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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I haven't used my algebra skills much for years and they seem to have atrophied significantly!

I'm having real trouble working out how to re-arrange a formula I've come across to get $x$ by itself on the left hand side. It looks like this:


I've tried every method I can remember but I can't get rid of those pesky square roots!

Any ideas?

share|cite|improve this question
Is it not just a question of squaring on both sides? – utdiscant Mar 12 '11 at 23:40
@utdiscant: Not "just" that; that will still leave a radical and a nested radical on the right hand side. You can probably go from there, though. – Arturo Magidin Mar 12 '11 at 23:46
There are so many square roots that I wind up with as many after squaring (due to cross terms) as there were before. – Ross Millikan Mar 12 '11 at 23:59
I've had Mathematica churning away at isolating $x$ for a few minutes now—since Mathematica didn't isolate the variable in a few seconds, I'm guessing that isolating it is not at all simple. – Isaac Mar 13 '11 at 0:44

I would start by multiplying the numerator and denomenator on the right by $E-\sqrt{F+G\sqrt{A^2-x^2}}$: $$\frac{x}{\sqrt{A^2 - x^2}} = \frac{\left(B + \sqrt{C + Dx}\right)\left(E - \sqrt{F + G\sqrt{A^2 - x^2}}\right)}{E^2-F+G\sqrt{A^2 - x^2}}$$

It may also help the manipulation to set $y = \sqrt{A^2 - x^2}$ for a while.

share|cite|improve this answer
@F.P.Adams: Note that the LHS side with that substitution becomes: $\frac{x^2}{y^2}$ – Orbling Mar 13 '11 at 0:38
"squaring the left side" sounds like an abusive way to treat an equality – yasmar Mar 13 '11 at 0:40
derp yeah sorry – leif Mar 13 '11 at 2:37

Your Answer


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