In this article we will analyze in detail the absolute value of a number. We will give various definitions of the modulus of a number, introduce notation and provide graphic illustrations. At the same time, let's consider various examples finding the modulus of a number by definition. After this, we will list and justify the main properties of the module. At the end of the article, we’ll talk about how a module is defined and located complex number.
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Number module - definition, notation and examples
First we introduce number modulus designation. We will write the modulus of the number a as , that is, to the left and right of the number we will put vertical dashes to form the modulus sign. Let's give a couple of examples. For example, module −7 can be written as ; module 4.125 is written as , and the module has a notation of the form .
The following definition of module applies to , and therefore to , and to integers, and to rational, and to irrational numbers, as to the constituent parts of the set of real numbers. We will talk about the modulus of a complex number in.
Definition.
Modulus of number a– this is either the number a itself, if a – positive number, or the number −a, the opposite of the number a, if a is a negative number, or 0, if a=0.
The voiced definition of the modulus of a number is often written in the following form 
, this entry means that if a>0 , if a=0 , and if a<0
.
The record can be presented in a more compact form 
. This notation means that if (a is greater than or equal to 0), and if a<0
.
There is also the entry 
. Here we should separately explain the case when a=0. In this case we have , but −0=0, since zero is considered a number that is opposite to itself.
Let's give examples of finding the modulus of a number using a stated definition. For example, let's find the modules of the numbers 15 and . Let's start by finding . Since the number 15 is positive, its modulus, by definition, is equal to this number itself, that is, . What is the modulus of a number? Since is a negative number, its modulus is equal to the number opposite to the number, that is, the number 
. Thus, .
To conclude this point, we present one conclusion that is very convenient to use in practice when finding the modulus of a number. From the definition of the modulus of a number it follows that the modulus of a number is equal to the number under the modulus sign without taking into account its sign, and from the examples discussed above this is very clearly visible. The stated statement explains why the module of a number is also called absolute value of the number. So the modulus of a number and the absolute value of a number are one and the same.
Modulus of a number as a distance
Geometrically, the modulus of a number can be interpreted as distance. Let's give determining the modulus of a number through distance.
Definition.
Modulus of number a– this is the distance from the origin on the coordinate line to the point corresponding to the number a.
This definition is consistent with the definition of the modulus of a number given in the first paragraph. Let's clarify this point. The distance from the origin to the point corresponding to a positive number is equal to this number. Zero corresponds to the origin, therefore the distance from the origin to the point with coordinate 0 is equal to zero (you do not need to set aside a single unit segment and not a single segment that makes up any fraction of a unit segment in order to get from point O to a point with coordinate 0). The distance from the origin to a point with a negative coordinate is equal to the number opposite to the coordinate of this point, since it is equal to the distance from the origin to the point whose coordinate is opposite number.
For example, the modulus of the number 9 is equal to 9, since the distance from the origin to the point with coordinate 9 is equal to nine. Let's give another example. The point with coordinate −3.25 is located at a distance of 3.25 from point O, so 
.
The stated definition of the modulus of a number is a special case of the definition of the modulus of the difference of two numbers.
Definition.
Modulus of the difference of two numbers a and b is equal to the distance between the points of the coordinate line with coordinates a and b.
That is, if points on the coordinate line A(a) and B(b) are given, then the distance from point A to point B is equal to the modulus of the difference between the numbers a and b. If we take point O (origin) as point B, then we get the definition of the modulus of a number given at the beginning of this paragraph.
Determining the modulus of a number using the arithmetic square root
Occasionally occurs determining modulus via arithmetic square root.
For example, let's calculate the moduli of the numbers −30 and based on this definition. We have. Similarly, we calculate the module of two thirds: 
.
The definition of the modulus of a number through the arithmetic square root is also consistent with the definition given in the first paragraph of this article. Let's show it. Let a be a positive number, and let −a be a negative number. Then 
And 
, if a=0 , then 
.
Module properties
The module has a number of characteristic results - module properties. Now we will present the main and most frequently used of them. When justifying these properties, we will rely on the definition of the modulus of a number in terms of distance.
Let's start with the most obvious property of the module - The modulus of a number cannot be a negative number. In literal form, this property has the form for any number a. This property is very easy to justify: the modulus of a number is a distance, and distance cannot be expressed as a negative number.
Let's move on to the next module property. The modulus of a number is zero if and only if this number is zero. The modulus of zero is zero by definition. Zero corresponds to the origin; no other point on the coordinate line corresponds to zero, since each real number is associated with a single point on the coordinate line. For the same reason, any number other than zero corresponds to a point different from the origin. And the distance from the origin to any point other than point O is not zero, since the distance between two points is zero if and only if these points coincide. The above reasoning proves that only the modulus of zero is equal to zero.
Go ahead. Opposite numbers have equal modules, that is, for any number a. Indeed, two points on the coordinate line, the coordinates of which are opposite numbers, are at the same distance from the origin, which means the modules of the opposite numbers are equal.
The following property of the module is: The modulus of the product of two numbers is equal to the product of the moduli of these numbers, that is, . By definition, the modulus of the product of numbers a and b is equal to either a·b if , or −(a·b) if . From the rules of multiplication of real numbers it follows that the product of the moduli of numbers a and b is equal to either a·b, , or −(a·b) if , which proves the property in question.
The modulus of the quotient of a divided by b is equal to the quotient of the modulus of a number divided by the modulus of b, that is, . Let us justify this property of the module. Since the quotient is equal to the product, then. By virtue of the previous property we have 
. All that remains is to use the equality , which is valid by virtue of the definition of the modulus of a number.
The following property of a module is written as an inequality: 
, a , b and c are arbitrary real numbers. The written inequality is nothing more than triangle inequality. To make this clear, let’s take points A(a), B(b), C(c) on the coordinate line, and consider a degenerate triangle ABC, whose vertices lie on the same line. By definition, the modulus of the difference is equal to the length of the segment AB, - the length of the segment AC, and - the length of the segment CB. Since the length of any side of a triangle does not exceed the sum of the lengths of the other two sides, then the inequality is true 
, therefore, the inequality is also true.
The inequality just proved is much more common in the form 
. The written inequality is usually considered as a separate property of the module with the formulation: “ The modulus of the sum of two numbers does not exceed the sum of the moduli of these numbers" But the inequality follows directly from the inequality if we put −b instead of b and take c=0.
Modulus of a complex number
Let's give definition of the modulus of a complex number. May it be given to us complex number, written in algebraic form, where x and y are some real numbers, representing, respectively, the real and imaginary parts of a given complex number z, and is the imaginary unit.
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 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 it right part 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 most easily explained 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 universal method solving equations with modulus. This is the interval method. But we will look at it later.
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Tochilkina Yulia
The work presents various methods for solving equations with a modulus.
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Municipal budgetary educational institution
"Average comprehensive school No. 59"
Equations with modulus
Abstract work
Performed 9A class student
MBOU "Secondary School No. 59" Barnaul
Tochilkina Yulia
Supervisor
Zakharova Lyudmila Vladimirovna,
mathematic teacher
MBOU "Secondary School No. 59" Barnaul
Barnaul 2015
Introduction
I'm in ninth grade. This academic year I will have to take the final certification for the basic school course. To prepare for the exam, we purchased the collection of D. A. Maltsev Mathematics. 9th grade. Looking through the collection, I discovered equations containing not only one, but also several modules. The teacher explained to me and my classmates that such equations are called “nested module” equations. This name seemed unusual to us, and the solution, at first glance, was quite complicated. This is how the topic for my work “Equations with modulus” appeared. I decided to study this topic more deeply, especially since it will be useful to me when taking exams at the end of the school year and I think it will be needed in grades 10 and 11. All of the above determines the relevance of the topic I have chosen.
Goal of the work :
- Consider various methods solving equations with modulus.
- Learn to solve equations containing a sign absolute value, various methods
To work on the topic, the following tasks were formulated:
Tasks:
- Explore theoretical material on the topic “Modulus of a real number”.
- Consider methods for solving equations and consolidate the acquired knowledge by solving problems.
- Apply the acquired knowledge when solving various equations containing the modulus sign in high school
Object of study:methods for solving equations with modulus
Subject of study:equations with modulus
Research methods:
Theoretical : study of literature on the research topic;
Internet - information.
Analysis information obtained from studying the literature; results obtained when solving equations with modulus different ways.
Comparison methods for solving equations is the subject of the rationality of their use when solving various equations with a modulus.
“We start thinking when we hit something.” Paul Valery.
1. Concepts and definitions.
The concept of “module” is widely used in many sections of the school mathematics course, for example, in the study of absolute and relative errors of an approximate number; in geometry and physics the concepts of a vector and its length (vector modulus) are studied. Module concepts applied in courses higher mathematics, physics and technical sciences studied in higher educational institutions.
The word “module” comes from the Latin word “modulus”, which means “measure”. This word has many meanings and is used not only in mathematics, physics and technology, but also in architecture, programming and other exact sciences.
It is believed that the term was proposed by Cotes, a student of Newton. The modulus sign was introduced in the 19th century by Weierstrass.
In architecture, a module is the initial unit of measurement established for a given architectural structure.
In technology, this is a term used in various fields of technology, used to designate various coefficients and quantities, for example, elastic modulus, engagement modulus...
In mathematics, modulus has several meanings, but I will consider it as the absolute value of a number.
Definition1: Modulus (absolute value) of a real number A this number itself is called if A ≥0, or the opposite number – and if A the modulus of zero is zero.
When solving equations with a modulus, it is convenient to use the properties of the modulus.
Let's consider the evidence of properties 5,6,7.
Statement 5. Equality │ a+b │=│ a │+│ b │ is true if av ≥ 0.
Proof. Indeed, after squaring both sides of this equality, we obtain │ a+b │²=│ a │²+2│ ab │+│ c │²,
a²+ 2 ab+b²=a²+ 2│ ab │+ b², from where │ ab │= ab
And the last equality will be true when av ≥0.
Statement 6. Equality │ a-c │=│ a │+│ c │ is true when av ≤0.
Proof. To prove it, it is sufficient in the equality
│ а+в │=│ а │+│ в │ replace в with - в, then а· (- в ) ≥0, whence ав ≤0.
Statement 7. Equality │ a │+│ b │= a+b performed at a ≥0 and b ≥0.
Proof . Having considered four cases a ≥0 and b ≥0; a ≥0 and c A in ≥0; A V a ≥0 and b ≥0.
(a-c) in ≥0.
Geometric interpretation
|a| - this is the distance on the coordinate line from the point with the coordinate A , to the origin.
|-a| |a|
A 0 a x
Geometric interpretation of the meaning of |a| clearly confirms that |-a|=|a|
If a 0, then on the coordinate line there are two points a and –a, equidistant from zero, the modules of which are equal.
If a=0, then on the coordinate line |a| represented by point 0.
Definition 2: An equation with a modulus is an equation containing a variable under the absolute value sign (under the modulus sign). For example: |x +3|=1
Definition 3: Solving an equation means finding all its roots, or proving that there are no roots.
2. Solution methods
From the definition and properties of a module, the main methods for solving equations with a module follow:
- "Expanding" a module (i.e. using a definition);
- Using the geometric meaning of the module (property 2);
- Graphic solution method;
- Using equivalent transformations (properties 4.6);
- Replacement of a variable (this uses property 5).
- Interval method.
I've decided enough a large number of examples, but in the work I present to your attention only a few, in my opinion, typical examples, solved in various ways, because the rest duplicate each other and in order to understand how to solve equations with a modulus there is no need to consider all the solved examples.
SOLVING EQUATIONS | f(x)| = a
Consider the equation | f(x)| = a, a R
An equation of this type can be solved by the definition of the modulus:
If A then the equation has no roots.
If a= 0, then the equation is equivalent to f(x)=0.
If a>0, then the equation is equivalent to the set
Example. Solve the equation |3x+2|=4.
Solution.
|3x+2|=4, then 3x+2=4,
3x+2= -4;
X=-2,
X=2/3
ANSWER: -2;2/3.
SOLVING EQUATIONS USING THE GEOMETRICAL PROPERTIES OF THE MODULE.
Example 1. Solve the equation /x-1/+/x-3/=6.
Solution.
Solving this equation means finding all such points on the numerical axis Ox, for each of which the sum of the distances from it to the points with coordinates 1 and 3 is equal to 6.
Not a single point from the segmentdoes not satisfy this condition, because the sum of the indicated distances is 2. Outside this segment there are two points: 5 and -1.
1 1 3 5
Answer: -1;5
Example 2. Solve equation |x 2 +x-5|+|x 2 +x-9|=10.
Solution.
Let us denote x 2 +x-5= a, then / a /+/ a-4 /=10. Let's find points on the Ox axis such that for each of them the sum of the distances to points with coordinates 0 and 4 is equal to 10. This condition is satisfied by -4 and 7.
3 0 4 7
So x 2 +x-5= 4 x 2 +x-5=7
X 2 +x-2=0 x 2 +x-12=0
X 1= 1, x 2= -2 x 1= -4, x 2= 3 Answer: -4;-2; 1; 3.
SOLVING EQUATIONS | f(x)| = | g(x)|.
- Since | a |=|in |, if a= in, then an equation of the form | f(x)| = | g(x )| equivalent to the totality
Example 1.
Solve the equation | x –2| = |3 – x |.
Solution.
This equation is equivalent to two equations:
x – 2 = 3 – x (1) and x – 2 = –3 + x (2)
2 x = 5 –2 = –3 – incorrect
X = 2.5 the equation has no solutions.
ANSWER: 2.5.
Example 2.
Solve equation |x 2 +3x-20|= |x 2 -3x+ 2|.
Solution.
Since both sides of the equation are non-negative, thensquaring is an equivalent transformation:
(x 2 +3x-20) 2 = (x 2 -3x+2) 2
(x 2 +3x-20) 2 - (x 2 -3x+2) 2 =0,
(x 2 +3x-20-x 2 +3x-2) (x 2 +3x-20+x 2 -3x+2)=0,
(6x-22)(2x 2 -18)=0,
6x-22=0 or 2x 2 -18=0;
X=22/6, x=3, x=-3.
X=11/3
Answer: -3; 3; 11/3.
SOLUTION OF EQUATIONS OF THE VIEW | f(x)| = g(x).
The difference between these equations and| f(x)| = a the fact that the right side is also a variable. And it can be both positive and negative. Therefore, you need to specially verify its non-negativity, because the modulus cannot be equal to negative number(property№1 )
1 way
Solution of the equation | f(x)| = g(x ) reduces to a set of solutions to the equationsand checking the fairness of inequality g(x )>0 for the found values of the unknown.
Method 2 (by module definition)
Since | f(x)| = g(x) if f(x) = 0; | f(x)| = - f(x) if f(x)
Example.
Solve equation |3 x –10| = x – 2.
Solution.
This equation is equivalent to the combination of two systems:
ANSWER: 3; 4.
SOLUTION OF EQUATIONS OF THE FORM |f 1 (x)|+|f 2 (x)|+…+|f n (x)|=g(x)
The solution of equations of this type is based on the definition of the modulus. For each function f 1 (x), f 2 (x), …, f n (x) it is necessary to find the domain of definition, its zeros and discontinuity points, dividing the general domain of definition into intervals, in each of which the functions f 1 (x), f 2 (x), …, f n (x) retain their sign. Next, using the definition of the module, for each of the found areas we obtain an equation that must be solved on this interval. This method is called "interval method»
Example.
Solve the equation |x-2|-3|x+4|=1.
Solution.
Let's find the points at which the submodular expressions are equal to zero
x-2=0, x+4=0,
x=2; x=-4.
Let's divide the number line into intervals x
Solving the equation comes down to solving three systems:
ANSWER: -15, -1.8.
GRAPHICAL METHOD FOR SOLVING EQUATIONS CONTAINING MODULE SIGN.
The graphical method of solving equations is approximate, since the accuracy depends on the selected unit segment, the thickness of the pencil, the angles at which the lines intersect, etc. But this method allows you to estimate how many solutions a given equation has.
Example. Solve graphically the equation |x - 2| + |x - 3| + |2x - 8| = 9
Solution. Let's construct graphs of functions in one coordinate system
y=|x - 2| + |x - 3| + |2x - 8| and y=9.
To construct a graph, you need to consider this function on each interval (-∞; 2); [ 3/2 ; ∞ )
Answer: (- ∞ ; 4/3] [ 3/2 ; ∞ )
We also used the method of equivalent transformations when solving the equations | f(x)| = | g(x)|.
EQUATIONS WITH A COMPLEX MODULE
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 various methods.
Example 1.
Solve the equation ||||x| – |–2| –1| –2| = 2.
Solution.
By definition of a module, we have:
Let's solve the first equation.
- ||| x |–2| –1| = 4
| x | – 2 = 5;
| x | = 7;
x = 7.
Let's solve the second equation.
- ||| x | –2| –1| = 0,
|| x | –2| = 1,
| x | –2 = 1,
| x | = 3 and | x | = 1,
x = 3; x = 1.
Answer: 1; 3; 7.
Example 2.
Solve the equation |2 – |x + 1|| = 3.
Solution.
Let's solve the equation by introducing a new variable.
Let | x + 1| = y, then |2 – y | = 3, from here
Let's do the reverse replacement:
(1) | x + 1| = –1 – no solutions.
(2) | x + 1| = 5
ANSWER: –6; 4.
Example3.
How many roots does the equation have | 2 | x | -6 | = 5 - x?
Solution. Let's solve the equation using equivalence schemes.
Equation | 2 | x | -6 | = 5 is equivalent to the system:
Solving equations and inequalities with modulus often causes difficulties. However, if you understand well what it is the absolute value of a number, And how to correctly expand expressions containing a modulus sign, then the presence in the equation expression under the modulus sign, ceases to be an obstacle to its solution.
A little theory. Each number has two characteristics: the absolute value of the number and its sign.
For example, the number +5, or simply 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 a number x is called the modulus of the number and is denoted by |x|.
As we 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 appear under the modulus 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 a module according to the module expansion rule.
Then our equation or inequality becomes 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 look at 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 received two numerical intervals: x≥3 and x<3.
Let us consider into which equations the original equation is transformed on each interval:
A) For x≥3 |x-3|=x-3, and our wounding has the form:
Attention! This equation exists only on the interval x≥3!
Let's open the brackets and present similar terms:
and solve this equation.
This equation has roots:
x 1 =0, x 2 =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 is satisfied only by 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 present similar terms. We get the equation:
x 1 =2, x 2 =3
Attention! since the equation 3-x=-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.
Among examples per module Often there are equations where you need to find module roots in a 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’s easier to look for a solution equations with modules graphically. Below is the method opening of double modules using examples common in practice. Understand the algorithm for calculating equations with modules well, so that you don’t have problems on quizzes, tests, and just to know.
Example 1. Solve the equation modulo |3|x|-5|=-2x-2.
Solution: Always start opening equations from the internal module
|x|=0 <->x=0.
At the point x=0, the equation with modulus is divided by 2.
At x< 0
подмодульная функция отрицательная, поэтому при раскрытии знак меняем на противоположный
|-3x-5|=-2x-2.
For x>0 or equal, expanding the module 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 limitation entirely belongs to the area in which we are solving. Let's move variables and constants to 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 moduli for positive variables
|3x-5|=-2x-2.
Expanding the module we get two systems of equations 
From the first equation, which is common to the 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 (there are no points of intersection). So the only roots of a module with a module are the values
x=-3; x=-1.4.
Example 2. Solve the equation with modulus ||x-1|-2|=3x-4.
Solution: Let's start by opening the internal module
|x-1|=0 <=>x=1.
A submodular function changes sign at one. For smaller values it is negative, for 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 modulus equation; it must be greater than zero.
3x-4>=0 -> x>=4/3.
This means that there is no need to solve the first equation, since it was 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 received two values, the first of which is rejected because it does not belong to the required interval. Finally, the equation has one solution x=7/4.
Example 3. Solve the equation with modulus ||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 number line into two intervals. Respectively, submodular function changes sign when passing through 2.5. Let us write down the condition for the solution on the right side of the equation with modulus.
x+3>=0 -> x>=-3.
So the solution can be values no less than (-3) . Let's expand the module for a negative value of the internal module
|-(2x-5)-1|=x+3;
|-2x+4|=x+3.
This module will also give 2 equations when expanded
-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 .
We reject the value x=7, since we were looking for a solution in the interval [-3;2.5]. Now we open the internal 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 get 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 modulus x=9, and there are two in total (x=1/3). By substitution you can check the correctness of the calculations performed
Answer: x=1/3; x=9.
Example 4. Find solutions to the double module ||3x-1|-5|=2x-3.
Solution: Let's expand the internal module of the equation
|3x-1|=0 <=>x=1/3.
The point x=2.5 divides the number line into two intervals, and given equation for two cases. We write down the condition for the solution based on the form of the equation on the right side
2x-3>=0 -> x>=3/2=1.5.
It follows that we are interested in values >=1.5. Thus modular equation consider on 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 moduli. Next, we will expand the module for x>2.5. We get the following equation
|3x-1-5|=2x-3;
|3x-6|=2x-3.
Expanding the module, we get 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 correspond to the condition x>2.5, we reject it.
Finally we have one root of the equation with moduli x=3.
Performing a check
||3*3-1|-5|=2*3-3 3=3
.
The root of the equation with the modulus was calculated correctly.
Answer: x=1/3; x=9.