X-number module. What is the modulus of a number in mathematics

Solving equations and inequalities with modulus is often difficult. However, if you understand well what is the absolute value of a numberand how to correctly expand expressions containing a modulus sign, then the presence in the equation the expression under the module sign, ceases to be an obstacle to its solution.

A bit of theory. Each number has two characteristics: the absolute value of the number, and its sign.

For example, the number +5, or just 5 has a "+" sign and an absolute value of 5.

The number -5 has a "-" sign and an absolute value of 5.

The absolute values \u200b\u200bof 5 and -5 are 5.

The absolute value of the number x is called the modulus of the number and is denoted by | x |.

As we can see, the absolute value of a number is equal to the number itself, if this number is greater than or equal to zero, and to this number with the opposite sign, if this number is negative.

The same applies to any expressions that are under the module sign.

The module expansion rule looks like this:

| f (x) | \u003d f (x) if f (x) ≥ 0, and

| f (x) | \u003d - f (x) if f (x)< 0

For example | x-3 | \u003d x-3 if x-3≥0 and | x-3 | \u003d - (x-3) \u003d 3-x if x-3<0.

To solve an equation containing an expression under the modulus sign, you must first expand the module according to the module expansion rule.

Then our equation or inequality transforms into two different equations existing at two different numerical ranges.

One equation exists on a numerical interval where the expression under the modulus sign is non-negative.

And the second equation exists on the interval where the expression under the modulus sign is negative.

Let's look at a simple example.

Let's solve the equation:

| x-3 | \u003d -x 2 + 4x-3

1. Let's expand the module.

| x-3 | \u003d x-3 if x-3≥0, i.e. if x≥3

| x-3 | \u003d - (x-3) \u003d 3-x if x-3<0, т.е. если х<3

2. We got two numerical ranges: x≥3 and x<3.

Consider the equations into which the original equation is transformed at each interval:

A) For x≥3 | x-3 | \u003d x-3, and our equation has the form:

Attention! This equation exists only on the interval x≥3!

Let's expand the brackets, we will give similar terms:

and solve this equation.

This equation has roots:

x 1 \u003d 0, x 2 \u003d 3

Attention! since the equation x-3 \u003d -x 2 + 4x-3 exists only on the interval x≥3, we are only interested in the roots that belong to this interval. Only x 2 \u003d 3 satisfies this condition.

B) For x<0 |x-3|=-(x-3) = 3-x, и наше уравнение приобретает вид:

Attention! This equation exists only on the interval x<3!

Let's expand the parentheses, we will give similar terms. We get the equation:

x 1 \u003d 2, x 2 \u003d 3

Attention! since the equation 3-x \u003d -x 2 + 4x-3 exists only on the interval x<3, нас интересуют только те корни, которые принадлежат этому промежутку. Этому условию удовлетворяет только х 1 =2.

So: from the first interval we take only the root x \u003d 3, from the second - the root x \u003d 2.

Modulus is the absolute value of an expression. In order to somehow denote a module, it is customary to use straight brackets. The value that is enclosed in straight brackets is the value that is taken modulo. The process of solving any module consists in expanding the very right brackets, which are called modular brackets in mathematical language. Their disclosure takes place according to a certain number of rules. Also, in the order of solving the modules, there are also the sets of values \u200b\u200bof those expressions that were in the module brackets. In most of all cases, a module is expanded in such a way that an expression that was submodular gets both positive and negative values, including the value zero. If we start from the established properties of the module, then in the process various equations or inequalities from the original expression are drawn up, which then need to be solved. Let's figure out how to solve modules.

Solution process

Solving a module begins by writing the original equation with a module. To answer the question of how to solve equations with a module, you need to expand it completely. To solve such an equation, the module is expanded. All modular expressions must be considered. It is necessary to determine at what values \u200b\u200bof the unknown quantities included in its composition, the modular expression in parentheses turns to zero. To do this, it is enough to equate the expression in modular brackets to zero, and then calculate the solution of the resulting equation. The found values \u200b\u200bmust be recorded. In the same way, it is also necessary to determine the value of all unknown variables for all modules in this equation. Next, you need to deal with the definition and consideration of all cases of existence of variables in expressions when they are different from the value zero. To do this, you need to write down some system of inequalities according to all modules in the original inequality. Inequalities should be designed so that they cover all available and possible values \u200b\u200bfor a variable that are found on the number line. Then you need to draw this very numerical line for visualization, on which in the future you will postpone all the obtained values.

Almost everything can now be done on the Internet. The module is no exception to the rule. You can solve it online on one of the many modern resources. All those values \u200b\u200bof the variable that are in the zero module will be a special constraint that will be used in the process of solving the modular equation. In the original equation, it is required to expand all available modular brackets, while changing the sign of the expression so that the values \u200b\u200bof the desired variable coincide with those values \u200b\u200bthat can be seen on the number line. The resulting equation must be solved. The value of the variable that will be obtained during the solution of the equation must be checked against the constraint set by the module itself. If the value of the variable fully satisfies the condition, then it is correct. All roots that will be obtained during the solution of the equation, but will not fit the constraints, must be discarded.

We don't choose mathher profession, and she chooses us.

Russian mathematician Yu.I. Manin

Equations with modulus

The most difficult to solve problems of school mathematics are equations containing variables under the modulus sign. To successfully solve such equations, you need to know the definition and basic properties of the module. Naturally, students should have the skills to solve equations of this type.

Basic concepts and properties

Modulus (absolute value) of a real number denoted and is defined as follows:

The simple properties of a module include the following relationships:

Note, that the last two properties are valid for any even degree.

In addition, if, where, then

More complex module properties, which can be effectively used to solve equations with modules, are formulated by means of the following theorems:

Theorem 1. For any analytic functions and the inequality holds

Theorem 2. Equality is equivalent to inequality.

Theorem 3. Equality tantamount to inequality.

Let's consider typical examples of solving problems on the topic "Equations, containing variables under the module sign ".

Solving equations with modulus

The most common method in school mathematics for solving equations with a module is the method, based on the expansion of modules. This method is versatile, however, in general, its application can lead to very cumbersome calculations. In this regard, students should know other, more effective methods and techniques for solving such equations. In particular, you need to have skills in applying theorems, given in this article.

Example 1.Solve the equation. (1)

Decision. Equation (1) will be solved by the "classical" method - the method of expanding the modules. To do this, we split the number axis points and into intervals and consider three cases.

1. If, then,,, and equation (1) takes the form. Hence it follows. However, here, therefore, the found value is not the root of equation (1).

2. If, then from equation (1) we obtain or .

Since then root of equation (1).

3. If, then equation (1) takes the form or . Note that.

Answer:,.

When solving subsequent equations with a module, we will actively use the properties of modules in order to increase the efficiency of solving such equations.

Example 2. Solve the equation.

Decision. Since and, then the equation implies... In this regard,,, and the equation takes the form... Hence we get... But , therefore, the original equation has no roots.

Answer: there are no roots.

Example 3. Solve the equation.

Decision. Since, then. If, then, and the equation takes the form.

From here we get.

Example 4. Solve the equation.

Decision.We rewrite the equation in an equivalent form. (2)

The resulting equation belongs to equations of the type.

Taking into account Theorem 2, it can be argued that equation (2) is equivalent to an inequality. From here we get.

Answer:.

Example 5. Solve the equation.

Decision. This equation has the form... Therefore , according to Theorem 3, here we have the inequality or .

Example 6. Solve the equation.

Decision. Let us assume that. Because , then the given equation takes the form of a quadratic equation, (3)

where ... Since equation (3) has a single positive root and then ... Hence, we obtain two roots of the original equation: and.

Example 7. Solve the equation. (4)

Decision. Since the equation is equivalent to a combination of two equations: and, then, when solving equation (4), it is necessary to consider two cases.

1. If, then or.

From here we get, and.

2. If, then or.

Since, then.

Answer:,,,.

Example 8. Solve the equation . (5)

Decision. Since and, then. From this and from Eq. (5) it follows that and, i.e. here we have the system of equations

However, this system of equations is inconsistent.

Answer: there are no roots.

Example 9. Solve the equation. (6)

Decision.If we denote, then and from equation (6) we obtain

Or . (7)

Since equation (7) has the form, this equation is equivalent to an inequality. From here we get. Since, then or.

Answer:.

Example 10. Solve the equation. (8)

Decision. According to Theorem 1, we can write

(9)

Taking into account equation (8), we conclude that both inequalities (9) turn into equalities, i.e. the system of equations holds

However, by Theorem 3, the above system of equations is equivalent to the system of inequalities

(10)

Solving the system of inequalities (10), we obtain. Since the system of inequalities (10) is equivalent to equation (8), the original equation has a single root.

Answer:.

Example 11. Solve the equation. (11)

Decision. Let and, then equality follows from equation (11).

Hence it follows that and. Thus, here we have a system of inequalities

The solution to this system of inequalities is and.

Answer:,.

Example 12. Solve the equation. (12)

Decision. Equation (12) will be solved by the method of sequential expansion of modules. To do this, consider several cases.

1. If, then.

1.1. If, then and,.

1.2. If, then. But , therefore, in this case, equation (12) has no roots.

2. If, then.

2.1. If, then and,.

2.2. If, then and.

Answer:,,,,.

Example 13. Solve the equation. (13)

Decision. Since the left-hand side of equation (13) is non-negative, then and. In this regard, and equation (13)

takes the form or.

It is known that the equation is equivalent to the combination of two equations and, solving which we get,. Because , then equation (13) has one root.

Answer:.

Example 14. Solve system of equations (14)

Decision. Since and, then and. Therefore, from the system of equations (14) we obtain four systems of equations:

The roots of the above systems of equations are the roots of the system of equations (14).

Answer: ,,,,,,,.

Example 15. Solve system of equations (15)

Decision. Since, then. In this regard, from the system of equations (15) we obtain two systems of equations

The roots of the first system of equations are and, and from the second system of equations we obtain and.

Answer:,,,.

Example 16. Solve system of equations (16)

Decision. From the first equation of system (16) it follows that.

Since then ... Consider the second equation of the system. Because thethen, and the equation takes the form, , or .

If you substitute the value into the first equation of system (16), then, or.

Answer:,.

For a deeper study of problem solving methods, related to solving equations, containing variables under the module sign, you can recommend tutorials from the list of recommended reading.

1. Collection of problems in mathematics for applicants to technical colleges / Ed. M.I. Skanavi. - M .: Peace and Education, 2013 .-- 608 p.

2. Suprun V.P. Mathematics for high school students: problems of increased complexity. - M .: CD "Librokom" / URSS, 2017 .-- 200 p.

3. Suprun V.P. Mathematics for high school students: non-standard problem solving methods. - M .: CD "Librokom" / URSS, 2017 .-- 296 p.

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And it is calculated in accordance with the following rules:

For brevity, use | a |... So, | 10 | \u003d 10; - 1/3 \u003d | 1/3 |; | -100 | \u003d 100, etc.

Any size x corresponds to a fairly accurate value | x|. And that means identity at= |x| sets at as some argument function x.

Schedulethis function presented below.

For x > 0 |x| = x, and for x< 0 |x|= -x; in this regard, the line y \u003d | x| at x\u003e 0 combined with a straight line y \u003d x(bisector of the first coordinate angle), and for x< 0 - с прямой y \u003d -x(bisector of the second coordinate angle).

Selected equations include unknowns under the sign module.

Arbitrary examples of such equations - | x— 1| = 2, |6 — 2x| =3x+ 1, etc.

Solving Equationscontaining the unknown under the modulus sign is based on the fact that if the absolute value of the unknown number x is equal to a positive number a, then this number x itself is equal to either a or -a.

for instance: if | x| \u003d 10, then or x\u003d 10, or x = -10.

Consider solving individual equations.

Let's analyze the solution of the equation | x- 1| = 2.

Let's expand the module then the difference x- 1 can be equal to either + 2 or - 2. If x - 1 \u003d 2, then x \u003d 3; if x - 1 \u003d - 2, then x \u003d - 1. We make a substitution and get that both of these values \u200b\u200bsatisfy the equation.

Answer.This equation has two roots: x 1 = 3, x 2 = - 1.

Let's analyze equation solution | 6 — 2x| = 3x+ 1.

After module expansionwe get: or 6 - 2 x= 3x+ 1, or 6 - 2 x= - (3x+ 1).

In the first case x \u003d 1, and in the second x= - 7.

Checking. When x= 1 |6 — 2x| = |4| = 4, 3x + 1 \u003d 4; it follows from the court, x = 1 - rootgiven equations.

When x = - 7 |6 — 2x| = |20| = 20, 3x+ 1 \u003d - 20; since 20 ≠ -20, then x \u003d - 7 is not a root of this equation.

Answer. Haveequations single root: x = 1.

Equations of this type can be solve and graphically.

So let's decide eg, graphically equation | x- 1| = 2.

Initially, we perform the construction function graphics at = |x- 1 |. Let's first draw a graph of the function at=x- 1:

That part of it graphicswhich is located above the axis x we will not change. For her x - 1\u003e 0 and therefore | x-1|=x-1.

The part of the graph that is located under the axis x, depict symmetrically about this axis. Since for this part x - 1 < 0 и соответственно |x - 1|= - (x - 1). The resulting line (solid line) and will function graph y \u003d | x—1|.

This line will cross with straight at \u003d 2 at two points: M 1 with abscissa -1 and M 2 with abscissa 3. And, accordingly, the equation | x- 1 | \u003d 2 there will be two roots: x 1 = - 1, x 2 = 3.

The module of numbers introduces a new concept in mathematics. Let's take a closer look at what a number module is and how to work with it?

Let's consider an example:

We left the house for the store. We walked 300 m, mathematically this expression can be written as +300, the meaning of the number 300 from the “+” sign will not change. Distance or modulus of a number in mathematics is one and the same can be written like this: | 300 | \u003d 300. The modulus of a number is indicated by two vertical lines.

And then we walked 200m in the opposite direction. Mathematically, we can write the return path as -200. But we do not say that “we have passed minus two hundred meters,” although we have returned, because the distance as a value remains positive. For this, the concept of a module was introduced in mathematics. You can write the distance or the modulus of the number -200 like this: | -200 | \u003d 200.

Module properties.

Definition:
Modulus of a number or absolute value of a number Is the distance from the starting point to the destination point.

The modulus of a non-zero integer, always a positive number.

The module is written like this:

1. The absolute value of a positive number is equal to the number itself.
| a | \u003da

2. The absolute value of a negative number is equal to the opposite number.
|- a | \u003da

3. The modulus of zero is zero.
|0|=0

4. Modules of opposite numbers are equal.
| a | \u003d | -a | \u003da

Questions on the topic:
What is the modulus of a number?
Answer: The module is the distance from the starting point to the destination point.

If you put a “+” in front of an integer, what happens?
Answer: the number will not change its meaning, for example, 4 \u003d + 4.

If you put a “-” in front of an integer, what happens?
Answer: The number will change to, for example, 4 and -4.

Which numbers have the same modulus?
Answer: positive numbers and zero will have the same modulus. For example, 15 \u003d | 15 |.

Which numbers have the opposite modulus?
Answer: for negative numbers, the modulus will be equal to the opposite number. For example, | -6 | \u003d 6.

Example # 1:
Find the modulus of numbers: a) 0 b) 5 c) -7?

Decision:
a) | 0 | \u003d 0
b) | 5 | \u003d 5
c) | -7 | \u003d 7

Example # 2:
Are there two different numbers whose absolute values \u200b\u200bare equal?

Decision:
|10|=10
|-10|=10

Modules of opposite numbers are equal.

Example # 3:
What are the two opposite numbers that have a module of 9?

Decision:
|9|=9
|-9|=9

Answer: 9 and -9.

Example # 4:
Follow the steps: a) | +5 | + | -3 | b) | -3 | + | -8 | c) | +4 | - | +1 |

Decision:
a) | +5 | + | -3 | \u003d 5 + 3 \u003d 8
b) | -3 | + | -8 | \u003d 3 + 8 \u003d 11
c) | +4 | - | +1 | \u003d 4 - 1 \u003d 3

Example # 5:
Find: a) modulus of number 2 b) modulus of number 6 c) modulus of number 8 d) modulus of number 1 e) modulus of number 0.
Decision:

a) the modulus of the number 2 is denoted as | 2 | or | +2 | It is the same.
|2|=2

b) the modulus of the number 6 is denoted as | 6 | or | +6 | It is the same.
|6|=6

c) the modulus of the number 8 is denoted as | 8 | or | +8 | It is the same.
|8|=8

d) the modulus of the number 1 is denoted as | 1 | or | +1 | It is the same.
|1|=1

e) the modulus of the number 0 is denoted as | 0 |, | +0 | or | -0 | It is the same.
|0|=0