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Solving a cubic sort of!

  1. Jul 7, 2011 #1
    Solving a cubic.... sort of!

    Hi,

    Can the equation [itex]\sqrt{Ax-x^3}+\sqrt{Bx-x^3}=C[/itex] be solved explicitly?

    All of MathLab, Maple and WolframAlpha seem to give an explicit solution but they dont show how they come to it. I'm afraid they may be missing other possible solutions.

    Thanks,
     
  2. jcsd
  3. Jul 7, 2011 #2

    lanedance

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    Re: Solving a cubic.... sort of!

    why not start by simplifying to
    [tex] (\sqrt{A}+\sqrt{B})\sqrt{x-x^3} = C[/tex]
     
  4. Jul 7, 2011 #3

    lanedance

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    Re: Solving a cubic.... sort of!

    note that if you square things you may need to account for the fact the argument of a square root cannot be zero
     
  5. Jul 7, 2011 #4
    Re: Solving a cubic.... sort of!

    Well, for starters, your factorization is incorrect...
     
  6. Jul 7, 2011 #5

    lanedance

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    Re: Solving a cubic.... sort of!

    good point, read it incorrectly, must be late ;)... will have another look
     
    Last edited: Jul 7, 2011
  7. Jul 7, 2011 #6

    lanedance

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    Re: Solving a cubic.... sort of!

    so i did a wolfram alpha check and the solution looks nasty... so there probably is a closed form method but its likely pretty involved & tedious, after looking at it closer, I can't see any easy way to simplify...
     
  8. Jul 7, 2011 #7

    lanedance

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    Re: Solving a cubic.... sort of!

    getting late, so i've probably made a mistake, but see what you think of this logic...
    assume x>0 and hopefully we don't divide by zero anywhere
    [tex] \sqrt{x}(\sqrt{A-x^2}+\sqrt{B-x^2})=C[/tex]
    multiply by the rational to get
    [tex] \sqrt{x}(A-B)=C(\sqrt{A-x^2}-\sqrt{B-x^2})[/tex]

    re-arrange both to get
    [tex] \sqrt{A-x^2}+\sqrt{B-x^2}=\frac{C}{\sqrt{x}}[/tex]
    [tex] \sqrt{A-x^2}-\sqrt{B-x^2}=\frac{\sqrt{x}(A-B)}{C}[/tex]

    adding and subtracting we get to
    [tex] 2\sqrt{A-x^2}=\frac{C}{\sqrt{x}}+\frac{\sqrt{x}(A-B)}{C}[/tex]
    [tex] 2\sqrt{B-x^2}= \frac{C}{\sqrt{x}}-\frac{\sqrt{x}(A-B)}{C}[/tex]

    square both
    [tex] 4(A-x^2)=\frac{C^2}{x}+2(A-B)+\frac{x(A-B)^2}{C^2}[/tex]
    [tex] 2(B-x^2)= \frac{C^2}{x}-2(A-B)+\frac{x(A-B)^2}{C^2}[/tex]

    which is starting to look a bit more tractable as in effect we can solve a cubic now...
     
  9. Jul 7, 2011 #8
    Re: Solving a cubic.... sort of!

    Here's another thought...

    In the original equation, solutions exist only for positive C, and both sides of the equation are trivially positive where solutions exist (in the real numbers).

    Given this, square both sides of the equation; since if f(x) = g(x) and both f(x) and g(x) are positive, then [f(x)]^2 = [g(x)]^2 iff f(x) = g(x).

    Rearrange terms so that only the radical term (only one remains after squaring) is on the right, and everything else is on the left of the equation. Clearly the right-hand side is positive, so solutions exist only when the left-hand side is positive... we must check our answers for x to ensure that this condition is satisfied. For now, square again.

    Combine all like terms, and you arrive (unless I goofed) at an equation like this:

    (4C^2)x^3 + (A^2 + B^2 - 2AB)x^2 - 2C^2(A + B)x + C^4 = 0.

    This is a cubic equation in standard form, and since there is a method to solve cubics, this can be solved. Simply check the answers against the conditions we have identified and that should contain all answers...
     
  10. Jul 7, 2011 #9

    disregardthat

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    Re: Solving a cubic.... sort of!

    [itex]\sqrt{Ax-x^3}+\sqrt{Bx-x^3}=C[/itex]

    so

    [itex]C(\sqrt{Ax-x^3}-\sqrt{Bx-x^3}) = Ax-x^3-(Bx-x^3) = (A-B)x[/itex]

    from the original equation

    [itex]\sqrt{Bx-x^3}=C-\sqrt{Ax-x^3}[/itex]

    which we can insert into our new equation:

    [itex](A-B)x = C(\sqrt{Ax-x^3}-(C-\sqrt{Ax-x^3})) = 2C\sqrt{Ax-x^3}-C^2[/itex]

    [itex]2C\sqrt{Ax-x^3} = (A-B)x+C^2[/itex]

    squaring yields

    [itex]2C(Ax-x^3) = (A-B)^2x^2+2C^2(A-B)x+C^4[/itex]

    i.e.

    [itex]2Cx^3+(A-B)^2x^2 + (2C^2(A-B) - 2AC)x+C^4 = 0[/itex]

    which can be solved using the formula for third-degree polynomials.

    If you want to solve it yourself without the formula, I suggest the following:
    - do a linear transformation of x to remove the coefficient of x^2
    - substitute x = y+a/y for a suitable constant a to transform it into a second-degree polynomial equation in y^2

    Note that we may have generated more solutions that there are by squaring, so make sure that you are finding the correct ones.
     
  11. Jul 7, 2011 #10

    Mute

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    Re: Solving a cubic.... sort of!

    You forgot to square the 2C in going to the second line there.
     
  12. Jul 7, 2011 #11
    Re: Solving a cubic.... sort of!

    ^ Does that mean his reduction to a cubic yields the same thing that I said in the post before his?
     
  13. Jul 8, 2011 #12

    lanedance

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    Re: Solving a cubic.... sort of!

    I think we've all done the same thing in essence - your post in #7 probably gives the clearest method
     
  14. Jul 8, 2011 #13

    disregardthat

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    Re: Solving a cubic.... sort of!

    Thanks, I will correct it.

    EDIT: apparantly I can't do it now.
     
  15. Jul 8, 2011 #14

    lanedance

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    Re: Solving a cubic.... sort of!

    yeah I think you have to get in within a certain time frame
     
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