt{ab}\ \ {\rm if}\ \ a+b=c)$

Bill Dubuque
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