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Tricky algebraic equation

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm helping a friend out with an optimization problem, and I wound up with a tricky equation in which I have to solve for x. Is there a particular way to go about doing this?

    [itex]4x^2 - 2x \sqrt{r^2 - x^2} - 2r^2 = 0[/itex],

    where [itex]r[/itex] is a positive constant.

    2. Relevant equations

    3. The attempt at a solution
    So far I've only typed the equation into WolframAlpha, but the solution they provide is beyond my understanding.
     
  2. jcsd
  3. Jun 19, 2012 #2
    Do you want to solve for x in terms of r?
     
  4. Jun 19, 2012 #3
    Put the sqrt on the other side of the eqn and square both sides, should be able to do it from there.
     
  5. Jun 19, 2012 #4
    Yes.
    Oh, I don't know why I hadn't considered that. Unfortunately, after simplifying it a bit, I'm still not sure how to solve for x:
    [itex]4x^2 - 2x\sqrt{r^2 - x^2} - 2r^2 = 0[/itex]
    [itex]2x^2 - x\sqrt{r^2 - x^2} - r^2 = 0[/itex]
    [itex]2x^2 - r^2 = x\sqrt{r^2 - x^2}[/itex]
    [itex](2x^2 - r^2)^2 = (x\sqrt{r^2 - x^2})^2[/itex]
    [itex]4x^4 - 4x^2r^2 + r^4 = x^2 (r^2 - x^2)[/itex]
    [itex]4x^4 - 4x^2r^2 + r^4 = x^2r^2 - x^4[/itex]
    [itex]5x^4 - 5x^2r^2 + r^4 = 0[/itex]
     
  6. Jun 19, 2012 #5

    Curious3141

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    Bring the term with the square root to the RHS. Square both sides. You should now have a quartic in x, which is actually a quadratic in x2. Solve for x2. At this stage, put the values of x2 you get back into the original equation to ensure they satisfy the original problem. Squaring equations may introduce redundant roots.

    Finally, take square roots to get x.
     
  7. Jun 19, 2012 #6

    Curious3141

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    As I wrote, quartic in x, but a quadratic in x2. To make it clearer, let x2 = y and solve for y.
     
  8. Jun 19, 2012 #7

    micromass

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    Put [itex]y=x^2[/itex] then you get a quadratic equation.
     
  9. Jun 19, 2012 #8

    Dick

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    Now substitute u=x^2 and you have a quadratic equation in u. You can solve that, right?
     
  10. Jun 19, 2012 #9
    Ah, thanks. Making that kind of substitution isn't as second-nature as it should be.
     
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