How to solve equations with multiple modules. Equations with modulus

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, quadratic equations are clicks like nuts for most children, but with such a far from complicated concept as a module 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, solving a 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 numerical 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 module properties.

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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Instructions

If the module is presented as a continuous function, then the value of its argument can be either positive or negative: | x | \u003d x, x ≥ 0; | x | \u003d - x, x

The modulus is zero, and the modulus of any positive number is to him. If the argument is negative, then after expanding the parentheses, its sign changes from minus to plus. This leads to the conclusion that the modules of the opposite are equal: | -х | \u003d | x | \u003d x.

The module of a complex number is found by the formula: | a | \u003d √b ² + c ², and | a + b | ≤ | a | + | b |. If the argument contains a positive number as a factor, then it can be moved outside the parenthesis, for example: | 4 * b | \u003d 4 * | b |.

If the argument is presented as a complex number, then for the convenience of calculations, the order of the members of the expression enclosed in square brackets is allowed: | 2-3 | \u003d | 3-2 | \u003d 3-2 \u003d 1 because (2-3) is less than zero.

The raised argument is simultaneously under the sign of the root of the same order - it is solved using: √a² \u003d | a | \u003d ± a.

If you are faced with a task that does not specify the condition for expanding the module brackets, then you do not need to get rid of them - this will be the final result. And if you want to reveal them, then you must indicate the ± sign. For example, you need to find the value of the expression √ (2 * (4-b)) ². His solution looks like this: √ (2 * (4-b)) ² \u003d | 2 * (4-b) | \u003d 2 * | 4-b |. Since the sign of the expression 4-b is unknown, it must be left in brackets. If you add an additional condition, for example, | 4-b | \u003e

The modulus of zero is zero, and the modulus of any positive number is to itself. If the argument is negative, then after expanding the parentheses, its sign changes from minus to plus. This leads to the conclusion that the absolute values \u200b\u200bof opposite numbers are equal: | -х | \u003d | x | \u003d x.

The module of a complex number is found by the formula: | a | \u003d √b ² + c ², and | a + b | ≤ | a | + | b |. If the argument contains a positive integer as a factor, then it can be moved outside the parenthesis, for example: | 4 * b | \u003d 4 * | b |.

The modulus cannot be negative, so any negative number is converted to positive: | -x | \u003d x, | -2 | \u003d 2, | -1/7 | \u003d 1/7, | -2.5 | \u003d 2.5.

If the argument is presented as a complex number, then for the convenience of calculations, it is allowed to change the order of the members of the expression enclosed in square brackets: | 2-3 | \u003d | 3-2 | \u003d 3-2 \u003d 1 because (2-3) is less than zero.

If you are faced with a task that does not specify the condition for expanding the module brackets, then you do not need to get rid of them - this will be the final result. And if you want to reveal them, then you must indicate the ± sign. For example, you need to find the value of the expression √ (2 * (4-b)) ². His solution looks like this: √ (2 * (4-b)) ² \u003d | 2 * (4-b) | \u003d 2 * | 4-b |. Since the sign of the expression 4-b is unknown, it must be left in brackets. If you add an additional condition, for example, | 4-b | \u003e 0, then the result will be 2 * | 4-b | \u003d 2 * (4 - b). A specific number can also be specified as the unknown element, which should be taken into account, since it will affect the sign of the expression.

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 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 program. 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 of coordinates to a given 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 size 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 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