In this lesson, we will continue to solve rational inequalities of increased complexity using the interval method. In the examples, more complex combined functions will be used and typical errors that arise when solving such inequalities will be considered.
Theme: Dietreal inequalities and their systems
Lesson: Solving Rational Inequalitiespovextreme complexity
1. Lesson topic, introduction
We solved rational inequalities form and the interval method was used to solve them. The function was either linear, or fractional linear, or a polynomial.
2. Problem solving
Let us consider inequalities of another type.
1. Solve the inequality
We transform the inequality using equivalent transformations.
Now we can explore the function 
Consider a function with no roots.
Let's schematically depict and read the graph of the function (Fig. 1).
The function is positive for any .
Since we have established that 
we can divide both sides of the inequality by this expression.
For a fraction to be positive, the numerator must have a positive denominator. 
Let's consider a function.
Let's schematically depict the graph of the function - a parabola, which means the branches are directed downwards (Fig. 2).
2. Solve the inequality
Consider the function 
1. Domain of definition
2. Function zeros
3. Select intervals of constancy.
4. Arranging the signs (Fig. 3).
If the parenthesis is in an odd degree, when passing through the root, the function changes sign. If the parenthesis is to an even power, the function does not change sign.
We made a typical mistake - we did not include the root in the answer. AT this case equality to zero is allowed, because the inequality is not strict.
In order to avoid such mistakes, it is necessary to remember that
Answer: 
We considered the interval method for complex inequalities and possible typical errors, as well as ways to eliminate them.
Let's consider one more example.
3. Solve the inequality
Let's factorize each bracket separately.
, so this factor can be ignored.
Now you can apply the interval method.
Consider 
We will not reduce the numerator and denominator by, this is a mistake.
1. Domain of definition
2. We already know the zeros of the function
It is not the zero of the function, because it is not included in the domain of definition - in this case, the denominator zero.
3. Determine the intervals of sign constancy.
4. We place signs on the intervals and select the intervals that satisfy our conditions (Fig. 4).
3. Conclusion
We have considered inequalities of increased complexity, but the interval method gives us the key to solving them, so we will use it in the future.
1. Mordkovich A. G. et al. Algebra 9th grade: Proc. For general education Institutions. - 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.
2. Mordkovich A. G. et al. Algebra 9th grade: Task book for students of educational institutions / A. G. Mordkovich, T. N. Mishustina et al. - 4th ed. — M.: Mnemosyne, 2002.-143 p.: ill.
3. Yu. N. Makarychev, Algebra. Grade 9: textbook. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. - 7th ed., Rev. and additional - M .: Mnemosyne, 2008.
4. Sh. A. Alimov, Yu. M. Kolyagin, and Yu. V. Sidorov, Algebra. Grade 9 16th ed. - M., 2011. - 287 p.
5. Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 12th ed., erased. — M.: 2010. — 224 p.: ill.
6. Algebra. Grade 9 At 2 hours. Part 2. Task book for students of educational institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. - 12th ed., Rev. — M.: 2010.-223 p.: ill.
1. Mordkovich A. G. et al. Algebra 9th grade: Task book for students of educational institutions / A. G. Mordkovich, T. N. Mishustina et al. - 4th ed. - M .: Mnemosyne, 2002.-143 p.: ill. No. 37; 45(a, c); 47(b, d); 49.
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To figure out how to solve quadratic equations, we need to figure out what a quadratic function is and what properties it has.
Surely you wondered why a quadratic function is needed at all? Where can we apply its graph (parabola)? Yes, you just have to look around, and you will notice that every day in Everyday life you face her. Have you noticed how a thrown ball flies in physical education? "In an arc"? The most correct answer would be "in a parabola"! And along what trajectory does the jet move in the fountain? Yes, also in a parabola! And how does a bullet or projectile fly? That's right, also in a parabola! Thus, knowing the properties quadratic function, it will be possible to solve many practical problems. For example, at what angle should you throw the ball to ensure the greatest flight range? Or where will the projectile end up if it is fired at a certain angle? etc.
quadratic function
So, let's figure it out.
For example, . What are equal here, and? Well, of course, and!
What if, i.e. less than zero? Well, of course, we are “sad”, which means that the branches will be directed downwards! Let's look at the chart.
This figure shows a graph of a function. Since, i.e. less than zero, the branches of the parabola point downwards. In addition, you probably already noticed that the branches of this parabola intersect the axis, which means that the equation has 2 roots, and the function takes both positive and negative values!
At the very beginning, when we gave the definition of a quadratic function, it was said that and are some numbers. Can they be equal to zero? Well, of course they can! I’ll even reveal an even bigger secret (which is not a secret at all, but it’s worth mentioning): no restrictions are imposed on these numbers (and) at all!
Well, let's see what happens to the graphs if and are equal to zero.
As you can see, the graphs of the considered functions (u) have shifted so that their vertices are now at the point with coordinates, that is, at the intersection of the axes and, this did not affect the direction of the branches. Thus, we can conclude that they are responsible for the "movement" of the parabola graph along the coordinate system.
The function graph touches the axis at a point. So the equation has one root. Thus, the function takes values greater than or equal to zero.
We follow the same logic with the graph of the function. It touches the x-axis at a point. So the equation has one root. Thus, the function takes values less than or equal to zero, that is.
Thus, to determine the sign of an expression, the first thing to do is to find the roots of the equation. This will be very useful to us.
Square inequality
Square inequality is an inequality consisting of a single quadratic function. Thus, all quadratic inequalities are reduced to the following four types:
When solving such inequalities, we will need the ability to determine where the quadratic function is greater, less, or equal to zero. I.e:
- if we have an inequality of the form, then in fact the problem is reduced to determining the numerical range of values for which the parabola lies above the axis.
- if we have an inequality of the form, then in fact the problem comes down to determining the numerical interval of x values for which the parabola lies below the axis.
If the inequalities are not strict (and), then the roots (the coordinates of the intersections of the parabola with the axis) are included in the desired numerical interval, with strict inequalities they are excluded.
This is all quite formalized, but do not despair and be afraid! Now let's look at examples, and everything will fall into place.
When solving quadratic inequalities, we will adhere to the above algorithm, and inevitable success awaits us!
| Algorithm | Example: |
| 1) Let's write the quadratic equation corresponding to the inequality (simply change the inequality sign to the equal sign "="). | |
| 2) Find the roots of this equation. | |
| 3) Mark the roots on the axis and schematically show the orientation of the branches of the parabola ("up" or "down") |  |
| 4) Let's place on the axis the signs corresponding to the sign of the quadratic function: where the parabola is above the axis, put "", and where below - "". |  |
| 5) We write out the interval (s) corresponding to "" or "", depending on the inequality sign. If the inequality is not strict, the roots are included in the interval; if it is strict, they are not included. |
Got it? Then fasten forward!
Well, did it work? If you have any difficulties, then understand the solutions.
Decision:
Let's write out the intervals corresponding to the sign " ", since the inequality sign is " ". The inequality is not strict, so the roots are included in the intervals:
We write the corresponding quadratic equation:
Let's find the roots of this quadratic equation:
We schematically mark the obtained roots on the axis and arrange the signs:
Let's write out the intervals corresponding to the sign " ", since the inequality sign is " ". The inequality is strict, so the roots are not included in the intervals:
We write the corresponding quadratic equation:
Let's find the roots of this quadratic equation:
this equation has one root
We schematically mark the obtained roots on the axis and arrange the signs:
Let's write out the intervals corresponding to the sign " ", since the inequality sign is " ". For any function takes non-negative values. Since the inequality is not strict, the answer is
We write the corresponding quadratic equation:
Let's find the roots of this quadratic equation:
Schematically draw a graph of a parabola and place the signs:
Let's write out the intervals corresponding to the sign " ", since the inequality sign is " ". For any, the function takes positive values, therefore, the solution to the inequality will be the interval:
SQUARE INEQUALITIES. MIDDLE LEVEL
Quadratic function.
Before talking about the topic of "square inequalities", let's remember what a quadratic function is and what its graph is.
| A quadratic function is a function of the form |
In other words, this second degree polynomial.
The graph of a quadratic function is a parabola (remember what that is?). Its branches are directed upwards if "a) the function takes only positive values for all, and in the second () - only negative:
In the case when the equation () has exactly one root (for example, if the discriminant is zero), this means that the graph touches the axis:
Then, similarly to the previous case, for , the function is nonnegative for all, and for , it is nonpositive.
So, after all, we have recently learned to determine where the quadratic function is greater than zero, and where it is less:
If the quadratic inequality is not strict, then the roots are included in the numerical interval, if strict, they are not.
If there is only one root, it's okay, there will be the same sign everywhere. If there are no roots, everything depends only on the coefficient: if, then the whole expression is greater than 0, and vice versa.
Examples (decide for yourself):
Answers:
There are no roots, so the entire expression on the left side takes the sign of the highest coefficient: for all. This means that there are no solutions to the inequality.
If the quadratic function on the left side is “incomplete”, the easier it is to find the roots:
SQUARE INEQUALITIES. BRIEFLY ABOUT THE MAIN
quadratic function is a function of the form:
The graph of a quadratic function is a parabola. Its branches are directed upwards if, and downwards if:
- If you want to find a number interval on which the square trinomial is greater than zero, then this is the number interval where the parabola lies above the axis.
- If you want to find a number interval on which the square trinomial is less than zero, then this is the number interval where the parabola lies below the axis.
Types of square inequalities:
All quadratic inequalities are reduced to the following four types:
Solution algorithm:
| Algorithm | Example: |
| 1) Let's write the quadratic equation corresponding to the inequality (simply change the inequality sign to the equal sign ""). | |
| 2) Find the roots of this equation. | |
| 3) Mark the roots on the axis and schematically show the orientation of the branches of the parabola ("up" or "down") | |
| 4) Let's place on the axis the signs corresponding to the sign of the quadratic function: where the parabola is above the axis, we put "", and where it is lower - "". | |
| 5) We write out the interval (s) corresponding to (s) "" or "", depending on the inequality sign. If the inequality is not strict, the roots are included in the interval; if the inequality is strict, they are not included. |
Well, the topic is over. If you are reading these lines, then you are very cool.
Because only 5% of people are able to master something on their own. And if you have read to the end, then you are in the 5%!
Now the most important thing.
You've figured out the theory on this topic. And, I repeat, it's ... it's just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough ...
For what?
For successful passing the exam, for admission to the institute on the budget and, MOST IMPORTANTLY, for life.
I will not convince you of anything, I will just say one thing ...
People who received a good education, earn much more than those who did not receive it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because much more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the exam and be ultimately ... happier?
FILL YOUR HAND, SOLVING PROBLEMS ON THIS TOPIC.
On the exam, you will not be asked theory.
You will need solve problems on time.
And, if you haven’t solved them (LOTS!), you will definitely make a stupid mistake somewhere or simply won’t make it in time.
It's like in sports - you need to repeat many times to win for sure.
Find a collection anywhere you want necessarily with solutions, detailed analysis and decide, decide, decide!
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If you don't like our tasks, find others. Just don't stop with theory.
“Understood” and “I know how to solve” are completely different skills. You need both.
Find problems and solve!
Introduction…………………………………………………………… 3
1. Classification of errors with examples…………………………… .…… …5
1.1. Classification by types of tasks…… ……………………… … ……….5
1.2. Classification by types of transformations…………………………………10
2. Tests……………………………………………….… .…………………….12
3. Protocols of decisions……………………….….……………………… 18
3.1. Protocols of incorrect solutions …………………………… ... 18
3.2. Answers (protocols of correct decisions)………………………………….34
3.3. Mistakes made in decisions…………………………………… 51
Appendix……………………….……………………………………………… 53
Literature………………………………………………………………………….56
INTRODUCTION
“They learn from mistakes,” says folk wisdom. But in order to learn from a negative experience, first of all, you need to see the error. Unfortunately, the student is often unable to detect it when solving a particular problem. As a result, the idea arose to conduct a study, the purpose of which is to identify typical mistakes made by students, as well as to classify them as fully as possible.
As part of this study, a large set of problems was considered and solved from the options for April testing, tests and written assignments for entrance exams at Omsk State University, various manuals and collections of tasks for applicants to universities, materials were carefully studied correspondence school at NOF OmGU. The data received was subjected to detailed analysis, while much attention was paid to the logic of decisions. Based on these data, the most frequently made mistakes, that is, typical ones, were identified.
Based on the results of this analysis, an attempt was made to systematize the characteristic errors and classify them by types of transformations and types of problems, among which the following were considered: quadratic inequalities, systems of inequalities, fractional-rational equations, equations with modulus, irrational equations, systems of equations, tasks for movement, tasks for work and labor productivity, trigonometric equations, systems trigonometric equations, planimetry.
The classification is accompanied by an illustration in the form of incorrect decision protocols, which makes it possible to help students develop the ability to check and control themselves, critically evaluate their activities, find errors and ways to eliminate them.
The next step was to work with tests. For each task, five answers were offered, of which one is correct, and the remaining four are incorrect, but they are not taken randomly, but correspond to the solution in which a specific error, standard for tasks of this type, was made. This provides a basis for predicting the degree of “roughness” of the error and the development of the main mental operations(analysis, synthesis, comparison, generalization). Tests have the following structure:
Error codes are divided into three types: OK - the correct answer, a numeric code - an error from the classification by task types, an alphabetic code - an error from the classification by types of transformations. Their decoding can be found in Chapter 1. Classification of errors with examples.
Further tasks were offered to find an error in the solution. These materials were used when working with students of the correspondence school at the NOF OmSU, as well as in the advanced training courses for teachers in Omsk and the Omsk region, conducted by the NOF OmSU.
In the future, on the basis of the work done, it is possible to create a system for monitoring and evaluating the level of knowledge and skills of the test person. It becomes possible to identify problem areas in the work, to fix successful methods and techniques, to analyze what training content it is advisable to expand. But for the greatest effectiveness of these methods, the interest of the student is necessary. For this purpose, together with Chubrik A.V. and a small software product was developed that generates incorrect solutions of linear and quadratic equations (theoretical basis and algorithms - me and Chuubrik A.V., assistance in implementation - student gr. MP-803 Filimonov M.V.). Working with this program gives the student the opportunity to act as a teacher, whose student is a computer.
The results obtained can serve as the beginning of a more serious study, which in the near and long term will be able to make the necessary adjustments to the system of teaching mathematics.
1. CLASSIFICATION OF ERRORS WITH EXAMPLES
1.1. Classification by task types
1. Algebraic equations and inequalities.
1.1. Square inequalities. Systems of inequalities:
1.1.1. Roots found wrong square trinomial: the Vieta theorem and the formula for finding the roots are incorrectly used;
1.1.2. The graph of a square trinomial is incorrectly depicted;
|
|
1.1.3. The argument values are incorrectly defined for which the inequality is satisfied;
1.1.4. Division by an expression containing an unknown value;
1.1.5. In systems of inequalities, the intersection of the solutions of all inequalities is incorrectly taken;
1.1.6. Incorrectly included or not included the ends of the intervals in the final answer;
1.1.7. Rounding.
1.2. Fractional-rational equations:
1.2.1. Incorrectly indicated or not indicated ODZ: it was not taken into account that the denominator of the fraction should not be equal to zero;
ODZ: .
1.2.2. Upon receipt of a response, the ODZ is not taken into account;
12. Dalinger V.A. Common math mistakes in entrance exams and how to avoid them. - Omsk: Publishing House of the Omsk IUU, 1991.
3. Dalinger V.A. Everything to ensure success in the final and entrance exams in mathematics. Issue 5. Exponential, logarithmic equations, inequalities and their systems: Tutorial. - Omsk: OmGPU Publishing House, 1996.
4. Dalinger V.A. Beginnings mathematical analysis: Typical errors, their causes and ways of prevention: Study guide. - Omsk: "Publisher-Polygraphist", 2002.
5. Dalinger V.A., Zubkov A.N. Handbook for passing the exam in mathematics: Analysis of the mistakes of applicants in mathematics and ways to prevent them. - Omsk: OmGPU Publishing House, 1991.
6. Kutasov A.D. Exponential and logarithmic equations, inequalities, systems: Teaching aid N7. - Publishing House of the Russian Open University, 1992.
The mistakes made by students when solving logarithmic equations and inequalities are very diverse: from incorrect design of the solution to logical errors. These and other errors will be discussed in this article.
1. The most typical mistake is that students, when solving equations and inequalities, without additional explanations, use transformations that violate equivalence, which leads to the loss of roots and the appearance of extraneous horses.
Let's look at specific examples of errors of this kind, but first we draw the reader's attention to the following thought: do not be afraid to acquire extraneous roots, they can be discarded by checking, be afraid to lose roots.
a) Solve the equation:
log3(5 - x) = 3 - log3(-1 - x).
Students often solve this equation in the following way.
log3(5 - x) = 3 - log3(-1 - x), log3(5 - x) + log3(-1 - x) = 3, log3((5 - x)(-1 - x)) = 3 , (5 - x)(-1 - x) = 33, x2 - 4x - 32 = 0,
x1 = -4; x2 = 8.
Students often, without additional reasoning, write down both numbers in response. But as the check shows, the number x = 8 is not the root of the original equation, since at x = 8 the left and right sides of the equation lose their meaning. The check shows that the number x = -4 is the root of the given equation.
b) Solve the equation
The domain of definition of the original equation is given by the system
To solve the given equation, we pass to the logarithm in base x, we obtain
We see that the left and right sides of this last equation at x = 1 are not defined, but this number is the root of the original equation (we can verify this by direct substitution). Thus, the formal transition to a new base led to the loss of the root. To avoid losing the root x = 1, you should specify that the new base must be a positive number other than one, and consider the case x = 1 separately.
2. A whole group of errors, or rather shortcomings, consists in the fact that students do not pay due attention to finding the domain of definition of equations, although in some cases it is precisely this domain that is the key to the solution. Let's take a look at an example in this regard.
solve the equation
Let's find the domain of definition of this equation, for which we solve the system of inequalities:
Whence we have x = 0. Let's check by direct substitution whether the number x = 0 is the root of the original equation
Answer: x = 0.
3. A typical mistake of students is that they do not know the definitions of concepts, formulas, formulations of theorems, and algorithms at the required level. Let's confirm what has been said with the following example.
solve the equation
Here is an erroneous solution to this equation:
Verification shows that x = -2 is not the root of the original equation.
The conclusion suggests itself that the given equation has no roots.
However, it is not. By substituting x = -4 into the given equation, we can verify that this is a root.
Let's analyze why the root was lost.
In the original equation, the expressions x and x + 3 can be both negative or both positive at the same time, but when passing to the equation, these same expressions can only be positive. Consequently, there was a narrowing of the domain of definition, which led to the loss of roots.
To avoid losing the root, you can proceed as follows: let's move in the original equation from the logarithm of the sum to the logarithm of the product. In this case, the appearance of extraneous roots is possible, but you can get rid of them by substitution.
4. Many mistakes made when solving equations and inequalities are the result of the fact that students very often try to solve problems according to a template, that is, in the usual way. Let's show this with an example.
Solve the inequality
An attempt to solve this inequality in the usual algorithmic ways will not lead to an answer. The solution here should consist in estimating the values of each term on the left side of the inequality on the domain of the inequality.
Find the domain of definition of the inequality:
For all x from the interval (9;10] the expression has positive values (values exponential function always positive).
For all x from the interval (9;10] the expression x - 9 has positive values, and the expression lg(x - 9) has negative values or zero, then the expression (- (x - 9) lg(x - 9) is positive or equal to zero.
Finally, we have x∈ (9;10]. Note that for such values of the variable, each term on the left side of the inequality is positive (the second term may be equal to zero), which means that their sum is always greater than zero. Therefore, the solution to the original inequality is interval (9;10].
5. One of the errors is related to the graphical solution of equations.
solve the equation
Our experience shows that students, solving this equation graphically (note that it cannot be solved by other elementary methods), receive only one root (it is the abscissa of a point lying on the line y = x), because the graphs of functions
These are graphs of mutually inverse functions.
In fact, the original equation has three roots: one of them is the abscissa of the point lying on the bisector of the first coordinate angle y \u003d x, the other root and the third root.
Note that equations of the form logax = ax at 0< a < e-e всегда имеют три действительных корня.
This example aptly illustrates the following output: graphic solution the equations f(x) = g(x) are “perfect” if both functions are different-monotone (one of them is increasing and the other is decreasing), and not sufficiently mathematically correct in the case of monotone functions (both either decrease or increase simultaneously).
6. A number of typical mistakes are due to the fact that students do not quite correctly solve equations and inequalities based on a functional approach. We will show typical errors of this kind.
a) Solve the equation xx = x.
The function on the left side of the equation is exponential-power, and if so, then the following restrictions should be imposed on the basis of the degree: x > 0, x ≠ 1. Let's take the logarithm of both parts of the given equation:
Whence we have x = 1.
The logarithm did not lead to a narrowing of the domain of definition of the original equation. But nevertheless we have lost two roots of the equation; by direct observation, we find that x = 1 and x = -1 are the roots of the original equation.
b) Solve the equation
As in the previous case, we have an exponential-power function, which means x > 0, x ≠ 1.
To solve the original equation, we take the logarithm of both parts of it in any base, for example, in base 10:
Given that the product of two factors is equal to zero when at least one of them is equal to zero, while the other makes sense, we have a set of two systems:
The first system has no solution; from the second system we get x = 1. Given the restrictions imposed earlier, the number x = 1 should not be the root of the original equation, although by direct substitution we make sure that this is not the case.
7. Consider some of the errors associated with the concept complex function kind. Let's show the error with an example.
Determine the type of monotonicity of the function .
Our practice shows that the vast majority of students determine monotonicity in this case only by the base of the logarithm, and since 0< 0,5 < 1, то отсюда следует ошибочный вывод - функция убывает.
Not! This function is increasing.
Conditionally for the view function, you can write:
Increasing (Descending) = Descending;
Increasing (Increasing) = Increasing;
Decreasing (Descending) = Increasing;
Decreasing (Increasing) = Decreasing;
8. Solve the equation
This task is taken from the third part of the Unified State Examination, which is assessed by points (the maximum score is 4).
Here is a solution that contains errors, which means that the maximum score will not be given for it.
We reduce the logarithms to base 3. The equation will take the form
By potentiating, we get
x1 = 1, x2 = 3.
Let's check to identify extraneous roots
, 1 = 1,
so x = 1 is the root of the original equation.
so x = 3 is not the root of the original equation.
Let us explain why this solution contains errors. The essence of the error is that the entry contains two gross errors. The first mistake: the record does not make sense at all. Second error: It is not true that the product of two factors, one of which is 0, is necessarily zero. Zero will be if and only if one factor is 0 and the second factor makes sense. Here, just, the second multiplier does not make sense.
9. Let us return to the error already commented on above, but at the same time we will give some new arguments.
When solving logarithmic equations, they pass to the equation. Each root of the first equation is also a root of the second equation. The converse, generally speaking, is not true, therefore, moving from equation to equation , it is necessary to check the roots of the latter by substitution into the original equation at the end. Instead of checking the roots, it is advisable to replace the equation with an equivalent system
If, when solving the logarithmic equation, the expressions
where n - even number, are transformed, respectively, according to the formulas , , , then, since in many cases the domain of definition of the equation is narrowed, some of its roots may be lost. Therefore, it is advisable to apply these formulas in the following form:
n is an even number.
Conversely, if when solving the logarithmic equation, the expressions , , , where n is an even number, are converted, respectively, into the expressions
then the domain of definition of the equation can expand, due to which it is possible to acquire extraneous roots. Keeping this in mind, in such situations it is necessary to monitor the equivalence of transformations and, if the domain of definition of the equation expands, check the resulting roots.
10. When solving logarithmic inequalities using substitution, we always first solve a new inequality with respect to a new variable, and only in its solution do we make a transition to the old variable.
Schoolchildren very often mistakenly make the reverse transition earlier, at the stage of finding the roots. rational function obtained on the left side of the inequality. This should not be done.
11. Let us give an example of another error related to the solution of inequalities.
Solve the inequality
.
Here is an erroneous solution that students very often offer.
Let's square both sides of the original inequality. Will have:
whence we obtain an incorrect numerical inequality , which allows us to conclude that the given inequality has no solutions.