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Deriving general cube roots

by AndyCav
Tags: cube, deriving, roots
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AndyCav
#1
Apr21-05, 10:57 AM
P: 7
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
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AndyCav
#2
Apr21-05, 11:00 AM
P: 7
PS Anyone know how some peole are making Latex code work in their messages?
AndyCav
#3
Apr21-05, 11:04 AM
P: 7
[tex]x_{0}=\sqrt{\frac{-c}{a}}[/tex]

Ha ha, I appear to have found out!

AndyCav
#4
Apr21-05, 11:18 AM
P: 7
Deriving general cube roots

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
AndyCav
#5
Apr21-05, 11:26 AM
P: 7
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.
CRGreathouse
#6
Apr21-05, 11:34 AM
Sci Advisor
HW Helper
P: 3,684
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.
AndyCav
#7
Apr21-05, 12:28 PM
P: 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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