Linear equations with modulus. How to solve modulo equations: basic rules

This online math calculator will help you solve an equation or inequality with moduli. Program for solving equations and inequalities with moduli not only gives the answer to the problem, it leads detailed solution with explanations

, i.e. displays the process of obtaining the result.

This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

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x^2 + 2|x-1| -6 = 0
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Solving the inequality \(|2x+7|

But the main way to solve equations and inequalities with moduli is associated with the so-called “revelation of the modulus by definition”:
if \(a \geq 0 \), then \(|a|=a \);
if \(a As a rule, an equation (inequality) with moduli is reduced to a set of equations (inequalities) that do not contain the modulus sign.

In addition to the above definition, the following statements are used:
1) If \(c > 0\), then the equation \(|f(x)|=c \) is equivalent to the set of equations: \(\left[\begin(array)(l) f(x)=c \\ f(x)=-c \end(array)\right.
2) If \(c > 0 \), then the inequality \(|f(x)| 3) If \(c \geq 0 \), then the inequality \(|f(x)| > c \) is equivalent to a set of inequalities : \(\left[\begin(array)(l) f(x) c \end(array)\right. \)
4) If both sides of the inequality \(f(x) EXAMPLE 1. Solve the equation \(x^2 +2|x-1| -6 = 0\).

If \(x-1 \geq 0\), then \(|x-1| = x-1\) and the given equation takes the form
\(x^2 +2(x-1) -6 = 0 \Rightarrow x^2 +2x -8 = 0 \).
If \(x-1 \(x^2 -2(x-1) -6 = 0 \Rightarrow x^2 -2x -4 = 0 \).
Thus, the given equation should be considered separately in each of the two indicated cases.
1) Let \(x-1 \geq 0 \), i.e. \(x\geq 1\). From the equation \(x^2 +2x -8 = 0\) we find \(x_1=2, \; x_2=-4\).
The condition \(x \geq 1 \) is satisfied only by the value \(x_1=2\).

2) Let \(x-1 Answer: \(2; \;\; 1-\sqrt(5) \)

EXAMPLE 2. Solve the equation \(|x^2-6x+7| = \frac(5x-9)(3)\). First way
(module expansion by definition).

Reasoning as in example 1, we come to the conclusion that the given equation needs to be considered separately if two conditions are met: \(x^2-6x+7 \geq 0 \) or \(x^2-6x+7
1) If \(x^2-6x+7 \geq 0 \), then \(|x^2-6x+7| = x^2-6x+7 \) and the given equation takes the form \(x^2 -6x+7 = \frac(5x-9)(3) \Rightarrow 3x^2-23x+30=0 \). Having solved this quadratic equation, we get: \(x_1=6, \; x_2=\frac(5)(3) \).
Let's find out whether the value \(x_1=6\) satisfies the condition \(x^2-6x+7 \geq 0\). To do this, substitute the indicated value into the quadratic inequality. We get: \(6^2-6 \cdot 6+7 \geq 0 \), i.e. \(7 \geq 0 \) is a true inequality.

2) If \(x^2-6x+7 Value \(x_3=3\) satisfies the condition \(x^2-6x+7 Value \(x_4=\frac(4)(3) \) does not satisfy the condition \ (x^2-6x+7 So, the given equation has two roots: \(x=6, \; x=3 \).

Second way. If the equation is given \(|f(x)| = h(x) \), then with \(h(x) \(\left[\begin(array)(l) x^2-6x+7 = \frac (5x-9)(3) \\ x^2-6x+7 = -\frac(5x-9)(3) \end(array)\right \)
Both of these equations were solved above (using the first method of solving the given equation), their roots are as follows: \(6,\; \frac(5)(3),\; 3,\; \frac(4)(3)\). The condition \(\frac(5x-9)(3) \geq 0 \) of these four values ​​is satisfied only by two: 6 and 3. This means that the given equation has two roots: \(x=6, \; x=3 \ ).

Third way(graphic).
1) Let's build a graph of the function \(y = |x^2-6x+7| \). First, let's construct a parabola \(y = x^2-6x+7\).
We have \(x^2-6x+7 = (x-3)^2-2 \). The graph of the function \(y = (x-3)^2-2\) can be obtained from the graph of the function \(y = x^2 \) by shifting it by 3 scale units to the right (along the x-axis) and by 2 scale units down ( along the y-axis).
The straight line x=3 is the axis of the parabola we are interested in. As control points for more accurate plotting, it is convenient to take point (3; -2) - the vertex of the parabola, point (0; 7) and point (6; 7) symmetrical to it relative to the axis of the parabola.

To now construct a graph of the function \(y = |x^2-6x+7| \), you need to leave unchanged those parts of the constructed parabola that lie not below the x-axis, and mirror that part of the parabola that lies below the x-axis relative to the x axis.

2) Let's build a graph of the linear function \(y = \frac(5x-9)(3)\). It is convenient to take points (0; –3) and (3; 2) as control points. It is important that the point x = 1.8 of the intersection of the straight line with the abscissa axis is located to the right of the left point of intersection of the parabola with the abscissa axis - this is the point \(x=3-\sqrt(2) \) (since \(3-\sqrt(2 ) 3) Judging by the drawing, the graphs intersect at two points - A(3; 2) and B(6; 7). Substituting the abscissas of these points x = 3 and x = 6 into the given equation, we are convinced that in both cases. In another value, the correct numerical equality is obtained. This means that our hypothesis was confirmed - the equation has two roots: x = 3 and x = 6. Answer: 3;

Comment

EXAMPLE 2. Solve the equation \(|x^2-6x+7| = \frac(5x-9)(3)\).
. The graphical method, for all its elegance, is not very reliable. In the example considered, it worked only because the roots of the equation are integers.

Consider the first interval: \((-\infty; \; -3) \).
If x Consider the second interval: \([-3; \; 2) \).
If \(-3 \leq x Consider the third interval: \(. Now we expand the internal module for x>2.5. We obtain an equation with one module
|2x-5-1|=x+3;
|2x-6|=x+3.
When expanding the module, we obtain the following linear equations
-2x+6=x+3 or 2x-6=x+3;
2x+x=6-3 or 2x-x=3+6;
3x=3; x=1 or x=9 .
The first value x=1 does not satisfy the condition x>2.5.
So on this interval we have one root of the equation with modulus x=9, and there are two in total (x=1/3). By substitution you can check the correctness of the calculations performed

Answer: x=1/3; x=9. Example 4.
Find solutions to the double module ||3x-1|-5|=2x-3.
Solution: Let's expand the internal module of the equation <=>|3x-1|=0
x=1/3.
The point x=2.5 divides the number line into two intervals and the given equation into two cases. We write down the condition for the solution based on the form of the equation on the right side 2x-3>=0
-> x>=3/2=1.5. It follows that we are interested in values ​​>=1.5. Thus modular equation
,
consider on two intervals
|-(3x-1)-5|=2x-3;
|-3x-4|=2x-3.
The resulting module, when expanded, is divided into 2 equations
-3x-4=2x-3 or 3x+4=2x-3;
2x+3x=-4+3 or 3x-2x=-3-4;
5x=-1; x=-1/5 or x=-7 .
Both values ​​do not fall into the interval, that is, they are not solutions to the equation with moduli. Next, we will expand the module for x>2.5. We get the following equation
|3x-1-5|=2x-3;.
|3x-6|=2x-3
Expanding the module, we get 2 linear equations 3x-6=2x-3 or
–(3x-6)=2x-3; 3x-2x=-3+6
or 2x+3x=6+3;
x=3 or 5x=9; x=9/5=1.8.
The second value found does not correspond to the condition x>2.5, we reject it.
Finally we have one root of the equation with modules x=3.
||3*3-1|-5|=2*3-3 3=3 .
Performing a check
So on this interval we have one root of the equation with modulus x=9, and there are two in total (x=1/3). By substitution you can check the correctness of the calculations performed

The root of the equation with the modulus was calculated correctly.

Tochilkina Yulia

The work presents various methods for solving equations with a modulus.

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Municipal budgetary educational institution

"Secondary school No. 59"

Equations with modulus

Abstract work Performed

9A class student

MBOU "Secondary School No. 59" Barnaul

Tochilkina Yulia

Supervisor

Zakharova Lyudmila Vladimirovna,

9A class student

mathematic teacher

Barnaul 2015

I'm in ninth grade. This academic year I will have to take the final certification for the basic school course. To prepare for the exam, we purchased the collection of D. A. Maltsev Mathematics. 9th grade. Looking through the collection, I discovered equations containing not only one, but also several modules. The teacher explained to me and my classmates that such equations are called “nested module” equations. This name seemed unusual to us, and the solution, at first glance, was quite complicated. This is how the topic for my work “Equations with modulus” appeared. I decided to study this topic more deeply, especially since it will be useful to me when taking exams at the end of the school year and I think it will be needed in grades 10 and 11. Everything said above determines the relevance of the topic I have chosen.

Goal of the work :

1. Consider various methods for solving equations with modulus.
2. Learn to solve equations containing an absolute value sign using various methods

To work on the topic, the following tasks were formulated:

Tasks:

1. Study theoretical material on the topic “Modulus of a real number.”
2. Consider methods for solving equations and consolidate the acquired knowledge by solving problems.
3. Apply the acquired knowledge when solving various equations containing the modulus sign in high school

Object of study:methods for solving equations with modulus

Subject of study:equations with modulus

Research methods:

Theoretical : study of literature on the research topic;

Internet - information.

Analysis information obtained from studying the literature; results obtained by solving equations with modulus in various ways.

Comparison methods for solving equations is the subject of the rationality of their use when solving various equations with a modulus.

“We start thinking when we hit something.” Paul Valery.

1. Concepts and definitions.

The concept of “module” is widely used in many sections of the school mathematics course, for example, in the study of absolute and relative errors of an approximate number; in geometry and physics the concepts of a vector and its length (vector modulus) are studied. The concepts of the module are 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”. This word has many meanings and is used not only in mathematics, physics and technology, but also in architecture, programming and other exact sciences.

It is believed that the term was proposed by Cotes, a student of Newton. The modulus sign was introduced in the 19th century by Weierstrass.

In architecture, a module is the initial unit of measurement established for a given architectural structure.

In technology, this is a term used in various fields of technology, used to designate various coefficients and quantities, for example, elastic modulus, engagement modulus...

In mathematics, modulus has several meanings, but I will consider it as the absolute value of a number.

Definition1: Modulus (absolute value) of a real number A this number itself is called if A ≥0, or the opposite number – and if A the modulus of zero is zero.

When solving equations with a modulus, it is convenient to use the properties of the modulus.

Let's consider the evidence of properties 5,6,7.

Statement 5. Equality │ a+b │=│ a │+│ b │ is true if av ≥ 0.

Proof. Indeed, after squaring both sides of this equality, we obtain │ a+b │²=│ a │²+2│ ab │+│ c │²,

a²+ 2 ab+b²=a²+ 2│ ab │+ b², from where │ ab │= ab

And the last equality will be true when av ≥0.

Statement 6. Equality │ a-c │=│ a │+│ c │ is true when av ≤0.

Proof. To prove it, it is sufficient in the equality

│ а+в │=│ а │+│ в │ replace в with - в, then а· (- в ) ≥0, whence ав ≤0.

Statement 7. Equality │ a │+│ b │= a+b performed at a ≥0 and b ≥0.

Proof . Having considered four cases a ≥0 and b ≥0; a ≥0 and c A in ≥0; A V a ≥0 and b ≥0.

(a-c) in ≥0.

Geometric interpretation

|a| is the distance on the coordinate line from the point with the coordinate A , to the origin.

|-a| |a|

A 0 a x

Geometric interpretation of the meaning of |a| clearly confirms that |-a|=|a|

If a 0, then on the coordinate line there are two points a and –a, equidistant from zero, 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 a module, the main methods for solving equations with a module follow:

1. "Expanding" a module (i.e. using a definition);
2. Using the geometric meaning of the module (property 2);
3. Graphic solution method;
4. Using equivalent transformations (properties 4.6);
5. Replacement of a variable (this uses property 5).
6. Interval method.

I solved quite a large number of examples, but in this work I present to your attention only a few, in my opinion, typical examples, solved in different ways, because the rest duplicate each other and in order to understand how to solve equations with a modulus there is no need to consider all the solved examples.

SOLVING EQUATIONS | f(x)| = a

Consider the equation | f(x)| = a, a R

An equation of this type can be solved by the definition of the modulus:

If A then the equation has no roots.

If a= 0, then the equation is equivalent to f(x)=0.

If a>0, then the equation is equivalent to the set

Example. Solve the equation |3x+2|=4.

Solution.

|3x+2|=4, then 3x+2=4,

3x+2= -4;

X=-2,

X=2/3

ANSWER: -2;2/3.

SOLVING EQUATIONS USING THE GEOMETRICAL PROPERTIES OF THE MODULE.

Example 1. Solve the equation /x-1/+/x-3/=6.

Solution.

Solving this equation means finding all such points on the numerical axis Ox, for each of which the sum of the distances from it to the points with coordinates 1 and 3 is equal to 6.

Not a single point from the segmentdoes not satisfy this condition, because the sum of the indicated distances is equal to 2. Outside this segment there are two points: 5 and -1.

1 1 3 5

Answer: -1;5

Example 2. Solve equation |x 2 +x-5|+|x 2 +x-9|=10.

Solution.

Let's denote x 2 +x-5= a, then / a /+/ a-4 /=10. Let's find points on the Ox axis such that for each of them the sum of the distances to points with coordinates 0 and 4 is equal to 10. This condition is satisfied by -4 and 7.

3 0 4 7

So x 2 +x-5= 4 x 2 +x-5=7

X 2 +x-2=0 x 2 +x-12=0

X 1= 1, x 2= -2 x 1= -4, x 2= 3 Answer: -4;-2; 1; 3.

SOLVING EQUATIONS | f(x)| = | g(x)|.

1. Since | a |=|in |, if a= in, then an equation of the form | f(x)| = | g(x )| equivalent to the totality

Example 1.

Solve the equation | x –2| = |3 – x |.

Solution.

This equation is equivalent to two equations:

x – 2 = 3 – x (1) and x – 2 = –3 + x (2)

2 x = 5 –2 = –3 – incorrect

X = 2.5 the equation has no solutions.

ANSWER: 2.5.

Example 2.

Solve equation |x 2 +3x-20|= |x 2 -3x+ 2|.

Solution.

Since both sides of the equation are non-negative, thensquaring is an equivalent transformation:

(x 2 +3x-20) 2 = (x 2 -3x+2) 2

(x 2 +3x-20) 2 - (x 2 -3x+2) 2 =0,

(x 2 +3x-20-x 2 +3x-2) (x 2 +3x-20+x 2 -3x+2)=0,

(6x-22)(2x 2 -18)=0,

6x-22=0 or 2x 2 -18=0;

X=22/6, x=3, x=-3.

X=11/3

Answer: -3; 3; 11/3.

SOLUTION OF EQUATIONS OF THE VIEW | f(x)| = g(x).

The difference between these equations and| f(x)| =a the fact that the right side is also a variable. And it can be both positive and negative. Therefore, you need to specifically verify its non-negativity, because the modulus cannot be equal to a negative number (property№1 )

1 way

Solution of the equation | f(x)| = g(x ) reduces to a set of solutions to the equationsand checking the fairness of inequality g(x )>0 for the found values ​​of the unknown.

Method 2 (by module definition)

Since | f(x)| = g(x) if f(x) = 0; | f(x)| = - f(x) if f(x)

Example.

Solve equation |3 x –10| = x – 2.

Solution.

This equation is equivalent to the combination of two systems:

ANSWER: 3; 4.

SOLUTION OF EQUATIONS OF THE FORM |f 1 (x)|+|f 2 (x)|+…+|f n (x)|=g(x)

The solution of equations of this type is based on the definition of the modulus. For each function f 1 (x), f 2 (x), …, f n (x) it is necessary to find the domain of definition, its zeros and discontinuity points, dividing the general domain of definition into intervals, in each of which the functions f 1 (x), f 2 (x), …, f n (x) retain their sign. Next, using the definition of the module, for each of the found areas we obtain an equation that must be solved on this interval. This method is called "interval method»

Example.

Solve the equation |x-2|-3|x+4|=1.

Solution.

Let's find the points at which the submodular expressions are equal to zero

x-2=0, x+4=0,

x=2; x=-4.

Let's divide the number line into intervals x

Solving the equation comes down to solving three systems:

ANSWER: -15, -1.8.

GRAPHICAL METHOD FOR SOLVING EQUATIONS CONTAINING MODULE SIGN.

The graphical method of solving equations is approximate, since the accuracy depends on the selected unit segment, the thickness of the pencil, the angles at which the lines intersect, etc. But this method allows you to estimate how many solutions a given equation has.

Example. Solve graphically the equation |x - 2| + |x - 3| + |2x - 8| = 9

Solution. Let's construct graphs of functions in one coordinate system

y=|x - 2| + |x - 3| + |2x - 8| and y=9.

To construct a graph, it is necessary to consider this function on each interval (-∞; 2);

[ 3/2 ; ∞ )

Answer: (- ∞ ; 4/3] [ 3/2 ; ∞ ) f(x)| = | g(x)|.

We also used the method of equivalent transformations when solving the equations |

EQUATIONS WITH A COMPLEX MODULE

Example 1.

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.

Solution.

Solve the equation ||||x| – |–2| –1| –2| = 2.

By definition of a module, we have:

1. Let's solve the first equation.

||| x |–2| –1| = 4

| x | – 2 = 5;

| x | = 7;

x = 7.

1. 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.

Example 2.

Answer: 1; 3; 7.

Solution.

Solve the equation |2 – |x + 1|| = 3.

Let's solve the equation by introducing a new variable.

Let | x + 1| = y, then |2 – y | = 3, from here

(1) | x + 1| = –1 – no solutions.

(2) | x + 1| = 5

ANSWER: –6; 4.

Example3.

How many roots does the equation have | 2 | x | -6 | = 5 - x?

Solution. Let's solve the equation using equivalence schemes.

Equation | 2 | x | -6 | = 5 is equivalent to the system: