Deriving general cube roots

  1. Apr 21, 2005 #1
    I've been trying to derive general solutions for cubic roots, i.e. the general solutions of (will Latex just work?)

    $ax^3+bx^2+cx+d=0$

    I do not want to be shown the solutions - but does anyone know what direction to go in to achieve this?

    I thought I'd found solutions at one point but they relied on one of the solutions (the simplest in form) being $\sqrt{-c/a}$ and that doesn't appear to be true!

    Andy
     
  2. jcsd
  3. Apr 21, 2005 #2
    PS Anyone know how some peole are making Latex code work in their messages?
     
  4. Apr 21, 2005 #3
    [tex]x_{0}=\sqrt{\frac{-c}{a}}[/tex]

    Ha ha, I appear to have found out!
     
  5. Apr 21, 2005 #4
    This is what I tried:

    [tex]f(x)=ax^3+bx^2+cx+d=0[/tex] is the general equation to solve, and intersects the y axis a distance d from the origin.

    So, consider [tex]f(x-x_0)=g(x)[/tex] where [tex]x_0[/tex] is one solution of [tex]f(x)=0[/tex]. As we have shifted the original function along the x-axis the new function [tex]g(x)[/tex] now passes through the origin.

    After substituting [tex]x-x_0[/tex] into f to get g, expand g to get it in the form [tex]g(x)=px^3+qx^2+rx+s[/tex]. As said above g passes through the origin so that s=0.

    Therefore, [tex]s=0=d-ax_0^3+bx_0^2-cx_0[/tex]. Rearranging for d in terms of [tex]x_0[/tex] and a, b and c and substituting back into the original form for f we get [tex]f(x)=a(x^3-x_0^3)+b(x^2-x_0^2)+c(x-x_0)[/tex].

    Now, as [tex]x_0[/tex] is by definition a solution of f; [tex]f(x_0)=0[/tex] we get:

    [tex]ax_0^3+cx_0=0[/tex] which gives the trivial solution [tex]x_0=0[/tex] (i.e. d=0) and the two solutions

    [tex]x_0=+-\sqrt{\frac{-c}{a}}[/tex].


    These don't seem to work. Am I doing something stupid?

    Andy
     
    Last edited: Apr 21, 2005
  6. Apr 21, 2005 #5
    There's meant to be a sqaure root sign over that -c/a but it's not coming out on my screen.

    [tex]x_0[/tex] is of course just one solution of the cubic.
     
  7. Apr 21, 2005 #6

    CRGreathouse

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    If you have one solution [tex]x_0[/tex] to the cubic, you can just divide by [tex]x-x_0[/tex] to reduce it to a quadratic and you're done.

    If you really want a general solution like Cardanno's, I think the first step would be to remove the [tex]bx^2[/tex] term by a change of variable.
     
  8. Apr 21, 2005 #7
    Well I'm trying to derive all solutions generally so I don't have [tex]x_0[/tex] to begin with. That's a fantastic idea about the change of variable to remove the even term though....! Cheers! I'll have dinner then have another try.
     
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