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 z
1, z
2, z
3, 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:
where
If we omit the intermediate calculations, the final formulas for the three real roots of the canonical equation can be represented as
where
Then the formulas for the roots of the original equation will look like:
x1 = y1 - , x2 = y2 - , x3 = y3 - .
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 - , x2 = y2 - , x3 = y3 - .
3 ) Δ=0 => 2 real roots:
Therefore, x
1 = y
1 -
, x
2 = y
2 -
.
If Δ=0 and p=q=0, then the canonical equation has only one root y
1=0. Accordingly, the original equation will have a single root
x = -.
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 x
3 + 6x
2 + 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, x
2 + 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 y
3 + 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:
x
1 = y
1 - 2 = 3 - 2 = 1,
x
2 = y
2 - 2 = -3 - 2 = -5,
x
3 = y
3 - 2 = 0 - 2 = -2.
Answer: -5, -2, 1.
Example 2.
Solve the equation x
3 + 3x
2 + 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 y
3 + 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:
x
1 = y
1 - 1 = 0 - 2 = -1,
x
2 = y
2 - 1 = i - 1 = i - 1,
x
3 = y
3 - 1 = -i - 1 = -i - 1.
Answer: -1, -1+i, -1-i.
Example 3.
Solve the equation x
3 + 12x
2 + 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 y
3 + 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:
x
1 = y
1 - 4 = -4 - 4 = -8,
x
2 = y
2 - 4 = 2 - 4 = -2.
Answer: -8, -2.
Example 4.
Solve the equation x
3 + 9x
2 + 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 y
3 + 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 x
3 + 18x
2 + 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 y
3 + 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 x
3 + 12x
2 + 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
<=> y
3 - 12y
2 + 48y - 64 + 12y
2 - 96y + 192 + 3y - 8 = 0 <=>
<=> y
3 - 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 t
3 and get a quadratic equation with respect to t
3: t
6 + 120t
3 + 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: