Examples of solving equations with a module. Number module (absolute value of a number), definitions, examples, properties. Number module - definition, notation and examples
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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 be aware of other, more effective methods and techniques for solving such equations. In particular, you must 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... However , 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. As , 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. However , 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 Eq. (13) is non-negative, 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,. As , 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. Insofar asthen, 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, can I advise tutorials from the list of recommended literature.
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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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. Based on the established properties of the module, then in the process various equations or inequalities from the original expression are compiled, which then need to be solved. Let's figure out how to solve modules.
Solution process
The solution to the module begins by writing the original equation with the 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 brackets 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.
One of the most challenging topics for students is solving equations that contain a variable under the modulus sign. Let's figure it out for a start, what is this connected with? Why, for example, do quadratic equations do most of the children clicks like nuts, but with such a far from complicated concept as a module, it has so many problems?
In my opinion, all these difficulties are associated with the lack of clearly formulated rules for solving equations with a modulus. So, deciding quadratic equation, the student knows for sure that he needs to first apply the discriminant formula, and then the formula for the roots of the quadratic equation. But what if there is a module in the equation? We will try to clearly describe the necessary action plan for the case when the equation contains an unknown under the modulus sign. Here are some examples for each case.
But first, let's remember module definition... So, the modulus of the number a this number itself is called if a non-negative and -aif the number a less than zero. You can write it like this:
| a | \u003d a if a ≥ 0 and | a | \u003d -a if a< 0
Speaking about the geometric sense of the module, it should be remembered that each real number corresponds to a certain point on the number axis - its k 
coordinate. So, the modulus or absolute value of a number is the distance from this point to the origin of the numerical axis. Distance is always specified as a positive number. Thus, the absolute value of any negative number is a positive number. By the way, even at this stage, many students begin to get confused. Any number can be in the module, but the result of applying the module is always a positive number.
Now let's go directly to solving the equations.
1. Consider an equation of the form | x | \u003d c, where c is a real number. This equation can be solved using the modulus definition.
We divide all real numbers into three groups: those that are greater than zero, those that are less than zero, and the third group is the number 0. Let's write the solution in the form of a diagram:
(± c if c\u003e 0
If | x | \u003d c, then x \u003d (0, if c \u003d 0
(no roots if with< 0
1) | x | \u003d 5, because 5\u003e 0, then x \u003d ± 5;
2) | x | \u003d -5, because -five< 0, то уравнение не имеет корней;
3) | x | \u003d 0, then x \u003d 0.
2. An equation of the form | f (x) | \u003d b, where b\u003e 0. To solve this equation, it is necessary to get rid of the modulus. We do it like this: f (x) \u003d b or f (x) \u003d -b. Now it is necessary to solve each of the obtained equations separately. If in the original equation b< 0, решений не будет.
1) | x + 2 | \u003d 4, because 4\u003e 0, then
x + 2 \u003d 4 or x + 2 \u003d -4
2) | x 2 - 5 | \u003d 11, because 11\u003e 0, then
x 2 - 5 \u003d 11 or x 2 - 5 \u003d -11
x 2 \u003d 16 x 2 \u003d -6
x \u003d ± 4 no roots
3) | x 2 - 5x | \u003d -8, because -8< 0, то уравнение не имеет корней.
3. An equation of the form | f (x) | \u003d g (x). Within the meaning of the module, such an equation will have solutions if its right-hand side is greater than or equal to zero, i.e. g (x) ≥ 0. Then we will have:
f (x) \u003d g (x)or f (x) \u003d -g (x).
1) | 2x - 1 | \u003d 5x - 10. This equation will have roots if 5x - 10 ≥ 0. This is where the solution of such equations begins.
1.O.D.Z. 5x - 10 ≥ 0
2. Solution:
2x - 1 \u003d 5x - 10 or 2x - 1 \u003d - (5x - 10)
3. We unite ODZ. and the solution is:
The root x \u003d 11/7 does not fit according to the O.D.Z., it is less than 2, and x \u003d 3 satisfies this condition.
Answer: x \u003d 3
2) | x - 1 | \u003d 1 - x 2.
1.O.D.Z. 1 - x 2 ≥ 0. Let us solve this inequality by the method of intervals:
(1 - x) (1 + x) ≥ 0
2. Solution:
x - 1 \u003d 1 - x 2 or x - 1 \u003d - (1 - x 2)
x 2 + x - 2 \u003d 0 x 2 - x \u003d 0
x \u003d -2 or x \u003d 1 x \u003d 0 or x \u003d 1
3. We combine the solution and ODZ:
Only the roots x \u003d 1 and x \u003d 0 are suitable.
Answer: x \u003d 0, x \u003d 1.
4. An equation of the form | f (x) | \u003d | g (x) |. Such an equation is equivalent to the following two equations f (x) \u003d g (x) or f (x) \u003d -g (x).
1) | x 2 - 5x + 7 | \u003d | 2x - 5 |. This equation is equivalent to the following two:
x 2 - 5x + 7 \u003d 2x - 5 or x 2 - 5x +7 \u003d -2x + 5
x 2 - 7x + 12 \u003d 0 x 2 - 3x + 2 \u003d 0
x \u003d 3 or x \u003d 4 x \u003d 2 or x \u003d 1
Answer: x \u003d 1, x \u003d 2, x \u003d 3, x \u003d 4.
5. Equations solved by the substitution method (variable substitution). This solution method is easiest to explain with a specific example. So, let a quadratic equation with a modulus be given:
x 2 - 6 | x | + 5 \u003d 0. By the property of the module x 2 \u003d | x | 2, so the equation can be rewritten as follows:
| x | 2 - 6 | x | + 5 \u003d 0. Let us replace | x | \u003d t ≥ 0, then we will have:
t 2 - 6t + 5 \u003d 0. Solving this equation, we get that t \u003d 1 or t \u003d 5. Let's return to the replacement:
| x | \u003d 1 or | x | \u003d 5
x \u003d ± 1 x \u003d ± 5
Answer: x \u003d -5, x \u003d -1, x \u003d 1, x \u003d 5.
Let's look at another example:
x 2 + | x | - 2 \u003d 0. By the property of the module x 2 \u003d | x | 2, therefore
| x | 2 + | x | - 2 \u003d 0. Let's make the replacement | x | \u003d t ≥ 0, then:
t 2 + t - 2 \u003d 0. Solving this equation, we obtain t \u003d -2 or t \u003d 1. Let's return to the replacement:
| x | \u003d -2 or | x | \u003d 1
No roots x \u003d ± 1
Answer: x \u003d -1, x \u003d 1.
6. Another type of equations - equations with a "complex" module. These equations include equations that have “modules in a module”. Equations of this kind can be solved using the properties of the module.
1) | 3 - | x || \u003d 4. We will proceed in the same way as in equations of the second type. Because 4\u003e 0, then we get two equations:
3 - | x | \u003d 4 or 3 - | x | \u003d -4.
Now we express the modulus x in each equation, then | x | \u003d -1 or | x | \u003d 7.
We solve each of the obtained equations. There are no roots in the first equation, because -1< 0, а во втором x = ±7.
The answer is x \u003d -7, x \u003d 7.
2) | 3 + | x + 1 || \u003d 5. We solve this equation in the same way:
3 + | x + 1 | \u003d 5 or 3 + | x + 1 | \u003d -5
| x + 1 | \u003d 2 | x + 1 | \u003d -8
x + 1 \u003d 2 or x + 1 \u003d -2. No roots.
Answer: x \u003d -3, x \u003d 1.
There is also a universal method for solving equations with a module. This is the spacing method. But we will consider it later.
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The term (module) literally translated from Latin means "measure". This concept was introduced into mathematics by the English scientist R. Cotes. And the German mathematician K. Weierstrass introduced the modulus sign - the symbol that denotes this concept when writing.
In contact with
For the first time, this concept is studied in mathematics in the 6th grade secondary school curriculum. According to one definition, modulus is the absolute value of a real number. In other words, to find out the absolute value of a real number, you must discard its sign.
Graphically absolute value and denoted as | a |.
The main distinguishing feature of this concept is that it is always a non-negative quantity.
Numbers that differ from each other only in sign are called opposite. If the value is positive, then its opposite will be negative, and zero is the opposite of itself.
Geometric meaning
If we consider the concept of a module from the point of view of geometry, then it will denote the distance, which is measured in unit segments from the origin to set point... This definition fully reveals the geometric meaning of the term under study.
This can be graphically expressed as follows: | a | \u003d OA.
Absolute magnitude properties
Below we will consider all the mathematical properties of this concept and methods of writing in the form of literal expressions:
Features of solving equations with a module
If we talk about solving mathematical equations and inequalities that contain module, then you need to remember that to solve them you need to open this sign.
For example, if the sign of an absolute value contains some mathematical expression, then before opening the module, it is necessary to take into account the current mathematical definitions.
| A + 5 | \u003d A + 5if, A is greater than or equal to zero.
5-Aif, and the value is less than zero.
In some cases, the sign can be expanded unambiguously for any values \u200b\u200bof the variable.
Let's take another example. Let's build a coordinate line, on which we mark all numerical values, the absolute value of which will be 5.
First, you need to draw a coordinate line, mark the origin of coordinates on it and set the size of a unit segment. In addition, the line must have a direction. Now on this straight line it is necessary to apply markings, which will be equal to the value of the unit segment.
Thus, we can see that on this coordinate line there will be two points of interest to us with values \u200b\u200bof 5 and -5.
The unit of a number is easy to find, and the theory behind it is important when solving problems.
The properties and rules of disclosure used in solving exercises and in exams will be useful to schoolchildren and students. Make money with your knowledge at https://teachs.ru!
What is a module in mathematics
The modulus of a number describes the distance on a number line from zero to a point, regardless of the direction in which the point lies from zero. Mathematical notation : | x |.
In other words, it is the absolute value of the number. The definition proves that the value is never negative.
Module properties
It is important to remember the following properties:
Complex number module
The absolute value of a complex number is the length of a directed segment drawn from the beginning of the complex plane to the point (a, b).
This directional line is also a vector representing a complex number a + bi, so the absolute value of a complex number is the same as the magnitude (or length) of the vector representing a + bi.
How to solve equations with a module
An equation with modulus is an equality that contains an absolute value expression. If, for a real number, it represents its distance from the origin on the number line, then modulus inequalities are a type of inequality that consists of absolute values.
Equations like | x | \u003d a
The equation | x | \u003d a has two answers x \u003d a and x \u003d –abecause both options are located on the coordinate line at a distance a from 0.
Equality with absolute value has no solution if the value is negative.
If | x |< a представляет собой расстояние чисел от начала координат, это значит, что нужно искать все числа, чье расстояние от начала координат меньше a.
Equations like | x | \u003d | y |
When there are absolute values \u200b\u200bon either side of the equations, you need to consider both possibilities for acceptable definitions - positive and negative expressions.
For example, for the equality | x - a | \u003d | x + b | there are two options: (x - a) \u003d - (x + b) or (x - a) \u003d (x + b).
Equations like | x | \u003d y
Equations of this kind contain the absolute value of the expression with a variable to the left of zero, and another unknown to the right. The y variable can be either greater than or less than zero.
To get an answer in such equality, you need to solve a system of several equations, in which you need to make sure that y is a non-negative value:
Solving inequalities with modulus
To better understand how to expand the module in different types of equalities and inequalities, you need to analyze examples.
Equations of the form | x | \u003d a
Example 1 (algebra grade 6). Solve: | x | + 2 \u003d 4.
Decision.
Such equations are solved in the same way as equalities without absolute values. This means that by moving the unknowns to the left and the constants to the right, the expression does not change.
After moving the constant to the right, we got: | x | \u003d 2.
Since unknowns are related to absolute value, this equality has two answers: 2 and −2 .
Answer: 2 and −2 .
Example 2(algebra grade 7). Solve the inequality | x + 2 | ≥ 1.
Decision.
The first thing to do is find the points where the absolute value changes. To do this, the expression equates to 0 ... Received: x \u003d –2.
It means that –2 - turning point.
Let's divide the interval into 2 parts:
- for x + 2 ≥ 0
[−1; + ∞).
- for x + 2< 0
The common answer for these two inequalities is the interval (−∞; –3].
Final decision – combining the answers of the individual parts:
x∈ (–∞; –3] ∪ [–1; + ∞).
Answer: x∈ (–∞; –3] ∪ [–1; + ∞) .
Equations of the form | x | \u003d | y |
Example 1 (algebra grade 8). Solve the equation with two modules: 2 * | x \u200b\u200b- 1 | + 3 \u003d 9 - | x - 1 |.
Decision:
Answer: x 1 \u003d 3; x 2 \u003d − 1.
Example 2 (algebra grade 8). Solve inequality:
Decision:
Equations of the form | x | \u003d y
Example 1 (algebra grade 10). Find x:
Decision:
It is very important to check the right side, otherwise you can write erroneous roots in response. From the system you can see what does not lie in the gap.
Answer: x \u003d 0.
Sum module
Difference modulus
The absolute value of the difference between two numbers x and y is equal to the distance between points with coordinates X and Y on the coordinate line.
Example 1.
Example 2.
Negative number modulus
To find the absolute value of a number that is less than zero, you need to know how far from zero it is. Since the distance is always positive (it is impossible to go through "negative" steps, they are just steps in the other direction), the result is always positive. I.e,
Simply put, the absolute value of a negative number has the opposite meaning.
Zero modulus
Known property:
This is why the absolute value cannot be said to be a positive number: zero is neither negative nor positive.
Squared module
The squared modulus is always equal to the squared expression:
Examples of graphs with a module
Often in tests and exams there are tasks that can be solved only by analyzing the graphs. Let's consider such tasks.
Example 1.
A function f (x) \u003d | x | is given. It is necessary to build a graph from - 3 to 3 with a step of 1.
Decision:
Explanation: the figure shows that the graph is symmetrical about the Y-axis.
Example 2... It is necessary to draw and compare the graphs of the functions f (x) \u003d | x – 2 | and g (x) \u003d | x | –2.
Decision:
Explanation: A constant within an absolute value moves the entire graph to the right if its value is negative, and to the left if it is positive. But the constant outside will move the graph up if the value is positive, and down if it is negative (like - 2 in function g (x)).
Vertex coordinate x (the point at which two lines connect, the top of the graph) is the number by which the graph is shifted to the left or right. And the coordinate y Is the value by which the graph moves up or down.
You can build such graphs using online plotting applications. With their help, you can visually see how constants affect functions.
Interval method in tasks with a module
The spacing method is one of the best ways to find the answer in module problems, especially if there are several in the expression.
To use the method, you need to do the following:
- Set each expression to zero.
- Find the values \u200b\u200bof variables.
- Apply on the numerical line the points obtained in step 2.
- Determine the sign of expressions (negative or positive value) in the intervals and draw the symbol - or +, respectively. The easiest way to determine the sign is using the substitution method (substituting any value from the interval).
- Solve inequalities with the resulting signs.
Example 1... Solve by the method of intervals.
Decision:
