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Deriving general cube roots 
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#1
Apr2105, 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 


#2
Apr2105, 11:00 AM

P: 7

PS Anyone know how some peole are making Latex code work in their messages?



#3
Apr2105, 11:04 AM

P: 7

[tex]x_{0}=\sqrt{\frac{c}{a}}[/tex]
Ha ha, I appear to have found out! 


#4
Apr2105, 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(xx_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 xaxis the new function [tex]g(x)[/tex] now passes through the origin. After substituting [tex]xx_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=dax_0^3+bx_0^2cx_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^3x_0^3)+b(x^2x_0^2)+c(xx_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 


#5
Apr2105, 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. 


#6
Apr2105, 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]xx_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. 


#7
Apr2105, 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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