# Rationalize the Denominator

• September 2nd 2012, 11:58 AM
Diesal
Rationalize the Denominator
http://i.imgur.com/psqwL.gif

I can't seem to solve this one. I tried reversing the signs, which didn't work for me. How do I solve this one?
• September 2nd 2012, 12:08 PM
Plato
Re: Rationalize the Denominator
Quote:

Originally Posted by Diesal

Multiply numerator and denominator by $(1-\sqrt{3}+\sqrt{5})$
• September 2nd 2012, 01:00 PM
Deveno
Re: Rationalize the Denominator
that won't quite do the job.

$(1 + \sqrt{3} - \sqrt{5})(1 - \sqrt{3} + \sqrt{5}) = 1^2 - (\sqrt{3} - \sqrt{5})^2$

$= 1 - (3 - 2\sqrt{15} + 5) = -7 + 2\sqrt{15}$, which certainly isn't rational.

i propose instead multiplying top and bottom by:

$7 + 3\sqrt{3} + \sqrt{5} + 2\sqrt{15}$, since:

$(1 + \sqrt{3} - \sqrt{5})(7 + 3\sqrt{3} + \sqrt{5} + 2\sqrt{15}) = 11$
• September 2nd 2012, 01:07 PM
Plato
Re: Rationalize the Denominator
You are correct. I entered is incorrectly into my CAS.
• September 2nd 2012, 07:51 PM
Prove It
Re: Rationalize the Denominator
Quote:

Originally Posted by Deveno
that won't quite do the job.

$(1 + \sqrt{3} - \sqrt{5})(1 - \sqrt{3} + \sqrt{5}) = 1^2 - (\sqrt{3} - \sqrt{5})^2$

$= 1 - (3 - 2\sqrt{15} + 5) = -7 + 2\sqrt{15}$, which certainly isn't rational.

Though you could then multiply top and bottom by \displaystyle \begin{align*} -7 - 2\sqrt{15} \end{align*}...