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One of the most difficult topics for students is solving equations containing a variable under the modulus sign. Let's first figure out what this is connected with? Why, for example, do most children crack quadratic equations like nuts, but have so many problems with such a far from complex concept as a module?

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 formulas for the roots of the quadratic equation. What to do if a modulus is found in the equation? We will try to clearly describe necessary plan actions in the case when the equation contains an unknown under the modulus sign. We will give several examples for each case.

But first, let's remember module definition. So, modulo the number a this number itself is called if a non-negative and -a, if number a less than zero. You can write it like this:

|a| = a if a ≥ 0 and |a| = -a if a< 0

Talking about geometric sense module, it should be remembered that each real number corresponds to a certain point on the number axis - its to coordinate. So, the module or absolute value of a number is the distance from this point to the origin of the numerical axis. The distance is always specified as a positive number. Thus, the modulus of any negative number is a positive number. By the way, even at this stage, many students begin to get confused. The module can contain any number, but the result of using the module is always a positive number.

Now let's move directly to solving the equations.

1. Consider an equation of the form |x| = 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. We write the solution in the form of a diagram:

(±c, if c > 0

If |x| = c, then x = (0, if c = 0

(no roots if with< 0

1) |x| = 5, because 5 > 0, then x = ±5;

2) |x| = -5, because -5< 0, то уравнение не имеет корней;

3) |x| = 0, then x = 0.

2. Equation of the form |f(x)| = b, where b > 0. To solve this equation it is necessary to get rid of the module. We do it this way: f(x) = b or f(x) = -b. Now you need to solve each of the resulting equations separately. If in the original equation b< 0, решений не будет.

1) |x + 2| = 4, because 4 > 0, then

x + 2 = 4 or x + 2 = -4

2) |x 2 – 5| = 11, because 11 > 0, then

x 2 – 5 = 11 or x 2 – 5 = -11

x 2 = 16 x 2 = -6

x = ± 4 no roots

3) |x 2 – 5x| = -8, because -8< 0, то уравнение не имеет корней.

3. An equation of the form |f(x)| = g(x). According to 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) = g(x) or f(x) = -g(x).

1) |2x – 1| = 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 = 5x – 10 or 2x – 1 = -(5x – 10)

3. We combine O.D.Z. and the solution, we get:

The root x = 11/7 does not fit the O.D.Z., it is less than 2, but x = 3 satisfies this condition.

Answer: x = 3

2) |x – 1| = 1 – x 2 .

1. O.D.Z. 1 – x 2 ≥ 0. Let’s solve this inequality using the interval method:

(1 – x)(1 + x) ≥ 0

2. Solution:

x – 1 = 1 – x 2 or x – 1 = -(1 – x 2)

x 2 + x – 2 = 0 x 2 – x = 0

x = -2 or x = 1 x = 0 or x = 1

3. We combine the solution and O.D.Z.:

Only roots x = 1 and x = 0 are suitable.

Answer: x = 0, x = 1.

4. Equation of the form |f(x)| = |g(x)|. Such an equation is equivalent to the following two equations f(x) = g(x) or f(x) = -g(x).

1) |x 2 – 5x + 7| = |2x – 5|. This equation is equivalent to the following two:

x 2 – 5x + 7 = 2x – 5 or x 2 – 5x +7 = -2x + 5

x 2 – 7x + 12 = 0 x 2 – 3x + 2 = 0

x = 3 or x = 4 x = 2 or x = 1

Answer: x = 1, x = 2, x = 3, x = 4.

5. Equations solved by the substitution method (variable replacement). This method solutions are easiest to explain in specific example. So, let us be given a quadratic equation with modulus:

x 2 – 6|x| + 5 = 0. By the modulus property x 2 = |x| 2, so the equation can be rewritten as follows:

|x| 2 – 6|x| + 5 = 0. Let's make the replacement |x| = t ≥ 0, then we will have:

t 2 – 6t + 5 = 0. Solving this equation, we find that t = 1 or t = 5. Let’s return to the replacement:

|x| = 1 or |x| = 5

x = ±1 x = ±5

Answer: x = -5, x = -1, x = 1, x = 5.

Let's look at another example:

x 2 + |x| – 2 = 0. By the modulus property x 2 = |x| 2, therefore

|x| 2 + |x| – 2 = 0. Let’s make the replacement |x| = t ≥ 0, then:

t 2 + t – 2 = 0. Solving this equation, we get t = -2 or t = 1. Let’s return to the replacement:

|x| = -2 or |x| = 1

No roots x = ± 1

Answer: x = -1, x = 1.

6. Another type of equations is equations with a “complex” modulus. Such equations include equations that have “modules within a module.” Equations of this type can be solved using the properties of the module.

1) |3 – |x|| = 4. We will act in the same way as in equations of the second type. Because 4 > 0, then we get two equations:

3 – |x| = 4 or 3 – |x| = -4.

Now let us express the modulus x in each equation, then |x| = -1 or |x| = 7.

We solve each of the resulting equations. There are no roots in the first equation, because -1< 0, а во втором x = ±7.

Answer x = -7, x = 7.

2) |3 + |x + 1|| = 5. We solve this equation in a similar way:

3 + |x + 1| = 5 or 3 + |x + 1| = -5

|x + 1| = 2 |x + 1| = -8

x + 1 = 2 or x + 1 = -2. No roots.

Answer: x = -3, x = 1.

There is also a universal method for solving equations with a modulus. This is the interval method. But we will look at it later.

website, when copying material in full or in part, a link to the source is required.

One of the most difficult topics for students is solving equations containing a variable under the modulus sign. Let's first figure out what this is connected with? Why, for example, do most children crack quadratic equations like nuts, but have so many problems with such a far from complex concept as a module?

In my opinion, all these difficulties are associated with the lack of clearly formulated rules for solving equations with a modulus. So, when solving a quadratic equation, the student knows for sure that he needs to first apply the discriminant formula, and then the formulas for the roots of the quadratic equation. What to do if a modulus is found 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. We will give several examples for each case.

But first, let's remember module definition. So, modulo the number a this number itself is called if a non-negative and -a, if number a less than zero. You can write it like this:

|a| = a if a ≥ 0 and |a| = -a if a< 0

Speaking about the geometric meaning of the module, it should be remembered that each real number corresponds to a certain point on the number axis - its coordinate. So, the module or absolute value of a number is the distance from this point to the origin of the numerical axis. The distance is always specified as a positive number. Thus, the modulus of any negative number is a positive number. By the way, even at this stage, many students begin to get confused. The module can contain any number, but the result of using the module is always a positive number.

Now let's move directly to solving the equations.

1. Consider an equation of the form |x| = 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. We write the solution in the form of a diagram:

(±c, if c > 0

If |x| = c, then x = (0, if c = 0

(no roots if with< 0

1) |x| = 5, because 5 > 0, then x = ±5;

2) |x| = -5, because -5< 0, то уравнение не имеет корней;

3) |x| = 0, then x = 0.

2. Equation of the form |f(x)| = b, where b > 0. To solve this equation it is necessary to get rid of the module. We do it this way: f(x) = b or f(x) = -b. Now you need to solve each of the resulting equations separately. If in the original equation b< 0, решений не будет.

1) |x + 2| = 4, because 4 > 0, then

x + 2 = 4 or x + 2 = -4

2) |x 2 – 5| = 11, because 11 > 0, then

x 2 – 5 = 11 or x 2 – 5 = -11

x 2 = 16 x 2 = -6

x = ± 4 no roots

3) |x 2 – 5x| = -8, because -8< 0, то уравнение не имеет корней.

3. An equation of the form |f(x)| = g(x). According to 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) = g(x) or f(x) = -g(x).

1) |2x – 1| = 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 = 5x – 10 or 2x – 1 = -(5x – 10)

3. We combine O.D.Z. and the solution, we get:

The root x = 11/7 does not fit the O.D.Z., it is less than 2, but x = 3 satisfies this condition.

Answer: x = 3

2) |x – 1| = 1 – x 2 .

1. O.D.Z. 1 – x 2 ≥ 0. Let’s solve this inequality using the interval method:

(1 – x)(1 + x) ≥ 0

2. Solution:

x – 1 = 1 – x 2 or x – 1 = -(1 – x 2)

x 2 + x – 2 = 0 x 2 – x = 0

x = -2 or x = 1 x = 0 or x = 1

3. We combine the solution and O.D.Z.:

Only roots x = 1 and x = 0 are suitable.

Answer: x = 0, x = 1.

4. Equation of the form |f(x)| = |g(x)|. Such an equation is equivalent to the following two equations f(x) = g(x) or f(x) = -g(x).

1) |x 2 – 5x + 7| = |2x – 5|. This equation is equivalent to the following two:

x 2 – 5x + 7 = 2x – 5 or x 2 – 5x +7 = -2x + 5

x 2 – 7x + 12 = 0 x 2 – 3x + 2 = 0

x = 3 or x = 4 x = 2 or x = 1

Answer: x = 1, x = 2, x = 3, x = 4.

5. Equations solved by the substitution method (variable replacement). This solution method is easiest to explain with a specific example. So, let us be given a quadratic equation with modulus:

x 2 – 6|x| + 5 = 0. By the modulus property x 2 = |x| 2, so the equation can be rewritten as follows:

|x| 2 – 6|x| + 5 = 0. Let's make the replacement |x| = t ≥ 0, then we will have:

t 2 – 6t + 5 = 0. Solving this equation, we find that t = 1 or t = 5. Let’s return to the replacement:

|x| = 1 or |x| = 5

x = ±1 x = ±5

Answer: x = -5, x = -1, x = 1, x = 5.

Let's look at another example:

x 2 + |x| – 2 = 0. By the modulus property x 2 = |x| 2, therefore

|x| 2 + |x| – 2 = 0. Let’s make the replacement |x| = t ≥ 0, then:

t 2 + t – 2 = 0. Solving this equation, we get t = -2 or t = 1. Let’s return to the replacement:

|x| = -2 or |x| = 1

No roots x = ± 1

Answer: x = -1, x = 1.

6. Another type of equations is equations with a “complex” modulus. Such equations include equations that have “modules within a module.” Equations of this type can be solved using the properties of the module.

1) |3 – |x|| = 4. We will act in the same way as in equations of the second type. Because 4 > 0, then we get two equations:

3 – |x| = 4 or 3 – |x| = -4.

Now let us express the modulus x in each equation, then |x| = -1 or |x| = 7.

We solve each of the resulting equations. There are no roots in the first equation, because -1< 0, а во втором x = ±7.

Answer x = -7, x = 7.

2) |3 + |x + 1|| = 5. We solve this equation in a similar way:

3 + |x + 1| = 5 or 3 + |x + 1| = -5

|x + 1| = 2 |x + 1| = -8

x + 1 = 2 or x + 1 = -2. No roots.

Answer: x = -3, x = 1.

There is also a universal method for solving equations with a modulus. This is the interval method. But we will look at it later.

blog.site, when copying material in full or in part, a link to the original source is required.

The modulus is the absolute value of the expression. To somehow indicate a module, it is customary to use straight brackets. The value that is enclosed in even brackets is the value that is taken modulo. The process of solving any module consists in opening those very straight brackets, which in mathematical language are called modular brackets. Their disclosure occurs according to a certain number of rules. Also, in the order of solving the modules, the sets of values ​​of those expressions that were in the modular brackets are found. In most cases, the module is expanded in such a way that the expression that was submodular receives 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 compiled, 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 the module. To answer the question of how to solve equations with a modulus, you need to open 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 ​​of the unknown quantities included in its composition, the modular expression in brackets becomes zero. In order to do this, it is enough to equate the expression in modular brackets to zero, and then calculate the solution to the resulting equation. The found values ​​must be recorded. In the same way, you also need to determine the value of all unknown variables for all modules in given equation. Next, you need to start defining and considering all cases of the 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 corresponding to all modules in the original inequality. Inequalities must be drawn up so that they cover all available and possible values for a variable that is found on the number line. Then you need to draw this same number line for visualization, on which to later plot 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 ​​of 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, you need to open all the available modular brackets, while changing the sign of the expression so that the values ​​of the desired variable coincide with those values ​​that are visible on the number line. The resulting equation must be solved. The value of the variable that will be obtained during solving the equation must be checked against the limitation that is specified 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 restrictions, must be discarded.