I have an equation a little like this;

$$2c^3 - 3oc^2 + oG = 0$$

In my equations, $o$ is always $> G$ . I'm trying to find a general solution rather than constantly estimating it graphically, and I got onto the Wikipedia article on cubic equations. If I've done my algebra correctly, the discriminant is $\operatorname{Dis} = 108(G^2\cdot o^2)(o^2 - G^2)$ which is always real and positive, meaning there should be three real solutions.

My problem is this however; there is a square root term that is $-27(2)\cdot \operatorname{Dis}$ ; this will always be negative, and always give a pair of complex roots; worse, I then have to take the cubic root of this at least twice; this is throwing me and I don't know how to explicitly solve the equation, or find a general form?

Can anyone help or point out any mistakes I'm making ? I'd be most grateful!



Check this out this paper Savante Janson on Roots of Polynomials of Degree 3 and 4. Also try the recently posted answer for Stack Exchange Question 145590. These may lead you in a good direction.

  • $\begingroup$ Thanks Tim - seems I'm stuck with an ugly looking expression to try and solve..I'll have at it for now! $\endgroup$
    – DRG
    Dec 13 '12 at 0:55

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