Solving Cubic Equations with Chebyshev Roots

[junglebeast]

Say I have a monic polynomial,

x^3 + ax^2 + bx + c

with a=-2.372282, b=1.862273, c=-0.483023

The discriminant is given by

a^2 b^2 - 4 b^3 - 4 a^3 c - 27 c^2 + 18 ab c

which is < 0, indicating 1 real root and 2 complex conjugates.

A method for solving a general cubic using the Chebyshev root is explained here,
http://www.statemaster.com/encyclopedia/Cubic-equation

but (a^3 - 3b) is negative, which means that "t" will be imaginary. But t is then used in the Chebyshev cubic root, which is only defined for real numbers [-2, inf].

 

[HallsofIvy]

The site you give does NOT say that the Chebyshev cubic root is "only defined for real numbers [-2, inf]". It gives simple formulas for the root in the case that the argument is in that interval, and then tells how to expand it to complex arguments.

 

[junglebeast]

EDIT - deleted previous response.

Let me clarify my remaining confusion from here:
http://www.exampleproblems.com/wiki/index.php/Cubic_equation

They define s = t^2, then say

If s < 0 then the reduction to Chebyshev polynomial form has given a t which is a pure imaginary number. In this case $$i C_{1\over3}(-it)-iC_{1\over3}(it)$$ is the sole real root. We are now evaluating a real root by means of a function of a purely imaginary argument; however we can avoid this by using the function

$$S_{1\over3}(t) = iC_{1\over3}(-it)-iC_{1\over3}(it) = 2 \operatorname{sinh}\left(\operatorname{arcsinh}\left({t\over2}\right)/3\right)$$,

which is a real function of a real variable with no singularities along the real axis.

Well, as they already pointed out t is imaginary, but then they say $$S_{1\over3}(t)$$ is a function of a real variable! But it's still defined in terms of t only! So how can I calculate $$S_{1\over3}(t)$$ given that I can't compute the imaginary t?