In general, an equation with a degree higher than 4 cannot be solved in radicals. But sometimes we can still find the roots of the polynomial on the left in the equation of the highest degree, if we represent it as a product of polynomials in degree at most 4. The solution to such equations is based on factoring a polynomial, so we advise you to repeat this topic before studying this article.
Most often one has to deal with equations of higher degrees with integer coefficients. In these cases, we can try to find the rational roots, and then factor the polynomial in order to then transform it into a lower-degree equation that is easy to solve. Within the framework of this material, we will consider just such examples.
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Equations of the highest degree with integer coefficients
All equations of the form a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 \u003d 0, we can reduce to an equation of the same degree by multiplying both sides by a n n - 1 and changing a variable of the form y \u003d a n x:
a n x n + a n - 1 x n - 1 +. ... ... + a 1 x + a 0 \u003d 0 ann xn + an - 1 ann - 1 xn - 1 +… + a 1 (an) n - 1 x + a 0 (an) n - 1 \u003d 0 y \u003d anx ⇒ yn + bn - 1 yn - 1 +… + b 1 y + b 0 \u003d 0
The resulting coefficients will also be whole. Thus, we will need to solve the reduced equation of the n-th degree with integer coefficients, which has the form x n + a n x n - 1 +… + a 1 x + a 0 \u003d 0.
Calculate the whole roots of the equation. If the equation has integer roots, you need to look for them among the divisors of the free term a 0. Let us write them down and substitute them into the original equality in turn, checking the result. Once we have obtained an identity and found one of the roots of the equation, we can write it in the form x - x 1 · P n - 1 (x) \u003d 0. Here x 1 is the root of the equation, and P n - 1 (x) is the quotient of dividing x n + a n x n - 1 +… + a 1 x + a 0 by x - x 1.
Substitute the rest of the divisors written out in P n - 1 (x) \u003d 0, starting with x 1, since the roots can be repeated. After obtaining the identity, the root x 2 is considered found, and the equation can be written as (x - x 1) (x - x 2) P n - 2 (x) \u003d 0. Here P n - 2 (x) will be the quotient of dividing P n - 1 (x) by x - x 2.
We continue to iterate over the divisors. Find all whole roots and denote their number as m. After that, the original equation can be represented as x - x 1 x - x 2 · ... · x - x m · P n - m (x) \u003d 0. Here P n - m (x) is a polynomial of degree n - m. It is convenient to use Horner's scheme for counting.
If our original equation has integer coefficients, we cannot end up with fractional roots.
As a result, we got the equation P n - m (x) \u003d 0, the roots of which can be found in any convenient way. They can be irrational or complex.
Let's show with a specific example how such a solution scheme is applied.
Example 1
Condition: find the solution to the equation x 4 + x 3 + 2 x 2 - x - 3 \u003d 0.
Decision
Let's start by finding whole roots.
We have a free term equal to minus three. It has divisors of 1, - 1, 3, and - 3. Let's substitute them in the original equation and see which of them will result in identities.
With x equal to one, we get 1 4 + 1 3 + 2 · 1 2 - 1 - 3 \u003d 0, which means that one will be the root of this equation.
Now we perform division of the polynomial x 4 + x 3 + 2 x 2 - x - 3 by (x - 1) in a column:
Hence, x 4 + x 3 + 2 x 2 - x - 3 \u003d x - 1 x 3 + 2 x 2 + 4 x + 3.
1 3 + 2 1 2 + 4 1 + 3 \u003d 10 ≠ 0 (- 1) 3 + 2 (- 1) 2 + 4 - 1 + 3 \u003d 0
We got an identity, which means that we have found another root of the equation, equal to - 1.
Divide the polynomial x 3 + 2 x 2 + 4 x + 3 by (x + 1) in a column:
We get that
x 4 + x 3 + 2 x 2 - x - 3 \u003d (x - 1) (x 3 + 2 x 2 + 4 x + 3) \u003d \u003d (x - 1) (x + 1) (x 2 + x + 3)
Substitute the next divisor into the equality x 2 + x + 3 \u003d 0, starting with - 1:
1 2 + (- 1) + 3 = 3 ≠ 0 3 2 + 3 + 3 = 15 ≠ 0 (- 3) 2 + (- 3) + 3 = 9 ≠ 0
The resulting equalities will be incorrect, which means that the equation no longer has integral roots.
The remaining roots will be the roots of the expression x 2 + x + 3.
D \u003d 1 2 - 4 1 3 \u003d - 11< 0
It follows from this that this square trinomial has no real roots, but has complex conjugate ones: x \u003d - 1 2 ± i 11 2.
Let's clarify that instead of long division, we can use Horner's scheme. This is done like this: after we have determined the first root of the equation, we fill in the table.
In the table of coefficients, we can immediately see the coefficients of the quotient of the division of polynomials, which means that x 4 + x 3 + 2 x 2 - x - 3 \u003d x - 1 x 3 + 2 x 2 + 4 x + 3.
After finding the next root equal to - 1, we get the following:
Answer: x \u003d - 1, x \u003d 1, x \u003d - 1 2 ± i 11 2.
Example 2
Condition: Solve the equation x 4 - x 3 - 5 x 2 + 12 \u003d 0.
Decision
The free term has divisors 1, - 1, 2, - 2, 3, - 3, 4, - 4, 6, - 6, 12, - 12.
We check them in order:
1 4 - 1 3 - 5 1 2 + 12 \u003d 7 ≠ 0 (- 1) 4 - (- 1) 3 - 5 (- 1) 2 + 12 \u003d 9 ≠ 0 2 4 2 3 - 5 2 2 + 12 \u003d 0
Hence, x \u003d 2 will be the root of the equation. Divide x 4 - x 3 - 5 x 2 + 12 by x - 2 using Horner's scheme:
As a result, we get x - 2 (x 3 + x 2 - 3 x - 6) \u003d 0.
2 3 + 2 2 - 3 2 - 6 \u003d 0
Hence, 2 will again be a root. Divide x 3 + x 2 - 3 x - 6 \u003d 0 by x - 2:
As a result, we get (x - 2) 2 (x 2 + 3 x + 3) \u003d 0.
It makes no sense to check the remaining divisors, since the equality x 2 + 3 x + 3 \u003d 0 is faster and more convenient to solve using the discriminant.
Let's solve the quadratic equation:
x 2 + 3 x + 3 \u003d 0 D \u003d 3 2 - 4 1 3 \u003d - 3< 0
We get a complex conjugate pair of roots: x \u003d - 3 2 ± i 3 2.
Answer: x \u003d - 3 2 ± i 3 2.
Example 3
Condition: find the real roots for the equation x 4 + 1 2 x 3 - 5 2 x - 3 \u003d 0.
Decision
x 4 + 1 2 x 3 - 5 2 x - 3 \u003d 0 2 x 4 + x 3 - 5 x - 6 \u003d 0
We perform multiplication 2 3 of both sides of the equation:
2 x 4 + x 3 - 5 x - 6 \u003d 0 2 4 x 4 + 2 3 x 3 - 20 2 x - 48 \u003d 0
Replace the variables y \u003d 2 x:
2 4 x 4 + 2 3 x 3 - 20 2 x - 48 \u003d 0 y 4 + y 3 - 20 y - 48 \u003d 0
As a result, we have a standard 4th degree equation that can be solved using the standard scheme. Let's check the divisors, divide and get in the end that it has 2 real roots y \u003d - 2, y \u003d 3 and two complex roots. We will not present the complete solution here. Due to the replacement, the real roots of this equation will be x \u003d y 2 \u003d - 2 2 \u003d - 1 and x \u003d y 2 \u003d 3 2.
Answer: x 1 \u003d - 1, x 2 \u003d 3 2
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Basic goals:
- To consolidate the concept of a whole rational equation of the th 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 by 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-seminar training system (lectures - explanation of new material, seminars - problem solving).
- Information and communication technologies (frontal survey, oral work with the class).
- Differentiated teaching, group and individual forms.
- The use of a research method in teaching aimed at developing the mathematical apparatus and thinking abilities of each specific student.
- Printed material - an individual short 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 - Study of a new topic (lecture). The purpose of the stage: to formulate the main methods for solving equations of higher degrees (n > 3)
- Summarizing.
The purpose of the stage: once again highlight the key points in the material studied in the lesson. - Homework.
The purpose of the stage: to formulate homework for the students.
Lesson summary
1. Organizational moment.
Formulation of the topic of the lesson: “Equations of the highest degrees. Methods for their solution ”.
2. Actualization of students' knowledge.
Theoretical survey - conversation. Repetition of some previously studied information from the theory. Students formulate basic definitions and formulate necessary theorems. Examples are given to demonstrate the level of knowledge acquired earlier.
- The concept of an equation in one variable.
- The concept of the root of an equation, solving an equation.
- The concept of a linear equation in one variable, the concept of a quadratic equation in one variable.
- The concept of equivalence of equations, equation-consequence (concept of extraneous roots), transition not by consequence (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 the whole rational equation. Reduced whole rational equation.
- Transition to a set of equations of lower degrees by factoring the original equation into factors.
- Polynomial concept n-th degree from x... Bezout's theorem. Consequences from Bezout's theorem. Root theorems ( Z-roots and Q-roots) of an entire rational equation with integer coefficients (reduced and non-reduced, respectively).
- Horner's scheme.
3. Studying a new topic.
We will consider the whole rational equation n-th degree 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 n-th degree from x, a n ≠ 0. If a a n \u003d 1 then such an equation is called the reduced whole rational equation n-th degree. Consider such equations for different values n and list the main methods for solving them.
n \u003d 1 - linear equation.
n \u003d 2 - quadratic equation. Discriminant formula. Formula for calculating roots. Vieta's theorem. Selecting a complete square.
n \u003d 3 - 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.
Reverse cubic equation of the form ax 3 + bx 2 + bx + a \u003d 0. 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 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. Equation given. Let us write down the divisors of the free term ( + 1; + 3; + 5; + 15). Let's apply Horner's scheme:
| x 3 | x 2 | x 1 | x 0 | output | |
| 1 | -9 | 23 | -15 | ||
| 1 | 1 | 1 x 1 - 9 \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 irreducible entire rational equation with integer coefficients.
Example: 9 x 3 + 27x 2 – x - 3 \u003d 0. The equation is not reduced. 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) Therefore, we will look for roots among the values \u200b\u200b( + 1; + ; + ; + 3). Let's apply Horner's scheme:
| x 3 | x 2 | x 1 | x 0 | output | |
| 9 | 27 | -1 | -3 | ||
| 1 | 9 | 1 x 9 + 27 \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 calculation when selecting Q -roots it can be convenient to make a change of variable, go to the reduced equation and select Z -roots.
- If the free term is 1
- If you can use a substitution of the form 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 Italian mathematicians Gerolamo Cardano (1501-1576), Nicolo Tartaglia (1500-1557), Scipione del Ferro (1465-1526). This formula is outside the scope of our course.
n \u003d 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 \u003d 0. Substitution y = x 2. From here y 1 = 4, y 2 \u003d -9. therefore x 1,2 = + 2 .
- Reverse equation of the fourth degree of the form ax 4 + bx 3 + c x 2 + bx + a = 0.
We solve by combining terms with the same coefficients by replacing the form
- ax 4 + bx 3 + cx 2 – bx + a = 0.
- Generalized fourth-degree return equation of the form ax 4 + bx 3 + cx 2 + kbx + k 2a \u003d 0.
- General view replacement. Some standard replacements.
Example 3 . Replacing the general view (follows from the form of a specific equation).
n = 3.
Equation with integer coefficients. Fitting Q-roots n = 3.
General formula. There is a universal method for solving equations of the fourth degree. This formula is associated with the name of Ludovico Ferrari (1522-1565). This formula is outside the scope of our course.
n > 5 - equations of the fifth and higher degrees.
Equation with integer coefficients. Selection of Z-roots based on the theorem. Horner's scheme. The algorithm is similar to that considered above for n = 3.
Equation with integer coefficients. Fitting Q-roots based on the theorem. Horner's scheme. The algorithm is similar to that considered above for n = 3.
Symmetric equations. Any return equation of odd degree has a root x \u003d -1 and after factoring it into factors we obtain that one factor has the form ( x + 1), and the second factor is the return equation of an even degree (its degree is one less than the degree of the original equation). Any recurrent 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 uniformity.
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 Evariste Galois (1811–1832 )).
- Let us recall again that in practice it is possible to use combinations the above methods. It is convenient to pass to a set of equations of lower degrees by factorization of the original equation.
- Widely used in practice remained outside the scope of our today's discussion. graphic 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 is reduced to showing that the equation has no roots. To prove this, we analyze the behavior of the functions under consideration on intervals of monotonicity. Example: equation x 8 – x 3 + 1 \u003d 0 has no roots.
- Using the monotonicity property of functions ... There are situations when the use of various properties of functions makes it possible to simplify the task.
Example 1: equation x 5 + 3x - 4 \u003d 0 has one root x\u003d 1. By the property of monotonicity of the analyzed functions, there are no other roots.
Example 2: equation x 4 + (x - 1) 4 \u003d 97 has roots x 1 \u003d -2 and x 2 \u003d 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 of 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 solution method in this case is the most effective, as well as correctly apply the chosen method.
5. Homework.
: p. 7, p. 164-174, No. 33-36, 39-44, 46.47.
: №№ 9.1–9.4, 9.6–9.8, 9.12, 9.14–9.16, 9.24–9.27.
Possible topics of reports or abstracts on this topic:
- Formula Cardano
- Graphical method for solving equations. Solution examples.
- Methods for the approximate solution of equations.
Analysis of the assimilation of the material and the interest of students in the topic:
Experience shows that students are primarily interested in the possibility of recruiting Z-roots and Q-roots of equations using a fairly simple algorithm using Horner's scheme. Students are also interested in various standard types of variable substitutions that can greatly simplify the problem. Graphical solution methods are usually of particular interest. In this case, you can additionally disassemble the tasks into a graphical method for solving equations; discuss the general view of the graph for a polynomial of 3, 4, 5 degrees; analyze how the number of roots of equations of 3, 4, 5 degrees is related to the type of the corresponding graph. Below is a list of books in which you can find additional information on this topic.
List of references:
- Vilenkin N.Ya. et al. “Algebra. A textbook for 9th grade students with in-depth study of mathematics ”- M., Education, 2007 - 367 p.
- Vilenkin N.Ya., Shibasov L.P., Shibasova Z.F. “Behind the pages of a mathematics textbook. Arithmetic. Algebra. 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. Textbook for grades 8-9 with in-depth study of mathematics ”- M., Education, 2008 - 301 p.
- Zvavich L.I. and others. “Algebra and the beginning of analysis. 8-11 cl. A manual for schools and classes with in-depth study of mathematics ”- M., Bustard, 1999 - 352 p.
- Zvavich L.I., Averyanov D.I., Pigarev B.P., Trushanina T.N. "Tasks in mathematics to prepare for the written exam in grade 9" - M., Education, 2007 - 112 p.
- Ivanov A.A., Ivanov A.P. “Thematic tests for the systematization of knowledge in mathematics” part 1 - M., Fizmatkniga, 2006 - 176 p.
- Ivanov A.A., Ivanov A.P. “Thematic tests for the systematization of 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 assignments in mathematics. Part 2-9 grade "- M., MCNMO, 2009 - 184 p.
- Makarychev Yu.N., Mindyuk N.G."Algebra. Additional chapters to the 9th grade school textbook. A textbook for students in schools and classes with advanced study of mathematics. " - M., Education, 2006 - 224 p.
- Mordkovich A.G. "Algebra. In-depth study. 8th grade. Textbook "- M., Mnemosina, 2006 - 296 p.
- Savin A.P. “Encyclopedic Dictionary of a Young Mathematician” - M., Pedagogy, 1985 - 352 p.
- Survillo G.S., Simonov A.S. “Didactic materials on algebra for grade 9 with in-depth study of mathematics” - M., Education, 2006 - 95 p.
- Chulkov P.V. “Equations and inequalities in the school mathematician course. Lectures 1–4 ”- M., September 1, 2006 - 88 p.
- Chulkov P.V. “Equations and inequalities in the school mathematician course. Lectures 5–8 ”- M., September 1, 2009 - 84 p.
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Introduction
Solving algebraic equations of higher degrees with one unknown is one of the most difficult and ancient mathematical problems. The most outstanding mathematicians of antiquity were engaged in these problems.
The solution of equations of the n-th degree is an important task for modern mathematics as well. Interest in them is quite large, since these equations are closely related to the search for the roots of equations that are not considered in the school curriculum in mathematics.
Problem: the lack of skills in solving equations of higher degrees in various ways among students prevents them from successfully preparing for the final certification in mathematics and mathematical olympiads, teaching in a specialized mathematical class.
The listed facts determined relevance of our work "Solving equations of higher degrees."
Possession of the simplest ways to solve equations of the n-th degree reduces the time to complete the task, on which the result of the work and the quality of the learning process depend.
Purpose of work: study of known methods for solving equations of higher degrees and identification of the most accessible of them for practical use.
Based on this goal, the work identified the following tasks:
Study literature and Internet resources on this topic;
Get acquainted with the historical facts regarding this topic;
Describe the different ways to solve higher-degree equations
compare the degree of complexity of each of them;
To acquaint classmates with methods of solving equations of higher degrees;
Create a set 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 - ways to solve equations of higher degrees.
Hypothesis:there is no general method and unified algorithm that allows finding solutions to equations of the n-th degree in a finite number of steps.
Research methods:
- bibliographic method (analysis of literature on the research topic);
- classification method;
- method of qualitative analysis.
Theoretical significanceresearch consists in the systematization of methods for solving equations of higher degrees and the description of their algorithms.
Practical significance - submitted material on this topic and the development of a textbook for students on this topic.
1 HIGHER DEGREE EQUATIONS
1.1 The concept of an 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, and , a 0 ≠ 0 .
Polynomial a 0 xⁿ + a 1 xn -1 + a 2 xⁿ - ² + ... + an -1 x + an is called a polynomial of degree n. The odds are distinguished by their names: a 0 - senior coefficient; an is a free member.
Definition 2. Solutions or roots for a given equation are all values \u200b\u200bof the variable x, which turn this equation into a 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 value of the variable xis also called the root of the polynomial. To solve an equation means to find all its roots or to establish that they do not exist.
If a a 0 \u003d 1, then such an equation is called the reduced whole rational equation n th degree.
For equations of the third and fourth degrees, there are Cardano and Ferrari formulas expressing 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 finding the roots of the equation is not an easy task. Nevertheless, in many special cases, this problem is solved to the end. Let's 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 scientists mastered the solution of a quadratic equation and solved systems of two equations, of which one is of the second degree. With the help of 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 that, being presented in ordinary language, may seem confusing and complicated ...
A universal formula for finding the roots of an algebraic equation nth degree no. Many, of course, came up with the tempting idea to find for any power n formulas that would express the roots of the equation in terms of its coefficients, that is, would solve the equation in radicals.
It was only in the 16th century that Italian mathematicians managed to advance further - to find formulas for n \u003d 3 and n \u003d 4. At the same time, Scipio, Dal, Ferro and his students Fiori and Tartaglia were engaged in the question of the general solution of equations of the third 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 of solving cubic equations were considered, as well as the method for solving equations of the 4th degree discovered by his student L. Ferrari.
F. Viet gave a complete exposition of questions related to the solution of equations of the third and fourth degrees.
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 in terms of radicals.
In the course of the study, it was revealed that modern science knows many ways to solve equations of the n-th 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 Vieta's theorem (for equations of degree n\u003e 2), Bezout's theorems, Horner's schemes, 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 became a discovery for us. These include - the method of indefinite coefficients, the selection of the full degree, symmetric equations.
2. SOLUTION OF INTEGER EQUATIONS OF HIGHER DEGREES WITH INTEGER COEFFICIENTS
2.1 Solving 3rd degree equations. Formula D. Cardano
Consider equations of the form x 3 + px + q \u003d 0.We transform the general equation to the form: x 3 + px 2 + qx + r \u003d 0. Let's write the formula for the sum cube; We add it to the original equality and replace it with y... We get the equation: y 3 + (q -) (y -) + (r - \u003d 0. After transformations, we have: y 2 + py + q \u003d 0. Now, again, write the formula for the sum cube:
(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 x, we get the equation x 3 - 3abx - (a 3 + b 3) = 0. Now you can see that the original equation is equivalent to the system: and Solving the system, we get:
We have obtained a formula for solving the reduced equation of the 3rd degree. She bears the name of the Italian mathematician Cardano.
Let's look at 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 cumbersome. Therefore, in practice, it is rarely used.
But the one who masters this formula can use it when solving third-degree equations on the exam.
2.2 Vieta's theorem
From the course of mathematics, we know this theorem for a quadratic equation, but few people know that it is also 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
; from this it follows that the following equalities can be written into the system:
The formulas derived by Viet for quadratic equations and demonstrated by us for equations of the third degree are also true for polynomials of higher degrees.
Let's solve the cubic equation:
Conclusion: this method is universal and easy enough for students to understand, since Vieta's theorem is familiar to them 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's 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 coefficients are integers, and the free term is nonzero, has an integer root, then this root is a divisor of the free term.
Considering that on the left side of the equation there is a polynomial of degree n, the theorem has a different interpretation.
Theorem. When dividing a polynomial of degree n with respect to x binomial x - a the remainder is equal to the value of the dividend at x \u003d a... (letter a can denote any real or imaginary number, i.e. any complex number).
Evidence: let be f (x) denotes an arbitrary n-th degree polynomial with respect to the variable x, and let, when dividing by the binomial ( x-a) happened in private q (x), and in the remainder R... It's obvious that q (x) there will be some polynomial (n - 1) th degree with respect to xand 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, then this would mean that the division is not satisfied. So, R from x does not depend. By the definition of division, we get the identity: f (x) \u003d (x-a) q (x) + R.
Equality is valid for any value of x, which means that it is also valid for x \u003d a, we 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 the value q (x) at x \u003d a.The remainder R remained the same as it 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 factor q (a) there is a certain number. Therefore, from the equality we get: f (a) \u003d R,h.t.d.
Example 1.Find the remainder of the division of a polynomial x 3 - 3x 2 + 6x-5 on a binomial
x-2. By Bezout's theorem : R \u003d f(2) = 23-322 + 62 -5 \u003d 3. Answer: R \u003d3.
Note that Bezout's theorem is important not so much in itself as in its consequences. (Attachment 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 Bezout's theorem, it is necessary:
Find all integer divisors of the free term;
Find at least one root of the equation from these divisors;
Divide the left side of the equation by (Ha);
Write down the product of the divisor and the quotient on the left side of the equation;
Solve the resulting equation.
Consider, for example, solving the equation x 3 + 4x 2 + x -6 = 0 .
Solution: find the divisors of the free term ± 1 ; ± 2; ± 3; ± 6. Let us calculate the values \u200b\u200bat x \u003d1, 1 3 + 41 2 + 1- 6 \u003d 0. Divide the left side of the equation by ( x-1). We will perform the division "with a corner", we get:
Conclusion: Bezout's theorem, one of the ways that we consider in our work, is studied in the program of optional classes. It is difficult to understand, because in order to own it, you need to know all the consequences from it, but at the same time Bezout's theorem is one of the main assistants of students on the exam.
2.4 Horner's scheme
To divide a polynomial by a binomial x-α you can use a special simple trick invented by English mathematicians of the 17th century, later called Horner's scheme. In addition to finding the roots of equations, according to Horner's scheme, it is easier to calculate their values. For this, it is necessary to substitute the value of the variable into the polynomial Pn (x) \u003d a 0 xn + a 1 x n-1 + a 2 xⁿ - ² +… ++ an -1 x + an. (one)
Consider the division of the polynomial (1) by the binomial x-α.
Let us express the coefficients of the incomplete quotient b 0 xⁿ - ¹+ b 1 xⁿ - ²+ b 2 xⁿ - ³+…+ bn -1 and the remainder r in terms of the coefficients of the polynomial Pn ( x) and the 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 .
Calculations according to Horner's scheme 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, Horner's scheme can be applied to the quotient b 0 x +b 1 x + ... +bn -1 according to the table. If in the column below bn -1 it turns out 0 again, which means that α is a multiple root.
Consider an example: Solve the equation x 3 + 4x 2 + x -6 = 0.
Apply the factorization of the polynomial on the left side of the equation, Horner's scheme to the left side of the equation.
Solution: find the divisors of the free term ± 1; ± 2; ±3; ± 6.
|
6 ∙ 1 + (-6) = 0 |
The quotients are numbers 1, 5, 6, and the remainder is r \u003d 0.
Hence, x 3 + 4x 2 + x - 6 = (x - 1) (x 2 + 5x + 6) = 0.
Hence: 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 have shown the use of two different methods of factoring polynomials. In our opinion, Horner's scheme is the most practical and economical.
2.5 Solving equations of the 4th degree. Ferrari method
Cardano's student Ludovic Ferrari discovered a way to solve a 4th degree equation. The Ferrari method consists of two stages.
Stage I: equations of the form are represented as a 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, but 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: Let us isolate the 4th degree, we get: For any dthe expression will be a perfect square. Add to both sides of the equation we get
On the left side there is a full square, you can pick up dso that the right-hand side (2) also becomes a complete square. Let's imagine that we have achieved this. Then our equation looks like this:
Finding the root afterwards will not be difficult. To choose the right one d it is necessary that the discriminant of the right-hand side (3) vanish, i.e.
So to find d, it is necessary to solve this equation of the 3rd degree. Such an auxiliary equation is called resolution.
We can easily find the whole root of the resolvent: d \u003d1
Substituting the equation into (1), we obtain
Conclusion: Ferrari's method is universal, but complicated and cumbersome. At the same time, if the solution algorithm is clear, then 4th degree equations can be solved by this method.
2.6 Method of undefined coefficients
The success of solving the equation of the 4th degree by the Ferrari method depends on whether we solve the resolvent - the equation of the 3rd degree, which, as we know, is not always possible.
The essence of the method of undefined coefficients is that the type of factors into which a given polynomial is decomposed is guessed, and the coefficients of these factors (also polynomials) are determined by multiplying the factors and equating the coefficients at the same degrees of the variable.
Example: Solve the equation:
Suppose that the left-hand side of our equation can be decomposed into two square trinomials with integer coefficients such that the identity
Obviously, the coefficients in front of the uni should be equal to 1, and the free terms should be equal to one + 1, the other has 1.
The coefficients in front of x... We denote them by and and and to determine them, we multiply both trinomials on 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 method of indefinite coefficients 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 the same degrees x.
2.7 Symmetric Equations
Definition.An equation of the form is called symmetric if the first coefficients on the left in the equation 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 the root x= - 1. Next, we can lower the degree of the equation by dividing it by ( x +one). It turns out that when the symmetric equation is divided by ( x +1) a symmetric equation of even degree is obtained. 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) \u003d 0.
Let us group the terms with symmetric
) + 3(x +. We denote at= x +, let's 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, let's go back to replacing 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 equations is not often encountered, but if you come across it, then it can be solved easily and simply without resorting to cumbersome calculations.
2.8 Isolation 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 get
Where is the only root.
RESULTS OF THE STUDY
Based on the results of the work, we came to the following conclusions:
Thanks to the studied theory, we got acquainted with various methods for solving entire equations of higher degrees;
D. Cardano's formula is difficult to apply and gives a high probability of making errors in the calculation;
- L. Ferrari's method makes it possible to reduce the solution of an equation of the fourth degree to a cubic one;
- Bezout's theorem can be applied both for cubic equations and for equations of the fourth degree; it is more understandable and visual when applied to the solution of 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, it is easier to calculate the values \u200b\u200bof the polynomials on the left side of the equation;
Of particular interest was the solution of equations by the method of undefined coefficients, the solution of symmetric equations.
In the course of the research work, it was found that students get acquainted with the simplest methods of solving equations of the highest degree in the 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 an increased interest in the subject of "mathematics".
The most popular methods for solving equations of higher degrees, which are found in solving Olympiad, competitive problems and as a result of student preparation for exams, are methods based on the application of Bezout's theorem, Horner's scheme and the introduction of a new variable.
Demonstration of the results of research work, i.e. ways to solve equations that are not studied in the school curriculum in mathematics, interested classmates.
Conclusion
Having studied educational and scientific literature, Internet resources in youth educational forums
Marina A. Trifanova
teacher of mathematics, MOU "Gymnasium No. 48 (multidisciplinary)", Talnakh
The triune goal of the lesson:
Educational:
systematization and generalization of knowledge on the solution of equations of higher degrees.
Developing:
to promote the development of logical thinking, the ability to work independently, skills of mutual control and self-control, the ability to speak and listen.
Educational:
developing a habit of constant employment, fostering responsiveness, hard work, accuracy.
Lesson type:
a lesson in the complex application of knowledge, skills and abilities.
Lesson form:
airing, physical training, various forms of work.
Equipment:
supporting notes, cards with assignments, lesson monitoring matrix.
DURING THE CLASSES
I. Organizational moment
- Communicating the lesson goal to students.
- Homework check (Appendix 1). Work with reference notes (Appendix 2).
Equations and answers for each of them are written on the board. Students check the answers and give a quick analysis of the solution to each equation or answer the teacher's questions (frontal survey). Self-control - students give themselves grades and hand over exercise books to the teacher to correct or approve grades. The grade school is written on the chalkboard:
“5+” - 6 equations;
“5” - 5 equations;
“4” - 4 equations;
“3” - 3 equations.
Teacher's homework questions:
1 equation
- What change of variables is made in the equation?
- What equation is obtained after changing variables?
2 equation
- What polynomial did both sides of the equation divide into?
- What change of variables was obtained?
3 equation
- Which polynomials need to be multiplied to simplify the solution of this equation?
4 equation
- Name the function f (x).
- How were the rest of the roots found?
Equation 5
- How many gaps were obtained to solve the equation?
6 equation
- In what ways could this equation be solved?
- Which solution is more rational?
II. Group work 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: "Deconstruct the proposed method for solving the equation and explain it using this example."
- Group work 15 minutes.
- Examples are written on the board (the board is divided into 4 parts).
- The group report takes 2 - 3 minutes.
- The teacher corrects the reports of the groups and helps in case of difficulty.
The group work continues on cards 5 - 8. Each equation is given 5 minutes for group discussion. Then there is a report on this equation at the board - a short analysis of the solution. The equation may not be completely solved - it is being finalized at home, but the sequence of its solution in the classroom is discussed all over.
III. Independent work.Appendix 4.
- Each student receives an individual assignment.
- Time work takes 20 minutes.
- 5 minutes before the end of the lesson, the teacher gives open-ended answers for each equation.
- Students change notebooks in a circle and check the answers from a friend. Give marks.
- Notebooks are handed over to the teacher for checking and correcting grades.
IV. Lesson summary.
Homework.
Checkout the solution to unfinished equations. Prepare for a control slice.
Grading.
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Slide captions:
Equations of higher degrees (roots of a polynomial in one variable).
Plan lectures. No. 1. Equations of the highest degrees in the school mathematics course. No. 2. Standard form of a polynomial. № 3. Integral roots of a polynomial. Horner's scheme. № 4. Fractional roots of a polynomial. № 5. Equations of the form: (x + a) (x + b) (x + c)… \u003d A № 6. Return equations. № 7. Homogeneous equations. № 8. The method of undefined coefficients. № 9. Functional - graphic method. № 10. Vieta's formulas for equations of higher degrees. № 11. Non-standard methods for solving equations of higher degrees.
Equations of higher degrees in the school mathematics course. 7th grade. Standard form of a polynomial. Actions with polynomials. Factoring a polynomial. In the regular class 42 hours, in the special class 56 hours. 8 special class. Integer roots of a polynomial, division of polynomials, recurrent equations, difference and sum of n - th powers of a binomial, method of undefined coefficients. Yu.N. Makarychev "Additional chapters to the school course of 8th grade algebra", MLGalitsky Collection of problems in algebra 8th - 9th grade. 9 special class. Rational roots of a polynomial. Generalized return equations. Vieta's formulas for equations of higher degrees. N. Ya. Vilenkin “Grade 9 algebra with in-depth study. 11 special class. Identity of polynomials. Polynomial in several variables. Functional - graphical method for solving equations of higher degrees.
Standard form of a polynomial. The polynomial P (x) \u003d a ⁿ x ⁿ + a n-1 x n-1 +… + a₂x ² + a₁x + a₀. It is called a standard polynomial. a n x ⁿ is the highest term of the polynomial and n is the coefficient of the highest term of the polynomial. For a n \u003d 1, P (x) is called a reduced polynomial. and ₀ is the free term of the polynomial P (x). n is the degree of the polynomial.
Integer roots of a polynomial. Horner's scheme. Theorem 1. If an integer a is a root of the polynomial P (x), then a is a divisor of the free term P (x). Example No. 1. Solve the equation. X⁴ + 2x³ \u003d 11x² - 4x - 4 Let us bring the equation to its standard form. X⁴ + 2x³ - 11x² + 4x + 4 \u003d 0. We have the polynomial P (x) \u003d x ⁴ + 2x³ - 11x² + 4x + 4 The divisions of the free term: ± 1, ± 2, ± 4. x \u003d 1 root of the equation because P (1) \u003d 0, x \u003d 2 is the root of the equation since P (2) \u003d 0 Bezout's theorem. The remainder of dividing the polynomial P (x) by the binomial (x - a) is equal to P (a). Consequence. If a is a root of the polynomial P (x), then P (x) is divisible by (x - a). In our equation, P (x) is divisible by (x - 1) and by (x - 2), and hence by (x - 1) (x - 2). When dividing P (x) by (x ² - 3x + 2) in the quotient, we get a trinomial x ² + 5x + 2 \u003d 0, which has roots x \u003d (- 5 ± √17) / 2
Fractional roots of a polynomial. Theorem # 2. If p / g is the root of the polynomial P (x), then p is the divisor of the free term, g is the divisor of the coefficient of the leading term P (x). Example # 2. Solve the equation. 6x³ - 11x² - 2x + 8 \u003d 0. Free term divisors: ± 1, ± 2, ± 4, ± 8. None of these numbers satisfy the equation. There are no intact roots. Natural divisors of the coefficient of the leading term P (x): 1, 2, 3, 6. Possible fractional roots of the equation: ± 2/3, ± 4/3, ± 8/3. By checking, we make sure that P (4/3) \u003d 0. X \u003d 4/3 is the root of the equation. According to Horner's scheme, we divide P (x) by (x - 4/3).
Examples for an independent solution. Solve the equations: 9x³ - 18x \u003d x - 2, x ³ - x ² \u003d x - 1, x ³ - 3x² -3x + 1 \u003d 0, X ⁴ - 2x³ + 2x - 1 \u003d 0, X⁴ - 3x² + 2 \u003d 0 , x ⁵ + 5x³ - 6x² \u003d 0, x ³ + 4x² + 5x + 2 \u003d 0, X⁴ + 4x³ - x ² - 16x - 12 \u003d 0 4x³ + x ² - x + 5 \u003d 0 3x⁴ + 5x³ - 9x² - 9x + 10 \u003d 0. Answers: 1) ± 1/3; 2 2) ± 1, 3) -1; 2 ± √3, 4) ± 1, 5) ± 1; ± √2, 6) 0; 1 7) -2; -1, 8) -3; -one; ± 2, 9) - 5/4 10) -2; - 5/3; one.
Equations of the form (x + a) (x + b) (x + c) (x + d)… \u003d A. Example №3. Solve the equation (x + 1) (x + 2) (x + 3) (x + 4) \u003d 24. a \u003d 1, b \u003d 2, c \u003d 3, d \u003d 4 a + d \u003d b + c. We multiply the first bracket with the fourth and the second with the third. (x + 1) (x + 4) (x + 20 (x + 3) \u003d 24. (x ² + 5x + 4) (x ² + 5x + 6) \u003d 24. Let x ² + 5x + 4 \u003d y , then y (y + 2) \u003d 24, y² + 2y - 24 \u003d 0 y₁ \u003d - 6, y₂ \u003d 4.x² + 5x + 4 \u003d -6 or x² + 5x + 4 \u003d 4.x² + 5x + 10 \u003d 0, D
Examples for an independent solution. (x + 1) (x + 3) (x + 5) (x + 7) \u003d -15, x (x + 4) (x + 5) (x + 9) + 96 \u003d 0, x (x + 3 ) (x + 5) (x + 8) + 56 \u003d 0, (x - 4) (x - 3) (x - 2) (x - 1) \u003d 24, (x - 3) (x -4) ( x - 5) (x - 6) \u003d 1680, (x ² - 5x) (x + 3) (x - 8) + 108 \u003d 0, (x + 4) ² (x + 10) (x - 2) + 243 \u003d 0 (x ² + 3x + 2) (x ² + 9x + 20) \u003d 4, Indication: x + 3x + 2 \u003d (x + 1) (x + 2), x ² + 9x + 20 \u003d (x + 4) (x + 5) Answers: 1) -4 ± √6; - 6; - 2. 6) - 1; 6; (5 ± √97) / 2 7) -7; -one; -4 ± √3.
Reflexive equations. Definition # 1. An equation of the form: ax⁴ + in ³ + cx ² + in + a \u003d 0 is called the return equation of the fourth degree. Definition number 2. An equation of the form: ah⁴ + bx ³ + cx ² + bx + k² a \u003d 0 is called a generalized return equation of the fourth degree. k² a: a \u003d k²; kv: v \u003d k. Example No. 6. Solve the equation x ⁴ - 7x³ + 14x² - 7x + 1 \u003d 0. Divide both sides of the equation by x ². x ² - 7x + 14 - 7 / x + 1 / x ² \u003d 0, (x ² + 1 / x ²) - 7 (x + 1 / x) + 14 \u003d 0. Let x + 1 / x \u003d y. We square both sides of the equality. x ² + 2 + 1 / x ² \u003d y², x ² + 1 / x ² \u003d y² - 2. We get the quadratic equation y² - 7y + 12 \u003d 0, y₁ \u003d 3, y₂ \u003d 4.x + 1 / x \u003d 3 or x + 1 / x \u003d 4. We get two equations: x ² - 3x + 1 \u003d 0, x ² - 4x + 1 \u003d 0. Example №7. 3x⁴ - 2x³ - 31x² + 10x + 75 \u003d 0. 75: 3 \u003d 25, 10: (- 2) \u003d -5, (-5) ² \u003d 25. The condition of the generalized return equation is fulfilled k \u003d -5. Solved similarly to example No. 6. Divide both sides of the equation by x ². 3x⁴ - 2x - 31 + 10 / x + 75 / x ² \u003d 0.3 (x ⁴ + 25 / x ²) - 2 (x - 5 / x) - 31 \u003d 0. Let x - 5 / x \u003d y, we square both sides of the equality x ² - 10 + 25 / x ² \u003d y², x ² + 25 / x ² \u003d y² + 10. We have the quadratic equation 3y² - 2y - 1 \u003d 0, y₁ \u003d 1, y₂ \u003d - 1 / 3. x - 5 / x \u003d 1 or x - 5 / x \u003d -1/3. We get two equations: x ² - x - 5 \u003d 0 and 3x² + x - 15 \u003d 0
Examples for an independent solution. 1.78x⁴ - 133x³ + 78x² - 133x + 78 \u003d 0, 2.x - 5x³ + 10x² - 10x + 4 \u003d 0, 3.x ⁴ - x ³ - 10x² + 2x + 4 \u003d 0, 4.6x⁴ + 5x³ - 38x² -10x + 24 \u003d 0.5 x ⁴ + 2x³ - 11x² + 4x + 4 \u003d 0.6 x ⁴ - 5x³ + 10x² -10x + 4 \u003d 0. Answers: 1) 2/3; 3/2, 2) 1; 2 3) -1 ± √3; (3 ± √17) / 2, 4) -1 ± √3; (7 ± √337) / 12 5) 1; 2; (-5 ± √17) / 2, 6) 1; 2.
Homogeneous equations. Definition. An equation of the form a₀ u³ + a₁ u² v + a₂ uv² + a₃ v³ \u003d 0 is called a homogeneous equation of the third degree with respect to u v. Definition. An equation of the form a₀ u⁴ + a₁ u³v + a₂ u²v² + a₃ uv³ + a₄ v⁴ \u003d 0 is called a homogeneous equation of the fourth degree with respect to u v. Example No. 8. Solve the equation (x ² - x + 1) ³ + 2x⁴ (x ² - x + 1) - 3x⁶ \u003d 0 Homogeneous equation of the third degree with respect to u \u003d x ²- x + 1, v \u003d x ². We divide both sides of the equation by x ⁶. We checked beforehand that x \u003d 0 is not the root of the equation. (x ² - x + 1 / x ²) ³ + 2 (x ² - x + 1 / x ²) - 3 \u003d 0. (x ² - x + 1) / x ²) \u003d y, y³ + 2y - 3 \u003d 0, y \u003d 1 root of the equation. We divide the polynomial P (x) \u003d y³ + 2y - 3 by y - 1 according to Horner's scheme. In the quotient, we get a trinomial that has no roots. Answer: 1.
Examples for an independent solution. 1.2 (x² + 6x + 1) ² + 5 (X² + 6X + 1) (X² + 1) + 2 (X² + 1) ² \u003d 0, 2. (X + 5) ⁴ - 13X² (X + 5) ² + 36X⁴ \u003d 0, 3.2 (X² + X + 1) ² - 7 (X - 1) ² \u003d 13 (X³ - 1), 4.2 (X -1) ⁴ - 5 (X² - 3X + 2) ² + 2 (x - 2) ⁴ \u003d 0, 5. (x ² + x + 4) ² + 3x (x ² + x + 4) + 2x² \u003d 0, Answers: 1) -1; -2 ± √3, 2) -5/3; -5/4; 5/2; 5 3) -1; -1/2; 2; 4 4) ± √2; 3 ± √2, 5) No roots.
The method of undefined coefficients. Theorem №3. Two polynomials P (x) and G (x) are identical if and only if they have the same degree and the coefficients of the same degrees of the variable in both polynomials are equal. Example No. 9. Factor the polynomial y⁴ - 4y³ + 5y² - 4y + 1.y⁴ - 4y³ + 5y² - 4y + 1 \u003d (y2 + wu + c) (y2 + v₁y + c₁) \u003d y ⁴ + y³ (b₁ + b) + y² (s₁ + s + b₁v) + y (sun ₁ + sv ₁) + ss ₁. According to theorem №3 we have a system of equations: s₁ + s \u003d -4, s₁ + s + s₁v \u003d 5, ss ₁ + sv св \u003d -4, ss ₁ \u003d 1. It is necessary to solve the system in integers. The last equation in integers can have solutions: c \u003d 1, c₁ \u003d 1; c \u003d -1, c₁ \u003d -1. Let с \u003d с ₁ \u003d 1, then from the first equation we have в₁ \u003d -4 –в. We substitute in the second equation of the system в² + 4в + 3 \u003d 0, в \u003d -1, в₁ \u003d -3 or в \u003d -3, в₁ \u003d -1. These values \u200b\u200bfit the third equation in the system. With c \u003d c ₁ \u003d -1 D
Example No. 10. Factor the polynomial y³ - 5y + 2. y³ -5y + 2 \u003d (y + a) (y² + wu + c) \u003d y³ + (a + b) y² + (ab + c) y + ac. We have a system of equations: a + b \u003d 0, ab + c \u003d -5, ac \u003d 2. Possible integer solutions of the third equation: (2; 1), (1; 2), (-2; -1), (-1 ; -2). Let a \u003d -2, c \u003d -1. From the first equation of the system в \u003d 2, which satisfies the second equation. Substituting these values \u200b\u200binto the desired equality, we get the answer: (y - 2) (y² + 2y - 1). Second way. У³ - 5у + 2 \u003d у³ -5у + 10 - 8 \u003d (у³ - 8) - 5 (у - 2) \u003d (у - 2) (у² + 2у -1).
Examples for an independent solution. Factor the polynomials: 1.y⁴ + 4y³ + 6y² + 4y -8, 2.y⁴ - 4y³ + 7y² - 6y + 2, 3. x ⁴ + 324, 4.y⁴ -8y³ + 24y² -32y + 15.5. Solve the equation using the factorization method: a) x ⁴ -3x² + 2 \u003d 0, b) x ⁵ + 5x³ -6x² \u003d 0. Answers: 1) (y² + 2y -2) (y² + 2y +4), 2) (y - 1) ² (y² -2y + 2), 3) (x ² -6x + 18) (x ² + 6x + 18), 4) (y - 1) (y - 3) (y² - 4y + 5), 5a) ± 1; ± √2, 5b) 0; one.
Functional - graphical method for solving equations of higher degrees. Example No. 11. Solve the equation x ⁵ + 5x -42 \u003d 0. Function y \u003d x ⁵ increasing, function y \u003d 42 - 5x decreasing (k
Examples for an independent solution. 1. Using the property of monotonicity of the function, prove that the equation has a single root, and find this root: a) x ³ \u003d 10 - x, b) x ⁵ + 3x³ - 11√2 - x. Answers: a) 2, b) √2. 2. Solve the equation using the functional-graphic method: a) x \u003d ³ √x, b) l x l \u003d ⁵ √x, c) 2 \u003d 6 - x, d) (1/3) \u003d x +4, d ) (x - 1) ² \u003d log₂ x, e) log \u003d (x + ½) ², g) 1 - √x \u003d ln x, h) √x - 2 \u003d 9 / x. Answers: a) 0; ± 1, b) 0; 1, c) 2, d) -1, e) 1; 2, f) 1, g) 1, h) 9.
Vieta's formulas for equations of higher degrees. Theorem 5 (Vieta's theorem). If the equation a x ⁿ + a x ⁿ +… + a₁x + a₀ has n different real roots x ₁, x ₂,…, x, then they satisfy the equalities: For a quadratic equation ax² + bx + c \u003d o: x ₁ + x ₂ \u003d -v / a, x₁x ₂ \u003d s / a; For the cubic equation a₃x ³ + a₂x ² + a₁x + a₀ \u003d o: x ₁ + x ₂ + x ₃ \u003d -a₂ / a₃; x₁x ₂ + x₁x ₃ + x₂x ₃ \u003d a₁ / a₃; х₁х₂х ₃ \u003d -а₀ / а₃; …, For the n-th degree equation: x ₁ + x ₂ +… x \u003d - a / a, x₁x ₂ + x₁x ₃ +… + x x \u003d a / a,…, x₁x ₂ ... · x \u003d (- 1 ) ⁿ a₀ / a. The converse theorem is also true.
Example No. 13. Write a cubic equation, the roots of which are inverse to the roots of the equation x ³ - 6x² + 12x - 18 \u003d 0, and the coefficient at x ³ is 2. 1. By Vieta's theorem for the cubic equation we have: x ₁ + x ₂ + x ₃ \u003d 6, x₁x ₂ + x₁x ₃ + x₂x ₃ \u003d 12, x₁x₂x ₃ \u003d 18. 2. We compose the reciprocals of the given roots and for them we apply the inverse Vieta theorem. 1 / x ₁ + 1 / x ₂ + 1 / x ₃ \u003d (x₂x ₃ + x₁x ₃ + x₁x ₂) / x₁x₂x ₃ \u003d 12/18 \u003d 2/3. 1 / x₁x ₂ + 1 / x₁x ₃ + 1 / x₂x ₃ \u003d (x ₃ + x ₂ + x ₁) / x₁x₂x ₃ \u003d 6/18 \u003d 1/3, 1 / x₁x₂x ₃ \u003d 1/18. We get the equation x ³ + 2 / 3x² + 1 / 3x - 1/18 \u003d 0 · 2 Answer: 2x³ + 4 / 3x² + 2 / 3x -1/9 \u003d 0.
Examples for an independent solution. 1. Write a cubic equation, the roots of which are inverse to the squares of the roots of the equation x ³ - 6x² + 11x - 6 \u003d 0, and the coefficient at x ³ is 8. Answer: 8x³ - 98 / 9x² + 28 / 9x -2/9 \u003d 0. Non-standard methods for solving equations of higher degrees. Example No. 12. Solve the x ⁴ -8x + 63 \u003d 0 equation. Factor the left side of the equation. Let's select the exact squares. X⁴ - 8x + 63 \u003d (x ⁴ + 16x² + 64) - (16x² + 8x + 1) \u003d (x² + 8) ² - (4x + 1) ² \u003d (x ² + 4x + 9) (x ² - 4x + 7) \u003d 0. Both discriminants are negative. Answer: no roots.
Example No. 14. Solve the equation 21x³ + x² - 5x - 1 \u003d 0. If the free term of the equation is ± 1, then the equation is transformed into the reduced equation by replacing x \u003d 1 / y. 21 / у³ + 1 / у² - 5 / у - 1 \u003d 0 · у³, у³ + 5у² -у - 21 \u003d 0. у \u003d -3 is the root of the equation. (y + 3) (y2 + 2y -7) \u003d 0, y \u003d -1 ± 2√2. x ₁ \u003d -1/3, x ₂ \u003d 1 / -1 + 2√2 \u003d (2√2 + 1) / 7, X₃ \u003d 1 / -1 -2√2 \u003d (1-2√2) / 7 ... Example No. 15. Solve the equation 4x³-10x² + 14x - 5 \u003d 0. Multiply both sides of the equation by 2. 8x³ -20x² + 28x - 10 \u003d 0, (2x) ³ - 5 (2x) ² + 14 · (2x) -10 \u003d 0. We introduce a new variable y \u003d 2x, we get the reduced equation y³ - 5y² + 14y -10 \u003d 0, y \u003d 1 is the root of the equation. (y - 1) (y² - 4y + 10) \u003d 0, D
Example No. 16. Prove that the equation x ⁴ + x ³ + x - 2 \u003d 0 has one positive root. Let f (x) \u003d x ⁴ + x ³ + x - 2, f ’(x) \u003d 4x³ + 3x² + 1\u003e o for x\u003e o. The function f (x) is increasing for x\u003e o, and the value f (o) \u003d -2. Obviously, the equation has one positive ch.t.d. root. Example No. 17. Solve the equation 8x (2x² - 1) (8x⁴ - 8x² + 1) \u003d 1. IF Sharygin "Optional course in mathematics for grade 11" .M. Enlightenment 1991 p90. 1. l x l 1 2x² - 1\u003e 1 and 8x⁴ -8x² + 1\u003e 1 2. Make the substitution x \u003d cozy, y € (0; n). For other values \u200b\u200bof y, the values \u200b\u200bof x are repeated, and the equation has no more than 7 roots. 2x² - 1 \u003d 2 cos²y - 1 \u003d cos2y, 8x⁴ - 8x² + 1 \u003d 2 (2x² - 1) ² - 1 \u003d 2 cos²2y - 1 \u003d cos4y. 3. The equation becomes 8 cosycos2ycos4y \u003d 1. Multiply both sides of the equation by siny. 8 sinycosycos2ycos4y \u003d siny. Applying the double angle formula 3 times, we get the equation sin8y \u003d siny, sin8y - siny \u003d 0
Completion of the solution of example No. 17. Apply the sine difference formula. 2 sin7y / 2 cos9y / 2 \u003d 0. Taking into account that y € (0; n), y \u003d 2nk / 3, k \u003d 1, 2, 3 or y \u003d n / 9 + 2nk / 9, k \u003d 0, 1, 2, 3. Returning to the variable x, we obtain answer: Cos2 n / 7, cos4 n / 7, cos6 n / 7, cos n / 9, ½, cos5 n / 9, cos7 n / 9. Examples for an independent solution. Find all values \u200b\u200bof a for which the equation (x ² + x) (x ² + 5x + 6) \u003d a has exactly three roots. The answer is 9/16. Hint: Graph the left side of the equation. F max \u003d f (0) \u003d 9/16. The straight line y \u003d 9/16 intersects the graph of the function at three points. Solve the equation (x ² + 2x) ² - (x + 1) ² \u003d 55. Answer: -4; 2. Solve the equation (x + 3) ⁴ + (x + 5) ⁴ \u003d 16. Answer: -5; -3. Solve the equation 2 (x ² + x + 1) ² -7 (x - 1) ² \u003d 13 (x ³ - 1). Answer: -1; -1/2, 2; 4 Find the number of real roots of the equation x ³ - 12x + 10 \u003d 0 on [-3; 3/2]. Hint: find the derivative and investigate for monot.
Examples for independent solution (continued). 6. Find the number of real roots of the equation x ⁴ - 2x³ + 3/2 \u003d 0. Answer: 2 7. Let x ₁, x ₂, x ₃ be the roots of the polynomial P (x) \u003d x ³ - 6x² -15x + 1. Find X₁² + x ₂² + x ₃². Answer: 66. Direction: Apply Vieta's theorem. 8. Prove that for a\u003e o and arbitrary real in the equation x ³ + ax + b \u003d o has only one real root. Hint: Carry out proof by contradiction. Apply Vieta's theorem. 9. Solve the equation 2 (x ² + 2) ² \u003d 9 (x ³ + 1). Answer: ½; one; (3 ± √13) / 2. Hint: reduce the equation to a homogeneous one using the equalities X² + 2 \u003d x + 1 + x ² - x + 1, x ³ + 1 \u003d (x + 1) (x ² - x + 1). 10. Solve the system of equations x + y \u003d x ², 3y - x \u003d y². Answer: (0; 0), (2; 2), (√2; 2 - √2), (- √2; 2 + √2). 11. Solve the system: 4y² -3xy \u003d 2x -y, 5x² - 3y² \u003d 4x - 2y. Answer: (o; o), (1; 1), (297/265; - 27/53).
Test. Option 1. 1. Solve the equation (x ² + x) - 8 (x ² + x) + 12 \u003d 0. 2. Solve the equation (x + 1) (x + 3) (x + 5) (x + 7) \u003d - 15 3. Solve the equation 12x² (x - 3) + 64 (x - 3) ² \u003d x ⁴. 4. Solve the equation x ⁴ - 4x³ + 5x² - 4x + 1 \u003d 0 5. Solve the system of equations: x² + 2y² - x + 2y \u003d 6, 1.5x² + 3y² - x + 5y \u003d 12.
Option 2 1. (x ² - 4x) ² + 7 (x ² - 4x) + 12 \u003d 0.2 x (x + 1) (x + 5) (x + 6) \u003d 24.3 x ⁴ + 18 (x + 4) ² \u003d 11x² (x + 4). 4.x ⁴ - 5x³ + 6x² - 5x + 1 \u003d 0. 5.x² - 2xy + y² + 2x²y - 9 \u003d 0, x - y - x²y + 3 \u003d 0.3 option. 1. (x ² + 3x) ² - 14 (x ² + 3x) + 40 \u003d 0 2. (x - 5) (x-3) (x + 3) (x + 1) \u003d - 35.3 x4 + 8x² (x + 2) \u003d 9 (x + 2) ². 4.x ⁴ - 7x³ + 14x² - 7x + 1 \u003d 0. 5.x + y + x ² + y ² \u003d 18, xy + x ² + y² \u003d 19.
Option 4. (x ² - 2x) ² - 11 (x ² - 2x) + 24 \u003d o. (x -7) (x-4) (x-2) (x + 1) \u003d -36. X⁴ + 3 (x -6) ² \u003d 4x² (6 - x). X⁴ - 6x³ + 7x² - 6x + 1 \u003d 0. X² + 3xy + y² \u003d - 1, 2x² - 3xy - 3y² \u003d - 4. Additional task: The remainder of dividing the polynomial P (x) by (x - 1) is 4, the remainder of division by (x + 1) is equal to 2, and when divided by (x - 2) is equal to 8. Find the remainder of division of P (x) by (x ³ - 2x² - x + 2).
Answers and instructions: option No. 1 No. 2. No. 3. No. 4. No. 5. 1. - 3; ± 2; 1 1; 2; 3. -five; -four; one; 2. Homogeneous equation: u \u003d x -3, v \u003d x² -2; -one; 3; 4. (2; 1); (2/3; 4/3). Indication: 1 · (-3) + 2 · 2 2. -6; -2; -4 ± √6. -3 ± 2√3; - four; - 2.1 ± √11; four; - 2. Homogeneous equation: u \u003d x + 4, v \u003d x² 1; 5; 3 ± √13. (2; 1); (0; 3); (- thirty). Indication: 2 · 2 + 1. 3. -6; 2; four; 12 -3; -2; four; 12 -6; -3; -one; 2. Homogeneous u \u003d x + 2, v \u003d x² -6; ± 3; 2 (2; 3), (3; 2), (-2 + √7; -2 - √7); (-2 - √7; -2 + √7). Hint: 2 -1. 4. (3 ± √5) / 2 2 ± √3 2 ± √3; (3 ± √5) / 2 (5 ± √21) / 2 (1; -2), (-1; 2). Guideline: 1 4 + 2.
Solution of an additional task. By Bezout's theorem: P (1) \u003d 4, P (-1) \u003d 2, P (2) \u003d 8. P (x) \u003d G (x) (x ³ - 2x² - x + 2) + ax² + inx + from. Substitute 1; - one; 2.P (1) \u003d G (1) · 0 + a + b + c \u003d 4, a + b + c \u003d 4.P (-1) \u003d a - b + c \u003d 2, P (2) \u003d 4a² + 2b + c \u003d 8. Solving the resulting system of three equations, we get: a \u003d b \u003d 1, c \u003d 2. Answer: x ² + x + 2.
Criterion No. 1 - 2 points. 1 point - one computational error. No. 2,3,4 - 3 points each. 1 point - led to a quadratic equation. 2 points - one computational error. No. 5. - 4 points. 1 point - expressed one variable through another. 2 points - got one of the solutions. 3 points - one computational error. Additional task: 4 points. 1 point - Bezout's theorem was applied for all four cases. 2 points - made up a system of equations. 3 points - one computational error.