TYPES OF EQUATIONS

Algebraic equations. Equations of the form f n= 0, where f n- a polynomial in one or more variables, are called algebraic equations. A polynomial is an expression of the form

f n = a 0 x i y j ... v k + a 1 x l y m ... v n +¼ + a s x p y q ... v r,

Where x, y, ..., v are variables, and i, j, ..., r are exponents (non-negative integers). A polynomial in one variable is written like this:

f(x) = a 0 x n + a 1 x n – 1 + ... + a n – 1 x + a n

or, in a particular case, 3 x 4 – x 3 + 2x 2 + 4x– 1. An algebraic equation with one unknown is any equation of the form f(x) = 0. If a 0 ¹ 0, then n is called the degree of the equation. For example, 2 x+ 3 = 0 – equation of the first degree; equations of the first degree are called linear, since the graph of the function y=ax+b looks like a straight line. Equations of the second degree are called quadratic, and equations of the third degree are called cubic. Equations of higher degrees have similar names.

Transcendental equations. Equations containing transcendental functions, such as logarithmic, exponential, or trigonometric functions, are called transcendental. The following equations are an example:

where lg is the base 10 logarithm.

Differential equations. So called equations containing one or more functions and their derivatives or differentials. Differential equations have proven to be an exceptionally valuable means of accurately formulating the laws of nature.

Integral equations. Equations containing an unknown function under the integral sign, for example, f (s) = ò K (s, t) f(t) dt, Where f(s) And K(s,t) are given, and f(t) is to be found.

Diophantine equations. A Diophantine equation is an algebraic equation in two or more unknowns with integer coefficients, the solution of which is sought in integer or rational numbers. For example equation 3 x – 5y= 1 has a solution x = 7, y= 4; in general, its solutions are integers of the form x = 7 + 5n, y = 4 + 3n.

SOLUTION OF ALGEBRAIC EQUATIONS

For all the types of equations listed above, there are no general solution methods. Yet in many cases, especially for algebraic equations of a certain type, there is a fairly complete theory of their solution.

Linear equations. These simple equations are solved by reducing them to an equivalent equation that directly shows the value of the unknown. For example, the equation x+ 2 = 7 can be reduced to the equivalent equation x= 5 by subtracting the number 2 from the right and left sides. The steps involved in reducing a simple equation, such as x+ 2 = 7, to the equivalent, are based on the use of four axioms.

1. If equal values ​​are increased by the same number, then the results will be equal.

2. If the same number is subtracted from equal values, then the results will be equal.

3. If equal values ​​are multiplied by the same number, then the results will be equal.

4. If equal values ​​are divided by the same number, then the results will be equal.

For example, to solve equation 2 x+ 5 = 15, we use Axiom 2 and subtract the number 5 from the right and left sides, resulting in the equivalent equation 2 x= 10. Then we use Axiom 4 and divide both sides of the resulting equation by 2, as a result of which the original equation reduces to the form x= 5, which is the desired solution.

Quadratic equations. Solutions to the general quadratic equation ax 2 + bx + c= 0 can be obtained using the formula

Thus, there are two solutions, which in a particular case may coincide.

Other algebraic equations. Explicit formulas, similar to the formula for solving a quadratic equation, can only be written for equations of the third and fourth degrees. But even these formulas are complex and do not always help to easily find the roots. As for the equations of the fifth degree or higher, for them, as N. Abel proved in 1824, it is impossible to indicate a general formula that would express the roots of the equation through its coefficients using radicals. In some special cases, equations of higher degrees can be easily solved by factorizing their left-hand side, i.e. factoring it out.

For example, the equation x 3 + 1 = 0 can be written in factorized form ( x + 1)(x 2 – x+ 1) = 0. We find solutions by setting each of the factors equal to zero:

So the roots are x= –1, , i.e. only 3 roots.

If the equation is not factorizable, then approximate solutions should be used. The main methods for finding approximate solutions were developed by Horner, Newton and Greffe. However, in all cases there is a strong belief that the solution exists: the algebraic equation n th degree has exactly n roots.

Systems of linear equations. Two linear equations with two unknowns can be written as

The solution of such a system is found using the determinants

It makes sense if If D= 0, then two cases are possible. (1) At least one of the determinants and is nonzero. In this case, there is no solution to the equations; the equations are inconsistent. A numerical example of such a situation is the system

(2) Both determinants are equal to zero. In this case, the second equation is simply a multiple of the first, and there are an infinite number of solutions.

General theory considers m linear equations with n variables:

If m = n and matrix ( aij) is non-degenerate, then the solution is unique and can be found by Cramer’s rule:

Where A ji– algebraic complement of an element aij in matrix ( aij). More generally, there are the following theorems. Let r is the rank of the matrix ( aij), s is the rank of the bordered matrix ( aij; b i), which is obtained from aij appending a column of numbers b i. Then: (1) if r = s, then exists n–r linearly independent solutions; (2) if r< s , then the equations are inconsistent and there are no solutions.

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Algebraic equations and methods for their solution

A.1 Polynomial and its roots

Consider a set of (n+1) real numbers , a polynomial (polynomial) of degree n with the above coefficients, an expression of the form is called:

https://pandia.ru/text/78/119/images/image003_38.gif" width="257" height="25 src="> (2)

is called an algebraic equation of degree n.

The roots of equation (2) are also called the roots of the polynomial.

Here are some facts about the roots of polynomials.

Fact 1. Any polynomial of odd degree has at least one real root.

Comment. Even knowing that the equation has a root, finding this root can be very difficult.

Example 1 The equation obviously has roots 0 and p.

Example 2 Establishing the roots of the equation, which certainly exist, is a rather difficult task.

Fact 2. If the coefficients of the polynomial are integers, then the rational roots of this equation (if any) have the form, where the numbers k and m are natural, and k is the divisor of the free term, m is the divisor of the main coefficient.

Example 3 https://pandia.ru/text/78/119/images/image010_16.gif" width="348" height="41 src="> (repetitive numbers have been shortened).

The check shows that the numbers 2, and are suitable.

The task of separating rational roots is greatly simplified if the leading coefficient in the polynomial is equal to one. In this case, the possible rational roots of the equation can only be integers that divide the free term of the polynomial.

Example 4 The polynomial has the following integer roots: . Checking possible roots (this can be done quite quickly with Horner's schemes) we make sure that the only integer root of the equation is 2.

Fact 3. If the number is the root of a polynomial, then this polynomial can be represented as a product a method of division by a "corner", very similar to that which is applied to ordinary numbers.

Let's take an example.

Example 5 Let's divide by:

https://pandia.ru/text/78/119/images/image021_6.gif" width="177" height="25">. Note that the first factor has a negative discriminant, so it (and the original polynomial) is larger than the roots does not have.

Fact 4.Any polynomial with real coefficients can be represented as:

https://pandia.ru/text/78/119/images/image023_6.gif" width="16 height=24" height="24"> - root multiplicity, - square trinomials that do not have real roots (they are called irreducible).

Comment. When solving equations and inequalities, one can reduce to irreducible trinomials.

P.2. Grouping as a way to find the roots of a polynomial

Unfortunately (and this has been proven), there is no universal algorithm that allows (like a square trinomial) to find the roots of any polynomial. There are special formulas for solving equations of the third and fourth degree, but they are laborious and are not studied in the school course. Therefore, other methods are often used, such as root separation (discussed in the first paragraph), the grouping method and its special case - the selection of full squares.

The essence of the grouping method is as follows: the members of the polynomial are divided into groups (hence the name) so that after reduction of similar ones, each group will be decomposed into factors, and one of the factors will be contained in each group. This common factor is taken out of brackets and the original polynomial is decomposed into the product of two lower degree polynomials.

Consider an example.

Example 6 Factorize polynomial by grouping method

https://pandia.ru/text/78/119/images/image027_3.gif" width="272" height="24 src=">

(https://pandia.ru/text/78/119/images/image029_3.gif" width="64" height="21">, we will include the first term in the first group, the second term in the third).

https://pandia.ru/text/78/119/images/image031_4.gif" width="51" height="24">, we find the expansion:

.

Both square trinomials have negative discriminants, so their further decomposition is impossible.

Example 7 Factorize the polynomial:

https://pandia.ru/text/78/119/images/image034_3.gif" width="35" height="21"> you need to dress a part that is a multiple of 14: for example, 70-1, 84-15, 98-29 or 42 + 27. The first option leads to a dead end. Consider the second option. We get:

https://pandia.ru/text/78/119/images/image036_2.gif" width="603" height="24">.

Thus,

P.3. Examples of solving the simplest algebraic equations

Polynomials are the simplest algebraic equations. In this subsection, we consider some examples of solving such equations.

Example 8 Find the roots of the equation

https://pandia.ru/text/78/119/images/image041_2.gif" width="89" height="19 src=">.

Let's start with the smallest number - three.

https://pandia.ru/text/78/119/images/image043_2.gif" width="40 height=23" height="23"> is one of the roots of the equation. To find the other roots, we divide the left side of the equation by :

https://pandia.ru/text/78/119/images/image046_2.gif" width="107" height="21">. Using, for example, Vieta's formulas, we obtain two other roots: .

Answer: https://pandia.ru/text/78/119/images/image049_2.gif" width="124" height="21 src=">.

Solution. The problem can be reduced to a biquadratic equation, but we will try to use factorization..gif" width="616" height="24 src=">.

The roots of the first factor: https://pandia.ru/text/78/119/images/image052_2.gif" width="63" height="41 src=">.

Next, consider an example of an equation that reduces to a rational one. A feature of such equations is the obligatory requirement to check the found roots of the region of admissible values. For example, at the Unified State Exam a few years ago, a “simple” task was proposed.

Example 10 solve the equation

DIV_ADBLOCK37">

P. 4. Fractional algebraic equations

The simplest fractional algebraic expression has the form:

https://pandia.ru/text/78/119/images/image055_2.gif" width="40" height="23 src=">.gif" width="111" height="41 src=">.

Solution: Let's bring the fractions to a common denominator:

https://pandia.ru/text/78/119/images/image059_2.gif" width="207" height="41">.

Both roots of the numerator are not roots of the denominator (verify this by directly substituting both roots into the denominator), so they are solutions to the equation considered.

If a fractional-rational equation contains many elementary expressions, then, after transformations, a rather cumbersome expression can form in the numerator, finding the roots of which will be very difficult. But in some cases it is possible to reduce a complex equation to a simpler one, using, for example, a change of variables. Consider an example.

Example 12. solve the equation

https://pandia.ru/text/78/119/images/image061_0.gif" width="81" height="41"> are mutually inverse (their product is equal to one). Let us introduce the following replacement: . The original equation will take view:

https://pandia.ru/text/78/119/images/image064_0.gif" height="16">, we get a quadratic equation:

https://pandia.ru/text/78/119/images/image066_0.gif" width="93" height="23">. Let's perform the reverse substitution. Get and solve the set of two equations: 2. Index, residential address , email (if any), telephone (home or mobile)

3. School data (for example: MBOU No. 1 Bikin village)

4. Surname, I. O. teacher of mathematics (for example: mathematic teacher)

M 10.2.1. Solve the equation by factoring the polynomial:

M 10.2.2. Solve fractional rational equation

a) https://pandia.ru/text/78/119/images/image082_0.gif" width="209" height="21 src=">. ( Note: multiply first the first factor with the fourth and the second with the third. Label the first piecey, the second product will then be represented as y+2. Solve the resulting quadratic equation and make the reverse substitution.)

c) https://pandia.ru/text/78/119/images/image084_0.gif" width="165" height="41 src=">. ( Note: try adding some number to the first two terms so that the sum turns out to be the reciprocal of the one in third place with a factor of -10. See examples 12 and 13 below..)

Algebraic equations. Definition

Let the functions f(x) and u(x) be defined on some set A. And let it be necessary to find the set X on which these functions take equal values, in other words, find all values ​​of x for which the equality holds: f(x)= c(x).

In this formulation, this equality is called an equation with unknown x.

An equation is called algebraic if only algebraic operations are performed on the unknown - addition, subtraction, multiplication, division, raising to a power and extracting a root with a natural exponent.

Algebraic equations contain only algebraic functions (whole, rational, irrational). An algebraic equation in general form can be represented by a polynomial of the nth degree with real coefficients:

For example,

The set A is called the set (region) of admissible values ​​of the unknown for the given equation.

The set X is called the set of solutions, and any of its solutions x=a is called the root of this equation. Solving an equation means finding the set of all its solutions or proving that there are none.

Methods for solving algebraic equations

In many scientific and engineering problems, it is required to solve an equation of the form

where f(x) is a given continuous non-linear function.

Analytically it is possible to find a solution only for the simplest equations. In most cases, it is necessary to solve an equation of the form (1) by numerical methods.

The numerical solution of equation (1) is usually carried out in two stages. At the first stage, you need to find such intervals of change of the variable x, where only one root is located. This problem is usually solved graphically. At the second stage, the individual roots are refined. Various methods are used for this.

Methods for solving nonlinear equations are divided into direct and iterative. Direct methods allow you to write the roots in the form of a formula. However, the equations encountered in practice are not always possible to solve. simple methods. To solve them, iterative methods are used, i.e. methods of successive approximations.

Direct methods - the solution is found in a previously known number of arithmetic operations, the solution is strict. Examples: Gauss method, square root method, Cramer's rule, etc.

Iterative methods are methods of successive approximations in which it is impossible to predict the number of arithmetic operations that will be required to solve an equation (system) with a given accuracy. Examples: the method of simple iterations, the Gauss-Seidel method, the method of dividing a segment in half, etc.

In this paper, we study and compare the method of simple iterations and the method of half division of a segment.

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1 Algebraic equations where Definition. Algebraic is an equation of the form 0, P () 0, some real numbers. 0 0 In this case, the variable is called the unknown, and the numbers 0 are the coefficients of the equation (), the order (or degree) of the equation. Definition. A number is called a solution (or root) of the equation () if, when the number is substituted into the equation 0 P instead, the correct equality 0 P is obtained. Depending on the coefficients, the equation () may have a single real root, several roots, or no real roots. To solve an equation means to find all its roots (only real solutions are considered in the school course) or to prove that the equation has no solutions. and We will consider the equation () at. For (cubic equation) there are formulas for the roots of the equation 0 P in radicals, known as Cordano's formulas. When equation () is unsolvable in radicals, i.e. the solution of the equation 0 P at cannot be expressed in terms of its coefficients 0, using a finite number of arithmetic operations (operations of addition, subtraction, multiplication, division and extracting the arithmetic root). The proof of this statement was first obtained by the Norwegian mathematician Abel in the year 6. In some cases, the solution of algebraic equations of higher degrees, including the third and fourth, can be found quite simply. Such a possibility is completely determined by the coefficients, 0, of the polynomial P. Corollary from Bezout's theorem. If is the root of the polynomial (P 0), then the polynomial P is divisible by the binomial without remainder, i.e. there exists a polynomial such that P F F. P

2 "corner". Equation () in this case is equivalent to a set of equations Dividing one polynomial Equation 0, F 0. P by another Q m, m, you can produce P degree can have no more than real roots, taking into account the multiplicity. Moreover, an equation of odd degree always has at least one real root. If the real numbers ..., are the roots of the equation 0, then the identity P holds, For equations of higher degrees (), the Vieta theorem is valid, which we formulate in the case of and. If the real numbers and are the roots of the cubic equation 0, 0, then they satisfy the conditions: b c d d, c, b. If the real numbers, and are the roots of the equation of the fourth degree 0, 0, then they satisfy the conditions: b c d e If the rational number is 0 e, d, b. p, where p q q c, an irreducible fraction, is the root of an equation with integer coefficients, then p must be a divisor of the constant term

3 , and q is a divisor of the coefficient 0 at the highest degree. In particular, the integer roots 0 p of the reduced equation 0 with integer coefficients are divisors of the free term. This statement follows from the last equality in (.7) If the sum of all coefficients of equation 0 has a root. P is zero, then the equation For example, the sum of the coefficients of the equation is zero, so it has a root. If in the equation the sum of the coefficients at odd powers is equal to the sum of the free term and the coefficients at even powers, then the equation has a root. For example, in the equation we have 6 7, so the root of this equation. Let us consider separate classes of algebraic equations of higher degrees and study methods for their solution. Biquadratic equations. Definition. A biquadratic equation is an equation of the form where 0. b c 0, () To solve this equation, a change of variables y is used, where y 0. In this case, a quadratic equation y by c 0 is obtained. Since the equation () is an equation of the fourth degree, it has no more than four real roots. If y and y are its solutions, then the original biquadratic equation will be equivalent to the set: The method of selecting the root (roots). 0 y y. If the given algebraic equation () with integer coefficients has integer roots, then they must be sought among the divisors of the free term

4 equations (). The rational roots p 0 of the equation () with integer coefficients q p should be sought among numbers such that p is a divisor of the free term, q and q is the divisor of the coefficient 0 at the highest degree in equation (). These properties underlie the method of selecting the roots of an algebraic equation. Example. Solve equation 0. Solution. This equation is reduced and has integer coefficients. Therefore, the integer roots of this equation (if any) are contained among the divisors of the free term:,. It is easy to see that is the root of the equation. To find the remaining roots, we divide the polynomial into a binomial “corner”: 0. For equation 0, we again find the root by selection, and then divide the polynomial into a binomial: 0, Equation 0 has no real roots. Thus, the

The 5th degree equation has two real roots. Answer.,. Change of variables method. If, when changing variables, the original equation is simplified (for example, its degree is reduced), then we boldly introduce a new variable. Example. Solve the equation. Solution. If you open the brackets and bring like terms, you get the equation 6 0, which is very difficult to solve. Although it is an equation with integer coefficients, but as we will see below, it does not have integer roots. Therefore, we will use another method: we introduce a new variable y and solve the quadratic equation y y. Its roots are y and y. Accordingly, the original equation will be equivalent to the combination of two equations. We solve the obtained quadratic equations.,. 0,D0,. or 0, D 7 0, there are no solutions. Thus, the original equation of the th degree has two roots and. Answer.,. Example. Find the largest negative root of equation 0. Solution. It is very difficult to find the roots of this equation, so we will use the following trick: multiply (or divide) this equation by a certain number so that the highest term of the equation becomes the cube of some expression

6 Note that, and introduce a new variable y. As a result, we obtain the equation y y y 6 0, which is equivalent to the original one. By selection, we find its roots y, y and y, which will correspond to the roots of the original equation, and. The largest negative root is . Answer. The largest negative root. You can introduce another variable and consider a quadratic equation with respect to one of the obtained ("old" or "new") variables. Example. Find the smallest root of the equation 6 0. Solution. Let's transform the original equation as follows: Let's introduce a new variable y 6 and get the equation 6 y y 0. Let's solve the resulting equation as a quadratic one with respect to y. y or y. D 6 y y 0, y, Let's return to the variable, we get two quadratic equations.

7 6, 9 0, D 0 0, 9 0, 9 0, 9 0 6, 0, D 9 We choose the smallest of them. Since 0 0, then 9., so the smallest solution. 9 0 Reply. Smallest Solution.. Return Equations Definition. Equations of the form 0 0 are called recurrent or symmetric, for which the coefficients in symmetrical positions are equal, that is, for k 0,. k k For example, it is recurrent, since 0, 9, 6. The following statements are true for reciprocal equations. A reciprocal equation of odd degree always has a root and, after division by a binomial, is reduced to a reciprocal equation of even degree. A reciprocal equation of even degree can be reduced to an equation of half the degree by introducing the variable y. Let us illustrate these statements with examples. Example. Solve the equation Solution. It is easy to see that this equation is reciprocal of an odd degree and, therefore, has a root. We divide the polynomial into a binomial:

8 It remains to solve the reciprocal equation of the th degree Since 0 is not the root of this equation, we can divide both parts of this equation by Let's make a change of variables i.e. y.. Get y. Then y, We get the equation y 0y 6 0 (the degree of the equation is halved!) Let's solve the quadratic equation y 0y 0. According to Vieta's theorem, the numbers y and y 6 are its roots. We have further

9 0.6 0, D 0.6 0.9,. Thus, the original equation of the th degree has roots:, and. Answer., and. D Using monotonicity of functions and other special techniques To solve non-standard algebraic equations, one has to use various techniques to transform the equation to an equivalent form, introduce new variables, study the function Solving equations of the form g f as part of the equation 0 f, etc. f is sometimes convenient to build on the use of the monotonicity property of functions. This technique is based on the following theorem. Theorem. Let the equation f g be defined on the set X R ; the function f is monotonically increasing (decreasing) on ​​X, and g is monotonically decreasing (increasing). If both E f, E g are ranges of f g on the set X and E f Eg, then there is a unique point 0 X such that g f, i.e. the equation 0 0 f g has a unique solution. This theorem is valid for any equations of the form g for algebraic ones. Example 6. Solve equation 96 E f. y f Eg 0 X g f, and not just Solution. The power function y, N, is defined on the entire real line and is a strictly increasing function on R. Therefore, the left side of the given

10 of the equation f is a strictly increasing function on R as the sum of two strictly increasing functions. Right part 96 g is identically constant. Therefore, in accordance with Theorem 6, the equation has a unique solution. It's easy to see what it is. Answer.. Example 7. Solve the equation. Solution Y. But Y for any R, and therefore the equation 0 Y, and hence the original (.), has no solution. Answer.

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Chapter 1. History of quadratic and higher-order equations 1.1 Equations in Ancient Babylon Algebra arose in connection with the solution of various problems using equations. Tasks usually require

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Typical variant "Complex numbers Polynomials and rational fractions" Task Given two complex numbers and cos sn Find and write the result in algebraic form write the result in trigonometric

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2.22. Take out the common factor (n is a natural number) out of brackets: 1) x n + 3 + x n ; 3) z 3n - z n ; 2) y n + 2 - y n - 2, n > 2; 4) 5n + 4 + 2 5n + 2-3 5n + 1. 2.23. Each number was assigned

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7 Trigonometric Equations and Inequalities Comment

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The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Equations have been used by man since ancient times and since then their use has only increased.

Equations containing the symbol \[\sqrtx\] are called equations with square root. The square root of a non-negative number \ is a non-negative number whose square is equal to \. \[(\sqrt a=x, x_2=a; x, a\pm0)\]. The number or expression under the root sign must always be non-negative.

Exist different ways solutions of such equations:

Squaring a number by multiplying the number by itself;

Simplifying the roots, if possible, by removing full roots from it;

The use of imaginary numbers to obtain the root of negative numbers;

Application of the division algorithm in a column;

And others.

For clarity, we solve the following equation with a square root:

\[\sqrt(x-5)=3\]

We multiply each side of the equation by itself to get rid of the radicals:

Now we have the simplest linear equation, which is solved as follows:

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