12
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ What happens if you now multiply by $\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{12}-\sqrt{5}}$? $\endgroup$
    – Henry
    Commented Mar 21, 2014 at 7:41
  • $\begingroup$ @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. $\endgroup$ Commented Mar 21, 2014 at 7:43
  • $\begingroup$ You say you have a denominator of $\sqrt{12}+\sqrt{5}$, so it seemed the obvious thing to do as $(\sqrt{12}+\sqrt{5})( \sqrt{12}-\sqrt{5})= \sqrt{12}^2-\sqrt{5}^2=12-5=7$ $\endgroup$
    – Henry
    Commented Mar 21, 2014 at 7:46
  • $\begingroup$ No no I mean the original case of $\sqrt{2} + \sqrt{3} + \sqrt{5}$ $\endgroup$ Commented Mar 21, 2014 at 7:47
  • 1
    $\begingroup$ In general you do what you have one: remove one of the square roots in the denominator and then remove the others $\endgroup$
    – Henry
    Commented Mar 21, 2014 at 7:48

4 Answers 4

11
$\begingroup$

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!

$\endgroup$
0
3
$\begingroup$

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

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

$\endgroup$
0
2
$\begingroup$

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

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

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

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

Can you see it at least now?

$\endgroup$
12
  • 1
    $\begingroup$ 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... $\endgroup$
    – 5xum
    Commented Mar 21, 2014 at 7:23
  • $\begingroup$ Also, how would I know by which fraction to multiply to rationalize a fraction with a trinomial denominator as in this case. $\endgroup$ Commented Mar 21, 2014 at 7:24
  • $\begingroup$ @5xum : That is more than a hint... That is OP's responsibility to fill the gap which can be easily done.... $\endgroup$
    – user87543
    Commented Mar 21, 2014 at 7:26
  • $\begingroup$ @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 $2\sqrt{6}$ $\endgroup$
    – user87543
    Commented Mar 21, 2014 at 7:28
  • $\begingroup$ @downvoter : Why down vote? $\endgroup$
    – user87543
    Commented Mar 21, 2014 at 7:29
1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .