To solve a system of linear equations with two variable method addition, you need:

1) multiply the left and right parts of one or both equations by a certain number so that the coefficients for one of the variables in the equations become opposite numbers;

2) fold term by term received equations and find the value of one of the variables;

3) substitute the found value of one variable into one of these equations and find the value of the second variable.

If in this system the coefficients for one variable are opposite numbers, then we will start solving the system immediately from point 2).

Examples. Solve a system of linear equations with two variables using the addition method.

Since the coefficients at y are opposite numbers (-1 and 1), we start the solution from point 2). We add the equations term by term and get the equation 8x = 24. The second equation of the system can be written any equation original system.

Find x and substitute its value into the 2nd equation.

We solve the 2nd equation: 9-y \u003d 14, hence y \u003d -5.

Let's do verification. Substitute the values ​​x = 3 and y = -5 into the original system of equations.

Note. The check can be done orally and not recorded if the check is not specified in the condition.

Answer: (3; -5).

If we multiply the 1st equation by (-2), then the coefficients for the variable x will become opposite numbers:

We add these equalities term by term.

We will obtain an equivalent system of equations, in which the 1st equation is the sum of two equations of the previous system, and the 2nd equation of the system we will write the 1st equation of the original system ( usually write the equation with smaller coefficients):

We find at from the 1st equation and the resulting value is substituted into the 2nd.

We solve the last equation of the system and get x = -2.

Answer: (-2; 1).

Let's make the coefficients for the variable at opposite numbers. To do this, we multiply all the terms of the 1st equation by 5, and all the terms of the 2nd equation by 2.

Substitute the value x=4 into the 2nd equation.

3 · 4 - 5y \u003d 27. Let's simplify: 12 - 5y \u003d 27, hence -5y \u003d 15, and y \u003d -3.

Answer: (4; -3).

To solve a system of linear equations with two variables using the substitution method, we proceed as follows:

1) we express one variable through another in one of the equations of the system (x through y or y through x);

2) we substitute the resulting expression into another equation of the system and get linear equation with one variable;

3) we solve the resulting linear equation with one variable and find the value of this variable;

4) the found value of the variable is substituted into expression (1) for another variable and we find the value of this variable.

Examples. Solve a system of linear equations using the substitution method.

Express X through y from the 1st equation. We get: x \u003d 7 + y. We substitute the expression (7 + y) instead of X into the 2nd equation of the system.

We got the equation: 3 · (7+y)+2y=16. This is a one variable equation at. We solve it. Let's open the brackets: 21+3y+2y=16. Collecting terms with a variable at on the left side, and the free terms on the right. When transferring a term from one part of the equality to another, we change the sign of the term to the opposite.

We get: 3y + 2y \u003d 16-21. We present like terms in each part of the equation. 5y=-5. We divide both sides of the equality by the coefficient of the variable. y=-5:5; y=-1. Substitute this value at into the expression x=7+y and find X. We get: x=7-1; x=6. A pair of variable values ​​x=6 and y=-1 is the solution to this system.

Write down: (6; -1). Answer: (6; -1). It is convenient to write these arguments as shown below, i.e., systems of equations - on the left under each other. On the right - calculations, necessary explanations, verification of the solution, etc.

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I. Ordinary differential equations

1.1. Basic concepts and definitions

A differential equation is an equation that relates an independent variable x, the desired function y and its derivatives or differentials.

Symbolically differential equation is written like this:

F(x,y,y")=0, F(x,y,y")=0, F(x,y,y",y",.., y (n))=0

A differential equation is called ordinary if the desired function depends on one independent variable.

By solving the differential equation is called such a function that turns this equation into an identity.

The order of the differential equation is the order of the highest derivative in this equation

Examples.

1. Consider the first order differential equation

The solution to this equation is the function y = 5 ln x. Indeed, by substituting y" into the equation, we get - an identity.

And this means that the function y = 5 ln x– is the solution of this differential equation.

2. Consider the second order differential equation y" - 5y" + 6y = 0. The function is the solution to this equation.

Really, .

Substituting these expressions into the equation, we get: , - identity.

And this means that the function is the solution of this differential equation.

Integration of differential equations is the process of finding solutions to differential equations.

General solution of the differential equation is called a function of the form , which includes as many independent arbitrary constants as the order of the equation.

Partial solution of the differential equation is called the solution obtained from the general solution for different numerical values ​​of arbitrary constants. The values ​​of arbitrary constants are found at certain initial values ​​of the argument and function.

The graph of a particular solution of a differential equation is called integral curve.

Examples

1. Find a particular solution to a first-order differential equation

xdx + ydy = 0, if y= 4 at x = 3.

Decision. Integrating both sides of the equation, we get

Comment. An arbitrary constant C obtained as a result of integration can be represented in any form convenient for further transformations. In this case, taking into account the canonical equation of the circle, it is convenient to represent an arbitrary constant С in the form .

is the general solution of the differential equation.

A particular solution of an equation that satisfies the initial conditions y = 4 at x = 3 is found from the general by substituting the initial conditions into the general solution: 3 2 + 4 2 = C 2 ; C=5.

Substituting C=5 into the general solution, we get x2+y2 = 5 2 .

This is a particular solution of the differential equation obtained from the general solution under given initial conditions.

2. Find the general solution of the differential equation

The solution of this equation is any function of the form , where C is an arbitrary constant. Indeed, substituting into the equations, we obtain: , .

Therefore, this differential equation has infinite set solutions, since for different values ​​of the constant C, the equality determines various solutions equations.

For example, by direct substitution, one can verify that the functions are solutions of the equation .

A problem in which it is required to find a particular solution to the equation y" = f(x, y) satisfying the initial condition y(x0) = y0, is called the Cauchy problem.

Equation solution y" = f(x, y), satisfying the initial condition, y(x0) = y0, is called a solution to the Cauchy problem.

The solution of the Cauchy problem has a simple geometric meaning. Indeed, according to these definitions, to solve the Cauchy problem y" = f(x, y) given that y(x0) = y0, means to find the integral curve of the equation y" = f(x, y) which goes through given point M0 (x0,y 0).

II. First order differential equations

2.1. Basic concepts

A first-order differential equation is an equation of the form F(x,y,y") = 0.

The first order differential equation includes the first derivative and does not include higher order derivatives.

The equation y" = f(x, y) is called a first-order equation solved with respect to the derivative.

A general solution of a first-order differential equation is a function of the form , which contains one arbitrary constant.

Example. Consider a first order differential equation.

The solution to this equation is the function .

Indeed, replacing in this equation with its value, we obtain

i.e 3x=3x

Therefore, the function is a general solution of the equation for any constant C.

Find a particular solution of this equation that satisfies the initial condition y(1)=1 Substituting initial conditions x=1, y=1 into the general solution of the equation , we obtain whence C=0.

Thus, we obtain a particular solution from the general one by substituting into this equation, the resulting value C=0 is a private decision.

2.2. Differential equations with separable variables

A differential equation with separable variables is an equation of the form: y"=f(x)g(y) or through differentials , where f(x) and g(y) are given functions.

For those y, for which , the equation y"=f(x)g(y) is equivalent to the equation in which the variable y is present only on the left side, and the variable x is present only on the right side. They say, "in the equation y"=f(x)g(y separating the variables.

Type equation is called a separated variable equation.

After integrating both parts of the equation on x, we get G(y) = F(x) + C is the general solution of the equation, where G(y) and F(x) are some antiderivatives, respectively, of functions and f(x), C arbitrary constant.

Algorithm for solving a first-order differential equation with separable variables

Example 1

solve the equation y" = xy

Decision. Derivative of a function y" replace with

we separate the variables

Let's integrate both parts of the equality:

Example 2

2yy" = 1- 3x 2, if y 0 = 3 at x0 = 1

This is a separated variable equation. Let's represent it in differentials. To do this, we rewrite this equation in the form From here

Integrating both parts of the last equality, we find

Substituting initial values x 0 = 1, y 0 = 3 find With 9=1-1+C, i.e. C = 9.

Therefore, the desired partial integral will be or

Example 3

Write an equation for a curve passing through a point M(2;-3) and having a tangent with a slope

Decision. According to the condition

This is a separable variable equation. Dividing the variables, we get:

Integrating both parts of the equation, we get:

Using the initial conditions, x=2 and y=-3 find C:

Therefore, the desired equation has the form

2.3. Linear differential equations of the first order

A first-order linear differential equation is an equation of the form y" = f(x)y + g(x)

where f(x) and g(x)- some given functions.

If a g(x)=0 then the linear differential equation is called homogeneous and has the form: y" = f(x)y

If then the equation y" = f(x)y + g(x) called heterogeneous.

General solution of a linear homogeneous differential equation y" = f(x)y given by the formula: where With is an arbitrary constant.

In particular, if C \u003d 0, then the solution is y=0 If linear homogeneous equation has the form y" = ky where k is some constant, then its general solution has the form: .

General solution of a linear inhomogeneous differential equation y" = f(x)y + g(x) given by the formula ,

those. is equal to the sum of the general solution of the corresponding linear homogeneous equation and the particular solution of this equation.

For a linear inhomogeneous equation of the form y" = kx + b,

where k and b- some numbers and a particular solution will be a constant function . Therefore, the general solution has the form .

Example. solve the equation y" + 2y +3 = 0

Decision. We represent the equation in the form y" = -2y - 3 where k=-2, b=-3 The general solution is given by the formula .

Therefore, where C is an arbitrary constant.

2.4. Solution of linear differential equations of the first order by the Bernoulli method

Finding a General Solution to a First-Order Linear Differential Equation y" = f(x)y + g(x) reduces to solving two differential equations with separated variables using the substitution y=uv, where u and v- unknown functions from x. This solution method is called the Bernoulli method.

Algorithm for solving a first-order linear differential equation

y" = f(x)y + g(x)

1. Enter a substitution y=uv.

2. Differentiate this equality y"=u"v + uv"

3. Substitute y and y" in given equation: u"v + uv" =f(x)uv + g(x) or u"v + uv" + f(x)uv = g(x).

4. Group the terms of the equation so that u take it out of brackets:

5. From the bracket, equating it to zero, find the function

This is a separable equation:

Divide the variables and get:

Where . .

6. Substitute the received value v into the equation (from item 4):

and find the function This is a separable equation:

7. Write the general solution in the form: , i.e. .

Example 1

Find a particular solution to the equation y" = -2y +3 = 0 if y=1 at x=0

Decision. Let's solve it with substitution y=uv,.y"=u"v + uv"

Substituting y and y" into this equation, we get

Grouping the second and third terms on the left side of the equation, we take out the common factor u out of brackets

We equate the expression in brackets to zero and, having solved the resulting equation, we find the function v = v(x)

We got an equation with separated variables. We integrate both parts of this equation: Find the function v:

Substitute the resulting value v into the equation We get:

This is a separated variable equation. We integrate both parts of the equation: Let's find the function u = u(x,c) Let's find a general solution: Let us find a particular solution of the equation that satisfies the initial conditions y=1 at x=0:

III. Higher order differential equations

3.1. Basic concepts and definitions

A second-order differential equation is an equation containing derivatives not higher than the second order. In the general case, the second-order differential equation is written as: F(x,y,y",y") = 0

The general solution of a second-order differential equation is a function of the form , which includes two arbitrary constants C1 and C2.

A particular solution of a second-order differential equation is a solution obtained from the general one for some values ​​of arbitrary constants C1 and C2.

3.2. Linear homogeneous differential equations of the second order with constant ratios.

Linear homogeneous differential equation of the second order with constant coefficients is called an equation of the form y" + py" + qy = 0, where p and q are constant values.

Algorithm for solving second-order homogeneous differential equations with constant coefficients

1. Write the differential equation in the form: y" + py" + qy = 0.

2. Compose its characteristic equation, denoting y" through r2, y" through r, y in 1: r2 + pr +q = 0

1. Substitution method: from any equation of the system we express one unknown in terms of another and substitute it into the second equation of the system.

Task. Solve the system of equations:

Decision. From the first equation of the system, we express at through X and substitute into the second equation of the system. Let's get the system equivalent to the original.

After bringing such terms, the system will take the form:

From the second equation we find: . Substituting this value into the equation at = 2 - 2X, we get at= 3. Therefore, the solution of this system is a pair of numbers .

2. Method algebraic addition : by adding two equations, get an equation with one variable.

Task. Solve the system equation:

Decision. Multiplying both sides of the second equation by 2, we get the system equivalent to the original. Adding the two equations of this system, we arrive at the system

After reducing similar terms, this system will take the form: From the second equation we find . Substituting this value into Equation 3 X + 4at= 5, we get , where . Therefore, the solution of this system is a pair of numbers .

3. Method for introducing new variables: we are looking for some repeated expressions in the system, which we will denote by new variables, thereby simplifying the form of the system.

Task. Solve the system of equations:

Decision. Let's write down this system otherwise:

Let be x + y = u, hu = v. Then we get the system

Let's solve it by the substitution method. From the first equation of the system, we express u through v and substitute into the second equation of the system. Let's get the system those.

From the second equation of the system we find v 1 = 2, v 2 = 3.

Substituting these values ​​into the equation u = 5 - v, we get u 1 = 3,
u 2 = 2. Then we have two systems

Solving the first system, we get two pairs of numbers (1; 2), (2; 1). The second system has no solutions.

Exercises for independent work

1. Solve systems of equations using the substitution method.

Equations and systems of equations of the first degree

Two numbers or some expressions connected by the sign "=" form equality. If the given numbers or expressions are equal for any values ​​of the letters, then such equality is called identity.

For example, when it is stated that for any a valid:

a + 1 = 1 + a, here equality is an identity.

Equation is called an equality containing unknown numbers marked with letters. These letters are called unknown. There can be more than one unknown in an equation.

For example, in equation 2 X + at = 7X– 3 two unknowns: X and at.

The expression on the left side of the equation (2 X + at) is called the left side of the equation, and the expression on the right side of the equation (7 X– 3) is called its right side.

The value of the unknown at which the equation becomes an identity is called decision or root equations.

For example, if in equation 3 X+ 7=13 instead of unknown X substitute the number 2, we get the identity. Therefore, the value X= 2 satisfies the given equation and the number 2 is the solution or root of the given equation.

The two equations are called equivalent(or equivalent), if all solutions of the first equation are solutions of the second and vice versa, all solutions of the second equation are solutions of the first. To equivalent equations also include equations that have no solutions.

For example Equations 2 X– 5 = 11 and 7 X+ 6 = 62 are equivalent since they have the same root X= 8; equations X + 2 = X+ 5 and 2 X + 7 = 2X are equivalent because both have no solutions.

Properties of equivalent equations

1. To both sides of the equation, you can add any expression that makes sense for all allowed values unknown; the resulting equation will be equivalent to the given one.

Example. Equation 2 X– 1 = 7 has a root X= 4. Adding 5 to both sides, we get equation 2 X– 1 + 5 = 7 + 5 or 2 X+ 4 = 12 which has the same root X = 4.

2. If both parts of the equation have the same terms, then they can be omitted.

Example. Equation 9 x + 5X = 18 + 5X has one root X= 2. Omitting in both parts 5 X, we get equation 9 X= 18 which has the same root X = 2.

3. Any term of the equation can be transferred from one part of the equation to another by changing its sign to the opposite.

Example. Equation 7 X - 11 = 3 has one root X= 2. If we transfer 11 to right side with opposite sign, we get equation 7 X= 3 + 11 which has the same solution X = 2.

4. Both parts of the equation can be multiplied by any expression (number) that makes sense and is non-zero for all admissible values ​​of the unknown, the resulting equation will be equivalent to this one.

Example. Equation 2 X - 15 = 10 – 3X has a root X= 5. Multiplying both sides by 3, we get the equation 3(2 X - 15) = 3(10 – 3X) or 6 X – 45 =30 – 9X, which has the same root X = 5.

5. The signs of all terms of the equation can be reversed (this is equivalent to multiplying both parts by (-1)).

Example. Equation - 3 x + 7 = - 8 after multiplying both parts by (-1) will take the form 3 X - 7 = 8. The first and second equations have a single root X = 5.

6. Both sides of the equation can be divided by the same number other than zero (that is, not equal to zero).

Example..gif" width="49 height=25" height="25">.gif" width="131" height="28"> is equivalent to this one because it has the same two roots: and https:/ /pandia.ru/text/78/105/images/image006_96.gif" width="125" height="48 src="> after multiplying both parts by 14, it will look like this:

https://pandia.ru/text/78/105/images/image009_71.gif" width="77 height=20" height="20">, where arbitrary numbers, X- unknown, called first degree equation with one unknown(or linear equation with one unknown).

Example. 2 X + 3 = 7 – 0,5X ; 0,3X = 0.

A first degree equation with one unknown always has one solution; a linear equation may not have solutions () or have an infinite number of them (https://pandia.ru/text/78/105/images/image013_59.gif" width="344 height=48" height="48">.

Decision. Multiply all terms in the equation by the least common multiple of the denominators, which is 12.

https://pandia.ru/text/78/105/images/image015_49.gif" width="183 height=24" height="24">.gif" width="371" height="20 src="> .

We group in one part (left) the terms containing the unknown, and in the other part (right) - the free terms:

https://pandia.ru/text/78/105/images/image019_34.gif" width="104" height="20">. Dividing both parts by (-22), we get X = 7.

Systems of two equations of the first degree with two unknowns

An equation like https://pandia.ru/text/78/105/images/image021_34.gif" width="87" height="24 src="> is called first degree equation with two unknowns x and at. If found general solutions two or more equations, then they say that these equations form a system, they are usually written one under the other and combined with a curly bracket, for example.

Each pair of unknowns that simultaneously satisfies both equations of the system is called system solution. Solve the system- this means finding all solutions of this system or showing that it does not have them. The two systems of equations are called equivalent (equivalent), if all solutions of one of them are solutions of the other, and vice versa, all solutions of the other are solutions of the first.

For example, the solution to the system is a pair of numbers X= 4 and at= 3. These numbers are also the only solution systems . Therefore, these systems of equations are equivalent.

Ways to solve systems of equations

1. Method of algebraic addition. If the coefficients for some unknown in both equations are equal in absolute value, then by adding both equations (or subtracting one from the other), you can get an equation with one unknown. By solving this equation, one unknown is determined, and by substituting it into one of the equations of the system, the second unknown is found.

Examples: Solve systems of equations: 1) .

Here the coefficients for at are equal in absolute value but opposite in sign. To get an equation with one unknown equation we add the systems term by term:

Received value X= 4 we substitute into some equation of the system, for example, into the first one, and find the value at: .

Answer: X = 4; at = 3.

2) https://pandia.ru/text/78/105/images/image029_23.gif" width="112" height="57 src=">.gif" width="220" height="87 src=" >

https://pandia.ru/text/78/105/images/image033_21.gif" width="103" height="47 src=">.

2. Substitution method. From any equation of the system, we express one of the unknowns in terms of the rest, and then we substitute the value of this unknown into the remaining equations. Consider this method with specific examples:

1) Let's solve the system of equations. Let us express one of the unknowns from the first equation, for example X: https://pandia.ru/text/78/105/images/image036_18.gif" width="483" height="24 src=">

Substitute at= 1 into the expression for X, we get .

Answer: https://pandia.ru/text/78/105/images/image039_18.gif" width="99" height="55 src=">. In this case, it is convenient to express at from the second equation:

https://pandia.ru/text/78/105/images/image041_16.gif" width="660" height="24">Substitute the value X= 5 into the expression for at, we get https://pandia.ru/text/78/105/images/image043_15.gif" width="96" height="24 src=">.

3) Let's solve the system of equations https://pandia.ru/text/78/105/images/image045_12.gif" width="205" height="48">. Substituting this value into the second equation, we get an equation with one unknown at: https://pandia.ru/text/78/105/images/image049_11.gif" width="128" height="48">

Answer: https://pandia.ru/text/78/105/images/image051_12.gif" width="95" height="108 src="> .

Let's rewrite the system as: . We replace the unknowns by setting , we get linear system ..gif" width="11 height=17" height="17"> into the second equation, we get an equation with one unknown:

Substituting the value v into the expression for t, we get: https://pandia.ru/text/78/105/images/image060_9.gif" width="92 height=51" height="51"> we find .

Answer: https://pandia.ru/text/78/105/images/image062_9.gif" width="120" height="57">, where are coefficients for unknowns, https://pandia.ru/text/ 78/105/images/image065_10.gif" width="67" height="52 src=">, then the system has the only thing decision.

B) If https://pandia.ru/text/78/105/images/image067_9.gif" width="105" height="52 src=">, then the system has infinite set solutions.

Example..gif" width="47" height="48 src=">), so the system has a unique solution.

Really, .

https://pandia.ru/text/78/105/images/image073_7.gif" width="115" height="48 src=">.

Example..gif" width="91 height=48" height="48"> or after reduction , hence the system has no solutions.

Example..gif" width="116 height=48" height="48"> or after shortening , so the system has an infinite number of solutions.

Equations Containing Modulus

When solving equations containing a module, the concept of a module is used real number. module (absolute value ) real number a the number itself is called if and opposite number (– a), if https://pandia.ru/text/78/105/images/image082_7.gif" width="20" height="28">.

So, https://pandia.ru/text/78/105/images/image084_8.gif" width="44" height="28 src=">, since the number 3 > 0; , since the number is 5< 0, поэтому ; , as (); , as .

Module properties:

1) https://pandia.ru/text/78/105/images/image093_7.gif" width="72" height="28 src=">

3) https://pandia.ru/text/78/105/images/image095_8.gif" width="123" height="56 src=">

5) https://pandia.ru/text/78/105/images/image097_7.gif" width="73" height="28 src=">.

Considering that the expression under the module can take two values ​​https://pandia.ru/text/78/105/images/image099_8.gif" width="68" height="20 src=">, then this equation reduces to solving two equations: and or and ..gif" width="52" height="20 src=">. Let's do a check by substituting each value X into the condition: if https://pandia.ru/text/78/105/images/image106_5.gif" width="165" height="28 src=">..gif" width="144" height="28 src=">.

Answer: https://pandia.ru/text/78/105/images/image104_6.gif" width="49" height="20 src=">.

Example..gif" width="408" height="55">

Answer: https://pandia.ru/text/78/105/images/image111_6.gif" width="41" height="20 src=">.

Example..gif" width="137" height="20"> and . Set aside the resulting values X on the numerical axis, breaking it into intervals:

If https://pandia.ru/text/78/105/images/image117_5.gif" width="144" height="24">, because in this interval, both expressions are under the module sign less than zero, and, removing the module, we must change the sign of the expression to the opposite. Let's solve the resulting equation:

Gif" width="75 height=24" height="24">. The boundary value can be included in both the first and second span, just as the value can be included in both the second and third. In the second interval, our equation will take the form: - this expression does not make sense, i.e., on this interval, the equation of solutions does not have solutions under the modulus sign, we equate them to zero. We find the roots of all expressions, with

Next spacing https://pandia.ru/text/78/105/images/image124_6.gif" width="225" height="20">..gif" width="52" height="20 src="> .gif" width="125" height="25">, where a, b, c are arbitrary numbers ( a≠ 0), and x is a variable called square. To solve this equation, you need to calculate the discriminant D = b 2 – 4ac. If a D> 0, then the quadratic equation has two solutions (roots): and .

If a D= 0, the quadratic equation obviously has two identical solutions(multiples of the root).

If a D< 0, квадратное уравнение не имеет real roots.

If one of the coefficients b or c zero, then the quadratic equation can be solved without calculating the discriminant:

1) https://pandia.ru/text/78/105/images/image131_5.gif" width="28" height="18 src="> x(ax+ b)=0

2)ax 2 + c = 0 ax 2 = – c; if https://pandia.ru/text/78/105/images/image135_3.gif" width="101" height="52">.

There are dependencies between the coefficients and roots of the quadratic equation, known as formulas or Vieta's theorem:

Bisquare equations are equations of the form https://pandia.ru/text/78/105/images/image138_4.gif" width="53" height="29">, then from the original equation we get a quadratic equation, from which we find at, and then X, according to the formula .

Example. solve the equation . We bring the expressions in both parts of the equality to common denominator..gif" width="212" height="29 src=">. We solve the resulting quadratic equation: , in this equation a= 1, b= –2,c= -15, then the discriminant is equal to: D = b 2 – 4ac= 64. Equation roots: , ..gif" width="130 height=25" height="25">. We make the replacement. Then the equation becomes is a quadratic equation, where a= 1, b= – 4,c= 3, its discriminant is: D = b 2 – 4ac = 16 – 12 = 4.

The roots of the quadratic equation are equal, respectively: and .

Roots of the original equation , , , ..gif" width="78" height="51">, where PN(x) and Pm(x) are polynomials of degrees n and m respectively. A fraction is zero if the numerator is zero and the denominator is not, but such a polynomial equation is mainly obtained only after lengthy transformations, transitions from one equation to another. In the process of solving, therefore, each equation is replaced by some new one, and the new one may have new roots. To follow these changes in the roots, to prevent the loss of roots and to be able to reject the extra ones is the task right decision equations.

It is clear that the best way- each time replace one equation with an equivalent one, then the roots of the last equation will be the roots of the original one. However, such perfect path difficult to implement in practice. As a rule, the equation is replaced by its consequence, which is not necessarily equivalent to it at all, while all the roots of the first equation are the roots of the second, i.e., the loss of roots does not occur, but extraneous ones may appear (or may not appear). In the case when at least once in the process of transformations the equation was replaced by an unequal one, we need mandatory check obtained roots.

So, if the decision was carried out without an analysis of the equivalence and sources of occurrence extraneous roots, check is obligatory part solutions. Without verification, the solution will not be considered complete, even if extraneous roots have not appeared. When they appeared and were not discarded, then this decision is simply wrong.

Here are some properties of a polynomial:

The root of the polynomial call the value x, for which the polynomial is equal to zero. Any polynomial of degree n has exactly n roots. If the polynomial equation is written as , then , where x 1, x 2,…, xn are the roots of the equation.

Any polynomial has even degree with real coefficients there is at least one real root, but in general it always has odd number real roots. A polynomial of even degree may not have real roots, and when they do, their number is even.

A polynomial under any circumstances can be decomposed into linear factors and square trinomials with negative discriminant. If we know its root x 1, then PN(x) = (x -x 1) Pn- 1(x).

If a PN(x) = 0 is an equation of even degree, then in addition to the method of factoring it, you can try to introduce a change of variable, with the help of which the degree of the equation will decrease.

Example. Solve the equation:

This equation of the third (odd) degree means that it is impossible to introduce an auxiliary variable that will lower the degree of the equation. It must be solved by factoring the left side, for which we first open the brackets, and then write it in standard form.

We get: x 3 + 5x – 6 = 0.

This is the reduced equation (coefficient at the highest degree equal to one), so we are looking for its roots among the factors of the free term - 6. These are the numbers ±1, ±2, ±3, ±6. Substituting x= 1 into the equation, we see that x= 1 is its root, so the polynomial x 3 + 5x–6 = 0 divided by ( x- 1) no residue. Let's do this division:

x 3 + 5x –6 = 0 x- 1

x 3 – x 2 x 2+x + 6

x 2 + 5x- 6

x 2– x

https://pandia.ru/text/78/105/images/image167_4.gif"> 6 x- 6

https://pandia.ru/text/78/105/images/image168_4.gif" width="50"> 6 x- 6

So x 3 + 5x –6 = 0; (x- 1)(x 2+ x + 6) = 0

The first equation gives the root x= 1, which is already selected, and in the second equation D< 0, it does not have real solutions. Since the ODZ of this equation , it is possible not to check.

Example..gif" width="52" height="21 src=">. If you multiply the first factor with the third, and the second with the fourth, then these products will have the same parts, which depend on x: (x 2 + 4x – 5)(x 2 + 4x – = 0.

Let be x 2 + 4x = y, then we write the equation in the form ( y – 5)(y- 21) – 297 = 0.

This quadratic equation has solutions: y 1 = 32, y 2 = - 6 ..gif" width="140" height="61 src=">; ODZ: x ≠ – 9.

If we reduce this equation to a common denominator, a polynomial of the fourth degree will appear in the numerator. So, it is allowed to change the variable, which will lower the degree of the equation. Therefore, it is not necessary to immediately reduce this equation to a common denominator. Here you can see that on the left is the sum of the squares. So, you can add it to full square sums or differences. In fact, subtract and add twice the product of the bases of these squares: https://pandia.ru/text/78/105/images/image179_3.gif" width="80" height="59 src=">, then y 2 + 18y– 40 = 0. According to the Vieta theorem y 1 = 2; y 2 = – 20. https://pandia.ru/text/78/105/images/image183_4.gif" width="108 height=32" height="32">, and in the second D< 0. Эти корни удовлетворяют ОДЗ

Answer: https://pandia.ru/text/78/105/images/image185_4.gif" width="191 height=51" height="51">.gif" width="73 height=48" height=" 48"> .gif" width="132" height="50 src=">.

We get a quadratic equation a(y 2 https://pandia.ru/text/78/105/images/image192_3.gif" width="213" height="31">.

Irrational equations

irrational called an equation in which the variable is contained under the sign of the radical (root ) or under the sign of elevation to fractional degree()..gif" width="120" height="32"> and have the same domain of definition of the unknown. When squaring the first and second equations, we get the same equation . The solutions of this equation are the solutions of both irrational equations.

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