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
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
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.
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.
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.