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Gerry wrote:
> Hi
>
> can someone help me to verify that the solutions of the cubic
> ax^3 + (a - 1)x^2 - (b + 1)x + 1 - 2b = 0
>
> can be defined as follows:
>
> say b=na+m,
> and 3y1^2+3y1y2+y2^2^2 (x1,x2,x3 are the roots of the cubic)
>
> then the following quadratic form in y1,2
> (2n+1)a^2+(3m+1)a+1=3y1^2+3y1y2+y2^2
> holds.
>
> Is it always possible to determin the roots x1,2,3 by solving the
> quadratic form if we know a,b,n,m?
If you make your substitutions y1=x2-x1,y2=x3-2x2+x1
in 3y1^2+3y1y2+y2^2, you simply get the symmetric function of the roots
(x1+x2+x3)^2 -3(x1*x2+x1*x3+x2*x3).
Any symmetric function of the roots can be expressed in terms of the
coefficients
of the equation, so all you obtain is
3y1^2+3y1y2+y2^2 =(x1+x2+x3)^2 -3(x1*x2+x1*x3+x2*x3) =
[(a-1)/a]^2 -3(b+1)/a
How would this help find x1,x2,x3 ?
Say a = 1, b= -1
x^3 + 3 =0
3y1^2+3y1y2+y2^2 = 0
y1/y2 = z
3 z^2 +3z +1 = 0
z = (-3 +/- sqrt( -3))/6
z = (x2-x1)/(x3-2x2+x1) = (w^2 -1)/3
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