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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 y1=x2-x1,y2=x3-2x2+x1 (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?
Hi, Gerry:
It is always possible to solve this (or any other) cubic equation
by solving an "auxiliary" quadratic equation. See, for example:
http://www.sosmath.com/algebra/factor/fac11/fac11.html
However you seem to be putting the cart before the horse by
defining y1, y2 in terms of x1,x2,x3 (at least if your ultimate
goal was to find roots x1,x2,x3).
What are you trying to do?
regards, chip
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