System of equations. Detailed theory with examples (2020). Solving simple linear equations Types of equations and methods for solving them

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Linear equation is an algebraic equation. In this equation, the total degree of its constituent polynomials is equal to one.

Linear equations are presented as follows:

In general form: a 1 x 1 + a 2 x 2 + … + a n x n + b = 0

In canonical form: a 1 x 1 + a 2 x 2 + … + a n x n = b.

Linear equation with one variable.

A linear equation with 1 variable is reduced to the form:

ax+ b=0.

For example:

2x + 7 = 0. Where a=2, b=7;

0.1x - 2.3 = 0. Where a=0.1, b=-2.3;

12x + 1/2 = 0. Where a=12, b=1/2.

The number of roots depends on a And b:

When a= b=0 , which means that the equation has an unlimited number of solutions, since .

When a=0 , b≠ 0 , which means the equation has no roots, since .

When a ≠ 0 , which means the equation has only one root.

Linear equation with two variables.

Equation with variable x is an equality of type A(x)=B(x), Where A(x) And B(x)- expressions from x. When substituting the set T values x into the equation we get a true numerical equality, which is called truth set this equation or solution of a given equation, and all such values ​​of the variable are roots of the equation.

Linear equations of 2 variables are presented in the following form:

In general form: ax + by + c = 0,

In canonical form: ax + by = -c,

In linear function form: y = kx + m, Where .

The solution or roots of this equation is the following pair of variable values (x;y), which turns it into an identity. A linear equation with 2 variables has an unlimited number of these solutions (roots). The geometric model (graph) of this equation is a straight line y=kx+m.

If an equation contains x squared, then the equation is called

After we have studied the concept of equalities, namely one of their types - numerical equalities, we can move on to another important type - equations. Within the framework of this material, we will explain what an equation and its root are, formulate basic definitions and give various examples of equations and finding their roots.

Concept of equation

Typically, the concept of an equation is taught at the very beginning of a school algebra course. Then it is defined like this:

Definition 1

Equation called an equality with an unknown number that needs to be found.

It is customary to denote unknowns in small Latin letters, for example, t, r, m, etc., but x, y, z are most often used. In other words, the equation is determined by the form of its recording, that is, equality will be an equation only when it is reduced to a certain form - it must contain a letter, the value that must be found.

Let us give some examples of the simplest equations. These can be equalities of the form x = 5, y = 6, etc., as well as those that include arithmetic operations, for example, x + 7 = 38, z − 4 = 2, 8 t = 4, 6: x = 3.

After the concept of brackets is learned, the concept of equations with brackets appears. These include 7 · (x − 1) = 19, x + 6 · (x + 6 · (x − 8)) = 3, etc. The letter that needs to be found can appear more than once, but several times, like, for example, in the equation x + 2 + 4 · x − 2 − x = 10 . Also, unknowns can be located not only on the left, but also on the right or in both parts at the same time, for example, x (8 + 1) − 7 = 8, 3 − 3 = z + 3 or 8 x − 9 = 2 (x + 17) .

Further, after students become familiar with the concepts of integers, reals, rationals, natural numbers, as well as logarithms, roots and powers, new equations appear that include all these objects. We have devoted a separate article to examples of such expressions.

In the 7th grade curriculum, the concept of variables appears for the first time. These are letters that can take on different meanings (for more details, see the article on numeric, letter and variable expressions). Based on this concept, we can redefine the equation:

Definition 2

The equation is an equality involving a variable whose value needs to be calculated.

That is, for example, the expression x + 3 = 6 x + 7 is an equation with the variable x, and 3 y − 1 + y = 0 is an equation with the variable y.

One equation can have more than one variable, but two or more. They are called, respectively, equations with two, three variables, etc. Let us write down the definition:

Definition 3

Equations with two (three, four or more) variables are equations that include a corresponding number of unknowns.

For example, an equality of the form 3, 7 · x + 0, 6 = 1 is an equation with one variable x, and x − z = 5 is an equation with two variables x and z. An example of an equation with three variables would be x 2 + (y − 6) 2 + (z + 0, 6) 2 = 26.

Root of the equation

When we talk about an equation, the need immediately arises to define the concept of its root. Let's try to explain what it means.

Example 1

We are given a certain equation that includes one variable. If we substitute a number for the unknown letter, the equation becomes a numerical equality - true or false. So, if in the equation a + 1 = 5 we replace the letter with the number 2, then the equality will become false, and if 4, then the correct equality will be 4 + 1 = 5.

We are more interested in precisely those values ​​with which the variable will turn into a true equality. They are called roots or solutions. Let's write down the definition.

Definition 4

Root of the equation They call the value of a variable that turns a given equation into a true equality.

The root can also be called a solution, or vice versa - both of these concepts mean the same thing.

Example 2

Let's take an example to clarify this definition. Above we gave the equation a + 1 = 5. According to the definition, the root in this case will be 4, because when substituted instead of a letter it gives the correct numerical equality, and two will not be a solution, since it corresponds to the incorrect equality 2 + 1 = 5.

How many roots can one equation have? Does every equation have a root? Let's answer these questions.

Equations that do not have a single root also exist. An example would be 0 x = 5. We can substitute an infinite number of different numbers into it, but none of them will turn it into a true equality, since multiplying by 0 always gives 0.

There are also equations that have several roots. They can have either a finite or an infinite number of roots.

Example 3

So, in the equation x − 2 = 4 there is only one root - six, in x 2 = 9 two roots - three and minus three, in x · (x − 1) · (x − 2) = 0 three roots - zero, one and two, there are infinitely many roots in the equation x=x.

Now let us explain how to correctly write the roots of the equation. If there are none, then we write: “the equation has no roots.” In this case, you can also indicate the sign of the empty set ∅. If there are roots, then we write them separated by commas or indicate them as elements of a set, enclosing them in curly braces. So, if any equation has three roots - 2, 1 and 5, then we write - 2, 1, 5 or (- 2, 1, 5).

It is allowed to write roots in the form of simple equalities. So, if the unknown in the equation is denoted by the letter y, and the roots are 2 and 7, then we write y = 2 and y = 7. Sometimes subscripts are added to letters, for example, x 1 = 3, x 2 = 5. In this way we point to the numbers of the roots. If the equation has an infinite number of solutions, then we write the answer as a numerical interval or use generally accepted notation: the set of natural numbers is denoted N, integers - Z, real numbers - R. Let's say, if we need to write that the solution to the equation will be any integer, then we write that x ∈ Z, and if any real number from one to nine, then y ∈ 1, 9.

When an equation has two, three roots or more, then, as a rule, we talk not about roots, but about solutions to the equation. Let us formulate the definition of a solution to an equation with several variables.

Definition 5

The solution to an equation with two, three or more variables is two, three or more values ​​of the variables that turn the given equation into a correct numerical equality.

Let us explain the definition with examples.

Example 4

Let's say we have the expression x + y = 7, which is an equation with two variables. Let's substitute one instead of the first, and two instead of the second. We will get an incorrect equality, which means that this pair of values ​​will not be a solution to this equation. If we take the pair 3 and 4, then the equality becomes true, which means we have found a solution.

Such equations may also have no roots or an infinite number of them. If we need to write down two, three, four or more values, then we write them separated by commas in parentheses. That is, in the example above, the answer will look like (3, 4).

In practice, you most often have to deal with equations containing one variable. We will consider the algorithm for solving them in detail in the article devoted to solving equations.

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An equation is a mathematical expression that is an equality and contains an unknown. If an equality is true for any admissible values ​​of the unknowns included in it, then it is called an identity; for example: a relation of the form (x – 1)2 = (x – 1)(x – 1) holds for all values ​​of x.

If an equation containing an unknown x holds only for certain values ​​of x and not for all values ​​of x, as in the case of an identity, then it may be useful to determine those values ​​of x for which the equation is valid. Such values ​​of x are called roots or solutions of the equation. For example, the number 5 is the root of the equation 2x + 7= 17.

In the branch of mathematics called equation theory, the main subject of study is methods for solving equations. In the school algebra course, much attention is paid to equations.

The history of the study of equations goes back many centuries. The most famous mathematicians who contributed to the development of the theory of equations were:

Archimedes (c. 287–212 BC) was an ancient Greek scientist, mathematician and mechanic. While studying a problem that reduced to a cubic equation, Archimedes discovered the role of the characteristic, which was later called the discriminant.

Francois Viet lived in the 16th century. He made great contributions to the study of various problems in mathematics. In particular, he introduced letter designations for the coefficients of the equation and established a connection between the roots of the quadratic equation.

Leonhard Euler (1707 – 1783) - mathematician, mechanic, physicist and astronomer. Author of St. 800 works on mathematical analysis, differential equations, geometry, number theory, approximate calculations, celestial mechanics, mathematics, optics, ballistics, shipbuilding, music theory, etc. He had a significant influence on the development of science. He derived formulas (Euler's Formulas) expressing trigonometric functions of the variable x through an exponential function.

Lagrange Joseph Louis (1736 - 1813), French mathematician and mechanic. He has carried out outstanding research, including research on algebra (the symmetric function of the roots of an equation, on differential equations (the theory of singular solutions, the method of variation of constants).

J. Lagrange and A. Vandermonde are French mathematicians. In 1771, a method for solving systems of equations (the substitution method) was first used.

Gauss Karl Friedrich (1777 -1855) - German mathematician. He wrote a book outlining the theory of equations for dividing a circle (i.e., the equations xn - 1 = 0), which in many ways was a prototype of Galois theory. In addition to general methods for solving these equations, he established a connection between them and the construction of regular polygons. For the first time since the ancient Greek scientists, he made a significant step forward in this matter, namely: he found all those values ​​of n for which a regular n-gon can be constructed with a compass and a ruler. I studied the method of addition. I concluded that systems of equations can be added, divided, and multiplied.

O. I. Somov - enriched various parts of mathematics with important and numerous works, among them the theory of certain algebraic equations of higher degrees.

Galois Evariste (1811-1832) - French mathematician. His main merit is the formulation of a set of ideas that he came to in connection with the continuation of research on the solvability of algebraic equations, begun by J. Lagrange, N. Abel, and others, and created the theory of algebraic equations of higher degrees with one unknown.

A. V. Pogorelov (1919 – 1981) - His work combines geometric methods with analytical methods of the theory of partial differential equations. His works also had a significant impact on the theory of nonlinear differential equations.

P. Ruffini - Italian mathematician. He devoted a number of works to proving the unsolvability of equations of degree 5, systematically using the closedness of the set of substitutions.

Despite the fact that scientists have been studying equations for a long time, science does not know how and when people needed to use equations. It is only known that people have been solving problems leading to the solution of the simplest equations since the time they became human. Another 3 - 4 thousand years BC. e. The Egyptians and Babylonians knew how to solve equations. The rule for solving these equations coincides with the modern one, but it is unknown how they got there.

In Ancient Egypt and Babylon, the false position method was used. An equation of the first degree with one unknown can always be reduced to the form ax + b = c, in which a, b, c are integers. According to the rules of arithmetic operations, ax = c - b,

If b > c, then c b is a negative number. Negative numbers were unknown to the Egyptians and many other later peoples (they began to be used in mathematics on an equal footing with positive numbers only in the seventeenth century). To solve problems that we now solve with equations of the first degree, the false position method was invented. In the Ahmes papyrus, 15 problems are solved by this method. The Egyptians had a special sign for an unknown number, which until recently was read "how" and translated as "heap" ("heap" or "unknown number" of units). Now they read a little less inaccurately: “yeah.” The solution method used by Ahmes is called the method of one false position. Using this method, equations of the form ax = b are solved. This method involves dividing each side of the equation by a. It was used by both the Egyptians and Babylonians. Different peoples used the method of two false positions. The Arabs mechanized this method and obtained the form in which it was transferred to the textbooks of European peoples, including Magnitsky’s Arithmetic. Magnitsky calls the solution a “false rule” and writes in the part of his book outlining this method:

This part is very cunning, because you can put everything with it. Not only what is in citizenship, but also the higher sciences in space, which are listed in the sphere of heaven, as the wise have needs.

The content of Magnitsky's poems can be briefly summarized as follows: this part of arithmetic is very tricky. With its help, you can calculate not only what is needed in everyday practice, but it also solves the “higher” questions that confront the “wise.” Magnitsky uses the “false rule” in the form that the Arabs gave it, calling it “the arithmetic of two errors” or the “method of scales.” Indian mathematicians often gave problems in verse. Lotus problem:

Above the quiet lake, half a measure above the water, the color of the lotus was visible. He grew up alone, and the wind, like a wave, Bent him to the side, and no longer

Flower over water. The fisherman's eye found him two meters from the place where he grew up. How deep is the lake water here? I'll ask you a question.

Types of equations

Linear equations

Linear equations are equations of the form: ax + b = 0, where a and b are some constants. If a is not equal to zero, then the equation has one single root: x = - b: a (ax + b; ax = - b; x = - b: a.).

For example: solve the linear equation: 4x + 12 = 0.

Solution: Since a = 4, and b = 12, then x = - 12: 4; x = - 3.

Check: 4 (- 3) + 12 = 0; 0 = 0.

Since 0 = 0, then -3 is the root of the original equation.

Answer. x = -3

If a is equal to zero and b is equal to zero, then the root of the equation ax + b = 0 is any number.

For example:

0 = 0. Since 0 is equal to 0, then the root of the equation 0x + 0 = 0 is any number.

If a is equal to zero and b is not equal to zero, then the equation ax + b = 0 has no roots.

For example:

0 = 6. Since 0 is not equal to 6, then 0x – 6 = 0 has no roots.

Systems of linear equations.

A system of linear equations is a system in which all equations are linear.

To solve a system means to find all its solutions.

Before solving a system of linear equations, you can determine the number of its solutions.

Let a system of equations be given: (a1x + b1y = c1, (a2x + b2y = c2.

If a1 divided by a2 is not equal to b1 divided by b2, then the system has one unique solution.

If a1 divided by a2 is equal to b1 divided by b2, but equal to c1 divided by c2, then the system has no solutions.

If a1 divided by a2 is equal to b1 divided by b2, and equal to c1 divided by c2, then the system has infinitely many solutions.

A system of equations that has at least one solution is called simultaneous.

A consistent system is called definite if it has a finite number of solutions, and indefinite if the set of its solutions is infinite.

A system that does not have a single solution is called inconsistent or contradictory.

Methods for solving linear equations

There are several ways to solve linear equations:

1) Selection method. This is the simplest way. It consists in selecting all valid values ​​of the unknown by enumeration.

For example:

Solve the equation.

Let x = 1. Then

4 = 6. Since 4 is not equal to 6, then our assumption that x = 1 was incorrect.

Let x = 2.

6 = 6. Since 6 is equal to 6, then our assumption that x = 2 was correct.

Answer: x = 2.

2) Simplification method

This method consists in transferring all terms containing the unknown to the left side, and the known ones to the right with the opposite sign, bringing similar ones, and dividing both sides of the equation by the coefficient of the unknown.

For example:

Solve the equation.

5x – 4 = 11 + 2x;

5x – 2x = 11 + 4;

3x = 15; : (3) x = 5.

Answer. x = 5.

3) Graphic method.

It consists in constructing a graph of the functions of a given equation. Since in a linear equation y = 0, the graph will be parallel to the y-axis. The point of intersection of the graph with the x-axis will be the solution to this equation.

For example:

Solve the equation.

Let y = 7. Then y = 2x + 3.

Let's plot the functions of both equations:

Methods for solving systems of linear equations

In seventh grade, they study three ways to solve systems of equations:

1) Substitution method.

This method consists in expressing one unknown in terms of another in one of the equations. The resulting expression is substituted into another equation, which then turns into an equation with one unknown, and then it is solved. The resulting value of this unknown is substituted into any equation of the original system and the value of the second unknown is found.

For example.

Solve the system of equations.

5x - 2y - 2 = 1.

3x + y = 4; y = 4 - 3x.

Let's substitute the resulting expression into another equation:

5x – 2(4 – 3x) -2 = 1;

5x – 8 + 6x = 1 + 2;

11x = 11; : (11) x = 1.

Let's substitute the resulting value into the equation 3x + y = 4.

3 1 + y = 4;

3 + y = 4; y = 4 – 3; y = 1.

Examination.

/3 1 + 1 = 4,

\5 · 1 – 2 · 1 – 2 = 1;

Answer: x = 1; y = 1.

2) Method of addition.

This method is that if a given system consists of equations that, when added term by term, form an equation with one unknown, then by solving this equation, we will obtain the value of one of the unknowns. The resulting value of this unknown is substituted into any equation of the original system and the value of the second unknown is found.

For example:

Solve the system of equations.

/3у – 2х = 5,

\5x – 3y = 4.

Let's solve the resulting equation.

3x = 9; : (3) x = 3.

Let's substitute the resulting value into the equation 3y – 2x = 5.

3у – 2 3 = 5;

3у = 11; : (3) y = 11/3; y = 3 2/3.

So x = 3; y = 3 2/3.

Examination.

/3 11/3 – 2 3 = 5,

\5 · 3 – 3 · 11/ 3 = 4;

Answer. x = 3; y = 3 2/3

3) Graphic method.

This method is based on the fact that equations are plotted in one coordinate system. If the graphs of an equation intersect, then the coordinates of the intersection point are the solution to this system. If the graphs of the equation are parallel lines, then this system has no solutions. If the graphs of the equations merge into one straight line, then the system has infinitely many solutions.

For example.

Solve the system of equations.

18x + 3y - 1 = 8.

2x - y = 5; 18x + 3y - 1 = 8;

Y = 5 - 2x; 3y = 9 - 18x; : (3) y = 2x - 5. y = 3 - 6x.

Let's build graphs of the functions y = 2x - 5 and y = 3 - 6x on the same coordinate system.

The graphs of the functions y = 2x - 5 and y = 3 - 6x intersect at point A (1; -3).

Therefore, the solution to this system of equations will be x = 1 and y = -3.

Examination.

2 1 - (- 3) = 5,

18 1 + 3 (-3) - 1 = 8.

18 - 9 – 1 = 8;

Answer. x = 1; y = -3.

Conclusion

Based on all of the above, we can conclude that equations are necessary in the modern world not only for solving practical problems, but also as a scientific tool. That is why so many scientists have studied this issue and continue to study it.

Types of algebraic equations and methods for solving them

For students interested in mathematics, when solving algebraic equations of higher degrees, an effective method for quickly finding roots, dividing with a remainder by the binomial x -  or by ax + b, is the Horner scheme.

Consider Horner's scheme.

Let us denote the incomplete quotient when dividing P(x) by x –  through

Q (x) = b 0 x n -1 + b 1 x n -2 + ... + b n -1, and the remainder is b n.

Since P(x) = Q (x)(x–) + b n, then the equality holds

a 0 x n + а 1 x n -1 + … + а n = (b 0 x n -1 + b 1 x n -2 + … + b n -1)(х– ) + b n

Let's open the brackets on the right side and compare the coefficients for the same powers of x on the left and right. We obtain that a 0 = b 0 and for 1  k  n the relations a k = b k -  b k -1 hold. It follows that b 0 = a 0 and b k = a k +  b k -1, 1  k  n.

We write the calculation of the coefficients of the polynomial Q (x) and the remainder b n in the form of a table:

a 0

a 1

a 2

A n-1

A n

b 0 = a 0

b 1 = a 1 +  b 0

b 2 = a 2 +  b 1

b n-1 = a n-1 +  b n-2

b n = a n +  b n-1

Example 1. Divide the polynomial 2x 4 – 7x 3 – 3x 2 + 5x – 1 by x + 1.

Solution. We use Horner's scheme.

When dividing 2x 4 – 7x 3 – 3x 2 + 5x – 1 by x + 1 we get 2x 3 – 9x 2 + 6x – 1

Answer: 2 x 3 – 9x 2 + 6x – 1

Example 2. Calculate P(3), where P(x) = 4x 5 – 7x 4 + 5x 3 – 2x + 1

Solution. Using Bezout's theorem and Horner's scheme, we obtain:

Answer: P(3) = 535

Exercise

Using Horner's diagram, divide the polynomial

4x 3 – x 5 + 132 – 8x 2 on x + 2;

2) Divide the polynomial

2x 2 – 3x 3 – x + x 5 + 1 on x + 1;

3) Find the value of the polynomial P 5 (x) = 2x 5 – 4x 4 – x 2 + 1 for x = 7.

1.1. Finding rational roots of equations with integer coefficients

The method for finding rational roots of an algebraic equation with integer coefficients is given by the following theorem.

Theorem: If an equation with integer coefficients has rational roots, then they are the quotient of dividing the divisor of the free term by the divisor of the leading coefficient.

Proof: a 0 x n + a 1 x n -1 + … + a n = 0

Let x = p/ q is a rational root, q, p are coprime.

Substituting the fraction p/q into the equation and freeing ourselves from the denominator, we get

a 0 r n + a 1 p n -1 q + … + a n -1 pq n -1 + a n q n = 0 (1)

Let's rewrite (1) in two ways:

a n q n = р(– а 0 р n -1 – а 1 р n -2 q – … – а n -1 q n -1) (2)

a 0 r n = q (– а 1 р n -1 –… – а n -1 рq n -2 – а n q n -1) (3)

From equality (2) it follows that a n q n is divisible by p, and since q n and p are coprime, then a n is divisible by p. Similarly, from equality (3) it follows that a 0 is divisible by q. The theorem has been proven.

Example 1. Solve the equation 2x 3 – 7x 2 + 5x – 1 = 0.

Solution. The equation does not have integer roots; we find the rational roots of the equation. Let the irreducible fraction p /q be the root of the equation, then p is found among the divisors of the free term, i.e. among the numbers  1, and q among the positive divisors of the leading coefficient: 1; 2.

Those. rational roots of the equation must be sought among the numbers  1,  1/2, denote P 3 (x) = 2x 3 – 7x 2 + 5x – 1, P 3 (1)  0, P 3 (–1)  0,

P 3 (1/2) = 2/8 – 7/4 + 5/2 – 1 = 0, 1/2 is the root of the equation.

2x 3 – 7x 2 + 5x – 1 = 2x 3 – x 2 – 6 x 2 + 3x + 2x – 1 = 0.

We get: x 2 (2x – 1) – 3x (2x – 1)+ (2x – 1) = 0; (2x – 1)(x 2 – 3x + 1) = 0.

Equating the second factor to zero and solving the equation, we get

Answer:
,

Exercises

Solve equations:

6x 3 – 25x 2 + 3x + 4 = 0;

6x 4 – 7x 3 – 6x 2 + 2x + 1 = 0;

3x 4 – 8x 3 – 2x 2 + 7x – 1 = 0;

1.2. Reciprocal equations and solution methods

Definition. An equation with integer powers with respect to an unknown is called reciprocal if its coefficients, equidistant from the ends of the left side, are equal to each other, i.e. equation of the form

A x n + bx n -1 + cx n -2 + … + cx 2 + bx + a = 0

Reciprocal equation of odd degree

A x 2 n +1 + bx 2 n + cx 2 n -1 + … + cx 2 + bx + a = 0

always has a root x = – 1. Therefore, it is equivalent to combining the equation x + 1 = 0 and  x 2 n +  x 2 n -1 + … +  x +  = 0. The last equation is a reciprocal equation of even degree. Thus, solving reciprocal equations of any degree is reduced to solving a reciprocal equation of even degree.

How to solve it? Let a reciprocal equation of even degree be given

A x 2 n + bx 2 n -1 + … + dx n +1 + ex n + dx n -1 + … + bx + a = 0

Note that x = 0 is not a root of the equation. Then we divide the equation by x n, we get

A x n + bx n -1 + … + dx + e + dx -1 + … + bx 1- n + аx -n = 0

We group the terms of the left side in pairs

A( x n + x - n ) + b (x n -1 + x -(n -1) + … + d(x + x -1 ) + e = 0

We make the replacement x + x -1 = y. After substituting the expressions x 2 + x -2 = y 2 – 2;

x 3 + x -3 = y 3 – 3y; x 4 + x -4 = y 4 – 4y + 2 into the equation we get the equation for atАу n + By n -1 +Cy n -2 + … + Ey + D = 0.

To solve this equation, you need to solve several quadratic equations of the form x + x -1 = y k, where k = 1, 2, ... n. Thus, we obtain the roots of the original equation.

Example 1. Solve the equation x 7 + x 6 – 5x 5 – 13x 4 – 13x 3 – 5x 2 + 2x + 1 = 0.

Solution. x = – 1 is the root of the equation. Let's apply Horner's scheme.

Our equation will take the form:

(x + 1)(x 6 + x 5 – 6x 4 – 7x 3 – 6x 2 + x + 1) = 0

1) x + 1 = 0, x = -1;

2) x 6 + x 5 – 6x 4 – 7x 3 – 6x 2 + x + 1 = 0 | : x 3  0; x 3 + x 2 – 6x – 7 – 6/x + 1/x 2 + 1/x 3 =0.

Grouping, we get: .

Let's introduce a replacement:
;
;
.

We get relatively at equation: y 3 – 3y + y 2 – 2 – 6y – 7 = 0;

y 3 + y 2 – 9y – 9 = 0; y 2 (y + 1) – 9 (y + 1) = 0; (y + 1)(y 2 – 9); y 1 = -1, y 2,3 =  3.

Solving equations
,
,
,

we get the roots:
,
,
,

Answer: x 1 = -1,
,

Exercises

Solve equations.

2x 5 + 5x 4 – 13x 3 – 13x 2 + 5x + 2 = 0;

2x 4 + 3x 3 – 16x 2 + 3x + 2 = 0;

15x 5 + 34x 4 + 15x 3 – 15x 2 – 34x – 15 = 0.

1.3. Variable replacement method for solving equations

The variable replacement method is the most common method. The art of making a variable change is to see which change makes the most sense and will lead to success more quickly.

If given the equation

F(f(x)) = 0, (1)

then by replacing the unknown y = f (x) it is first reduced to the equation

F(y) = 0, (2)

and then after finding all the solutions to equation (2) y 1, y 2, ..., y n, ... is reduced to solving the set of equations f (x) = y 1, f (x) = y 2,..., f (x) = y 2,...

The main ways to implement the variable replacement method are:

using the basic property of a fraction;

highlighting the square of the binomial;

transition to a system of equations;

opening brackets in pairs;

opening parentheses in pairs and dividing both sides of the equation;

decreasing the degree of the equation;

double replacement.

1.3.1. Reducing the power of an equation

Solve the equation (x 2 + x + 2)(x 2 + x + 3) = 6 (3)

Solution. Let's denote x 2 + x + 2 = y, then let's take y (y + 1) = 6, solving the latter, we get y 1 = 2, y 2 = -3. This equation (3) is equivalent to the set of equations x 2 + x + 2 = 2

x 2 + x + 2 = -3

Solving the first, we get x 1 = 0, x 2 = -1. Solving the second, we get
,

Answer: x 1 = 0, x 2 = -1,

1.3.2. Fourth degree equation of the form (x + a)(x +b )(x + c )(x + d ) = m , where a + b = c + d, or a + c = b + d, or a + d = b + c.

Example. Solve the equation (x - 1)(x - 7)(x -4)(x + 2) = 40

Solution. – 1- 4 = - 7 + 2, - 5 = - 5, multiplying these pairs of brackets, we get the equation (x 2 - 5x - 14)(x 2 - 5x + 4) = 40

Let's introduce the replacement: x 2 - 5x – 14 = y, we get the equation y(y + 18) = 40, y 2 + 18y = 40, y 2 + 18y – 40 = 0. y 1 = -20, y 2 = 2. Returning to the original variable, we solve a set of equations:

X 2 - 5x – 14 = - 20 x 1 = 2; x 2 = 3

x 2 - 5x – 14 = 2 x 3.4 =

Answer: x 1 = 2; x 2 = 3 x 3.4 =

1.3.3. An equation of the form (x + a)(x + b)(x + c)(x + d) = Ex 2,

Where ab = cd, or ac =bd, or ad = bc. Open the brackets in pairs and divide both parts by x 2  0.

Example. (x - 1)(x - 2)(x - 8)(x - 4) = 4x 2

Solution. The product of the numbers in the first and third and in the second and fourth brackets are equal, i.e. – 8 (- 1) = (- 2)(- 4). Let's multiply the indicated pairs of brackets and write the equation (x 2 - 9x + 8)(x 2 - 6x + 8) = 4x 2.

Since x = 0 is not a root of the equation, we divide both sides of the equation by x 2  0, we get:
, replacement:
, the original equation will take the form:t(t+3) =4, t 2 + 3 t=4, t 2 + 3 t – 4=0, t 1 =1, t 2 = - 4.

Let's return to the original variable:

x 2 - 10x + 8 = 0

x 2 - 5x + 8 = 0

We solve the first equation, we get x 1,2 = 5 

The second equation has no roots.

Answer: x 1.2 = 5 

1.3.4. Equation of the fourth type (ax 2 + b 1 x + c)(a x 2 + b 2 x + c) = A x 2

Equation (ax 2 + b 1 x+ c)(a x 2 + b 2 x + c) = A x 2, where c  0, A  2
, which after replacing the unknown
can be rewritten as a square and can be easily solved.

Example. (x 2 + x+ 2)(x 2 + 2x + 2) = 2x 2

Solution. It is easy to see that x = 0 is not a root of this equation by dividing this equation by x 2 , we get the equation

replacement
, we get the equation (y+1)(y+2) = 2, solving it, we have roots y 1 = 0; at 2 = - 3, therefore the original equation is equivalent to the set of equations

solving, we get x 1 = -1; x 2 = -2.

Answer: x 1 = -1; x 2 = -2

1.3.5. Equation of the form: a (cx 2 + p 1 x + q) 2 + b (cx 2 + p 2 x + q) 2 = Ax 2

The equation a(cx 2 + p 1 x + q) 2 + b(cx 2 + p 2 x + q) 2 = Ax 2 where a, b, c, q, A are such that q 0, A 0, c 0, a 0, b 0 has no root x = 0, so dividing the equation by x 2 , we obtain an equivalent equation
, which after replacement
can be rewritten as a quadratic equation that can be easily solved. + 1)( x 2 – 14x + 15 = 0

x 2 – 7 x + 15 = 0

Answer: