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operation on radicals

bergausstein

Active member
Jul 30, 2013
191
can you tell me what is the proper conjugate of the denominator.
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!
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
I'd first multiply by the conjugate with respect to $\sqrt{5}$ and then multiply by the conjugate with respect to $\sqrt{3}$.
 

bergausstein

Active member
Jul 30, 2013
191
what do you mean by "with resperct to $\sqrt{5}$ and $\sqrt{3}$?
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
First multiply the denominator by $2 + \sqrt{3} + \sqrt{5}$ to get $\left (2 + \sqrt{3}\right )^2 - 5 = 2 + 4\sqrt{3}$ and then multiply by $2 - 4\sqrt{3}$
 

bergausstein

Active member
Jul 30, 2013
191
my answer is

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

is this correct?
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
No. Can you check your calculations?
 
Last edited:

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
Don't forget you also need to multiply the numerator

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

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

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

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
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
You can also do it "the other way":

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

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

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

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

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

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).