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Chip Eastham wrote:
>
> 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
Hi Chip,
well for one i was hoping that there would be a method to solve the
quadratic form in two variables in case the cubic has integer
coefficients.(this would be helpfull as a solution to factor coprimes)
Putting the cart up front seems to be a general approach for higher
degrees.
The roots of a polynomial of degree n are always related to some of the
many solutions of a quadratic form with n-1 variables.
Thanks
Gerry
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