This page is based on: R.W.D. Nickalls, *A new approach to solving the cubic:
Cardan's solution revealed*, Mathematical Gazette 77 (1993), 354-359.
You can download
this article as a PDF (Acrobat) file.

Note that this page will use a little bit of calculus, but you can read (and hopefully understand) most of it without knowing calculus.

Tartaglia's first step was to depress the cubic by shifting the graph of the cubic horizontally by the quantity

Let the roots be denoted by *x*_{1},*x*_{2} and *x*_{3}.
The cubic then has the form

Multiplying out we obtain:

Thus setting

A cubic (in black) and its depressed counter part (in blue). Note that the roots of the depressed cubic add up to 0. |

Note that this cubic is odd. Since (and its third derivative is not 0), it will have its only inflection point at the origin. What can we say about the relative extrema? If it has any, it will have one local minimum and one local maximum: Since , the extrema will be located at

This quantity will play a major role in what follows, we set

The quantity tells us how many extrema the cubic will have: If , the cubic has one local minimum and one local maximum, if , the cubic has no extrema.

The meaning of and

Here comes the most crucial observation of this page: It is now obvious to see that the cubic will have one real root when |

The relative size of |d| and |h| determines the number of real roots. |

Let's now follow Cardano's approach and see where the quantity *d*^{2}-*h*^{2} enters the picture.

We want to solve an equation of the form

Using Tartaglia's substitution

we obtain the equation

Here d is a quantity depending on a, b, c, and d

which leads to the triquadratic equation for

which yields

Remembering that , we can write this as

There it is:

(No secrets: You can check that the roots are .) In this case

Cardano's approach yields

Discarding the solution with the negative sign we obtain

Now it helps tremendously to notice that

Indeed, you can check that

(Note that this is cheating, though; there is no purely algebraic way to figure this out: Try it!). Consequently

and finally the real root is given by

*Remark*: I have presented an example, where
.
If, on the other hand, ,
the cubic function will have no extreme points, and thus exactly one real root. You can check that the algebra still works the same if :
We are never using
or *h* in our calculations, only
and *h*^{2}. Thus all calculations will still only deal with **real** numbers.

(No secrets again: You can check that the roots are , and 2.)

Our goal is to find one real root; the other two real roots can then be found by polynomial division and the quadratic formula.

In this case

Cardano's approach yields

I hope you are not surprised by ; we were expecting that! Consequently, discarding the solution with the minus sign, we obtain

Thus

and one of the solutions to the cubic is given by

A paradox: Even though the solution is

The *Casus Irreducibilis* historically led to the study of complex numbers. You might be surprised that complex numbers did not enter the picture via the quadratic formula, as they usually do nowadays in the school curriculum. The reason: There was no "paradox"! The complex solutions could not be seen, they were "imaginary".

One of the great algebraists of the 20th century, B.L. van der Waerden observes in his book *Algebra I*, that the *Casus Irreducibilis* is unavoidable. There will never be an algebraic improvement of the cubic formula, which avoids the usage of complex numbers.

Let's consider the depressed cubic equation

again. We will use the

Substituting in the cubic yields

Using , we can write this as

Next observe that by deMoivre's formula

Finally remembering that , it follows that we can write the cubic equation as

We will only obtain solutions for , if the right hand side is bounded by 1 in absolute value, i.e., if

Thus this method will only work when the cubic has three real roots, in the

In this case,

is one solution of the trigonometric equation, yielding as a solution to the cubic equation

It is not hard to see that the other two solutions are given by

Let's try this for our example

Here,

Consequently

This yields the solutions

as

By "inspection", we can see that is a root of this polynomial; in fact, the polynomial has an elementary factorization:

Thus the cubic equation has the roots , and .

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