Cubic equation
A cubic equation has the general form:
$$ \large ax^3 + bx^2 + cx + d = 0 $$
$$ \large a \neq 0 $$
Here \( \large a, b, c, d \) are constants, and \( \large x \) is the unknown. Since the highest power is three, we call it a cubic equation.
How to solve a cubic equation?
There is no simple solution formula as for quadratic equations. Instead, different methods are used depending on the form of the equation.
1. Find an obvious root
Sometimes one can guess a solution by trying simple numbers, e.g. \( \large x = -2, -1, 0, 1, 2 \). If it gives 0, we have found a root.
Example:
$$ \large x^3 - x = 0 $$
Here we can factor out \( \large x \):
$$ \large x(x^2 - 1) = 0 $$
This gives three solutions:
$$ \large x = 0 $$
$$ \large x = 1 $$
$$ \large x = -1 $$
2. Polynomial division
If one finds a root, one can use polynomial division to reduce the cubic equation to a quadratic equation. The quadratic can then be solved with the discriminant and the quadratic formula.
Example:
$$ \large x^3 - 6x^2 + 11x - 6 = 0 $$
We guess \( \large x = 1 \) and insert:
$$ \large 1^3 - 6 \cdot 1^2 + 11 \cdot 1 - 6 \quad \Leftrightarrow $$
$$ \large 1 - 6 + 11 - 6 \quad \Leftrightarrow $$
$$ \large 0 $$
Thus \( \large x = 1 \) is a root. We divide the polynomial by \( \large (x-1) \):
$$ \large x^3 - 6x^2 + 11x - 6 \quad : \quad (x - 1) \quad \Leftrightarrow $$
$$ \large x^2 - 5x + 6 $$
Now we have a quadratic equation:
$$ \large x^2 - 5x + 6 = 0 $$
The discriminant is:
$$ \large D = (-5)^2 - 4 \cdot 1 \cdot 6 \quad \Leftrightarrow $$
$$ \large D = 25 - 24 \quad \Leftrightarrow $$
$$ \large D = 1 $$
The solutions are:
$$ \large x = \frac{5 \pm \sqrt{1}}{2} \quad \Leftrightarrow $$
$$ \large x = \frac{5 \pm 1}{2} $$
Thus:
$$ \large x = 2 $$
$$ \large x = 3 $$
Together with \( \large x = 1 \) we have three solutions.
3. Graphical method
One can always draw the graph of \( \large f(x) = ax^3 + bx^2 + cx + d \) and read where it intersects the x-axis. This corresponds to finding the roots.
A cubic graph can have one or three real roots. It always has at least one, because it goes to \( -\infty \) on one side and \( +\infty \) on the other.
Cardano’s method
There actually exists a general solution formula for cubic equations. It was discovered in the 16th century by the Italians Tartaglia and Cardano.
The idea is first to eliminate the \( \large x^2 \)-term. This is done with a substitution:
$$ \large x = y - \frac{b}{3a} $$
Afterwards one obtains a reduced cubic equation of the form:
$$ \large y^3 + py + q = 0 $$
Cardano’s formula in practice
We consider the equation:
$$ \large x^3 - 6x - 9 = 0 $$
It is already in reduced form, so we set \( \large p = -6 \), \( \large q = -9 \).
Discriminant:
$$ \large \Delta = \Bigl(\frac{q}{2}\Bigr)^2 + \Bigl(\frac{p}{3}\Bigr)^3 \quad \Leftrightarrow $$
$$ \large \Delta = \Bigl(-\frac{9}{2}\Bigr)^2 + (-2)^3 \quad \Leftrightarrow $$
$$ \large \Delta = \frac{81}{4} - 8 \quad \Leftrightarrow $$
$$ \large \Delta = \frac{49}{4} $$
\( \large \Delta > 0 \), so there is one real solution.
Cardano’s formula:
$$ \large y = \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} \;+\; \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}} $$
Insert \( \large q = -9 \), \( \large \sqrt{\Delta} = \frac{7}{2} \):
$$ \large y = \sqrt[3]{\frac{9}{2} + \frac{7}{2}} + \sqrt[3]{\frac{9}{2} - \frac{7}{2}} \quad \Leftrightarrow $$
$$ \large y = \sqrt[3]{8} + \sqrt[3]{1} \quad \Leftrightarrow $$
$$ \large y = 2 + 1 \quad \Leftrightarrow $$
$$ \large y = 3 $$
Thus \( \large x = 3 \) is a real root.
The remaining roots:
$$ \large x^3 - 6x - 9 = (x-3)(x^2 + 3x + 3) $$
The quadratic factor gives the discriminant:
$$ \large D = 3^2 - 4 \cdot 1 \cdot 3 \quad \Leftrightarrow $$
$$ \large D = 9 - 12 \quad \Leftrightarrow $$
$$ \large D = -3 $$
So the two remaining roots are complex:
$$ \large x = \frac{-3 \pm i\sqrt{3}}{2} $$
Altogether the solutions are:
$$ \large x = 3 \quad \text{(real)} $$
$$ \large x = \frac{-3 \pm i\sqrt{3}}{2} \quad \text{(not real)} $$
Note: If there are three real solutions, Cardano’s formula can still be used, but the calculation goes through complex numbers even though the final answers are real. This is called casus irreducibilis.
Casus irreducibilis – three real roots
When \( \large \Delta < 0 \), there are three real roots. In this case, one can use a trigonometric method with cosine.
Example:
$$ \large x^3 - 3x + 1 = 0 $$
Here \( \large p = -3 \), \( \large q = 1 \).
We set:
$$ \large x = 2\sqrt{\frac{-p}{3}} \cos(\theta) $$
and determine the angle from:
$$ \large \cos(3\theta) = \frac{-q}{2} \cdot \sqrt{\frac{-27}{p^3}} $$
Insert \( \large p=-3 \), \( \large q=1 \):
$$ \large \cos(3\theta) = \frac{-1}{2} \cdot \sqrt{\frac{-27}{(-3)^3}} \quad \Leftrightarrow $$
$$ \large \cos(3\theta) = \frac{-1}{2} \cdot \sqrt{1} \quad \Leftrightarrow $$
$$ \large \cos(3\theta) = -\frac{1}{2} $$
Thus:
$$ \large 3\theta = 120^\circ, \; 240^\circ, \; 480^\circ $$
Hence:
$$ \large \theta = 40^\circ, \; 80^\circ, \; 160^\circ $$
Now insert in:
$$ \large x = 2\sqrt{\frac{-p}{3}} \cos(\theta) $$
Since \( \large p=-3 \), we get \( \large 2\sqrt{1} = 2 \), so:
$$ \large x_1 = 2\cos(40^\circ) \approx 1.53 $$
$$ \large x_2 = 2\cos(80^\circ) \approx 0.35 $$
$$ \large x_3 = 2\cos(160^\circ) \approx -1.88 $$
Here we clearly see three real solutions without using complex numbers.
Summary
- A cubic equation has the form \( \large ax^3 + bx^2 + cx + d = 0 \).
- Solutions can be found by guessing a root, factoring, or using polynomial division.
- Graphically, one can always see that there is at least one real root.
- Cardano’s formula provides a general method but is technical and rarely used in practice.
- If \( \large \Delta > 0 \): one real root. If \( \large \Delta = 0 \): multiple roots coincide. If \( \large \Delta < 0 \): three real roots (casus irreducibilis).
- In casus irreducibilis, one can use a trigonometric cosine formula to find all three roots.