Irrational inequalities. Theory and examples Methods for solving inequalities with roots

Etc. Ivanova

METHODS FOR SOLVING IRRATIONAL INEQUALITIES

CDO and NIT SRPTL

UDC 511 (O75.3)

BBK 22.1Ya72

Compiled by T. D. Ivanova

Reviewer: M. I. Baisheva - Candidate of Pedagogical Sciences, Associate Professor of the Department

mathematical analysis of the Faculty of Mathematics

Institute of Mathematics and Informatics of Yakutsk

state university

Methods for solving irrational inequalities: Methodological guide

M 34 for students in grades 9-11 / comp. Ivanova T.D. from Suntar Suntarsky ulus

RS (Y): CDO NIT SRPTL, 2007, - 56 p.

The manual is addressed to high school students of secondary schools, as well as those entering universities as a methodological guide to solving irrational inequalities. In the manual, the main methods for solving irrational inequalities are discussed in detail, examples of solving irrational inequalities with parameters are given, and examples for an independent solution are proposed. Teachers can use the manual as a didactic material for conducting independent work, when reviewing the topic "Irrational inequalities".

The manual reflects the teacher's experience in studying the topic "Irrational inequalities" with students.

Problems are taken from materials of entrance exams, methodological newspapers and magazines, training manuals, a list of which is given at the end of the manual

UDC 511 (O75.3)

BBK 22.1Ya72

 T.D. Ivanova, comp., 2006.

 CDO NIT SRPTL, 2007.

Foreword 5

Introduction 6

Section I. Examples of solving the simplest irrational inequalities 7

Section II Inequalities of the Form
> g (x), g (x), g (x) 9

Section III. Inequalities of the form
;
;

;
13

Section IV. Inequalities with multiple roots of even degree 16

Section V. Replacement Method (Introducing a New Variable) 20

Section VI. Inequalities of the Form f (x)
0; f (x) 0;

Section VII. Inequalities of the form
25

Section VIII. Using radical expression transformations

in irrational inequalities 26

Section IX. Graphical solution of irrational inequalities 27

Section X. Mixed Inequalities 31

Section XI. Using the monotonicity property of a function 41

Section XII. Function Replacement Method 43

Section XIII. Examples of solving inequalities directly

by the method of intervals 45

Section XIV. Examples of solving irrational inequalities with parameters 46

Literature 56

REVIEW

This teaching aid is intended for students in grades 10-11. As practice shows, schoolchildren and applicants experience special difficulties in solving irrational inequalities. This is due to the fact that in school mathematics this section is considered insufficiently, various methods for solving such inequalities are not considered, in a more extended way. Also, school teachers feel a lack of methodological literature, which manifests itself in a limited amount of problem material with an indication of various approaches, methods of solution.

The tutorial discusses methods for solving irrational inequalities. Ivanova T.D. at the beginning of each section introduces students to the main idea of ​​the method, then examples are shown with explanations, and also offers problems for independent solution.

The compiler uses the most "effective" methods for solving irrational inequalities that are encountered when entering higher educational institutions with increased requirements for students' knowledge.

Students can gain invaluable experience and skill in solving complex irrational inequalities by reading this manual. I believe that this manual will also be useful for mathematics teachers working in specialized classes, as well as for developers of elective courses.

Candidate of Pedagogy, Associate Professor of the Department of Mathematical Analysis of the Mathematical Faculty of the Institute of Mathematics and Informatics, Yakutsk State University

Baisheva M.I.

FOREWORD

The manual is addressed to high school students of secondary schools, as well as those entering universities as a methodological guide to solving irrational inequalities. In the manual, the main methods for solving irrational inequalities are analyzed in detail, approximate examples of how to solve irrational inequalities are given, examples of solving irrational inequalities with parameters are given, and examples for an independent solution are proposed, for some of them short answers and instructions are given.

When analyzing examples, self-solving inequalities, it is assumed that the student is able to solve linear, square and other inequalities, owns various methods for solving inequalities, in particular, the method of intervals. It is proposed to solve the inequality in several ways.

Teachers can use the manual as a didactic material for conducting independent work, when reviewing the topic "Irrational inequalities".

The manual reflects the teacher's experience in studying the topic "Irrational inequalities" with students.

Problems are selected from materials of entrance exams to higher educational institutions, methodological newspapers and magazines on mathematics "September 1", "Mathematics at school", "Quant", textbooks, a list of which is given at the end of the manual.

INTRODUCTION

Irrational inequalities are called inequalities in which variables or a function of a variable are included under the root sign.

The main standard method for solving irrational inequalities is the sequential raising of both sides of the inequality to a power in order to get rid of the root. But this operation often leads to the appearance of extraneous roots or, even, to the loss of roots, i.e. leads to inequality that is not equal to the original one. Therefore, one must very carefully monitor the equivalence of transformations and consider only those values ​​of the variable for which the inequality makes sense:

if the root is even, then the radical expression must be non-negative and the value of the root is also a non-negative number.

if the root of the degree is an odd number, then the radical expression can take any real number and the sign of the root coincides with the sign of the radical expression.

it is possible to raise both parts of the inequality to an even power only after making sure that they are non-negative;

raising both sides of the inequality to the same odd power is always an equivalent transformation.

ChapterI... Examples of solving the simplest irrational inequalities

Examples 1- 6:

Solution:

1.a)
.

b)
.

2.a)

b)

3.a)
.

b)
.

4.a)

b)

5.a)
.

b)

6.a)
.

b)
.

7.

8.a)
.

b)

9.a)
.

b)

11.

12. Find the smallest positive integer x that satisfies the inequality

13.a) Find the midpoint of the interval of the solution to the inequality

b) Find the arithmetic mean of all integer values ​​x for which the inequality has a solution 4

14. Find the smallest negative solution to the inequality

15.a)
;

b)

Section II. Inequalities of the form> g (x), g (x),g (x)

Similarly, as in the solution of examples 1-4, we argue when solving inequalities of the indicated form.

Example 7 : Solve inequality
> NS + 1

Solution: ODZ inequality: NS-3. There are two possible cases for the right side:

a) NS+ 10 (the right side is non-negative) or b) NS + 1

Consider a) If NS+10, i.e. NS- 1, then both sides of the inequality are non-negative. We square both parts: NS + 3 >NS+ 2NS+ 1. We obtain the square inequality NS+ NS – 2 x x - 1, we get -1

Consider b) If NS+1 x x -3

Combining the solutions of cases a) -1 and b) NS-3, we write down the answer: NS
.

It is convenient to write all the reasoning when solving Example 7 as follows:

The original inequality is equivalent to a set of systems of inequalities
.

NS

Answer: .

Reasoning when solving inequalities of the form

1.> g(x); 2. g(x); 3. g(x); 4. g(x) can be briefly written in the form of the following schemes:

I. > g(x)

2. g(x)

3. g(x)

4. g(x)
.

Example 8 :
NS.

Solution: The original inequality is equivalent to the system

x> 0

Answer: NS
.

Tasks for an independent solution:

b)

b)
.

b)

b)

20.a)
x

b)

21. a)

In this lesson we will consider the solution of irrational inequalities, we will give various examples.

Topic: Equations and Inequalities. Systems of equations and inequalities

Lesson:Irrational inequalities

When solving irrational inequalities, it is often necessary to raise both sides of the inequality to some degree, this is a rather important operation. Let us recall the features.

Both sides of the inequality can be squared if both of them are non-negative, only then we get the correct inequality from the true inequality.

Both sides of the inequality can be cube in any case, if the original inequality was true, then when we cube, we get the correct inequality.

Consider an inequality of the form:

The radical expression must be non-negative. The function can take on any value, there are two cases to consider.

In the first case, both sides of the inequality are non-negative, we have the right to square. In the second case, the right-hand side is negative, and we have no right to square. In this case, it is necessary to look at the meaning of inequality: here a positive expression (square root) is greater than a negative expression, which means that inequality is always satisfied.

So, we have the following solution scheme:

In the first system, we do not protect the radical expression separately, since when the second inequality of the system is satisfied, the radical expression must automatically be positive.

Example 1 - Solve Inequality:

According to the scheme, we pass to the equivalent set of two systems of inequalities:

Let's illustrate:

Rice. 1 - illustration of the solution of example 1

As we can see, when getting rid of irrationality, for example, when squaring, we get a set of systems. Sometimes this complex design can be simplified. In the resulting set, we have the right to simplify the first system and get an equivalent set:

As an independent exercise, it is necessary to prove the equivalence of these populations.

Consider an inequality of the form:

Similar to the previous inequality, we consider two cases:

In the first case, both sides of the inequality are non-negative, we have the right to square. In the second case, the right-hand side is negative, and we have no right to square. In this case, it is necessary to look at the meaning of inequality: here a positive expression (square root) is less than a negative expression, which means that inequality is contradictory. There is no need to consider the second system.

We have an equivalent system:

Sometimes an irrational inequality can be solved graphically. This method is applicable when the corresponding graphs can be easily built and their intersection points can be found.

Example 2 - Solve inequalities graphically:

a)

b)

We have already solved the first inequality and we know the answer.

To solve inequalities graphically, you need to plot the function on the left side and the function on the right side.

Rice. 2. Graphs of functions and

To plot the function graph, it is necessary to transform the parabola into a parabola (mirror it about the y-axis), shift the resulting curve by 7 units to the right. The graph confirms that this function decreases monotonically in its domain of definition.

The function graph is a straight line, it is easy to plot it. The y-intercept is (0; -1).

The first function decreases monotonically, the second increases monotonically. If the equation has a root, then it is the only one, it is easy to guess from the graph:.

When the argument is less than the root, the parabola is above the straight line. When the argument is between three and seven, the line is above the parabola.

We have the answer:

An effective method for solving irrational inequalities is the method of intervals.

Example 3 - Solve inequalities using the interval method:

a)

b)

according to the method of intervals, it is necessary to temporarily move away from inequality. To do this, transfer everything in the given inequality to the left side (get zero on the right) and introduce a function equal to the left side:

now it is necessary to examine the resulting function.

ODZ:

We have already solved this equation graphically, so we do not dwell on determining the root.

Now it is necessary to select intervals of constancy and determine the sign of the function at each interval:

Rice. 3. Intervals of constancy for example 3

Recall that to determine the signs on an interval, it is necessary to take a sample point and substitute it into the function; the resulting sign will be retained by the function throughout the entire interval.

Let's check the value at the boundary point:

The obvious answer is:

Consider the following type of inequalities:

First, we write down the ODZ:

The roots exist, they are non-negative, we can square both parts. We get:

We got an equivalent system:

The resulting system can be simplified. When the second and third inequalities are satisfied, the first is automatically true. We have:

Example 4 - Solve Inequality:

We act according to the scheme - we get an equivalent system.

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Any inequality that includes a function under the root is called irrational... There are two types of such inequalities:

In the first case, the root is less than the function g (x); in the second, it is larger. If g (x) - constant, inequality is drastically simplified. Please note: outwardly, these inequalities are very similar, but their solution schemes are fundamentally different.

Today we will learn how to solve irrational inequalities of the first type - they are the simplest and most understandable. The inequality sign can be strict or non-strict. The following statement is true for them:

Theorem. Any irrational inequality of the form

Equivalent to the system of inequalities:

Not weak? Let's take a look at where such a system comes from:

1. f (x) ≤ g 2 (x) - everything is clear here. This is the original squared inequality;
2. f (x) ≥ 0 is the ODZ of the root. Let me remind you: the arithmetic square root exists only from non-negative numbers;
3. g (x) ≥ 0 is the range of the root. By squaring inequality, we burn the cons. As a result, extra roots may arise. The inequality g (x) ≥ 0 cuts them off.

Many students "get stuck" on the first inequality of the system: f (x) ≤ g 2 (x) - and completely forget the other two. The result is predictable: wrong decision, lost points.

Since irrational inequalities are a rather complex topic, we will analyze 4 examples at once. From elementary to really complex. All tasks are taken from the entrance exams of the Moscow State University. M.V. Lomonosov.

Examples of problem solving

Task. Solve the inequality:

Before us is the classic irrational inequality: f (x) = 2x + 3; g (x) = 2 is a constant. We have:

By the end of the solution, only two of the three inequalities remain. Because the inequality 2 ≥ 0 always holds. We intersect the remaining inequalities:

So, x ∈ [−1,5; 0.5]. All dots are filled because inequalities are not strict.

Task. Solve the inequality:

We apply the theorem:

We solve the first inequality. To do this, we open the square of the difference. We have:

2x 2 - 18x + 16< (x − 4) 2 ;
2x 2 - 18x + 16< x 2 − 8x + 16:
x 2 - 10x< 0;
x (x - 10)< 0;
x ∈ (0; 10).

Now let's solve the second inequality. There too square trinomial:

2x 2 - 18x + 16 ≥ 0;
x 2 - 9x + 8 ≥ 0;
(x - 8) (x - 1) ≥ 0;
x ∈ (−∞; 1] ∪∪∪∪)