One of the most difficult topics for students is solving equations containing a variable under the modulus sign. Let's see for a start what is it connected with? Why, for example, quadratic equations most children click like nuts, but with such a far from the most complex 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, 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. But what if a module is encountered in the equation? We will try to clearly describe the necessary plan of action in the case when the equation contains an unknown under the modulus sign. We give several examples for each case.
But first, let's remember module definition. So, the modulus of the number a the number itself is called if a non-negative and -a if the 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 to 
coordinate. So, the module or the absolute value of a number is the distance from this point to the origin of the numerical axis. The distance is always given 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. Any number can be in the module, but the result of applying the module is always a positive number.
Now let's move on 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 definition of the modulus.
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 -five< 0, то уравнение не имеет корней;
3) |x| = 0, then x = 0.
2. An equation of the form |f(x)| = b, where b > 0. To solve this equation, it is necessary to get rid of the modulus. We do it like this: f(x) = b or f(x) = -b. Now it is necessary to solve separately each of the obtained equations. 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 side is greater than or equal to zero, i.e. g(x) ≥ 0. Then we 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. Combine O.D.Z. and the solution, we get:
The root x \u003d 11/7 does not fit according to O.D.Z., it is less than 2, and x \u003d 3 satisfies this condition.
Answer: x = 3
2) |x – 1| \u003d 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 \u003d 1 - x 2 or x - 1 \u003d - (1 - x 2)
x 2 + x - 2 = 0 x 2 - x = 0
x = -2 or x = 1 x = 0 or x = 1
3. Combine solution and O.D.Z.:
Only the roots x = 1 and x = 0 are suitable.
Answer: x = 0, x = 1.
4. An 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 (change of variable). 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 = 0. By the property of the module x 2 = |x| 2 , so the equation can be rewritten as follows:
|x| 2–6|x| + 5 = 0. Let's make the change |x| = 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| = 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 property of the module x 2 = |x| 2 , so
|x| 2 + |x| – 2 = 0. Let's make the change |x| = t ≥ 0, then:
t 2 + t - 2 \u003d 0. Solving this equation, we get, t \u003d -2 or t \u003d 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's express the module 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 -one< 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. There are no roots.
Answer: x = -3, x = 1.
There is also a universal method for solving equations with a modulus. This is the spacing method. But we will consider it further.
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The absolute value of a number a is the distance from the origin to the point BUT(a).
To understand this definition, we substitute instead of a variable a any number, for example 3 and try to read it again:
The absolute value of a number 3 is the distance from the origin to the point BUT(3 ).
It becomes clear that the module is nothing more than the usual distance. Let's try to see the distance from the origin to point A( 3 )
The distance from the origin of coordinates to point A( 3 ) is equal to 3 (three units or three steps).
The modulus of a number is indicated by two vertical lines, for example:
The modulus of the number 3 is denoted as follows: |3|
The modulus of the number 4 is denoted as follows: |4|
The modulus of the number 5 is denoted as follows: |5|
We looked for the modulus of the number 3 and found out that it is equal to 3. So we write:
Reads like: "The modulus of three is three"
Now let's try to find the modulus of the number -3. Again, we return to the definition and substitute the number -3 into it. Only instead of a dot A use new point B. point A we have already used in the first example.
The modulus of the number is 3 call the distance from the origin to the point B(—3 ).
The distance from one point to another cannot be negative. Therefore, the modulus of any negative number, being a distance, will also not be negative. The module of the number -3 will be the number 3. The distance from the origin to the point B(-3) is also equal to three units:
Reads like: "The modulus of a number minus three is three"
The modulus of the number 0 is 0, since the point with coordinate 0 coincides with the origin, i.e. distance from origin to point O(0) equals zero:
"The modulus of zero is zero"
We draw conclusions:
- The modulus of a number cannot be negative;
- For a positive number and zero, the modulus is equal to the number itself, and for a negative one, to the opposite number;
- Opposite numbers have equal modules.
Opposite numbers
Numbers that differ only in signs are called opposite. For example, the numbers −2 and 2 are opposites. They differ only in signs. The number −2 has a minus sign, and 2 has a plus sign, but we don’t see it, because plus, as we said earlier, is traditionally not written.
More examples of opposite numbers:
Opposite numbers have equal modules. For example, let's find modules for −2 and 2
The figure shows that the distance from the origin to the points A(−2) And B(2) equal to two steps.
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Solving Equations and Inequalities with Modulus often causes problems. However, if you understand well what is the absolute value of a number, And how to correctly expand expressions containing the modulo sign, then the presence in the equation 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 of the numbers 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 modulus 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)|= f(x) if f(x) ≥ 0, and
|f(x)|= - f(x) if f(x)< 0
For example |x-3|=x-3 if x-3≥0 and |x-3|=-(x-3)=3-x if x-3<0.
To solve an equation containing an expression under the modulus sign, you must first expand module by module expansion rule.
Then our equation or inequality is transformed into two different equations existing on two different numerical intervals.
One equation exists on a numerical interval on which the expression under the modulus sign is non-negative.
And the second equation exists on the interval on which the expression under the modulus sign is negative.
Let's consider a simple example.
Let's solve the equation:
|x-3|=-x 2 +4x-3
1. Let's open the module.
|x-3|=x-3 if x-3≥0, i.e. if x≥3
|x-3|=-(x-3)=3-x if x-3<0, т.е. если х<3
2. We got two numerical intervals: x≥3 and x<3.
Consider what equations the original equation is transformed into on each interval:
A) For x≥3 |x-3|=x-3, and our equation looks like:
Attention! This equation exists only on the interval x≥3!
Let's open the brackets, 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=-x 2 +4x-3 exists only on the interval x≥3, we are only interested in those roots that belong to this interval. This condition satisfies only x 2 =3.
B) At 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 open the brackets and give like 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=3, from the second - the root x=2.
We don't choose math her profession, and she chooses us.
Russian mathematician Yu.I. Manin
Modulo Equations
The most difficult problems to solve in school mathematics are equations containing variables under the module sign. To successfully solve such equations, it is necessary 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 the module include the following relationships:
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 with modules, are formulated by means of the following theorems:
Theorem 1.For any analytic functions And the inequality
Theorem 2. Equality is the same as inequality.
Theorem 3. Equality is equivalent to the inequality.
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 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. In this regard, students should also be aware of other, more efficient methods and techniques for solving such equations. In particular, need to have the skills to apply theorems, given in this article.
Example 1 Solve the equation. (one)
Solution. Equation (1) will be solved by the "classical" method - the module expansion method. To do this, we break the numerical axis dots and intervals and consider three cases.
1. If , then , , , and equation (1) takes the form . It follows from here. However, here , so the found value is not the root of equation (1).
2. If , then from equation (1) we obtain or .
Since then the root of equation (1).
3. If , then equation (1) takes the form or . Note that .
Answer: , .
When solving the following 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.
Solution. Since and then it follows from the equation. In this regard, , , and the equation becomes. From here we get. But , so the original equation has no roots.
Answer: no roots.
Example 3 solve the equation.
Solution. Since , then . If , then , and the equation becomes.
From here we get .
Example 4 solve the equation.
Solution.Let us rewrite the equation in an equivalent form. (2)
The resulting equation belongs to equations of the type .
Taking into account Theorem 2, we can state that equation (2) is equivalent to the inequality . From here we get .
Answer: .
Example 5 Solve the equation.
Solution. This equation has the form. That's why , according to Theorem 3, here we have the inequality or .
Example 6 solve the equation.
Solution. Let's 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 . From here we get two roots of the original equation: And .
Example 7 solve the equation. (4)
Solution. Since the equationis equivalent to the 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 8solve the equation . (5)
Solution. Since and , then . From here and from Eq. (5) it follows that and , i.e. here we have a system of equations
However, this system of equations is inconsistent.
Answer: no roots.
Example 9 solve the equation. (6)
Solution. If we designate and from equation (6) we obtain
Or . (7)
Since equation (7) has the form , this equation is equivalent to the inequality . From here we get . Since , then or .
Answer: .
Example 10solve the equation. (8)
Solution.According to Theorem 1, we can write
(9)
Taking into account equation (8), we conclude that both inequalities (9) turn into equalities, i.e. there is a system of equations
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)
Solution. Let and , then the equation (11) implies the equality .
From this it follows that and . Thus, here we have a system of inequalities
The solution to this system of inequalities are And .
Answer: , .
Example 12.solve the equation. (12)
Solution. Equation (12) will be solved by the method of successive 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 13solve the equation. (13)
Solution. Since the left 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 a system of equations (14)
Solution. 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 a system of equations (15)
Solution. 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 a system of equations (16)
Solution. It follows from the first equation of system (16) that .
Since then . Consider the second equation of the system. Insofar as, then , and the equation becomes, , or .
If we substitute the valueinto the first equation of system (16), then , or .
Answer: , .
For a deeper study of problem solving methods, related to the solution of equations, containing variables under the module sign, you can advise tutorials from the list of recommended literature.
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: tasks of increased complexity. - M .: KD "Librocom" / URSS, 2017. - 200 p.
3. Suprun V.P. Mathematics for high school students: non-standard methods for solving problems. - M .: KD "Librocom" / URSS, 2017. - 296 p.
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Among examples per module often there are equations where you need to find module roots in module, that is, an equation of the form
||a*x-b|-c|=k*x+m .
If k=0 , that is, the right side is equal to a constant (m) then it is easier to look for a solution equations with modules graphically. Below is the methodology deployment of double modules on common practice examples. Understand well the algorithm for calculating equations with modules, so as not to have problems on control, tests, and just to know.
Example 1 Solve the equation module in module |3|x|-5|=-2x-2.
Solution: Always start expanding equations from the internal module
|x|=0 <->x=0.
At the point x=0, the equation with modulus is divided by 2 .
For x< 0
подмодульная функция отрицательная, поэтому при раскрытии знак меняем на противоположный
|-3x-5|=-2x-2.
For x>0 or equal, expanding the modulus we get
|3x-5|=-2x-2 .
Let's solve the equation for negative variables (x< 0)
. Оно разлагается на две системы уравнений. Первое уравнение получаем из условия, что функция после знака равенства неотрицательна. Второе - раскрывая модуль в одной системе принимаем, что подмодульная функция положительная, в иной отрицательная - меняем знак правой или левой части (зависит от методики преподавания).
From the first equation, we get that the solution should not exceed (-1) , i.e.
This restriction belongs entirely to the area in which we are solving. Let's move the variables and constants on opposite sides of equality in the first and second systems
and find a solution 
Both values belong to the interval that is being considered, that is, they are roots.
Consider an equation with modules for positive variables
|3x-5|=-2x-2.
Expanding the module, we obtain two systems of equations 
From the first equation, which is common for two systems, we obtain the familiar condition
which, in intersection with the set on which we are looking for a solution, gives an empty set (no intersection points). So the only roots of module with module are the values
x=-3; x=-1.4.
Example 2 Solve the equation with modulo ||x-1|-2|=3x-4.
Solution: Let's start by expanding the inner module
|x-1|=0 <=>x=1.
A submodule function changes sign at one. At smaller values it is negative, at larger values it is positive. In accordance with this, when expanding the internal module, we obtain two equations with the module
x |-(x-1)-2|=3x-4;
x>=1 -> |x-1-2|=3x-4.
Be sure to check the right side of the equation with the modulus, it must be greater than zero.
3x-4>=0 -> x>=4/3.
This means that there is no need to solve the first of the equations, since it is written for x< 1,
что не соответствует найденному условию. Раскроем модуль во втором уравнении
|x-3|=3x-4 ->
x-3=3x-4 or x-3=4-3x;
4-3=3x-x or x+3x=4+3;
2x=1 or 4x=7;
x=1/2 or x=7/4.
We got two values, the first of which is rejected, because it does not belong to the desired interval. The final equation has one solution x=7/4.
Example 3 Solve the equation with modulo ||2x-5|-1|=x+3.
Solution: Let's open the internal module
|2x-5|=0 <=>x=5/2=2.5.
The point x=2.5 splits the numerical axis into two intervals. Respectively, submodule function changes sign when passing through 2.5. Let us write the condition on the solution on the right side of the equation with modulus.
x+3>=0 -> x>=-3.
So the solution can be values not less than (-3) . Let's expand the modulus for the negative value of the internal modulus
|-(2x-5)-1|=x+3;
|-2x+4|=x+3.
This module will also, when expanded, give 2 equations
-2x+4=x+3 or 2x-4=x+3;
2x+x=4-3 or 2x-x=3+4;
3x=1; x=1/3 or x=7 .
The value x=7 is rejected, since we were looking for a solution on the interval [-3;2.5]. Now expand the inner module for x>2.5 . We get an equation with one module
|2x-5-1|=x+3;
|2x-6|=x+3.
When expanding the module, we obtain the following linear equations
-2x+6=x+3 or 2x-6=x+3;
2x+x=6-3 or 2x-x=3+6;
3x=3; x=1 or x=9 .
The first value x=1 does not satisfy the condition x>2.5. So on this interval we have one root of the equation with module x=9, and there are only two of them (x=1/3). By substitution, you can check the correctness of the calculations performed
Answer: x=1/3; x=9.
Example 4 Find solutions of the double module ||3x-1|-5|=2x-3.
Solution: Expand the inner module of the equation
|3x-1|=0 <=>x=1/3.
The point x=2.5 divides the numerical axis into two intervals, and the given equation into two cases. We write down the condition for the solution, based on the type of equation on the right side
2x-3>=0 -> x>=3/2=1.5.
It follows that we are interested in values >=1.5 . In this way modular equation look at two intervals
,
|-(3x-1)-5|=2x-3;
|-3x-4|=2x-3.
The resulting module, when expanded, is divided into 2 equations
-3x-4=2x-3 or 3x+4=2x-3;
2x+3x=-4+3 or 3x-2x=-3-4;
5x=-1; x=-1/5 or x=-7 .
Both values do not fall into the interval , that is, they are not solutions to the equation with modules. Next, expand the modulus for x>2.5 . We get the following equation
|3x-1-5|=2x-3;
|3x-6|=2x-3.
Expanding the module, we obtain 2 linear equations
3x-6=2x-3 or –(3x-6)=2x-3;
3x-2x=-3+6 or 2x+3x=6+3;
x=3 or 5x=9; x=9/5=1.8.
The second value found does not meet the x>2.5 condition, we reject it.
Finally we have one root of the equation with modules x=3 .
We perform a check
||3*3-1|-5|=2*3-3 3=3
.
The root of the equation with the modulus calculated correctly.
Answer: x=1/3; x=9.