How to remove the module in the inequality. Modulo inequalities

There are several ways to solve inequalities containing a modulus. Let's consider some of them.

1) Solving the inequality using the geometric property of the module.

Let me remind you what the geometric property of the module is: the module of the number x is the distance from the origin to the point with the x coordinate.

In the course of solving inequalities in this way, 2 cases may arise:

1. |x| ≤ b,

And the inequality with modulus obviously reduces to a system of two inequalities. Here the sign can be strict, in which case the points in the picture will be “punched out”.

2. |x| ≥ b, then the picture of the solution looks like this:

And the inequality with the modulus obviously reduces to the set of two inequalities. Here the sign can be strict, in which case the points in the picture will be “punched out”.

Example 1

Solve the inequality |4 – |x|| ≥ 3.

Solution.

This inequality is equivalent to the following set:

U [-1;1] U

Example 2

Solve the inequality ||x+2| – 3| ≤ 2.

Solution.

This inequality is equivalent to the following system.

(|x + 2| – 3 ≥ -2
(|x + 2| – 3 ≤ 2,
(|x + 2| ≥ 1
(|x + 2| ≤ 5.

We solve separately the first inequality of the system. It is equivalent to the following set:

U[-1; 3].

2) Solving inequalities using the definition of the module.

Let me remind you to start module definition.

|a| = a if a ≥ 0 and |a| = -a if a< 0.

For example, |34| = 34, |-21| = -(-21) = 21.

Example 1

Solve the inequality 3|x – 1| ≤ x + 3.

Solution.

Using the module definition, we get two systems:

(x – 1 ≥ 0
(3(x – 1) ≤ x + 3

(x - 1< 0
(-3(x - 1) ≤ x + 3.

Solving the first and second systems separately, we get:

(x ≥ 1
(x ≤ 3,

(x< 1
(x ≥ 0.

The solution to the original inequality will be all solutions of the first system and all solutions of the second system.

Answer: x€.

3) Solving inequalities by squaring.

Example 1

Solve the inequality |x 2 – 1|< | x 2 – x + 1|.

Solution.

Let's square both sides of the inequality. I note that squaring both sides of the inequality is possible only if they are both positive. In this case, we have modules on both the left and right, so we can do this.

(|x 2 – 1|) 2< (|x 2 – x + 1|) 2 .

Now let's use the following module property: (|x|) 2 = x 2 .

(x 2 - 1) 2< (x 2 – x + 1) 2 ,

(x 2 - 1) 2 - (x 2 - x + 1) 2< 0.

(x 2 - 1 - x 2 + x - 1) (x 2 - 1 + x 2 - x + 1)< 0,

(x - 2)(2x 2 - x)< 0,

x(x - 2)(2x - 1)< 0.

We solve by the interval method.

Answer: x € (-∞; 0) U (1/2; 2)

4) Solving inequalities by the change of variables method.

Example.

Solve the inequality (2x + 3) 2 – |2x + 3| ≤ 30.

Solution.

Note that (2x + 3) 2 = (|2x + 3|) 2 . Then we get the inequality

(|2x + 3|) 2 – |2x + 3| ≤ 30.

Let's make the change y = |2x + 3|.

Let us rewrite our inequality taking into account the replacement.

y 2 – y ≤ 30,

y 2 – y – 30 ≤ 0.

We factorize the square trinomial on the left.

y1 = (1 + 11) / 2,

y2 = (1 - 11) / 2,

(y - 6)(y + 5) ≤ 0.

We solve by the interval method and get:

Back to replacement:

5 ≤ |2x + 3| ≤ 6.

This double inequality is equivalent to the system of inequalities:

(|2x + 3| ≤ 6
(|2x + 3| ≥ -5.

We solve each of the inequalities separately.

The first is equivalent to the system

(2x + 3 ≤ 6
(2x + 3 ≥ -6.

Let's solve it.

(x ≤ 1.5
(x ≥ -4.5.

The second inequality obviously holds for all x, since the modulus is, by definition, a positive number. Since the solution of the system is all x that simultaneously satisfy both the first and second inequality of the system, then the solution of the original system will be the solution of its first double inequality (after all, the second is true for all x).

Answer: x € [-4.5; 1.5].

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Maths is a symbol of the wisdom of science,

an example of scientific rigor and simplicity,

the standard of perfection and beauty in science.

Russian philosopher, professor A.V. Voloshinov

Modulo inequalities

The most difficult problems to solve in school mathematics are the inequalities, containing variables under the module sign. To successfully solve such inequalities, it is necessary to know the properties of the module well and have the skills to use them.

Basic concepts and properties

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

The simple properties of the module include the following relationships:

AND .

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

Also, if , where , then and

More complex module properties, which can be effectively used in solving equations and inequalities with modules, are formulated by means of the following theorems:

Theorem 1.For any analytic functions and the inequality.

Theorem 2. Equality is equivalent to the inequality.

Theorem 3. Equality is equivalent to the inequality.

The most common inequalities in school mathematics, containing unknown variables under the modulo sign, are inequalities of the form and where some positive constant.

Theorem 4. Inequality is equivalent to a double inequality, and the solution to the inequalityreduces to solving the set of inequalities and .

This theorem is a particular case of Theorems 6 and 7.

More complex inequalities, containing the module are inequalities of the form, and .

Methods for solving such inequalities can be formulated using the following three theorems.

Theorem 5. Inequality is equivalent to the combination of two systems of inequalities

AND (1)

Proof. Since then

This implies the validity of (1).

Theorem 6. Inequality is equivalent to the system of inequalities

Proof. Because , then from the inequality follows that . Under this condition, the inequalityand in this case the second system of inequalities (1) turns out to be inconsistent.

The theorem has been proven.

Theorem 7. Inequality is equivalent to the combination of one inequality and two systems of inequalities

AND (3)

Proof. Since , then the inequality always executed, if .

Let , then the inequalitywill be tantamount to inequality, from which the set of two inequalities follows and .

The theorem has been proven.

Consider typical examples of solving problems on the topic “Inequalities, containing variables under the module sign.

Solving inequalities with modulus

The simplest method for solving inequalities with modulus is the method, based on module expansion. This method is generic, however, in the general case, its application can lead to very cumbersome calculations. Therefore, students should also know other (more efficient) methods and techniques for solving such inequalities. In particular, need to have the skills to apply theorems, given in this article.

Example 1Solve the inequality

. (4)

Solution.Inequality (4) will be solved by the "classical" method - the moduli expansion method. To this end, we break the numerical axis dots and intervals and consider three cases.

1. If , then , , , and inequality (4) takes the form or .

Since the case is considered here, , is a solution to inequality (4).

2. If , then from inequality (4) we obtain or . Since the intersection of intervals and is empty, then there are no solutions to inequality (4) on the considered interval.

3. If , then inequality (4) takes the form or . It's obvious that is also a solution to inequality (4).

Answer: , .

Example 2 Solve the inequality.

Solution. Let's assume that . Because , then the given inequality takes the form or . Since , then and hence follows or .

However , therefore or .

Example 3 Solve the inequality

. (5)

Solution. Because , then inequality (5) is equivalent to the inequalities or . From here, according to Theorem 4, we have a set of inequalities and .

Answer: , .

Example 4Solve the inequality

. (6)

Solution. Let's denote . Then from inequality (6) we obtain the inequalities , , or .

From here, using the interval method, we get . Because , then here we have a system of inequalities

The solution to the first inequality of system (7) is the union of two intervals and , and the solution of the second inequality is the double inequality. This implies , that the solution to the system of inequalities (7) is the union of two intervals and .

Answer: ,

Example 5Solve the inequality

. (8)

Solution. We transform inequality (8) as follows:

Or .

Applying the interval method, we obtain a solution to inequality (8).

Answer: .

Note. If we put and in the condition of Theorem 5, then we obtain .

Example 6 Solve the inequality

. (9)

Solution. From inequality (9) it follows. We transform inequality (9) as follows:

Or

Since , then or .

Answer: .

Example 7Solve the inequality

. (10)

Solution. Since and , then or .

In this connection and inequality (10) takes the form

Or

. (11)

It follows from this that or . Since , then inequality (11) also implies or .

Answer: .

Note. If we apply Theorem 1 to the left side of inequality (10), then we get . From here and from inequality (10) it follows, that or . Because , then inequality (10) takes the form or .

Example 8 Solve the inequality

. (12)

Solution. Since then and inequality (12) implies or . However , therefore or . From here we get or .

Answer: .

Example 9 Solve the inequality

. (13)

Solution. According to Theorem 7, the solutions to inequality (13) are or .

Let now. In this case and inequality (13) takes the form or .

If we combine intervals and , then we obtain a solution to inequality (13) of the form.

Example 10 Solve the inequality

. (14)

Solution. Let us rewrite inequality (14) in an equivalent form: . If we apply Theorem 1 to the left side of this inequality, then we obtain the inequality .

From here and from Theorem 1 it follows, that inequality (14) is satisfied for any values.

Answer: any number.

Example 11. Solve the inequality

. (15)

Solution. Applying Theorem 1 to the left side of inequality (15), we get . From here and from inequality (15) follows the equation, which looks like.

According to Theorem 3, the equation is equivalent to the inequality. From here we get.

Example 12.Solve the inequality

. (16)

Solution. From inequality (16), according to Theorem 4, we obtain the system of inequalities

When solving the inequalitywe use Theorem 6 and obtain the system of inequalitiesfrom which follows.

Consider the inequality. According to Theorem 7, we obtain a set of inequalities and . The second population inequality holds for any real.

Consequently , the solution of inequality (16) are.

Example 13Solve the inequality

. (17)

Solution. According to Theorem 1, we can write

(18)

Taking into account inequality (17), we conclude that both inequalities (18) turn into equalities, i.e. there is a system of equations

By Theorem 3, this system of equations is equivalent to the system of inequalities

or

Example 14Solve the inequality

. (19)

Solution. Since , then . Let us multiply both parts of inequality (19) by the expression , which for any values ​​takes only positive values. Then we obtain an inequality that is equivalent to inequality (19), of the form

From here we get or , where . Since and then the solutions to inequality (19) are and .

Answer: , .

For a deeper study of methods for solving inequalities with a module, it is advisable to refer to tutorials, listed in the list of recommended readings.

1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.

2. Suprun V.P. Mathematics for high school students: methods for solving and proving inequalities. – M.: Lenand / URSS, 2018. - 264 p.

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

Do you have any questions?

To get the help of a tutor - register.

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modulo number this number itself is called if it is non-negative, or the same number with the opposite sign if it is negative.

For example, the modulus of 6 is 6, and the modulus of -6 is also 6.

That is, the modulus of a number is understood as an absolute value, the absolute value of this number without taking into account its sign.

Denoted as follows: |6|, | X|, |a| etc.

(For more details, see the "Module of Number" section).

Modulo Equations.

Example 1 . solve the equation|10 X - 5| = 15.

Solution.

In accordance with the rule, the equation is equivalent to the combination of two equations:

10X - 5 = 15
10X - 5 = -15

We decide:

10X = 15 + 5 = 20
10X = -15 + 5 = -10

X = 20: 10
X = -10: 10

X = 2
X = -1

Answer: X 1 = 2, X 2 = -1.

Example 2 . solve the equation|2 X + 1| = X + 2.

Solution.

Since the modulus is a non-negative number, then X+ 2 ≥ 0. Accordingly:

X ≥ -2.

We make two equations:

2X + 1 = X + 2
2X + 1 = -(X + 2)

We decide:

2X + 1 = X + 2
2X + 1 = -X - 2

2X - X = 2 - 1
2X + X = -2 - 1

X = 1
X = -1

Both numbers are greater than -2. So both are roots of the equation.

Answer: X 1 = -1, X 2 = 1.

Example 3 . solve the equation

|X + 3| - 1
————— = 4
X - 1

Solution.

The equation makes sense if the denominator is not equal to zero - so if X≠ 1. Let's take this condition into account. Our first action is simple - we don’t just get rid of the fraction, but we transform it in such a way as to get the module in its purest form:

|X+ 3| - 1 = 4 ( X - 1),

|X + 3| - 1 = 4X - 4,

|X + 3| = 4X - 4 + 1,

|X + 3| = 4X - 3.

Now we have only the expression under the modulus on the left side of the equation. Move on.
The modulus of a number is a non-negative number - that is, it must be greater than or equal to zero. Accordingly, we solve the inequality:

4X - 3 ≥ 0

4X ≥ 3

X ≥ 3/4

Thus, we have a second condition: the root of the equation must be at least 3/4.

In accordance with the rule, we compose a set of two equations and solve them:

X + 3 = 4X - 3
X + 3 = -(4X - 3)

X + 3 = 4X - 3
X + 3 = -4X + 3

X - 4X = -3 - 3
X + 4X = 3 - 3

X = 2
X = 0

We received two responses. Let's check if they are the roots of the original equation.

We had two conditions: the root of the equation cannot be equal to 1, and it must be at least 3/4. That is X ≠ 1, X≥ 3/4. Both of these conditions correspond to only one of the two answers received - the number 2. Hence, only it is the root of the original equation.

Answer: X = 2.

Inequalities with the modulus.

Example 1 . Solve the inequality| X - 3| < 4

Solution.

The module rule says:

|a| = a, if a ≥ 0.

|a| = -a, if a < 0.

The modulus can have both a non-negative and a negative number. So we have to consider both cases: X- 3 ≥ 0 and X - 3 < 0.

1) When X- 3 ≥ 0 our original inequality remains as it is, only without the modulo sign:
X - 3 < 4.

2) When X - 3 < 0 в исходном неравенстве надо поставить знак минус перед всем подмодульным выражением:

-(X - 3) < 4.

Opening the brackets, we get:

-X + 3 < 4.

Thus, from these two conditions, we have come to the union of two systems of inequalities:

X - 3 ≥ 0
X - 3 < 4

X - 3 < 0
-X + 3 < 4

Let's solve them:

X ≥ 3
X < 7

X < 3
X > -1

So, in our answer we have the union of two sets:

3 ≤ X < 7 U -1 < X < 3.

Determine the smallest and largest values. These are -1 and 7. At the same time X greater than -1 but less than 7.
Besides, X≥ 3. Hence, the solution to the inequality is the entire set of numbers from -1 to 7, excluding these extreme numbers.

Answer: -1 < X < 7.

Or: X ∈ (-1; 7).

Add-ons.

1) There is a simpler and shorter way to solve our inequality - graphical. To do this, draw a horizontal axis (Fig. 1).

Expression | X - 3| < 4 означает, что расстояние от точки X to point 3 less than four units. We mark the number 3 on the axis and count 4 divisions to the left and right of it. On the left we will come to point -1, on the right - to point 7. Thus, the points X we just saw without calculating them.

Moreover, according to the inequality condition, -1 and 7 themselves are not included in the set of solutions. Thus, we get the answer:

1 < X < 7.

2) But there is another solution that is even simpler than the graphical way. To do this, our inequality must be presented in the following form:

4 < X - 3 < 4.

After all, this is how it is according to the rule of the module. The non-negative number 4 and the similar negative number -4 are the boundaries of the solution to the inequality.

4 + 3 < X < 4 + 3

1 < X < 7.

Example 2 . Solve the inequality| X - 2| ≥ 5

Solution.

This example differs significantly from the previous one. The left side is greater than 5 or equal to 5. From a geometric point of view, the solution to the inequality is all the numbers that are at a distance of 5 units or more from point 2 (Fig. 2). The graph shows that these are all numbers that are less than or equal to -3 and greater than or equal to 7. So, we have already received the answer.

Answer: -3 ≥ X ≥ 7.

Along the way, we solve the same inequality by rearranging the free term to the left and right with the opposite sign:

5 ≥ X - 2 ≥ 5

5 + 2 ≥ X ≥ 5 + 2

The answer is the same: -3 ≥ X ≥ 7.

Or: X ∈ [-3; 7]

Example solved.

Example 3 . Solve the inequality 6 X 2 - | X| - 2 ≤ 0

Solution.

Number X can be positive, negative or zero. Therefore, we need to take into account all three circumstances. As you know, they are taken into account in two inequalities: X≥ 0 and X < 0. При X≥ 0, we simply rewrite our original inequality as is, only without the modulo sign:

6x 2 - X - 2 ≤ 0.

Now for the second case: if X < 0. Модулем отрицательного числа является это же число с противоположным знаком. То есть пишем число под модулем с обратным знаком и опять же освобождаемся от знака модуля:

6X 2 - (-X) - 2 ≤ 0.

Expanding the brackets:

6X 2 + X - 2 ≤ 0.

Thus, we have received two systems of equations:

6X 2 - X - 2 ≤ 0
X ≥ 0

6X 2 + X - 2 ≤ 0
X < 0

We need to solve inequalities in systems - which means we need to find the roots of two quadratic equations. To do this, we equate the left-hand sides of the inequalities to zero.

Let's start with the first one:

6X 2 - X - 2 = 0.

How to solve a quadratic equation - see the "Quadric Equation" section. We will immediately name the answer:

X 1 \u003d -1/2, x 2 \u003d 2/3.

From the first system of inequalities, we get that the solution to the original inequality is the entire set of numbers from -1/2 to 2/3. We write the union of solutions for X ≥ 0:
[-1/2; 2/3].

Now let's solve the second quadratic equation:

6X 2 + X - 2 = 0.

Its roots:

X 1 = -2/3, X 2 = 1/2.

Conclusion: when X < 0 корнями исходного неравенства являются также все числа от -2/3 до 1/2.

Let's combine the two answers and get the final answer: the solution is the whole set of numbers from -2/3 to 2/3, including these extreme numbers.

Answer: -2/3 ≤ X ≤ 2/3.

Or: X ∈ [-2/3; 2/3].