Cardano's formula for solving cubic equations

A method of solving a third-degree equation using Cardano's formula.

In the general case, one can find the roots of a third-degree equation in the domain of complex numbers using the Cardano formula for a cubic equation in canonical form, named after the Italian mathematician D. Cardano.
General a third-degree equation ax³ + bx² + cx + d = 0, a ≠ 0, by replacing a variable x = y - b/(3a), is reduced to the canonical three - term form y³ + py + q = 0, where

Another variable replacement y = t - p/(3t) leads a three-term cubic equation to the form
Multiplying both parts of the equation by t³, we get The last equation is a quadratic equation with respect to t³, its roots can be written out explicitly:

Hence
Therefore, the roots of a three-term cubic equation are equal Although the expressions for y₁ and y₂ look different, they are the same numbers. Let's transform them as follows. In the expression for y₁ we multiply both the numerator and the denominator of the fraction by t₂, and in the expression for y₂ we multiply both the numerator and the denominator of the fraction by t₁. We get


Similarly, for y₂

Therefore, y₁ = y₂ = t₁ + t₂, and the Cardano formula for the roots of the canonical cubic equation has the form:



The expression is called the discriminant of the third-degree equation y³ + py + q = 0.

As is known, the n-th root of a complex number z, z = r *(cosφ + isinφ)
has n complex values where Therefore, has n values z₁, z₂, z₃, where
There are two cubic roots in the Cardano formula, and their values must be combined according to the following rule: for each of the three values of the first cubic root the value of the second cubic root is taken so that the ratio is fulfilled z_i*z_j = -.
To avoid such a combination of values of different cubic roots, you can use the formula


or, what is the same thing,


Each value of y found by the Cardano formula corresponds to the solution of the original equation x = y - b/(3a).
Depending on the value of the discriminant Δ a cubic equation can have either 3 real roots, (Δ <0), or 1 real root and two complex conjugates (Δ > 0), or 2 real roots (Δ=0) or one real root (Δ=0, p=q=0). Let's consider all these cases.

1) Δ <0   =>   3 real roots:


If we omit the intermediate calculations, the final formulas for the three real roots of the canonical equation can be represented as


Then the formulas for the roots of the original equation will look like:
x₁ = y₁ - 

b/3a

,   x₂ = y₂ - 

b/3a

,   x₃ = y₃ - 

b/3a

.   
2 ) Δ > 0   =>   1 real root and two complex conjugates:


The formulas for the roots of the original equation are the same as in the previous case
x₁ = y₁ - 

b/3a

,   x₂ = y₂ - 

b/3a

,   x₃ = y₃ - 

b/3a

.   
3 ) Δ=0   =>   2 real roots:

Therefore, x₁ = y₁ - ,   x₂ = y₂ - .   
If Δ=0 and p=q=0, then the canonical equation has only one root y₁=0. Accordingly, the original equation will have a single root
x = -

b/3a

.
If a cubic equation has an integer or rational root, then, of course, the easiest way is to find this root by selection, then reduce the original equation to a quadratic one by division. If there are no rational roots, then only the Cardano formula can help find a solution.
The practical use of the Cardano formula for solving cubic equations is extremely difficult due to cumbersome calculations. But in special cases, it is quite easy to do this, for example, for the first case (Δ < 0) when q = 0 to find three real roots, or for the third case. (Δ = 0). For the second case, when Δ > 0 , the formulas for the roots of the cubic equation can always be written out. Thus, the application of Cardano's formula is justified if the equation does not have rational roots.

Let us consider the application of Cardano's formula to solving cubic equations using examples.

Examples.

Example 1.
Solve the equation x³ + 6x² + 3x - 10 = 0.
Solution.
This equation can be easily solved without using the Cardano formula. It is easy to pick up the root x = 1. By dividing the left side of the equation by x - 1 using Horner's scheme, we obtain



Therefore, x² + 7x + 10 = 0. Solving this quadratic equation, we get


Now let's find the roots of the original equation using Cardano's formula. For this equation a = 1, b = 6, c =3, d = -10. Replacing a variable x = y - b/(3a) = y - 6/3 = y - 2 brings the original equation to the form y³ + py + q = 0, where


Let's calculate the discriminant of this equation


Since Δ <0   =>   the canonical equation has 3 real roots. Since q = 0   =>   φ =    =>   


Then for the roots of the original equation we get:
x₁ = y₁ - 2 = 3 - 2 = 1,
x₂ = y₂ - 2 = -3 - 2 = -5,
x₃ = y₃ - 2 = 0 - 2 = -2.
Answer: -5, -2, 1.

Example 2.
Solve the equation x³ + 3x² + 4x + 2 = 0.

Solution.
For this equation a = 1, b = 3, c =4, d = 2. Replacing the variable x = y - b/(3a) = y - 3/3 = y - 1 brings the original equation to the form y³ + py + q = 0, где


Let's calculate the discriminant of this equation

Since Δ >0   =>   the canonical equation has 1 real root and two complex conjugates:
Then for the roots of the original equation we get:
x₁ = y₁ - 1 = 0 - 2 = -1,
x₂ = y₂ - 1 = i - 1 = i - 1,
x₃ = y₃ - 1 = -i - 1 = -i - 1.
Answer: -1, -1+i, -1-i.

Example 3.
Solve the equation x³ + 12x² + 36x + 32 = 0.

Solution. For this equation a = 1, b = 12, c =36, d = 32. Replacing the variable x = y - b/(3a) = y - 12/3 = y - 4 brings the original equation to the form y³ + py + q = 0, где



Let's calculate the discriminant of this equation

Since Δ = 0   =>   the equation has 2 real roots:

Then for the roots of the original equation we get:
x₁ = y₁ - 4 = -4 - 4 = -8,
x₂ = y₂ - 4 = 2 - 4 = -2.
Answer: -8, -2.

Example 4.
Solve the equation x³ + 9x² + 9x - 137 = 0.

Solution.
For this equation a = 1, b = 9, c =9, d = -137. Replacing the variable x = y - b/(3a) = y - 9/3 = y - 3 brings the original equation to the form y³ + py + q = 0, где


Let's calculate the discriminant of this equation


Since Δ >0   =>   the canonical equation has 1 real root and two complex conjugates:


Then for the roots of the original equation we get:


Answer:

Example 5.
Solve the equation x³ + 18x² + 90x + 50 = 0.

Solution.
For this equation a = 1, b = 18, c =90, d = 50. Replacing the variable x = y - b/(3a) = y - 18/3 = y - 6 brings the original equation to the form y³ + py + q = 0, где


Let's calculate the discriminant of this equation


Так как Δ > 0   =>   the canonical equation has 1 real root and two complex conjugates:

Then for the roots of the original equation we get:


Answer:

Since it is quite difficult to remember the intermediate formulas for finding the roots of a cubic equation using Cardano's formula, you can simply repeat the derivation of Cardano's formula for this equation. Let's consider the corresponding example.

Example.
Find the real roots of the equation x³ + 12x² + 3x + 4 = 0.

Solution.
For this equation a = 1, b = 12, c = 3, d = 4. Let's replace the variable x = y - b/(3a) = y - 12/3 = y - 4:
(y - 4)³ + 12(y - 4)² + 3(y - 4) + 4 = 0    <=>   y³ - 12y² + 48y - 64 + 12y² - 96y + 192 + 3y - 8 = 0   <=>   
 <=>  y³ - 45y + 120 = 0.
Therefore, p = -45, q = 120, Δ = (60)² - (15)³ = 225 >0. So the original equation has one real root.
Now let's make the following variable substitution y = t - p/(3t) = t + 45/(3t) = t + 15/t. We get


We multiply this equation by t³ and get a quadratic equation with respect to t³: t⁶ + 120t³ + 3375 = 0.
Hence,
Then Now you can find y by the formula y = t + .
Instead of t, you can substitute either or the result will be the same:

Or

Thus, the real root of the original equation is equal to
Answer: