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Tochilkina Julia
The paper presents various methods for solving equations with a modulus.
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Municipal budgetary educational institution
"Secondary school No. 59"
Modulo Equations
Abstract work
Performed 9th grade student
MBOU "Secondary School No. 59", Barnaul
Tochilkina Julia
Supervisor
Zakharova Ludmila Vladimirovna,
mathematic teacher
MBOU "Secondary School No. 59", Barnaul
Barnaul 2015
Introduction
I am in ninth grade. This academic year I have to pass the final certification for the course of the basic school. To prepare for the exam, we purchased a collection of D. A. Maltsev Mathematics. Grade 9 Looking through the collection, I found equations containing not only one, but also several modules. The teacher explained to me and my classmates that such equations are called "nested modules" equations. This name seemed unusual for us, and the solution at first glance, rather complicated. This is how the topic for my work “Equations with a modulus” appeared. I decided to study this topic in more depth, especially since it will come in handy for me when passing exams at the end of the school year and I think that I will need it in grades 10 and 11. All of the above determines the relevance of the topic I have chosen.
Objective :
- Consider various methods solution of equations with modulus.
- Learn to solve equations containing the sign of the absolute value using various methods
To work on the topic, the following tasks were formulated:
Tasks:
- To study theoretical material on the topic "The modulus of a real number."
- Consider methods for solving equations and consolidate the knowledge gained by solving problems.
- Apply the acquired knowledge in solving various equations containing the sign of the modulus in high school
Object of study:methods for solving equations with a modulus
Subject of study:modulo equations
Research methods:
Theoretical : study of literature on the research topic;
Internet - information.
Analysis information obtained in the study of literature; results obtained when solving equations with modulus different ways.
Comparison ways of solving equations, the subject of rationality of their use in solving various equations with a module.
“We start thinking when we bump into something.” Paul Valerie.
1. Concepts and definitions.
The concept of "module" is widely used in many sections school course mathematics, for example, in the study of the absolute and relative errors of an approximate number; in geometry and physics, the concepts of a vector and its length (vector modulus) are studied. The concept of a module is used in courses of higher mathematics, physics and technical sciences studied in higher educational institutions.
The word "module" comes from the Latin word "modulus", which means "measure" in translation. 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 to be used by Kots, a student of Newton. The module sign was introduced in the 19th century by Weierstrass.
In architecture, a module is the initial unit of measure established for a given architectural structure.
In engineering, this is a term used in various areas technique, which serves to designate various coefficients and quantities, for example, the modulus of elasticity, the modulus of engagement ...
In mathematics, a modulus has several meanings, but I will treat it as the absolute value of a number.
Definition1 : Modulus (absolute value) of a real number a the number itself is called if a ≥0, or the opposite number - what if a the modulus of zero is zero.
When solving equations with a module, it is convenient to use the properties of the module.
Consider the proofs of 5,6,7 properties.
Statement 5. Equality │ is true if av ≥ 0.
Proof. Indeed, after squaring both parts of this equality, we get, │ a+v │²=│ a │²+2│ ab │+│ to │²,
a² + 2 av + b² \u003d a² + 2│ av │ + b², from where │ av │ = av
And the last equality will be true for av ≥0.
Statement 6. Equality │ a-c │=│ a │+│ c │ is true when av ≤0.
Proof. To prove it, it suffices in the equality
│ a + in │=│ a │+│ in │ replace in with - in, then a (- in) ≥0, whence av ≤0.
Statement 7. Equality │ a │+│ in │= a + in performed at a ≥0 and b ≥0.
Proof . Considering four cases a ≥0 and b ≥0; a ≥0 and b a at ≥0; a in a ≥0 and b ≥0.
(a-c) in ≥0.
Geometric interpretation
|a| is the distance on the coordinate line from the point with coordinate a , to the origin of coordinates.
|-a| |a|
A 0 a x
Geometric interpretation of the meaning |a| clearly confirms that |-a|=|a|
If a 0, then on the coordinate line there are two points a and -a, equidistant from zero, whose modules 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 the 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);
- Graphical solution method;
- Use of equivalent transformations (properties 4.6);
- Variable substitution (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.
SOLUTION OF EQUATIONS | f(x)| = a
Consider the equation | f(x)| = a, and R
An equation of this kind can be solved by defining the modulus:
If a 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.
SOLUTION OF EQUATIONS USING THE GEOMETRIC PROPERTIES OF THE MODULE.
Example 1 Solve the equation /x-1/+/x-3/=6.
Solution.
To solve this equation means to find 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.
None of the points on the segmentdoes not satisfy this condition, because the sum of the specified 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.
Denote x 2 + x-5 \u003d a, then / a / + / a-4 /=10. Let's find points on the x-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 \u003d 4 x 2 + x-5 \u003d 7
X 2 + x-2 \u003d 0 x 2 + x-12 \u003d 0
X 1 \u003d 1, x 2 \u003d -2 x 1 \u003d -4, x 2 \u003d 3 Answer: -4; -2; one; 3.
SOLUTION OF EQUATIONS | f(x)| = | g(x)|.
- Since | a |=|b |, if a=b, then an equation of the form | f(x)| = | g(x )| is tantamount to an aggregate
Example1.
Solve equation | x–2| = |3 - x |.
Solution.
This equation is equivalent to two equations:
x - 2 \u003d 3 - x (1) and x - 2 \u003d -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 the equivalent transformation:
(x 2 + 3x-20) 2 \u003d (x 2 -3x + 2) 2
(x 2 + 3x-20) 2 - (x 2 -3x + 2) 2 \u003d 0,
(x 2 + 3x-20-x 2 + 3x-2) (x 2 + 3x-20 + x 2 -3x + 2) \u003d 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 in that the right side is also a variable. And it can be both positive and negative. Therefore, it must be specially verified that it is non-negative, because the modulus cannot be equal to negative number(property№1 )
1 way
Equation solution | f(x)| = g(x ) is reduced to the set of solutions to the equationsand checking the validity of the inequality g(x )>0 for the found values of the unknown.
2 way (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:
O t e t: 3; four.
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 module. 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 that break general area definitions into intervals, in each of which the functions f 1 (x), f 2 (x), …, f n (x) keep their sign. Further, using the definition of the module, for each of the found areas we obtain an equation that must be solved on a given interval. This method is called "interval method»
Example.
Solve the equation |x-2|-3|x+4|=1.
Solution.
Let's find the points where the submodule expressions are equal to zero
x-2=0, x+4=0,
x=2; x=-4.
Let's break the number line into intervals x
The solution of the equation is reduced to the solution of three systems:
Answer: -15, -1.8.
GRAPHIC METHOD FOR SOLVING EQUATIONS CONTAINING MODULE SIGN.
The graphical way of solving the equations is approximate, since the accuracy depends on the selected single 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 particular equation has.
Example. Solve graphically the equation |x - 2| + |x - 3| + |2x - 8| = 9
Solution. Let us construct graphs of functions in one coordinate system
y=|x - 2| + |x - 3| + |2x - 8| and y=9.
To build a graph, it is necessary to consider this function on each interval (-∞; 2); [ 3/2 ; ∞)
Answer: (- ∞ ; 4/3] [ 3/2 ; ∞ )
We also used the method of equivalent transformations in solving the equations | f(x)| = | g(x)|.
EQUATIONS WITH "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 the 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.
O n e t: 1; 3; 7.
Example 2
Solve equation |2 – |x + 1|| = 3.
Solution.
Let's solve the equation by introducing a new variable.
Let | x + 1| = y , then |2 – y | = 3, hence
Let's do the reverse substitution:
(1) | x + 1| = -1 - no solutions.
(2) | x + 1| = 5
A n e t: -6; four.
Example3 .
How many roots does the equation | 2 | x | -6 | = 5 - x?
Solution. Let's solve the equation using equivalence schemes.
Equation | 2 | x | -6 | = 5 -x is equivalent to the system:
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The term (module) in literal translation from Latin means "measure". This concept was introduced into mathematics by the English scientist R. Cotes. And the German mathematician K. Weierstrass introduced the module sign - a symbol by which this concept is denoted when writing.
First this concept studied in mathematics under the 6th grade program high school. According to one definition, the modulus is the absolute value of a real number. In other words, to find out the modulus of a real number, you must discard its sign.
Graphically absolute value a denoted as |a|.
The main distinguishing feature of this concept is that it is always a non-negative value.
Numbers that differ from each other only in sign are called opposite numbers. If the value is positive, then its opposite is negative, and zero is its own opposite.
geometric value
If we consider the concept of the module from the standpoint of geometry, then it will denote the distance, which is measured in unit segments from the origin to given point. This definition fully reveals geometric meaning the term being studied.
Graphically, this can be expressed as follows: |a| = O.A.
Absolute value properties
Below we will consider all the mathematical properties of this concept and ways of writing in the form of literal expressions:
Features of solving equations with a modulus
If we talk about solving mathematical equations and inequalities that contain module, then you need to remember that in order to solve them, you will need to open this sign.
For example, if the sign of the absolute value contains some mathematical expression, then before opening the module, it is necessary to take into account the current mathematical definitions.
|A + 5| = A + 5 if A is greater than or equal to zero.
5-A if A is less than zero.
In some cases, the sign can be unambiguously expanded for any value of the variable.
Let's consider one more example. Let's construct a coordinate line, on which we mark all numerical values, the absolute value of which will be 5.
First you need to draw a coordinate line, designate the origin of coordinates on it and set the size of a single segment. In addition, the line must have a direction. Now on this straight line it is necessary to apply markings that will be equal to the value of a single segment.
Thus, we can see that on this coordinate line there will be two points of interest to us with values 5 and -5.
Solving Equations and Inequalities with Modulus often causes problems. However, if you understand well what is the absolute value of a number, and how to correctly expand expressions containing the modulo sign, then the presence in the equation expression under the module sign ceases to be an obstacle to its solution.
A bit of theory. Each number has two characteristics: the absolute value of the number, and its sign.
For example, the number +5, or just 5, has a "+" sign and an absolute value of 5.
The number -5 has a "-" sign and an absolute value of 5.
The absolute values of the numbers 5 and -5 are 5.
The absolute value of the number x is called the modulus of the number and is denoted by |x|.
As we can see, the modulus of a number is equal to the number itself, if this number is greater than or equal to zero, and to this number with the opposite sign, if this number is negative.
The same applies to any expressions that are under the module sign.
The module expansion rule looks like this:
|f(x)|= f(x) if f(x) ≥ 0, and
|f(x)|= - f(x) if f(x)< 0
For example |x-3|=x-3 if x-3≥0 and |x-3|=-(x-3)=3-x if x-3<0.
To solve an equation containing an expression under the modulus sign, you must first expand module by module expansion rule.
Then our equation or inequality is transformed into two different equations existing on two different numerical intervals.
One equation exists on a numerical interval on which the expression under the modulus sign is non-negative.
And the second equation exists on the interval on which the expression under the modulus sign is negative.
Let's consider a simple example.
Let's solve the equation:
|x-3|=-x 2 +4x-3
1. Let's open the module.
|x-3|=x-3 if x-3≥0, i.e. if x≥3
|x-3|=-(x-3)=3-x if x-3<0, т.е. если х<3
2. We got two numerical intervals: x≥3 and x<3.
Consider what equations the original equation is transformed into on each interval:
A) For x≥3 |x-3|=x-3, and our equation looks like:
Attention! This equation exists only on the interval x≥3!
Let's open the brackets, give similar terms:
and solve this equation.
This equation has roots:
x 1 \u003d 0, x 2 \u003d 3
Attention! since the equation x-3=-x 2 +4x-3 exists only on the interval x≥3, we are only interested in those roots that belong to this interval. This condition satisfies only x 2 =3.
B) At x<0 |x-3|=-(x-3) = 3-x, и наше уравнение приобретает вид:
Attention! This equation exists only on the interval x<3!
Let's open the brackets and give like terms. We get the equation:
x 1 \u003d 2, x 2 \u003d 3
Attention! since the equation 3-x \u003d -x 2 + 4x-3 exists only on the interval x<3, нас интересуют только те корни, которые принадлежат этому промежутку. Этому условию удовлетворяет только х 1 =2.
So: from the first interval we take only the root x=3, from the second - the root x=2.
We don't choose math her profession, and she chooses us.
Russian mathematician Yu.I. Manin
Modulo Equations
The most difficult problems to solve in school mathematics are equations containing variables under the module sign. To successfully solve such equations, it is necessary to know the definition and basic properties of the module. Naturally, students should have the skills to solve equations of this type.
Basic concepts and properties
Modulus (absolute value) of a real number denoted and is defined as follows:
The simple properties of the module include the following relationships:
Note, that the last two properties hold for any even degree.
Also, if , where , then and
More complex module properties, which can be effectively used in solving equations with modules, are formulated by means of the following theorems:
Theorem 1.For any analytic functions and the inequality
Theorem 2. Equality is the same as inequality.
Theorem 3. Equality is equivalent to the inequality.
Consider typical examples of solving problems on the topic “Equations, containing variables under the module sign.
Solving Equations with Modulus
The most common method in school mathematics for solving equations with a modulus is the method, based on module expansion. This method is generic, however, in the general case, its application can lead to very cumbersome calculations. In this regard, students should also be aware of other, more effective methods and methods for solving such equations. In particular, need to have the skills to apply theorems, given in this article.
Example 1 Solve the equation. (one)
Solution. Equation (1) will be solved by the "classical" method - the module expansion method. To do this, we break the numerical axis dots and intervals and consider three cases.
1. If , then , , , and equation (1) takes the form . It follows from here. However, here , so the found value is not the root of equation (1).
2. If , then from equation (1) we obtain or .
Since then the root of equation (1).
3. If , then equation (1) takes the form or . Note that .
Answer: , .
When solving the following equations with a module, we will actively use the properties of modules in order to increase the efficiency of solving such equations.
Example 2 solve the equation.
Solution. Since and then it follows from the equation. In this regard, , , and the equation becomes. From here we get. However , so the original equation has no roots.
Answer: no roots.
Example 3 solve the equation.
Solution. Since , then . If , then , and the equation becomes.
From here we get .
Example 4 solve the equation.
Solution.Let us rewrite the equation in an equivalent form. (2)
The resulting equation belongs to equations of the type .
Taking into account Theorem 2, we can state that equation (2) is equivalent to the inequality . From here we get .
Answer: .
Example 5 Solve the equation.
Solution. This equation has the form. That's why , according to Theorem 3, here we have the inequality or .
Example 6 solve the equation.
Solution. Let's assume that . Because , then the given equation takes the form of a quadratic equation, (3)
where . Since equation (3) has a single positive root and , then . From here we get two roots of the original equation: and .
Example 7 solve the equation. (4)
Solution. Since the equationis equivalent to the combination of two equations: and , then when solving equation (4) it is necessary to consider two cases.
1. If , then or .
From here we get , and .
2. If , then or .
Since , then .
Answer: , , , .
Example 8solve the equation . (5)
Solution. Since and , then . From here and from Eq. (5) it follows that and , i.e. here we have a system of equations
However, this system of equations is inconsistent.
Answer: no roots.
Example 9 solve the equation. (6)
Solution. If we designate and from equation (6) we obtain
Or . (7)
Since equation (7) has the form , this equation is equivalent to the inequality . From here we get . Since , then or .
Answer: .
Example 10solve the equation. (8)
Solution.According to Theorem 1, we can write
(9)
Taking into account equation (8), we conclude that both inequalities (9) turn into equalities, i.e. there is a system of equations
However, by Theorem 3, the above system of equations is equivalent to the system of inequalities
(10)
Solving the system of inequalities (10) we obtain . Since the system of inequalities (10) is equivalent to equation (8), the original equation has a single root .
Answer: .
Example 11. solve the equation. (11)
Solution. Let and , then the equation (11) implies the equality .
From this it follows that and . Thus, here we have a system of inequalities
The solution to this system of inequalities are and .
Answer: , .
Example 12.solve the equation. (12)
Solution. Equation (12) will be solved by the method of successive expansion of modules. To do this, consider several cases.
1. If , then .
1.1. If , then and , .
1.2. If , then . However , therefore, in this case, equation (12) has no roots.
2. If , then .
2.1. If , then and , .
2.2. If , then and .
Answer: , , , , .
Example 13solve the equation. (13)
Solution. Since the left side of equation (13) is non-negative, then and . In this regard, , and equation (13)
takes the form or .
It is known that the equation is equivalent to the combination of two equations and , solving which we get, . Because , then equation (13) has one root.
Answer: .
Example 14 Solve a system of equations (14)
Solution. Since and , then and . Therefore, from the system of equations (14) we obtain four systems of equations:
The roots of the above systems of equations are the roots of the system of equations (14).
Answer: ,, , , , , , .
Example 15 Solve a system of equations (15)
Solution. Since , then . In this regard, from the system of equations (15) we obtain two systems of equations
The roots of the first system of equations are and , and from the second system of equations we obtain and .
Answer: , , , .
Example 16 Solve a system of equations (16)
Solution. It follows from the first equation of system (16) that .
Since then . Consider the second equation of the system. Because the, then , and the equation becomes, , or .
If we substitute the valueinto the first equation of system (16), then , or .
Answer: , .
For a deeper study of problem solving methods, related to the solution of equations, containing variables under the module sign, you can advise tutorials from the list of recommended literature.
1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.
2. Suprun V.P. Mathematics for high school students: tasks of increased complexity. - M .: KD "Librocom" / URSS, 2017. - 200 p.
3. Suprun V.P. Mathematics for high school students: non-standard methods for solving problems. - M .: KD "Librocom" / URSS, 2017. - 296 p.
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