Most recent change of CubicEquation

Edit made on February 11, 2013 by ColinWright at 22:31:04

Deleted text in red / Inserted text in green

The next stage up from the quadratic equation, the general cubic
equation has the form EQN:ax^3+bx^2+cx+d=0.

Just as there is a solution to the quadratic equation, a general
solution to the cubic was found by Tartaglia and Ferro.

The problem is, even when there are three real solutions, the
intermediate calculations use the roots of negative numbers,
and this led directly to the acceptance of the complex numbers.

In practice, numerical solutions are found rather than using a
closed form formula. Newton's Method is an efficient way to find
solutions, although one must be aware that the basins of attraction
have fractal boundaries.

Also related: quartic equation
!! Outline of the solution
Start with a general cubic. Since the leading coefficient can't
be zero we can divide through by that, so our general cubic is:
* EQN:x^3+ax^2+bx+c=0
Substitute EQN:x=t-a/3 giving
* EQN:(t-a/3)^3+a(t-a/3)^2+b(t-a/3)+c=0
That has eliminated Expand that and gather the quadratic term, so we have /t/ terms together and you get
* EQN:t^3+pt+q=0
for suitable /p/ and /q./ Notice that there is no quadratic term.

Now substitute EQN:t=u-\frac{p}{3u} and multiply by EQN:u^3 to get
* EQN:u^6+qu^3+\frac{p^3}{27}=0

Now we have a quadratic in EQN:u^3 so using the quadratic equation
formula we get two possible answers for EQN:u^3. Each of these
gives three cube roots, giving six possible answers. These occur
in pairs of identical solutions, giving three in all as required.

The Wikipedia and MathWorld articles have more information.