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

p, q Another variable replacement y = t - p/(3t) leads a three-term cubic equation to the form variable substitution
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:
t1, t2
Hence t1, t2
Therefore, the roots of a three-term cubic equation are equalthe roots of a three-term cubic equation 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
p/3t1
by t2, and in the expression for y2 we multiply both the numerator and the denominator of the fraction
p/3t2
by t1. We get

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

Cardano's formula for the roots of the canonical cubic equation

The expression the discriminant of the cubic equation y^3+ py+q=0 is called the discriminant of the third-degree equation y3 + py + q = 0.

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As is known, the n-th root of a complex number z, z = r *(cosφ + isinφ)
has n complex values the root of the n-th degree has n complex values where z Therefore, z has n values z1, z2, z3, where z1 z2 z3
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 zi the value of the second cubic root is taken so zj that the ratio is fulfilled zi*zj = -
p/3
.
To avoid such a combination of values of different cubic roots, you can use the formula

y
or, what is the same thing,

y
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:

3 real roots where 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

3 real roots of the canonical equation where 3 real roots of the canonical equation 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:

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:
2 real roots
Therefore, x1 = y1 - 
b/3a
,   x2 = y2 - 
b/3a
.   
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 of using the Cardano formula to solve cubic equations


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

Division by x-1 of the left side of the equation using Horner's scheme

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

The roots of the quadratic equation x^2+7x+10 =0
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

calculation of p q
Let's calculate the discriminant of this equation

discriminant of the equation
Since Δ <0   =>   the canonical equation has 3 real roots. Since q = 0   =>   φ = 
π/2
   =>   

roots of the equation
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

calculation of p q
Let's calculate the discriminant of this equation
discriminant of the equation
Since Δ >0   =>   the canonical equation has 1 real root and two complex conjugates:
roots of the equation 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

calculation of p q

Let's calculate the discriminant of this equation
discriminant of the equation
Since Δ = 0   =>   the equation has 2 real roots:
roots of the equation
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

calculation of p q
Let's calculate the discriminant of this equation

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

roots of the equation
Then for the roots of the original equation we get:

roots of the equation x^3+9x^2+9x-137 =0
Answer:roots of the equation x^3+9x^2+9x-137 =0

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

calculation of p q
Let's calculate the discriminant of this equation

discriminant of the equation
Так как Δ > 0   =>   the canonical equation has 1 real root and two complex conjugates:
roots of the equation
Then for the roots of the original equation we get:

roots of the equation x^3+9x^2+9x-137 =0
Answer:roots of the equation x^3+18x^2+90x+50 =0

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

replacing a variable
We multiply this equation by t3 and get a quadratic equation with respect to t3: t6 + 120t3 + 3375 = 0.
Hence, t
Then t1 t2 Now you can find y by the formula y = t + 
15/t
.
Instead of t, you can substitute either t1 or t2 the result will be the same:
y
Or
y
Thus, the real root of the original equation is equal to x
Answer: root of the equation

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