Class: 9

Basic goals:

1. Reinforce the concept of an entire rational equation of the th degree.
2. Formulate the basic methods for solving equations of higher degrees (n > 3).
3. Teach basic methods for solving higher-order equations.
4. Learn to use the type of equation to determine the most effective way to solve it.

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

• Lecture-seminar teaching system (lectures - explanation of new material, seminars - problem solving).
• Information and communication technologies (frontal survey, oral work with the class).
• Differentiated learning, group and individual forms.
• Using a research method in teaching aimed at developing the mathematical apparatus and thinking abilities of each individual student.
• Printed material – an individual brief summary of the lesson (basic concepts, formulas, statements, lecture material condensed in the form of diagrams or tables).

Lesson plan:

1. Organizing time.
   The purpose of the stage: to include students in educational activities, to determine the content of the lesson.
2. Updating students' knowledge.
   The purpose of the stage: to update students’ knowledge on previously studied related topics
3. Studying a new topic (lecture). Goal of the stage: to formulate the basic methods for solving equations of higher degrees (n > 3)
4. Summarizing.
   The purpose of the stage: to once again highlight the key points in the material studied in the lesson.
5. Homework.
   The purpose of the stage: to formulate homework for students.

Lesson summary

1. Organizational moment.

Formulation of the lesson topic: “Equations of higher powers. Methods for solving them.”

2. Updating students' knowledge.

Theoretical survey - conversation. Repetition of some previously studied information from the theory. Students formulate basic definitions and formulate the necessary theorems. Give examples to demonstrate the level of previously acquired knowledge.

• The concept of an equation with one variable.
• The concept of the root of an equation, the solution of an equation.
• The concept of a linear equation with one variable, the concept of a quadratic equation with one variable.
• The concept of equivalence of equations, equations-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 a whole rational equation n th degree. Standard form of a whole 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. Corollaries from Bezout's theorem. Root theorems ( Z-roots and Q-roots) of an entire rational equation with integer coefficients (reduced and unreduced, respectively).
• Horner's scheme.

3. Studying a new topic.

We will consider the whole rational equation n-th power of 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 integer rational equation n th degree. Let us consider such equations for various values n and list the main methods for solving them.

n= 1 – linear equation.

n= 2 – quadratic equation. Discriminant formula. Formula for calculating roots. Vieta's theorem. Selecting a complete square.

n= 3 – cubic equation.

Grouping method.

Example: x 3 – 4x 2 – x+ 4 = 0 (x – 4)(x 2– 1) = 0 x 1 = 4 , x 2 = 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 search in this case is finite, and we select the roots using a certain algorithm in accordance with the theorem about Z-roots of the given whole rational equation with integer coefficients.

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

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 search in this case is finite and we select the roots using a certain algorithm in accordance with the theorem about Q-roots of an unreduced integer rational equation with integer coefficients.

Example: 9 x 3 + 27x 2 – x– 3 = 0. The equation is unreduced. Let us write down the divisors of the free term ( + 1; + 3). Let us write down the divisors of the coefficient at the highest power of the unknown. ( + 1; + 3; + 9) Consequently, we will look for roots among the values ​​( + 1; + ; + ; + 3). Let's apply Horner's scheme:

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

For ease of calculation when selecting Q -roots It can be convenient to make a change of variable, go to the given equation and select Z -roots.

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

n= 4 – 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. Replacement y = x 2. From here y 1 = 4, y 2 = -9. That's why x 1,2 = + 2 .

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

• Generalized recurrent equation of the fourth degree of the form ax 4 + bx 3 + cx 2 + kbx + k 2 a = 0.

Example 3 . General view replacement(follows from the type of specific equation).

n = 3.

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

General formula. There is a universal method for solving fourth degree equations. This formula is associated with the name of Ludovico Ferrari (1522–1565). This formula is beyond 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 that 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 that discussed above for n = 3.

Symmetric equations. Any reciprocal equation of odd degree has a root x= -1 and after factoring it into factors we find 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 species. 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 equations of the fifth degree (this was shown by the Italian mathematician Paolo Ruffini (1765–1822) and the Norwegian mathematician Niels Henrik Abel (1802–1829)) and higher degrees (this was shown by the French mathematician Evariste Galois (1811–1832) )).

• Let us recall once 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 factoring the original equation.
• Outside the scope of our discussion today are those widely used in practice. graphical methods solving equations and approximate solution methods equations of higher degrees.
• There are situations when the equation does not have R-roots.
Then the solution comes down to showing that the equation has no roots. To prove this, we analyze the behavior of the functions under consideration on monotonicity intervals. Example: equation x 8 – x 3 + 1 = 0 has no roots.
• Using the property of monotonicity of functions
. There are situations when using various properties of functions allows you to simplify the task.
Example 1: Equation x 5 + 3x– 4 = 0 has one root x= 1. Due to the property of monotonicity of the analyzed functions, there are no other roots.
Example 2: Equation x 4 + (x– 1) 4 = 97 has roots x 1 = -2 and x 2 = 3. Having analyzed the behavior of the corresponding functions on intervals of monotonicity, we conclude that there are no other roots.

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 algorithms listed above. Depending on the type of equation, we will have to learn to determine which method of solution in a given case is the most effective, as well as correctly apply the chosen method.

5. Homework.

: paragraph 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 for reports or abstracts on this topic:

• Cardano formula
• Graphical method for solving equations. Examples of solutions.
• Methods for approximate solution of equations.

Analysis of student learning and interest in the topic:

Experience shows that students’ interest is primarily aroused by 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 substitution of variables, which can significantly simplify the type of problem. Graphical solution methods are usually of particular interest. In this case, you can additionally analyze problems using a graphical method for solving equations; discuss the general form of the graph for a polynomial of degree 3, 4, 5; analyze how the number of roots of equations of degree 3, 4, 5 is related to the appearance of the corresponding graph. Below is a list of books where you can find additional information on this topic.

Bibliography:

1. Vilenkin N.Ya. and others. “Algebra. Textbook for 9th grade students with in-depth study of mathematics” - M., Prosveshchenie, 2007 - 367 p.
2. Vilenkin N.Ya., Shibasov L.P., Shibasova Z.F.“Behind the pages of a mathematics textbook. Arithmetic. Algebra. 10-11 grades” – M., Education, 2008 – 192 p.
3. Vygodsky M.Ya.“Handbook of Mathematics” – M., AST, 2010 – 1055 p.
4. Galitsky M.L.“Collection of problems in algebra. Textbook for grades 8-9 with in-depth study of mathematics” - M., Prosveshchenie, 2008 - 301 p.
5. Zvavich L.I. and others. “Algebra and the beginnings of analysis. 8–11 grades A manual for schools and classes with in-depth study of mathematics” - M., Drofa, 1999 - 352 p.
6. Zvavich L.I., Averyanov D.I., Pigarev B.P., Trushanina T.N.“Mathematics assignments for preparing for the written exam in 9th grade” - M., Prosveshchenie, 2007 - 112 p.
7. Ivanov A.A., Ivanov A.P.“Thematic tests for systematizing knowledge in mathematics” part 1 – M., Fizmatkniga, 2006 – 176 p.
8. Ivanov A.A., Ivanov A.P.“Thematic tests for systematizing knowledge in mathematics” part 2 – M., Fizmatkniga, 2006 – 176 p.
9. Ivanov A.P.“Tests and tests in mathematics. Tutorial". – M., Fizmatkniga, 2008 – 304 p.
10. Leibson K.L.“Collection of practical tasks in mathematics. Part 2–9 grades” – M., MTSNM, 2009 – 184 p.
11. Makarychev Yu.N., Mindyuk N.G."Algebra. Additional chapters for the 9th grade school textbook. A textbook for students in schools and classes with in-depth study of mathematics.” – M., Education, 2006 – 224 p.
12. Mordkovich A.G."Algebra. In-depth study. 8th grade. Textbook” – M., Mnemosyne, 2006 – 296 p.
13. Savin A.P.“Encyclopedic Dictionary of a Young Mathematician” - M., Pedagogy, 1985 - 352 p.
14. Survillo G.S., Simonov A.S.“Didactic materials on algebra for grade 9 with in-depth study of mathematics” - M., Prosveshchenie, 2006 - 95 p.
15. Chulkov P.V.“Equations and inequalities in the school mathematics course. Lectures 1–4” – M., September 1st, 2006 – 88 p.
16. Chulkov P.V.“Equations and inequalities in the school mathematics course. Lectures 5–8” – M., September 1, 2009 – 84 p.

Let's consider solving equations with one variable of degree higher than the second.

The degree of the equation P(x) = 0 is the degree of the polynomial P(x), i.e. the greatest of the powers of its terms with a coefficient not equal to zero.

So, for example, the equation (x 3 – 1) 2 + x 5 = x 6 – 2 has the fifth degree, because after the operations of opening the brackets and bringing similar ones, we obtain the equivalent equation x 5 – 2x 3 + 3 = 0 of the fifth degree.

Let us recall the rules that will be needed to solve equations of degree higher than two.

Statements about the roots of a polynomial and its divisors:

1. A polynomial of nth degree has a number of roots not exceeding n, and roots of multiplicity m occur exactly m times.

2. A polynomial of odd degree has at least one real root.

3. If α is the root of P(x), then P n (x) = (x – α) · Q n – 1 (x), where Q n – 1 (x) is a polynomial of degree (n – 1).

4.

5. The reduced polynomial with integer coefficients cannot have fractional rational roots.

6. For a third degree polynomial

P 3 (x) = ax 3 + bx 2 + cx + d one of two things is possible: either it is decomposed into the product of three binomials

Р 3 (x) = а(х – α)(х – β)(х – γ), or decomposes into the product of a binomial and a square trinomial Р 3 (x) = а(х – α)(х 2 + βх + γ ).

7. Any polynomial of the fourth degree can be expanded into the product of two square trinomials.

8. A polynomial f(x) is divisible by a polynomial g(x) without a remainder if there is a polynomial q(x) such that f(x) = g(x) · q(x). To divide polynomials, the “corner division” rule is used.

9. For the polynomial P(x) to be divisible by a binomial (x – c), it is necessary and sufficient that the number c be the root of P(x) (Corollary of Bezout’s theorem).

10. Vieta's theorem: If x 1, x 2, ..., x n are real roots of the polynomial

P(x) = a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + … + x n = -a 1 /a 0,

x 1 x 2 + x 1 x 3 + … + x n – 1 x n = a 2 /a 0,

x 1 x 2 x 3 + … + x n – 2 x n – 1 x n = -a 3 / a 0,

x 1 · x 2 · x 3 · x n = (-1) n a n / a 0 .

Solving Examples

Example 1.

Find the remainder of division P(x) = x 3 + 2/3 x 2 – 1/9 by (x – 1/3).

Solution.

By corollary to Bezout’s theorem: “The remainder of a polynomial divided by a binomial (x – c) is equal to the value of the polynomial of c.” Let's find P(1/3) = 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R = 0.

Example 2.

Divide with a “corner” 2x 3 + 3x 2 – 2x + 3 by (x + 2). Find the remainder and incomplete quotient.

Solution:

2x 3 + 3x 2 – 2x + 3| x + 2

2x 3 + 4 x 2 2x 2 – x

X 2 – 2 x

Answer: R = 3; quotient: 2x 2 – x.

Basic methods for solving higher degree equations

1. Introduction of a new variable

The method of introducing a new variable is already familiar from the example of biquadratic equations. It consists in the fact that to solve the equation f(x) = 0, a new variable (substitution) t = x n or t = g(x) is introduced and f(x) is expressed through t, obtaining a new equation r(t). Then solving the equation r(t), the roots are found:

(t 1, t 2, …, t n). After this, a set of n equations q(x) = t 1 , q(x) = t 2 , … , q(x) = t n is obtained, from which the roots of the original equation are found.

Example 1.

(x 2 + x + 1) 2 – 3x 2 – 3x – 1 = 0.

Solution:

(x 2 + x + 1) 2 – 3(x 2 + x) – 1 = 0.

(x 2 + x + 1) 2 – 3(x 2 + x + 1) + 3 – 1 = 0.

Substitution (x 2 + x + 1) = t.

t 2 – 3t + 2 = 0.

t 1 = 2, t 2 = 1. Reverse substitution:

x 2 + x + 1 = 2 or x 2 + x + 1 = 1;

x 2 + x - 1 = 0 or x 2 + x = 0;

Answer: From the first equation: x 1, 2 = (-1 ± √5)/2, from the second: 0 and -1.

2. Factorization by grouping and abbreviated multiplication formulas

The basis of this method is also not new and consists in grouping terms in such a way that each group contains a common factor. To do this, sometimes it is necessary to use some artificial techniques.

Example 1.

x 4 – 3x 2 + 4x – 3 = 0.

Solution.

Let's imagine - 3x 2 = -2x 2 – x 2 and group:

(x 4 – 2x 2) – (x 2 – 4x + 3) = 0.

(x 4 – 2x 2 +1 – 1) – (x 2 – 4x + 3 + 1 – 1) = 0.

(x 2 – 1) 2 – 1 – (x – 2) 2 + 1 = 0.

(x 2 – 1) 2 – (x – 2) 2 = 0.

(x 2 – 1 – x + 2)(x 2 – 1 + x - 2) = 0.

(x 2 – x + 1)(x 2 + x – 3) = 0.

x 2 – x + 1 = 0 or x 2 + x – 3 = 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 = (-1 ± √13)/2.

3. Factorization by the method of undetermined coefficients

The essence of the method is that the original polynomial is factorized with unknown coefficients. Using the property that polynomials are equal if their coefficients are equal at the same powers, the unknown expansion coefficients are found.

Example 1.

x 3 + 4x 2 + 5x + 2 = 0.

Solution.

A polynomial of degree 3 can be expanded into the product of linear and quadratic factors.

x 3 + 4x 2 + 5x + 2 = (x – a)(x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 = x 3 +bx 2 + cx – ax 2 – abx – ac,

x 3 + 4x 2 + 5x + 2 = x 3 + (b – a)x 2 + (cx – ab)x – ac.

Having solved the system:

(b – a = 4,
(c – ab = 5,
(-ac = 2,

(a = -1,
(b = 3,
(c = 2, i.e.

x 3 + 4x 2 + 5x + 2 = (x + 1)(x 2 + 3x + 2).

The roots of the equation (x + 1)(x 2 + 3x + 2) = 0 are easy to find.

Answer: -1; -2.

4. Method of selecting a root using the highest and free coefficient

The method is based on the application of theorems:

1) Every integer root of a polynomial with integer coefficients is a divisor of the free term.

2) In order for the irreducible fraction p/q (p is an integer, q is a natural number) to be the root of an equation with integer coefficients, it is necessary that the number p be an integer divisor of the free term a 0, and q be a natural divisor of the leading coefficient.

Example 1.

6x 3 + 7x 2 – 9x + 2 = 0.

Solution:

6: q = 1, 2, 3, 6.

Therefore, p/q = ±1, ±2, ±1/2, ±1/3, ±2/3, ±1/6.

Having found one root, for example – 2, we will find other roots using corner division, the method of indefinite coefficients or Horner’s scheme.

Answer: -2; 1/2; 1/3.

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In the 16th century, mathematicians stumbled upon complex numbers almost by accident (see Chapter 11). By the 18th century, complex numbers were considered an extension of the field of real numbers, but working with them still led to parity errors, as 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, producing several additional proofs over the following years. In 1849, he reworked the first version, this time using complex numbers. In 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 long-standing question of whether solving high-order polynomial equations requires generating numbers of a higher order than complex ones.

One of the most thorny problems in algebra of that time was the question of whether the fifth-order polynomial, the quintic, could be solved by algebraic methods, that is, using a finite number of algebraic steps. Nowadays in school they teach 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 not a single method was found for quintics. The fundamental theorem of algebra may seem to hold the prospect 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 then). 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 kind 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 a scientific article in which he stated that the quintic is not solvable by algebraic means, as, indeed, is any polynomial of a higher order. Abel believed that this article would serve as his ticket 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 did not read the article. In 1826, the Norwegian government finally provided funds for Abel to travel around Europe. Fearing that personal communication with Gauss would not bring him much joy, the mathematician decided not to visit Göttingen and instead went to Berlin. There he became friends with August Leopold Krelle (1780–1855), a mathematician, architect and engineer who advised the Prussian Ministry of Education on matters of mathematics. Krell intended to found the Journal of Pure and Applied Mathematics. So Abel got the opportunity to disseminate 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 undecidable by algebraic methods. And then he left for Paris. This trip greatly upset Abel, because he practically did not receive the support he needed from French mathematicians. He became close to Augustin Louis Cauchy (1789–1857), who was at that time the leading luminary of mathematical analysis, but had a very complex character. As Abel himself put it, "Cauchy is mad and nothing can be done about it, although at present he is the only one who is capable of anything in mathematics." If we try to justify the manifestations of disrespect and neglect emanating from Gauss and Cauchy, we can say that the quintic achieved a certain fame and attracted the attention of both respected mathematicians and originalists. Abel returned to Norway, where he suffered increasingly from tuberculosis. He continued to send his work to Crelle, but died in 1829, unaware of how much his reputation had become established 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 the fourth order cannot be solved using radicals such as square roots, cube roots, or higher order ones. 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 unforgiving towards those whom he considered less talented than himself, and at the same time he hated 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 devoured the rest of the works of Legendre and, later, Abel. His enthusiasm, self-confidence and intolerance led to truly terrible consequences in his relationships with teachers and examiners. Galois took part in a competition to enter the École Polytechnique, the cradle of French mathematics, but failed the exam due to lack of preparation. For 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's second failure to enter the École Polytechnique has become part of mathematical folklore. He was so used to constantly holding complex mathematical ideas in his head that he was infuriated by the petty nagging of the examiners. Since the examiners had difficulty understanding his explanations, he threw a dry erase cloth from the board at one of them in the face. Soon after this, his father died, committing suicide as a result of church intrigue. A riot practically broke out at his funeral. In February 1830, Galois wrote the following three papers, sending them to the Academy of Sciences for the Grand Prix in Mathematics. Joseph Fourier, then secretary of the academy, died without reading them, and after his death the articles were not found among his papers. Such a torrent of disappointment would have overwhelmed anyone. Galois rebelled against those in power because he felt that they did not recognize his merits and destroyed his father. He plunged headlong into politics, becoming an ardent republican - not the wisest decision in France in 1830. In a last-ditch attempt, he sent a scientific paper to the famous French physicist and mathematician Simeon Denis Poisson (1781–1840), who responded by demanding further 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 rebellion! This time he was sentenced to six months in prison on trumped-up charges of illegally wearing the uniform of the disbanded artillery battalion to which he had joined. Released on parole, he took up a task that disgusted him as much as everything else in life. In his letters to his devoted friend Chevalier, his disappointment is felt. On May 29, 1832, he accepted a challenge to a duel, the reasons for which are not fully understood. “I fell victim to a dishonest coquette. My life is extinguished in a miserable quarrel,” he writes in “Letter to All Republicans.” Galois's most famous work was sketched the night before the fatal duel. Scattered in the margins are complaints: “I don’t have time anymore, I don’t have time anymore.” He was forced to leave to others the detailed exposition of intermediate steps that were not essential for understanding the main idea. He needed to put on paper the basis of his discoveries - the origins of what is now called Galois's theorem. He ended his will by asking Chevalier to "appeal to Jacobi and Gauss to give their public opinion, not as to the correctness, but as to the importance of these theorems." Early in the morning, Galois went to meet his rival. They had to shoot from a distance of 25 steps. Galois was wounded and died in hospital the next morning. He was only twenty years old.

Galois built 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 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 having 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's 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 through the other. These relations were formalized into a 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 Galois theory are very complex, and, most likely, partly as a result of this, they were initially difficult to understand. But once the level of abstraction allowed him to move from algebraic solutions of equations to the algebraic structure of their associated groups, Galois was able to predict the solvability of an equation based on the properties of such groups. Moreover, his theory also provided a method by which these roots themselves could be found. As for the 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 using 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 his compatriot's article be found. Cauchy eventually found the article, only to have it lost again by the academy's editors! That same year, Abel was awarded the Grand Prix in Mathematics (shared 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 the introduction expressed regret that the academy had initially rejected Galois's work because of its complexity - "clarity of presentation is indeed necessary when the author leads the reader off the beaten path into uncharted wild territories." He continues: “Galois is no more! Let us not fall into useless criticism. Let's put aside the shortcomings and look at the advantages! " The fruits of Galois' short life fit into just sixty pages. The editor of a mathematical journal for candidates to the École Normale and the École Polytechnique commented on the Galois case as follows: “A candidate of high intelligence was eliminated 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 social status, titles and elegies in honor of some stingy prince, whose wallet will be opened with the help of these incense - with the threat of closing it when the praise ends. You will not see here reverent eulogies, written in letters three times larger than the text itself, addressed to those of high position in science, to some wise patron - something obligatory (I would say inevitable) for someone at the age of twenty who wants to write something. I am not telling anyone here that I owe their advice and support for all the good that comes out of my work. I don't say this because it would be a lie. If I were to mention any of the greats in society or in science (the distinction between these two classes of people is almost imperceptible at the present time), I swear it would not be a sign of gratitude. I owe it to them that I published the first of these two articles so late, and that I wrote all this in prison - 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 shut. castle in relation to stupid and evil zoiles. I think I can use the word "zoiles" without fear of being accused of impropriety, since that is what I call 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 more often than not simply got lost in the files of gentlemen members of the academy, although, in truth, I cannot imagine such indiscretion on the part of people who are responsible for Abel's death. In my opinion, anyone would like to be compared with 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 extracts from it were sent in February 1829, but none of this was printed, and even the manuscript turned out to be impossible to return.

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First, let's remember the basic formulas of powers and their properties.

Product of a number a occurs 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 = 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 indicator.

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

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

Let's take a simple equation:

2 x = 2 3

This example can be solved even in your head. 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 to formalize this decision:

2 x = 2 3
x = 3

In order to solve such an equation, we removed identical grounds(that is, twos) and wrote down what was left, these are degrees. We got the answer we were looking for.

Now let's summarize our decision.

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

Now let's look at a few examples:

Let's start with something 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 powers.

x+2=4 The simplest equation is obtained.
x=4 – 2
x=2
Answer: x=2

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

3 3x - 9 x+8 = 0

First, move 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 = (3 2) x+8

We get 9 x+8 =(3 2) x+8 =3 2x+16

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

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

Let's look at the following example:

2 2x+4 - 10 4 x = 2 4

First of all, we look at the bases, bases two and four. And we need them to be the same. We transform the four using 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 bother us. What to do with them? If you look closely you can see that on the left side we have 2 2x repeated, 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 entire equation by 6:

Let's imagine 4=2 2:

2 2x = 2 2 bases are the same, we discard them and equate the degrees.
2x = 2 is the simplest equation. Divide it by 2 and 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

Our bases are the same, equal to three. In this example, you can see that the first three has a degree twice (2x) than the second (just x). In this case, you can solve replacement method. We replace the number with the smallest degree:

Then 3 2x = (3 x) 2 = t 2

We replace all x powers in the equation with t:

t 2 - 12t+27 = 0
We get a quadratic equation. Solving through the discriminant, we get:
D=144-108=36
t 1 = 9
t2 = 3

Returning to the variable x.

Take t 1:
t 1 = 9 = 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 = 3 = 3 x
3 x = 3 1
x 2 = 1
Answer: x 1 = 2; x 2 = 1.

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