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Basic methods for solving equations. Lesson “Equivalence of equations Checking roots. Loss of roots when solving equations Reasons for the appearance of extraneous roots when solving equations

Basic methods for solving equations. Lesson “Equivalence of equations Checking roots. Loss of roots when solving equations Reasons for the appearance of extraneous roots when solving equations

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Subject trigonometric equations begins with a school lecture, which is structured in the form of a heuristic conversation. The lecture discusses theoretical material and examples of solving all typical problems according to the plan:

The simplest trigonometric equations.
Basic methods for solving trigonometric equations.
Homogeneous equations.

In the following lessons, independent skills development begins, based on the application of the principle of joint activity between teacher and student. First, goals for students are set, i.e. it is determined who wants to know no more than what is required by the state standard, and who is ready to do more.

The final diagnosis is created taking into account level differentiation, which allows students to consciously determine the minimum knowledge that is necessary to receive a grade of “3”. Based on this, multi-level materials are selected to diagnose students’ knowledge. Such work allows for an individual approach to students, including everyone in conscious learning activities, developing self-organization and self-learning skills, and ensuring a transition to active, independent thinking.

The seminar is conducted after practicing the basic skills of solving trigonometric equations. Several lessons before the seminar, students are given questions that will be discussed during the seminar.

The seminar consists of three parts.

1. The introductory part covers all the theoretical material, including an introduction to the problems that will arise when solving complex equations.

2. The second part discusses the solution of equations of the form:

and cosx + bsinx = c.
a (sinx + cosx) + bsin2x + c = 0.
equations solvable by reducing the degree.

These equations use universal substitution, degree reduction formulas, and the auxiliary argument method.

3. The third part discusses the problems of root loss and acquisition extraneous roots. Shows how to select roots.

Pupils work in groups. To solve the examples, well-trained guys are called in who can show and explain the material.

The seminar is designed for a well-prepared student, because... it addresses issues somewhat beyond the scope of the program material. It includes equations of a more complex form, and especially addresses problems encountered in solving complex trigonometric equations.

The seminar was held for students in grades 10–11. Each student had the opportunity to expand and deepen their knowledge on this topic, to compare the level of their knowledge not only with the requirements for a school graduate, but also with the requirements for those entering V.U.Z.

SEMINAR

Subject:"Solving Trigonometric Equations"

Goals:

Generalize knowledge on solving trigonometric equations of all types.
Focus on problems: loss of roots;

extraneous roots; root selection.

DURING THE CLASSES.

I. Introductory part

1. Basic methods for solving trigonometric equations
Factorization.
Introduction of a new variable.

Functional-graphic method.

2. Some types of trigonometric equations.

Equations that reduce to quadratic equations for cos x = t, sin x = t.

Asin 2 x + Bcosx + C = 0; Acos 2 x + Bsinx + C = 0.

They are solved by introducing a new variable.

Homogeneous equations of the first and second degree First degree equation:

Asinx + Bcosx = 0 divide by cos x, we get Atg x + B = 0 Second degree equation:

Asin 2 x + Bsinx cosx + Сcos 2 x = 0 divide by cos 2 x, we get Atg 2 x + Btgx + C = 0

They are solved by factorization and by introducing a new variable.

All methods apply.

Downgrade:

1). Аcos2x + Вcos 2 x = C; Acos2x + Bsin 2 x = C.

Solved by factorization method.

2). Asin2x + Bsin 2 x = C; Asin2x + Bcos 2 x = C. Equation of the form:

A(sinx + cosx) + Bsin2x + C = 0.

Reduced to square with respect to t = sinx + cosx;

sin2x = t 2 – 1.

3. Formulas.
x + 2n; Checking is required!

Decreasing power: cos 2 x = (1 + cos2x): 2; sin 2 x = (1 – cos 2x): 2

Auxiliary argument method.

Let us replace Acosx + Bsinx with Csin (x + ), where sin = a/C; cos=v/c;

If you see a square, lower the degree.
If you see a piece, make an amount.
If you see the amount, do the work.

5. Loss of roots, extra roots.

Loss of roots: divide by g(x); dangerous formulas (universal substitution). With these operations we narrow the scope of definition.

Excess roots: raised to an even power; multiply by g(x) (get rid of the denominator).

With these operations we expand the scope of definition.

II. Examples of trigonometric equations

1) 1. Equations of the form Asinx + Bcosx = C

Universal substitution.O.D.Z. x – any.

3 sin 2x + cos 2x + 1= 0.

tgx = u. x/2 + n;

u = – 1/3.

tan x = –1/3, x = arctan (–1/3) + k, k Z. Examination:

3sin( + 2n) + cos( + 2n) + 1= 3 sin + cos + 1 = 0 – 1 + 1 = 0.

x = /2 + n, n e Z. Is the root of the equation. Answer:

2) x = arctan(–1/3) + k, k Z. x = /2 + n, n Z.

Functional-graphic method. O.D.Z. x – any.
Sinx – cosx = 1

Sinx = cosx + 1.

x = /2 + n, n e Z. Is the root of the equation. Let's plot the functions: y = sinx, y = cosx + 1.

3) x = /2 + 2 n, Z; x = + 2k, k Z.

Introduction of an auxiliary argument. O.D.Z.: x – any.

8cosx + 15 sinx = 17.

8/17 cosx + 15/17 sinx = 1, because (8/17) 2 + (15/17) 2 = 1, then there exists such that sin = 8/17,

x = /2 + n, n e Z. Is the root of the equation. cos = 15/17, which means sin cosx + sinx cos = 1; = arcsin 8/17.

x = /2 + 2n – , x = /2 + 2n – arcsin 8/17, n Z.

2. Reducing the order: Acos2x + Bsin2x = C. Acos2x + Bcos2x = C.

1). sin 2 3x + sin 2 4x + sin 2 6x + sin 2 7x = 2. O.D.Z.: x – any.
1 – cos 6x + 1 – cos 8x + 1 – cos 12x + 1 – cos 14x = 4
cos 6x + cos 8x + cos 12x + cos 14x = 0
2cos10x cos 4x + 2cos 10x cos 2x = 0
2cos 10x(cos 4x + cos 2x) = 0
2cos10x 2cos3x cosx = 0

x = /2 + n, n e Z. Is the root of the equation. cos10x = 0, cos3x = 0, cosx = 0.

x = /20 + n/10, n Z. x = /6 + k/3, k Z, x = /2 + m, m Z.

At
k = 1 and m = 0

k = 4 and m = 1.

the series are the same.

3. Reduction to homogeneity. Asin2x + Bsin 2 x = C, Asin2x + Bcos 2 x = C.
1) 5 sin 2 x + 3 sinx cosx + 6 cos 2 x = 5. ODZ: x – any.
5 sin 2 x + 3 sinx cosx + 6cos 2 x – 5 sin 2 x – 5 cos 2 x = 0
3 sinxcosx + cos 2 x = 0 (1) cannot be divided by cos 2 x, since we lose roots.
cos 2 x = 0 satisfies the equation.
cosx (3 sinx + cosx) = 0
cosx = 0. 3 sinx + cosx = 0.

x = /2 + n, n e Z. Is the root of the equation. x = /2 + k, k Z. tgx = –1/3, x = –/6 + n, n Z.

x = /2 + k, k Z. , x = –/6 + n, n Z

4. Equation of the form: A(sinx + cosx) + B sin2x + C = 0.
1). 4 + 2sin2x – 5(sinx + cosx) = 0. O.D.Z.: x – any.
sinx + cosx = t, sin2x = t 2 – 1. < 2
4 + 2t 2 – 2 – 5t = 0, | t |
2 t 2 – 5t + 2 = 0. t 1 = 2, t 2 = S.
sinx + cosx = S. cosx = sin(x + /2),
sinx +sin(x + /2) = 1/2,
2sin(x + /4) cos(–/4) = 1/2
x +/4 = (–1) k arcsin(1/2 O 2) + k, k Z.

x = /2 + n, n e Z. Is the root of the equation. x = (–1) k arcsin(1/22) – /4 + k, k Z.

5. Factorization.

1) cos 2 x – 2 cosx = 4 sinx – sin2x
cosx(cosx – 2) = 2 sinx (2 – cosx),
(cosx – 2)(cosx + 2 sinx) = 0.

1) cosx = 2, no roots.
2) cosx + 2 sinx = 0
2tgx + 1 = 0

x = /2 + n, n e Z. Is the root of the equation. x = arctan(1/2) + n, n Z.

III. Problems arising when solving trigonometric equations

1. Loss of roots: divide by g(x); We use dangerous formulas.

1) Find the error.

1 – cosx = sinx *sinx/2,
1 – cosx = 2sin 2 x/2 formula.
2 sin 2 x/2 = 2 sinx/2* сosx/2* sinx/2 divide by 2 sin 2 x/2,
1 = cosx/2
x/2 = 2n, x = 4n, n" Z.
Lost roots sinx/2 = 0, x = 2k, k Z.

Correct solution: 2sin 2 x/2(1 – cosx/2) = 0.

sin 2 x/2 = 0
x = 2k, k Z.

1 – cosx /2 = 0
x = 4p n, n Z.

2. Extraneous roots: we get rid of the denominator; raise to an even power.

1). (sin4x – sin2x – сos3x + 2sinx – 1) : (2sin2x – 3) = 0. O.D.Z.: sin2x 3 / 2.

2cos3x sinx – cos3x + 2sinx – 1 = 0
(cos3x + 1)(2sinx – 1) = 0

1). сos3x + 1 = 0
x = /3 + 2n/3, n Z.

2). 2sinx – 1 = 0
x = (–1) k /6 + k, k Z.

I. x = /3 + 2n/3
1. n = 0
sin 2 /3 = 3 / 2
do not satisfy. O.D.Z.

2. n = 1
sin 2= 0
satisfy O.D.Z.

3. n = 2
sin 2/ 3 = –3 / 2
satisfy O.D.Z.

II. x = (–1) k /6 + k, k Z
1.k = 0
sin 2/6 = 3 / 2
do not satisfy O.D.Z.
2. k = 1
sin 2*5/6 = –3 / 2
satisfy O.D.Z.

x = /2 + n, n e Z. Is the root of the equation. x = + 2k, x = 5/3 + 2k, x = 5/6 + 2k, k Z. t = 5 sin3x = 0

Loss of roots and extraneous roots when solving equations

Municipal educational institution "Secondary school No. 2 with in-depth study individual items"city of Vsevolozhsk. Research work prepared by a student of grade 11B: Vasily Vasiliev. Project manager: Egorova Lyudmila Alekseevna.

Equation First, consider various ways solutions to this equation sinx+cosx =- 1

Solution No. 1 sinx+cosx =-1 i Y x 0 1 sin(x+)=- 1 sin(x+)=- x+ =- +2 x+ = +2 + x=- +2 x= +2 Answer: +2

Solution No. 2 sinx+cosx =- 1 i Answer: +2 y x 0 1 2sin cos + - + + = 0 sin cos + = 0 cos (cos + sin)= 0 cos =0 cos + sin =1 = + m tg =-1 = + m =- + x=- +2 x= +2

Solution No. 3 I y x 0 1 sinx+cosx =- 1 2 = x= x+ x sin2x=0 2x= x= Answer:

sinx+cosx =-1 Solution No. 4 i y x 0 1 + =- 1 2tg +1- =-1- 2tg =- 2 =- + n x= - + 2 n Answer: - + 2 n

Let's compare solutions Correct solutions Let's figure out in what cases extraneous roots may appear and why No. 2 Answer: +2 No. 3 Answer: No. 4 Answer: + 2 n No. 1 Answer: +2

Checking the solution Is it necessary to check? Check the roots just in case, to be on the safe side? This is of course useful when it’s easy to substitute, but mathematicians are rational people and don’t do unnecessary things. Let's look at different cases and remember when verification is really needed.

1. The simplest ready-made formulas c osx =a x=a =a s inx =a t gx =a In cases where the roots are found using the simplest, ready-made formulas, the check does not need to be done. However, when using such formulas, you should remember the conditions under which they can be used. For example, the formula = can be used under the condition a 0, -4ac 0 And the answer x= arccos2+2 for the equation cosx =2 is considered a gross error, since the formula x= arccos a +2 can only be used for the roots of the equation cosx =a, where | a | 1

2. Transformations More often, when solving equations, you have to carry out a lot of transformations. If an equation is replaced by a new one that has all the roots of the previous one, and it is transformed so that no loss or acquisition of roots occurs, then such equations are called equivalent. 1. When transferring the components of an equation from one part to another. 2. When adding the same number to both sides. 3. When both sides of an equation are multiplied by the same non-zero number. 4 . When applying identities that are true on the set of all real numbers. However, verification is not required!

However, not every equation can be solved by equivalent transformations. More often it is necessary to apply unequal transformations. Often such transformations are based on the use of formulas that are not valid for all real values. In this case, in particular, the domain of definition of the equation changes. This error is found in solution #4. Let's look at the error, but first let's look again at solution No. 4. sinx+cosx=-1 + =-1 2tg +1- =-1- 2tg =-2 =- + n x = - + 2 n The error lies in the formula sin2x= This formula can be used, but you should additionally check whether the roots are numbers of the form + for which tg is not defined. Now it is clear that the solution is the loss of roots. Let's see it through to the end.

Solution No. 4 i y x 0 1 Let's check the numbers = + n by substitution: x= + 2 n sin(+ 2 n)+ cos (+ 2 n)=sin + cos =0+(-1)=- 1 So x= +2 n is the root of the equation Answer: +2 sinx+cosx =-1 + =- 1 2tg +1- =-1- 2tg =- 2 =- + n x= - + 2 n

We looked at one of the ways to lose roots; there are a great many of them in mathematics, so you need to solve carefully, remembering all the rules. Just as you can lose the roots of an equation, you can also acquire extra ones in the course of solving it. Let's consider solution No. 3 in which such an error was made.

Solution #3 I y x 0 1 2 2 and extra roots! Extraneous roots could appear when both sides of the equation were squared. In this case, it is necessary to check. For n=2k we have sin k+cos k=-1; cos k=-1 for k=2m-1 , Then n=2(2m+1)=4m+2 , x= = +2 m , Answer: +2 For n=2k+1 we have sin +cos =- 1 sin(+ k)+ cos (+ k)=- 1 cos k-sin k=- 1 cos k=-1 at k=2m+1 n=2(2m+1)+ 1=2m+3 x= (4m+3)= +2 m=- +2 sinx+cosx =- 1 = x= x+ x sin2x=0 2x= x=

So, we looked at a couple of possible cases, of which there are a great many. Try not to waste your time and make stupid mistakes.

§ 1. LOST AND EXTRAINED ROOTS WHEN SOLVING EQUATIONS (BY EXAMPLES)

REFERENCE MATERIAL

1. Two theorems in § 3 of Chapter VII talked about what actions on equations do not violate their equivalence.

2. Let us now consider such operations on equations that can lead to a new equation that is unequal to the original equation. Instead of general considerations, we will limit ourselves to considering only specific examples.

3. Example 1. Given an equation. Open the brackets in this equation, move all the terms to the left side and solve quadratic equation. Its roots are

If we reduce both sides of the equation by common multiplier then you get an equation that is unequal to the original one, since it has only one root

Thus, reducing both sides of the equation by a factor containing the unknown may result in the roots of the equation being lost.

4. Example 2. Given the equation This equation has a single root. Let us square both sides of this equation, and we get. Solving this equation, we find two roots:

We see that the new equation is not equivalent to the original equation. The root is the root of the equation which, after squaring both sides, leads to the equation

5. Extraneous roots can also appear when both sides of the equation are multiplied by a factor containing an unknown, if this factor vanishes for real values of x.

Example 3. If we multiply both sides of the equation by then we get a new equation which, after transferring the term from the right side to the left and factoring it, gives an equation from either

The root does not satisfy an equation that has only one root

From here we conclude: when squaring both sides of the equation (in general to an even power), as well as when multiplying by a factor containing an unknown and vanishing at real values of the unknown, extraneous roots may appear.

All considerations expressed here on the issue of the loss and appearance of extraneous roots of an equation apply equally to any equations (algebraic, trigonometric, etc.).

6. An equation is called algebraic if only algebraic operations are performed on the unknown - addition, subtraction, multiplication, division, exponentiation and root extraction. natural indicator(and the number of such operations is finite).

So, for example, the equations

are algebraic, and the equations

The following transformations are most often used when solving equations:

Other transformations

In the list presented in the previous paragraph, we deliberately did not include such transformations as raising both sides of the equation to the same natural degree, logarithm, potentiation of both sides of the equation, extraction of the root of one degree from both sides of the equation, release from the external function and others. The fact is that these transformations are not so general: transformations from the above list are used to solve equations of all types, and the transformations just mentioned are used to solve certain types of equations (irrational, exponential, logarithmic, etc.). They are discussed in detail within the framework of the corresponding methods for solving the corresponding types of equations. Here are links to their detailed descriptions:

Raising both sides of an equation to the same natural power.
Taking logarithms of both sides of the equation.
Potentiating both sides of the equation.
Extracting the root of the same power from both sides of an equation.
Replacing an expression corresponding to one of the parts of the original equation with an expression from another part of the original equation.

The links provided contain comprehensive information on the listed transformations. Therefore, we will no longer dwell on them in this article. All subsequent information applies to transformations from the list of basic transformations.

What is the result of transforming the equation?

Carrying out all the above transformations can give either an equation that has the same roots as the original equation, or an equation whose roots contain all the roots of the original equation, but which may also have other roots, or an equation whose roots will not include all roots of the transformed equation. In the following paragraphs we will analyze which of these transformations, under which conditions, lead to which equations. This is extremely important to know for successfully solving equations.

Equivalent transformations of equations

Of particular interest are transformations of equations that result in equivalent equations, that is, equations that have the same set of roots as the original equation. Such transformations are called equivalent transformations. In school textbooks, the corresponding definition is not given explicitly, but it is easy to read from the context:

Definition

Equivalent transformations of equations are transformations that give equivalent equations.

So why are equivalent transformations interesting? The fact is that if with their help it is possible to come from the equation being solved to a fairly simple equivalent equation, then solving this equation will give the desired solution to the original equation.

Of the transformations listed in the previous paragraph, not all are always equivalent. Some transformations are equivalent only if certain conditions. Let's make a list of statements that determine which transformations and under what conditions are equivalent transformations of the equation. To do this, we will take the above list as a basis, and to the transformations that are not always equivalent, we will add conditions that give them equivalence. Here is the list:

Replacing an expression on the left or right side of an equation with an expression that does not change the variables for the equation is an equivalent transformation of the equation.

Let us explain why this is so. To do this, we take an equation with one variable (similar reasoning can be carried out for equations with several variables) of the form A(x)=B(x), we denoted the expressions on its left and right sides as A(x) and B(x), respectively . Let the expression C(x) be identically equal to the expression A(x), and the ODZ of the variable x of the equation C(x)=B(x) coincides with the ODZ of the variable x for the original equation. Let us prove that the transformation of the equation A(x)=B(x) into the equation C(x)=B(x) is an equivalent transformation, that is, we will prove that the equations A(x)=B(x) and C(x) =B(x) are equivalent.

To do this, it is enough to show that any root of the original equation is a root of the equation C(x)=B(x), and any root of the equation C(x)=B(x) is a root of the original equation.

Let's start with the first part. Let q be the root of the equation A(x)=B(x), then when we substitute it for x we will get the correct numerical equality A(q)=B(q). Since the expressions A(x) and C(x) are identically equal and the expression C(q) makes sense (this follows from the condition that the OD for the equation C(x)=B(x) coincides with the OD for the original equation) , then the numerical equality A(q)=C(q) is true. Next we use the properties of numerical equalities. Due to the symmetry property, the equality A(q)=C(q) can be rewritten as C(q)=A(q) . Then, due to the transitivity property, the equalities C(q)=A(q) and A(q)=B(q) imply the equality C(q)=B(q). This proves that q is the root of the equation C(x)=B(x) .

The second part, and with it the entire statement as a whole, is proved in an absolutely analogous way.

The essence of the analyzed equivalent transformation is as follows: it allows you to work separately with expressions on the left and right sides of the equations, replacing them with identically equal expressions on the original ODZ of variables.

The most common example: we can replace the sum of numbers on the right side of the equation x=2+1 with its value, which will result in an equivalent equation of the form x=3. Indeed, we replaced the expression 2+1 with the identically equal expression 3, and the ODZ of the equation did not change. Another example: on the left side of the equation 3·(x+2)=7·x−2·x+4−1 we can, and on the right – , which will lead us to the equivalent equation 3·x+6=5·x+ 3. The resulting equation is indeed equivalent, since we replaced the expressions with identically equal expressions and at the same time obtained an equation that has an OD that coincides with the OD for the original equation.

Adding the same number to both sides of an equation or subtracting the same number from both sides of an equation is an equivalent transformation of the equation.

Let us prove that adding the same number c to both sides of the equation A(x)=B(x) gives the equivalent equation A(x)+c=B(x)+c and that subtracting from both sides of the equation A(x) =B(x) of the same number c gives the equivalent equation A(x)−c=B(x)−c.

Let q be the root of the equation A(x)=B(x), then the equality A(q)=B(q) is true. The properties of numerical equalities allow us to add to both sides of a true numerical equality or subtract the same number from its parts. Let us denote this number as c, then the equalities A(q)+c=B(q)+c and A(q)−c=B(q)−c are valid. From these equalities it follows that q is the root of the equation A(x)+c=B(x)+c and the equation A(x)−c=B(x)−c.

Now back. Let q be the root of the equation A(x)+c=B(x)+c and the equation A(x)−c=B(x)−c, then A(q)+c=B(q)+c and A (q)−c=B(q)−c . We know that subtracting the same number from both sides of a true numerical equality produces a true numerical equality. We also know that adding the correct numerical equality to both sides gives the correct numerical equality. Let us subtract the number c from both sides of the correct numerical equality A(q)+c=B(q)+c, and add the number c to both sides of the equality A(x)−c=B(x)−c. This will give us the correct numerical equalities A(q)+c−c=B(q)+c−c and A(q)−c+c=B(q)+c−c, from which we conclude that A(q) =B(q) . From the last equality it follows that q is the root of the equation A(x)=B(x) .

This proves the original statement as a whole.

Let us give an example of such a transformation of equations. Let's take the equation x−3=1, and transform it by adding the number 3 to both sides, after which we get the equation x−3+3=1+3, which is equivalent to the original one. It is clear that in the resulting equation you can perform operations with numbers, as we discussed in the previous item on the list, as a result we have the equation x=4. So, performing equivalent transformations, we accidentally solved the equation x−3=1, its root is the number 4. The considered equivalent transformation is very often used to get rid of identical numerical terms located in different parts of the equation. For example, in both the left and right sides of the equation x 2 +1=x+1 there is the same term 1, subtracting the number 1 from both sides of the equation allows us to move on to the equivalent equation x 2 +1−1=x+1−1 and further to equivalent equation x 2 =x, and thereby get rid of these identical terms.

Adding to both sides of the equation or subtracting from both sides of the equation an expression for which the ODZ is not narrower than the ODZ for the original equation is an equivalent transformation.

Let's prove this statement. That is, we prove that the equations A(x)=B(x) and A(x)+C(x)=B(x)+C(x) are equivalent, provided that the ODZ for the expression C(x) is not already , than ODZ for the equation A(x)=B(x) .

First we prove one auxiliary point. Let us prove that, under the specified conditions, the OD equations before and after the transformation are the same. Indeed, the ODZ for the equation A(x)+C(x)=B(x)+C(x) can be considered as the intersection of the ODZ for the equation A(x)=B(x) and the ODZ for the expression C(x) . From this and from the fact that the ODZ for the expression C(x) is not narrower by condition than the ODZ for the equation A(x)=B(x), it follows that the ODZ for the equations A(x)=B(x) and A (x)+C(x)=B(x)+C(x) are the same.

Now we will prove the equivalence of the equations A(x)=B(x) and A(x)+C(x)=B(x)+C(x), provided that the ranges of acceptable values for these equations are the same. We will not give a proof of the equivalence of the equations A(x)=B(x) and A(x)−C(x)=B(x)−C(x) under the specified condition, since it is similar.

Let q be the root of the equation A(x)=B(x), then the numerical equality A(q)=B(q) is true. Since the ODZ of the equations A(x)=B(x) and A(x)+C(x)=B(x)+C(x) are the same, then the expression C(x) makes sense at x=q, which means C(q) is some number. If we add C(q) to both sides of the correct numerical equality A(q)=B(q) , this will give the correct numerical inequality A(q)+C(q)=B(q)+C(q) , from which it follows that q is the root of the equation A(x)+C(x)=B(x)+C(x) .

Back. Let q be the root of the equation A(x)+C(x)=B(x)+C(x), then A(q)+C(q)=B(q)+C(q) is a true numerical equality. We know that subtracting the same number from both sides of a true numerical equality produces a true numerical equality. Subtract C(q) from both sides of the equality A(q)+C(q)=B(q)+C(q) , this gives A(q)+C(q)−C(q)=B(q)+C(q)−C(q) and further A(q)=B(q) . Therefore, q is the root of the equation A(x)=B(x) .

Thus, the statement in question is completely proven.

Let's give an example of this transformation. Let's take the equation 2 x+1=5 x+2. We can add to both sides, for example, the expression −x−1. Adding this expression will not change the ODZ, which means that such a transformation is equivalent. As a result of this, we obtain the equivalent equation 2 x+1+(−x−1)=5 x+2+(−x−1). This equation can be transformed further: open the brackets and reduce similar terms on its left and right sides (see the first item in the list). After performing these actions, we obtain the equivalent equation x=4·x+1. The transformation of equations under consideration is often used to get rid of identical terms that are simultaneously on the left and right sides of the equation.

If you move a term in an equation from one part to another, changing the sign of this term to the opposite, you will get an equation equivalent to the given one.

This statement is a consequence of the previous ones.

Let us show how this equivalent transformation of the equation is carried out. Let's take the equation 3·x−1=2·x+3. Let's move the term, for example, 2 x from the right side to the left, changing its sign. In this case, we obtain the equivalent equation 3·x−1−2·x=3. You can also move minus one from the left side of the equation to the right, changing the sign to plus: 3 x−2 x=3+1. Finally, bringing similar terms leads us to the equivalent equation x=4.

Multiplying or dividing both sides of an equation by the same non-zero number is an equivalent transformation.

Let's give a proof.

Let A(x)=B(x) be some equation and c be some number different from zero. Let us prove that multiplying or dividing both sides of the equation A(x)=B(x) by the number c is an equivalent transformation of the equation. To do this, we prove that the equations A(x)=B(x) and A(x) c=B(x) c, as well as the equations A(x)=B(x) and A(x):c= B(x):c - equivalent. This can be done this way: prove that any root of the equation A(x)=B(x) is a root of the equation A(x) c=B(x) c and a root of the equation A(x):c=B(x) :c , and then prove that any root of the equation A(x) c=B(x) c , like any root of the equation A(x):c=B(x):c, is a root of the equation A(x) =B(x) . Let's do it.

Let q be the root of the equation A(x)=B(x) . Then the numerical equality A(q)=B(q) is true. Having studied the properties of numerical equalities, we learned that multiplying or dividing both sides of a true numerical equality by the same number other than zero leads to a true numerical equality. Multiplying both sides of the equality A(q)=B(q) by c, we obtain the correct numerical equality A(q) c=B(q) c, from which it follows that q is the root of the equation A(x) c= B(x)·c . And dividing both sides of the equality A(q)=B(q) by c, we obtain the correct numerical equality A(q):c=B(q):c, from which it follows that q is the root of the equation A(x):c =B(x):c .

Now in the other direction. Let q be the root of the equation A(x) c=B(x) c. Then A(q)·c=B(q)·c is a true numerical equality. Dividing both parts of it by a non-zero number c, we obtain the correct numerical equality A(q)·c:c=B(q)·c:c and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) . If q is the root of the equation A(x):c=B(x):c . Then A(q):c=B(q):c is a true numerical equality. Multiplying both parts of it by a non-zero number c, we obtain the correct numerical equality A(q):c·c=B(q):c·c and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) .

The statement has been proven.

Let's give an example of this transformation. With its help, you can, for example, get rid of fractions in the equation. To do this, you can multiply both sides of the equation by 12. The result is an equivalent equation of the form , which can then be transformed into the equivalent equation 7 x−3=10, which does not contain fractions in its notation.

Multiplying or dividing both sides of an equation by the same expression, the OD for which is not narrower than the OD for the original equation and does not vanish by the OD for the original equation, is an equivalent transformation.

Let's prove this statement. To do this, we prove that if the ODZ for the expression C(x) is not narrower than the ODZ for the equation A(x)=B(x), and C(x) does not vanish on the ODZ for the equation A(x)=B( x) , then the equations A(x)=B(x) and A(x) C(x)=B(x) C(x), as well as the equations A(x)=B(x) and A( x):C(x)=B(x):C(x) - equivalent.

Let q be the root of the equation A(x)=B(x) . Then A(q)=B(q) is a true numerical equality. From the fact that the ODZ for the expression C(x) is not the same ODZ for the equation A(x)=B(x), it follows that the expression C(x) makes sense when x=q. This means that C(q) is some number. Moreover, C(q) is nonzero, which follows from the condition that the expression C(x) does not vanish. If we multiply both sides of the equality A(q)=B(q) by a non-zero number C(q), this will give the correct numerical equality A(q)·C(q)=B(q)·C(q) , from which it follows that q is the root of the equation A(x)·C(x)=B(x)·C(x) . If we divide both sides of the equality A(q)=B(q) by a non-zero number C(q), this will give the correct numerical equality A(q):C(q)=B(q):C(q) , from which it follows that q is the root of the equation A(x):C(x)=B(x):C(x) .

Back. Let q be the root of the equation A(x)·C(x)=B(x)·C(x) . Then A(q)·C(q)=B(q)·C(q) is a true numerical equality. Note that the ODZ for the equation A(x) C(x)=B(x) C(x) is the same as the ODZ for the equation A(x)=B(x) (we justified this in one of the previous paragraphs current list). Since C(x) by condition does not vanish on the ODZ for the equation A(x)=B(x), then C(q) is a nonzero number. Dividing both sides of the equality A(q) C(q)=B(q) C(q) by a non-zero number C(q) we obtain the correct numerical equality A(q)·C(q):C(q)=B(q)·C(q):C(q) and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) . If q is the root of the equation A(x):C(x)=B(x):C(x) . Then A(q):C(q)=B(q):C(q) is a true numerical equality. Multiplying both sides of the equality A(q):C(q)=B(q):C(q) by a non-zero number C(q) we obtain the correct numerical equality A(q):C(q)·C(q)=B(q):C(q)·C(q) and further A(q)=B(q) . It follows that q is the root of the equation A(x)=B(x) .

The statement has been proven.

For clarity, we give an example of carrying out a disassembled transformation. Let's divide both sides of the equation x 3 ·(x 2 +1)=8·(x 2 +1) by the expression x 2 +1. This transformation is equivalent, since the expression x 2 +1 does not vanish on the OD for the original equation and the OD of this expression is not narrower than the OD for the original equation. As a result of this transformation, we obtain the equivalent equation x 3 ·(x 2 +1):(x 2 +1)=8·(x 2 +1):(x 2 +1), which can be further transformed to the equivalent equation x 3 =8.

Transformations leading to corollary equations

In the previous paragraph, we examined which transformations from the list of basic transformations and under what conditions are equivalent. Now let's see which of these transformations and under what conditions lead to corollary equations, that is, to equations that contain all the roots of the transformed equation, but in addition to them may also have other roots - extraneous roots for the original equation.

Transformations leading to corollary equations are in demand no less than equivalent transformations. If with their help it is possible to obtain an equation that is quite simple in terms of solution, then its solution and subsequent elimination of extraneous roots will give a solution to the original equation.

Note that all equivalent transformations can be considered special cases of transformations that lead to corollary equations. This is understandable, because there is an equivalent equation special case consequence equations. But from a practical point of view, it is more useful to know that the transformation under consideration is precisely equivalent, and not leading to a corollary equation. Let us explain why this is so. If we know that the transformation is equivalent, then the resulting equation will definitely not have roots extraneous to the original equation. And the transformation leading to the corollary equation may be the cause of the appearance of extraneous roots, which obliges us in the future to carry out an additional action - sifting out extraneous roots. Therefore, in this section of the article we will focus on transformations, as a result of which extraneous roots may appear for the original equation. And it is really important to be able to distinguish such transformations from equivalent transformations in order to clearly understand when it is necessary to filter out extraneous roots, and when this is not necessary.

Let's analyze the entire list of basic transformations of equations given in the second paragraph of this article in order to search for transformations, as a result of which extraneous roots may appear.

Replacing expressions on the left and right sides of the equation with identically equal expressions.

We have proven that this transformation is equivalent if its implementation does not change the OD. And if the DL changes, what will happen? The narrowing of the ODZ can lead to the loss of roots; this will be discussed in more detail in the next paragraph. And with the expansion of the ODZ, extraneous roots may appear. It is not difficult to justify this. Let us present the corresponding reasoning.

Let the expression C(x) be such that it is identically equal to the expression A(x) and the OD for the equation C(x)=B(x) is wider than the OD for the equation A(x)=B(x). Let us prove that the equation C(x)=B(x) is a consequence of the equation A(x)=B(x), and that among the roots of the equation C(x)=B(x) there may be roots that are foreign to the equation A( x)=B(x) .

Let q be the root of the equation A(x)=B(x) . Then A(q)=B(q) is a true numerical equality. Since the ODZ for the equation C(x)=B(x) is wider than the ODZ for the equation A(x)=B(x), then the expression C(x) is defined at x=q. Then, taking into account the identical equality of the expressions C(x) and A(x) , we conclude that C(q)=A(q) . From the equalities C(q)=A(q) and A(q)=B(q), due to the transitivity property, the equality C(q)=B(q) follows. From this equality it follows that q is the root of the equation C(x)=B(x) . This proves that under the specified conditions the equation C(x)=B(x) is a consequence of the equation A(x)=B(x) .

It remains to prove that the equation C(x)=B(x) can have roots different from the roots of the equation A(x)=B(x). Let us prove that any root of the equation C(x)=B(x) from the ODZ for the equation A(x)=B(x) is a root of the equation A(x)=B(x). Path p is the root of the equation C(x)=B(x), belonging to the ODZ for the equation A(x)=B(x). Then C(p)=B(p) is a true numerical equality. Since p belongs to the ODZ for the equation A(x)=B(x), then the expression A(x) is defined for x=p. From this and from the identical equality of the expressions A(x) and C(x) it follows that A(p)=C(p) . From the equalities A(p)=C(p) and C(p)=B(p), due to the transitivity property, it follows that A(p)=B(p), which means p is the root of the equation A(x)= B(x) . This proves that any root of the equation C(x)=B(x) from the ODZ for the equation A(x)=B(x) is a root of the equation A(x)=B(x). In other words, on the ODZ for the equation A(x)=B(x) there cannot be roots of the equation C(x)=B(x), which are extraneous roots for the equation A(x)=B(x). But according to the condition, the ODZ for the equation C(x)=B(x) is wider than the ODZ for the equation A(x)=B(x). And this allows the existence of a number r that belongs to the ODZ for the equation C(x)=B(x) and does not belong to the ODZ for the equation A(x)=B(x), which is the root of the equation C(x)=B(x). That is, the equation C(x)=B(x) may have roots that are foreign to the equation A(x)=B(x), and all of them will belong to the set to which the ODZ for the equation A(x)=B is extended (x) when replacing the expression A(x) in it with the identically equal expression C(x).

So, replacing the expressions on the left and right sides of the equation with identically equal expressions, as a result of which the ODZ is expanded, in the general case leads to a corollary equation (that is, it can lead to the appearance of extraneous roots) and only in a particular case leads to equivalent equation (in the event that the resulting equation does not have roots foreign to the original equation).

Let us give an example of carrying out a parsed transformation. Replacing the expression on the left side of the equation identically equal to it by the expression x·(x−1) leads to the equation x·(x−1)=0, in this case the expansion of the ODZ occurs - the number 0 is added to it. The resulting equation has two roots 0 and 1, and substituting these roots into the original equation shows that 0 is an extraneous root for the original equation, and 1 is the root of the original equation. Indeed, substituting zero into the original equation gives the meaningless expression , since it contains division by zero, and substituting one gives the correct numerical equality , which is the same as 0=0 .

Note that a similar transformation of a similar equation into the equation (x−1)·(x−2)=0, as a result of which the ODZ also expands, does not lead to the appearance of extraneous roots. Indeed, both roots of the resulting equation (x−1)·(x−2)=0 - numbers 1 and 2, are roots of the original equation, which is easy to verify by checking by substitution. With these examples, we once again wanted to emphasize that replacing an expression on the left or right side of the equation with an identically equal expression, which expands the ODZ, does not necessarily lead to the appearance of extraneous roots. But it can also lead to their appearance. So, if such a transformation took place in the process of solving the equation, then it is necessary to carry out a check in order to identify and filter out extraneous roots.

Most often, the ODZ of an equation can expand and extraneous roots may appear due to the replacement by zero of the difference of identical expressions or the sum of expressions with opposite signs, due to the replacement by zero of products with one or more zero factors, due to the reduction of fractions and due to the use of properties roots, powers, logarithms, etc.

Adding the same number to both sides of an equation or subtracting the same number from both sides of an equation.

We showed above that this transformation is always equivalent, that is, leading to an equivalent equation. Go ahead.

Adding the same expression to both sides of an equation or subtracting the same expression from both sides of an equation.

In the previous paragraph, we added a condition that the ODZ for the expression being added or subtracted should not be narrower than the ODZ for the equation being transformed. This condition made the transformation in question equivalent. Here there are arguments similar to those given at the beginning of this paragraph of the article regarding the fact that an equivalent equation is a special case of a corollary equation and that knowledge about the equivalence of a transformation is practically more useful than knowledge about the same transformation, but from the standpoint of the fact that it leads to corollary equation.

Is it possible, as a result of adding the same expression or subtracting the same expression from both sides of an equation, to obtain an equation that, in addition to all the roots of the original equation, will have some other roots? No, he can not. If the ODZ for the expression being added or subtracted is not narrower than the ODZ for the original equation, then as a result of the addition or subtraction an equivalent equation will be obtained. If the ODZ for the expression being added or subtracted is narrower than the ODZ for the original equation, then this can lead to the loss of roots, and not to the appearance of extraneous roots. We'll talk more about this in the next paragraph.

Transferring a term from one part of the equation to another with the sign changed to the opposite.

This transformation of the equation is always equivalent. Therefore, it makes no sense to consider it as a transformation leading to an equation-consequence, for the reasons stated above.

Multiplying or dividing both sides of an equation by the same number.

In the previous paragraph, we proved that if the multiplication or division of both sides of the equation is carried out by a non-zero number, then this is an equivalent transformation of the equation. Therefore, again, there is no point in talking about it as a transformation leading to a corollary equation.

But here it is worth paying attention to the disclaimer about the difference from zero of the number by which both sides of the equation are multiplied or divided. For division this clause is clear - with primary classes we realized that You can't divide by zero. Why this clause for multiplication? Let's think about what multiplying both sides of the equation by zero results in. For clarity, let's take a specific equation, for example, 2 x+1=x+5. This is a linear equation that has a single root, which is the number 4. Let's write down the equation that will be obtained by multiplying both sides of this equation by zero: (2 x+1) 0=(x+5) 0. Obviously, the root of this equation is any number, because when you substitute any number into this equation instead of the variable x, you get the correct numerical equality 0=0. That is, in our example, multiplying both sides of the equation by zero led to a corollary equation, which caused the appearance of an infinite number of extraneous roots for the original equation. Moreover, it is worth noting that in this case the usual methods of screening out extraneous roots do not cope with their task. This means that the transformation performed is useless for solving the original equation. And this is a typical situation for the transformation under consideration. This is why a transformation such as multiplying both sides of an equation by zero is not used to solve equations. We still have to look at this transformation and other transformations that should not be used to solve equations in the last paragraph.

Multiplying or dividing both sides of an equation by the same expression.

In the previous paragraph, we proved that this transformation is equivalent if two conditions are met. Let's remind them. The first condition: the OD for this expression should not be narrower than the OD for the original equation. The second condition: the expression by which the multiplication or division is carried out must not vanish on the ODZ for the original equation.

Let's change the first condition, that is, we will assume that the OD for the expression by which we plan to multiply or divide both sides of the equation is narrower than the OD for the original equation. As a result of such a transformation, an equation will be obtained for which the ODZ will be narrower than the ODZ for the original equation. Such transformations can lead to the loss of roots; we will talk about them in the next paragraph.

What will happen if we remove the second condition about the non-zero values of the expression by which both sides of the equation are multiplied or divided by the ODZ for the original equation?

Dividing both sides of the equation by the same expression, which vanishes by the OD for the original equation, will result in an equation whose OD is narrower than the OD for the original equation. Indeed, numbers will fall out of it, turning the expression by which the division was carried out to zero. This can lead to root loss.

What about multiplying both sides of the equation by the same expression, which vanishes on the ODZ for the original equation? It can be shown that when both sides of the equation A(x)=B(x) are multiplied by the expression C(x), for which the ODZ is not narrower than the ODZ for the original equation, and which vanishes by the ODZ for the original equation, the equation is obtained is a consequence that, in addition to all the roots of the equation A(x)=B(x), it can also have other roots. Let's do this, especially since this paragraph of the article is precisely devoted to transformations leading to corollary equations.

Let the expression C(x) be such that the ODZ for it is not narrower than the ODZ for the equation A(x)=B(x), and it vanishes on the ODZ for the equation A(x)=B(x) . Let us prove that in this case the equation A(x)·C(x)=B(x)·C(x) is a consequence of the equation A(x)=B(x) .

Let q be the root of the equation A(x)=B(x) . Then A(q)=B(q) is a true numerical equality. Since the ODZ for the expression C(x) is not narrower than the ODZ for the equation A(x)=B(x), then the expression C(x) is defined at x=q, which means that C(q) is a certain number. Multiplying both sides of a true numerical equality by any number gives a true numerical equality, therefore, A(q)·C(q)=B(q)·C(q) is a true numerical equality. This means q is the root of the equation A(x)·C(x)=B(x)·C(x) . This proves that any root of the equation A(x)=B(x) is a root of the equation A(x) C(x)=B(x) C(x), which means that the equation A(x) C (x)=B(x)·C(x) is a consequence of the equation A(x)=B(x) .

Note that under the specified conditions, the equation A(x)·C(x)=B(x)·C(x) may have roots that are foreign to the original equation A(x)=B(x). They are all numbers from the ODZ for the original equation that turn the expression C(x) to zero (all numbers that turn the expression C(x) to zero are the roots of the equation A(x) C(x)=B(x) C(x) , since their substitution into the indicated equation gives the correct numerical equality 0=0 ), but which are not roots of the equation A(x)=B(x) . The equations A(x)=B(x) and A(x)·C(x)=B(x)·C(x) under the specified conditions will be equivalent when all numbers from the ODZ for the equation A(x)=B (x) , which make the expression C(x) vanish, are the roots of the equation A(x)=B(x) .

So, multiplying both sides of the equation by the same expression, the ODZ for which is not narrower than the ODZ for the original equation, and which vanishes by the ODZ for the original equation, in the general case leads to a corollary equation, that is, it can lead to the appearance of foreign roots.

Let's give an example to illustrate. Let's take the equation x+3=4. Its only root is the number 1. Let's multiply both sides of this equation by the same expression, which vanishes by the ODZ for the original equation, for example, by x·(x−1) . This expression vanishes at x=0 and x=1. Multiplying both sides of the equation by this expression gives us the equation (x+3) x (x−1)=4 x (x−1). The resulting equation has two roots: 1 and 0. The number 0 is an extraneous root for the original equation that appeared as a result of the transformation.

Transformations that may lead to loss of roots

Some conversions from under certain conditions can lead to loss of roots. For example, when dividing both sides of the equation x·(x−2)=x−2 by the same expression x−2, the root is lost. Indeed, as a result of such a transformation, the equation x=1 is obtained with a single root, which is the number 1, and the original equation has two roots 1 and 2.

It is necessary to clearly understand when roots are lost as a result of transformations, so as not to lose roots when solving equations. Let's figure this out.

As a result of these transformations, loss of roots can occur if and only if the ODZ for the transformed equation turns out to be narrower than the ODZ for the original equation.

To prove this statement, two points need to be substantiated. First, it is necessary to prove that if, as a result of the indicated transformations of the equation, the ODZ is narrowed, then a loss of roots may occur. And, secondly, it is necessary to justify that if, as a result of these transformations, the roots are lost, then the ODZ for the resulting equation is narrower than the ODZ for the original equation.

If the ODZ for the equation obtained as a result of the transformation is narrower than the ODZ for the original equation, then, naturally, not a single root of the original equation located outside the ODZ for the resulting equation can be the root of the equation obtained as a result of the transformation. This means that all these roots will be lost when moving from the original equation to an equation for which the ODZ is narrower than the ODZ for the original equation.

Now back. Let us prove that if, as a result of these transformations, the roots are lost, then the ODZ for the resulting equation is narrower than the ODZ for the original equation. This can be done by the opposite method. The assumption that as a result of these transformations, the roots are lost, but the ODZ is not narrowed, contradicts the statements proven in the previous paragraphs. Indeed, from these statements it follows that if, when carrying out the indicated transformations, the ODZ is not narrowed, then either equivalent equations or corollary equations are obtained, which means that loss of roots cannot occur.

So, the reason for the possible loss of roots when carrying out basic transformations of equations is the narrowing of the ODZ. It is clear that when solving equations, we should not lose roots. Here, naturally, the question arises: “What should we do to avoid losing roots when transforming equations?” We will answer it in the next paragraph. Now let's go through the list of basic transformations of equations to see in more detail which transformations can lead to the loss of roots.

Replacing expressions on the left and right sides of the equation with identically equal expressions.

If you replace the expression on the left or right side of the equation with an identically equal expression, the OD for which is narrower than the OD for the original equation, this will lead to a narrowing of the OD, and because of this, roots may be lost. Most often, replacement of expressions on the left or right side of equations with identically equal expressions, carried out on the basis of some properties of roots, powers, logarithms and some trigonometric formulas. For example, replacing the expression on the left side of the equation with an identically equal expression narrows the ODZ and leads to the loss of the root −16. Similarly, replacing the expression on the left side of the equation with an identically equal expression leads to an equation for which the ODZ is narrower than the ODZ for the original equation, which entails the loss of the root −3.

Adding the same number to both sides of an equation or subtracting the same number from both sides of an equation.

This transformation is equivalent, therefore, roots cannot be lost during its implementation.

Adding the same expression to both sides of an equation or subtracting the same expression from both sides of an equation.

If you add or subtract an expression whose OD is narrower than the OD for the original equation, this will lead to a narrowing of the OD and, as a consequence, to a possible loss of roots. It's worth keeping this in mind. But here it is worth noting that in practice it is usually necessary to resort to adding or subtracting expressions that are present in the recording of the original equation, which does not lead to a change in the ODZ and does not entail the loss of roots.

Transferring a term from one part of the equation to another with the sign changed to the opposite.

This transformation of the equation is equivalent, therefore, as a result of its implementation, the roots are not lost.

Multiplying or dividing both sides of an equation by the same number other than zero.

This transformation is also equivalent, and because of it, the loss of roots does not occur.

Multiplying or dividing both sides of an equation by the same expression.

This transformation can lead to a narrowing of the OD in two cases: when the OD for the expression by which the multiplication or division is carried out is narrower than the OD for the original equation, and when division is carried out by an expression that becomes zero on the OD for the original equation. Note that in practice it is usually not necessary to resort to multiplying and dividing both sides of the equation by an expression with a narrower VA. But you have to deal with division by an expression that turns into zero for the original equation. There is a method that allows you to cope with the loss of roots during such division, we will talk about it in the next paragraph of this article.

How to avoid root loss?

If you use only transformations from to transform equations and at the same time do not allow narrowing of the ODZ, then the loss of roots will not occur.

Does this mean that no other transformations of the equations can be made? No, it doesn't mean that. If you come up with some other transformation of the equation and fully describe it, that is, indicate when it leads to equivalent equations, when to corollary equations, and when it can lead to the loss of roots, then it could well be adopted.

Should we completely abandon reforms that would narrow DPD? Should not be doing that. It would not hurt to keep in your arsenal transformations in which a finite number of numbers drop out of the ODZ for the original equation. Why shouldn’t such transformations be abandoned? Because there is a method to avoid root loss in such cases. It consists of a separate check of the numbers falling out of the ODZ to see if there are roots of the original equation among them. You can check this by substituting these numbers into the original equation. Those of them that, when substituted, give the correct numerical equality, are the roots of the original equation. They need to be included in the answer. After such a check, you can safely carry out the planned transformation without fear of losing your roots.

A typical transformation in which the ODZ for an equation is narrowed down to several numbers is to divide both sides of the equation by the same expression, which becomes zero at several points from the ODZ for the original equation. This transformation is the basis of the solution method reciprocal equations. But it is also used to solve other types of equations. Let's give an example.

The equation can be solved by introducing a new variable. To introduce a new variable, you need to divide both sides of the equation by 1+x. But with such a division, a loss of root may occur, since although the ODZ for the expression 1+x is not narrower than the ODZ for the original equation, the expression 1+x becomes zero at x=−1, and this number belongs to the ODZ for the original equation. This means that the root −1 may be lost. To eliminate the loss of a root, you should separately check whether −1 is a root of the original equation. To do this, you can substitute −1 into the original equation and see what equality you get. In our case, the substitution gives the equality, which is the same as 4=0. This equality is false, which means −1 is not the root of the original equation. After such a check, you can carry out the intended division of both sides of the equation by 1 + x, without fear that loss of roots may occur.

At the end of this paragraph, let us once again turn to the equations from the previous paragraph and. Transformation of these equations based on identities and leads to a narrowing of the ODZ, and this entails the loss of roots. At this point, we said that in order not to lose our roots, we need to abandon reforms that narrow the DZ. This means that these transformations must be abandoned. But what should we do? It is possible to carry out transformations not based on identities and , due to which the ODZ is narrowed, and on the basis of identities and . As a result of the transition from the original equations and to the equations and there is no narrowing of the ODZ, which means that the roots will not be lost.

Here we especially note that when replacing expressions with identically equal expressions, you must carefully ensure that the expressions are exactly identically equal. For example, in Eq. it is impossible to replace the expression x+3 with an expression in order to simplify the appearance of the left side to , since the expressions x+3 and are not identically equal, because their values do not coincide at x+3<0 . В нашем примере такое преобразование приводит к потере корня. А в общем случае замена выражения не тождественно равным выражением приводит к уравнению, которое не позволяет получить решение исходного уравнения.

Transformations of equations that should not be used

The transformations mentioned in this article are usually sufficient for practical needs. That is, you shouldn’t be too bothered about coming up with any other transformations; it’s better to focus on the correct use of the already proven ones.

Literature

Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. In 2 hours. Part 1. Textbook for students of general education institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.
Algebra and the beginning of mathematical analysis. 10th grade: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. - 3rd ed. - M.: Education, 2010.- 368 p.: ill.-ISBN 978-5-09-022771-1.