Mathematics is a symbol of the wisdom of science,
a model of scientific rigor and simplicity,
the standard of excellence and beauty in science.
Russian philosopher, professor A.V. Voloshinov
Inequalities with modulus
The most difficult problems to solve in school mathematics are inequalities, containing variables under the modulus sign. To successfully solve such inequalities, you must have a good knowledge of the properties of the module and have the skills to use them.
Basic concepts and properties
Modulus (absolute value) of a real number denoted by and is defined as follows:
The simple properties of a module include the following relationships:
AND .
Note, that the last two properties are valid for any even degree.
Moreover, if, where, then and
More complex module properties, which can be effectively used when solving equations and inequalities with moduli, are formulated through the following theorems:
Theorem 1.For any analytical functions And inequality is true.
Theorem 2. Equality tantamount to inequality.
Theorem 3. Equality tantamount to inequality.
The most common inequalities in school mathematics, containing unknown variables under the modulus sign, are inequalities of the form and where some positive constant.
Theorem 4. Inequality is equivalent to double inequality, and the solution to the inequalityreduces to solving a set of inequalities And .
This theorem is a special case of Theorems 6 and 7.
More complex inequalities, containing a 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
I (1)
Proof. Since then
This implies the validity of (1).
Theorem 6. Inequality is equivalent to the system of inequalities
Proof. Because , then from inequality follows that . Under this condition, the inequalityand in this case the second system of inequalities (1) will turn 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
I (3)
Proof. Since , then the inequality always executed, If .
Let , then inequalitywill be equivalent to inequality, from which follows a set of two inequalities And .
The theorem has been proven.
Let's look at typical examples of solving problems on the topic “Inequalities, containing variables under the modulus sign."
Solving inequalities with modulus
The simplest method for solving inequalities with modulus is the method, based on module expansion. This method is universal, however, in the general case, its use can lead to very cumbersome calculations. Therefore, students should know other (more effective) methods and techniques for solving such inequalities. In particular, it is necessary to have skills in applying theorems, given in this article.
Example 1.Solve inequality
. (4)
Solution.We will solve inequality (4) using the “classical” method – the method of revealing modules. For this purpose, we divide the number axis dots and into intervals and consider three cases.
1. If , then , , , and inequality (4) takes the form or .
Since the case is considered here, it is a solution to inequality (4).
2. If, then from inequality (4) we obtain or . Since the intersection of intervals And is empty, then on the interval of solutions under consideration there is no inequality (4).
3. If, then inequality (4) takes the form or . It's obvious that is also a solution to inequality (4).
Answer: , .
Example 2. Solve inequality.
Solution. Let's assume that. Because , then the given inequality takes the form or . Since then and from here it follows or .
However, therefore or.
Example 3. Solve 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 4.Solve 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 to 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 5.Solve inequality
. (8)
Solution. Let us transform inequality (8) as follows:
Or .
Using the interval method, we obtain a solution to inequality (8).
Answer: .
Note. If we put and in the conditions of Theorem 5, we obtain .
Example 6. Solve inequality
. (9)
Solution. From inequality (9) it follows. Let us transform inequality (9) as follows:
Or
Since , then or .
Answer: .
Example 7.Solve inequality
. (10)
Solution. Since and , then or .
In this regard and inequality (10) takes the form
Or
. (11)
It follows 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 this and inequality (10) it follows, what or . Because , then inequality (10) takes the form or .
Example 8. Solve inequality
. (12)
Solution. Since then and from inequality (12) it follows or . However, therefore or. From here we get or .
Answer: .
Example 9. Solve inequality
. (13)
Solution. According to Theorem 7, the solution to inequality (13) is or .
Let it be now. In this case and inequality (13) takes the form or .
If you combine the intervals And , then we obtain a solution to inequality (13) of the form.
Example 10. Solve inequality
. (14)
Solution. Let us rewrite inequality (14) in an equivalent form: . If we apply Theorem 1 to the left side of this inequality, 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 inequality
. (15)
Solution. Applying Theorem 1 to the left side of inequality (15), we get . This and inequality (15) yield the equation, which has the form.
According to Theorem 3, the equation tantamount to inequality. From here we get.
Example 12.Solve inequality
. (16)
Solution. From inequality (16), according to Theorem 4, we obtain a system of inequalities
When solving the inequalityLet us use Theorem 6 and obtain a system of inequalitiesfrom which it follows.
Consider the inequality. According to Theorem 7, we obtain a set of inequalities And . The second population inequality is valid for any real.
Hence , the solution to inequality (16) is.
Example 13.Solve 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 14.Solve inequality
. (19)
Solution. Since, then. Let us multiply both sides of inequality (19) by the expression , which takes only positive values for any 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 solution to inequality (19) is And .
Answer: , .
For a more in-depth study of methods for solving inequalities with a modulus, we recommend turning to textbooks, given in the list of recommended literature.
1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.
2. Suprun V.P. Mathematics for high school students: methods of 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.: CD “Librocom” / URSS, 2017. – 296 p.
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Equations and inequalities with modulus.
Explanatory note.
This course is devoted to a systematic presentation of educational material related to the concept of the modulus of a number and aspects of its application. It discusses various methods for solving equations and inequalities with modulus based on its definition, properties and graphical interpretation.
The course is characterized by a practical orientation. Its main content consists of educational tasks. Some of them are given with a complete solution illustrating one or another method. Others are included for independent work. The presentation of practical solutions is accompanied by the necessary theoretical information.
The course is aimed at developing in students a broader understanding of the module. In addition, the tasks of the unified exam in mathematics require the ability to operate with the module. Thus, the main role of the course is to prepare students to successfully pass the Unified State Exam.
Educational and thematic plan
Material for classes
Lesson 1. Determining the modulus of a number and its application in solving equations.
Definition. Modulus of a non-negative real number X call this number itself: | x | = X; The modulus of a negative real number x is the opposite number: | x | = - x .
In short it is written like this:
|x | =
The term “module” (from the Latin modulus - measure) was introduced by the English mathematician R. Cortes (1682-1716), and the modulus sign was introduced by the German mathematician K. Weierstrass (1815-1897) in 1841. Using the above definition, you can solve equations and inequalities that contain a modulus. Now let's look at some simple examples.
Example 1. Solve the equation |3-3x|= -1.
Solution. By property of the module, the expression | 3-3x | is non-negative, so it can never be equal to (-1).
Answer. No solutions.
Example 2. Solve the equation | 3x -x 2 -2 | = 3x -x 2
Solution. We will not solve this equation in traditional ways, but note that it has the following form:
|A | = A.
Note that, by definition of the module, this equality is necessarily satisfied when A>0, and when A<0 оно не может быть верным. Поэтому исходное уравнение равносильно квадратному неравенству 3х – х 2 - 2 >0, which we already know how to solve.
Answer. .
Example 3. Solve the equation | x + 2 | = | 2x – 1 |.
Solution. Let's square both sides of the equation. This can be done because both sides of the original equation are non-negative. We get
| x + 2 | 2 = | 2x – 1 | 2.
Obviously, in this equation we can remove the modules and write the equivalent quadratic equation
(x + 2) 2 = (2x – 1) 2,
Transforming which, we get
x 2 + 4x + 4 = 4x 2 – 4x + 1, 3x 2 – 8x – 3 = 0.
Answer. ( -1/3 .3).
Now let's move on to more traditional tasks.
The main technique for solving equations and inequalities containing the expression |f (x)| is to reveal the module by definition, namely, the entire range of permissible values M is divided into two subsets M 1 and M 2 such that
f (x)>0 for all x M 1, then |f (x)| =f(x)
f(x)<0 для всех х ∊ М 2 ,тогда |f (x )| = - f (x )
Example 4. Solve the equation | 2x – 3 | = 3x – 7.
Solution. Consider cases: 1. 2x – 3 >0, 2x – 3 = 3x – 7, x = 4
2. 2x - 3<0, -2х + 3 = 3х- 7, х=2-не является корнем, т.к. при х=2 2х-3>0. Answer: 4.
This method is not the only one. When solving an equation of the form
| f(x) | = g(x)
The following two methods are most widely used.
The first, standard, is based on the disclosure of the module, based on its definition, and consists in the transition to an equivalent set of systems
| f(x) | = g(x)
The second method is to move from the original equation to an equivalent system
| f(x) | = g(x)
The first method should be used in the case of a complex expression for the function g (x) and not very complex for the function f (x); the second, on the contrary, is better to use if the expression for g (x) is simple.
Example 5. Solve the equation |x | = x - √2x +1 + 1 (Using the first method)
Example 6. Solve the equation 3|x 2 -2x -1| = 5x +1 (Using the second method)
Inequality of the form | f(x) |< g (x ) гораздо удобнее решать, перейдя двойному неравенству или к равносильной ему системе двух неравенств
| f(x) | g(x) -g(x) f(x) g(x)
Similarly, an inequality of the form
| f(x) | g(x)
Solve equations
3|y 2 – 6y + 7| = 5y – 9 |x | - |x – 1| = 1 |x 2 – 1| = (x – 1)
x 2 + |x – 1| = 1 |x 2 + 2x – 3| = x 2 + x – 20
Solve inequalities
|2x – 5|< 3 |x 2 – 2x – 3| < 3x - 3
x 2 – 6 > |x | |3 - |x – 2| |< 1
Lesson 2. Interval method for solving equations and inequalities containing a modulus.
Solve the equation | x -2| + |2x -3| = 5. Expanding sequentially the modules included in the equation under consideration, we will have to consider four systems and a obviously unsuitable case. And if there are three or more modules in the equation, the number of systems will increase even more. Therefore, to solve problems that involve two or more modules, it is more rational to use the interval method.
To apply the interval method when solving equations with modules, the number line must be divided into intervals in such a way that on each of them all submodular expressions retain constant signs and, therefore, on each interval all modules are revealed in a certain way.
Example 1. Solve the equation | 3x +4| + 2|x -3| = 16
Let us mark the points x = - 4/3 and x = 3 on the numerical axis, at which the submodular expressions vanish. Let us determine the signs of submodular expressions on the three resulting intervals.
Case 1. When x>3, both modules open with a “+” sign. We get the system
x >3,
3x+4+2(x-3) = 16 x=18/5 (18/5>3)
Case 2. At -4/3 4/3 3x+4+2(-x+3) = 16. The equation of this system has a root x=6, which does not satisfy the inequality of the system, therefore, it is not the root of the given equation. Case 3. At x< -4/3 оба модуля раскрываются со знаком «-«, получаем x< -4/3, 3x-4+(-x+3) = 16. This system has a unique solution x = -14/5. Answer:(-14/5; 18/5). The solution to inequalities containing a modulus is, in most cases, constructed similarly to the solution to the corresponding equations. The main difference is that after freeing ourselves from modules, we need to solve, naturally, not an equation, but an inequality. There is one more difference. If when solving equations one can widely use verification of the obtained solutions, then in the case of inequalities it can be difficult to discard extraneous solutions by verification. This means that when solving inequalities, they try to use mainly equivalent transitions. Example 2. Solve the inequality |x – 4| + |x + 1|<7 Solution. On the number line it is necessary to mark the numbers x=-1 and x=4, at which the expressions under the moduli signs turn to zero. Then we place expression signs on the three resulting intervals (x-4) and (x+1). __________________________ The resulting sets of signs indicate to us which cases need to be considered. As a result of expanding the modules in these three cases, we obtain three systems. Solving these systems and combining the answers, we get Answer: (-2;5). Exercises for independent work Solve the equations: | x – 1| + |x – 2| + |x – 3| = 4 |6 – 2x | + |3x + 7| - 2|4x + 11| = x – 3 | |3x – 1| - |2x + 1| | = 1 Solve the inequalities: |x – 1| + |x + 2|< 3
|x – 1|< |2x
– 3| - |x
– 2| |x
2 – 3| + x
2 + x
< 7. Lesson 4. Solving equations and inequalities with modules on the coordinate line. When studying the distance between two points A(x 1) and B(x 2) on a coordinate line, a formula is derived according to which AB = | x 1 - x 2 |. Using this formula, you can solve equations and inequalities of the form |x – a | = b , |x – a | = |x – b |, |x – a | | x – a |>|x – b |, as well as equations and inequalities that can be reduced to them. Example 1. Solve the equation |x – 3| = 1. Solution. Translating this equation into the “language of distances,” we get the sentence “the distance from a point with coordinate x to a point with coordinate 3 is 1.” Consequently, solving the equation comes down to finding points distant from the point with coordinate 3 by a distance of 1. Let us turn to the geometric illustration. _______________________________________________________ The roots of the equation are the numbers 2 and 4. Example 2. Solve the equation | 2x + 1 | = 3 Reducing this equation to the form | x – (-1/2) | = 3/2, use the distance formula. Answer: -2;1. Example 3: Solve the equation |x + 2| = |x – 1|. Solution. Let's write this equation in the form |x – (-2)| = |x – 1|. Based on geometric considerations, it is not difficult to understand that the root of the last equation is the coordinate of a point equidistant from the points with coordinates 1 and -2. Answer: -0.5. Example 4: Solve the inequality |x – 1|<2. Solution. Based on geometric concepts, we come to the conclusion that the solutions to this inequality are the coordinates of points located at a distance less than 2 from the point with coordinate 1. Answer: (-1;3) Exercises for independent work | x – 2| = 0.4 | 10 – x |< 7 | x
+ 4 | = | x
– 4 | | x + 3 | = 0.7 | x + 1 | > 1 | x + 2.5| = | x - 3.3| | x – 2.5|< 0,5 | x
+ 8 | >0.7 | x | > | x – 2 | | x – 5 |< | x – 1 |
. This online math calculator will help you solve an equation or inequality with moduli. Program for solving equations and inequalities with moduli not only gives the answer to the problem, it leads detailed solution with explanations, i.e. displays the process of obtaining the result. This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions. In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases. Enter an equation or inequality with moduli It was discovered that some scripts necessary to solve this problem were not loaded, and the program may not work. Because There are a lot of people willing to solve the problem, your request has been queued. If you noticed an error in the solution, then you can write about this in the Feedback Form. Our games, puzzles, emulators: In a basic school algebra course, you may encounter the simplest equations and inequalities with moduli. To solve them, you can use a geometric method based on the fact that \(|x-a| \) is the distance on the number line between points x and a: \(|x-a| = \rho (x;\; a) \). For example, to solve the equation \(|x-3|=2\) you need to find points on the number line that are distant from point 3 at a distance of 2. There are two such points: \(x_1=1\) and \(x_2=5\) . Solving the inequality \(|2x+7| But the main way to solve equations and inequalities with moduli is associated with the so-called “revelation of the modulus by definition”: In addition to the above definition, the following statements are used: If \(x-1 \geq 0\), then \(|x-1| = x-1\) and the given equation takes the form EXAMPLE 2. Solve the equation \(|x^2-6x+7| = \frac(5x-9)(3)\). First way(module expansion by definition). 1) If \(x^2-6x+7 \geq 0 \), then \(|x^2-6x+7| = x^2-6x+7 \) and the given equation takes the form \(x^2 -6x+7 = \frac(5x-9)(3) \Rightarrow 3x^2-23x+30=0 \). Having solved this quadratic equation, we get: \(x_1=6, \; x_2=\frac(5)(3) \). 2) If \(x^2-6x+7 Value \(x_3=3\) satisfies the condition \(x^2-6x+7 Value \(x_4=\frac(4)(3) \) does not satisfy the condition \ (x^2-6x+7 So, the given equation has two roots: \(x=6, \; x=3 \). Second way. If the equation \(|f(x)| = h(x) \) is given, then with \(h(x) \(\left[\begin(array)(l) x^2-6x+7 = \frac (5x-9)(3) \\ x^2-6x+7 = -\frac(5x-9)(3) \end(array)\right. \) Third way(graphic). It is important that the point x = 1.8 of the intersection of the straight line with the abscissa axis is located to the right of the left point of intersection of the parabola with the abscissa axis - this is the point \(x=3-\sqrt(2)\) (since \(3-\sqrt(2 ) 3) Judging by the drawing, the graphs intersect at two points - A(3; 2) and B(6; 7).Substituting the abscissas of these points x = 3 and x = 6 into the given equation, we are convinced that both In another value, the correct numerical equality is obtained. This means that our hypothesis was confirmed - the equation has two roots: x = 3 and x = 6. Answer: 3; 6. Comment. The graphical method, for all its elegance, is not very reliable. In the example considered, it worked only because the roots of the equation are integers. EXAMPLE 3. Solve the equation \(|2x-4|+|x+3| = 8\) First way Consider the first interval: \((-\infty; \; -3) \). In simple terms, a modulus is a “number without a minus.” And it is precisely in this duality (in some places you don’t have to do anything with the original number, but in others you will have to remove some kind of minus) that is where the whole difficulty lies for beginning students. There is also a geometric definition. It is also useful to know, but we will turn to it only in complex and some special cases, where the geometric approach is more convenient than the algebraic one (spoiler: not today). Definition. Let point $a$ be marked on the number line. Then the module $\left| x-a \right|$ is the distance from point $x$ to point $a$ on this line. If you draw a picture, you will get something like this: One way or another, from the definition of a module its key property immediately follows: the modulus of a number is always a non-negative quantity. This fact will be a red thread running through our entire narrative today. Now let's look at the inequalities. There are a great many of them, but our task now is to be able to solve at least the simplest of them. Those that reduce to linear inequalities, as well as to the interval method. I have two big lessons on this topic (by the way, very, VERY useful - I recommend studying them): If you know all this, if the phrase “let’s move from inequality to equation” does not make you have a vague desire to hit yourself against the wall, then you are ready: welcome to hell to the main topic of the lesson. :) This is one of the most common problems with modules. It is required to solve an inequality of the form: \[\left| f\right| \ltg\] The functions $f$ and $g$ can be anything, but usually they are polynomials. Examples of such inequalities: \[\begin(align) & \left| 2x+3 \right| \lt x+7; \\ & \left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0; \\ & \left| ((x)^(2))-2\left| x \right|-3 \right| \lt 2. \\\end(align)\] All of them can be solved literally in one line according to the following scheme: \[\left| f\right| \lt g\Rightarrow -g \lt f \lt g\quad \left(\Rightarrow \left\( \begin(align) & f \lt g, \\ & f \gt -g \\\end(align) \right.\right)\] It is easy to see that we get rid of the module, but in return we get a double inequality (or, which is the same thing, a system of two inequalities). But this transition takes into account absolutely all possible problems: if the number under the modulus is positive, the method works; if negative, it still works; and even with the most inadequate function in place of $f$ or $g$, the method will still work. Naturally, the question arises: couldn’t it be simpler? Unfortunately, it's not possible. This is the whole point of the module. However, enough with the philosophizing. Let's solve a couple of problems: Task. Solve the inequality: \[\left| 2x+3 \right| \lt x+7\] Solution. So, we have before us a classic inequality of the form “the modulus is less” - there’s even nothing to transform. We work according to the algorithm: \[\begin(align) & \left| f\right| \lt g\Rightarrow -g \lt f \lt g; \\ & \left| 2x+3 \right| \lt x+7\Rightarrow -\left(x+7 \right) \lt 2x+3 \lt x+7 \\\end(align)\] Do not rush to open the parentheses preceded by a “minus”: it is quite possible that due to your haste you will make an offensive mistake. \[-x-7 \lt 2x+3 \lt x+7\] \[\left\( \begin(align) & -x-7 \lt 2x+3 \\ & 2x+3 \lt x+7 \\ \end(align) \right.\] \[\left\( \begin(align) & -3x \lt 10 \\ & x \lt 4 \\ \end(align) \right.\] \[\left\( \begin(align) & x \gt -\frac(10)(3) \\ & x \lt 4 \\ \end(align) \right.\] The problem was reduced to two elementary inequalities. Let us note their solutions on parallel number lines: The intersection of these sets will be the answer. Answer: $x\in \left(-\frac(10)(3);4 \right)$ Task. Solve the inequality: \[\left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0\] Solution. This task is a little more difficult. First, let’s isolate the module by moving the second term to the right: \[\left| ((x)^(2))+2x-3 \right| \lt -3\left(x+1 \right)\] Obviously, we again have an inequality of the form “the module is smaller”, so we get rid of the module using the already known algorithm: \[-\left(-3\left(x+1 \right) \right) \lt ((x)^(2))+2x-3 \lt -3\left(x+1 \right)\] Now attention: someone will say that I'm a bit of a pervert with all these parentheses. But let me remind you once again that our key goal is correctly solve the inequality and get the answer. Later, when you have perfectly mastered everything described in this lesson, you can pervert it yourself as you wish: open parentheses, add minuses, etc. To begin with, we’ll simply get rid of the double minus on the left: \[-\left(-3\left(x+1 \right) \right)=\left(-1 \right)\cdot \left(-3 \right)\cdot \left(x+1 \right) =3\left(x+1 \right)\] Now let's open all the brackets in the double inequality: Let's move on to the double inequality. This time the calculations will be more serious: \[\left\( \begin(align) & ((x)^(2))+2x-3 \lt -3x-3 \\ & 3x+3 \lt ((x)^(2))+2x -3 \\ \end(align) \right.\] \[\left\( \begin(align) & ((x)^(2))+5x \lt 0 \\ & ((x)^(2))-x-6 \gt 0 \\ \end( align)\right.\] Both inequalities are quadratic and can be solved by the interval method (that’s why I say: if you don’t know what this is, it’s better not to take on modules yet). Let's move on to the equation in the first inequality: \[\begin(align) & ((x)^(2))+5x=0; \\ & x\left(x+5 \right)=0; \\ & ((x)_(1))=0;((x)_(2))=-5. \\\end(align)\] As you can see, the output is an incomplete quadratic equation, which can be solved in an elementary way. Now let's look at the second inequality of the system. There you will have to apply Vieta's theorem: \[\begin(align) & ((x)^(2))-x-6=0; \\ & \left(x-3 \right)\left(x+2 \right)=0; \\& ((x)_(1))=3;((x)_(2))=-2. \\\end(align)\] We mark the resulting numbers on two parallel lines (separate for the first inequality and separate for the second): Answer: $x\in \left(-5;-2 \right)$ I think that after these examples the solution scheme is extremely clear: A similar algorithm exists for inequalities of the following type, when the modulus is greater than the function. However, there are a couple of serious “buts”. We’ll talk about these “buts” now. They look like this: \[\left| f\right| \gtg\] Similar to the previous one? It seems. And yet such problems are solved in a completely different way. Formally, the scheme is as follows: \[\left| f\right| \gt g\Rightarrow \left[ \begin(align) & f \gt g, \\ & f \lt -g \\\end(align) \right.\] In other words, we consider two cases: In this case, the options are combined with a square bracket, i.e. We have before us a combination of two requirements. Please note again: this is not a system, but a totality, therefore in the answer the sets are combined rather than intersecting. This is a fundamental difference from the previous point! In general, many students are completely confused with unions and intersections, so let’s sort this issue out once and for all: To make it even easier to remember, just draw legs to these signs to make glasses (just don’t now accuse me of promoting drug addiction and alcoholism: if you are seriously studying this lesson, then you are already a drug addict): Translated into Russian, this means the following: the union (totality) includes elements from both sets, therefore it is in no way less than each of them; but the intersection (system) includes only those elements that are simultaneously in both the first set and the second. Therefore, the intersection of sets is never larger than the source sets. So it became clearer? That is great. Let's move on to practice. Task. Solve the inequality: \[\left| 3x+1 \right| \gt 5-4x\] Solution. We proceed according to the scheme: \[\left| 3x+1 \right| \gt 5-4x\Rightarrow \left[ \begin(align) & 3x+1 \gt 5-4x \\ & 3x+1 \lt -\left(5-4x \right) \\\end(align) \ right.\] We solve each inequality in the population: \[\left[ \begin(align) & 3x+4x \gt 5-1 \\ & 3x-4x \lt -5-1 \\ \end(align) \right.\] \[\left[ \begin(align) & 7x \gt 4 \\ & -x \lt -6 \\ \end(align) \right.\] \[\left[ \begin(align) & x \gt 4/7\ \\ & x \gt 6 \\ \end(align) \right.\] We mark each resulting set on the number line, and then combine them: It is quite obvious that the answer will be $x\in \left(\frac(4)(7);+\infty \right)$ Answer: $x\in \left(\frac(4)(7);+\infty \right)$ Task. Solve the inequality: \[\left| ((x)^(2))+2x-3 \right| \gt x\] Solution. Well? Nothing - everything is the same. We move from an inequality with a modulus to a set of two inequalities: \[\left| ((x)^(2))+2x-3 \right| \gt x\Rightarrow \left[ \begin(align) & ((x)^(2))+2x-3 \gt x \\ & ((x)^(2))+2x-3 \lt -x \\\end(align) \right.\] We solve every inequality. Unfortunately, the roots there will not be very good: \[\begin(align) & ((x)^(2))+2x-3 \gt x; \\ & ((x)^(2))+x-3 \gt 0; \\&D=1+12=13; \\ & x=\frac(-1\pm \sqrt(13))(2). \\\end(align)\] The second inequality is also a bit wild: \[\begin(align) & ((x)^(2))+2x-3 \lt -x; \\ & ((x)^(2))+3x-3 \lt 0; \\&D=9+12=21; \\ & x=\frac(-3\pm \sqrt(21))(2). \\\end(align)\] Now you need to mark these numbers on two axes - one axis for each inequality. However, you need to mark the points in the correct order: the larger the number, the further the point moves to the right. And here a setup awaits us. If everything is clear with the numbers $\frac(-3-\sqrt(21))(2) \lt \frac(-1-\sqrt(13))(2)$ (the terms in the numerator of the first fraction are less than the terms in the numerator of the second , so the sum is also less), with the numbers $\frac(-3-\sqrt(13))(2) \lt \frac(-1+\sqrt(21))(2)$ there will also be no difficulties (positive number obviously more negative), then with the last couple everything is not so clear. Which is greater: $\frac(-3+\sqrt(21))(2)$ or $\frac(-1+\sqrt(13))(2)$? The placement of points on the number lines and, in fact, the answer will depend on the answer to this question. So let's compare: \[\begin(matrix) \frac(-1+\sqrt(13))(2)\vee \frac(-3+\sqrt(21))(2) \\ -1+\sqrt(13)\ vee -3+\sqrt(21) \\ 2+\sqrt(13)\vee \sqrt(21) \\\end(matrix)\] We isolated the root, got non-negative numbers on both sides of the inequality, so we have the right to square both sides: \[\begin(matrix) ((\left(2+\sqrt(13) \right))^(2))\vee ((\left(\sqrt(21) \right))^(2)) \ \ 4+4\sqrt(13)+13\vee 21 \\ 4\sqrt(13)\vee 3 \\\end(matrix)\] I think it’s a no brainer that $4\sqrt(13) \gt 3$, so $\frac(-1+\sqrt(13))(2) \gt \frac(-3+\sqrt(21)) (2)$, the final points on the axes will be placed like this: Let me remind you that we are solving a set, so the answer will be a union, not an intersection of shaded sets. Answer: $x\in \left(-\infty ;\frac(-3+\sqrt(21))(2) \right)\bigcup \left(\frac(-1+\sqrt(13))(2 );+\infty \right)$ As you can see, our scheme works great for both simple and very tough problems. The only “weak point” in this approach is that you need to correctly compare irrational numbers (and believe me: these are not only roots). But a separate (and very serious) lesson will be devoted to comparison issues. And we move on. Now we get to the most interesting part. These are inequalities of the form: \[\left| f\right| \gt\left| g\right|\] Generally speaking, the algorithm that we will talk about now is correct only for the module. It works in all inequalities where there are guaranteed non-negative expressions on the left and right: What to do with these tasks? Just remember: In inequalities with non-negative “tails”, both sides can be raised to any natural power. There will be no additional restrictions. First of all, we will be interested in squaring - it burns modules and roots: \[\begin(align) & ((\left(\left| f \right| \right))^(2))=((f)^(2)); \\ & ((\left(\sqrt(f) \right))^(2))=f. \\\end(align)\] Just don’t confuse this with taking the root of a square: \[\sqrt(((f)^(2)))=\left| f \right|\ne f\] Countless mistakes were made when a student forgot to install a module! But this is a completely different story (these are, as it were, irrational equations), so we will not go into this now. Let's solve a couple of problems better: Task. Solve the inequality: \[\left| x+2 \right|\ge \left| 1-2x \right|\] Solution. Let's immediately notice two things: Therefore, we can square both sides of the inequality to get rid of the modulus and solve the problem using the usual interval method: \[\begin(align) & ((\left(\left| x+2 \right| \right))^(2))\ge ((\left(\left| 1-2x \right| \right) )^(2)); \\ & ((\left(x+2 \right))^(2))\ge ((\left(2x-1 \right))^(2)). \\\end(align)\] At the last step, I cheated a little: I changed the sequence of terms, taking advantage of the evenness of the module (in fact, I multiplied the expression $1-2x$ by −1). \[\begin(align) & ((\left(2x-1 \right))^(2))-((\left(x+2 \right))^(2))\le 0; \\ & \left(\left(2x-1 \right)-\left(x+2 \right) \right)\cdot \left(\left(2x-1 \right)+\left(x+2 \ right)\right)\le 0; \\ & \left(2x-1-x-2 \right)\cdot \left(2x-1+x+2 \right)\le 0; \\ & \left(x-3 \right)\cdot \left(3x+1 \right)\le 0. \\\end(align)\] We solve using the interval method. Let's move from inequality to equation: \[\begin(align) & \left(x-3 \right)\left(3x+1 \right)=0; \\ & ((x)_(1))=3;((x)_(2))=-\frac(1)(3). \\\end(align)\] We mark the found roots on the number line. Once again: all points are shaded because the original inequality is not strict! Let me remind you for those who are especially stubborn: we take the signs from the last inequality, which was written down before moving on to the equation. And we paint over the areas required in the same inequality. In our case it is $\left(x-3 \right)\left(3x+1 \right)\le 0$. OK it's all over Now. The problem is solved. Answer: $x\in \left[ -\frac(1)(3);3 \right]$. Task. Solve the inequality: \[\left| ((x)^(2))+x+1 \right|\le \left| ((x)^(2))+3x+4 \right|\] Solution. We do everything the same. I won't comment - just look at the sequence of actions. Square it: \[\begin(align) & ((\left(\left| ((x)^(2))+x+1 \right| \right))^(2))\le ((\left(\left | ((x)^(2))+3x+4 \right| \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))\le ((\left(((x)^(2))+3x+4 \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))-((\left(((x)^(2))+3x+4 \ right))^(2))\le 0; \\ & \left(((x)^(2))+x+1-((x)^(2))-3x-4 \right)\times \\ & \times \left(((x) ^(2))+x+1+((x)^(2))+3x+4 \right)\le 0; \\ & \left(-2x-3 \right)\left(2((x)^(2))+4x+5 \right)\le 0. \\\end(align)\] Interval method: \[\begin(align) & \left(-2x-3 \right)\left(2((x)^(2))+4x+5 \right)=0 \\ & -2x-3=0\ Rightarrow x=-1.5; \\ & 2((x)^(2))+4x+5=0\Rightarrow D=16-40 \lt 0\Rightarrow \varnothing . \\\end(align)\] There is only one root on the number line: Answer: $x\in \left[ -1.5;+\infty \right)$. A small note about the last task. As one of my students accurately noted, both submodular expressions in this inequality are obviously positive, so the modulus sign can be omitted without harm to health. But this is a completely different level of thinking and a different approach - it can conditionally be called the method of consequences. About it - in a separate lesson. Now let’s move on to the final part of today’s lesson and look at a universal algorithm that always works. Even when all previous approaches were powerless. :) What if all these techniques don't help? If the inequality cannot be reduced to non-negative tails, if it is impossible to isolate the module, if in general there is pain, sadness, melancholy? Then the “heavy artillery” of all mathematics comes onto the scene—the brute force method. In relation to inequalities with modulus it looks like this: So how? Weak? Easily! Only for a long time. Let's see in practice: Task. Solve the inequality: \[\left| x+2 \right| \lt \left| x-1 \right|+x-\frac(3)(2)\] Solution. This crap doesn't boil down to inequalities like $\left| f\right| \lt g$, $\left| f\right| \gt g$ or $\left| f\right| \lt \left| g \right|$, so we act ahead. We write out submodular expressions, equate them to zero and find the roots: \[\begin(align) & x+2=0\Rightarrow x=-2; \\ & x-1=0\Rightarrow x=1. \\\end(align)\] In total, we have two roots that divide the number line into three sections, within which each module is revealed uniquely: Let's look at each section separately. 1. Let $x \lt -2$. Then both submodular expressions are negative, and the original inequality will be rewritten as follows: \[\begin(align) & -\left(x+2 \right) \lt -\left(x-1 \right)+x-1.5 \\ & -x-2 \lt -x+1+ x-1.5 \\ & x \gt 1.5 \\\end(align)\] We got a fairly simple limitation. Let's intersect it with the initial assumption that $x \lt -2$: \[\left\( \begin(align) & x \lt -2 \\ & x \gt 1.5 \\\end(align) \right.\Rightarrow x\in \varnothing \] Obviously, the variable $x$ cannot simultaneously be less than −2 and greater than 1.5. There are no solutions in this area. 1.1. Let us separately consider the borderline case: $x=-2$. Let's just substitute this number into the original inequality and check: is it true? \[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1.5 \right|)_(x=-2) ) \\ & 0 \lt \left| -3\right|-2-1.5; \\ & 0 \lt 3-3.5; \\ & 0 \lt -0.5\Rightarrow \varnothing . \\\end(align)\] It is obvious that the chain of calculations has led us to an incorrect inequality. Therefore, the original inequality is also false, and $x=-2$ is not included in the answer. 2. Let now $-2 \lt x \lt 1$. The left module will already open with a “plus”, but the right one will still open with a “minus”. We have: \[\begin(align) & x+2 \lt -\left(x-1 \right)+x-1.5 \\ & x+2 \lt -x+1+x-1.5 \\& x \lt -2.5 \\\end(align)\] Again we intersect with the original requirement: \[\left\( \begin(align) & x \lt -2.5 \\ & -2 \lt x \lt 1 \\\end(align) \right.\Rightarrow x\in \varnothing \] And again, the set of solutions is empty, since there are no numbers that are both less than −2.5 and greater than −2. 2.1. And again a special case: $x=1$. We substitute into the original inequality: \[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1.5 \right|)_(x=1)) \\ & \left| 3\right| \lt \left| 0\right|+1-1.5; \\ & 3 \lt -0.5; \\ & 3 \lt -0.5\Rightarrow \varnothing . \\\end(align)\] Similar to the previous “special case”, the number $x=1$ is clearly not included in the answer. 3. The last piece of the line: $x \gt 1$. Here all modules are opened with a plus sign: \[\begin(align) & x+2 \lt x-1+x-1.5 \\ & x+2 \lt x-1+x-1.5 \\ & x \gt 4.5 \\ \end(align)\] And again we intersect the found set with the original constraint: \[\left\( \begin(align) & x \gt 4.5 \\ & x \gt 1 \\\end(align) \right.\Rightarrow x\in \left(4.5;+\infty \right)\] Finally! We have found an interval that will be the answer. Answer: $x\in \left(4,5;+\infty \right)$ Finally, one remark that may save you from stupid mistakes when solving real problems: Solutions to inequalities with moduli usually represent continuous sets on the number line - intervals and segments. Isolated points are much less common. And even less often, it happens that the boundary of the solution (the end of the segment) coincides with the boundary of the range under consideration. Consequently, if boundaries (the same “special cases”) are not included in the answer, then the areas to the left and right of these boundaries will almost certainly not be included in the answer. And vice versa: the border entered into the answer, which means that some areas around it will also be answers. Keep this in mind when reviewing your solutions.
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Do not forget indicate which task you decide what enter in the fields.A little theory.
Equations and inequalities with moduli
if \(a \geq 0 \), then \(|a|=a \);
if \(a As a rule, an equation (inequality) with moduli is reduced to a set of equations (inequalities) that do not contain the modulus sign.
1) If \(c > 0\), then the equation \(|f(x)|=c \) is equivalent to the set of equations: \(\left[\begin(array)(l) f(x)=c \\ f(x)=-c \end(array)\right. \)
2) If \(c > 0 \), then the inequality \(|f(x)| 3) If \(c \geq 0 \), then the inequality \(|f(x)| > c \) is equivalent to a set of inequalities : \(\left[\begin(array)(l) f(x) c \end(array)\right. \)
4) If both sides of the inequality \(f(x) EXAMPLE 1. Solve the equation \(x^2 +2|x-1| -6 = 0\).
\(x^2 +2(x-1) -6 = 0 \Rightarrow x^2 +2x -8 = 0 \).
If \(x-1 \(x^2 -2(x-1) -6 = 0 \Rightarrow x^2 -2x -4 = 0 \).
Thus, the given equation should be considered separately in each of the two indicated cases.
1) Let \(x-1 \geq 0 \), i.e. \(x\geq 1\). From the equation \(x^2 +2x -8 = 0\) we find \(x_1=2, \; x_2=-4\). The condition \(x \geq 1 \) is satisfied only by the value \(x_1=2\).
2) Let \(x-1 Answer: \(2; \;\; 1-\sqrt(5) \)
Reasoning as in example 1, we come to the conclusion that the given equation needs to be considered separately if two conditions are met: \(x^2-6x+7 \geq 0 \) or \(x^2-6x+7
Let's find out whether the value \(x_1=6\) satisfies the condition \(x^2-6x+7 \geq 0\). To do this, substitute the indicated value into the quadratic inequality. We get: \(6^2-6 \cdot 6+7 \geq 0 \), i.e. \(7 \geq 0 \) is a true inequality. This means that \(x_1=6\) is the root of the given equation.
Let's find out whether the value \(x_2=\frac(5)(3)\) satisfies the condition \(x^2-6x+7 \geq 0\). To do this, substitute the indicated value into the quadratic inequality. We get: \(\left(\frac(5)(3) \right)^2 -\frac(5)(3) \cdot 6 + 7 \geq 0 \), i.e. \(\frac(25)(9) -3 \geq 0 \) is an incorrect inequality. This means that \(x_2=\frac(5)(3)\) is not the root of the given equation.
Both of these equations were solved above (using the first method of solving the given equation), their roots are as follows: \(6,\; \frac(5)(3),\; 3,\; \frac(4)(3)\). The condition \(\frac(5x-9)(3) \geq 0 \) of these four values is satisfied by only two: 6 and 3. This means that the given equation has two roots: \(x=6, \; x=3 \ ).
1) Let's build a graph of the function \(y = |x^2-6x+7| \). First, let's construct a parabola \(y = x^2-6x+7\). We have \(x^2-6x+7 = (x-3)^2-2 \). The graph of the function \(y = (x-3)^2-2\) can be obtained from the graph of the function \(y = x^2\) by shifting it 3 scale units to the right (along the x-axis) and 2 scale units down ( along the y-axis). The straight line x=3 is the axis of the parabola we are interested in. As control points for more accurate plotting, it is convenient to take point (3; -2) - the vertex of the parabola, point (0; 7) and point (6; 7) symmetrical to it relative to the axis of the parabola.
To now construct a graph of the function \(y = |x^2-6x+7| \), you need to leave unchanged those parts of the constructed parabola that lie not below the x-axis, and mirror that part of the parabola that lies below the x-axis relative to the x axis.
2) Let's build a graph of the linear function \(y = \frac(5x-9)(3)\). It is convenient to take points (0; –3) and (3; 2) as control points. 
The expression 2x–4 becomes 0 at the point x = 2, and the expression x + 3 becomes 0 at the point x = –3. These two points divide the number line into three intervals: \(x
If x Consider the second interval: \([-3; \; 2) \).
If \(-3 \leq x Consider the third interval: \(
Graphical module definition Solving inequalities. Interval method
1. Inequalities of the form “Modulus is less than function”
Intersection of many 
Again, since we are solving a system of inequalities, we are interested in the intersection of the shaded sets: $x\in \left(-5;-2 \right)$. This is the answer. 2. Inequalities of the form “Modulus is greater than function”
Difference between intersection and union of sets
Union of sets 
A case of ugly roots 3. Inequalities with non-negative “tails”
Getting rid of the modulus sign 
The answer is a whole interval 4. Method of enumeration of options
Partitioning the number line by zeros of submodular functions