Solution of non-linear equations. The theory of finding the roots of a nonlinear equation

MINISTRY OF EDUCATION OF THE RUSSIAN FEDERATION

EAST SIBERIAN STATE

UNIVERSITY OF TECHNOLOGY

Abstract on the topic: “Solution of nonlinear equations

simple iteration method"

Completed:. Bubeev B.M.

Checked by: Shirapov D.Sh.

Introduction

Nonlinear equations can be divided into 2 classes - algebraic and transcendental. Algebraic equations are called equations containing only algebraic functions (whole, rational, irrational). In particular, a polynomial is an entire algebraic function. Equations containing other functions (trigonometric, exponential, logarithmic, and others) are called transcendent.

Methods for solving nonlinear equations are divided into two groups:

precise methods;

iterative methods.

Exact Methods allow us to write the roots in the form of some finite relation (formula). From the school algebra course, such methods are known for solving trigonometric, logarithmic, exponential, as well as the simplest algebraic equations.

As is known, many equations and systems of equations do not have analytical solutions. First of all, this applies to most transcendental equations. It is also proved that it is impossible to construct a formula by which it would be possible to solve an arbitrary algebraic equation of degree higher than the fourth. In addition, in some cases the equation contains coefficients that are known only approximately, and, therefore, the very problem of accurately determining the roots of the equation loses its meaning. To solve them, we use iterative methods with a given degree of accuracy.

Let the equation

Function f(x) is continuous on the interval [ a, b] together with their 1st and 2nd order derivatives.

Values f(x) at the ends of the segment have different signs ( f(a)  f(b) < 0).

First and second derivatives f"(x) And f""(x) retain a certain sign throughout the segment.

Conditions 1) and 2) guarantee that on the interval [ a, b] there is at least one root, and it follows from 3) that f(x) is monotonic on this interval and therefore the root will be unique.

Solve Equation (1) iterative method means to establish whether it has roots, how many roots and find the values ​​of the roots with the required accuracy.

Any value that reverses a function f(x) to zero, i.e. such that:

called root equations(1) or zero functions f(x).

The problem of finding the root of the equation f(x) = 0 by the iterative method consists of two stages:

root separation- finding the approximate value of the root or the segment containing it;

refinement of approximate roots- bringing them to a given degree of accuracy.

The process of separating the roots begins with the establishment of the signs of the function f(x) in the boundary x=a And x=b points in the area of ​​its existence.

Example 1 . Separate the roots of the equation:

Therefore, equation (2) has three real roots lying in the intervals [-3, -1], and .

Approximate values ​​of the roots ( initial approximations) can also be known from the physical meaning of the problem, from solving a similar problem with different initial data, or can be found graphically.

Common in engineering practice graphic way determination of approximate roots.

Taking into account that the real roots of equation (1) are the intersection points of the graph of the function f(x) with the x-axis, it is enough to graph the function f(x) and mark the intersection points f(x) with axle Oh, or mark on axis Oh segments containing one root. Plotting can often be greatly simplified by replacing equation (1) equivalent him with the equation:

where functions f 1 (x) And f 2 (x) - simpler than the function f(x). Then, plotting the graphs of the functions at=f 1 (x) And at = f 2 (x), we obtain the desired roots as the abscissas of the intersection points of these graphs.

Figure 2.

Example 2 . Graphically separate the roots of the equation (Figure 2):

x lg x= 1.

Equation (4) can be conveniently rewritten as an equality:

Hence it is clear that the roots of equation (4) can be found as the abscissas of the intersection points of the logarithmic curve y= log x and hyperbole y = . Having constructed these curves, we will approximately find the only root of equation (4) or determine its containing segment .

The iterative process consists in successive refinement of the initial approximation X 0 . Each such step is called iteration. As a result of iterations, a sequence of approximate values ​​of the root is found X 1 , X 2 , ..., X n . If these values ​​with an increase in the number of iterations n approaching true value root, then we say that the iterative process converges.

Simple iteration method

To use the iteration method, the original non-linear equation f(X) = 0 is replaced by the equivalent equation

Geometrically, the iteration method can be explained as follows. Let's build on a plane hoy function graphs y = x And y= (X). Each real root of equation (8) is the abscissa of the intersection point M crooked y= (X) with a straight line y = x(Figure 6, A).

Figure 6

Departing from some point A 0 [x 0 ,  (x 0)], we build a broken line A 0 IN 1 A 1 IN 2 A 2 ... (“ladder”), the links of which are alternately parallel to the axis Oh and axes OU, vertices A 0 , A 1 , A 2 , ... lie on a curve y= (X), and the vertices IN 1 , IN 2 , IN 3 , …, - on a straight line y = x. General abscissas of points A 1 and IN 1 , A 2 and IN 2 , ... are obviously respectively successive approximations X 1 , X 2 , ... root .

Another kind of broken line is also possible A 0 IN 1 A 1 IN 2 A 2 ... - “spiral” (Figure 6, b). A solution in the form of a “ladder” is obtained if the derivative " ( X) is positive, and the solution is in the form of a “spiral” if " ( X) is negative.

In Figure 6, a, b curve at =  (X) in the vicinity of the root - gently sloping, that is<1, и процесс итерации сходится. Однако, если рассмотреть случай, где >1, then the iteration process can be divergent (Figure 7).

Figure 7

Therefore, for the practical application of the iteration method, it is necessary to find out sufficient conditions for the convergence of the iterative process.

Theorem:Let the function  (X) defined and differentiable on the segment [a, b], and all its values  (X) [a,b].

Then if there is a proper fraction q such that

for a<x<b, That: 1)iteration process

converges regardless of the initial value I X 0  [a,b];

2) limit value is the only root of the equation x = (X) on the segment [a, b].

Example 5 . The equation

f(x)  x 3 - x - 1 = 0

has a root , because f(1) = - 1 < 0 и f(2) = 5 > 0.

Equation (10) can be written as

X = X 3 - 1.

 (X) = X 3 - 1 and " ( X) = 3X 2 ;

" (X) 3 at 1 X 2

and, consequently, the conditions for the convergence of the iteration process are not met.

If we write equation (10) in the form

then we will have:

.

From here at 1 X 2 and, therefore, the iteration process for Eq. (12) converges quickly. equations method dividing the segment in half ... in memory in the form simple variables. The result of this... iteration) of type Real; d is the discriminant of the Real type; x1 - first root equations found method solutions square equations ...

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Let a function be given that is continuous together with its several derivatives. It is required to find all or some real roots of the equation

This task is divided into several subtasks. First, it is necessary to determine the number of roots, to investigate their nature and location. Second, find the approximate values ​​of the roots. Thirdly, to choose from them the roots of interest to us and calculate them with the required accuracy. The first and second tasks are solved, as a rule, by analytical or graphical methods. In the case when only the real roots of equation (1) are sought, it is useful to compile a table of function values. If the function has different signs in two neighboring nodes of the table, then between these nodes lies an odd number of roots of the equation (at least one). If these nodes are close, then most likely there is only one root between them.

The found approximate values ​​of the roots can be refined using various iterative methods. Let's consider three methods: 1) the method of dichotomy (or dividing the segment in half); 2) simple iteration method; and 3) Newton's method.

Methods for solving the problem

Bisection method

The simplest method for finding the root of the nonlinear equation (1) is the half division method.

Let a continuous function be given on the segment. If the values ​​of the function at the ends of the segment have different signs, i.e. then this means that there is an odd number of roots inside the given segment. Let, for definiteness, have only one root. The essence of the method is to halve the length of the segment at each iteration. We find the middle of the segment (see Fig. 1) Calculate the value of the function and select the segment on which the function changes its sign. Divide the new segment in half again. And we continue this process until the length of the segment is equal to the predetermined error in calculating the root. The construction of several successive approximations according to formula (3) is shown in Figure 1.

So, the algorithm of the dichotomy method:

1. Set the interval and error.

2. If f(a) and f(b) have the same signs, issue a message about the impossibility of finding the root and stop.

Fig.1.

3. Otherwise calculate c=(a+b)/2

4. If f(a) and f(c) have different signs, put b=c, otherwise a=c.

5. If the length of the new segment, then calculate the value of the root c=(a+b)/2 and stop, otherwise go to step 3.

Since in N steps the length of the segment is reduced by 2 N times, the given error in finding the root will be reached in iterations.

As can be seen, the rate of convergence is low, but the advantages of the method include simplicity and unconditional convergence of the iterative process. If the segment contains more than one root (but an odd number), then one will always be found.

Comment. To determine the interval in which the root lies, an additional analysis of the function is required, based either on analytical estimates or on the use of a graphical solution method. It is also possible to organize a search of function values ​​at different points until the function sign-changing condition is met

Finding the roots of a nonlinear equation

coursework

Computer science, cybernetics and programming

Flowcharts implementing numerical methods - for the dichotomy method: Flowchart for the chord method: Flowchart for the Newton method: Program listing unit Unit1; interfce uses Windows Messges SysUtils Vrints Clsses Grphics Controls Forms Dilogs TeEngine Series ExtCtrls TeeProcs Chrt Menus OleCtnrs StdCtrls xCtrls OleCtrls VCF1 Mth; type TForm1 = clssTForm GroupBox1: TGroupBox; OleContiner2: TOleContiner; MinMenu1: TMinMenu; N1: TMenuItem; Chrt1: TChrt; Series1:...

RUSSIAN STATE UNIVERSITY OF OIL AND GAS them. THEM. GUBKINA

Department of Informatics

Course work

in the discipline "Informatics".

Subject: " Finding the roots of a nonlinear equation "

Done: student

Manepova A. M

group: GI-12-05

Checked:

Moscow 2013

Assignment for course work. Theory of finding the roots of a nonlinear equation. Description of the numerical methods used. 1. Method of half division (dichotomy) 2. Method of chords 3. Newton's method Calculations in the mathematical package Mat lab Report on the results of calculating the approximate value of the root of the equation in MS Excel. Calculation results using Parameter Selection Calculation results using Search for Solutions Description of the application created in the Delphi environment. Block schemes implementing numerical methods Program listing Application window image Analysis of the results Literature.

Assignment for course work.

1. calculation done in math package Matlab (Mathematica 5 .) (function file for describing a non-linear equation, graph, solution in symbolic and numerical form).
2. Finding roots of a non-linear equation in spreadsheets MS Excel (type of a non-linear equation, a graph for finding the roots of a non-linear equation, find the root of a non-linear equation using conditional analysis tools: “Parameter selection”, “Search for a solution”).
3. Create an application to find the roots of the nonlinear equation in Delphi environment (a type of non-linear equation, a graph on a given interval, for each method: the results of tabulating a function on a given interval with a given step, for each method of the numerical method, a user subroutine with parameter transfer). The results are displayed on the form in the form of a table and in a file. Provide for a change in the accuracy of the value (E<= 0 , 001).
4. kind of equation

Theory of finding the roots of a nonlinear equation. Description of the numerical methods used.

Let the function , which is continuous along with its several derivatives. It is required to find all or some real roots of the equation

.
This task is divided into several subtasks. First, you need to determine the number of roots, examine them character and location. Second, find the approximate values ​​of the roots. Thirdly, choose the roots of interest to us from them and calculate them with the required accuracy e. The first and second tasks are solved, as a rule, by analytical or graphical methods. In the case when only the real roots of the equation are searched, it is useful to make a table of values functions . If at two neighboring nodes tables function has different signs, then between these nodes lies an odd number of roots of the equation (at least one). If these nodes are close, then most likely there is only one root between them.
The found approximate values ​​of the roots can be refined using various iterative methods.

Consider three methods: 1) the dichotomy method (or dividing the segment in half); 2) simple iteration method and 3) method Newton.

1. Method of half division (dichotomy)

Let a continuous function be given on the intervalIf the function values ​​at the ends of the segment have different signs, i.e.then this means that there is an odd number of roots inside the given segment. Let, for definiteness, have only one root. The essence of the method is to halve the length of the segment at each iteration. We find the midpoint of the segment according to the formula:We calculate the value of the functionand choose the segment on which the function changes its sign . Divide the new segment in half again. And this one process we continue until the length of the segment is equal to the predetermined error in calculating the root E.

2. Method of chords

When solving a non-linear equation by the chord method, the intervals , on which there is only one solution, and the accuracy Ɛ are specified. Then, through two points with coordinates (a, F (a)) and (b, F (b)) we draw a straight line segment (chord) and determine the point of intersection of this line with the abscissa axis. At the same time, they ate F (a) * F (b)<0, то праву границу интервала пееносиим в точку x (b=x). Если указанное условие не выполняется, то в точку x the left border of the interval is transferred (a=x). The search for a solution stops when the specified accuracy |F(x)|>Ɛ is reached. Calculations are carried out until the inequality is fulfilled:. The iterative formula of the chord method has the form:

3. Newton's method

To solve the equation numericallysimple iteration method, it must be reduced to the following form:, where contraction mapping.

For best convergence method at the next approximation pointthe condition must be met. The solution of this equation is sought in the form, Then:

Assuming that the approach point is "close enough" to the root, and that the given function continuous , the final formula for is:

With this in mind, the function is defined by the expression:

This function, in a neighborhood of the root, performs a contraction mapping, and an algorithm for finding a numerical solution to the equationreduces to an iterative calculation procedure:

Calculations in a mathematical package mat lab

In the mathematical package, according to the condition of the assignment, a graph of the function was built and the root of the equation was found using a symbolic solution ( solve ) and numerically using built-in functions: fzero and fsolve . To describe my function, I used a function file.

The following figure shows the graph of the function:

Used to write commands M-file:

In the command window, the following results were obtained:

r 1 =

r2 =

r 3 =

r4 =

8.0000

r5 =

7.9979 -8.0000

Report on the results of calculating the approximate value of the root of the equation in MS Excel.

MS Excel the calculation of the approximate value of the root of the equation was carried out using the built-in options "Selection of parameters" and "Search for solutions". To select the initial approximation, I previously built a diagram.

Calculation results using Parameter Selection

x =-9 (based on chart)

As a result of using Parameter Selection, the root was found x = -8.01.

Calculation results using Search for Solutions

We chose as the initial approximation x =-9 (based on chart)

After execution, the following result was obtained:

Finding a solution gave me meaning x = -8.00002

Description of the application created in the Delphi environment.

When creating an application in the environment Delphi the interface provided the output of the form of the function and the graph. Finding the root of a non-linear equation was implemented using three methods: Dichotomy Method, Chord Method and Newton's Method. In contrast to the calculation excel , where the roots were found using the selection of parameters and the search for a solution, the program provides for the input of the calculation accuracy by the user. The calculation results are displayed both in the application window and in a text file.

Block schemes implementing numerical methods

Flowchart for the dichotomy method:

Flowchart for chord method:

Flowchart for Newton's method:

Program listing

unit Unit1;

interface

uses

Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,

Dialogs, TeEngine, Series, ExtCtrls, TeeProcs, Chart, Menus, OleCtnrs,

StdCtrls, AxCtrls, OleCtrls, VCF1, Math;

type

TForm1 = class(TForm)

GroupBox1: TGroupBox;

OleContainer2: TOleContainer;

MainMenu1: TMainMenu;

N1: TMenuItem;

Chart1: TChart;

Series1: TPointSeries;

N2: TMenuItem;

N3: TMenuItem;

N4: TMenuItem;

N5: TMenuItem;

Label1: T Label;

Edit1: TEdit;

GroupBox2: TGroupBox;

GroupBox3: TGroupBox;

GroupBox4: TGroupBox;

Label2: T Label;

Label3: T Label;

Edit2: TEdit;

Edit3: TEdit;

Edit4: TEdit;

Label4: T Label;

Edit5: TEdit;

Label5: T Label;

Edit7: TEdit;

Label7: T Label;

F1Book1: TF1Book;

F1Book2: TF1Book;

F1Book3: TF1Book;

F1Book4: TF1Book;

Procedure N1Click(Sender: TObject);

Procedure N3Click(Sender: TObject);

Procedure FormCreate(Sender: TObject);

Procedure N4Click(Sender: TObject);

Procedure N5Click(Sender: TObject);

Private

(Private declarations)

Public

(Public declarations)

end;

const

xmin:real=-20;

xmax:real=20;

Form1: TForm1;

X,y,t,a,b,cor:real;

I,n:integer;

Fail:textfile;

implementation

($R *.dfm)

function f(x:real):real;

begin

f:=(8+x)/(x*sqrt(sqr(x)-4));

end;

function f1(x:real):real;

begin

f1:=(-power(x,3)-16*x*x+32)/(x*X*sqrt(power(x*x-4,3)));

end;

procedure metoddix(ta,tb,eps:real;var xk:real;var kolvo: integer);

begin

number:=0;

repeat

xk:=(ta+tb)/2;

kolvo:=kolvo+1;

Form1.F1book1.NumberRC:=xk;

Form1.F1book1.NumberRC:=f(xk);

if f(ta)*f(xk)<0 then tb:=xk

elseta:=xk;

until (abs(f(xk))<=eps);

end;

procedure metodhord(ta,tb,eps:real;var xk:real;var kolvo: integer);

begin

number:=0;

repeat

xk:= ta-f(ta)*(ta-tb)/(f(ta)-f(tb));

kolvo:=kolvo+1;

Form1.F1book2.NumberRC:=xk;

Form1.F1book2.NumberRC:=f(xk);

if f(ta)*f(xk)<0 then tb:=xk

elseta:=xk;

until (abs(f(xk))<=eps);

end;

procedure metodnyutona(ta,eps:real;var xk:real;var kolvo: integer);

begin

number:=0;

repeat

xk:= ta-f(ta)/f1(ta);

ta:=xk;

kolvo:=kolvo+1;

Form1.F1book3.NumberRC:=xk;

Form1.F1book3.NumberRC:=f(xk);

until (abs(f(xk))<=eps);

end;

procedure TForm1.N1Click(Sender: TObject);

begin

x:=xmin;

i:=0;

while x<=xmax do

begin

if abs(x)>5 then

Begin

I:=i+1;

Y:=f(x);

Series1.Addxy(x,y);

F1book4.NumberRC:=x;

F1book4.NumberRC:=y;

end;

x:=x+0.5;

end;

end;

procedure TForm1.N3Click(Sender: TObject); // Calculation of the root by the method of half division

begin

F1book1.ClearRange(1,1,100,2,3);

t:=strtofloat(Edit1.Text);

a:=strtofloat(Edit2.Text);

b:=strtofloat(Edit3.Text);

metoddix(a, b, t, cor, n);

F1book4.TextRC:=" dichotomy ";

F1book4.TextRC:="root=";

F1book4.NumberRC:=cor;

F1book4.TextRC:="y=";

F1book4.NumberRC:=f(cor);

F1book4.TextRC:=" number of iterations =";

F1book4.NumberRC:=n;

Append(fail);

writeln(fail);

Writeln(fail," Dichotomy calculation ");

closefile(fail);

end;

procedure TForm1.FormCreate(Sender: TObject);

begin

Assignfile(fail," report .txt");

Rewrite(fail);

closefile(fail);

end;

procedure TForm1.N4Click(Sender: TObject); // Calculation of the root by the method of chords

begin

F1book2.ClearRange(1,1,100,2,3);

t:=strtofloat(Edit1.Text);

a:=strtofloat(Edit5.Text);

b:=strtofloat(Edit4.Text);

metodhord(a, b, t, cor, n);

F1book4.TextRC:=" chords ";

F1book4.TextRC:="root=";

F1book4.NumberRC:=cor;

F1book4.TextRC:="y=";

F1book4.NumberRC:=f(cor);

F1book4.TextRC:=" number of iterations =";

F1book4.NumberRC:=n;

Assignfile(fail," report .txt");

Append(fail);

writeln(fail);

Writeln(fail," Calculation by the method of chords ");

writeln(fail,"Calculation accuracy = ",t:10:7);

Writeln(fail,"Initial guess:a = ",a:8:3," b = ",b:8:3);

writeln(fail, "Root found : x = ",cor:8:3, " y=f(x)= ",f(cor):8:6);

writeln(fail, "Number of iterations = ",n);

closefile(fail);

end;

procedure TForm1.N5Click(Sender: TObject); // Calculation of the root by Newton's method

begin

F1book3.ClearRange(1,1,100,2,3);

t:=strtofloat(Edit1.Text);

a:=strtofloat(Edit7.Text);

methododnyutona(a,t,cor,n);

F1book4.TextRC:="Newton";

F1book4.TextRC:="root=";

F1book4.NumberRC:=cor;

F1book4.TextRC:="y=";

F1book4.NumberRC:=f(cor);

F1book4.TextRC:=" number of iterations =";

F1book4.NumberRC:=n;

Assignfile(fail," report .txt");

Append(fail);

writeln(fail);

Writeln(fail," Calculation by Newton's method ");

writeln(fail,"Calculation accuracy = ",t:10:7);

Writeln(fail,"Initial guess:a = ",a:8:3," b = ",b:8:3);

writeln(fail, "Root found : x = ",cor:8:3, " y=f(x)= ",f(cor):8:6);

writeln(fail, "Number of iterations = ",n);

closefile(fail);

end;

end.

Application window image

The initial interface looks like this:

After performing the calculations for E<= 0,001:

As a report, the file “Report. txt":

Analysis of the results

In accordance with the task for the course work in the mathematical package, I found the root of a non-linear equation ( x =-8) and plotted.

In spreadsheets, the root of the equation was found using two built-in options "Parameter selection" and "Search for a solution", while "Search for a solution" still gave a more accurate value. The results almost coincided with those in matlab.

To find the root in the environment Delphi the user has the ability to enter the accuracy of the calculation from the keyboard. Testing the program showed that for the same given calculation accuracy, Newton's method finds the desired value with a smaller number of iterations.

Thus, the calculations showed that it is possible to solve a nonlinear equation in different media. The most laborious calculation turned out to be in the environment Delphi.

Literature.

1. Amosov A.A. and other computational methods for engineers, Moscow, Vysshaya Shkola, 1994.
2. Faronov V.V. Delphi. Programming in a high-level language

3 . Walkenbach D . Microsoft Office Excel 2007.user bible

Volkov V.B. Easy-to-follow Excel 2010 Tutorial

Consider the problem of finding the roots of the nonlinear equation

The roots of equation (1) are those values ​​of x that, when substituting, turn it into an identity. Only for the simplest equations it is possible to find a solution in the form of formulas, i.e. analytical form. More often it is necessary to solve equations by approximate methods, the most widespread among which, in connection with the advent of computers, are numerical methods.

The algorithm for finding roots by approximate methods can be divided into two stages. At the first, the location of the roots is studied and their separation is carried out. There is an area in which there is a root of the equation or an initial approximation to the root x 0 . The simplest way to solve this problem is to study the graph of the function f(x) . In the general case, to solve it, it is necessary to involve all means of mathematical analysis.

The existence on the found interval of at least one root of equation (1) follows from the Bolzano condition:

f(a)*f(b)<0 (2)

It is also assumed that the function f(x) is continuous on the given segment. However, this condition does not answer the question about the number of roots of the equation on a given interval. If the requirement of continuity of the function is supplemented with the requirement of its monotonicity, and this follows from the sign-constancy of the first derivative, then we can assert the existence of a unique root on a given segment.

When localizing roots, it is also important to know the basic properties of this type of equation. For example, recall some properties of algebraic equations:

where are real coefficients.

• a) An equation of degree n has n roots, among which there can be both real and complex ones. Complex roots form complex conjugate pairs and, therefore, the equation has an even number of such roots. For an odd value of n, there is at least one real root.
• b) The number of positive real roots is less than or equal to the number of variable signs in the sequence of coefficients. Replacing x with -x in equation (3) allows you to estimate the number of negative roots in the same way. iteration newton dichotomy non-linear

At the second stage of solving equation (1), using the obtained initial approximation, an iterative process is constructed that makes it possible to refine the value of the root with some predetermined accuracy. The iterative process consists in successive refinement of the initial approximation. Each such step is called an iteration. As a result of the iteration process, a sequence of approximate values ​​of the roots of the equation is found. If this sequence approaches the true value of the root x as n grows, then the iterative process converges. An iterative process is said to converge to at least order m if the following condition is satisfied:

where С>0 is some constant. If m=1 , then one speaks of first-order convergence; m=2 - about quadratic, m=3 - about cubic convergence.

The iterative cycles end if, for a given permissible error, the criteria for absolute or relative deviations are met:

or the smallness of the residual:

This work is devoted to the study of an algorithm for solving nonlinear equations using Newton's method.

Method idea. An equation is chosen in which one of the variables is most simply expressed in terms of the other variables. The resulting expression for this variable is substituted into the remaining equations of the system.

1. b) Combination with other methods.

Method idea. If the direct substitution method is not applicable at the initial stage of the solution, then equivalent system transformations are used (term by term addition, subtraction, multiplication, division), and then direct substitution is carried out directly.

2) Method of independent solution of one of the equations.

Method idea. If the system contains an equation in which there are mutually inverse expressions, then a new variable is introduced and the equation is solved with respect to it. The system then breaks down into several simpler systems.

Solve a system of equations

Consider the first equation of the system:

Making the substitution , where t ≠ 0, we obtain

Whence t 1 = 4, t 2 = 1/4.

Returning to the old variables, consider two cases.

The roots of the equation 4y 2 - 15y - 4 \u003d 0 are y 1 \u003d 4, y 2 \u003d - 1/4.

The roots of the equation 4x 2 + 15x - 4 \u003d 0 are x 1 \u003d - 4, x 2 \u003d 1/4.

3) Reduction of the system to the union of simpler systems.

1. a) Factorization by taking out the common factor.

Method idea. If one of the equations has a common factor, then this equation is decomposed into factors and, taking into account the equality of the expression to zero, they proceed to solving simpler systems.

1. b) Factoring through the solution of the homogeneous equation.

Method idea. If one of the equations is a homogeneous equation (, then, having solved it with respect to one of the variables, we factor it, for example: a (x-x 1) (x-x 2) and, given the equality of the expression to zero, we proceed to solving simpler systems.

Let's solve the first system

1. c) Using homogeneity.

Method idea. If the system has an expression that is a product of variables, then using the method of algebraic addition, a homogeneous equation is obtained, and then the factorization method is used through the solution of a homogeneous equation.

4) Method of algebraic addition.

Method idea. In one of the equations, we get rid of one of the unknowns, for this we equalize the modules of the coefficients for one of the variables, then we perform either term-by-term addition of equations, or subtraction.

5) Method of multiplication of equations.

Method idea. If there are no such pairs (x; y) for which both parts of one of the equations vanish simultaneously, then this equation can be replaced by the product of both equations of the system.

Let's solve the second equation of the system.

Let = t, then 4t 3 + t 2 -12t -12 = 0. Applying the corollary from the polynomial root theorem, we have t 1 = 2.

Р(2) = 4∙2 3 + 2 2 - 12∙2 - 12 = 32 + 4 - 24 - 12 = 0. We lower the degree of the polynomial using the method of indefinite coefficients.

4t 3 + t 2 -12t -12 = (t - 2) (at 2 + bt + c).

4t 3 + t 2 -12t -12 = at 3 + bt 2 + ct - 2at 2 -2bt - 2c.

4t 3 + t 2 - 12t -12 = at 3 + (b - 2a) t 2 + (c -2b) t - 2c.

We get the equation 4t 2 + 9t + 6 = 0, which has no roots, since D = 9 2 - 4∙4∙6 = -15<0.

Returning to the variable y, we have = 2, whence y = 4.

Answer. (1;4).

6) The method of division of equations.

Method idea. If there are no such pairs (x; y) for which both parts of one of the equations vanish simultaneously, then this equation can be replaced by an equation that is obtained by dividing one equation of the system by another.

7) The method of introducing new variables.

Method idea. Some expressions from the original variables are taken as new variables, which leads to a simpler system than the original one from these variables. After the new variables are found, it is necessary to find the values ​​of the original variables.

Returning to the old variables, we have:

We solve the first system.

8) Application of the Vieta theorem.

Method idea. If the system is composed in this way, one of the equations is presented as a sum, and the second as a product of some numbers that are the roots of some quadratic equation, then using the Vieta theorem we compose a quadratic equation and solve it.

Answer. (1;4), (4;1).

Substitution is used to solve symmetric systems: x + y = a; xy = in. When solving symmetric systems, the following transformations are used:

x 2 + y 2 \u003d (x + y) 2 - 2xy \u003d a 2 - 2c; x 3 + y 3 \u003d (x + y) (x 2 - xy + y 2) \u003d a (a 2 -3c);

x 2 y + xy 2 \u003d xy (x + y) \u003d av; (x + 1) ∙ (y + 1) \u003d xy + x + y + 1 \u003d a + b + 1;

Answer. (1;1), (1;2), (2;1).

10) "Boundary problems".

Method idea. The solution of the system is obtained by logical reasoning related to the structure of the domain of definition or the set of values ​​of functions, the study of the sign of the discriminant of the quadratic equation.

The peculiarity of this system is that the number of variables in it is greater than the number of equations. For nonlinear systems, such a feature is often a sign of a "boundary problem". Based on the type of equations, we will try to find the set of values ​​of the function that occurs in both the first and second equations of the system. Since x 2 + 4 ≥ 4, it follows from the first equation that

The answer is (0;4;4), (0;-4;-4).

11) Graphical method.

Method idea. Build graphs of functions in one coordinate system and find the coordinates of their intersection points.

1) Having rewritten the first equation of the systems in the form y \u003d x 2, we come to the conclusion: the graph of the equation is a parabola.

2) Having rewritten the second equation of the systems in the form y \u003d 2 / x 2, we come to the conclusion: the graph of the equation is a hyperbola.

3) The parabola and the hyperbola intersect at point A. There is only one point of intersection, since the right branch of the parabola serves as a graph of an increasing function, and the right branch of the hyperbola is a decreasing one. Judging by the constructed geometric model, point A has coordinates (1; 2). Verification shows that the pair (1;2) is a solution to both equations of the system.