Solving equations of higher degrees by various methods

Basic goals:

To fix the concept of the whole rational equation of degree.
Formulate the main methods for solving equations of higher degrees (n > 3).
To teach the basic methods of solving equations of higher degrees.
To teach the type of equation to determine the most effective way to solve it.

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

Lecture and 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.
Using the research method in teaching aimed at developing the mathematical apparatus and mental abilities of each individual student.
Printed material is an individual brief 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 educational activities, to determine 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 basic methods for solving equations of higher degrees (n > 3)
Summarizing.
The purpose of the stage: once again highlight 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 theme of the lesson: “Equations of higher degrees. Methods for solving them. ”

2. Updating students' knowledge.

Theoretical survey - conversation. Repetition of some previously studied information from theory. Students formulate basic definitions and formulate 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.
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 of consequence (the concept of extraneous roots), the 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 ndegree. The standard form of the whole rational equation. The given whole rational equation.
Transition to a set of equations of lower degrees by factoring the original equation.
Polynomial concept ndegree from x. Bezout theorem. Corollaries from Bezout's theorem. Root Theorems ( Z-roots and Q-roots) of the whole rational equation with integer coefficients (respectively reduced and non-reduced).
Horner's scheme.

3. Learning a new topic.

We will consider the whole rational equation ndegree of the standard form with one unknown variable x: P n (x) \u003d 0, where P n (x) \u003d a n x n + a n-1 x n-1 + a 1 x + a 0 - polynomial ndegree from x, a n ≠ 0. If a a n \u003d 1 then such an equation is called a reduced integer rational equation ndegree. We consider such equations for various values n and list the main methods for solving them.

n \u003d 1 is a linear equation.

n \u003d 2 is the quadratic equation. The discriminant formula. Formula for calculating the roots. Vieta theorem. Full square selection.

n \u003d 3 is the cubic equation.

Grouping Method.

Example: x 3 - 4x 2 - x+ 4 \u003d 0 (x - 4) (x 2– 1) = 0 x 1 = 4 , x 2 = 1, x 3 = -1.

Return cubic equation of the form ax 3 + bx 2 + bx + a \u003d 0. We solve by combining the 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 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 \u003d 0. The equation is given. We write out the divisors of the free term ( + 1; + 3; + 5; + 15). We apply the Horner scheme:

	x 3	x 2	x 1	x 0	output
	1	-9	23	-15
1	1	1 x 1 - 9 \u003d -8	1 x (-8) + 23 \u003d 15	1 x 15 - 15 \u003d 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 search 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 \u003d 0. The equation is not reduced. We write out the divisors of the free term ( + 1; + 3). We write out the divisors of the coefficient with the highest degree of the unknown. ( + 1; + 3; + 9) Therefore, we will search for the roots among the values \u200b\u200b( + 1; + ; + ; + 3). We apply the Horner scheme:

	x 3	x 2	x 1	x 0	output
	9	27	-1	-3
1	9	1 x 9 + 27 \u003d 36	1 x 36 - 1 \u003d 35	1 x 35 - 3 \u003d 32 ≠ 0	1 - not root
-1	9	-1 x 9 + 27 \u003d 18	-1 x 18 - 1 \u003d -19	-1 x (-19) - 3 \u003d 16 ≠ 0	-1 - not root
	9	x 9 + 27 \u003d 30	x 30 - 1 \u003d 9	x 9 - 3 \u003d 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 calculating the selection of Q roots it is convenient to make a change of variable, go to the above equation and choose Z roots.

If the free term is 1

.

If you can use a view replacement y \u003d 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 is beyond the scope of our course.

n \u003d 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 \u003d 0. Replacement y = x 2. From here y 1 = 4, y 2 \u003d -9. therefore x 1,2 = + 2 .

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

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

ax 4 + bx 3 + cx 2 – bx + a = 0.

The generalized return equation of the fourth degree of the form ax 4 + bx 3 + cx 2 + kbx + k 2a \u003d 0.

General replacement. Some standard replacements.

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

n = 3.

Equation with integer coefficients. Q-root selection 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 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 considered above for n = 3.

Equation with integer coefficients. Q-root selection based on the theorem. Horner's scheme. The algorithm is similar to that considered above for n = 3.

Symmetric equations. Every odd return equation has a root x \u003d -1 and after factoring it, we get that one factor has the form ( x + 1), and the second factor is the reverse equation of even degree (its degree is one less than the degree of the original equation). Any return equation of even degree together with a root of the form x \u003d φ 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. Using 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 Nils Henrik Abel (1802–1829)) and higher degrees (this was shown by the French mathematician Evarist Galois (1811-1832 )).

Recall again that in practice it is possible to use combinations the methods listed above. It is convenient to go to the set of equations of lower degrees by factorization of the original equation.
Outside the scope of our discussion today, we have remained widely used in practice. graphical methods solving equations and approximate methods equations of higher degrees.

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

Using the monotonicity property of functions

4. Summarizing.

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 to determine which method of solution in this case is the most effective, and also to apply the chosen method correctly.

5. Homework.

: p. 7, p. 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 essays on this topic:

Formula Cardano
Graphic method for solving equations. Examples of solutions.
Methods for the approximate solution of equations.

Analysis of the assimilation of material and student interest in the topic:

Experience shows that the interest of students in the first place is the possibility of selection Z-roots and Q-roots of equations using a fairly simple algorithm using the Horner scheme. Students are also interested in various standard types of variable substitution, which can significantly simplify the type of task. Of particular interest are graphic solutions. In this case, you can additionally parse the tasks into a graphical method for solving equations; discuss the general view 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 form of the corresponding graph. Below is a list of books where you can find additional information on this topic.

List of references:

Vilenkin N.Ya. et al. “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 math textbook. Arithmetic. Algebra. Grade 10-11 ”- M., Education, 2008 - 192 p.
Vygodsky M.Ya. “Handbook of mathematics” - M., AST, 2010 - 1055 p.
Galitsky M.L.“Collection of problems in algebra. A textbook for grades 8-9 with an in-depth study of mathematics ”- M., Education, 2008 - 301 p.
Zvavich L.I. et al. “Algebra and the beginning of analysis. 8-11 cl. A manual for schools and classes with an in-depth study of mathematics ”- M., Drofa, 1999 - 352 p.
Zvavich L.I., Averyanov D.I., Pigarev B.P., Trushanina T.N. “Tasks in mathematics to prepare for a written exam in grade 9” - M., Education, 2007 - 112 p.
Ivanov A.A., Ivanov A.P. “Thematic tests for systematization of knowledge in mathematics”, part 1 - M., Fizmatkniga, 2006 - 176 p.
Ivanov A.A., Ivanov A.P. “Thematic tests for systematizing knowledge in mathematics” part 2 - M., Fizmatkniga, 2006 - 176 p.
Ivanov A.P. “Tests and tests in mathematics. Tutorial". - M., Fizmatkniga, 2008 - 304 p.
Leibson K.L. “Collection of practical tasks in mathematics. Part 2–9 grade ”- M., ICMMO, 2009 - 184 p.
Makarychev Yu.N., Mindyuk N.G."Algebra. Additional chapters to the school textbook of grade 9. A textbook for students in schools and classes with an in-depth study of mathematics. ” - M., Education, 2006 - 224 p.
Mordkovich A.G. "Algebra. In-depth study. 8th grade. Textbook ”- M., Mnemozina, 2006 - 296 p.
Savin A.P. “Encyclopedic Dictionary of the Young Mathematician” - M., Pedagogy, 1985 - 352 p.
Survillo G.S., Simonov A.S. “Didactic materials on algebra for grade 9 with an in-depth study of mathematics” - M., Education, 2006 - 95 p.
Chulkov P.V. “Equations and inequalities in the school course of mathematicians. Lectures 1–4 ”- M., First of September, 2006 - 88 p.
Chulkov P.V. “Equations and inequalities in the school course of mathematicians. Lectures 5–8 ”- M., First of September, 2009 - 84 p.

Trifanova Marina Anatolyevna
teacher of mathematics, Municipal Educational Establishment "Gymnasium No. 48 (multidisciplinary)", Talnakh

The triune goal of the lesson:

Educational:
systematization and generalization of knowledge on solving equations of higher degrees.
Developing:
to promote the development of logical thinking, the ability to work independently, the skills of mutual control and self-control, the ability to speak and listen.
Educating:
developing a habit of permanent employment; fostering responsiveness, industriousness, and accuracy.

Lesson type:

lesson of the integrated application of knowledge, skills.

Lesson form:

airing, physical education, various forms of work.

Equipment:

reference notes, task cards, lesson monitoring matrix.

DURING THE CLASSES

I. Organizational moment

Report the purpose of the lesson to students.
Checking homework (Appendix 1). Work with a supporting abstract (Appendix 2).

Equations and answers for each of them are written on the board. Students check the answers and give a brief analysis of the solution to each equation or answer the teacher's questions (frontal survey). Self-control - students give themselves grades and pass notebooks to the teacher for verification to correct grades or confirm them. The grade school is written on the board:

“5+” - 6 equations;
“5” - 5 equations;
“4” - 4 equations;
“3” - 3 equations.

Homework teacher questions:

1 equation

What is the change of variables made in the equation?
What equation is obtained after changing variables?

2 equation

Which polynomial divided both sides of the equation?
What variable substitution was received?

3 equation

What polynomials must be multiplied to simplify the solution of this equation?

4 equation

Name the function f (x).
How were the other roots found?

5 equation

How many gaps were obtained to solve the equation?

6 equation

What methods could solve this equation?
Which solution is more rational?

II. Work in groups is the main part of the lesson.

The class is divided into 4 groups. Each group is given a card with theoretical and practical (Appendix 3) questions: “Parse the proposed method for solving the equation and explain it with this example.”

Group work 15 minutes.
Examples are written on the board (the board is divided into 4 parts).
The group report takes 2 to 3 minutes.
The teacher corrects group reports and helps with difficulty.

Work in groups continues on cards No. 5 - 8. For each equation, 5 minutes are given for discussion in the group. Then a report on the given equation comes near the board - a brief analysis of the solution. The equation can not be solved to the end - it is being finalized at home, but the whole sequence of its solution in the class is discussed.

III. Independent work.Appendix 4.

Each student receives an individual assignment.
Work on time takes 20 minutes.
5 minutes before the end of the lesson, the teacher gives open answers for each equation.
Students change in a circle of notebooks and check the answers with a friend. Give grades.
Notebooks are handed over to the teacher for verification and adjustment of grades.

IV. Lesson summary.

Homework.

To issue a solution to unfinished equations. Prepare for a control slice.

Grading.

Consider solving equations with one variable of degree higher than the second.

The degree of the equation P (x) \u003d 0 is the degree of the polynomial P (x), i.e. the largest of the degrees of its members with a coefficient not equal to zero.

So, for example, the equation (x 3 - 1) 2 + x 5 \u003d x 6 - 2 has a fifth power, because after the operations of opening the brackets and bringing down similar ones, we obtain the equivalent equation x 5 - 2x 3 + 3 \u003d 0 of the fifth degree.

Recall the rules that will be needed to solve equations of degree higher than the second.

Statements about the roots of the polynomial and its divisors:

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

2. An odd degree polynomial has at least one real root.

3. If α is the root of P (x), then P n (x) \u003d (x - α) · Q n - 1 (x), where Q n - 1 (x) is a polynomial of degree (n - 1).

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

6. For a polynomial of the third degree

P 3 (x) \u003d ax 3 + bx 2 + cx + d one of the two is possible: either it decomposes into the product of three binomials

P 3 (x) \u003d a (x - α) (x - β) (x - γ), or decomposes into the product of the binomial and the square trinomial P 3 (x) \u003d a (x - α) (x 2 + βx + γ )

7. Any fourth-degree polynomial decomposes into a product of two square trinomials.

8. The polynomial f (x) is divisible by the polynomial g (x) without remainder if there exists a polynomial q (x) such that f (x) \u003d g (x) · q (x). For division of polynomials the rule "division by corner" is applied.

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

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

P (x) \u003d a 0 x n + a 1 x n - 1 + ... + a n, then the following equalities hold:

x 1 + x 2 + ... + x n \u003d -a 1 / a 0,

x 1 · x 2 + x 1 · x 3 + ... + x n - 1 · x n \u003d a 2 / a 0,

x 1 · x 2 · x 3 + ... + x n - 2 · x n - 1 · x n \u003d -a 3 / a 0,

x 1 · x 2 · x 3 · x n \u003d (-1) n a n / a 0.

Solution Examples

Example 1

Find the remainder of the division of P (x) \u003d x 3 + 2/3 x 2 - 1/9 by (x - 1/3).

Decision.

By the corollary of Bezout’s theorem: “The remainder of dividing a polynomial by a binomial (x - c) is equal to the value of the polynomial in c”. We find P (1/3) \u003d 0. Therefore, the remainder is 0 and the number 1/3 is the root of the polynomial.

Answer: R \u003d 0.

Example 2

Divide the "corner" 2x 3 + 3x 2 - 2x + 3 by (x + 2). Find the remainder and the partial quotient.

Decision:

2x 3 + 3x 2 - 2x + 3 | x + 2

2x 3 + 4 x 2 2x 2 - x

X 2 - 2 x

Answer: R \u003d 3; quotient: 2x 2 - x.

Basic methods for solving equations of higher degrees

1. Introduction of a new variable

The method of introducing a new variable is already familiar with the example of biquadratic equations. It consists in the fact that to solve the equation f (x) \u003d 0, a new variable (substitution) t \u003d x n or t \u003d g (x) is introduced and f (x) is expressed in terms of t, obtaining a new equation r (t). Solving then the equation r (t), find the roots:

(t 1, t 2, ..., t n). After that, a set of n equations q (x) \u003d t 1, q (x) \u003d t 2, ..., q (x) \u003d 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 \u003d 0.

Decision:

(x 2 + x + 1) 2 - 3 (x 2 + x) - 1 \u003d 0.

(x 2 + x + 1) 2 - 3 (x 2 + x + 1) + 3 - 1 \u003d 0.

Replacement (x 2 + x + 1) \u003d t.

t 2 - 3t + 2 \u003d 0.

t 1 \u003d 2, t 2 \u003d 1. Reverse replacement:

x 2 + x + 1 \u003d 2 or x 2 + x + 1 \u003d 1;

x 2 + x - 1 \u003d 0 or x 2 + x \u003d 0;

Answer: From the first equation: x 1, 2 \u003d (-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 the terms in such a way that each group contains a common factor. To do this, sometimes it is necessary to apply some artificial techniques.

Example 1

x 4 - 3x 2 + 4x - 3 \u003d 0.

Decision.

Imagine - 3x 2 \u003d -2x 2 - x 2 and group:

(x 4 - 2x 2) - (x 2 - 4x + 3) \u003d 0.

(x 4 - 2x 2 +1 - 1) - (x 2 - 4x + 3 + 1 - 1) \u003d 0.

(x 2 - 1) 2 - 1 - (x - 2) 2 + 1 \u003d 0.

(x 2 - 1) 2 - (x - 2) 2 \u003d 0.

(x 2 - 1 - x + 2) (x 2 - 1 + x - 2) \u003d 0.

(x 2 - x + 1) (x 2 + x - 3) \u003d 0.

x 2 - x + 1 \u003d 0 or x 2 + x - 3 \u003d 0.

Answer: There are no roots in the first equation, from the second: x 1, 2 \u003d (-1 ± √13) / 2.

3. Factorization by the method of indefinite 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 degrees, unknown expansion coefficients are found.

Example 1

x 3 + 4x 2 + 5x + 2 \u003d 0.

Decision.

A polynomial of degree 3 can be decomposed into a product of linear and square factors.

x 3 + 4x 2 + 5x + 2 \u003d (x - a) (x 2 + bx + c),

x 3 + 4x 2 + 5x + 2 \u003d x 3 + bx 2 + cx - ax 2 - abx - ac,

x 3 + 4x 2 + 5x + 2 \u003d x 3 + (b - a) x 2 + (cx - ab) x - ac.

Having solved the system:

(b - a \u003d 4,
(c - ab \u003d 5,
(-ac \u003d 2,

(a \u003d -1,
(b \u003d 3,
(c \u003d 2, i.e.

x 3 + 4x 2 + 5x + 2 \u003d (x + 1) (x 2 + 3x + 2).

The roots of the equation (x + 1) (x 2 + 3x + 2) \u003d 0 are easily found.

The answer is -1; -2.

4. Method of root selection by senior and free coefficient

The method relies 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 positive integer) to be the root of the equation with integer coefficients, it is necessary that the number p be the integer divisor of the free term and 0, and q is the natural divisor of the highest coefficient.

Example 1

6x 3 + 7x 2 - 9x + 2 \u003d 0.

Decision:

6: q \u003d 1, 2, 3, 6.

Therefore, p / q \u003d ± 1, ± 2, ± 1/2, ± 1/3, ± 2/3, ± 1/6.

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

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

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Introduction

Solving algebraic equations of higher degrees with one unknown is one of the most difficult and oldest mathematical problems. The most outstanding mathematicians of antiquity were engaged in these tasks.

Solving equations of degree n is an important task for modern mathematics. The interest in them is quite large, since these equations are closely related to the search for the roots of equations that are not considered by the school program in mathematics.

Problem: lack of skills in solving equations of higher degrees in various ways among students prevents them from successfully preparing for final certification in mathematics and mathematical olympiads, teaching in a specialized mathematical class.

The listed facts identified relevance our work "Solving equations of higher degrees."

Possession of the simplest methods for solving equations of the nth degree reduces the time to complete the task, on which the result of the work and the quality of the learning process depend.

Objective: the study of known methods for solving equations of higher degrees and identifying the most accessible of them for practical use.

Based on the goal, the following tasks:

To study the literature and Internet resources on this topic;

Get acquainted with historical facts related to this topic;

Describe the various ways to solve equations of higher degrees

compare the degree of difficulty of each of them;

To acquaint classmates with methods for solving equations of higher degrees;

Create a selection of equations for the practical application of each of the considered methods.

Object of study - equations of higher degrees with one variable.

Subject of study - methods for solving equations of higher degrees.

Hypothesis:there is no general method and a single algorithm that allows finding solutions of equations of the nth power in a finite number of steps.

Research Methods:

- bibliographic method (analysis of literature on the research topic);

- classification method;

- qualitative analysis method.

Theoretical significanceresearch consists in systematizing methods for solving equations of higher degrees and a description of their algorithms.

Practical significance - Presented material on this topic and the development of a textbook for students on this topic.

1. EQUATIONS OF THE HIGHEST DEGREES

1.1 the Concept of the equation of the n-th degree

Definition 1.An equation of the nth degree is an equation of the form

a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an \u003d 0, where the coefficients a 0, a 1, a 2…, an -1, an- any real numbers, moreover , a 0 ≠ 0 .

Polynomial a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an is called an nth degree polynomial. Odds are distinguished by name: a 0 - senior coefficient; an is a free member.

Definition 2. The solutions or roots for this equation are all variable values xwhich convert this equation to true numerical equality or, in which the polynomial a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an vanishes. This variable value xalso called the root of the polynomial. To solve the equation means to find all its roots or to establish that they are not.

If a a 0 \u003d 1, then such an equation is called a reduced integer rational equation n th degrees.

For equations of the third and fourth degree, there are Cardano and Ferrari formulas that express the roots of these equations in terms of radicals. It turned out that in practice they are rarely used. Thus, if n ≥ 3, and the coefficients of the polynomial are arbitrary real numbers, then the search for the roots of the equation is not an easy task. Nevertheless, in many special cases this problem is solved to the end. Let us dwell on some of them.

1.2 Historical facts of solving equations of higher degrees

Already in ancient times, people realized how important it is to learn how to solve algebraic equations. About 4000 years ago, Babylonian scholars mastered the solution of a quadratic equation and solved a system of two equations, of which one is of the second degree. Using equations of higher degrees, various problems of land surveying, architecture and military affairs were solved, many and various questions of practice and natural science were reduced to them, since the exact language of mathematics allows you to simply express facts and relationships, which, when set out in ordinary language, may seem confusing and complex .

A universal formula for finding the roots of an algebraic equation nth no degree. Many, of course, had the tempting idea to find formulas for any degree n that express the roots of the equation in terms of its coefficients, that is, solve the equation in radicals.

Only in the 16th century did Italian mathematicians advance further by finding formulas for n \u003d 3 and n \u003d 4. At the same time, Scipio, Dahl, Ferro and his students Fiori and Tartaglia dealt with the general solution of equations of the 3rd degree.

In 1545, the book of the Italian mathematician D. Cardano, “The Great Art, or the Rules of Algebra,” was published, where, along with other questions of algebra, general methods for solving cubic equations are considered, as well as a method for solving equations of the 4th degree, discovered by his student L. Ferrari.

A complete statement of the questions connected with the solution of equations of the 3rd and 4th degrees was given by F. Viet.

In the 20s of the 19th century, the Norwegian mathematician N. Abel proved that the roots of equations of the fifth degree cannot be expressed through radicals.

The study revealed that modern science knows many ways to solve equations of the nth degree.

The search for methods for solving equations of higher degrees that cannot be solved by the methods considered in the school curriculum resulted in methods based on the application of the Vieta theorem (for equations of degree n\u003e 2), Bezout's theorem, Horner's scheme, as well as the Cardano and Ferrari formula for solving cubic equations and equations of the fourth degree.

The paper presents methods for solving equations and their types, which have become a discovery for us. These include - the method of uncertain coefficients, the selection of the full degree, symmetric equations.

2. SOLUTION OF THE WHOLE EQUATIONS OF THE HIGHEST DEGREES WITH WHOLE COEFFICIENTS

2.1 Solution of equations of the 3rd degree. Formula D. Cardano

We consider equations of the form x 3 + px + q \u003d 0.We transform the equation of the general form to the form: x 3 + px 2 + qx + r \u003d 0. We write the formula for the cube of the sum; Add to the original equality and replace it with y. We get the equation: y 3 + (q -) (y -) + (r - \u003d 0. After the transformations, we have: y 2 + py + q \u003d 0. Now, again, write down the formula for the cube of the sum:

(a + b) 3 \u003d a 3 + 3a 2 b + 3ab 2 + b 3 \u003d a 3 + b 3 + 3ab (a + b), replace ( a + b)on the xwe get the equation x 3 - 3abx - (a 3 + b 3) = 0. Now it is clear that the original equation is equivalent to the system: and Solving the system, we obtain:

We got a formula for solving the reduced equation of the 3rd degree. She bears the name of the Italian mathematician Cardano.

Consider an example. Solve the equation:.

We have r \u003d 15 and q \u003d 124, then using the Cardano formula, we calculate the root of the equation

Conclusion: this formula is good, but not suitable for solving all cubic equations. However, it is bulky. Therefore, in practice it is rarely used.

But those who master this formula can use it when solving equations of the third degree on the exam.

2.2 Vieta Theorem

From a course in mathematics we know this theorem for a quadratic equation, but few people know that it is used to solve equations of higher degrees.

Consider the equation:

factor the left side of the equation, divide by ≠ 0.

We transform the right side of the equation to the form

; it follows that we can write the following equalities into the system:

The formulas deduced by Viet for quadratic equations and demonstrated by us for equations of degree 3 are also true for polynomials of higher degrees.

We solve the cubic equation:

Conclusion: this method is universal and easy enough for students to understand, since they know the Vieta theorem from the school curriculum for n = 2. At the same time, in order to find the roots of equations using this theorem, one must have good computational skills.

2.3 Bezout Theorem

This theorem is named after the French mathematician of the 18th century J. Bezout.

Theorem.If the equation a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an \u003d 0, in which all the coefficients are integers, and the free term is non-zero, has an integer root, then this root is a divisor of the free term.

Considering that the polynomial of the nth power is on the left side of the equation, the theorem has a different interpretation.

Theorem. When dividing a polynomial of degree n with respect to x on the binomial x - a the remainder is the value of the dividend for x \u003d a. (letter a can mean any real or imaginary number, i.e. any complex number).

Evidence: let be f (x) denotes an arbitrary polynomial of degree n with respect to the variable x and let it be divided by the binomial ( x-a) it turned out in private q (x), but in the remainder R. It's obvious that q (x) there will be some polynomial (n - 1st) degree relative x, and the remainder R will be a constant value, i.e. independent of x.

If the remainder R was a polynomial of the first degree with respect to x, this would mean that the division was not performed. So, R from x does not depend. By the definition of division, we obtain the identity: f (x) \u003d (x-a) q (x) + R.

Equality is valid for any value of x; therefore, it is also valid for x \u003d awe get: f (a) \u003d (a-a) q (a) + R. Symbol f (a) denotes the value of the polynomial f (x) at x \u003d a, q (a) denotes a value q (x) at x \u003d a.The remainder R stayed the way he was before since R from x does not depend. Composition ( x-a) q (a) \u003d 0, since the factor ( x-a) \u003d 0, and the multiplier q (a) there is a certain number. Therefore, from the equality we obtain: f (a) \u003d R,t.d.

Example 1Find the remainder of a polynomial division x 3 - 3x 2 + 6x-5 on the binomial

x-2. By the Bezout theorem : R \u003d f(2) = 23-322 + 62 -5 \u003d 3. Answer: R \u003d3.

We note that Bezout's theorem is not so important in itself as it is by its consequences. (Appendix 1)

Let us dwell on some methods of applying Bezout's theorem to solving practical problems. It should be noted that when solving equations using the Bezout theorem, it is necessary:

Find all integer divisors of a free member;

From these divisors find at least one root of the equation;

The left side of the equation is divided into (Ha);

Write the product of the divisor and the quotient on the left side of the equation;

Solve the resulting equation.

Consider the example of solving the equation x 3 + 4x 2 + x -6 = 0 .

Solution: find the terms of the free term ± 1 ; ± 2; ± 3; ± 6. We calculate the values \u200b\u200bfor x \u003d1, 1 3 + 41 2 + 1-6 \u003d 0. Divide the left side of the equation by ( x-1). The division is performed by the “corner”, we get:

Conclusion: Bezout's theorem is one of the methods that we consider in our work that is studied in the program of elective classes. It is difficult to understand, because in order to own it, you need to know all the consequences of it, but Bezu’s theorem is one of the main assistants to students on the exam.

2.4 Horner Scheme

To divide a polynomial into a binomial x-α you can use a special simple technique invented by English mathematicians of the XVII century, later called Horner's scheme. In addition to finding the roots of the equations, according to Horner's scheme, one can more easily calculate their values. To do this, substitute the value of the variable in the polynomial Pn (x) \u003d a 0 xn + a 1 x n-1 + a 2 xⁿ - ² + ... ++ an -1 x + an (1)

Consider the division of polynomial (1) by the binomial x-α.

Express the coefficients of the partial quotient b 0 xⁿ - ¹+ b 1 xⁿ - ²+ b 2 xⁿ - ³+…+ bn -1 and the remainder r through the coefficients of the polynomial Pn ( x) and number α. b 0 \u003d a 0 , b 1 = α b 0 + a 1 , b 2 = α b 1 + a 2 …, bn -1 =

= α bn -2 + an -1 = α bn -1 + an .

Horner calculations are presented in the form of the following table:

	and 0	a 1	a 2 ,
	b 0 \u003d a 0	b 1 = α b 0 + a 1	b 2 = α b 1 + a 2		r \u003d αb n-1 + an

Insofar as r \u003d Pn (α), then α is the root of the equation. In order to check whether α is a multiple root, the Horner scheme can already be applied to the quotient b 0 x +b 1 x + ... +bn -1 according to the table. If in the column under bn -1 it turns out again 0, which means that α is a multiple root.

Consider an example: solve the equation x 3 + 4x 2 + x -6 = 0.

We apply to the left side of the equation the factorization of the polynomial on the left side of the equation, the Horner scheme.

Solution: find the free term divisors ± 1; ± 2; ±3; ± 6.


				6 ∙ 1 + (-6) = 0

The coefficients of the quotient are numbers 1, 5, 6, and the remainder is r \u003d 0.

Means x 3 + 4x 2 + x - 6 = (x - 1) (x 2 + 5x + 6) = 0.

From here: x - 1 \u003d 0 or x 2 + 5x + 6 = 0.

x = 1, x 1 = -2; x 2 = -3. Answer:1,- 2, - 3.

Conclusion: thus, on one equation, we showed the application of two different methods of factoring polynomials. In our opinion, the Horner scheme is the most practical and economical.

2.5 Solution of equations of the 4th degree. Ferrari Method

Pupil Cardano Louis Ferrari discovered a way to solve the equation of the 4th degree. The Ferrari method consists of two stages.

Stage I: equations of the form are represented as the product of two square trinomials this follows from the fact that the equation is of the 3rd degree and at least one solution.

Stage II: the resulting equations are solved using factorization, however, in order to find the required factorization, it is necessary to solve cubic equations.

The idea is to represent the equations in the form A 2 \u003d B 2, where A \u003d x 2 + s

B-linear function of x. Then it remains to solve the equations A \u003d ± B.

For clarity, consider the equation: Separate the 4th degree, we get: For any dthe expression will be a full square. Add to both sides of the equation we get

On the left side is a full square, you can pick up dso that the right-hand side of (2) becomes a full square. Imagine that we have achieved this. Then our equation looks like this:

Finding the root afterwards will not be difficult. To choose the right d it is necessary that the discriminant of the right-hand side of (3) becomes zero, i.e.

So, to find d, it is necessary to solve this equation of the 3rd degree. Such an auxiliary equation is called resolvent.

We easily find the whole root of the resolvent: d \u003d1

Substituting the equation in (1), we obtain

Conclusion: the Ferrari method is universal, but complex and cumbersome. However, if the solution algorithm is clear, then equations of the 4th degree can be solved by this method.

2.6 Method of uncertain coefficients

The success of solving the 4th degree equation by the Ferrari method depends on whether we solved the resolvent — the 3rd degree equation, which, as we know, is not always successful.

The essence of the method of uncertain coefficients is that the form of the factors into which the given polynomial is decomposed is guessed, and the coefficients of these factors (also polynomials) are determined by multiplying the factors and equating the coefficients for the same degrees of the variable.

Example: solve the equation:

Suppose that the left side of our equation can be decomposed into two square trinomials with integer coefficients such that the identity

Obviously, the coefficients in front of them must be equal to 1, and the free terms in one + 1, the other has 1.

The coefficients facing x. We denote them by and and and to determine them, we multiply both trinomials of the right side of the equation.

As a result, we get:

Equating the coefficients at the same degrees xon the left and right sides of equality (1), we obtain a system for finding and

Having solved this system, we will have

So, our equation is equivalent to the equation

Having solved it, we get the following roots:.

The uncertain coefficient method is based on the following statements: any fourth-degree polynomial in the equation can be decomposed into the product of two second-degree polynomials; two polynomials are identically equal if and only if their coefficients are equal at equal degrees x

2.7 Symmetric equations

DefinitionAn equation of the form is called symmetric if the first coefficients in the equation on the left are equal to the first coefficients on the right.

We see that the first coefficients on the left are equal to the first coefficients on the right.

If such an equation has an odd degree, then it has a root x= - 1. Next, we can lower the degree of the equation by dividing it by ( x +1). It turns out that when dividing the symmetric equation by ( x +1) we obtain a symmetric equation of even degree. The proof of the symmetry of the coefficients is presented below. (Appendix 6) Our task is to learn how to solve symmetric equations of even degree.

For example: (1)

We solve equation (1), divide by x 2 (medium grade) \u003d 0.

We group the terms with symmetric

) + 3(x +. Denote at= x +, square both sides, hence \u003d at 2 So 2 ( at 2 or 2 at 2 + 3 solving the equation, we get at = , at \u003d 3. Next, we return to the replacement x + \u003d and x + \u003d 3. We get the equations and the first has no solution, and the second has two roots. Answer:.

Conclusion: this type of equation is not common, but if you come across it, you can solve it easily and simply without resorting to cumbersome calculations.

2.8 Allocation of the full degree

Consider the equation.

The left side is the cube of the sum (x + 1), i.e.

We extract the root of the third degree from both parts:, then we obtain

Where is the only root.

RESEARCH RESULTS

Based on the results of the work, we came to the following conclusions:

Thanks to the theory we studied, we became acquainted with various methods for solving entire equations of higher degrees;

Formula D. Cardano is difficult to use and makes it more likely to make mistakes in the calculation;

- the method of L. Ferrari allows us to reduce the solution of the equation of the fourth degree to cubic;

- Bezout's theorem can be applied both to cubic equations and to equations of the fourth degree; it is more understandable and visual in application to solving equations;

Horner's scheme helps to significantly reduce and simplify calculations in solving equations. In addition to finding the roots, according to Horner's scheme, one can more simply calculate the values \u200b\u200bof the polynomials on the left side of the equation;

Of particular interest were the solution of equations by the method of indefinite coefficients, the solution of symmetric equations.

In the course of the research work, it was found that students learn the simplest ways to solve equations of the highest degree in elective classes in mathematics, starting from the 9th or 10th grade, as well as in special courses of visiting mathematical schools. This fact was established as a result of a survey of mathematics teachers at MBOU “Secondary School No. 9” and students showing increased interest in the subject “mathematics”.

The most popular methods for solving equations of higher degrees that are encountered in solving olympiad, competitive problems and as a result of students preparing for exams are methods based on the application of Bezout's theorem, Horner's scheme, and the introduction of a new variable.

Demonstration of research results, i.e. methods of solving equations that are not studied in the school curriculum in mathematics, interested classmates.

Conclusion

Having studied the educational and scientific literature, Internet resources in youth educational forums

When solving algebraic equations, one often has to factor the polynomial. Factoring a polynomial means representing it as a product of two or more polynomials. We use some methods of decomposing polynomials quite often: taking out a common factor, applying reduced multiplication formulas, extracting a full square, grouping. Let's consider some more methods.

Sometimes, when factoring a polynomial, the following statements are useful:

1) if a polynomial with integer coefficients has a rational root (where is an irreducible fraction, then is the divisor of the free term and the divisor of the highest coefficient:

2) If in some way we choose the root of the polynomial of degree, then the polynomial can be represented in the form where the polynomial of degree

The polynomial can be found either by dividing the polynomial into a binomial “column”, or by correspondingly grouping the terms of the polynomial and extracting a factor from them, or by the method of indefinite coefficients.

Example. Factor polynomial

Decision. Since the coefficient at x4 is 1, the rational roots of this polynomial exist, they are divisors of 6, that is, they can be integers ± 1, ± 2, ± 3, ± 6. Denote this polynomial by P4 (x). Since P P4 (1) \u003d 4 and P4 (-4) \u003d 23, the numbers 1 and -1 are not the roots of the polynomial PA (x). Since P4 (2) \u003d 0, then x \u003d 2 is the root of the polynomial P4 (x), and therefore, this polynomial is divided by the two-member x - 2. Therefore, x4 -5x3 + 7x2 -5x +6 x-2 x4 -2x3 x3 -3x2 + x-3

3x3 + 7x2 -5x +6

3x3 + 6x2 x2 - 5x + 6 x2 - 2x

Consequently, P4 (x) \u003d (x - 2) (x3 - 3x2 + x - 3). Since xz - 3x2 + x - 3 \u003d x2 (x - 3) + (x - 3) \u003d (x - 3) (x2 + 1), then x4 - 5x3 + 7x2 - 5x + 6 \u003d (x - 2) (x - 3) (x2 + 1).

Parameter Input Method

Sometimes when factoring a polynomial, the method of introducing a parameter helps. We explain the essence of this method in the following example.

Example. x3 - (√3 + 1) x2 + 3.

Decision. Consider a polynomial with the parameter a: x3 - (a + 1) x2 + a2, which for a \u003d √3 turns into a given polynomial. We write this polynomial as a quadratic trinomial with respect to a: ar - ax2 + (x3 - x2).

Since the roots of this quadratic trinomial with respect to a are a1 \u003d x and a2 \u003d x2 - x, then the equality a2 - ax2 + (xs - x2) \u003d (a - x) (a - x2 + x) is valid. Therefore, the polynomial x3 - (√3 + 1) x2 + 3 is factorized by √3 - x and √3 - x2 + x, i.e.

x3 - (√3 + 1) x2 + 3 \u003d (x-√3) (x2-x-√3).

Method for introducing a new unknown

In some cases, by replacing the expression f (x) in the polynomial Pn (x), we can obtain a polynomial with respect to y through y, which can already be easily factorized. Then, after changing y to f (x), we obtain the factorization of the polynomial Pn (x).

Example. Factor the polynomial x (x + 1) (x + 2) (x + 3) -15.

Decision. We transform this polynomial as follows: x (x + 1) (x + 2) (x + 3) -15 \u003d [x (x + 3)] [(x + 1) (x + 2)] - 15 \u003d (x2 + 3x) (x2 + 3x + 2) - 15.

Denote x2 + 3x by y. Then we have y (y + 2) - 15 \u003d y2 + 2y - 15 \u003d y2 + 2y + 1 - 16 \u003d (y + 1) 2 - 16 \u003d (y + 1 + 4) (y + 1-4) \u003d ( y + 5) (y - 3).

Therefore, x (x + 1) (x + 2) (x + 3) - 15 \u003d (x2 + 3x + 5) (x2 + 3x - 3).

Example. Factor polynomial (x-4) 4+ (x + 2) 4

Decision. Denote x - 4 + x + 2 \u003d x - 1 by y.

(x - 4) 4 + (x + 2) 2 \u003d (y - 3) 4 + (y + 3) 4 \u003d y4 - 12y3 + 54y3 - 108y + 81 + y4 + 12y3 + 54y2 + 108y + 81 \u003d

2y4 + 108y2 + 162 \u003d 2 (y4 + 54y2 + 81) \u003d 2 [(yy + 27) 2 - 648] \u003d 2 (y2 + 27 - √b48) (y2 + 27 + √b48) \u003d

2 ((x-1) 2 + 27-√b48) ((x-1) 2 + 27 + √b48) \u003d 2 (x2-2x + 28--18√2) (x2-2x + 28 + 18√ 2 )

Combination of different methods

Often when factoring a polynomial, several of the methods discussed above must be applied sequentially.

Example. Factor the polynomial x4 - 3x2 + 4x-3.

Decision. Using the grouping, we rewrite the polynomial in the form x4 - 3x2 + 4x - 3 \u003d (x4 - 2x2) - (x2 -4x + 3).

Applying the full square selection method to the first bracket, we have x4 - 3x3 + 4x - 3 \u003d (x4 - 2 · 1 · x2 + 12) - (x2 -4x + 4).

Using the full square formula, we can now write that x4 - 3x2 + 4x - 3 \u003d (x2 -1) 2 - (x - 2) 2.

Finally, applying the formula of the difference of the squares, we obtain )

§ 2. Symmetric equations

1. Symmetric equations of the third degree

Equations of the form ax3 + bx2 + bx + a \u003d 0, and ≠ 0 (1) are called symmetric equations of the third degree. Since ax3 + bx2 + bx + a \u003d a (x3 + 1) + bx (x + 1) \u003d (x + 1) (ax2 + (b-a) x + a), then equation (1) is equivalent to the set of equations x + 1 \u003d 0 and ax2 + (b-a) x + a \u003d 0, which is not difficult to solve.

Example 1. Solve the equation

3x3 + 4x2 + 4x + 3 \u003d 0. (2)

Decision. Equation (2) is a symmetric equation of the third degree.

Since 3x3 + 4xg + 4x + 3 \u003d 3 (x3 + 1) + 4x (x + 1) \u003d (x + 1) (3x2 - 3x + 3 + 4x) \u003d (x + 1) (3x2 + x + 3) , then equation (2) is equivalent to the set of equations x + 1 \u003d 0 and 3x3 + x + 3 \u003d 0.

The solution to the first of these equations is x \u003d -1, the second equation has no solutions.

Answer: x \u003d -1.

2. Symmetric equations of the fourth degree

Equation of the form

(3) is called a symmetric equation of the fourth degree.

Since x \u003d 0 is not the root of equation (3), then, dividing both sides of equation (3) by x2, we obtain an equation equivalent to the original (3):

We rewrite equation (4) in the form:

In this equation we make a replacement, then we obtain the quadratic equation

If equation (5) has 2 roots y1 and y2, then the original equation is equivalent to the set of equations

If equation (5) has one root y0, then the original equation is equivalent to the equation

Finally, if equation (5) has no roots, then the original equation also has no roots.

Example 2. Solve the equation

Decision. This equation is a symmetric equation of the fourth degree. Since x \u003d 0 is not its root, then, dividing equation (6) by x2, we obtain the equation equivalent to it:

Having grouped the terms, we rewrite equation (7) in the form or in the form

Putting, we obtain an equation having two roots y1 \u003d 2 and y2 \u003d 3. Therefore, the original equation is equivalent to a set of equations

The solution to the first equation of this population is x1 \u003d 1, and the solution to the second is and.

Therefore, the original equation has three roots: x1, x2 and x3.

Answer: x1 \u003d 1 ,.

§3. Algebraic equations

1. The decrease in the degree of equation

Some algebraic equations by replacing a certain polynomial in them with one letter can be reduced to algebraic equations whose degree is less than the degree of the original equation and the solution of which is simpler.

Example 1. Solve the equation

Decision. We denote by, then equation (1) can be rewritten in the form The last equation has roots and Therefore, equation (1) is equivalent to the set of equations and. There is a solution to the first equation of this set, and there are solutions to the second equation

The solutions of equation (1) are

Example 2. Solve the equation

Decision. Multiplying both sides of the equation by 12 and denoting by,

We get the equation Rewrite this equation in the form

(3) and denoting by rewriting equation (3) in the form The last equation has roots, and therefore we find that equation (3) is equivalent to the combination of two equations and there are solutions to this set of equations, i.e., equation (2) is equivalent to the combination of equations and ( 4)

The solutions of (4) are and, they are the solutions of equation (2).

2. Equations of the form

The equation

(5) where are given numbers, can be reduced to a biquadratic equation by replacing the unknown, i.e., replacing

Example 3. Solve the equation

Decision. Denote by, t. e. make a change of variables or Then equation (6) can be rewritten in the form or, using the formula, in the form

Since the roots of the quadratic equation are, and then the solutions of equation (7) are the solutions of the set of equations and. This set of equations has two solutions and, therefore, the solutions of equation (6) are and

3. Equations of the form

The equation

(8) where the numbers α, β, γ, δ, and Α are such that α

Example 4. Solve the equation

Decision. We replace the unknowns, i.e., y \u003d x + 3 or x \u003d y - 3. Then equation (9) can be rewritten in the form

(y-2) (y-1) (y + 1) (y + 2) \u003d 10, i.e., in the form

(y2-4) (y2-1) \u003d 10 (10)

Biquadratic equation (10) has two roots. Therefore, equation (9) also has two roots:

4. Equations of the form

Equation, (11)

Where does not have a root x \u003d 0, therefore, dividing equation (11) by x2, we obtain the equation equivalent to it

Which, after replacing the unknown, will be rewritten in the form of a quadratic equation, the solution of which is not difficult.

Example 5. Solve the equation

Decision. Since h \u003d 0 is not the root of equation (12), dividing it by x2, we obtain the equation equivalent to it

Making the substitution unknown, we obtain the equation (y + 1) (y + 2) \u003d 2, which has two roots: y1 \u003d 0 and y1 \u003d -3. Therefore, the original equation (12) is equivalent to the set of equations

This set has two roots: x1 \u003d -1 and x2 \u003d -2.

Answer: x1 \u003d -1, x2 \u003d -2.

Comment. Equation of the form

In which, one can always bring to form (11) and, moreover, assuming α\u003e 0 and λ\u003e 0 to form.

5. Equations of the form

The equation

, (13) where the numbers, α, β, γ, δ, and Α are such that αβ \u003d γδ ≠ 0, we can rewrite it by multiplying the first bracket with the second, and the third with the fourth, in the form, i.e., equation (13) now written in the form (11), and its solution can be carried out in the same way as the solution of equation (11).

Example 6. Solve the equation

Decision. Equation (14) has the form (13), so we rewrite it in the form

Since x \u003d 0 is not a solution to this equation, dividing both its parts by x2, we obtain an equivalent initial equation. Making a change of variables, we obtain a quadratic equation, the solution of which is and. Therefore, the original equation (14) is equivalent to the set of equations and.

The solution to the first equation of this set is

The second equation of this set of solutions has no. So, the original equation has roots x1 and x2.

6. Equations of the form

The equation

(15) where the numbers a, b, c, q, A are such that, does not have a root x \u003d 0, therefore, dividing equation (15) by x2. we obtain an equation equivalent to it, which, after replacing the unknown, will be rewritten in the form of a quadratic equation, the solution of which is not difficult.

Example 7. The solution of the equation

Decision. Since x \u003d 0 is not the root of equation (16), then, dividing both its parts by x2, we obtain the equation

, (17) equivalent to equation (16). Having made the replacement unknown, we rewrite equation (17) in the form

The quadratic equation (18) has 2 roots: y1 \u003d 1 and y2 \u003d -1. Therefore, equation (17) is equivalent to a combination of equations and (19)

The set of equations (19) has 4 roots:,.

They will be the roots of equation (16).

§4. Rational equations

Equations of the form \u003d 0, where H (x) and Q (x) are polynomials, are called rational.

Having found the roots of the equation H (x) \u003d 0, then we need to check which of them are not the roots of the equation Q (x) \u003d 0. These roots and only they will be solutions to the equation.

Consider some methods for solving an equation of the form \u003d 0.

1. Equations of the form

The equation

(1) under certain conditions on numbers, it can be solved as follows. Grouping the terms of equation (1) in two and summing each pair, you need to get polynomials of the first or zero degree in the numerator, differing only in numerical factors, and in the denominators - trinomials with the same two terms containing x, then after changing the variables, the equation will either have also the form (1), but with a smaller number of terms, either will be equivalent to a combination of two equations, one of which will be the first degree, and the second will be an equation of the form (1), but with a smaller number of terms.

Example. Solve the equation

Decision. Grouping on the left side of equation (2) the first term with the last, and the second with the penultimate, we rewrite equation (2) in the form

Summing up the terms in each bracket, we rewrite equation (3) in the form

Since there is no solution to equation (4), dividing this equation by, we obtain the equation

, (5) equivalent to equation (4). We make a replacement for the unknown, then equation (5) can be rewritten in the form

Thus, the solution of equation (2) with five terms on the left side is reduced to the solution of equation (6) of the same form, but with three terms on the left side. Summing up all the terms on the left side of equation (6), we rewrite it in the form

The solutions to the equation are and. None of these numbers vanish the denominator of the rational function on the left side of equation (7). Therefore, equation (7) has these two roots, and therefore, the original equation (2) is equivalent to the set of equations

The solutions to the first equation of this set are

The solutions of the second equation from this set are

Therefore, the original equation has roots

2. Equations of the form

The equation

(8) under certain conditions on numbers it is possible to solve this way: it is necessary to select the integer part in each of the fractions of the equation, i.e., replace equation (8) with equation

Reduce it to the form (1) and then solve it by the method described in the previous paragraph.

Example. Solve the equation

Decision. We write equation (9) in the form or in the form

Summing up the terms in brackets, we rewrite equation (10) in the form

Making a replacement for the unknown, we rewrite equation (11) in the form

Summing the terms on the left side of equation (12), we rewrite it in the form

It is easy to see that equation (13) has two roots: and. Therefore, the original equation (9) has four roots:

3) Equations of the form.

An equation of the form (14) under certain conditions on numbers can be solved as follows: by expanding (if it is, of course, possible) each of the fractions on the left side of equation (14) into a sum of simple fractions

Reduce equation (14) to form (1), then, after conveniently rearranging the terms of the resulting equation, solve it by the method described in paragraph 1).

Example. Solve the equation

Decision. Since and, then, multiplying the numerator of each fraction in equation (15) by 2 and noting that equation (15) can be written as

Equation (16) has the form (7). Regrouping the terms in this equation, we rewrite it in the form or in the form

Equation (17) is equivalent to a combination of equations and

To solve the second equation of the set (18), we make the replacement of the unknown Then it will be rewritten in the form or in the form

Summing up all the terms on the left side of equation (19), rewrite it as

Since the equation has no roots, then equation (20) also does not have them.

The first equation of the population (18) has a single root Since this root is included in the ODZ of the second equation of the population (18), it is the only root of the population (18), and hence the original equation.

4. Equations of the form

The equation

(21) under certain conditions on the numbers and A, after presenting each term on the left side in the form, it can be reduced to the form (1).

Example. Solve the equation

Decision. We rewrite equation (22) in the form or in the form

Thus, equation (23) is reduced to the form (1). Now, grouping the first term with the last, and the second with the third, we rewrite equation (23) in the form

This equation is equivalent to the set of equations and. (24)

The last equation of the set (24) can be rewritten in the form

There are solutions to this equation, and since it is included in the ODZ of the second equation of the set (30), then the set (24) has three roots :. All of them are solutions to the original equation.

5. Equations of the form.

Equation of the form (25)

Under certain conditions, by replacing the unknown, the numbers can be reduced to an equation of the form

Example. Solve the equation

Decision. Since it is not a solution to equation (26), then dividing the numerator and denominator of each fraction on the left side by, we rewrite it in the form

After changing the variables, we rewrite equation (27) in the form

Solving equation (28) is and. Therefore, equation (27) is equivalent to the set of equations and. (29)