One of the most difficult topics for students, this 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 with this one it’s far from the best? complex concept How does the module have 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. 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
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, module or absolute value number is the distance from this point to the origin of the number axis. The distance is always specified as a positive number. Thus, the module 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.
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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, 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.
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MBOU Secondary School No. 17, Ivanovo
« Equations with modulus"
Methodological development
Compiled
math teacher
Lebedeva N.V.20010
Explanatory note
Chapter 1. Introduction
Section 2. Basic properties Section 3. Geometric interpretation of the concept of modulus of a number Section 4. Graph of the function y = |x| Section 5. ConventionsChapter 2. Solving equations containing a modulus
Section 1. Equations of the form |F(x)| = m (simplest) Section 2. Equations of the form F(|x|) = m Section 3. Equations of the form |F(x)| = G(x) Section 4. Equations of the form |F(x)| = ± F(x) (most beautiful) Section 5. Equations of the form |F(x)| = |G(x)| Section 6. Examples of solving non-standard equations Section 7. Equations of the form |F(x)| + |G(x)| = 0 Section 8. Equations of the form |a 1 x ± b 1 | ± |a 2 x ± in 2 | ± …|a n x ± in n | = m Section 9. Equations containing several modulesChapter 3. Examples of solving various equations with modulus.
Section 1. Trigonometric equations Section 2. Exponential equations Section 3. Logarithmic equations Section 4. Irrational equations Section 5. Assignments increased complexity Answers to the exercises BibliographyExplanatory note.
The concept of absolute value (modulus) of a real number is one of its essential characteristics. This concept is widespread in various sections of physical, mathematical and technical sciences. In the practice of teaching mathematics courses in high school in accordance with the Program of the Ministry of Defense of the Russian Federation, the concept of “absolute value of a number” appears repeatedly: in the 6th grade, the definition of a module is introduced, its geometric meaning; in the 8th grade, the concept of absolute error is formed, the solution of the simplest equations and inequalities containing a modulus is considered, and the properties of the arithmetic square root are studied; in 11th grade the concept is found in the section “Root n-th degree." Teaching experience shows that students often encounter difficulties when solving tasks that require knowledge of this material, and often they skip it without starting to implement it. In the texts exam tasks Similar tasks are also included for the 9th and 11th grade courses. In addition, the requirements that universities place on school graduates differ, namely, more high level than the requirements of the school curriculum. modern society It is very important to develop a mathematical thinking style, which manifests itself in certain mental skills. In the process of solving problems with modules, the ability to use techniques such as generalization and specification, analysis, classification and systematization, and analogy is required. Solving such tasks allows you to test your knowledge of the main sections school course, level logical thinking, initial research skills. this work is devoted to one of the sections - solving equations containing a module. It consists of three chapters. The first chapter introduces basic concepts and the most important theoretical considerations. The second chapter proposes nine main types of equations containing a module, discusses methods for solving them, and examines examples different levels difficulties. The third chapter offers more complex and non-standard equations (trigonometric, exponential, logarithmic and irrational). For each type of equation there are exercises for independent decision(answers and instructions are attached). The main purpose of this work is to provide methodological assistance to teachers in preparing for lessons and in organizing elective courses. The material can also be used as teaching aid for high school students. The tasks proposed in the work are interesting and not always easy to solve, which allows you to learning motivationstudents to become more aware, test their abilities, and improve the level of preparation of school graduates for entering universities. A differentiated selection of the proposed exercises involves a transition from the reproductive level of mastering the material to the creative one, as well as the opportunity to teach how to apply your knowledge when solving non-standard problems.
Chapter 1. Introduction. .
Section 1. Determination of absolute value : Definition The absolute value (modulus) of a real number A The absolute value (modulus) of a real number a non-negative number is called: or -A. │ The absolute value (modulus) of a real number │ Designation:│ The entry reads as follows: “modulus of the number a” or “absolute value of the number a”
│a, if a > 0
│ a│ = │ 0, if a = 0 (1)- and, if a 1) │2,5│ = 2,5 2) │-7│ = 7 3) │1 - √2│ = √2 – 1
- Examples:
a) │x - 8│, if x > 12 b) │2x + 3│, if x ≤ -2 │x – 8│= x – 8 │ 2x + 3│= - 2x – 3
Section 2. Basic properties. Let's consider the basic properties of absolute value. Property #1: Opposite numbers have equal modules, i.e.│а│=│- а│ Let us show that the equality is true. Let's write down the definition of the number : │- A= (2) - a│ The absolute value (modulus) of a real number Let's compare sets (1) and (2). Obviously, the definitions of the absolute values of numbers Let us show that the equality is true. Let's write down the definition of the number match up. Hence, Opposite numbers have equal modules, i.e.When considering the following properties, we will limit ourselves to their formulation, since their proof is given in Property #2: The absolute value of the sum of a finite number of real numbers does not exceed the sum of the absolute values of the terms: │а 1 + а 2 +…+ а n │ ≤│а 1 │+│а 2 │+ … + │а n │ Property #3: The absolute value of the difference between two real numbers does not exceed the sum of their absolute values: │а - в│ ≤│а│+│в│ Property #4: The absolute value of the product of a finite number of real numbers is equal to the product of the absolute values of the factors: │а·в│=│а│·│в│ Property #5: The absolute value of the quotient of real numbers is equal to the quotient of their absolute values:
Section 3. Geometric interpretation of the concept of modulus of a number.
Each real number can be associated with a point on the number line, which will be a geometric image of this real number. Each point on the number line corresponds to its distance from the origin, i.e. the length of the segment from the origin to a given point. This distance is always considered as a non-negative value. Therefore, the length of the corresponding segment will be the geometric interpretation of the absolute value of a given real number
The presented geometric illustration clearly confirms property No. 1, i.e. the moduli of opposite numbers are equal. From here the validity of the equality is easily understood: │х – а│= │а – x│. The solution to the equation │х│= m, where m ≥ 0, namely x 1.2 = ± m, also becomes more obvious. - and, if a 1) │х│= 4 x 1.2 = ± 4 2) │х - 3│= 1
x 1.2 = 2; 4 Section 4. Graph of the function y = │х│
The domain of this function is all real numbers.Section 5. Conventions.
In the future, when considering examples of solving equations, the following will be used symbols: ( - sign of the system [ - sign of the totality When solving a system of equations (inequalities), the intersection of solutions of the equations (inequalities) included in the system is found. When solving a set of equations (inequalities), the union of solutions included in the set of equations (inequalities) is found.Chapter 2. Solving equations containing a modulus.
In this chapter we will look at algebraic methods for solving equations containing one or more modules.Section 1. Equations of the form │F (x)│= m
An equation of this type is called the simplest. It has a solution if and only if m ≥ 0. By definition of the modulus, the original equation is equivalent to a set of two equations: │ F(x)│=m
- and, if a
№1. Solve the equation: │7х - 2│= 9
Answer: x 1
= - 1; X 2
= 1
4
/
7
№2
│x 2 + 3x + 1│= 1
x 2 + 3x + 2 = 0 x 2 +3x = 0 x 1 = -1; x 2 = -2 x (x + 3) = 0 x 1 = 0; x 2 = -3 Answer: the sum of the roots is - 2.№3
│x 4 -5x 2 + 2│= 2 x 4 – 5x 2 = 0 x 4 – 5x 2 + 4 = 0 x 2 (x 2 – 5) = 0 let’s denote x 2 = m, m ≥ 0 x = 0 ; ±√5 m 2 – 5m + 4 = 0 m = 1; 4 – both values satisfy the condition m ≥ 0 x 2 = 1 x 2 = 4 x = ± 1 x = ± 2 Answer: number of roots of equation 7. Exercises:
№1. Solve the equation and indicate the sum of the roots: │х - 5│= 3 №2 . Solve the equation and indicate the smaller root: │x 2 + x│= 0 №3 . Solve the equation and indicate the larger root: │x 2 – 5x + 4│= 4 №4 .Solve the equation and indicate the whole root: │2x 2 – 7x + 6│= 1 №5 .Solve the equation and indicate the number of roots: │x 4 – 13x 2 + 50│= 14
Section 2. Equations of the form F(│х│) = m
The function argument on the left side is under the modulus sign, and the right side is independent of the variable. Let's consider two ways to solve equations of this type. 1 way: By definition of absolute value, the original equation is equivalent to the combination of two systems. In each of which a condition is imposed on a submodular expression. F(│х│) =m
Since the function F(│x│) is even throughout the entire domain of definition, the roots of the equations F(x) = m and F(- x) = m are pairs of opposite numbers. Therefore, it is enough to solve one of the systems (when considering examples in this way, the solution to one system will be given). Method 2: Application of the method of introducing a new variable. In this case, the notation │x│= a is introduced, where a ≥ 0. This method less voluminous in design. - and, if a №1 . Solve the equation: 3x 2 – 4│x│= - 1 Let’s use the introduction of a new variable. Let us denote │x│= a, where a ≥ 0. We obtain the equation 3a 2 - 4a + 1 = 0 D = 16 – 12 = 4 a 1 = 1 a 2 = 1 / 3 Return to the original variable: │x│=1 and │х│= 1/3. Each equation has two roots. Answer: x 1 = 1; X 2 = - 1; X 3 = 1 / 3 ; X 4 = - 1 / 3 . №2. Solve the equation: 5x 2 + 3│x│- 1 = 1 / 2 │x│ + 3x 2
Let's find the solution to the first system of the population: 4x 2 + 5x – 2 =0 D = 57 x 1 = -5+√57 / 8 x 2 = -5-√57 / 8 Note that x 2 does not satisfy the condition x ≥ 0. Solution the second system will be the number opposite to the value x 1. Answer: x 1
=
-5+√57
/
8
; X 2
=
5-√57
/
8
.№3
.
Solve the equation: x 4 – │х│= 0 Let us denote │х│= a, where a ≥ 0. We get the equation a 4 – a = 0 a · (a 3 – 1) = 0 a 1 = 0 a 2 = 1 Return to the original variable: │х│=0 and │х│= 1 x = 0; ± 1 Answer: x 1
= 0; X 2
= 1; X 3
= - 1.
Exercises: №6. Solve the equation: 2│х│ - 4.5 = 5 – 3 / 8 │х│ №7 . Solve the equation, indicate the number of roots in your answer: 3x 2 - 7│x│ + 2 = 0 №8 . Solve the equation, indicate integer solutions in your answer: x 4 + │x│ - 2 = 0
Section 3. Equations of the form │F(x)│ = G(x)
The right-hand side of an equation of this type depends on a variable and, therefore, has a solution if and only if the right-hand side is a function G(x) ≥ 0. The original equation can be solved in two ways: 1 way: Standard, based on the disclosure of a module based on its definition and consists of an equivalent transition to a combination of two systems. │ F(x)│ =G(X)
This method can be rationally used in the case of a complex expression for the function G(x) and a less complex one for the function F(x), since it is assumed that inequalities with the function F(x) will be solved. Method 2: Consists in the transition to an equivalent system in which a condition is imposed on the right side. │ F(x)│=
G(x)
This method is more convenient to use if the expression for the function G(x) is less complex than for the function F(x), since the solution to the inequality G(x) ≥ 0 is assumed. In addition, in the case of several modules, it is recommended to use the second option. - and, if a
№1.
Solve the equation: │x + 2│= 6 -2x 
(1 way) Answer: x = 1 1
/
3
№2.
│x 2 – 2x - 1│= 2 (x + 1)
(2 way) Answer: The product of roots is 3.№3. Solve the equation and indicate the sum of the roots in your answer:
│x - 6│= x 2 - 5x + 9
Answer: the sum of the roots is 4.
Exercises: №9. │x + 4│= - 3x №10. Solve the equation, indicate the number of solutions in your answer:│x 2 + x - 1│= 2x – 1 №11 . Solve the equation, indicate the product of the roots in your answer:│x + 3│= x 2 + x – 6
Section 4. Equations of the form │F(x)│= F(x) and │F(x)│= - F(x)
Equations of this type are sometimes called “the most beautiful.” Since the right-hand side of the equations depends on the variable, solutions exist if and only if the right-hand side is non-negative. Therefore, the original equations are equivalent to the inequalities:│F(x)│= F(x) F(x) ≥ 0 and │F(x)│= - F(x) F(x) - and, if a №1 . Solve the equation, indicate the smaller whole root in your answer: │5x - 3│= 5x – 3 5x – 3 ≥ 0 5x ≥ 3 x ≥ 0.6 Answer: x = 1№2. Solve the equation, indicate the length of the interval in your answer: │х 2 - 9│= 9 – x 2 x 2 – 9 ≤ 0 (x – 3) (x + 3) ≤ 0 [- 3; 3] Answer: the length of the gap is 6.№3 . Solve the equation and indicate the number of integer solutions in your answer: │2 + x – x 2 │= 2 + x – x 2 2 + x – x 2 ≥ 0 x 2 – x – 2 ≤ 0 [- 1; 2] Answer: 4 whole solutions.№4 . Solve the equation and indicate the largest root in your answer:
│4 – x -
│= 4 – x – 
x 2 – 5x + 5 = 0 D = 5 x 1.2 = 
≈ 1,4Answer: x = 3.
Exercises:
№12.
Solve the equation, indicate the whole root in your answer: │x 2 + 6x + 8│= x 2 + 6x + 8 №13.
Solve the equation, indicate the number of integer solutions in your answer: │13x – x 2 - 36│+ x 2 – 13x + 36 = 0 №14.
Solve the equation; in your answer, indicate an integer that is not the root of the equation: 
Section 5. Equations of the form │F(x)│= │G(x)│
Since both sides of the equation are non-negative, the solution involves considering two cases: submodular expressions are equal or opposite in sign. Therefore, the original equation is equivalent to the combination of two equations: │ F(x)│= │ G(x)│
- and, if a
№1.
Solve the equation, indicate the whole root in your answer: │x + 3│=│2x - 1│ 
Answer: whole root x = 4.№2.
Solve the equation: │
x – x 2 - 1│=│2x – 3 – x 2 │ 
Answer: x = 2.№3
.
Solve the equation and indicate the product of the roots in your answer: 
Root equations 4x 2 + 2x – 1 = 0 x 1.2 = - 1±√5 / 4 Answer: the product of the roots is – 0.25. Exercises:
№15
. Solve the equation and indicate the whole solution in your answer: │x 2 – 3x + 2│= │x 2 + 6x - 1│ №16.
Solve the equation, indicate the smaller root in your answer:│5x - 3│=│7 - x│ №17
. Solve the equation and indicate the sum of the roots in your answer: 
Section 6. Examples of solving non-standard equations
In this section we will look at examples of non-standard equations, when solving which the absolute value of the expression is revealed by definition. - and, if a№1.
Solve the equation, indicate the sum of the roots in your answer: x · │x│- 5x – 6 = 0 
Answer: the sum of the roots is 1 №2.
.
Solve the equation, indicate the smaller root in your answer: x 2 - 4x · 
- 5 = 0 
Answer: smaller root x = - 5. №3.
Solve the equation: 
Answer: x = -1. Exercises:
№18.
Solve the equation and indicate the sum of the roots: x · │3x + 5│= 3x 2 + 4x + 3
№19.
Solve the equation: x 2 – 3x = 
№20.
Solve the equation: 
Section 7. Equations of the form │F(x)│+│G(x)│=0
It is easy to notice that on the left side of the equation of this type is the sum of non-negative quantities. Therefore, the original equation has a solution if and only if both terms are equal to zero at the same time. The equation is equivalent to the system of equations: │ F(x)│+│ G(x)│=0
- and, if a
№1
. Solve the equation: 
Answer: x = 2. №2.
Solve the equation: Answer: x = 1. Exercises:
№21.
Solve the equation: №22
. Solve the equation and indicate the sum of the roots in your answer: №23
. Solve the equation and indicate the number of solutions in your answer: Section 8. Equations of the form │a 1 x + b 1 │±│a 2 x + b 2 │± … │a n x +b n │= m
To solve equations of this type, the interval method is used. If we solve it by sequential expansion of modules, we get n sets of systems, which is very cumbersome and inconvenient. Let's consider the interval method algorithm: 1). Find variable values X, for which each module is equal to zero (zeros of submodular expressions):
2). Mark the found values on a number line, which is divided into intervals (the number of intervals is respectively equal to n+1
) 3). Determine with what sign each module is revealed at each of the resulting intervals (when formulating a solution, you can use a number line, marking the signs on it) 4). The original equation is equivalent to the aggregate n+1
systems, in each of which the variable’s membership is indicated X one of the intervals. - and, if a
№1
. Solve the equation and indicate the largest root in your answer: 
1). Let's find the zeros of the submodular expressions: x = 2; x = -3 2). Let's mark the found values on the number line and determine with what sign each module is revealed on the resulting intervals: x – 2 x – 2 x – 2 - - + - 3 2 x 2x + 6 2x + 6 2x + 6 - + + 3)
- no solutions The equation has two roots. Answer: the largest root x = 2. №2.
Solve the equation and provide the whole root in your answer: 
1). Let's find the zeros of the submodular expressions: x = 1.5; x = - 1 2). Let's mark the found values on the number line and determine with what sign each module is revealed on the resulting intervals: x + 1 x + 1 x + 1 - + + -1 1.5 x 2x – 3 2x – 3 2x – 3 - - +
3).
The last system has no solutions, therefore the equation has two roots. When solving the equation, you should pay attention to the “-” sign in front of the second module. Answer: whole root x = 7. №3.
Solve the equation, indicate the sum of the roots in your answer: 1). Let's find the zeros of the submodular expressions: x = 5; x = 1; x = - 2 2). Let's mark the found values on the number line and determine with what sign each module is revealed at the resulting intervals: x – 5 x – 5 x – 5 x – 5 - - - + -2 1 5 x x – 1 x – 1 x – 1 x – 1 - - + + x + 2 x + 2 x + 2 x + 2 - + + +
3).
The equation has two roots x = 0 and 2. Answer: the sum of the roots is 2. №4
.
Solve the equation: 1). Let's find the zeros of the submodular expressions: x = 1; x = 2; x = 3. 2). Let us determine with what sign each module is revealed on the resulting intervals. 3). 
Let's combine the solutions of the first three systems. Answer: ; x = 5.Exercises: №24. Solve the equation:
№25.
Solve the equation and indicate the sum of the roots in your answer: №26.
Solve the equation and indicate the smaller root in your answer: №27.
Solve the equation and indicate the larger root in your answer: Section 9. Equations containing several modules
Equations containing multiple modules assume the presence of absolute values in submodular expressions. The basic principle for solving equations of this type is the sequential disclosure of modules, starting with the “external” one. During the solution, the techniques discussed in sections No. 1, No. 3 are used.- and, if a
№1.
Solve the equation: 
Answer: x = 1; - eleven. №2.
Solve the equation:
Answer: x = 0; 4; - 4. №3.
Solve the equation and indicate the product of the roots in your answer: 
Answer: the product of the roots is – 8. №4.
Solve the equation: 
Let us denote the equations of the population (1)
Let's compare sets (1) and (2). Obviously, the definitions of the absolute values of numbers (2)
and consider the solution to each of them separately for ease of design. Since both equations contain more than one module, it is more convenient to carry out an equivalent transition to sets of systems. (1)
(2)
Answer:
Exercises:
№36.
Solve the equation, indicate the sum of the roots in your answer: 5 │3x-5│ = 25 x №37.
Solve the equation, if there is more than one root, indicate the sum of the roots in your answer: │x + 2│ x – 3x – 10 = 1 №38.
Solve the equation: 3 │2x -4│ = 9 │x│ №39.
Solve the equation and indicate the number of roots in your answer: 2 │ sin x│ = √2 №40
. Solve the equation and indicate the number of roots in your answer:
Section 3. Logarithmic equations.
Before solving the following equations, it is necessary to review the properties of logarithms and the logarithmic function. - and, if a №1. Solve the equation, indicate the product of the roots in your answer: log 2 (x+1) 2 + log 2 │x+1│ = 6 O.D.Z. x+1≠0 x≠ - 1Case 1: if x ≥ - 1, then log 2 (x+1) 2 + log 2 (x+1) = 6 log 2 (x+1) 3 = log 2 2 6 (x+1) 3 = 2 6 x+1 = 4 x = 3 – satisfies the condition x ≥ - 1 2 case: if x log 2 (x+1) 2 + log 2 (-x-1) = 6 log 2 (x+1) 2 + log 2 (-(x+1)) = 6 log 2 (-(x+1) 3) = log 2 2 6- (x+1) 3 = 2 6- (x+1) = 4 x = - 5 – satisfies condition x - 1
Answer: the product of the roots is – 15.
№2.
Solve the equation, indicate the sum of the roots in your answer: lg 
O.D.Z. 
Answer: the sum of the roots is 0.5.
№3.
Solve the equation: log 5 
O.D.Z. 
Answer: x = 9. №4.
Solve the equation: │2 + log 0.2 x│+ 3 = │1 + log 5 x│ O.D.Z. x > 0 Let's use the formula for moving to another base. │2 - log 5 x│+ 3 = │1 + log 5 x│
│2 - log 5 x│- │1 + log 5 x│= - 3 Let's find the zeros of the submodular expressions: x = 25; x = These numbers divide the range of acceptable values into three intervals, so the equation is equivalent to a set of three systems. 
Answer:
Another important fact: modulus is never negative. Whatever number we take - be it positive or negative - its modulus always turns out to be positive (or, in extreme cases, zero). This is why the modulus is often called the absolute value of a number.
In addition, if we combine the definition of the modulus for a positive and negative number, we obtain a global definition of the modulus for all numbers. Namely: the modulus of a number is equal to the number itself if the number is positive (or zero), or equal to the opposite number if the number is negative. You can write this as a formula:
There is also a modulus of zero, but it is always equal to zero. In addition, zero singular, which has no opposite.
Thus, if we consider the function $y=\left| x \right|$ and try to draw its graph, you will get something like this:
Modulus graph and example of solving the equation
From this picture it is immediately clear that $\left| -m \right|=\left| m \right|$, and the modulus graph never falls below the x-axis. But that’s not all: the red line marks the straight line $y=a$, which, for positive $a$, gives us two roots at once: $((x)_(1))$ and $((x)_(2)) $, but we'll talk about that later. :)
In addition to the purely algebraic definition, there is a geometric one. Let's say there are two points on the number line: $((x)_(1))$ and $((x)_(2))$. In this case, the expression $\left| ((x)_(1))-((x)_(2)) \right|$ is simply the distance between the specified points. Or, if you prefer, the length of the segment connecting these points:
Modulus is the distance between points on a number line This definition also implies that the modulus is always non-negative. But enough definitions and theory - let's move on to real equations. :)
Basic formula
Okay, we've sorted out the definition. But that didn’t make it any easier. How to solve equations containing this very module?
Calm, just calm. Let's start with the simplest things. Consider something like this:
\[\left| x\right|=3\]
So the modulus of $x$ is 3. What could $x$ be equal to? Well, judging by the definition, we are quite happy with $x=3$. Really:
\[\left| 3\right|=3\]
Are there other numbers? Cap seems to be hinting that there is. For example, $x=-3$ is also $\left| -3 \right|=3$, i.e. the required equality is satisfied.
So maybe if we search and think, we will find more numbers? But break it off: more numbers No. Equation $\left| x \right|=3$ has only two roots: $x=3$ and $x=-3$.
Now let's complicate the task a little. Let the function $f\left(x \right)$ hang out under the modulus sign instead of the variable $x$, and put an arbitrary number $a$ in place of the triple on the right. We get the equation:
\[\left| f\left(x \right) \right|=a\]
So how can we solve this? Let me remind you: $f\left(x \right)$ is an arbitrary function, $a$ is any number. Those. Anything at all! For example:
\[\left| 2x+1 \right|=5\]
\[\left| 10x-5 \right|=-65\]
Let's pay attention to the second equation. You can immediately say about him: he has no roots. Why? Everything is correct: because it requires that the modulus be equal to a negative number, which never happens, since we already know that the modulus is always a positive number or, in extreme cases, zero.
But with the first equation everything is more fun. There are two options: either there is a positive expression under the modulus sign, and then $\left| 2x+1 \right|=2x+1$, or this expression is still negative, and then $\left| 2x+1 \right|=-\left(2x+1 \right)=-2x-1$. In the first case, our equation will be rewritten as follows:
\[\left| 2x+1 \right|=5\Rightarrow 2x+1=5\]
And suddenly it turns out that the submodular expression $2x+1$ is really positive - it is equal to the number 5. That is we can safely solve this equation - the resulting root will be a piece of the answer:
Those who are particularly distrustful can try to substitute the found root into the original equation and make sure that there really is a positive number under the modulus.
Now let's look at the case of a negative submodular expression:
\[\left\( \begin(align)& \left| 2x+1 \right|=5 \\& 2x+1 \lt 0 \\\end(align) \right.\Rightarrow -2x-1=5 \Rightarrow 2x+1=-5\]
Oops! Again, everything is clear: we assumed that $2x+1 \lt 0$, and as a result we got that $2x+1=-5$ - indeed, this expression is less than zero. We solve the resulting equation, while already knowing for sure that the found root will suit us:
In total, we again received two answers: $x=2$ and $x=3$. Yes, the amount of calculations turned out to be a little larger than in the very simple equation $\left| x \right|=3$, but nothing fundamentally has changed. So maybe there is some universal algorithm?
Yes, such an algorithm exists. And now we will analyze it.
Getting rid of the modulus sign
Let us be given the equation $\left| f\left(x \right) \right|=a$, and $a\ge 0$ (otherwise, as we already know, there are no roots). Then you can get rid of the modulus sign using the following rule:
\[\left| f\left(x \right) \right|=a\Rightarrow f\left(x \right)=\pm a\]
Thus, our equation with a modulus splits into two, but without a modulus. That's all the technology is! Let's try to solve a couple of equations. Let's start with this
\[\left| 5x+4 \right|=10\Rightarrow 5x+4=\pm 10\]
Let’s consider separately when there is a ten plus on the right, and separately when there is a minus. We have:
\[\begin(align)& 5x+4=10\Rightarrow 5x=6\Rightarrow x=\frac(6)(5)=1,2; \\& 5x+4=-10\Rightarrow 5x=-14\Rightarrow x=-\frac(14)(5)=-2.8. \\\end(align)\]
That's all! We got two roots: $x=1.2$ and $x=-2.8$. The entire solution took literally two lines.
Ok, no question, let's look at something a little more serious:
\[\left| 7-5x\right|=13\]
Again we open the module with plus and minus:
\[\begin(align)& 7-5x=13\Rightarrow -5x=6\Rightarrow x=-\frac(6)(5)=-1,2; \\& 7-5x=-13\Rightarrow -5x=-20\Rightarrow x=4. \\\end(align)\]
A couple of lines again - and the answer is ready! As I said, there is nothing complicated about modules. You just need to remember a few rules. Therefore, we move on and begin with truly more complex tasks.
The case of a right-hand side variable
Now consider this equation:
\[\left| 3x-2 \right|=2x\]
This equation is fundamentally different from all previous ones. How? And the fact that to the right of the equal sign is the expression $2x$ - and we cannot know in advance whether it is positive or negative.
What to do in this case? First, we must understand once and for all that if the right side of the equation turns out to be negative, then the equation will have no roots- we already know that the module cannot be equal to a negative number.
And secondly, if the right part is still positive (or equal to zero), then you can act in exactly the same way as before: simply open the module separately with a plus sign and separately with a minus sign.
Thus, we formulate a rule for arbitrary functions $f\left(x \right)$ and $g\left(x \right)$ :
\[\left| f\left(x \right) \right|=g\left(x \right)\Rightarrow \left\( \begin(align)& f\left(x \right)=\pm g\left(x \right ), \\& g\left(x \right)\ge 0. \\\end(align) \right.\]
In relation to our equation we get:
\[\left| 3x-2 \right|=2x\Rightarrow \left\( \begin(align)& 3x-2=\pm 2x, \\& 2x\ge 0. \\\end(align) \right.\]
Well, we will somehow cope with the requirement $2x\ge 0$. In the end, we can stupidly substitute the roots that we get from the first equation and check whether the inequality holds or not.
So let’s solve the equation itself:
\[\begin(align)& 3x-2=2\Rightarrow 3x=4\Rightarrow x=\frac(4)(3); \\& 3x-2=-2\Rightarrow 3x=0\Rightarrow x=0. \\\end(align)\]
Well, which of these two roots satisfies the requirement $2x\ge 0$? Yes both! Therefore, the answer will be two numbers: $x=(4)/(3)\;$ and $x=0$. That's the solution. :)
I suspect that some of the students are already starting to get bored? Well, let's look at an even more complex equation:
\[\left| ((x)^(3))-3((x)^(2))+x \right|=x-((x)^(3))\]
Although it looks evil, in fact it is still the same equation of the form “modulus equals function”:
\[\left| f\left(x \right) \right|=g\left(x \right)\]
And it is solved in exactly the same way:
\[\left| ((x)^(3))-3((x)^(2))+x \right|=x-((x)^(3))\Rightarrow \left\( \begin(align)& ( (x)^(3))-3((x)^(2))+x=\pm \left(x-((x)^(3)) \right), \\& x-((x )^(3))\ge 0. \\\end(align) \right.\]
We will deal with inequality later - it is somehow too evil (in fact, it is simple, but we will not solve it). For now, it’s better to deal with the resulting equations. Let's consider the first case - this is when the module is expanded with a plus sign:
\[((x)^(3))-3((x)^(2))+x=x-((x)^(3))\]
Well, it’s a no brainer that you need to collect everything from the left, bring similar ones and see what happens. And this is what happens:
\[\begin(align)& ((x)^(3))-3((x)^(2))+x=x-((x)^(3)); \\& 2((x)^(3))-3((x)^(2))=0; \\\end(align)\]
We take the common factor $((x)^(2))$ out of brackets and get a very simple equation:
\[((x)^(2))\left(2x-3 \right)=0\Rightarrow \left[ \begin(align)& ((x)^(2))=0 \\& 2x-3 =0 \\\end(align) \right.\]
\[((x)_(1))=0;\quad ((x)_(2))=\frac(3)(2)=1.5.\]
Here we used important property product, for the sake of which we factored the original polynomial: the product is equal to zero when at least one of the factors is equal to zero.
Now let’s deal with the second equation in exactly the same way, which is obtained by expanding the module with a minus sign:
\[\begin(align)& ((x)^(3))-3((x)^(2))+x=-\left(x-((x)^(3)) \right); \\& ((x)^(3))-3((x)^(2))+x=-x+((x)^(3)); \\& -3((x)^(2))+2x=0; \\& x\left(-3x+2 \right)=0. \\\end(align)\]
Again the same thing: the product is equal to zero when at least one of the factors is equal to zero. We have:
\[\left[ \begin(align)& x=0 \\& -3x+2=0 \\\end(align) \right.\]
Well, we got three roots: $x=0$, $x=1.5$ and $x=(2)/(3)\;$. Well, which of this set will go into the final answer? To do this, remember that we have an additional constraint in the form of inequality:
How to take this requirement into account? Let’s just substitute the found roots and check whether the inequality holds for these $x$ or not. We have:
\[\begin(align)& x=0\Rightarrow x-((x)^(3))=0-0=0\ge 0; \\& x=1.5\Rightarrow x-((x)^(3))=1.5-((1.5)^(3)) \lt 0; \\& x=\frac(2)(3)\Rightarrow x-((x)^(3))=\frac(2)(3)-\frac(8)(27)=\frac(10) (27)\ge 0; \\\end(align)\]
Thus, the root $x=1.5$ does not suit us. And in response there will be only two roots:
\[((x)_(1))=0;\quad ((x)_(2))=\frac(2)(3).\]
As you can see, even in this case there was nothing complicated - equations with modules are always solved using an algorithm. You just need to have a good understanding of polynomials and inequalities. Therefore, we move on to more complex tasks - there will already be not one, but two modules.
Equations with two modules
Until now, we have studied only the simplest equations - there was one module and something else. We sent this “something else” to another part of the inequality, away from the module, so that in the end everything would be reduced to an equation of the form $\left| f\left(x \right) \right|=g\left(x \right)$ or even simpler $\left| f\left(x \right) \right|=a$.
But kindergarten ended - it's time to consider something more serious. Let's start with equations like this:
\[\left| f\left(x \right) \right|=\left| g\left(x \right) \right|\]
This is an equation of the form “modulus equals modulus”. Fundamentally important point is the absence of other terms and factors: only one module on the left, one more module on the right - and nothing more.
Someone will now think that such equations are more difficult to solve than what we have studied so far. But no: these equations are even easier to solve. Here's the formula:
\[\left| f\left(x \right) \right|=\left| g\left(x \right) \right|\Rightarrow f\left(x \right)=\pm g\left(x \right)\]
All! We simply equate submodular expressions by putting a plus or minus sign in front of one of them. And then we solve the resulting two equations - and the roots are ready! No additional restrictions, no inequalities, etc. Everything is very simple.
Let's try to solve this problem:
\[\left| 2x+3 \right|=\left| 2x-7 \right|\]
Elementary Watson! Expanding the modules:
\[\left| 2x+3 \right|=\left| 2x-7 \right|\Rightarrow 2x+3=\pm \left(2x-7 \right)\]
Let's consider each case separately:
\[\begin(align)& 2x+3=2x-7\Rightarrow 3=-7\Rightarrow \emptyset ; \\& 2x+3=-\left(2x-7 \right)\Rightarrow 2x+3=-2x+7. \\\end(align)\]
The first equation has no roots. Because when is $3=-7$? At what values of $x$? “What the hell is $x$? Are you stoned? There’s no $x$ there at all,” you say. And you'll be right. We have obtained an equality that does not depend on the variable $x$, and at the same time the equality itself is incorrect. That's why there are no roots. :)
With the second equation, everything is a little more interesting, but also very, very simple:
As you can see, everything was solved literally in a couple of lines - we didn’t expect anything else from a linear equation. :)
As a result, the final answer is: $x=1$.
So how? Difficult? Of course not. Let's try something else:
\[\left| x-1 \right|=\left| ((x)^(2))-3x+2 \right|\]
Again we have an equation of the form $\left| f\left(x \right) \right|=\left| g\left(x \right) \right|$. Therefore, we immediately rewrite it, revealing the modulus sign:
\[((x)^(2))-3x+2=\pm \left(x-1 \right)\]
Perhaps someone will now ask: “Hey, what nonsense? Why does “plus-minus” appear on the right-hand expression and not on the left?” Calm down, I’ll explain everything now. Indeed, in a good way we should have rewritten our equation as follows:
Then you need to open the brackets, move all the terms to one side of the equal sign (since the equation, obviously, will be square in both cases), and then find the roots. But you must admit: when “plus-minus” appears before three terms (especially when one of these terms is a quadratic expression), it somehow looks more complicated than the situation when “plus-minus” appears before only two terms.
But nothing prevents us from rewriting the original equation as follows:
\[\left| x-1 \right|=\left| ((x)^(2))-3x+2 \right|\Rightarrow \left| ((x)^(2))-3x+2 \right|=\left| x-1 \right|\]
What happened? Nothing special: they just swapped the left and right sides. A little thing that will ultimately make our life a little easier. :)
In general, we solve this equation, considering options with a plus and a minus:
\[\begin(align)& ((x)^(2))-3x+2=x-1\Rightarrow ((x)^(2))-4x+3=0; \\& ((x)^(2))-3x+2=-\left(x-1 \right)\Rightarrow ((x)^(2))-2x+1=0. \\\end(align)\]
The first equation has roots $x=3$ and $x=1$. The second is generally an exact square:
\[((x)^(2))-2x+1=((\left(x-1 \right))^(2))\]
Therefore, it has only one root: $x=1$. But we have already obtained this root earlier. Thus, only two numbers will go into the final answer:
\[((x)_(1))=3;\quad ((x)_(2))=1.\]
Mission Complete! You can take a pie from the shelf and eat it. There are 2 of them, yours is the middle one. :)
Important Note. The presence of identical roots for different options expansion of the modulus means that the original polynomials are factorized, and among these factors there will definitely be a common one. Really:
\[\begin(align)& \left| x-1 \right|=\left| ((x)^(2))-3x+2 \right|; \\& \left| x-1 \right|=\left| \left(x-1 \right)\left(x-2 \right) \right|. \\\end(align)\]
One of the module properties: $\left| a\cdot b \right|=\left| a \right|\cdot \left| b \right|$ (i.e. the modulus of the product is equal to the product of the moduli), so the original equation can be rewritten as follows:
\[\left| x-1 \right|=\left| x-1 \right|\cdot \left| x-2 \right|\]
As you can see, we really have a common factor. Now, if you collect all the modules on one side, you can take this factor out of the bracket:
\[\begin(align)& \left| x-1 \right|=\left| x-1 \right|\cdot \left| x-2 \right|; \\& \left| x-1 \right|-\left| x-1 \right|\cdot \left| x-2 \right|=0; \\& \left| x-1 \right|\cdot \left(1-\left| x-2 \right| \right)=0. \\\end(align)\]
Well, now remember that the product is equal to zero when at least one of the factors is equal to zero:
\[\left[ \begin(align)& \left| x-1 \right|=0, \\& \left| x-2 \right|=1. \\\end(align) \right.\]
Thus, the original equation with two modules has been reduced to the two simplest equations that we talked about at the very beginning of the lesson. Such equations can be solved literally in a couple of lines. :)
This remark may seem unnecessarily complex and inapplicable in practice. However, in reality, you may encounter much more complex problems than those we are looking at today. In them, modules can be combined with polynomials, arithmetic roots, logarithms, etc. And in such situations, the ability to lower the overall degree of the equation by taking something out of brackets can be very, very useful. :)
Now I would like to look at another equation, which at first glance may seem crazy. Many students get stuck on it, even those who think they have a good understanding of the modules.
However, this equation is even easier to solve than what we looked at earlier. And if you understand why, you'll get another trick for quickly solving equations with moduli.
So the equation is:
\[\left| x-((x)^(3)) \right|+\left| ((x)^(2))+x-2 \right|=0\]
No, this is not a typo: it is a plus between the modules. And we need to find at what $x$ the sum of two modules is equal to zero. :)
What's the problem anyway? But the problem is that each module is a positive number, or, in extreme cases, zero. What happens if you add two positive numbers? Obviously a positive number again:
\[\begin(align)& 5+7=12 \gt 0; \\& 0.004+0.0001=0.0041 \gt 0; \\& 5+0=5 \gt 0. \\\end(align)\]
The last line might give you an idea: the only time the sum of the modules is zero is if each module is zero:
\[\left| x-((x)^(3)) \right|+\left| ((x)^(2))+x-2 \right|=0\Rightarrow \left\( \begin(align)& \left| x-((x)^(3)) \right|=0, \\& \left| ((x)^(2))+x-2 \right|=0. \\\end(align) \right.\]
And when is the module equal to zero? Only in one case - when the submodular expression is equal to zero:
\[((x)^(2))+x-2=0\Rightarrow \left(x+2 \right)\left(x-1 \right)=0\Rightarrow \left[ \begin(align)& x=-2 \\& x=1 \\\end(align) \right.\]
Thus, we have three points at which the first module is reset to zero: 0, 1 and −1; as well as two points at which the second module is reset to zero: −2 and 1. However, we need both modules to be reset to zero at the same time, so among the found numbers we need to choose those that are included in both sets. Obviously, there is only one such number: $x=1$ - this will be the final answer.
Cleavage method
Well, we've already covered a bunch of problems and learned a lot of techniques. Do you think that's all? But no! Now we will look at the final technique - and at the same time the most important. We will talk about splitting equations with modulus. What will we even talk about? Let's go back a little and look at some simple equation. For example this:
\[\left| 3x-5 \right|=5-3x\]
In principle, we already know how to solve such an equation, because it is a standard construction of the form $\left| f\left(x \right) \right|=g\left(x \right)$. But let's try to look at this equation from a slightly different angle. More precisely, consider the expression under the modulus sign. Let me remind you that the modulus of any number can be equal to the number itself, or it can be opposite to this number:
\[\left| a \right|=\left\( \begin(align)& a,\quad a\ge 0, \\& -a,\quad a \lt 0. \\\end(align) \right.\]
Actually, this ambiguity is the whole problem: since the number under the modulus changes (it depends on the variable), it is not clear to us whether it is positive or negative.
But what if you initially require that this number be positive? For example, we require that $3x-5 \gt 0$ - in this case we are guaranteed to get a positive number under the modulus sign, and we can completely get rid of this very modulus:
Thus, our equation will turn into a linear one, which can be easily solved:
True, all these thoughts make sense only under the condition $3x-5 \gt 0$ - we ourselves introduced this requirement in order to unambiguously reveal the module. Therefore, let's substitute the found $x=\frac(5)(3)$ into this condition and check:
It turns out that for the specified value of $x$ our requirement is not met, because the expression turned out to be equal to zero, and we need it to be strictly greater than zero. Sad. :(
But it's okay! After all, there is another option $3x-5 \lt 0$. Moreover: there is also the case $3x-5=0$ - this also needs to be considered, otherwise the solution will be incomplete. So, consider the case $3x-5 \lt 0$:
Obviously, the module will open with a minus sign. But then a strange situation arises: both on the left and on the right in the original equation the same expression will stick out:
I wonder at what $x$ the expression $5-3x$ will be equal to the expression $5-3x$? Even Captain Obviousness would choke on his saliva from such equations, but we know: this equation is an identity, i.e. it is true for any value of the variable!
This means that any $x$ will suit us. However, we have a limitation:
In other words, the answer will not be a single number, but a whole interval:
Finally, there is one more case left to consider: $3x-5=0$. Everything is simple here: under the modulus there will be zero, and the modulus of zero is also equal to zero (this follows directly from the definition):
But then the original equation $\left| 3x-5 \right|=5-3x$ will be rewritten as follows:
We already obtained this root above when we considered the case of $3x-5 \gt 0$. Moreover, this root is a solution to the equation $3x-5=0$ - this is the limitation that we ourselves introduced to reset the module. :)
Thus, in addition to the interval, we will also be satisfied with the number lying at the very end of this interval:
Combining roots in modulo equations Total final answer: $x\in \left(-\infty ;\frac(5)(3) \right]$ It’s not very common to see such crap in the answer to a fairly simple (essentially linear) equation with modulus , really? Well, get used to it: the difficulty of the module is that the answers in such equations can be completely unpredictable.
Something else is much more important: we have just analyzed a universal algorithm for solving an equation with a modulus! And this algorithm consists of the following steps:
- Equate each modulus in the equation to zero. We get several equations;
- Solve all these equations and mark the roots on the number line. As a result, the straight line will be divided into several intervals, at each of which all modules are uniquely revealed;
- Solve the original equation for each interval and combine your answers.
That's all! There is only one question left: what to do with the roots obtained in step 1? Let's say we have two roots: $x=1$ and $x=5$. They will split the number line into 3 pieces:
Splitting the number line into intervals using points So what are the intervals? It is clear that there are three of them:
- The leftmost one: $x \lt 1$ — the unit itself is not included in the interval;
- Central: $1\le x \lt 5$ - here one is included in the interval, but five is not included;
- Rightmost: $x\ge 5$ - five is only included here!
I think you already understand the pattern. Each interval includes the left end and does not include the right.
At first glance, such an entry may seem inconvenient, illogical and generally some kind of crazy. But believe me: after a little practice, you will find that this approach is the most reliable and does not interfere with unambiguously opening the modules. It’s better to use such a scheme than to think every time: give the left/right end to the current interval or “throw” it into the next one.
Instructions
If a module is represented as a continuous function, then the value of its argument can be either positive or negative: |x| = x, x ≥ 0; |x| = - x, x
z1 + z2 = (x1 + x2) + i(y1 + y2);
z1 - z2 = (x1 - x2) + i(y1 - y2);
It is easy to see that addition and subtraction of complex numbers follows the same rule as addition and .
The product of two complex numbers is equal to:
z1*z2 = (x1 + iy1)*(x2 + iy2) = x1*x2 + i*y1*x2 + i*x1*y2 + (i^2)*y1*y2.
Since i^2 = -1, the final result is:
(x1*x2 - y1*y2) + i(x1*y2 + x2*y1).
The operations of exponentiation and root extraction for complex numbers are defined in the same way as for real numbers. However, in the complex region, for any number, there are exactly n numbers b such that b^n = a, that is, n roots of the nth degree.
In particular, this means that any algebraic equation of degree n with one variable has exactly n complex roots, some of which may be .
Video on the topic
Sources:
- Lecture "Complex numbers" in 2019
A root is an icon that represents mathematical operation finding a number whose raising to the power indicated before the root sign should give the number indicated under this very sign. Often, to solve problems that involve roots, it is not enough to just calculate the value. It is necessary to carry out additional operations, one of which is entering a number, variable or expression under the root sign.
Instructions
Determine the root exponent. An exponent is an integer that indicates the power to which the result of calculating the root must be raised in order to obtain the radical expression (the number from which this root is extracted). The root exponent as a superscript before the root icon. If this one is not specified, it is Square root, the degree of which is two. For example, the exponent of the root √3 is two, the exponent of ³√3 is three, the exponent of the root ⁴√3 is four, etc.
Raise the number you want to enter under the sign of the root to a power equal to the exponent of this root, determined by you in the previous step. For example, if you need to enter the number 5 under the root sign ⁴√3, then the index of the root degree is four and you need the result of raising 5 to the fourth power 5⁴=625. You can do this in any way convenient for you - in your head, using a calculator or the corresponding services hosted.
Enter the value obtained in the previous step under the root sign as a multiplier of the radical expression. For the example used in the previous step with adding ⁴√3 5 (5*⁴√3) under the root, this action can be done like this: 5*⁴√3=⁴√(625*3).
Simplify the resulting radical expression if possible. For an example from the previous steps, you just need to multiply the numbers under the root sign: 5*⁴√3=⁴√(625*3)=⁴√1875. This completes the operation of entering the number under the root.
If the problem contains unknown variables, then the steps described above can be done in general view. For example, if you need to enter an unknown variable x under the fourth root root, and the radical expression is 5/x³, then the entire sequence of actions can be written as follows: x*⁴√(5/x³)=⁴√(x⁴*5/x³)= ⁴√(x*5).
Sources:
- what is the root sign called?
Real numbers are not enough to solve any quadratic equation. The simplest of quadratic equations, having no roots among the real numbers - this is x^2+1=0. When solving it, it turns out that x=±sqrt(-1), and according to the laws of elementary algebra, extract the root even degree from negative numbers it is forbidden.
