EntranceEducational portal. Equations with modulus absolute value definition of modulus Section VI
Introduction
1. Absolute value in a high school course
1.1 Definitions and main theorems
1.2 Geometric interpretation of the concept |a|
2. Methods for solving equations and inequalities
2.1 Solving equations and inequalities using the definition of absolute value (modulus)
2.2 Method of solution using dependencies between numbers a and b, their modules and the squares of these numbers
2.3 Interval method
2.4 Graphical method
2.5 Method of sequential expansion of the module
2.6 Types of equations and inequalities and their solution
3. Additional ways to solve equations and inequalities
3.1 solving equations and inequalities containing a modulus using identities
3.2 Solving equations containing moduli of non-negative expressions
3.3 Solving equations using geometric interpretation
3.4 Solving equations by going to the consequence
3.5 Typical problems containing a variable under the modulus sign
3.6 Teachers’ advice on the sequence of studying equations and inequalities with a module in a school mathematics course
4. Equations and inequalities with a module in the Unified National Testing (UNT)
Conclusion
List of used literature
Introduction
Relevance of the topic is due to the fact that the module is widely used in various sections of the school course in mathematics, physics and technical sciences. For example, in the theory of approximate calculations, the concepts of absolute and relative errors of an approximate number are used, the concept of a vector and its length (vector modulus) is used in geometry and mechanics, in mathematical analysis the concept of modulus is contained in the definitions of limits, a bounded function. I believe that this topic requires more in-depth research, since it can be seen in various tasks of increased complexity that the authors of didactic materials offer students, in problems of mathematical Olympiads, UNT tasks and exams for admission to universities.
In the practice of teaching mathematics in secondary schools, the concept of the absolute value of a number (module) is encountered more than once.
In grade 6, in the topic of approximate calculations, the concept of the absolute value of a number is formed, while understanding the absolute error of an approximate number.
In the second half of the 6th grade, the definition of the absolute value of a number (module) is introduced and with the help of this concept the rules of operations on rational numbers are formulated.
In grade 8, when considering the properties of the arithmetic square root, the concept of the absolute value of a number finds its new application:
;
, where others.
In grade 9, when studying the limit of a sequence, students encounter expressions of the form:
The concept of the absolute value of a number receives its further development in 10th grade when studying the limit of a function, when studying a function for boundedness, when studying complex numbers.
In grade 11, the topic “Exponent with a rational exponent” examines the properties of roots n th degree, where the concept of absolute magnitude of a number is also used; so, for example,
=
Thus, in all classes, in accordance with curriculum, exercises containing the sign of the absolute value of a number should be included and considered.
In 6th grade you can solve equations of the form:
In grade 7, there is the opportunity to examine solutions to equations of the form: etc., systems of equations of the form:
As well as constructing graphs of functions: ; ; etc.
In grade 8, the concepts of absolute value extend to quadratic equations, the graph of a quadratic trinomial, etc. You can solve equations of the form: 
;
;
Novelty thesis : solved all the equations and inequalities with moles found in the UNT test tasks and examined the main mistakes made by students when solving them.
Target carrying out our research - to analyze the educational and methodological material, identify all the methods for solving equations and inequalities with a module and combine them in this work.
To achieve the goal, it is necessary to solve the following tasks:
Study the main theorems and definitions;
Describe the basic methods for solving equations and inequalities with modulus;
Identify non-standard methods for solving equations and inequalities with modulus.
Object of study: the process of teaching equations and inequalities at school.
Subject of research: methods for solving equations and inequalities containing a modulus sign in a school mathematics course.
Practical significance thesis is that this thesis presents all the methods and techniques for solving equations and inequalities that can be used in a school mathematics course.
The main research methods in my thesis are:
analytical,
comparative,
study of monographic publications and articles,
concrete historical,
generalization method.
This diploma is based on the following works: “Absolute value” Gaidukov I.I., “Equations and inequalities with modules and methods for solving them” Sevryukov P.F., Smolyakov A.N., “Algebra and principles of analysis. Equations and inequalities 10 - 11kl" Olehnik, Potapov, Pasichenko.
The first chapter discusses the theoretical side of the problem, the main theorems and concepts necessary for further research on this topic. equation problem inequality
In the second chapter of the thesis, we combined methods for solving equations and inequalities with a module that is included in the school curriculum.
In the third chapter, we presented non-standard techniques for solving equations and inequalities containing a module, studied in additional classes and used in solving Olympiad problems. Also considered here are typical tasks for solving equations and inequalities and tasks for test versions of the Unified National Testing (UNT).
When solving equations containing the sign of the absolute value, we will be based on the definition of the modulus of a number and the properties of the absolute value of a number.
1. Absolute value in a high school course
1.1 Definitions and main theorems
Let's consider the concept of the absolute magnitude of a number, or, which is the same, the modulus of a number for real numbers.
Definition 1.1.1 The absolute value (modulus) of a real number a is a non-negative number taken from two numbers a or - A.
The absolute value of the number a denote | A| and read “the absolute value of the number a”, or "modulus of number a".
From the definition it follows:
From the definition it follows that for any real number a, ≥0.
Examples 1.1.1:
;
Theorem 1.1.1 Opposite numbers have equal absolute values, i.e. = .
In fact, by definition of absolute value, we have:
=
=
Hence,
1.2 Geometric interpretation of the concept
It is known that everyone real number you can associate a point on a number line, then this point will be a geometric image of a given real number. Each point on the number line corresponds to its distance from the origin of the reference, and the length of a segment, the beginning of which is at the origin of the reference, and the end at a given point. This distance, or the length of the segment, is always considered as a non-negative quantity.
At the same time, each point on the number line can be associated with a directed segment (vector), which is characterized by length and direction.
The set of real numbers corresponds to the set of points on an oriented line, i.e. such a straight line on which, in addition to the origin and scale, a positive direction is established.
Then we can assume that the geometric interpretation of a real number is a vector coming from the origin and having an end at the point representing the given number. The length of this vector will be a geometric interpretation of the absolute value of a given real number.
The geometric interpretation of the meaning clearly confirms that =.
Examples 1.2.1:
If = 5, then A 1 =5 and A 2 =-5, or a =±5.
Therefore, this equality is satisfied by two numbers that correspond to two points on the number line.
If ˃10, then 
Where A˃10 and A˂ -10, or
Consequently, this inequality is satisfied by the set of numbers of two intervals: (-∞;-10) and (10;∞), and on the number line - two intervals corresponding to these intervals.
Translating an algebraic problem into geometric language is a convenient and powerful method for solving problems. As another example, let’s look at the block of Olympiad problems:
Example 1.2.2:
Given function: 
.
Solution: Let's build a graph of the function. To do this, note that , and then we can first build a graph of the function and then reflect it relative to the coordinate axis. Let's transform the expression defining the function:
Since this system defines an upper semicircle of radius 2 with a center at point , the graph of the original function is a union of the semicircles indicated in the figure.
Now solving problems is not difficult:
With) At there are no solutions, at the equation has three solutions, at - four solutions, at - two solutions.
b) The inequality holds for all of the segment .
a) the root of the equation is the abscissa of the point of intersection of the line with the graph of the function. Let's find it geometrically: the right triangle shaded in the figure is isosceles (the angular coefficient of the line is -1), its hypotenuse is the radius of the circle, its length is 2. Then the length of the leg lying on the abscissa is , and the required abscissa is equal to .
Geomet r logical meaning of the module r difference of quantities is the distance between them. For example, the geometric meaning of the expression |x– A| - the length of the segment of the coordinate axis connecting the points with the abscissas a and x. Translating an algebraic problem into geometric language often allows one to avoid cumbersome solutions.
Example 1.2.3: Let's solve the equation |x–1|+|x–2|=1 using the geometric interpretation of the modulus.
We will reason as follows: based on the geometric interpretation of the module, the left side of the equation is the sum of distances from some abscissa point x to two fixed points with abscissas 1 and 2. Then it is obvious that all points with abscissas from the segment have the required property, and points located outside this segment - no. Hence the answer: the set of solutions to the equation is the segment.
Answer: x
Example 1.2.4: Let's solve the equation |x – 1| - |x – 2|=1 1 using the geometric interpretation of the module.
We will reason similarly to the previous example, and we will find that the difference in distances to points with abscissas 1 and 2 is equal to one only for points located on the coordinate axis to the right of the number 2. Therefore, the solution to this equation will not be the segment enclosed between points 1 and 2, and the ray emerging from point 2 and directed in the positive direction of the OX axis.
Answer: x sum of distances d+f equal to the length of the segment AB, i.e. 7. It is also easy to set what for points X<2 or x>5 sum of distances d+f>7. Therefore, the solution to the equation is the interval.
b) Let's expand the modulus sign. To do this, plot the points -2 and 5 on the number line. These points divide it into three intervals. Let us consider the signs of the modules in each of the intervals.
In interval 1 (X<-2) we get: -(x–5)–(x+2)=7 or –x+5–x–2=7 or - 2x+3=7, from where we get: x=-2. But this point is not included in the considered interval. That's why x=-2 is not a solution.
In interval 2: X we get: -(x–5)+(x+2)=7 or 7=7. Since the equality is correct, any point in this interval is a solution to this equation.
In interval 3 (x>5) we get: (x-5)+(x+2)=7 or 2x-3=7, where x=5. Dot x=5 is not included in the interval under consideration and is not a solution to the equation.
So, the solution to this equation is: -2x5.
Exercises for independent work:
Solve equations:
Lesson No. 3. Solving quadratic equations with modulus.
Let's consider solving quadratic equations with modules using examples:
No. 1. Solve the equation
Let's introduce the replacement =y, then at y 0 the equation takes the form:
y 2 –6у+8=0, from where y 1 = 2 and y 2 = 4. a x= 2 or -2; 4 or -4.
No. 2. Solve the equation:
The equation is equivalent to the system: From where X=1.
No. 3. Solve the equation:
2X – 1.
The equation has a solution provided that 2 X–10, and equality is possible provided: the meanings of the expressions x 2 + x–1 and 2 X–1 are the same or opposite. That. we have: x0.5. Let's make up the equations: x 2 + x–1=2X–1 or x 2+X–1=-(2X–1); solving which, we get
No. 4. Find the roots of the equation: 
.
Let's present this equation in the form: = X 2 – 1, from where:
x – 1 = x 2 – 1,
or x – 1 = - (x 2 – 1).
x 2 – 1 at x - 1 And x 1.Solving the equations, we obtain from the first: x=0 And x=1, from the second: x=-2 And x=1.
Answer: x=1; x=-2.
No. 5. Find the whole roots of the equation: = .
Using the definition of a module, we come to the conclusion that equality is possible if the values of the expressions x–x 2 –1 And 2x+3–x 2 equal or opposite, i.e. this equation is equivalent to a set of two equations:
Solving the set, we obtain the roots of this equation: x=-4;-0.5;2. Integers among them: -4 and 2.
No. 6. Solve the equation: =2x 2 –3x+1.
Let us denote the expression 3x-1-2x 2 letter A. Then this equation will take the form: =-a. Based on the analytical notation of the definition of the modulus, we can conclude that this equation is equivalent to the inequality: 3x–1-2x 2 0, solving which, we get the answer: x0.5 And x1.
Exercises for independent work.
Solve the equation:
No. 1.=x 2 + x–20.
No. 2. + 3x -5=0,
No. 3. =(x–1)(x+1),
No. 4. x 2 –6+5=0,
No. 5. x 2 +8=9,
No. 6.=x 2 –6x+6,
No. 7. x = -8.
Lesson No. 4. Solving equations containing absolute value with parameters.
Let's consider an example: solve an equation with a parameter
Let's build function graphs y=3–x And y=. Schedule y=3–x is fixed and does not depend on the parameter. Schedule y= obtained from the graph of the function y=, depends on parameter A. Therefore, let's consider 3 cases:
This case, as can be seen from the figure, will be when A<3 . The graphs of these functions intersect at a single point B. Consider triangle ABC, in which angle A is equal to angle B and equal to 45 0, draw the altitude VD in this triangle. Because triangle ABC is isosceles, then BD is also the median of this triangle. Therefore, the abscissa of point D X=(a + 3)/2.
This case occurs when A=3. Then the graphs of the functions coincide along the segment AB and the abscissa of any point on this ray is a solution to this equation, i.e. X<3.
In this case A>3. It can be seen that the graphs of the functions do not intersect, i.e. have no common points. Therefore, the equation has no solution.
Exercises for independent work:
Solve the equations:
No. 3. (a–2)=a–2,
No. 4. a 2 x 2 + a = 0.
Lesson No. 5. Solving linear inequalities with moduli.
Inequalities containing a variable under the modulus sign are solved in various ways; Let's look at a fairly simple example:
No. 1.Solve inequality:
First method: We have: >4,
Geometrically, the expression means the distance on the coordinate line between points X and 2.5. This means we need to find all such points X, which are more than 2 away from point 2.5, are points from the intervals X<0,5 And x>4.5.
Second method: Since both sides of the given inequality are non-negative, we square both sides of this inequality: 2 >4 2.
(2x–5) 2 >4 2 ,
(2x–5) 2 –16>0,
(2x–5–4)(2x–5+4)>0,
2(x–4.5) 2(x–0.5)>0,
(x–4.5)(x–0.5)>0.
Applying the interval method, we get: X<0 ,5 and x>4.5.
Third way: Expression 2x–5 may be non-negative or negative. Those. we have a combination of two systems:
Where: X<0,5 And x>4.5.
Let's look at a few more examples.
Example No. 2. Solve inequality:<3.
This inequality is equivalent to the combination of two systems:
From the first system we get 2x<5 , from the second -1<х<2 . Combining these two solutions we get: -1<х<5 .
Example No. 3. Solve inequality: 3 x+3.
This inequality is equivalent to the double inequality -x-33x–3x+3 or system 
We have : 0x3.
Exercises for independent work:
Solve inequalities:
№1. <3х+1,
№3. ->-2.
Lesson No. 6. Solving quadratic inequalities with moduli.
Let's look at example No. 1. Solve the inequality: +x–2<0 .
This inequality can be solved using the interval method. Let's consider another solution based on the following statement: for any value of a, the inequality is equivalent to the system of inequalities: ,and inequalityis equivalent to a set of inequalities.
Therefore, our inequality is equivalent to a system of inequalities: 
solving which, we get: 
Let's write down the answer: (1-;2-).
Example No. 2. Find integer solutions to the inequality: 2x–x 2. The problem comes down to solving a set of two systems of inequalities:
Let us solve the first system: from the first inequality we have: x1; x2.
from the second: 2x 2 –5x+20, or 0.5x2.
Having noted the found solutions to the first and second inequalities of the first system on the coordinate line, we find the intersection of the solutions.
That. 0.5x1 And x=2. This is the solution of the first system.
Let us solve the second system: from the first inequality we have: 1<х<2 , from the second: -(x 2 -3x+2)2x–x 2, or – x 2 +3x–2–2x+ x 2 0, or x2.
Noting the found solutions to the first and second inequalities of the second system on the coordinate line, we obtain: 1<х<2 . This is the solution of the second system.
Combining the found solutions to systems of inequalities 0.5x1; x=2; 1
Exercises for independent work:
Solve the inequalities:
№3. <3х–3,
No. 4. x 2 -3+2>0,
No. 5. x 2 x<3,
No. 6. x 2 -6x+7-<0,
No. 7. 3+x 2 –7>0,
№8. >.
Lesson No. 7. Solving inequalities containing absolute value with parameters.
Example. At what values A the inequality is true: ah 2 +4+a+3<0 ?
At x0 we have ah 2 +4x+a+3<0 . Senior coefficient A must be negative, the discriminant must be less than zero.
A<0, Д=16–4a(a+3)<0; 16-4а 2 -12а<0; а 2 +3а-4>0; A<-4 And a>1;
abscissa of the vertex of a parabola x 0 = -b/2a=- 4/2a=-2/a 0, where A<-4 .
At X<0 we have ah 2 –4x+a+3<0 . Arguing similarly, we get: A<-4 .
Answer: when A<-4 this inequality holds for all real values of x.
Exercises for independent work:
Solve inequalities with parameters:
No. 2. (Ha)<0,
No. 3. Are there values of a for which the inequality ah 2 >2+5 has no solutions?
Lessons No. 8 - 9. Interval method for solving equations and inequalities containing a modulus.
Let's consider the interval method using the example of solving the equation
-+3-2=x+2.
To solve this inequality, it is necessary to expand the modules. To do this, we select intervals, at each of which the expressions under the modulus sign take only positive or negative values. Finding such intervals is based on the theorem: if on the interval (a; b) the function f is continuous and does not vanish, then it retains a constant sign on this interval.
To highlight intervals of constant sign, we find the points at which the expressions written under the modulus become zero:
x+1=0, x=-1; x=0; x–1=0, x=1; x–2=0, x=2.
The resulting points will divide the line into the required intervals. Let's define the signs of expressions
x+1, x, x–1, x–2 on these intervals:
Taking into account the signs, we will expand the modules. As a result, we obtain a set of systems equivalent to this equation:
The last set is reduced to the form:
The solution to the set of systems and this equation: -2; X 2.
The technique used is called interval method. It is also used in solving inequalities.
Solve inequality: +x–2<0.
1) Find the zeros of the expression: x 2 -3x.
x 1 =0, x 2 =3.
2) Let's divide the coordinate line into intervals and set the sign of the expression x 2 -3x at each interval:
3) Let's expand the module: 
Solution of the first system: , solution of the second. The solution to this inequality: 
.
Exercises for independent work:
№3
Lesson No. 10 - 11. Solving inequalities of the form , through equivalent transitions.
Let us consider inequalities of the form and . Let us accept the following theorem without proof: for any value of a the inequalityis equivalent to a system of inequalities and the inequalityis equivalent to a set of inequalities
Let's look at an example: solve the inequality: 
>x+2.
Using the formulated theorem, let us move on to the set of inequalities:
System and inequality 0x>2 have no solutions. Therefore, the solution to the population (and to this inequality) is X.
Exercises for independent work:
Lesson No. 12. Application of the properties of absolute value in solving equations and inequalities.
When solving some tasks, the properties of the module are used. (If necessary, repeat them, see lesson No. 1).
Let us illustrate the use of module properties in solving the following examples.
Lesson objectives:
educational:
- repetition of various methods for solving equations containing the modulus sign;
- solving equations in various ways;
- solving equations offered in the entrance exams at Moscow State University;
- solving equations containing a modulus sign and a parameter;
educational:
- development of attention;
- developing the ability to correctly and clearly write down a decision;
- developing the ability to listen to classmates’ explanations;
- developing the ability to check one’s own decision;
developing:
- developing the ability to find the most rational way to solve;
- development of mathematical thinking;
- developing the ability to justify your decision;
- development of the ability to generalize acquired knowledge;
- developing the ability to solve equations with a parameter;
Equipment:
- blackboard;
- handouts with task conditions for working in groups;
- computer;
- projector;
- screen.
Knowledge, abilities, skills.
As a result of the lesson, students must repeat the basic techniques for solving equations containing a modulus sign, learn to solve similar equations at the level of school finals and competitive exams, learn to understand and be able to find solutions to equations containing a parameter.
PROGRESS OF THE LESSON
1) Repeat the definition of the modulus of a number and how to expand it depending on the sign of the argument.
2) Repeat the basic methods of solving equations containing expression modules:
a) solving equations by opening the module externally;
b) solving equations by opening the module from the inside;
c) solving equations containing modules using the variable change method;
d) solving equations containing several modules;
e) solving equations containing modules and parameters simultaneously.
3) Solving equations using various methods (work in groups).
4) Solving competitive exam equations (using computer).
5) Solving equations containing modules and parameters simultaneously (using a blackboard, computer and projector).
6) Summing up the lesson, grading.
Materials for the lesson:
1. For each of the indicated equations, select a solution method and solve it (solution on the board and in notebooks).
a) | 5 - 4x | = 1
b) | 6x2 _ 5x + 1 | = 5x - 6x2 - 1
c) x2 + 3|x+1| - 1 = 0
d) | x - 2| + |x + 4| = 8
e) 2|x + 2| + 3 = (x + 2)2
Answers: a) 1; 1.5; b) ; c) -1; d) -5; 3; e) -5; 1.
2. Work in groups (each group receives an envelope with an assignment and a card for grading and self-evaluation of the work performed).
Type of grading card. (Appendix 2)
Grading criteria:
“5” - solved 5 equations in different ways independently;
“4” - solved 5 equations in different ways and received one consultation from group members;
“3” - solved 5 equations in different ways and received two or three consultations from group members;
“2” - had difficulty solving equations and constantly consulted with group members;
The grade is given by the group after discussion and by the student himself, the final grade is given by the teacher.
CARD No. 1
a) | 3x-3 | = 6;
b) | x 2 - 3x - 10 | = 3x - x 2 + 10;
c) 1/|x| + 1/(x + 1) = 2;
d) | x 2 - 9 | + | x - 2| = 5;
e) | x - 1| + | x - 2| + | x - 3| = x.
CARD No. 2
a) | 3-2x | = 4;
b) | x 2 - 3x + 2 | = 3x - x 2 - 2;
c) 2/|x - 1| + 4/(x + 3) = 3;
d) | x 2 - 8x | - 9 = 0;
e) | x - 3 | + | x + 2 | - | x - 4 | = 3.
CARD No. 3
a) | 5x-4 | = 6;
b) x 2 + 2| x - 1 | - 1 = 0;
c) | x 2 - 2x | - 3 = 0;
d) (x - 3.5) 2 + 2| x - 3.5 | = 1.25;
e) | x + 2 | - | x - 3 | + | x - 1 | = 1.
3. Solving competitive exam equations.
a) Solve the equation: |||| x -3 | - 1 | + 2 | - 3| = 1
Let's expand the module externally and obtain a set of two equations:
||| x - 3 | - 1 | + 2 | - 3 = 1 and ||| x - 3 | - 1 | + 2 | - 3 = -1, transforming which we get:
||| x- 3 | - 1 | + 2 | = 4 and ||| x - 3 | - 1 | + 2 | = 2.
Let us expand the module externally again and obtain a set of four equations:
|| x - 3 | - 1 | + 2 = 4; || x - 3 | - 1 | + 2 = -4; || x - 3 | - 1 | + 2 = 2 and
|| x - 3 | - 1 | + 2 = -2.
Let us transform the resulting equations again:
|| x - 3 | - 1 | = 2; || x - 3 | - 1 | = -6; || x- 3 | - 1 | = 0 and || x - 3 | - 1 | = -4.
It is easy to see that the second and fourth of the resulting equations have no solution, since the modulus cannot take negative values.
Further expansion of the modules leads to the answer: x = 0; 2; 4; 6.
b) As homework, it is proposed to solve the following equations:
|| x - 2 | - 4 | = 3;
|||| x + 1| - 5 | + 1| - 2 | = 2;
|||| x + 3| - 2 | + 1 | - 3| = 3;
|| 2x - 7 | - x | = 7 - x;
|| x - 1 | - x - 3 | + x = 4;
|| 2x - 1 | - x - 3 | = 4 - x.
4. Solving equations with a parameter.
It is proposed to determine the number of roots of the equation depending on the value of the parameter A and solve this equation in two ways: analytical and graphical:
| x 2 - 2x - 3 | = A.
a) Graphical method for solving the equation:
To solve this equation, it is necessary to plot graphs of the following functions: y 1 = |x 2 - 2x - 3| and y 2 = A. The graph of the first function is a parabola, in which the region of negative values of the function is mapped to the region of positive values of the variable at relative to the axis X. The graph of the second function is a straight line parallel to the axis X.
It is easy to see that when A‹0 the resulting graphs do not intersect, which indicates the absence of solutions to this equation. At a = 0we have two points of intersection of the graphs, and, therefore, two solutions: x = -1 and x = 3. At 0
At a = There are three solutions: x 1 = 1 – 22 and x 2 = 1 + 22, and x 3 = x 4 = 1.
At A›4 solutions, as well as points of intersection of graphs, there are two left:
b) Analytical method for solving the equation:
The first conclusion can be drawn immediately: A> 0 because the modulus cannot take negative values. Thus, when a‹0 there are no solutions. At A= 0, we solve the quadratic equation: x 2 - 2x - 3 = 0, the solution of which is x 1 = -1 and x 2 = 3. When A›0 we solve two equations separately:
x 2 - 2x - 3 = A(1) and x 2 - 2x - 3 = - A (2).
Equation (1) has two solutions for any parameter values A> 0. Equation (2) has two solutions only at 0‹ A‹4, for these parameter values, the discriminant of the quadratic equation (2) is positive, and the roots of the equation are similar to x 3 and x 4 found with a graphical solution. At A= 4, the discriminant of equation (2) is equal to 0, the solution to equation (2) is one and equal to 1.
As a result of solving in any way, the following answer was obtained:
At A‹0 no solutions;
At A= 0x = -1; 3.
At 0
At a = 4: x 1 = 1 - 22 and x 2 = 1 + 22, and x 3 = x 4 = 1.
At A›4:
c) As homework, it is proposed to determine the number of roots of the equation depending on the values of parameter a:
1) | 5 + 2x - x 2 | = A; 2) x 2 - 6|x| + 5 = A; 3) x 2 - 3|x| = A.
5. Summing up the lesson, grading.