Equations of higher degrees. Power or exponential equations 5th degree equation

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First, let's recall the basic formulas of degrees and their properties.

Product of a number a happens on itself n times, we can write this expression as a a … a=a n

1. a 0 = 1 (a ≠ 0)

3. a n a m = a n + m

4. (a n) m = a nm

5. a n b n = (ab) n

7. a n / a m \u003d a n - m

Power or exponential equations- these are equations in which the variables are in powers (or exponents), and the base is a number.

Examples of exponential equations:

In this example, the number 6 is the base, it is always at the bottom, and the variable x degree or measure.

Let us give more examples of exponential equations.
2 x *5=10
16x-4x-6=0

Now let's look at how exponential equations are solved?

Let's take a simple equation:

2 x = 2 3

Such an example can be solved even in the mind. It can be seen that x=3. After all, in order for the left and right sides to be equal, you need to put the number 3 instead of x.
Now let's see how this decision should be made:

2 x = 2 3
x = 3

To solve this equation, we removed same grounds(that is, deuces) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our solution.

Algorithm for solving the exponential equation:
1. Need to check the same whether the bases of the equation on the right and on the left. If the grounds are not the same, we are looking for options to solve this example.
2. After the bases are the same, equate degree and solve the resulting new equation.

Now let's solve some examples:

Let's start simple.

The bases on the left and right sides are equal to the number 2, which means we can discard the base and equate their degrees.

x+2=4 The simplest equation has turned out.
x=4 - 2
x=2
Answer: x=2

In the following example, you can see that the bases are different, these are 3 and 9.

3 3x - 9 x + 8 = 0

To begin with, we transfer the nine to the right side, we get:

Now you need to make the same bases. We know that 9=3 2 . Let's use the power formula (a n) m = a nm .

3 3x \u003d (3 2) x + 8

We get 9 x + 8 \u003d (3 2) x + 8 \u003d 3 2 x + 16

3 3x \u003d 3 2x + 16 now it is clear that the bases on the left and right sides are the same and equal to three, which means we can discard them and equate the degrees.

3x=2x+16 got the simplest equation
3x-2x=16
x=16
Answer: x=16.

Let's look at the following example:

2 2x + 4 - 10 4 x \u003d 2 4

First of all, we look at the bases, the bases are different two and four. And we need to be the same. We transform the quadruple according to the formula (a n) m = a nm .

4 x = (2 2) x = 2 2x

And we also use one formula a n a m = a n + m:

2 2x+4 = 2 2x 2 4

Add to the equation:

2 2x 2 4 - 10 2 2x = 24

We gave an example for the same reasons. But other numbers 10 and 24 interfere with us. What to do with them? If you look closely, you can see that on the left side we repeat 2 2x, here is the answer - we can put 2 2x out of brackets:

2 2x (2 4 - 10) = 24

Let's calculate the expression in brackets:

2 4 — 10 = 16 — 10 = 6

We divide the whole equation by 6:

Imagine 4=2 2:

2 2x \u003d 2 2 bases are the same, discard them and equate the degrees.
2x \u003d 2 turned out to be the simplest equation. We divide it by 2, we get
x = 1
Answer: x = 1.

Let's solve the equation:

9 x - 12*3 x +27= 0

Let's transform:
9 x = (3 2) x = 3 2x

We get the equation:
3 2x - 12 3 x +27 = 0

The bases are the same for us, equal to three. In this example, it can be seen that the first triple has a degree twice (2x) than the second (just x). In this case, you can decide substitution method. The number with the smallest degree is replaced by:

Then 3 2x \u003d (3 x) 2 \u003d t 2

We replace all degrees with x's in the equation with t:

t 2 - 12t + 27 \u003d 0
We get a quadratic equation. We solve through the discriminant, we get:
D=144-108=36
t1 = 9
t2 = 3

Back to Variable x.

We take t 1:
t 1 \u003d 9 \u003d 3 x

That is,

3 x = 9
3 x = 3 2
x 1 = 2

One root was found. We are looking for the second one, from t 2:
t 2 \u003d 3 \u003d 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 \u003d 2; x 2 = 1.

On the site you can in the section HELP DECIDE to ask questions of interest, we will definitely answer you.

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Judging by the beginning of the publication, which we will omit here, the text was written by Yuri Ignatievich. And it is well written, and the problems are topical, that's just how to call Russia, as Mukhin does ...

No matter how anyone treats the anti-people government, Russia is above it and does not deserve insults. Even from the talented deceiver of the American agency NASA.

Appeal to comrade. Mukhin Yu.I.

Dear Yuri Ignatievich! I know that you visit these pages. Therefore, I am addressing you directly.

We all appreciate your selfless work in the field of exposing the lies of the West, the lies of America, the lies of pseudo-scientists, the lies of liberals. With pleasure and benefit for ourselves and society, we think about serious topics that you throw at us from time to time, be it meritocracy or metaphysics, love for national history or restoration of justice.

However, your definitions of our common Motherland with you are perplexing and greatly upset.

However, judge for yourself: how would you characterize a person who began to insult his mother who fell ill and because of this temporarily stopped working?

But Russia, no matter how it is called, and no matter how good or disgusting the power may be, Russia is our Motherland. Motherland. For her, our grandfathers shed their blood and laid down their lives.

Therefore, to put it on a par with power is to lower the spiritual sublime to the level of the material, and even low. Those. you are comparing completely different categories. A thing unacceptable for any sane person.

I beg you, dear comrade. Mukhin, think seriously about this.

... And with the equations (I did not know this), the situation is as follows. How to find roots quadratic equation guessed back in ancient Egypt.

How to find the roots of a cubic equation and a fourth degree equation was found in the sixteenth century, but they could not find the roots of a fifth degree equation until 2016. And far from ordinary people tried.

In the sixteenth century, the founder of symbolic algebra François Viet tried to find the roots of the equation of the fifth degree; in the nineteenth century, the founder of modern higher algebra, the French mathematician Evariste Galois, tried to do this; after him, the Norwegian mathematician Niels Henrik Abel tried to find the roots of the equations of the fifth degree, who, in the end, gave up and proved the impossibility of solving the equation of the fifth degree in a general form.

We read in Wikipedia about the merits of Abel: “Abel completed a brilliant study of an ancient problem:proved the impossibility of solving in a general form (in radicals) an equation of the 5th degree ...

In algebra, Abel found necessary condition so that the root of the equation is expressed "in radicals" through the coefficients of this equation. The sufficient condition was soon discovered by Galois, whose achievements were based on the work of Abel.

Abel gave concrete examples of a 5th degree equation whose roots cannot be expressed in radicals, and thus largely closed the ancient problem.

As you can see, if they tried to prove the Poincaré theorem all the time and Perelman turned out to be more successful than other mathematicians, then after Abel they did not take mathematics to the equations of the fifth degree.

And in 2014 mathematician from Tomsk Sergey Zaikov, which can be judged from the photo that he is already in years, and according to the data from the article about him, that he is a graduate of the Faculty of Applied Mathematics and Cybernetics of Tomsk state university, in the course of his work he received equations of the fifth degree. Dead end? Yes, dead end! But Sergey Zaikov undertook to break through it.

And in 2016, he found ways to solve equations of the fifth degree in a general form! He did what the mathematicians Galois and Abel proved the impossibility of.

I tried to find information about Sergei Zaikov on Wikipedia, but fuck you! About the mathematician Sergey Zaikov and about finding the solution of equations of the fifth degree by him no information!

The piquancy of the matter is also given by the fact that for mathematicians there is an analogue of the Nobel Prize - Abel Prize(Nobel forbade giving the prize to mathematicians and now it is given for mathematical bowel movements, calling them "physics").

This mathematical prize is in honor of the same Abel who proved the impossibility of what Zaikov did. However, self-nomination for this award is not allowed. But Zaikov is a lone mathematician and there are no organizations that could nominate him for this award.

True, we have the Academy of Sciences, but after all, academicians sit there not for the development of mathematics, but “to cut the loot”. Who needs this Zaikov there?

Well, for news agencies, Zaikov is not Perelman for you! Therefore, the discovery of Zaikov for the media is not a sensation.

Here is the fact that Poroshenko made a mistake with the door - yes! This is a real sensation!

Tomsk mathematician solved a problem that could not be solved for two hundred years

With the advent of algebra, its main task was considered to be the solution of algebraic equations. The solution of the equation of the second degree was already known in Babylon and Ancient Egypt. We go through these equations in school. Remember the equation x2 + ax + b = 0, and the discriminant?

Sergei Zaikov with a book

The solution of algebraic equations of the third and fourth degree was found in the sixteenth century. But it was not possible to solve the equation of the fifth degree. The reason was found by Lagrange. He showed that the solution of equations of the third and fourth degree became possible because they can be reduced to equations that had already been solved before. A third-degree equation can be reduced to a second-degree equation, and a fourth-degree equation can be reduced to a third-degree equation. But the equation of the fifth degree is reduced to the equation of the sixth, i.e., more complex, so the traditional methods of solution are not applicable.

The question of solving a fifth-degree equation moved forward only two hundred years ago, when Abel proved that not all fifth-degree equations can be solved in radicals, that is, in square, cubic and other roots known to us from school. And Galois soon, that is, two hundred years ago, found a criterion for determining which equations of the fifth degree can be solved in radicals and which cannot. It lies in the fact that the Galois group of radically solvable equations of the fifth degree must be either cyclic or metacyclic. But Galois did not find a way to solve in radicals those fifth-degree equations that are solvable in radicals. Galois theory is very famous, many books have been written about it.

So far, only particular solutions have been found for fifth-degree equations solvable in radicals. And only this year, the Tomsk mathematician Sergey Zaikov solved a problem that could not be solved for two hundred years. He published the book "How Algebraic Equations of the Fifth Degree Are Solved in Radicals", in which he indicated a method for solving any equations of the fifth degree that are solvable in radicals. Zaikov is a graduate of the Faculty of Applied Mathematics and Cybernetics of Tomsk State University. We were able to interview him.

— Sergey, why did you start solving this problem?

— I needed a solution to a fifth-degree equation to solve a problem from another branch of mathematics. I started figuring out how to find it and found out that not all of them are solved in radicals. Then I tried to find in the scientific literature a way to solve those equations that are solvable in radicals, but I found only a criterion by which one can determine which are solvable and which are not. I am not an algebraist, but, of course, as a graduate of the FPMK, I can also apply algebraic methods. Therefore, since 2014, I seriously began to look for a solution and found it myself.

The method was found by me two years ago, I prepared a book in which not only it was described, but also methods for solving some equations of degrees greater than the fifth. But I didn't have the money to publish it. This year I decided that it would be easier to publish only part of this work, and took only half of it, devoted to a method for solving a fifth-degree equation in radicals.

I made it my goal to publish something like a guide to solving this problem, understandable for mathematicians who need to solve a specific equation. Therefore, I simplified it by removing a lot of long formulas and a significant part of the theory, cutting it down by more than half, leaving only the necessary. Therefore, I got something like a book "for dummies", according to which mathematicians who are not familiar with Galois theory can solve the equation they need.

- For this, many thanks to Vladislav Beresnev, whom we have known for many years. He sponsored the publication of the book.

Is it possible for you to receive any prize in mathematics for solving this problem? For example, you mentioned Abel. But there is an Abel Prize in mathematics, which is considered an analogue of the Nobel Prize?

“That possibility cannot be completely ruled out. But you should not hope for it either.

For example, applications for candidates for the 2019 Abel Prize are due by September 15th. Moreover, self-nomination is not allowed. I'm a lone mathematician. There are no organizations or well-known mathematicians who will nominate me. Therefore, it will not be considered regardless of whether my work deserves this prize, and whether it is in the spirit of this prize that it be awarded to those who continue the work of Abel. But even if it is presented, everything also depends on the level of work of other candidates.

The book is intended for those who are not familiar with the Galois theory. The fundamentals of the Galois theory are given only in the part in which they are necessary for solving the equation, the solution method is described in detail, and techniques that simplify the solution are shown. A significant part of the book is devoted to an example of solving a particular equation. The reviewers of the book are Doctor of Technical Sciences Gennady Petrovich Agibalov and Doctor of Physics. mat. Sciences, Professor Petr Andreevich Krylov.

PREPARED ANASTASIA SKIRNEVSKAYA

In general, an equation that has a degree higher than 4 cannot be solved in radicals. But sometimes we can still find the roots of the polynomial on the left in the equation of the highest degree, if we represent it as a product of polynomials in a degree of no more than 4. The solution of such equations is based on the decomposition of the polynomial into factors, so we advise you to review this topic before studying this article.

Most often you have to deal with equations higher degrees with integer coefficients. In these cases, we can try to find rational roots, and then factoring the polynomial to then convert it to an equation of a lower degree, which will be easy to solve. In the framework of this material, we will consider just such examples.

Higher degree equations with integer coefficients

All equations of the form a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = 0 , we can reduce to an equation of the same degree by multiplying both sides by a n n - 1 and changing the variable of the form y = a n x:

a n x n + a n - 1 x n - 1 + . . . + a 1 x + a 0 = 0 ann xn + an - 1 ann - 1 xn - 1 + … + a 1 (an) n - 1 x + a 0 (an) n - 1 = 0 y = anx ⇒ yn + bn - 1 yn - 1 + … + b 1 y + b 0 = 0

The resulting coefficients will also be integers. Thus, we will need to solve the above equation nth degree with integer coefficients, having the form x n + a n x n - 1 + … + a 1 x + a 0 = 0 .

We calculate the integer roots of the equation. If the equation has integer roots, you need to look for them among the divisors of the free term a 0. Let's write them down and substitute them into the original equality one by one, checking the result. Once we have obtained an identity and found one of the roots of the equation, we can write it in the form x - x 1 · P n - 1 (x) = 0 . Here x 1 is the root of the equation, and P n - 1 (x) is the quotient of x n + a n x n - 1 + ... + a 1 x + a 0 divided by x - x 1 .

Substitute the remaining divisors in P n - 1 (x) = 0 , starting with x 1 , since the roots can be repeated. After obtaining the identity, the root x 2 is considered found, and the equation can be written as (x - x 1) (x - x 2) P n - 2 (x) \u003d 0. Here P n - 2 (x) will be quotient from dividing P n - 1 (x) by x - x 2 .

We continue to sort through the divisors. Find all integer roots and denote their number as m. After that, the original equation can be represented as x - x 1 x - x 2 · … · x - x m · P n - m (x) = 0 . Here P n - m (x) is a polynomial of n - m -th degree. For calculation it is convenient to use Horner's scheme.

If our original equation has integer coefficients, we cannot end up with fractional roots.

As a result, we got the equation P n - m (x) = 0, the roots of which can be found in any convenient way. They can be irrational or complex.

Let us show on a specific example how such a solution scheme is applied.

Example 1

Condition: find the solution of the equation x 4 + x 3 + 2 x 2 - x - 3 = 0 .

Solution

Let's start with finding integer roots.

We have an intercept equal to minus three. It has divisors equal to 1 , - 1 , 3 and - 3 . Let's substitute them into the original equation and see which of them will give identities as a result.

For x equal to one, we get 1 4 + 1 3 + 2 1 2 - 1 - 3 = 0, so one will be the root given equation.

Now let's divide the polynomial x 4 + x 3 + 2 x 2 - x - 3 by (x - 1) into a column:

So x 4 + x 3 + 2 x 2 - x - 3 = x - 1 x 3 + 2 x 2 + 4 x + 3.

1 3 + 2 1 2 + 4 1 + 3 = 10 ≠ 0 (- 1) 3 + 2 (- 1) 2 + 4 - 1 + 3 = 0

We got an identity, which means we found another root of the equation, equal to - 1.

We divide the polynomial x 3 + 2 x 2 + 4 x + 3 by (x + 1) in a column:

We get that

x 4 + x 3 + 2 x 2 - x - 3 = (x - 1) (x 3 + 2 x 2 + 4 x + 3) = = (x - 1) (x + 1) (x 2 + x + 3)

We substitute the next divisor into the equation x 2 + x + 3 = 0, starting from - 1:

1 2 + (- 1) + 3 = 3 ≠ 0 3 2 + 3 + 3 = 15 ≠ 0 (- 3) 2 + (- 3) + 3 = 9 ≠ 0

The resulting equalities will be incorrect, which means that the equation no longer has integer roots.

The remaining roots will be the roots of the expression x 2 + x + 3 .

D \u003d 1 2 - 4 1 3 \u003d - 11< 0

It follows from this that this square trinomial there are no real roots, but there are complex conjugate ones: x = - 1 2 ± i 11 2 .

Let us clarify that instead of dividing into a column, Horner's scheme can be used. This is done like this: after we have determined the first root of the equation, we fill in the table.

In the table of coefficients, we can immediately see the coefficients of the quotient from the division of polynomials, which means x 4 + x 3 + 2 x 2 - x - 3 = x - 1 x 3 + 2 x 2 + 4 x + 3.

After finding the next root, equal to - 1 , we get the following:

Answer: x \u003d - 1, x \u003d 1, x \u003d - 1 2 ± i 11 2.

Example 2

Condition: solve the equation x 4 - x 3 - 5 x 2 + 12 = 0.

Solution

The free member has divisors 1 , - 1 , 2 , - 2 , 3 , - 3 , 4 , - 4 , 6 , - 6 , 12 , - 12 .

Let's check them in order:

1 4 - 1 3 - 5 1 2 + 12 = 7 ≠ 0 (- 1) 4 - (- 1) 3 - 5 (- 1) 2 + 12 = 9 ≠ 0 2 4 2 3 - 5 2 2 + 12 = 0

So x = 2 will be the root of the equation. Divide x 4 - x 3 - 5 x 2 + 12 by x - 2 using Horner's scheme:

As a result, we get x - 2 (x 3 + x 2 - 3 x - 6) = 0 .

2 3 + 2 2 - 3 2 - 6 = 0

So 2 will again be a root. Divide x 3 + x 2 - 3 x - 6 = 0 by x - 2:

As a result, we get (x - 2) 2 (x 2 + 3 x + 3) = 0 .

Checking the remaining divisors does not make sense, since the equality x 2 + 3 x + 3 = 0 is faster and more convenient to solve using the discriminant.

Let's solve the quadratic equation:

x 2 + 3 x + 3 = 0 D = 3 2 - 4 1 3 = - 3< 0

We get a complex conjugate pair of roots: x = - 3 2 ± i 3 2 .

Answer: x = - 3 2 ± i 3 2 .

Example 3

Condition: find the real roots for the equation x 4 + 1 2 x 3 - 5 2 x - 3 = 0.

Solution

x 4 + 1 2 x 3 - 5 2 x - 3 = 0 2 x 4 + x 3 - 5 x - 6 = 0

We perform the multiplication 2 3 of both parts of the equation:

2 x 4 + x 3 - 5 x - 6 = 0 2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0

We replace the variables y = 2 x:

2 4 x 4 + 2 3 x 3 - 20 2 x - 48 = 0 y 4 + y 3 - 20 y - 48 = 0

As a result, we got a standard equation of the 4th degree, which can be solved according to the standard scheme. Let's check the divisors, divide and in the end we get that it has 2 real roots y \u003d - 2, y \u003d 3 and two complex ones. We will not present the entire solution here. By virtue of the replacement, the real roots of this equation will be x = y 2 = - 2 2 = - 1 and x = y 2 = 3 2 .

Answer: x 1 \u003d - 1, x 2 \u003d 3 2

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In the 16th century, mathematicians stumbled upon complex numbers almost by accident (see Chapter 11). TO XVIII century complex numbers were considered an extension of the domain real numbers, but working with them still led to a parity error, as in Leonard E's great work on number theory "Arithmetical Investigations" (1801) avoided the use of so-called "imaginary numbers". It seems to me that the most important part of this work is the first proof of the fundamental theorem of algebra. Gauss realized how important this theorem was, creating several additional proofs over the following years. In 1849 he revised the first version, this time using complex numbers. Using modern terms, we can say that for any finite polynomial equation with real or complex coefficients, all its roots will be real or complex numbers. Thus, we get a negative answer to the age-old question of whether the solution of polynomial equations requires high order create numbers of a higher order than complex ones.

One of the most thorny problems in algebra at that time was the question of whether algebraic methods, that is, using a finite number of algebraic steps, the fifth-order polynomial is a quintic. Now the school teaches the formula for solving quadratic equations, and since the 16th century, similar methods have been known for solving equations of the third and fourth degree (Chapter 11). But no method has been found for quintics. It may seem that the fundamental theorem of algebra contains the promise of a positive answer, but in fact it simply guarantees that solutions exist, it does not say anything about the existence of formulas that give exact solutions (approximate numerical and graphical methods already existed by that time). And then two mathematical geniuses with a tragic fate appeared.

Niels Henrik Abel (1802–1829) was born into a large poor family living in a small village in Norway, a country devastated by long years of war with England and Sweden. The teacher, who was friendly to the boy, gave him private lessons, but after the death of his father, at the age of eighteen, despite his young age and fragile health, Abel was forced to support his family. In 1824 he published scientific article, in which he stated that the quintic is not soluble by algebraic means, as, indeed, any polynomial of a higher order. Abel believed that this article would serve as his pass to the scientific world, and sent it to Gauss at the University of Göttingen. Unfortunately, Gauss never got around to cutting the pages with a knife (any reader had to do that in those days) and didn't read the article. In 1826, the Norwegian government finally provided funds for Abel to travel around Europe. Fearing that personal contact with Gauss would not bring him great joy, the mathematician decided not to visit Göttingen and instead went to Berlin. There he befriended August Leopold Crelle (1780–1855), a mathematician, architect and engineer who advised the Prussian Ministry of Education on matters of mathematics. Krell was going to found the Journal of Pure and Applied Mathematics. So Abel got the opportunity to distribute his work and published a lot, especially in the early issues of the Journal, which immediately began to be considered a very prestigious and authoritative scientific publication. The Norwegian published there an extended version of his proof that the quintic is unsolvable by algebraic methods. And then he went to Paris. This trip upset Abel very much, because he practically did not receive the support of French mathematicians that he needed so much. He became close to Augustin Louis Cauchy (1789-1857), who at that time was the main luminary mathematical analysis but was very complex. As Abel himself put it, "Cauchy is insane and nothing can be done about it, although at present he is the only one capable of anything in mathematics." If one tries to justify the disrespect and disdain emanating from Gauss and Cauchy, one can say that the Quintic achieved a certain fame and attracted the attention of both respected mathematicians and originals. Abel returned to Norway, where he was increasingly suffering from tuberculosis. He continued to send his papers to Crelle, but died in 1829, unaware of the extent to which his reputation had grown in the scientific world. Two days after his death, Abel received an offer to take a scientific position in Berlin.

Abel showed that any polynomial above fourth order cannot be solved with radicals like square roots, cube roots, or higher. However, the explicit conditions under which, in special cases, these polynomials could be solved, and the method for solving them, were formulated by Galois. Évariste Galois (1811–1832) lived a short and eventful life. He was an incredibly gifted mathematician. Galois was relentless towards those whom he considered less talented than himself, and at the same time he could not stand social injustice. He showed no aptitude for mathematics until he read Legendre's Elements of Geometry (published in 1794, this book was the main textbook for the next hundred years). Then he literally swallowed the rest of the works of Legendre and, later, Abel. His enthusiasm, self-confidence and intolerance led to truly dire consequences in his dealings with teachers and examiners. Galois took part in the competition for admission to the Polytechnic School - the cradle of French mathematics, but due to unpreparedness he failed the exam. Some time after meeting a new teacher who recognized his talent, he managed to keep his temper under control. In March 1829, Galois published his first paper on continued fractions, which he considered his most significant work. He sent a message about his discoveries to the Academy of Sciences, and Cauchy promised to present them, but forgot. Moreover, he simply lost the manuscript.

Galois' second failure to enter the Polytechnic School entered mathematical folklore. He was so accustomed to constantly having complex mathematical ideas in his head that he was infuriated by the petty nitpicking of the examiners. As the examiners had difficulty understanding his explanations, he threw an erasing cloth in the face of one of them. Soon after, his father died, having committed suicide as a result of church intrigues. A riot almost broke out at his funeral. In February 1830, Galois wrote the following three articles, sending them to the Academy of Sciences for the Grand Prix in mathematics. Joseph Fourier, then secretary of the academy, died without having read them, and after his death the articles were not found among his papers. Such a stream of disappointments would bring down anyone. Galois rebelled against those in power because he felt that they did not recognize his merits and killed his father. He plunged headlong into politics, becoming an ardent republican - not the wisest decision in France in 1830. In a last desperate attempt, he sent a scientific paper to the famous French physicist and mathematician Siméon Denis Poisson (1781–1840), who in reply demanded additional evidence.

This was the last straw. In 1831, Galois was arrested twice - first for allegedly calling for the assassination of King Louis Philippe, and then in order to protect him - the authorities feared a republican revolt! This time, he was sentenced to six months' imprisonment on trumped-up charges of illegally wearing the uniform of the disbanded artillery battalion he joined. Released on parole, he took up a business that disgusted him as much as everything else in life. In letters to devoted friend Chevalier feels his disappointment. On May 29, 1832, he accepted a challenge to a duel, the reasons for which are not fully understood. “I fell victim to a dishonorable coquette. My life is fading away in a miserable quarrel,” he writes in a Letter to All Republicans. Galois' most famous work was sketched the night before the fatal duel. Complaints are scattered in the margins: "I have no more time, I have no more time." He had to leave to others the details of the intermediate steps that were not essential to understanding the main idea. He needed to throw out on paper the basis of his discoveries - the origins of what is now called the Galois theorem. He ended his will by asking Chevalier to "request Jacobi and Gauss to give their opinion publicly, not on the correctness, but on the importance of these theorems." Early in the morning, Galois went to meet his rival. They had to shoot from a distance of 25 paces. Galois was wounded and died in the hospital the next morning. He was only twenty years old.

Galois relied on the work of Lagrange and Cauchy, but he developed a more general method. This was an extremely important achievement in the field of solving quintics. The scientist paid less attention to the original equations or graphical interpretation, and thought more about the nature of the roots themselves. To simplify, Galois considered only the so-called irreducible quintics, that is, those that could not be factorized in the form of polynomials of a lower order (as we said, for any polynomial equations up to the fourth order there are formulas for finding their roots). In general, an irreducible polynomial with rational coefficients is a polynomial that cannot be decomposed into simpler polynomials that have rational coefficients. For example, (x 5 - 1) can be factorized (x-1)(x 4 + x 3 + x 2 + x + 1), whereas (x 5 - 2) irreducible. Galois' goal was to determine the conditions under which all solutions of a general irreducible polynomial equation can be found in terms of radicals.

The key to the solution is that the roots of any irreducible algebraic equation are not independent, they can be expressed one in terms of the other. These relations were formalized into the group of all possible permutations, the so-called root symmetry group - for a quintic this group contains 5! = 5 x 4 x 3 x 2 x 1 = 120 elements. The mathematical algorithms of the Galois theory are very complex, and, most likely, partly because of this, they were initially understood with great difficulty. But after the level of abstraction made it possible to move from the algebraic solutions of equations to the algebraic structure of the groups associated with them, Galois was able to predict the solvability of the equation based on the properties of such groups. Moreover, his theory also provided a method by which these roots themselves could be found. With regard to quintics, the mathematician Joseph Liouville (1809–1882), who in 1846 published most of Galois's work in his Journal of Pure and Applied Mathematics, noted that the young scientist had proved a "beautiful theorem", and in order "to If an irreducible equation of the original degree is solvable in terms of radicals, it is necessary and sufficient that all its roots be rational functions of any two of them. Since this is impossible for a quintic, it cannot be solved with radicals.

In three years, the mathematical world has lost two of its brightest new stars. Mutual accusations and soul-searching followed, and Abel and Galois achieved well-deserved recognition, but only posthumously. In 1829, Carl Jacobi, through Legendre, learned of Abel's "lost" manuscript, and in 1830 a diplomatic scandal erupted when the Norwegian consul in Paris demanded that the article of his compatriot be found. In the end, Cauchy found the article, only to have it lost again in the academy editors! In the same year, Abel was awarded the Grand Prix in Mathematics (together with Jacobi) - but he was already dead. In 1841 his biography was published. In 1846, Liouville edited some of Galois's manuscripts for publication, and in his introduction expressed regret that the academy had initially rejected Galois's work because of its complexity - "indeed, clarity of presentation is necessary when the author leads the reader off the beaten path into uncharted wild territories." He continues: “Galois is no more! Let's not fall into useless criticism. Let's discard the flaws and look at the virtues! Fruit brief life Galois fit in only sixty pages. The editor of the mathematical journal for candidates for the École Normale and the Ecole Polytechnique commented on the Galois case as follows: “A candidate with a high intelligence was weeded out by an examiner with a lower level of thinking. Barbarus hic ego sum, quia non intelligor illis."

First of all, the second page of this work is not burdened with names, surnames, descriptions of the position in society, titles and elegies in honor of some miserly prince, whose purse will be opened with these incense - with the threat of closing it when the praises are over. You won't see here respectful praises, written in letters three times the size of the text itself, addressed to those who have a high position in science, to some wise patron - something mandatory (I would say inevitable) for someone at the age of twenty who wants to write something. I am not telling anyone here that I am indebted to their advice and support for all the good things in my work. I don't say this because it would be a lie. If I had to mention one of the greats in society or in science (at present the difference between these two classes of people is almost imperceptible), I swear it would not be a token of gratitude. I owe it to them that I published the first of these two papers so late, and that I wrote all this in prison - in a place that can hardly be considered suitable for scientific reflection, and I am often amazed at my restraint and ability to keep my mouth on castle in relation to stupid and vicious Zoils. It seems to me that I can use the word "Zoils" without fear of being accused of indecency, since that is how I refer to my opponents. I am not going to write here about how and why I was sent to prison, but I must say that my manuscripts were most often simply lost in the folders of gentlemen of the academy, although, to tell the truth, I cannot imagine such indiscretion on the part of people on whose conscience the death of Abel. In my opinion, anyone would like to be compared to this brilliant mathematician. Suffice it to say that my article on the theory of equations was sent to the Academy of Sciences in February 1830, that the extracts from it were sent in February 1829, and yet none of it was printed, and even the manuscript turned out to be impossible to return.

Galois, unpublished preface, 1832

Class: 9

Basic goals:

To consolidate the concept of an integer rational equation of the th degree.
Formulate the main methods for solving equations of higher degrees (n > 3).
To teach the basic methods for solving equations of higher degrees.
To teach by the form of the equation to determine the most effective method his decisions.

Forms, methods and pedagogical techniques that are used by the teacher in the classroom:

Lecture-seminar training system (lectures - explanation of new material, seminars - problem solving).
Information and communication technologies (frontal survey, oral work with the class).
Differentiated training, group and individual forms.
The use of the research method in training aimed at development mathematical apparatus and mental abilities of each individual student.
Printed material - an individual summary of the lesson (basic concepts, formulas, statements, lecture material is compressed in the form of diagrams or tables).

Lesson plan:

Organizing time.
The purpose of the stage: to include students in learning activities define the content of the lesson.
Updating students' knowledge.
The purpose of the stage: to update the knowledge of students on previously studied related topics
Learning a new topic (lecture). The purpose of the stage: to formulate the main methods for solving equations of higher degrees (n > 3)
Summarizing.
The purpose of the stage: to once again highlight the key points in the material studied in the lesson.
Homework.
The purpose of the stage: to formulate homework for students.

Lesson summary

1. Organizational moment.

The wording of the topic of the lesson: “Equations of higher degrees. Methods for their solution”.

2. Actualization of students' knowledge.

Theoretical survey - conversation. Repetition of some previously studied information from the theory. Students formulate basic definitions and give statements of necessary theorems. Examples are given, demonstrating the level of previously acquired knowledge.

The concept of an equation with one variable.
The concept of the root of the equation, the solution of the equation.
concept linear equation with one variable, the concept of a quadratic equation with one variable.
The concept of equivalence of equations, equation-consequences (the concept of extraneous roots), transition not by consequence (the case of loss of roots).
The concept of a whole rational expression with one variable.
The concept of an entire rational equation n th degree. The standard form of an entire rational equation. Reduced whole rational equation.
Transition to a set of equations of lower degrees by factoring the original equation.
The concept of a polynomial n th degree from x. Bezout's theorem. Consequences from Bezout's theorem. Root theorems ( Z-roots and Q-roots) of an entire rational equation with integer coefficients (reduced and nonreduced, respectively).
Horner's scheme.

3. Learning a new topic.

We will consider the whole rational equation n th power of the standard form with one unknown variable x:Pn(x)= 0 , where P n (x) = a n x n + a n-1 x n-1 + a 1 x + a 0– polynomial n th degree from x, a n ≠ 0 . If a n = 1 then such an equation is called a reduced whole rational equation n th degree. Let us consider such equations for different values n and list the main methods of their solution.

n= 1 is a linear equation.

n= 2 is a quadratic equation. Discriminant formula. Formula for calculating roots. Vieta's theorem. Selection of a full square.

n = 3 – cubic equation.

grouping method.

Example: x 3 – 4x 2 – x+ 4 = 0 (x - 4) (x 2– 1) = 0 x 1 = 4 , x2 = 1,x 3 = -1.

Reciprocal cubic equation of the form ax 3 + bx 2 + bx + a= 0. We solve by combining terms with the same coefficients.

Example: x 3 – 5x 2 – 5x + 1 = 0 (x + 1)(x 2 – 6x + 1) = 0 x 1 = -1, x 2 = 3 + 2, x 3 = 3 – 2.

Selection of Z-roots based on the theorem. Horner's scheme. When applying this method, it is necessary to emphasize that the enumeration in this case is finite, and we select the roots according to a certain algorithm in accordance with the theorem on Z-roots of the reduced whole rational equation with integer coefficients.

Example: x 3 – 9x 2 + 23x– 15 = 0. The equation is reduced. We write out the divisors of the free term ( + 1; + 3; + 5; + 15). Let's apply Horner's scheme:

	x 3	x 2	x 1	x 0	output
	1	-9	23	-15
1	1	1 x 1 - 9 = -8	1 x (-8) + 23 = 15	1 x 15 - 15 = 0	1 - root
	x 2	x 1	x 0

We get ( x – 1)(x 2 – 8x + 15) = 0 x 1 = 1, x 2 = 3, x 3 = 5.

Equation with integer coefficients. Selection of Q-roots based on the theorem. Horner's scheme. When applying this method, it is necessary to emphasize that the enumeration in this case is finite and we select the roots according to a certain algorithm in accordance with the theorem on Q-roots of an unreduced whole rational equation with integer coefficients.

Example: 9 x 3 + 27x 2 – x– 3 = 0. The equation is not reduced. We write out the divisors of the free term ( + 1; + 3). Let us write out the divisors of the coefficient at the highest power of the unknown. ( + 1; + 3; + 9) Therefore, we will look for roots among the values ( + 1; + ; + ; + 3). Let's apply Horner's scheme:

	x 3	x 2	x 1	x 0	output
	9	27	-1	-3
1	9	1 x 9 + 27 = 36	1 x 36 - 1 = 35	1 x 35 - 3 = 32 ≠ 0	1 is not a root
-1	9	-1 x 9 + 27 = 18	-1 x 18 - 1 = -19	-1 x (-19) - 3 = 16 ≠ 0	-1 is not a root
	9	x9 + 27 = 30	x 30 - 1 = 9	x 9 - 3 = 0	root
	x 2	x 1	x 0

We get ( x – )(9x 2 + 30x + 9) = 0 x 1 = , x 2 = - , x 3 = -3.

For the convenience of calculation when choosing Q -roots it can be convenient to make a change of variable, go to the above equation and adjust Z -roots.

If the intercept is 1

.

If it is possible to use the substitution of the form y=kx

.

Formula Cardano. There is a universal method for solving cubic equations - this is the Cardano formula. This formula is associated with the names of the Italian mathematicians Gerolamo Cardano (1501–1576), Nicolo Tartaglia (1500–1557), Scipio del Ferro (1465–1526). This formula lies outside the scope of our course.

n= 4 is an equation of the fourth degree.

grouping method.

Example: x 4 + 2x 3 + 5x 2 + 4x – 12 = 0 (x 4 + 2x 3) + (5x 2 + 10x) – (6x + 12) = 0 (x + 2)(x 3 + 5x- 6) = 0 (x + 2)(x– 1)(x 2 + x + 6) = 0 x 1 = -2, x 2 = 1.

Variable replacement method.

Biquadratic equation of the form ax 4 + bx 2+s = 0 .

Example: x 4 + 5x 2 - 36 = 0. Substitution y = x 2. From here y 1 = 4, y 2 = -9. That's why x 1,2 = + 2 .

Reciprocal equation of the fourth degree of the form ax 4 + bx 3+c x 2 + bx + a = 0.

We solve by combining terms with the same coefficients by replacing the form

ax 4 + bx 3 + cx 2 – bx + a = 0.

Generalized backward equation of the fourth degree of the form ax 4 + bx 3 + cx 2 + kbx + k2 a = 0.

General replacement. Some standard substitutions.

Example 3 . General view replacement(follows from the form of a particular equation).

n = 3.

Equation with integer coefficients. Selection of Q-roots n = 3.

General formula. There is a universal method for solving equations of the fourth degree. This formula is associated with the name of Ludovico Ferrari (1522-1565). This formula lies outside the scope of our course.

n > 5 - equations of the fifth and higher degrees.

Equation with integer coefficients. Selection of Z-roots based on the theorem. Horner's scheme. The algorithm is similar to the one discussed above for n = 3.

Equation with integer coefficients. Selection of Q-roots based on the theorem. Horner's scheme. The algorithm is similar to the one discussed above for n = 3.

Symmetric equations. Any reciprocal equation of odd degree has a root x= -1 and after decomposing it into factors, we get that one factor has the form ( x+ 1), and the second factor is a reciprocal equation of even degree (its degree is one less than the degree of the original equation). Any reciprocal equation of even degree together with a root of the form x = φ also contains the root of the form . Using these statements, we solve the problem by lowering the degree of the equation under study.

Variable replacement method. Use of homogeneity.

There is no general formula for solving entire fifth-degree equations (this was shown by the Italian mathematician Paolo Ruffini (1765–1822) and the Norwegian mathematician Nils Henrik Abel (1802–1829)) and higher powers (this was shown by the French mathematician Evariste Galois (1811–1832) )).

Recall again that in practice it is possible to use combinations the methods listed above. It is convenient to pass to a set of equations of lower degrees by factorization of the original equation.
Outside the scope of our today's discussion, there are widely used in practice graphic methods solving equations and approximate solution methods equations of higher degrees.

There are situations when the equation does not have R-roots.

Using the monotonicity property of functions

4. Summing up.

Summary: Now we have mastered the basic methods for solving various equations of higher degrees (for n > 3). Our task is to learn how to effectively use the above algorithms. Depending on the type of equation, we will have to learn how to determine which solution method is the most effective in this case, as well as correctly apply the chosen method.

5. Homework.

: item 7, pp. 164–174, nos. 33–36, 39–44, 46,47.

: №№ 9.1–9.4, 9.6–9.8, 9.12, 9.14–9.16, 9.24–9.27.

Possible topics of reports or abstracts on this topic:

Formula Cardano
Graphical method for solving equations. Solution examples.
Methods for approximate solution of equations.

Analysis of the assimilation of the material and students' interest in the topic:

Experience shows that the interest of students in the first place is the possibility of selecting Z-roots and Q-roots of equations using a fairly simple algorithm using Horner's scheme. Students are also interested in various standard types of variable substitution, which can significantly simplify the type of problem. Graphical methods of solution are usually of particular interest. In this case, you can additionally parse the tasks into a graphical method for solving equations; discuss general form graphics for a polynomial of 3, 4, 5 degrees; analyze how the number of roots of equations of 3, 4, 5 degrees is related to the type of the corresponding graph. Below is a list of books where you can find additional information on this topic.

Bibliography:

Vilenkin N.Ya. etc. “Algebra. A textbook for students in grades 9 with an in-depth study of mathematics ”- M., Education, 2007 - 367 p.
Vilenkin N.Ya., Shibasov L.P., Shibasova Z.F.“Behind the pages of a mathematics textbook. Arithmetic. Algebra. Grades 10-11” – M., Enlightenment, 2008 – 192 p.
Vygodsky M.Ya."Handbook of mathematics" - M., AST, 2010 - 1055 p.
Galitsky M.L.“Collection of problems in algebra. Tutorial for grades 8-9 with in-depth study of mathematics ”- M., Education, 2008 - 301 p.
Zvavich L.I. et al. “Algebra and the Beginnings of Analysis. 8–11 cells A manual for schools and classes with in-depth study of mathematics ”- M., Drofa, 1999 - 352 p.
Zvavich L.I., Averyanov D.I., Pigarev B.P., Trushanina T.N.“Assignments in mathematics to prepare for a written exam in grade 9” - M., Education, 2007 - 112 p.
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Ivanov A.P.“Tests and test papers mathematics. Tutorial". - M., Fizmatkniga, 2008 - 304 p.
Leibson K.L.“Collection of practical tasks in mathematics. Part 2–9 class” – M., MTsNMO, 2009 – 184 p.
Makarychev Yu.N., Mindyuk N.G."Algebra. Additional chapters for the 9th grade school textbook. Textbook for students of schools and classes with in-depth study of mathematics.” - M., Education, 2006 - 224 p.
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Savin A.P. “encyclopedic Dictionary young mathematician” – M., Pedagogy, 1985 – 352 p.
Survillo G.S., Simonov A.S. “Didactic materials in Algebra for Grade 9 with in-depth study of mathematics” – M., Enlightenment, 2006 – 95 p.
Chulkov P.V.“Equations and inequalities in the school course of mathematics. Lectures 1–4” – M., First of September, 2006 – 88 p.
Chulkov P.V.“Equations and inequalities in the school course of mathematics. Lectures 5–8” – M., First of September, 2009 – 84 p.