Sections: Maths
Class: 9
A mandatory learning outcome is the ability to solve an inequality of the form:
ax 2 + bx + c ><0
based on a schematic diagram quadratic function.
Most often, students make mistakes when solving square inequalities with a negative first coefficient. The textbook proposes in such cases to replace the inequality with an equivalent one with a positive coefficient at x 2 (example No. 3). It is important that students understand that they need to “forget” about the original inequality, for the solution it is necessary to depict a parabola with branches pointing upwards. You can argue otherwise.
Suppose we need to solve the inequality:
-x 2 + 2x -5<0
First, find out if the graph of the function y=-x 2 +2x-5 crosses the OX axis. To do this, we solve the equation:
The equation has no roots, therefore, the graph of the function y \u003d -x 2 + 2x-5 is entirely below the X axis and the inequality -x 2 + 2x-5<0 выполняется при любых значения Х. Необходимо показать учащимся оба способа решения и разрешить пользоваться любым из них.
The ability to solve is practiced at No. 111 and No. 119. It is imperative to consider such inequalities x 2 +5>0, -x 2 -3≤0; 3x 2 >0 etc.
Of course, when solving such inequalities, you can use a parabola. However, strong students should give answers immediately, without resorting to a drawing. In this case, it is necessary to require explanations, for example: x 2 ≥0 and x 2 +7>0 for any values of x. Depending on the level of preparation of the class, you can limit yourself to these numbers or use No. 120 No. 121. It is necessary to perform simple identical transformations in them, so there will be a repetition of the material covered here. These numbers are designed for strong students. If a good result is achieved and the solution of square inequalities does not cause any problems, then students can be invited to solve a system of inequalities in which one or both inequalities are square (exercise 193, 194).
It is interesting not only to solve quadratic inequalities, but also where else this solution can be applied: to find the domain of the function of studying a quadratic equation with parameters (Exercise 122-124). For the most advanced students, you can consider quadratic inequalities with parameters of the form:
Ax2+Bx+C>0 (≥0)
Ax 2+Bx+C<0 (≤0)
Where A,B,C are expressions depending on the parameters, A≠0,x are unknown.
Inequality Ax 2 +Bx+C>0
Investigated according to the following schemes:
1)If A=0, then we have a linear inequality Bx+C>0
2) If A≠0 and discriminant D>0, then we can factorize the square trinomial and get the inequality
A(x-x1) (x-x2)>0
x 1 and x 2 are the roots of the equation Ax 2 +Bx+C=0
3)If A≠0 and D<0 то если A>0 the solution will be the set of real numbers R; at A<0 решений нет.
The rest of the inequalities can be studied similarly.
Can be used when solving square inequalities, hence the property of a square trinomial
1) If A>0 and D<0 то Ax2+Bx+C>0- for all x.
2) If A<0 и D<0 то Ax2+Bx+C<0 при всех x.
When solving a quadratic inequality, it is more convenient to use a schematic representation of the graph of the function y=Ax2+Bx+C
Example: For all parameter values, solve the inequality
X 2 +2(b+1)x+b 2 >0
D=4(b+1) 2 -4b 2 =4b 2 +8b+4-4b 2
1) D<0 т.е. 2b+1<0
The coefficient in front of x 2 is equal to 1>0, then the inequality holds for all x, i.e. Х є R
2) D=0 => 2b+1=0
Then x 2 +x+0>0
x ½(-∞;-½) U (-½;∞)
3) D>0 =>2b+1>0
The roots of a square trinomial have the form:
X 1 \u003d-b-1-√2b+1
X 2 \u003d -b-1 + √2b + 1
The inequality takes the form
(x-x 1) (x-x 2)>0
Using the interval method, we get
x є(-∞;x 1) U (x 2 ;∞)
For an independent solution, give the following inequality
As a result of solving inequalities, the student should understand that in order to solve inequalities of the second degree, it is proposed to abandon the excessive detailing of the method of constructing a graph, from finding the coordinates of the vertices of the parabola, observing the scale, one can limit oneself to the image of a sketch of a graph of a quadratic function.
In the senior level, the solution of square inequalities is practically not an independent task, but acts as a component of the solution of another equation or inequality (logarithmic, exponential, trigonometric). Therefore, it is necessary to teach students how to quickly solve quadratic inequalities. You can refer to three theorems borrowed from the textbook by A.A. Kiseleva.
Theorem 1. Let a square trinomial ax 2 +bx+c be given, where a>0, having 2 different real roots(D>0).
Then: 1) For all values of the variable x that are less than the smaller root and greater than the larger root, the square trinomial is positive
2) For values of x between square roots, the trinomial is negative.
Theorem 2. Let a square trinomial ax 2 +bx+c be given, where a>0 having 2 identical real roots (D=0). Then for all values of x different from the roots of the square trinomial, the square trinomial is positive.
Theorem3. Let a square trinomial ax 2 +bx+c be given where a>0 has no real roots (D<0).Тогда при всех значениях x квадратный трехчлен положителен. Доказательство этих теорем приводить не надо.
For example: solve the inequality:
D=1+288=289>0
The solution is
X≤-4/3 and x≥3/2
Answer (-∞; -4/3] U
Answers are placed on the reverse side, you can see them after the allotted time has passed. It is most convenient to carry out this work at the beginning of the lesson at the signal of the teacher. (Attention, get ready, start). On command "Stop" the work is interrupted.
Working hours are determined depending on the level of preparation of the class. The increase in speed is an indicator of the student's work.
The ability to solve quadratic inequalities will be useful for students when passing the exam. In group B problems, there are more and more tasks related to the ability to solve quadratic inequalities.
For example:
A stone is thrown vertically upwards. Until the stone has fallen, the height at which it is located is described by the formula
(h is the height in meters, t is the time in seconds elapsed since the throw).
Find how many seconds the stone was at a height of at least 9 meters.
To solve it, you need to write an inequality:
5t2+18t-9≥0
Answer: 2.4 s
Starting to give students examples from the exam already in the 9th grade at the stage of studying the material, we are already preparing for the exam, solving square inequalities containing a parameter makes it possible to solve problems from group C.
A non-formal approach to studying the topic in grade 9 facilitates the assimilation of the material in the course “Algebra and the beginning of analysis” on such topics as “Application of the derivative” “Solving inequalities by the interval method” “Solving logarithmic and exponential inequalities” “Solving irrational inequalities”.
Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")
What's happened "square inequality"? Not a question!) If you take any quadratic equation and change the sign in it "=" (equal) to any inequality icon ( > ≥ < ≤ ≠ ), we get a quadratic inequality. For example:
1. x2 -8x+12 ≥ 0
2. -x 2 +3x > 0
3. x2 ≤ 4
Well, you get the idea...)
I knowingly linked equations and inequalities here. The fact is that the first step in solving any square inequality - solve the equation from which this inequality is made. For this reason - the inability to solve quadratic equations automatically leads to a complete failure in inequalities. Is the hint clear?) If anything, look at how to solve any quadratic equations. Everything is detailed there. And in this lesson we will deal with inequalities.
The inequality ready for solution has the form: left - square trinomial ax 2 +bx+c, on the right - zero. The inequality sign can be absolutely anything. The first two examples are here are ready for a decision. The third example still needs to be prepared.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)
you can get acquainted with functions and derivatives.
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 of a 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 instance, . 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 ahead!
Well, did it work? If you have any difficulties, then understand the solutions.
Solution:
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. AVERAGE 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 delivery Unified State Examination, 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? Do not 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!
Before you figure it out how to solve quadratic inequality, let's consider what inequality is called square.
Remember!
The inequality is called square, if the highest (greatest) power of the unknown "x" is equal to two.
Let's practice determining the type of inequality using examples.
How to solve a quadratic inequality
In previous lessons, we discussed how to solve linear inequalities. But unlike linear inequalities square ones are solved in a completely different way.
Important!
It is impossible to solve a quadratic inequality in the same way as a linear one!
To solve a quadratic inequality, a special method is used, which is called interval method.
What is the interval method
interval method called a special way of solving quadratic inequalities. Below we will explain how to use this method and why it is so named.
Remember!
To solve a quadratic inequality using the interval method, you need:
We understand that the rules described above are difficult to perceive only in theory, so we will immediately consider an example of solving a quadratic inequality using the algorithm above.
It is required to solve a quadratic inequality.
Now, as said in , draw "arches" over the intervals between the marked points.
Let's put signs inside the intervals. From right to left, alternating, starting with "+", we note the signs.
We just have to execute , that is, select the desired intervals and write them down in response. Let's return to our inequality.
Since in our inequality x 2 + x − 12 ", so we need negative intervals. Let's shade all negative areas on a numerical axis and we will write out them in the answer.
Only one interval turned out to be negative, which is between the numbers " −3" and "4", so we write it in response as a double inequality
"-3".
Let's write down the answer of the quadratic inequality.
Answer: -3
By the way, it is precisely because we consider the intervals between numbers when solving a quadratic inequality that the method of intervals got its name.
After receiving the answer, it makes sense to check it to make sure the solution is correct.
Let's choose any number that is in the shaded area of the received answer " −3" and substitute it instead of "x" in the original inequality. If we get the correct inequality, then we have found the answer to the quadratic inequality is correct.
Take, for example, the number "0" from the interval. Substitute it into the original inequality "x 2 + x − 12".
X 2 + x − 12
0 2 + 0 − 12 −12 (correct)
We got the correct inequality when substituting a number from the solution area, which means that the answer was found correctly.
Brief notation of the solution by the method of intervals
Abbreviated record of the solution of the quadratic inequality " x 2 + x − 12 ” method of intervals will look like this:
X 2 + x − 12
x2 + x − 12 = 0
x 1 =
|
x 2 =
|
||||||||||||||||||||
x 1 =
|
x 2 =
 Answer: x ≤ 0 ; x ≥
Consider an example where there is a negative coefficient in front of "x 2" in a square inequality. 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. The Beginnings of Mathematical Analysis: Typical Errors, Their Causes and Ways of Prevention: Textbook. - 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 is that 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 the 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
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 deciding logarithmic equation 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: where do we get the wrong numerical inequality, which allows us to conclude that the given inequality has no solutions. Read also:
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, 1 = 1,
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