## Most recent change of CubicEquation

Edit made on November 02, 2008 by derekcouzens at 13:43:50

Deleted text in red / Inserted text in green

WW
The next stage up from the quadratic equation, the general cubic
equation has the form EQN:y=ax^3+bx^2+cx+d. The question of
"solving" the cubic equation is to find values of /x/ for which
/y=0./
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
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 the quadratic term, so we have
* EQN:t^3+pt+q=0
for suitable /p/ and /q./

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.