Ferrari and cardano formulas. Cardano formula for solving cubic equation

Let's look again at the sum cube formula, but write it differently:

Compare this entry with equation (13) and try to establish a connection between them. Even with a hint it's not easy. We must pay tribute to the mathematicians of the Renaissance who solved the cubic equation without knowing alphabetic symbolism. Let's substitute into our formula:

It is now clear: in order to find the root of equation (13), it is enough to solve the system of equations

and take as the amount and . By replacing , this system is reduced to a very simple form:

Then you can act in different ways, but all “roads” will lead to the same quadratic equation. For example, according to Vieta’s theorem, the sum of the roots of the reduced quadratic equation is equal to the coefficient with a minus sign, and the product is equal to the free term. It follows that and are the roots of the equation

Let's write down these roots:

The variables and are equal to the cubic roots of and , and the desired solution to the cubic equation (13) is the sum of these roots:

This formula is known as Cardano formula.

Trigonometric solution

by substitution it is reduced to an “incomplete” form

, , . (14)

The roots , , of the “incomplete” cubic equation (14) are equal

, ,

Let the “incomplete” cubic equation (14) be valid.

a) If (the “irreducible” case), then

(b) If , , then

, .

, ,

, .

In all cases, the actual value of the cube root is taken.

Biquadratic equation

Algebraic equation of the fourth degree.

where a, b, c are some real numbers, called biquadratic equation. By substitution the equation is reduced to a quadratic equation followed by solving two binomial equations and ( and are the roots of the corresponding quadratic equation).

If and , then the biquadratic equation has four real roots:

If , ), then the biquadratic equation has two real roots and imaginary conjugate roots:

If and , then the biquadratic equation has four purely imaginary pairwise conjugate roots:

, .

Fourth degree equations

A method for solving fourth degree equations was found in the 16th century. Ludovico Ferrari, student of Gerolamo Cardano. That's what it's called - the method. Ferrari.

As in solving cubic and quadratic equations, in a fourth-degree equation

you can get rid of the term by substitution. Therefore, we will assume that the coefficient of the cube of the unknown is zero:

Ferrari's idea was to represent the equation in the form , where the left side is the square of the expression , and the right side is the square of a linear equation of , the coefficients of which depend on . After this, it remains to solve two quadratic equations: and . Of course, such a representation is only possible with a special choice of the parameter. It is convenient to take it in the form , then the equation will be rewritten as follows:

The right side of this equation is the quadratic trinomial of . It will be a complete square when its discriminant is equal to zero, i.e.

, or

This equation is called resolvent (i.e. "permissive"). It is relatively cubic, and Cardano’s formula allows us to find some of its roots. When the right side of equation (15) takes the form

and the equation itself is reduced to two quadratic ones:

Their roots give all the solutions to the original equation.

For example, let’s solve the equation

Here it will be more convenient to use not ready-made formulas, but the very idea of the solution. Let's rewrite the equation in the form

and add the expression to both sides so that a complete square is formed on the left side:

Now let’s equate the discriminant of the right side of the equation to zero:

or, after simplification,

One of the roots of the resulting equation can be guessed by sorting out the divisors of the free term: . After substituting this value we get the equation

where . The roots of the resulting quadratic equations are And . Of course, in the general case complex roots can also be obtained.

Any cubic equation with real coefficients has at least one real root, the other two are either also real or are a complex conjugate pair.

Let's start the review with the simplest cases - binomial And returnable equations. Then we move on to finding rational roots (if any). Let's finish with an example of finding the roots of a cubic equation using Cardano's formula for the general case.

Page navigation.

Solving a two-term cubic equation.

The binomial cubic equation has the form .

This equation is reduced to the form by dividing by a coefficient A that is different from zero. Next, apply the formula for abbreviated multiplication sum of cubes:

From the first bracket we find , and the square trinomial has only complex roots.

Example.

Find the real roots of the cubic equation.

Solution.

We apply the formula for abbreviated multiplication of difference of cubes:

From the first bracket we find that the square trinomial in the second bracket has no real roots, since its discriminant is negative.

Answer:

Solving the reciprocal cubic equation.

The reciprocal cubic equation has the form , where A and B are coefficients.

Let's group:

Obviously, x = -1 is the root of such an equation, and the roots of the resulting quadratic trinomial are easily found through the discriminant.

Example.

Solve cubic equation .

Solution.

This is a reciprocal equation. Let's group:

Obviously x = -1 is the root of the equation.

Finding the roots of a quadratic trinomial:

Answer:

Solving cubic equations with rational roots.

Let's start with the simplest case, when x=0 is the root of the cubic equation.

In this case, the free term D is equal to zero, that is, the equation has the form .

If you take x out of brackets, then a square trinomial will remain in brackets, the roots of which can be easily found either through the discriminant or by Vieta’s theorem .

Example.

Find the real roots of the equation .

Solution.

x=0 is the root of the equation. Let's find the roots of the quadratic trinomial.

Since its discriminant is less than zero, the trinomial has no real roots.

Answer:

x=0.

If the coefficients of a cubic equation are integers, then the equation can have rational roots.

When , multiply both sides of the equation by and change variables y = Ax:

We arrived at the given cubic equation. It can have whole roots, which are divisors of the free term. So we write down all the divisors and begin to substitute them into the resulting equation until we obtain an identical equality. The divisor at which the identity is obtained is the root of the equation. Therefore, the root of the original equation is .

Example.

Find the roots of the cubic equation.

Solution.

Let's transform the equation to the above: multiply by both sides and change the variable y = 2x.

The free term is 36. Let's write down all its divisors: .

We substitute them one by one into equality until identity is obtained:

So y = -1 is the root. It corresponds to .

Let's divide on, using:

We get

All that remains is to find the roots of the quadratic trinomial.

It's obvious that , that is, its multiple root is x=3.

Answer:

Comment.

This algorithm can be used to solve reciprocal equations. Since -1 is the root of any reciprocal cubic equation, we can divide the left side of the original equation by x+1 and find the roots of the resulting quadratic trinomial.

In the case when the cubic equation does not have rational roots, other solution methods are used, for example, specific methods.

Solving cubic equations using the Cardano formula.

In general, the roots of a cubic equation are found using the Cardano formula.

For the cubic equation the values are found . Next we find And .

We substitute the resulting p and q into the Cardano formula:

Dispute

FormulaCardano

Mostovoy

Odessa

Dispute

Disputes in the Middle Ages always presented an interesting spectacle, attracting idle townspeople, young and old. The topics of the debates were varied, but always scientific. At the same time, science was understood to be what was included in the list of the so-called seven liberal arts, which was, of course, theology. Theological disputes were the most frequent. They argued about everything. For example, about whether to associate a mouse with the holy spirit if it eats the sacrament, whether the Cumae Sibyl could have predicted the birth of Jesus Christ, why the brothers and sisters of the Savior are not canonized, etc.

About the dispute that was supposed to take place between the famous mathematician and the no less famous doctor, only the most general guesses were made, since no one really knew anything. They said that one of them deceived the other (it is unknown who exactly and to whom). Almost all those who gathered in the square had the most vague ideas about mathematics, but everyone was looking forward to the start of the debate. It was always interesting, you could laugh at the loser, regardless of whether he was right or wrong.

When the town hall clock struck five, the gates swung wide open and the crowd rushed inside the cathedral. On either side of the center line connecting the entrance to the altar, two high pulpits were erected near the two side columns, intended for debaters. Those present made a loud noise, not paying any attention to the fact that they were in the church. Finally, in front of the iron grille that separated the iconostasis from the rest of the central nave, a town crier in a black and purple cloak appeared and proclaimed: “Illustrious citizens of the city of Milan! Now the famous mathematician Niccolo Tartaglia from Brenia will speak to you. His opponent was supposed to be the mathematician and physician Geronimo Cardano. Niccolò Tartaglia accuses Cardano of being the last to publish in his book “Ars magna” a method for solving an equation of the 3rd degree, which belongs to him, Tartaglia. However, Cardano himself could not come to the debate and therefore sent his student Luige Ferrari. So, the debate is declared open, its participants are invited to the departments.” An awkward man with a hooked nose and a curly beard climbed onto the pulpit to the left of the entrance, and a young man in his twenties with a handsome, self-confident face ascended to the opposite pulpit. His entire demeanor reflected complete confidence that his every gesture and every word would be received with delight.

Tartaglia began.

Dear Sirs! You know that 13 years ago I managed to find a way to solve an equation of the 3rd degree and then, using this method, I won the dispute with Fiori. My method attracted the attention of your fellow citizen Cardano, and he used all his cunning art to find out the secret from me. He did not stop either from deception or outright forgery. You also know that 3 years ago Cardano’s book on the rules of algebra was published in Nuremberg, where my method, so shamelessly stolen, was made available to everyone. I challenged Cardano and his student to a competition. I proposed to solve 31 problems, the same number was proposed to me by my opponents. A deadline was set for solving problems - 15 days. In 7 days I managed to solve most of the problems that were compiled by Cardano and Ferrari. I printed them and sent them by courier to Milan. However, I had to wait a full five months until I received answers to my tasks. They were resolved incorrectly. This gave me grounds to challenge both of them to a public debate.

Tartaglia fell silent. The young man, looking at the unfortunate Tartaglia, said:

Dear Sirs! My worthy opponent allowed himself, in the very first words of his speech, to express so much slander against me and against my teacher; his argument was so unfounded that it would hardly take me any trouble to refute the first and show you the inconsistency of the second. First of all, what kind of deception can we talk about if Niccolo Tartaglia completely voluntarily shared his method with both of us? And this is how Geronimo Cardano writes about the role of my opponent in the discovery of the algebraic rule. He says that it is not he, Cardano, “but my friend Tartaglia who has the honor of discovering something so beautiful and amazing, surpassing human wit and all the talents of the human spirit. This discovery is truly a heavenly gift, such a wonderful proof of the power of the mind that comprehended it, that nothing can be considered unattainable for him.”

My opponent accused me and my teacher of allegedly giving the wrong solution to his problems. But how can the root of an equation be incorrect if by substituting it into the equation and performing all the actions prescribed in this equation, we arrive at identity? And if Senor Tartaglia wants to be consistent, then he should have responded to the remark why we, who stole, but in his words, his invention and used it to solve the proposed problems, received the wrong solution. We - my teacher and I - do not consider Signor Tartaglia's invention to be of little importance. This invention is wonderful. Moreover, relying largely on it, I found a way to solve an equation of the 4th degree, and in the Ars Magna my teacher talks about this. What does Senor Tartaglia want from us? What is he trying to achieve with the dispute?

Gentlemen, gentlemen,” Tartaglia shouted, “I ask you to listen to me!” I do not deny that my young opponent is very strong in logic and eloquence. But this cannot replace a true mathematical proof. The problems that I gave to Cardano and Ferrari were not solved correctly, but I will prove this too. Indeed, let us take, for example, an equation from among those solved. It is known...

An unimaginable noise arose in the church, completely absorbing the end of the sentence begun by the hapless mathematician. He was not allowed to continue. The crowd demanded that he shut up and that Ferrari should take the turn. Tartaglia, seeing that continuing the argument was completely useless, hastily descended from the pulpit and went out through the northern porch into the square. The crowd wildly greeted the “winner” of the dispute, Luigi Ferrari.

...This is how this dispute ended, which continues to cause more and more new disputes. Who actually owns the method for solving a 3rd degree equation? We're talking now - Niccolo Tartaglie. He discovered it, and Cardano tricked him into making the discovery. And if now we call the formula representing the roots of an equation of the 3rd degree through its coefficients the Cardano formula, then this is a historical injustice. However, is it unfair? How to calculate the degree of participation of each mathematician in the discovery? Maybe over time someone will be able to answer this question absolutely accurately, or maybe it will remain a mystery...

Cardano formula

Using modern mathematical language and modern symbolism, the derivation of Cardano's formula can be found using the following extremely elementary considerations:

Let us be given a general equation of the 3rd degree:

ax 3 +3bx 2 +3cx+d=0 (1)

If you put

, then we give the equation (1) to mind

Let's introduce a new unknown U using equality

By introducing this expression into (2) , we get

hence

If the numerator and denominator of the second term are multiplied by the expression and taken into account, the resulting expression for u turns out to be symmetrical with respect to the signs “+” and “-”, then we finally get

(The product of cubic radicals in the last equality must equal p).

This is the famous Cardano formula. If you go from y back to x, then we obtain a formula that determines the root of a general equation of the 3rd degree.

The young man who treated Tartaglia so mercilessly understood mathematics as easily as he understood the rights of unpretentious secrecy. Ferrari finds a way to solve a 4th degree equation. Cardano included this method in his book. What is this method?

Let (1)

- general equation of 4th degree.

If you put

then the equation (1) can be brought to mind

Where p,q,r- some coefficients depending on a,b,c,d,e. It is easy to see that this equation can be written as follows:

In fact, it is enough to open the brackets, then all terms containing t, cancels out, and we return to the equation (2) .

Let's select a parameter t so that the right side of the equation (3) was a perfect square relative to y. As is known, a necessary and sufficient condition for this is the vanishing of the discriminant of the coefficients of the trinomial (with respect to y) standing on the right:

We have obtained a complete cubic equation, which we can now solve. Let's find any of its roots and add it to the equation (3) , will now take the form

This is a quadratic equation. Solving it, you can find the root of the equation (2) , and therefore (1) .

4 months before his death, Cardano finished his autobiography, which he wrote intensely throughout the last year and which was supposed to sum up his difficult life. He felt death approaching. According to some reports, his own horoscope linked his death with his 75th birthday. He died on September 21, 1576. 2 days before the anniversary. There is a version that he committed suicide in anticipation of imminent death or even to confirm his horoscope. In any case, Cardano the astrologer took the horoscope seriously.

A note about Cardano's formula

Let us analyze the formula for solving the equation in the real domain. So,

When calculating x we have to take the square root first and then the cubic root. We can take the square root while remaining in the real region if . Two square root values that differ in sign appear in different terms for x. The values of the cube root in the real domain are unique and the result is a unique real root x at . By examining the graph of the cubic trinomial, it is easy to verify that it actually has a single real root at . At there are three real roots. When there is a double real root and a single root, and when there is a triple root x=0.

Let's continue studying the formula for . Turns out. What if an equation with integer coefficients has an integer root, when calculating it using the formula, intermediate irrationalities may arise. For example, the equation has a single root (real) - x=1. Cardano's formula gives for this single real root the expression

But virtually any proof involves using the fact that this expression is the root of the equation. If you do not guess this, indestructible cubic radicals will appear during the transformation.

The Cardano-Tartaglia problem was soon forgotten. The formula for solving the cubic equation was associated with the “Great Art” and gradually began to be called formula Cardano.

Many had a desire to restore the true picture of events in a situation where their participants undoubtedly did not tell the whole truth. For many, it was important to establish the extent of Cardano's guilt. By the end of the 19th century, some of the discussions began to take on the character of serious historical and mathematical research. Mathematicians realized what a big role Cardano's work played at the end of the 16th century. It became clear what Leibniz had noted even earlier: “Cardano was a great man with all his shortcomings; without them he would be perfect.”

MUNICIPAL VII STUDENT SCIENTIFIC AND PRACTICAL CONFERENCE “YOUTH: CREATIVITY, SEARCH, SUCCESS”

Anninsky municipal district

Voronezh region

Section:MATHEMATICS

Subject:"Cardano Formula: History and Application"

MKOU Anninskaya secondary school No. 3, 9 “B” class

Niccolò Fontana Tartaglia (Italian: NiccolòFontanaTartaglia, 1499-1557) - Italian mathematician.

In general, history tells that the formula was initially discovered by Tartaglia and handed over to Cardano in finished form, but Cardano himself denied this fact, although he did not deny Tartaglia’s involvement in the creation of the formula.

The name “Cardano’s formula” is firmly rooted behind the formula, in honor of the scientist who actually explained and presented it to the public.

When the town hall clock struck five, the gates swung wide open and the crowd rushed inside the cathedral. On either side of the center line connecting the entrance to the altar, two high pulpits were erected near the two side columns, intended for debaters. Those present made a loud noise, not paying any attention to the fact that they were in the church. Finally, in front of the iron grille that separated the iconostasis from the rest of the central nave, a town crier in a black and purple cloak appeared and proclaimed: “Illustrious citizens of the city of Milan! Now the famous mathematician Niccolo Tartaglia from Brenia will speak to you. His opponent was supposed to be the mathematician and physician Geronimo Cardano. Niccolò Tartaglia accuses Cardano of the fact that the latter, in his book “Arsmagna,” published a method for solving an equation of the 3rd degree, which belongs to him, Tartaglia. However, Cardano himself could not come to the debate and therefore sent his student Luige Ferrari. So, the debate is declared open, its participants are invited to the departments.” An awkward man with a hooked nose and a curly beard ascended to the pulpit to the left of the entrance, and a young man in his twenties with a handsome, self-confident face ascended to the opposite pulpit. His entire demeanor showed complete confidence that his every gesture and every word would be received with delight.

Tartaglia began.

Tartaglia fell silent. The young man, looking at the unfortunate Tartaglia, said:

My opponent accused me and my teacher of allegedly giving the wrong solution to his problems. But how can the root of an equation be incorrect if by substituting it into the equation and performing all the actions prescribed in this equation, we arrive at identity? And if Senor Tartaglia wants to be consistent, then he should have responded to the remark why we, who, in his words, stole his invention and used it to solve the proposed problems, received the wrong solution. We - my teacher and I - do not consider Signor Tartaglia's invention to be of little importance. This invention is wonderful. Moreover, relying largely on it, I found a way to solve an equation of the 4th degree, and in Arsmagna my teacher talks about this. What does Senor Tartaglia want from us? What is he trying to achieve with the dispute?

Gentlemen, gentlemen,” Tartaglia shouted, “I ask you to listen to me!” I do not deny that my young opponent is very strong in logic and eloquence. But this cannot replace a true mathematical proof. The problems I gave to Cardano and Ferrari were solved incorrectly, but I will prove it too. Indeed, let us take, for example, an equation from among those solved. It is known...

Thus ended this dispute, which continues to cause more and more new disputes. Who actually owns the method for solving a 3rd degree equation? We're talking now - Niccolo Tartaglie. He discovered it, and Cardano tricked him into making the discovery. And if now we call the formula representing the roots of an equation of the 3rd degree through its coefficients the Cardano formula, then this is a historical injustice. However, is it unfair? How to calculate the degree of participation of each mathematician in the discovery? Maybe over time someone will be able to answer this question absolutely accurately, or maybe it will remain a mystery...

Using modern mathematical language and modern symbolism, the derivation of Cardano's formula can be found using the following extremely elementary considerations:

Let us be given a general equation of the 3rd degree:

x 3 + ax 2 + bx + c = 0,

(1)

Wherea, b, c – arbitrary real numbers.

Let us replace the variable in equation (1)X to a new variable yaccording to the formula:

x 3 +ax 2 +bx+c = (y ) 3 + a(y ) 2 + b(y ) + c = y 3 3y 2 + 3y+ a(y 2 2y+ by = y 3 y 3 + (b

then equation (1) will take the formy 3 + ( b

If we introduce the notationp = b, q = ,

then the equation will take the formy 3 + py + q = 0.

This is the famous Cardano formula.

Roots of a cubic equationy 3 + py + q = 0 depend on the discriminant

IfD> 0, thena cubic polynomial has three different real roots.

IfD< 0, то a cubic polynomial has one real root and two complex roots (which are complex conjugate).

IfD = 0, it has a multiple root (either one root of multiplicity 2 and one root of multiplicity 1, both of which are real; or one single real root of multiplicity 3).

2.4. Examples of universal methods for solving cubic equations

Let's try to apply Cardan's formula to solving specific equations.

Example 1: x 3 +15 x+124 = 0

Herep = 15; q = 124.

Answer:X