Cubic equation

To solve a cubic equation

Cardano's method

Gerolamo Cardano published a method to solve a cubic equation in 1545. There is a description of this method on Wikipedia. But it is not too detailed and on the German Wikipedia even the last part is missing and so it’s impossible to implement a working algorithm by the aid of this. So I thought I try to do a more detailed explanation :-)

The method of Cardano is based on the reduction of the equation polynomia

to the form

And some variable substitutions as follows:

First of all the entire equation is divided by A to get the form

With a = B/A, b = C/A and c = D/A

Now Cardano makes a variable substitution and replaces x by the term y + α. With this the equation becomes:

This expanded becomes

Given that α = -a/3, the y² part becomes 0 and we get to the equation

With α replaced by -a/3

Now there is another substitution: y will be substituted by u + v and we get

This equation expanded gives

And with some brackets

Given that 3uv = - p the part in brackets becomes 0 and for q = - u³ – v³ the entire left part is 0.

This condition for q we put into square now and the one for p divided by 3 to the power of 3 and multiply it by 4.

By adding both conditions together we get

Let’s say

And we have

From this two equations we can derive

And with the substitution above

y = u + v

and (not to be forgotten :-)

)

x = y – a/3 = u + v – a/3

As we are looking for x and not for y.

Now there are the 3 possibilities D >= 0 , D = 0 and D < 0:

D > 0

If D > 0 the root of D is real and we get one real solution and two complex solutions for u and v:

As real solution.

As the cubic root of a real number can have two complex solutions. One at the complex angle of 120° and the other one at 240°. The two complex solutions are of the same length as u0 and v0, but one has the complex angle 120° and the other one 240°. That means with

for the complex angles we get

and

3 possible solutions for u and 3 possible solutions for v and if we combine them, we get 9 possible solutions. That’s pretty much and not all of them are valid :-)

.

Further above we had the clause u*v = - p/3 and this is not the case for all 9 combinations of u and v. Principally it’s valid for the u₀ * v₀, u₁ * v₂ and u₂ * v₁. Let’s have look at u₁ * v₂. That’s

If we separate

and

That only works for ε₁ * ε₂. If we combine ε₁ * ε₁, there remains an imaginary part and that can’t give - p/3. So, given that D > 0 there are the 3 solutions. But only the first one is real.

y₁ = u₀ + v₀
y₂ = u₁ + v₂
y₃ = u₂ + v₁

(and don’t forget x = y – a/3 :-)

)

D = 0

If D = 0, we have the same 3 solutions but as D = 0, u₀ = v₀ and

and y₃ = y₂ we get only 2 solutions.

y₁ = u₀ + v₀
y₂ = -(u₀ + v₀)/2

D < 0

The case D < 0 is the so called “Casus irreducibilis”. Here the root of D becomes complex and the cubic roots for u and v become complex too.

To solve this complex root we have to interpret the complex number as number of the length r and complex angle α.

With the substitution from further above

We get the same length for u and v:

The angle becomes for u:

The angle becomes for v:

To get the cubic root out of this now we have to calculate the cubic root of r and basically divide α by 3 that’s it. At least for the first solution :-)

. The cubic root of a complex number has not just 1 but 3 solutions. The angle α we get once by multiplying α/3 by 3, twice by multiplying (2π + α)/3 by 3 and thirdly by multiplying (4π + α)/3 by 3. That makes 3 solutions for u and 3 solutions for v. They look like:

And as αv = 2π - α

That would give 9 possible solutions again. But as we are interested in real solutions, we can reduce that to 3 real solutions. For the 3 combinations

u₀ + v₂
u₁ + v₁
u₂ + v₀

the imaginary part becomes 0 and the solutions are

That’s it

There is a small online solver in JavaScript for this on Cardano solver).

C# Demo Project Cardano

Cardano.zip