Maths is a symbol of the wisdom of science,
an example of scientific rigor and simplicity,
the standard of perfection and beauty in science.
Russian philosopher, professor A.V. Voloshinov
Modulo inequalities
The most difficult problems to solve in school mathematics are the inequalities, containing variables under the module sign. To successfully solve such inequalities, it is necessary to know the properties of the module well and have the skills to use them.
Basic concepts and properties
Modulus (absolute value) of a real number denoted and is defined as follows:
The simple properties of the module include the following relations:
AND .
Note, that the last two properties hold for any even degree.
Also, if , where , then and
More complex properties module, which can be effectively used in solving equations and inequalities with modules, are formulated by means of the following theorems:
Theorem 1.For any analytic functions and the inequality.
Theorem 2. Equality is equivalent to the inequality.
Theorem 3. Equality is equivalent to the inequality.
The most common inequalities in school mathematics, containing unknown variables under the modulo sign, are inequalities of the form and where some positive constant.
Theorem 4. Inequality is equivalent to a double inequality, and the solution to the inequalityreduces to solving the set of inequalities and .
This theorem is a particular case of Theorems 6 and 7.
More complex inequalities, containing the module are inequalities of the form, and .
Methods for solving such inequalities can be formulated using the following three theorems.
Theorem 5. Inequality is equivalent to the combination of two systems of inequalities
AND (1)
Proof. Since then
This implies the validity of (1).
Theorem 6. Inequality is equivalent to the system of inequalities
Proof. Because , then from the inequality follows that . Under this condition, the inequalityand in this case the second system of inequalities (1) turns out to be inconsistent.
The theorem has been proven.
Theorem 7. Inequality is equivalent to the combination of one inequality and two systems of inequalities
AND (3)
Proof. Since , then the inequality always executed, if .
Let , then the inequalitywill be tantamount to inequality, from which the set of two inequalities follows and .
The theorem has been proven.
Consider typical examples of solving problems on the topic “Inequalities, containing variables under the module sign.
Solving inequalities with modulus
The simplest method for solving inequalities with modulus is the method, based on module expansion. This method is generic, however, in the general case, its application can lead to very cumbersome calculations. Therefore, students should also know other (more efficient) methods and techniques for solving such inequalities. In particular, need to have the skills to apply theorems, given in this article.
Example 1Solve the inequality
. (4)
Solution.Inequality (4) will be solved by the "classical" method - the moduli expansion method. To this end, we break the numerical axis dots and intervals and consider three cases.
1. If , then , , , and inequality (4) takes the form or .
Since the case is considered here, , is a solution to inequality (4).
2. If , then from inequality (4) we obtain or . Since the intersection of intervals and is empty, then there are no solutions to inequality (4) on the considered interval.
3. If , then inequality (4) takes the form or . It's obvious that is also a solution to inequality (4).
Answer: , .
Example 2 Solve the inequality.
Solution. Let's assume that . Because , then the given inequality takes the form or . Since , then and hence follows or .
However , therefore or .
Example 3 Solve the inequality
. (5)
Solution. Because , then inequality (5) is equivalent to the inequalities or . From here, according to Theorem 4, we have a set of inequalities and .
Answer: , .
Example 4Solve the inequality
. (6)
Solution. Let's denote . Then from inequality (6) we obtain the inequalities , , or .
From here, using the interval method, we get . Because , then here we have a system of inequalities
The solution to the first inequality of system (7) is the union of two intervals and , and the solution of the second inequality is the double inequality. This implies , that the solution to the system of inequalities (7) is the union of two intervals and .
Answer: ,
Example 5Solve the inequality
. (8)
Solution. We transform inequality (8) as follows:
Or .
Applying the interval method, we obtain a solution to inequality (8).
Answer: .
Note. If we put and in the condition of Theorem 5, then we obtain .
Example 6 Solve the inequality
. (9)
Solution. From inequality (9) it follows. We transform inequality (9) as follows:
Or
Since , then or .
Answer: .
Example 7Solve the inequality
. (10)
Solution. Since and , then or .
In this connection and inequality (10) takes the form
Or
. (11)
It follows from this that or . Since , then inequality (11) also implies or .
Answer: .
Note. If we apply Theorem 1 to the left side of inequality (10), then we get . From here and from inequality (10) it follows, that or . Because , then inequality (10) takes the form or .
Example 8 Solve the inequality
. (12)
Solution. Since then and inequality (12) implies or . However , therefore or . From here we get or .
Answer: .
Example 9 Solve the inequality
. (13)
Solution. According to Theorem 7, the solutions to inequality (13) are or .
Let now. In this case and inequality (13) takes the form or .
If we combine intervals and , then we obtain a solution to inequality (13) of the form.
Example 10 Solve the inequality
. (14)
Solution. Let us rewrite inequality (14) in an equivalent form: . If we apply Theorem 1 to the left side of this inequality, then we obtain the inequality .
From here and from Theorem 1 it follows, that inequality (14) is satisfied for any values.
Answer: any number.
Example 11. Solve the inequality
. (15)
Solution. Applying Theorem 1 to the left side of inequality (15), we get . From here and from inequality (15) follows the equation, which looks like.
According to Theorem 3, the equation is equivalent to the inequality. From here we get.
Example 12.Solve the inequality
. (16)
Solution. From inequality (16), according to Theorem 4, we obtain the system of inequalities
When solving the inequalitywe use Theorem 6 and obtain the system of inequalitiesfrom which follows.
Consider the inequality. According to Theorem 7, we obtain a set of inequalities and . The second population inequality holds for any real.
Consequently , the solution of inequality (16) are.
Example 13Solve the inequality
. (17)
Solution. According to Theorem 1, we can write
(18)
Taking into account inequality (17), we conclude that both inequalities (18) turn into equalities, i.e. there is a system of equations
By Theorem 3, this system of equations is equivalent to the system of inequalities
or
Example 14Solve the inequality
. (19)
Solution. Since , then . Let us multiply both parts of inequality (19) by the expression , which for any values takes only positive values. Then we obtain an inequality that is equivalent to inequality (19), of the form
From here we get or , where . Since and then the solutions to inequality (19) are and .
Answer: , .
For a deeper study of methods for solving inequalities with a module, it is advisable to refer to tutorials, listed in the list of recommended readings.
1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.
2. Suprun V.P. Mathematics for high school students: methods for solving and proving inequalities. – M.: Lenand / URSS, 2018. - 264 p.
3. Suprun V.P. Mathematics for high school students: non-standard methods for solving problems. - M .: KD "Librocom" / URSS, 2017. - 296 p.
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Thomas Aquinas
The interval method allows you to solve any equations containing the modulus. The essence of this method is to split the numerical axis into several sections (intervals), and it is necessary to split the axis with the zeros of the expressions in the modules. Then, on each of the resulting sections, any submodule expression is either positive or negative. Therefore, each of the modules can be expanded either with a minus sign or with a plus sign. After these actions, it remains only to solve each of the obtained simple equations on the considered interval and combine the received answers.
Let's consider this method on a specific example.
|x + 1| + |2x – 4| – |x + 3| = 2x - 6.
1) Find the zeros of the expressions in the modules. To do this, we equate them to zero, and solve the resulting equations.
x + 1 = 0 2x – 4 = 0 x + 3 = 0
x = -1 2x = 4 x = -3
2) Arrange the resulting points in the desired order on the coordinate line. They will break the entire axis into four sections.
3) Let's determine on each of the resulting sections the signs of the expressions in the modules. To do this, we substitute in them any numbers from the intervals of interest to us. If the result of the calculation is a positive number, then we put "+" in the table, and if the number is negative, then we put "-". This can be pictured like this: 
4) Now we will solve the equation on each of the four intervals, opening the modules with the signs that are in the table. So, consider the first interval:
I interval (-∞; -3). On it, all modules are opened with a "-" sign. We get the following equation:
-(x + 1) - (2x - 4) - (-(x + 3)) \u003d 2x - 6. We present similar terms, having previously opened the brackets in the resulting equation:
X - 1 - 2x + 4 + x + 3 = 2x - 6
The answer received is not included in the considered interval, so it is not necessary to write it in the final answer.
II interval [-3; -one). At this interval in the table there are signs "-", "-", "+". This is how we reveal the modules of the original equation:
-(x + 1) - (2x - 4) - (x + 3) = 2x - 6. Simplify by expanding the brackets:
X - 1 - 2x + 4 - x - 3 \u003d 2x - 6. We present in the resulting equation the following:
x = 6/5. The resulting number does not belong to the interval under consideration, so it is not the root of the original equation.
III interval [-1; 2). We open the modules of the original equation with the signs that are in the figure in the third column. We get:
(x + 1) - (2x - 4) - (x + 3) = 2x - 6. Get rid of the brackets, move the terms containing the variable x to the left side of the equation, and not containing x to the right. Will have:
x + 1 - 2x + 4 - x - 3 = 2x - 6
The number 2 is not included in the considered interval.
IV interval) - they will automatically consider this an incorrect answer. Also, when testing, if a non-strict inequality with modules is specified, then among the solutions, look for areas with square brackets.
On the interval (-3; 0), expanding the module, we change the sign of the function to the opposite 
Taking into account the scope of the inequality disclosure, the solution will have the form
Together with the previous area, this will give two half-intervals
Example 5. Find a solution to the inequality
9x^2-|x-3|>=9x-2 
Solution:
A non-strict inequality is given, the submodule function of which is equal to zero at the point x=3. At smaller values it is negative, at larger values it is positive. We expand the module on the interval x<3.
Finding the discriminant of the equation
and roots 
Substituting the zero point, we find out that on the interval [-1/9; 1] the quadratic function is negative, therefore the interval is a solution. Next, open the module for x>3 
MOU "Khvastovichskaya secondary school»
"Method of intervals for solving equations and inequalities with several modules"
Research work in mathematics
Performed:
student of 10 "b" class
Golysheva Evgeniya
Supervisor:
mathematic teacher
Shapenskaya E.N.
Introduction……………………………………………………………………………… … ….3 Chapter 1. Methods for solving problems with several modules………………… …............4 1.1. Definition of the module. Solving by definition……………………………………………………………………………4 1.2 Solving equations with several modules using the method of intervals…………………………5 1.3 . Tasks with multiple modules. Solution methods……………………………....7 1.4. The method of intervals in problems with modules…………………………………………......9 Chapter 2. Equations and inequalities containing modules…………………………….…. 11 2.1 Solving equations with multiple modules using the interval method..….11 2.2 Solving inequalities with multiple modules using the interval method.…13 Conclusion……………………………………………………… ………………………...15 Literature………………………………………………………………….……….….16
Introduction
The concept of absolute value is one of the most important characteristics of a number, both in the field of real and in the field complex numbers. This concept is widely used not only in various sections of the school mathematics course, but also in the courses of higher mathematics, physics and technical sciences studied at universities. Problems related to absolute values are often found at mathematical Olympiads, entrance exams in universities and on the exam.
Topic:"Interval Method for Solving Equations and Inequalities with Multiple Modules by the Interval Method".
Objective area: maths.
Object of study: solution of equations and inequalities with the module.
Subject of study: interval method for multi-module solution.
Purpose of the study: reveal the efficiency of solving equations and inequalities with several modules by the interval method.
Hypothesis: if you use the interval method to solve inequalities and equations with several modules, you can greatly facilitate your work.
Working methods: collection of information and its analysis.
Tasks:
Study the literature on this topic.
Consider solutions of inequalities and equations with several modules.
Reveal the most effective method solutions.
Practical orientation of the project:
This work can be used as study guide for students and methodological manual for the teacher.
Chapter 1.
1.1 Definition of the module. Solution by definition.
By definition, the modulus, or absolute value, of a non-negative number a is the same as the number itself, and the modulus of a negative number is opposite number, that is - a:
The modulus of a number is always non-negative. Consider examples.
Example 1 Solve the equation |–x| = -3.
Here, there is no need to analyze cases, because the absolute value of the number is always non-negative, which means that this equation has no solutions.
Let us write the solution of these simplest equations in general view:
Example 2 Solve the equation |x| = 2 – x.
Solution. For x 0 we have the equation x = 2 – x, i.e. x = 1. Since 1 0, x = 1 is the root of the original equation. In the second case (x
Answer: x = 1.
Example 3 Solve equation 3|x – 3| + x = -1.
Solution. Here the division into cases is determined by the sign of the expression x – 3. For x – 3 ³ 0 we have 3x – 9 + x = –1 Û x = 2. But 2 – 3 0.
Answer: The equation has no roots.
Example 4 Solve the equation |x – 1| = 1 – x.
Solution. Since 1 - x \u003d - (x - 1), it follows directly from the definition of the module that those and only those x for which x - 1 0 satisfy the equation. This equation has been reduced to an inequality, and the answer is a whole interval (ray).
Answer: x 1.
1.2. Solving equations with a module using systems.
The examples analyzed earlier allow us to formulate the rules for exemption from the modulus sign in equations. For equations of the form |f(x)| = g(x) there are two such rules:
1st rule: |f(x)| = g(x) w (1)
2nd rule: |f(x)| = g(x) Û (2)
Let us explain the notation used here. Curly brackets denote systems, and square brackets denote collections.
Solutions to a system of equations are the values of a variable that simultaneously satisfy all the equations of the system.
The solutions of the set of equations are all values of the variable, each of which is the root of at least one of the equations of the set.
Two equations are equivalent if any solution to each of them is also a solution to the other, in other words, if the sets of their solutions are the same.
If the equation contains several modules, then you can get rid of them in turn, using the above rules. But there are usually shortcuts. We will get acquainted with them later, but now we will consider the solution of the simplest of these equations:
|f(x)| = |g(x)| Û
This equivalence follows from the fact obvious fact that if the modules of two numbers are equal, then the numbers themselves are either equal or opposite.
Example 1. Solve the equation |x 2 – 7x + 11| = x + 1.
Solution. Let's get rid of the module in two ways described above:
1 way: 2 way:
As you can see, in both cases it is necessary to solve the same two quadratic equations, but in the first case they are accompanied by quadratic inequalities, and in the second - linear. Therefore, the second method for this equation is simpler. Solving quadratic equations, we find the roots of the first , both roots satisfy the inequality . The discriminant of the second equation is negative, therefore, the equation has no roots.
Answer: .
Example 2. Solve the equation |x 2 – x – 6| = |2x2 + x – 1|.
Solution. We already know that it is not necessary to consider (as many as 4) variants of the distribution of signs of expressions under modules: this equation is equivalent to the combination of two quadratic equations without any additional inequalities: Which is equivalent to: The first equation has no set of solutions (its discriminant is negative), the second equation has two roots.
1.3. Tasks with multiple modules. Solution methods.
Sequential expansion of modules.
There are two main approaches to solving equations and inequalities containing several modules. You can call them "serial" and "parallel". Now let's get acquainted with the first of them.
His idea is that first one of the modules is isolated in one part of the equation (or inequality) and revealed by one of the methods described earlier. Then the same thing is repeated with each of the resulting equations with modules, and so on until we get rid of all the modules.
Example1. Solve the equation: +
Solution. We isolate the second module and open it using the first method, that is, simply by determining the absolute value:
We apply the second method of liberation from the module to the obtained two equations:
Finally, we solve the resulting four linear equations and select those roots that satisfy the corresponding inequalities. As a result, only two values remain: x = –1 and .
Answer: -1; .
Parallel expansion of modules.
You can remove all modules at once in an equation or inequality and write out all possible combinations of signs of submodule expressions. If there are n modules in the equation, then there will be 2 n options, because each of the n expressions under the module, when removing the module, can receive one of two signs - plus or minus. Basically, we need to solve all 2 n equations (or inequalities) freed from modules. But their solutions will also be solutions of the original problem only if they lie in areas where the corresponding equation (inequality) coincides with the original one. These areas are defined by expression signs under modules. We have already solved the following inequality, so you can compare different approaches to the solution.
Example 2.+
Solution.
Let's consider 4 possible character sets of expressions under modules.
Only the first and third of these roots satisfy the corresponding inequalities, and hence the original equation.
Answer: -1; .
Similarly, you can solve any problems with several modules. But, like any universal method, this solution method is far from always optimal. Below we will see how it can be improved.
1.4. The method of intervals in problems with modules
Looking more closely at the conditions different variants distribution of signs of submodule expressions in the previous solution, we will see that one of them, 1 - 3x
Imagine we are solving an equation that has three modulus of linear expressions; for example, |x – a| + |x – b| + |x – c| = m.
The first modulus is x - a for x ³ a and a - x for x b and x
They form four gaps. On each of them, each of the expressions under the modules retains its sign, therefore, the equation as a whole, after expanding the modules, has the same form on each interval. So, out of 8 theoretically possible options for opening modules, only 4 turned out to be enough for us!
You can also solve any problem with several modules. Namely, the numerical axis is divided into intervals of constant sign of all expressions under the modules, and then on each of them the equation or inequality is solved, into which the given problem turns on this interval. In particular, if all expressions under modules are rational, then it suffices to mark their roots on the axis, as well as the points where they are not defined, that is, the roots of their denominators. Marked points and set the required intervals of sign constancy. In the same way, we act when solving rational inequalities by the method of intervals. And the method we have described for solving problems with modules has the same name.
Example 1. Solve the equation.
Solution. Find the zeros of the function , whence . We solve the problem on each interval:
So this equation has no solutions.
Example 2. Solve the equation.
Solution. Find the zeros of the function . We solve the problem on each interval:
1) (no solutions);
Example 3. Solve the equation.
Solution. The expressions under the absolute value sign vanish at . Accordingly, we need to consider three cases:
2) - the root of the equation;
3) is the root of this equation.
Chapter 2. Equations and inequalities containing modules.
2.1 Solutions of equations with several modules using the method of intervals.
Example 1
Solve the equation:
|x+2| = |x-1|+x-3
-(x+2) = -(x-1) + x-3
X-2=-x+1+x-3
x=2 - does not satisfy
condition x
no solutions
2. If -2≤x
x+2 = -(x-1)+x-3
satisfies
condition -2
3. If x≥1, then
Answer: x=6
Example 2
Solve the equation:
1) Find zeros of submodule expressions
Zeros of submodule expressions break the numerical axis into several intervals. Arrange the signs of submodule expressions on these intervals.
At each interval, we open the modules and solve the resulting equation. After finding the root, we check that it belongs to the interval on which we are in this moment we are working.
1. :
- fits.
2. :
- does not fit.
3. :
– fits.
4. :
- does not fit. Answer:
2.2 Solving inequalities with multiple modules using the interval method.
Example 1
Solve the inequality:
|x-1| + |x-3| four
-(x-1) - (x-3) 4
2. If 1≤x
x-1– (x-3) 4
24 is wrong
no solutions
3. If x≥3, then
Answer: xЄ (-∞; 0) U (4; + ∞)
Example 2
Let's solve the inequality
Solution. The dots and (the roots of the expressions under the module) divide the entire numerical axis into three intervals, on each of which the modules should be expanded.
1) When is satisfied, and the inequality has the form , that is, . In this case, the answer is .
2) When , the inequality has the form , that is, . This inequality is true for any values of the variable , and, given that we solve it on the set , we get the answer in the second case .
3) When , the inequality is transformed to , and the solution in this case is . General solution of the inequality --- an association three responses received.
Thus, to solve equations and inequalities containing several modules, it is convenient to use the interval method. To do this, you need to find the zeros of the milestones of the submodule functions, denote them by odz equations and inequalities.
Conclusion
AT recent times in mathematics, methods are widely used to simplify the solution of problems, in particular the interval method, which makes it possible to significantly speed up calculations. Therefore, the study of the interval method for solving equations and inequalities with several modules is relevant.
In the process of working on the topic “Solving equations and inequalities containing the unknown under the modulus sign by the interval method”, I: studied the literature on this issue, got acquainted with the algebraic and graphical approach to solving equations and inequalities containing the unknown under the modulus sign, and came to the conclusion:
In some cases, when solving equations with a modulus, it is possible to solve equations according to the rules, and sometimes it is more convenient to use the interval method.
When solving equations and inequalities containing a modulus, the interval method is more visual and relatively simpler.
In the course of writing research work I have revealed many problems that can be solved using the interval method. The most important task is to solve equations and inequalities with multiple modules.
In the course of my work on solving inequalities and equations with several modules using the interval method, I found that the speed of solving problems doubled. This allows you to significantly speed up the workflow and reduce time costs. Thus, my hypothesis “if you use the interval method to solve inequalities and equations with several modules, you can greatly facilitate your work” was confirmed. In the process of working on the study, I gained experience in solving equations and inequalities with several modules. I think that the knowledge I have gained will allow me to avoid mistakes when solving.
Literature
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Olehnik S.N. Potapov M.K. Equations and inequalities. Non-standard solution methods. M.: Publishing House Factorial, 1997. - 219p.
Sevryukov P.F., Smolyakov A.N. Equations and inequalities with modules and methods for their solution. M.: Publishing House of Enlightenment 2005. - 112 p.
Sadovnichiy Yu.V. USE. Practicum in mathematics. Solution of equations and inequalities. Transformation of algebraic expressions. Moscow: Legion Publishing House 2015 - 128 p.
Shevkin A.V. Quadratic inequalities. interval method. M.: LLC " Russian word– educational book”, 2003. – 32 p.
http://padabum.com