In this special case, if you let A = ((x^2 + 10*y^2)/10^(1/3) (assuming I did
the algebra right, note there are 3 such A's, but only 1 real one) then you get
sage: A = var("A"); solve(2* x^2 + y^2 + z^2 - 1- A*z==0,z)
[z == (A - sqrt(A^2 - 4*y^2 - 8*x^2 + 4))/2, z == (sqrt(A^2 - 4*y^2 -
8*x^2 +
ok thanks. Do you no any other way how I could solve it
2008/6/21 David Joyner <[EMAIL PROTECTED]>:
>
> This is a 6th degree polynomial in z. There is no general formula for
> algebraically
> solving for roots of polynomials of degree 5 or higher.
>
>
> On Fri, Jun 20, 2008 at 8:20 PM, Max <[EMAI
This is a 6th degree polynomial in z. There is no general formula for
algebraically
solving for roots of polynomials of degree 5 or higher.
On Fri, Jun 20, 2008 at 8:20 PM, Max <[EMAIL PROTECTED]> wrote:
>
> Hi,
> i am looking for a way to solve this equation "10* (2* x^2 + y^2 + z^2
> - 1)^3- x
