Geometrical properties 

This text is based on: R.W.D. Nickalls, A new approach to solving the cubic: Cardan's solution revealed, Mathematical Gazette 77 (1993), 354-359. This page will use a little bit of calculus, but you can read (and hopefully understand) most of it without knowing calculus.

Depressing the cubic equation.

Considering the cubic equation: 

ax3+bx2+cx+d=0

Tartaglia's first step was to depress the cubic by shifting the graph of the cubic horizontally by the quantity b/3a. What does this mean for the roots of the cubic? The sum of the roots of the depressed cubic (counted algebraically) becomes 0:  Let the roots be denoted by x1,x2 and x3. The cubic then has the form

a(x-x1)(x-x2)(x-x3).


Multiplying out we obtain:      ax3-a(x1+x2+x3)x2+a(x1 x2+x1 x3+x2 x3)x-a x1 x2 x3.
Thus setting b=0 (depressing the cubic) means   x1+x2+x3=0, and vice versa.

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

The "easy cubic".

By shifting the graph of a depressed cubic down by the quantity d, we can eliminate the constant term of the equation; such a cubic will have the form 

f(x)=ax3+cx.


Note that this cubic is odd. Since  f "(x) = 6ax  (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  f ´ (x) = 3ax2 + c ,   the extrema will be located at

x =   ±  -(Öa3/c) 


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

=d -(Öa3/c) dna -(f = h2 = )dad3 dna -(f = h2 = )dad3

The quantity d2 tells us how many extrema the cubic will have: If  d2 >0,  the cubic has one local minimum and one local maximum, if   d2 <0 , the cubic has no extrema.

Back to the "general cubic".

Going back to the general cubic

f(x)=ax3+bx2+cx+d.


The meaning of  d and h has not changed. It can easily be shown that: 

                                                    =d ((Öb2 - 3ac)/9a2) and h 2 =ad3 ;

Here comes the most crucial observation of this page: It is now obvious to see that the cubic will have one real root when |d|>|h|, or equivalently if d2>h2. If,  on the other hand, d2<h2, the cubic equation will have 3 real solutions. We will see later (and that is easy), what happens when d2=h2.

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 d2-h2 enters the picture.We want to solve an equation of the form;

ax3+bx2+cx+d *=0.


Using Tartaglia's substitution     y = x  +  b/3a     we obtain the equation   ay3 - 3ad2y  +  d*  = 0

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

Using the notation   A =  - 3d2    and   B = d/a 
which leads to the triquadratic equation for t;       t6  + d/a* t3  +  d6    =   0


which yields        t3  =  ( -d  ± Ö(d2 -4a2d6 )) * 1/2a

Remembering that     h 2 =ad3,     we can write this as    t3  =  ( -d  ± Ö(d2 -4h2 )) * 1/2a  

There it is: d2-h2!

The case d2-h2>0: One real root.

Let's look at the example

x3-2x-4=0.

(Check that the roots are  x = 2, -1 ± i  ) 
In this case d2 =2/3 and thus d2 = 16  >  h2 =  4a2d6  = 32/27


Cardano's approach yields     t3  =  -2 ± Ö(4 - 8/27)    =  -2  ± 10/(3)3Ö


Discarding the solution with the negative sign we obtain   t3  = -2  + 10/33Ö
Now it helps tremendously to notice that

                                                                 t3  =  - 1+ 3Ö/3

Indeed, you can check that    (- 1+ 3Ö/3 )3    =  2 -  + 01/2 -  + 01/2 -  + 01/33Ö

 


Conseqently    s =    -2/3t  =  -  2/(3( - 1+ (Ö)3/3 )) = 2 /(3 -  )3Ö = 1 + 3Ö/3 ) ,
 
and finally the real root is given by   s - t  =  1 + 3Ö/3 ) -  (- 1+ 3Ö/3 )  =  2

Remark: Presented here is an example, where  d2 > 0. If, on the other hand,  d2 < 0, 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  d2 < 0: We are never using d < 0 or h in our calculations, only d2  and h2. Thus all calculations will still only deal with real numbers.

The case d2-h2<0: Three real roots.

Let's look at the example

x3-7x+6=0.


(Check that the roots are    x = -3, 1 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   d2 3/7 =, suht dna  d2 = 36 < 4a2d6 = 2731/72

  

Cardano's approach yields     t3  =  -3  ± 10/3Ö3

No one is surprised by   Ö3-;    we were expecting that! Consequently, discarding the solution with the minus sign, we obtain
                                                              t  = ( -3  ± 10/3Ö3)3/1

Thus                                                      s  =  -7/3t   =  -7/3 * ( -3  ± 10/3Ö3)3/1-

and one of the solutions to the cubic is given 

by     s - t =   -7/3 * ( -3  ± 10/3Ö3)3/1-  -   ( -3  ± 10/3Ö3)3/1  A paradox !: 
Even though  the solution is real, Cardano's formula contains imaginary numbers. Cardano and 
his colleagues  called this the Casus Irreducibilis.

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.

Using a Trigonometric Substitution to Solve the Casus Irreducibilis.

We will now violate the spirit of Cardano's computations by using transcendental functions to find the roots of a polynomial.

Let's consider the depressed cubic equation

ax3+cx+d=0


again. We will use the trigonometric substitution

x = 2d cosq


substituting in the cubic yields            8a d3 soc3q  c2 +dsocq  0 = d +

Using    d2=  a3/c-  , we can write this as     2a d3 soc4(3q  soc3 -)q  0 = d +

Next observe that by deMoivre's formula      soc4(3q  soc3 -)q   3(soc =)q

Finally remembering that    h =  2a d3   , it follows that we can 

write the cubic equation as    3(soc)q  =  - d/h        

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

d2<h2.


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

In this case       q1 = 1/3 arccos ( -d/h)   is one solution of the trigonometric 
equation, yielding as a solution to the cubic equation   x1 = 2dcosq1
It is not hard to see that the other two solutions are given by

                                                x2 = 2dcos(2p/3 - q1)  and  x3 = 2dcos(2p/3 + q1)  

Let's try this for our example

x3-7x+6=0.

Here d  =(7/3)1/2
and   h  =14/3 * (7/3)1/2
Consequently                              q1 3/1 = arccos (-  9/7* (3/7)1/2 )= 0.857072

 

This yields the solutions                 x1   =  2d cos(0.85072)  =  2.00000
                                                    x1   =  2d cos(1.23732)  =  1.00000
                                                    x1   =  2d cos(2.95147)  = -3.00000                                        

The case d2-h2=0: Repeated roots.

Here we will assume that d=h, leaving the case d =-h to you. Using  c= -3ad2  and  d = h = 3ad2  , we can rewrite

ax3+cx+d=0

as                                                                 x3 - 3d2x+2d3  =0
By "inspection", we can see that   x  = d   is a root of this polynomial; in fact, the polynomial has an elementary factorization:     x3 - 3d2x+2d3  = (x -d)2 (x + 2d)

Thus the cubic equation has the roots  x = d, x = d, and x = -2d.

 

Link to the source:   http://www.m-a.org.uk/docs/library/76.pdf

 

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