Lesson Objectives

Introduce students to this mathematical concept, as modulus of a number;
To teach schoolchildren the skills of finding modules of numbers;
Consolidate the studied material by performing various tasks;

Tasks

Consolidate children's knowledge about the modulus of number;
With Solution test items to check how the students learned the studied material;
Continue to instill interest in mathematics lessons;
To educate students in logical thinking, curiosity and perseverance.

Lesson Plan

1. General concepts and determination of the modulus of the number.
2. geometric sense module.
3. The modulus of the number of its properties.
4. Solving equations and inequalities that contain the modulus of a number.
5. History reference about the term "modulus of a number".
6. Task to consolidate knowledge of the topic covered.
7. Homework.

General concepts about the modulus of a number

The modulus of a number is usually called the number itself, if it does not have a negative value, or the same number is negative, but with the opposite sign.

That is, the modulus of a non-negative real number a is the number itself:

And, the modulus of a negative real number x will be the opposite number:

In writing, it will look like this:

For a better understanding, let's take an example. So, for example, the modulus of the number 3 is 3, and also the modulus of the number -3 is 3.

It follows from this that the modulus of a number means an absolute value, that is, its absolute value, but without taking into account its sign. To put it even more simply, it is necessary to discard the sign from the number.

The modulus of a number can be designated and look like this: |3|, |x|, |a| etc.

So, for example, the modulus of the number 3 is denoted by |3|.

Also, remember that the modulus of a number is never negative: |a|≥ 0.

|5| = 5, |-6| = 6, |-12.45| = 12.45 etc.

The geometric meaning of the module

The modulus of a number is the distance, which is measured in unit segments from the origin to the point. This definition expands the module with geometric point vision.

Let's take a coordinate line and denote two points on it. Let these points correspond to numbers such as -4 and 2.

Now let's take a look at this picture. We see that the point A indicated on the coordinate line corresponds to the number -4, and if you look closely, you will see that this point is located at a distance of 4 unit segments from the reference point 0. It follows that the length of the segment OA is equal to four units. In this case, the length of the segment OA, that is, the number 4 will be the modulus of the number -4.

In this case, the modulus of the number is denoted and written as follows: |−4| = 4.

Now take, and on the coordinate line, denote the point B.

This point B will correspond to the number +2, and, as we can see, it is located at a distance of two unit segments from the origin. It follows from this that the length of the segment OB is equal to two units. In this case, the number 2 will be the modulus of the number +2.

In writing it will look like this: |+2| = 2 or |2| = 2.

And now let's sum it up. If we take some unknown number a and denote it on the coordinate line by point A, then in this case the distance from point A to the origin, that is, the length of the segment OA, is precisely the modulus of the number "a".

In writing it will look like this: |a| = O.A.

Modulus of the number of its properties

And now let's try to highlight the properties of the module, consider all possible cases and write them using literal expressions:

First, the modulus of a number is a non-negative number, which means that the modulus of a positive number is equal to the number itself: |a| = a if a > 0;

Secondly, modules that consist of opposite numbers are equal: |a| = |–a|. That is, this property tells us that opposite numbers always have equal modules, that is, on the coordinate line, although they have opposite numbers, they are at the same distance from the reference point. It follows from this that the modules of these opposite numbers are equal.

Thirdly, the modulus of zero is equal to zero if this number is zero: |0| = 0 if a = 0. Here we can say with certainty that the modulus of zero is zero by definition, since it corresponds to the origin of the coordinate line.

The fourth property of the modulus is that the modulus of the product of two numbers is equal to the product of the modules of these numbers. Now let's take a closer look at what this means. If you follow the definition, then you and I know that the modulus of the product of numbers a and b will be equal to a b, or − (a b), if, a in ≥ 0, or - (a c), if, a in is greater than 0. In records it will look like this: |a b| = |a| |b|.

The fifth property is that the modulus of the quotient of numbers is equal to the ratio of the modules of these numbers: |a: b| = |a| : |b|.

And the following properties of the module of the number:

Solving equations and inequalities that contain the modulus of a number

When starting to solve problems that have a number modulus, it should be remembered that in order to solve such a task, it is necessary to reveal the sign of the module using knowledge of the properties that this task corresponds to.

Exercise 1

So, for example, if under the module sign there is an expression that depends on a variable, then the module should be expanded in accordance with the definition:

Of course, when solving problems, there are cases when the module is unambiguously revealed. If, for example, we take

, here we see that such an expression under the modulus sign is non-negative for any values ​​of x and y.

Or, for example, take

, we see that this modulus expression is not positive for any values ​​of z.

Task 2

In front of you is a coordinate line. On this line, it is necessary to mark the numbers, the modulus of which will be equal to 2.

Decision

First of all, we must draw a coordinate line. You already know that for this, first on the straight line you need to select the origin, direction and unit segment. Next, we need to put points from the origin that are equal to the distance of two unit segments.

As you can see, there are two such points on the coordinate line, one of which corresponds to the number -2, and the other to the number 2.

Historical information about the modulus of the number

The term "modulus" comes from the Latin name modulus, which means the word "measure" in translation. The term was coined by the English mathematician Roger Cotes. But the module sign was introduced thanks to the German mathematician Karl Weierstrass. When writing, a module is denoted with the following symbol: | |.

Questions to consolidate knowledge of the material

In today's lesson, we got acquainted with such a concept as the modulus of a number, and now let's check how you learned this topic by answering the questions posed:

1. What is the name of the number that is the opposite of a positive number?
2. What is the name of the number that is the opposite of a negative number?
3. Name the number that is the opposite of zero. Does such a number exist?
4. Name the number that cannot be the module of the number.
5. Define the modulus of a number.

Homework

1. Before you are numbers that you need to arrange in descending order of modules. If you complete the task correctly, you will recognize the name of the person who first introduced the term “module” into mathematics.

2. Draw a coordinate line and find the distance from M (-5) and K (8) to the origin.

Subjects > Mathematics > Mathematics Grade 6

This lesson will introduce the concept of the modulus of a real number and introduce some of its basic definitions, followed by examples that demonstrate the application of various of these definitions.

Subject:Real numbers

Lesson:Real Number Modulus

1. Module definitions

Consider such a concept as the modulus of a real number, it has several definitions.

Definition 1. The distance from a point on a coordinate line to zero is called modulus of number, which is the coordinate of the given point (Fig. 1).

Example 1 . Note that the modules of opposite numbers are equal and non-negative, since this is a distance, and it cannot be negative, and the distance from numbers symmetrical about zero to the origin are equal.

Definition 2. .

Example 2. Consider one of the tasks posed in the previous example to demonstrate the equivalence of the introduced definitions. , as we see, with a negative number under the module sign, adding one more minus in front of it provides a non-negative result, as follows from the definition of the module.

Consequence. The distance between two points with coordinates on the coordinate line can be found as follows regardless relative position points (Fig. 2).

2. Basic properties of the module

1. The modulus of any number is non-negative

2. The module of the product is the product of the modules

3. Module private - this is private modules

3. Problem solving

Example 3. Solve the equation.

Decision. Let's use the second module definition: and write our equation as a system of equations for various options module expansion.

Example 4. Solve the equation.

Decision. Similarly to the solution of the previous example, we obtain that .

Example 5. Solve the equation.

Decision. Let's solve through the corollary from the first definition of the module: . Let's depict this on the numerical axis, taking into account the fact that the desired root will be at a distance of 2 from point 3 (Fig. 3).

Based on the figure, we obtain the roots of the equation: , because the points with these coordinates are at a distance of 2 from point 3, as required in the equation.

Answer. .

Example 6. Solve the equation.

Decision. Compared to the previous problem, there is only one complication - this is that there is no complete similarity with the formulation of the corollary about the distance between numbers on the coordinate axis, since the plus sign is under the module sign, not the minus sign. But it is not difficult to bring it to the required form, which we will do:

Let's depict this on the numerical axis similarly to the previous solution (Fig. 4).

Equation roots .

Answer. .

Example 7. Solve the equation.

Decision. This equation is a little more complicated than the previous one, because the unknown is in second place and with a minus sign, in addition, it is also with a numerical factor. To solve the first problem, we use one of the properties of the module and get:

To solve the second problem, we will perform a change of variables: , which will lead us to the simplest equation . According to the second definition of the module . We substitute these roots into the replacement equation and obtain two linear equations:

Answer. .

4. Square root and modulus

Quite often, in the course of solving problems with roots, modules arise, and one should pay attention to the situations in which they arise.

At first glance at this identity, questions may arise: “why is the module there?” and “why is the identity false?”. It turns out that one can give a simple counterexample to the second question: if then must be true, which is equivalent, and this is not an identity.

After that, the question may arise: “Does such an identity solve the problem”, but there is also a counterexample for this proposal. If then must be true, what is equivalent, and this is a wrong identity.

Accordingly, if we remember that Square root from a non-negative number is a non-negative number, and the value of the modulus is non-negative, it becomes clear why the above statement is true:

.

Example 8. Calculate the value of the expression .

Decision. In such tasks, it is important not to get rid of the root thoughtlessly immediately, but to use the above identity, since .

As a special number, it has no sign.

Examples of writing numbers: + 36 , 6 ; − 273 ; 142. (\displaystyle +36(,)6;\ (-)273;\ 142.) last number has no sign and is therefore positive.

Note that plus and minus indicate the sign for numbers, but not for literal variables or algebraic expressions. For example, in formulas −t; a + b − (a 2 + b 2) (\displaystyle -t;\ a+b;\ -(a^(2)+b^(2))) the plus and minus symbols do not specify the sign of the expression they precede, but the sign arithmetic operation, so that the sign of the result can be anything, it is determined only after the expression has been evaluated.

In addition to arithmetic, the concept of a sign is used in other branches of mathematics, including for non-numeric mathematical objects (see below). The concept of a sign is also important in those branches of physics where physical quantities are divided into two classes, conditionally called positive and negative - for example, electric charges, positive and negative feedback, various forces of attraction and repulsion.

Number sign

Positive and negative numbers

Zero is not assigned any sign, that is + 0 (\displaystyle +0) and − 0 (\displaystyle -0) is the same number in arithmetic. In mathematical analysis, the meaning of symbols + 0 (\displaystyle +0) and − 0 (\displaystyle -0) may vary, see about it Negative and positive zero ; in computer science, the computer encoding of two zeros (integer type) may differ, see direct code.

In connection with the above, a few more useful terms are introduced:

• Number non-negative if it is greater than or equal to zero.
• Number non-positive if it is less than or equal to zero.
• Positive numbers without zero and negative numbers without zero are sometimes (to emphasize that they are non-zero) called "strictly positive" and "strictly negative" respectively.

The same terminology is sometimes used for real functions. For example, the function is called positive if all its values ​​are positive, non-negative, if all its values ​​are non-negative, etc. They also say that the function is positive/negative on a given interval of its definition..

For an example of using the function, see the article Square root#Complex numbers .

Modulus (absolute value) of a number

If the number x (\displaystyle x) drop the sign, the resulting value is called module or absolute value numbers x (\displaystyle x), it is denoted | x | . (\displaystyle |x|.) Examples: | 3 | = 3; | − 3 | = 3. (\displaystyle |3|=3;\ |(-3)|=3.)

For any real numbers a , b (\displaystyle a,b) the following properties hold.

Sign of non-numeric objects

Angle sign

The value of the angle on the plane is considered positive if it is measured counterclockwise, otherwise it is negative. Two cases of rotation are similarly classified:

• rotation on a plane - for example, rotation by (–90°) is clockwise;
• a rotation in space around an oriented axis is generally considered positive if the “gimlet rule” is satisfied, otherwise it is considered negative.

direction sign

In analytic geometry and physics, advances along a given straight line or curve are often conditionally divided into positive and negative ones. Such a division may depend on the formulation of the problem or on the chosen coordinate system. For example, when calculating the length of an arc of a curve, it is often convenient to assign a minus sign to this length in one of two possible directions.

Sign in computing

most significant bit
0	1	1	1	1	1	1	1	=	127
0	1	1	1	1	1	1	0	=	126
0	0	0	0	0	0	1	0	=	2
0	0	0	0	0	0	0	1	=	1
0	0	0	0	0	0	0	0	=	0
1	1	1	1	1	1	1	1	=	−1
1	1	1	1	1	1	1	0	=	−2
1	0	0	0	0	0	0	1	=	−127
1	0	0	0	0	0	0	0	=	−128
To represent the sign of an integer, most computers use

Today, friends, there will be no snot and sentiment. Instead, I will send you into battle with one of the most formidable opponents in the 8th-9th grade algebra course without further questions.

Yes, you understood everything correctly: we are talking about inequalities with a modulus. We will look at four basic techniques with which you will learn to solve about 90% of these problems. What about the other 10%? Well, we'll talk about them in a separate lesson. :)

However, before analyzing any tricks there, I would like to recall two facts that you already need to know. Otherwise, you risk not understanding the material of today's lesson at all.

What you already need to know

Captain Evidence, as it were, hints that in order to solve inequalities with a modulus, you need to know two things:

1. How are inequalities resolved?
2. What is a module.

Let's start with the second point.

Module Definition

Everything is simple here. There are two definitions: algebraic and graphic. Let's start with the algebra:

Definition. The module of the number $x$ is either the number itself, if it is non-negative, or the number opposite to it, if the original $x$ is still negative.

It is written like this:

\[\left| x \right|=\left\( \begin(align) & x,\ x\ge 0, \\ & -x,\ x \lt 0. \\\end(align) \right.\]

talking plain language, the modulus is "a number without a minus". And it is in this duality (somewhere you don’t need to do anything with the original number, but somewhere you have to remove some minus there) and all the difficulty for novice students lies.

Is there some more geometric definition. It is also useful to know it, but we will refer 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 the point $a$ be marked on the real line. Then the module $\left| x-a \right|$ is the distance from the point $x$ to the point $a$ on this line.

If you draw a picture, you get something like this:

Graphic definition module

One way or another, its key property immediately follows from the definition of the module: the modulus of a number is always a non-negative value. This fact will be a red thread running through our entire story today.

Solution of inequalities. Spacing Method

Now let's deal with inequalities. There are a great many of them, but our task now is to be able to solve at least the simplest of them. The ones that come down to linear inequalities, as well as to the method of intervals.

I have two big tutorials on this topic (by the way, very, VERY useful - I recommend studying):

1. The interval method for inequalities(especially watch the video);
2. Fractional-rational inequalities- a very voluminous lesson, but after it you will not have any questions at all.

If you know all this, if the phrase "let's move from inequality to equation" does not make you vaguely want to kill yourself against the wall, then you are ready: welcome to hell to the main topic of the lesson. :)

1. Inequalities of the form "Module less than function"

This is one of the most frequently encountered tasks with modules. It is required to solve an inequality of the form:

\[\left| f\right| \ltg\]

Anything can act as functions $f$ and $g$, but usually they are polynomials. Examples of such inequalities:

\[\begin(align) & \left| 2x+3\right| \ltx+7; \\ & \left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0; \\ & \left| ((x)^(2))-2\left| x \right|-3 \right| \lt 2. \\\end(align)\]

All of them are solved literally in one line according to the 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 instead 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 module 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: is it not easier? Unfortunately, you can't. This is the whole point of the module.

But enough of the philosophizing. Let's solve a couple of problems:

Task. Solve the inequality:

\[\left| 2x+3\right| \ltx+7\]

Decision. So, we have a classical inequality of the form “the module is less than” - there is 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 brackets that are preceded by a “minus”: it is quite possible that because of the 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 has been reduced to two elementary inequalities. We note their solutions on parallel real lines:

Intersection of many

The intersection of these sets will be the answer.

Answer: $x\in \left(-\frac(10)(3);4 \right)$

Task. Solve the inequality:

\[\left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0\]

Decision. This task is a little more difficult. To begin with, we 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 less”, so we get rid of the module according to 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 brackets. But once again I remind you that our key goal is correctly solve the inequality and get the answer. Later, when you have perfectly mastered everything that is described in this lesson, you can pervert yourself as you like: open brackets, add minuses, etc.

And for starters, we just 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 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 square and are solved by the interval method (that's why I say: if you don't know what it is, it's better not to take on modules yet). We pass 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 turned out to be incomplete. quadratic equation, which is solved elementarily. Now let's deal with the second inequality of the system. There you 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 obtained numbers on two parallel lines (separate for the first inequality and separate for the second):

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.

Answer: $x\in \left(-5;-2 \right)$

I think after these examples the solution scheme is very clear:

1. Isolate the module by moving all other terms to the opposite side of the inequality. Thus we get an inequality of the form $\left| f\right| \ltg$.
2. Solve this inequality by getting rid of the module as described above. At some point, it will be necessary to move from a double inequality to a system of two independent expressions, each of which can already be solved separately.
3. Finally, it remains only to cross the solutions of these two independent expressions - and that's it, we will get the final answer.

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 will talk about these “buts” now.

2. Inequalities of the form "Module is greater than function"

They look like this:

\[\left| f\right| \gt g\]

Similar to the previous one? Seem to be. Nevertheless, such tasks 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:

1. First, we simply ignore the module - we solve the usual inequality;
2. Then, in fact, we open the module with the minus sign, and then we multiply both parts of the inequality by −1, with a sign.

In this case, the options are combined with a square bracket, i.e. We have a combination of two requirements.

Pay attention again: before us is not a system, but an aggregate, therefore in the answer, the sets are combined, not intersected. This is a fundamental difference from the previous paragraph!

In general, many students have a lot of confusion with unions and intersections, so let's look into this issue once and for all:

• "∪" is a concatenation sign. In fact, this is a stylized letter "U", which came to us from in English and is an abbreviation for "Union", i.e. "Associations".
• "∩" is the intersection sign. This crap didn't come from anywhere, but just appeared as an opposition to "∪".

To make it even easier to remember, just add legs to these signs to make glasses (just don’t accuse me of promoting drug addiction and alcoholism now: if you are seriously studying this lesson, then you are already a drug addict):

Difference between intersection and union of sets

Translated into Russian, this means the following: the union (collection) includes elements from both sets, therefore, no less than each of them; but the intersection (system) includes only those elements that are both in the first set and in the second. Therefore, the intersection of sets is never greater than the source sets.

So it became clearer? That is great. Let's move on to practice.

Task. Solve the inequality:

\[\left| 3x+1 \right| \gt 5-4x\]

Decision. We act 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 population inequality:

\[\left[ \begin(align) & 3x+4x \gt 5-1 \\ & 3x-4x \lt -5-1 \\ \end(align) \right.\]

\[\left[ \begin(align) & 7x \gt 4 \\ & -x \lt -6 \\ \end(align) \right.\]

\[\left[ \begin(align) & x \gt 4/7\ \\ & x \gt 6 \\ \end(align) \right.\]

We mark each resulting set on the number line, and then combine them:

Union of sets

Obviously the answer is $x\in \left(\frac(4)(7);+\infty \right)$

Answer: $x\in \left(\frac(4)(7);+\infty \right)$

Task. Solve the inequality:

\[\left| ((x)^(2))+2x-3 \right| \gtx\]

Decision. Well? No, it's all the same. We pass 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 each inequality. Unfortunately, the roots will not be very good there:

\[\begin(align) & ((x)^(2))+2x-3 \gt x; \\ & ((x)^(2))+x-3 \gt 0; \\ &D=1+12=13; \\ & x=\frac(-1\pm \sqrt(13))(2). \\\end(align)\]

In the second inequality, there is also a bit of game:

\[\begin(align) & ((x)^(2))+2x-3 \lt -x; \\ & ((x)^(2))+3x-3 \lt 0; \\ &D=9+12=21; \\ & x=\frac(-3\pm \sqrt(21))(2). \\\end(align)\]

Now we need to mark these numbers on two axes - one axis for each inequality. However, points must be marked in right order: how more number, the further we shift the point to the right.

And here we are waiting for a setup. 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 smaller), with the numbers $\frac(-3-\sqrt(13))(2) \lt \frac(-1+\sqrt(21))(2)$ there will also be no difficulty (a positive number obviously more negative), but with the last couple, everything is not so simple. Which is larger: $\frac(-3+\sqrt(21))(2)$ or $\frac(-1+\sqrt(13))(2)$? The arrangement 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 $\frac(-1+\sqrt(13))(2) \gt \frac(-3+\sqrt(21)) (2)$, finally the points on the axes will be arranged like this:

Case of ugly roots

Let me remind you that we are solving a collection, so the answer will be the union, and not the intersection of the 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 tasks, and for very rigid ones. The only “weak spot” 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 questions of comparison. And we move on.

3. Inequalities with non-negative "tails"

So we got to the most interesting. These are inequalities of the form:

\[\left| f\right| \gt\left| g\right|\]

Generally speaking, the algorithm that we are going to talk about now is true 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 degree. 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 do not confuse this with taking the root of the square:

\[\sqrt(((f)^(2)))=\left| f \right|\ne f\]

Countless mistakes were made when a student forgot to install a module! But that's a completely different story (it's like irrational equations), so we won't go into that now. Let's better solve a couple of problems:

Task. Solve the inequality:

\[\left| x+2 \right|\ge \left| 1-2x \right|\]

Decision. We immediately notice two things:

1. This is a non-strict inequality. Points on the number line will be punched out.
2. Both sides of the inequality are obviously non-negative (this is a property of the module: $\left| f\left(x \right) \right|\ge 0$).

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, using the parity of the modulus (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 by 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!

Getting rid of the module sign

Let me remind you for the 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, this is $\left(x-3 \right)\left(3x+1 \right)\le 0$.

That's it. Problem solved.

Answer: $x\in \left[ -\frac(1)(3);3 \right]$.

Task. Solve the inequality:

\[\left| ((x)^(2))+x+1 \right|\le \left| ((x)^(2))+3x+4 \right|\]

Decision. We do everything the same. I will not comment - just look at the sequence of actions.

Let's square it:

\[\begin(align) & ((\left(\left| ((x)^(2))+x+1 \right| \right))^(2))\le ((\left(\left | ((x)^(2))+3x+4 \right| \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))\le ((\left(((x)^(2))+3x+4 \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))-((\left(((x)^(2))+3x+4 \ right))^(2))\le 0; \\ & \left(((x)^(2))+x+1-((x)^(2))-3x-4 \right)\times \\ & \times \left(((x) ^(2))+x+1+((x)^(2))+3x+4 \right)\le 0; \\ & \left(-2x-3 \right)\left(2((x)^(2))+4x+5 \right)\le 0. \\\end(align)\]

Spacing 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:

The answer is a whole range

Answer: $x\in \left[ -1.5;+\infty \right)$.

A small note about the last task. As one of my students accurately noted, both submodule expressions in this inequality are obviously positive, so the modulus sign can be omitted without harm to health.

But this is already a completely different level of thinking and a different approach - it can be conditionally called the method of consequences. About him - in a separate lesson. And now let's move on to the final part of today's lesson and consider universal algorithm which always works. Even when all previous approaches were powerless. :)

4. Method of enumeration of options

What if all these tricks don't work? If inequality cannot be reduced to non-negative tails, if it is impossible to isolate the module, if at all pain-sadness-longing?

Then the “heavy artillery” of all mathematics enters the scene - the enumeration method. With regard to inequalities with the modulus, it looks like this:

1. Write out all submodule expressions and equate them to zero;
2. Solve the resulting equations and mark the found roots on one number line;
3. The straight line will be divided into several sections, within which each module has a fixed sign and therefore unambiguously expands;
4. Solve the inequality on each such section (you can separately consider the boundary roots obtained in paragraph 2 - for reliability). Combine the results - this will be the answer. :)

Well, how? Weak? Easily! Only for a long time. Let's see in practice:

Task. Solve the inequality:

\[\left| x+2 \right| \lt\left| x-1 \right|+x-\frac(3)(2)\]

Decision. 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 let's go ahead.

We write out submodule 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, inside which each module is revealed uniquely:

Splitting the number line by zeros of submodular functions

Let's consider each section separately.

1. Let $x \lt -2$. Then both submodule expressions are negative, and the original inequality is 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 constraint. Let's intersect it with the original 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 but greater than 1.5. There are no solutions in this area.

1.1. Let's separately consider the boundary case: $x=-2$. Let's just substitute this number into the original inequality and check: does it hold?

\[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1,5 \right|)_(x=-2) ) \\ & 0 \lt \left| -3 \right|-2-1.5; \\ & 0 \lt 3-3.5; \\ & 0 \lt -0,5\Rightarrow \varnothing . \\\end(align)\]

Obviously, the chain of calculations has led us to the wrong inequality. Therefore, the original inequality is also false, and $x=-2$ is not included in the answer.

2. Now let $-2 \lt x \lt 1$. The left module will already open with a "plus", but the right one is still 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 empty set of solutions, since there are no numbers that are both less than −2.5 and greater than −2.

2.1. And again special case: $x=1$. We substitute into the original inequality:

\[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1,5 \right|)_(x=1)) \\ & \left| 3\right| \lt\left| 0 \right|+1-1.5; \\ & 3 \lt -0.5; \\ & 3 \lt -0,5\Rightarrow \varnothing . \\\end(align)\]

Similarly 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 expanded 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 the interval, which will be the answer.

Answer: $x\in \left(4,5;+\infty \right)$

Finally, one note that may save you from stupid mistakes when solving real problems:

Solutions of inequalities with modules are usually continuous sets on the number line - intervals and segments. Isolated points are much rarer. And even less often, it happens that the boundaries of the solution (the end of the segment) coincide with the boundary of the range under consideration.

Consequently, if the boundaries (those same “special cases”) are not included in the answer, then the areas to the left-right of these boundaries will almost certainly not be included in the answer either. And vice versa: the border entered in response, which means that some areas around it will also be responses.

Keep this in mind when you check your solutions.