Solution of inequality х2 9. Inequalities. Types of inequalities. What is the spacing method

First, a little lyrics to get a feel for the problem that the spacing method solves. Let's say we need to solve the following inequality:

(x - 5) (x + 3)> 0

What are the options? The first thing that comes to mind for most students is the rules "plus for plus equals plus" and "minus for minus equals plus". Therefore, it suffices to consider the case when both parentheses are positive: x - 5> 0 and x + 3> 0. Then we also consider the case when both parentheses are negative: x - 5< 0 и x + 3 < 0. Таким образом, наше неравенство свелось к совокупности двух систем, которая, впрочем, легко решается:

More advanced students will remember (maybe) that there is a quadratic function on the left, the graph of which is a parabola. Moreover, this parabola intersects the OX axis at the points x = 5 and x = −3. For further work, you need to open the brackets. We have:

x 2 - 2x - 15> 0

Now it is clear that the branches of the parabola are directed upward, because coefficient a = 1> 0. Let's try to draw a diagram of this parabola:

The function is greater than zero where it passes above the OX axis. In our case, these are the intervals (−∞ −3) and (5; + ∞) - this is the answer.

Please note: the picture shows exactly function diagram rather than her schedule. Because for a real graph, you need to count the coordinates, calculate the offsets and other crap that we don't need at all right now.

Why are these methods ineffective?

So, we looked at two solutions to the same inequality. Both of them turned out to be quite cumbersome. The first solution appears - just think about it! - a set of systems of inequalities. The second solution is also not particularly easy: you need to remember the parabola graph and a bunch of other small facts.

It was a very simple inequality. It has only 2 multipliers. Now imagine that the factors will not be 2, but at least 4. For example:

(x - 7) (x - 1) (x + 4) (x + 9)< 0

How can this inequality be addressed? Going through all the possible combinations of pros and cons? Yes, we will fall asleep faster than we can find a solution. Drawing a graph is also not an option, since it is not clear how such a function behaves on the coordinate plane.

For such inequalities, a special solution algorithm is needed, which we will consider today.

What is the spacing method

The interval method is a special algorithm designed to solve complex inequalities of the form f (x)> 0 and f (x)< 0. Алгоритм состоит из 4 шагов:

Solve the equation f (x) = 0. Thus, instead of inequality, we get an equation that is much easier to solve;
Mark all obtained roots on the coordinate line. Thus, the line is divided into several intervals;
Find out the sign (plus or minus) of the function f (x) on the rightmost interval. To do this, it is enough to substitute in f (x) any number that will be to the right of all marked roots;
Mark signs on the remaining intervals. To do this, it is enough to remember that when passing through each root, the sign changes.

That's all! After that, it remains only to write out the intervals that interest us. They are marked with a "+" sign if the inequality had the form f (x)> 0, or a "-" sign if the inequality has the form f (x)< 0.

At first glance, it might seem that the spacing method is some kind of tin. But in practice, everything will be very simple. It is worth a little practice - and everything will become clear. Take a look at the examples and see for yourself:

Task. Solve the inequality:

(x - 2) (x + 7)< 0

We work according to the method of intervals. Step 1: replace the inequality with the equation and solve it:

(x - 2) (x + 7) = 0

The product is equal to zero if and only if at least one of the factors is equal to zero:

x - 2 = 0 ⇒ x = 2;
x + 7 = 0 ⇒ x = −7.

We got two roots. Go to step 2: mark these roots on the coordinate line. We have:

Now step 3: find the sign of the function on the rightmost interval (to the right of the marked point x = 2). To do this, you need to take any number that is greater than the number x = 2. For example, take x = 3 (but no one forbids taking x = 4, x = 10 and even x = 10,000). We get:

f (x) = (x - 2) (x + 7);
x = 3;
f (3) = (3 - 2) (3 + 7) = 110 = 10;

We get that f (3) = 10> 0, so we put a plus sign in the rightmost interval.

Moving on to the last point - it is necessary to mark the signs on the remaining intervals. Remember that when passing through each root, the sign must change. For example, to the right of the root x = 2 there is a plus (we made sure of this in the previous step), so there must be a minus on the left.

This minus extends to the entire interval (−7; 2), so there is a minus to the right of the root x = −7. Therefore, there is a plus to the left of the root x = −7. It remains to mark these signs on the coordinate axis. We have:

Let's go back to the original inequality, which looked like:

(x - 2) (x + 7)< 0

So the function must be less than zero. Hence, we are interested in the minus sign, which appears only on one interval: (−7; 2). This will be the answer.

Task. Solve the inequality:

(x + 9) (x - 3) (1 - x)< 0

Step 1: set the left side to zero:

(x + 9) (x - 3) (1 - x) = 0;
x + 9 = 0 ⇒ x = −9;
x - 3 = 0 ⇒ x = 3;
1 - x = 0 ⇒ x = 1.

Remember: the product is zero when at least one of the factors is zero. That is why we have the right to equate each individual parenthesis to zero.

Step 2: mark all the roots on the coordinate line:

Step 3: find out the sign of the rightmost gap. We take any number that is greater than x = 1. For example, we can take x = 10. We have:

f (x) = (x + 9) (x - 3) (1 - x);
x = 10;
f (10) = (10 + 9) (10 - 3) (1 - 10) = 19 7 (−9) = - 1197;
f (10) = −1197< 0.

Step 4: arrange the rest of the signs. Remember that when passing through each root, the sign changes. As a result, our picture will look like this:

That's all. It remains only to write down the answer. Take another look at the original inequality:

(x + 9) (x - 3) (1 - x)< 0

This is an inequality of the form f (x)< 0, т.е. нас интересуют интервалы, отмеченные знаком минус. А именно:

x ∈ (−9; 1) ∪ (3; + ∞)

This is the answer.

A note on function signs

Practice shows that the greatest difficulties in the method of intervals arise at the last two steps, i.e. when placing signs. Many students begin to get confused: what numbers should be taken and where to put the signs.

To finally understand the method of intervals, consider two notes on which it is built:

A continuous function changes sign only at those points where it is zero... Such points break the coordinate axis into pieces, inside which the sign of the function never changes. That is why we solve the equation f (x) = 0 and mark the found roots on the line. The numbers found are "borderline" points that separate the pluses from the minuses.
To find out the sign of a function on any interval, it is enough to substitute any number from this interval into the function. For example, for the interval (−5; 6), we have the right to take x = −4, x = 0, x = 4 and even x = 1.29374, if we want. Why is it important? Because many students begin to gnaw doubts. Like, what if for x = −4 we get a plus, and for x = 0 - a minus? And nothing - this will never happen. All points on the same interval give the same sign. Remember this.

That's all there is to know about the spacing method. Of course, we have analyzed it in its simplest form. There are more complex inequalities - lax, fractional, and repeating-root inequalities. For them, you can also use the spacing method, but this is a topic for a separate large lesson.

Now I would like to analyze an advanced technique that dramatically simplifies the spacing method. More precisely, the simplification affects only the third step - the calculation of the sign on the rightmost piece of the straight line. For some reason, this technique does not work in schools (at least, no one explained this to me). But in vain - in fact, this algorithm is very simple.

So, the sign of the function is on the right side of the number axis. This piece has the form (a; + ∞), where a is the largest root of the equation f (x) = 0. In order not to blow up the brain, consider a specific example:

(x - 1) (2 + x) (7 - x)< 0;
f (x) = (x - 1) (2 + x) (7 - x);
(x - 1) (2 + x) (7 - x) = 0;
x - 1 = 0 ⇒ x = 1;
2 + x = 0 ⇒ x = −2;
7 - x = 0 ⇒ x = 7;

We got 3 roots. Let's list them in ascending order: x = −2, x = 1, and x = 7. Obviously, the largest root is x = 7.

For those who find it easier to reason graphically, I will mark these roots on the coordinate line. Let's see what happens:

It is required to find the sign of the function f (x) on the rightmost interval, i.e. on (7; + ∞). But as we have already noted, to determine the sign, you can take any number from this interval. For example, you can take x = 8, x = 150, etc. And now - the very technique that is not used in schools: let's take infinity as a number. More precisely, plus infinity, i.e. + ∞.

“What are you, you got stoned? How can you substitute infinity in the function? " - you might ask. But think about it: we do not need the value of the function itself, we only need the sign. Therefore, for example, the values f (x) = −1 and f (x) = −938 740 576 215 mean the same thing: the function is negative on this interval. Therefore, all that is required of you is to find the sign that arises at infinity, and not the value of the function.

In fact, substituting infinity is very simple. Let's go back to our function:

f (x) = (x - 1) (2 + x) (7 - x)

Imagine x is a very large number. A billion or even a trillion. Now let's see what happens in each parenthesis.

First bracket: (x - 1). What happens if you subtract one from a billion? The result is a number that is not very different from a billion, and this number will be positive. Likewise with the second bracket: (2 + x). If we add a billion to two, we get a billion and a penny - this is a positive number. Finally, the third parenthesis: (7 - x). Here there will be a minus one billion, from which they "chewed off" a pitiful piece in the form of a seven. Those. the resulting number will not differ much from minus billion - it will be negative.

It remains to find the sign of the entire work. Since we had a plus in the first brackets, and a minus in the last, we get the following construction:

(+) · (+) · (−) = (−)

The final sign is a minus! It doesn't matter what the value of the function itself is equal to. The main thing is that this value is negative, i.e. the rightmost interval has a minus sign. It remains to perform the fourth step of the spacing method: arrange all the signs. We have:

The original inequality was as follows:

(x - 1) (2 + x) (7 - x)< 0

Therefore, we are interested in the intervals marked with a minus sign. We write out the answer:

x ∈ (−2; 1) ∪ (7; + ∞)

That's the whole trick I wanted to tell you. In conclusion - one more inequality, which is solved by the method of intervals with the involvement of infinity. To visually shorten the solution, I will not write step numbers and extended comments. I will only write what you really need to write when solving real problems:

Task. Solve the inequality:

x (2x + 8) (x - 3)> 0

We replace the inequality with the equation and solve it:

x (2x + 8) (x - 3) = 0;
x = 0;
2x + 8 = 0 ⇒ x = −4;
x - 3 = 0 ⇒ x = 3.

We mark all three roots on the coordinate line (immediately with signs):

On the right side of the coordinate axis there is a plus, because the function looks like:

f (x) = x (2x + 8) (x - 3)

And if we substitute infinity (for example, a billion), we get three positive parentheses. Since the original expression must be greater than zero, we are only interested in the pluses. It remains to write out the answer:

x ∈ (−4; 0) ∪ (3; + ∞)

Today, friends, there will be no snot and sentimentality. Instead, I will send you into battle with one of the most formidable opponents in the 8-9 grade algebra course without any questions.

Yes, you understood everything correctly: we are talking about inequalities with a modulus. We will look at four basic techniques with which you will learn how to solve about 90% of such problems. What about the other 10%? Well, we'll talk about them in a separate lesson. :)

However, before analyzing any of the techniques, I would like to remind you of two facts that you already need to know. Otherwise, you risk not understanding the material of today's lesson at all.

What you need to know already

Captain Obvious is kind of hinting that two things need to be known in order to solve inequalities with modulus:

How inequalities are resolved;
What is a module.

Let's start with the second point.

Module definition

Everything is simple here. There are two definitions: algebraic and graphical. For starters - algebraic:

Definition. The modulus of the number $ x $ is either this number itself, if it is non-negative, or the number opposite to it, if the original $ x $ is still negative.

It is written like this:

\ [\ left | x \ right | = \ left \ (\ begin (align) & x, \ x \ ge 0, \\ & -x, \ x \ lt 0. \\\ end (align) \ right. \]

In simple terms, a module is a "number without a minus". And it is precisely in this duality (somewhere with the initial number nothing needs to be done, but somewhere you have to remove some kind of minus) that the whole difficulty for novice students lies.

There is also a geometric definition. It is also useful to know it, but we will refer to it only in complex and some special cases, where the geometric approach is more convenient than the algebraic one (spoiler: not today).

Definition. Let the point $ a $ be marked on the number line. Then the module $ \ left | x-a \ right | $ is the distance from the point $ x $ to the point $ a $ on this line.

If you draw a picture, you get something like this:

Graphical module definition

One way or another, its key property immediately follows from the definition of a module: the modulus of a number is always non-negative... This fact will be a red thread throughout our entire story today.

Solving inequalities. Spacing method

Now let's deal with inequalities. There are a great many of them, but our task now is to be able to solve at least the simplest of them. Those that reduce to linear inequalities, as well as the method of intervals.

On this topic, I have two great lessons (by the way, very, VERY useful - I recommend studying):

Spacing method for inequalities (especially watch the video);
Fractional rational inequalities are a huge lesson, but after that you won't have any questions at all.

If you know all this, if the phrase "let's move from inequality to an equation" does not make you vaguely want to kill yourself against the wall, then you are ready: welcome to hell to the main topic of the lesson. :)

1. Inequalities of the form "Module less than function"

This is one of the most common tasks with modules. It is required to solve an inequality of the form:

\ [\ left | f \ right | \ lt g \]

The functions $ f $ and $ g $ can be anything, but usually they are polynomials. Examples of such inequalities:

All of them are solved literally in one line according to the scheme:

\ [\ left | f \ right | \ lt g \ Rightarrow -g \ lt f \ lt g \ quad \ left (\ Rightarrow \ left \ (\ begin (align) & f \ lt g, \\ & f \ gt -g \\\ end (align) \ right. \ right) \]

It is easy to see that we get rid of the module, but instead we get a double inequality (or, which is the same, a system of two inequalities). But this transition takes into account absolutely all possible problems: if the number under the module is positive, the method works; if negative, it still works; and even with the most inadequate function in place of $ f $ or $ g $, the method will still work.

Naturally, the question arises: couldn't it be easier? Unfortunately, you can't. This is the whole feature of the module.

However, stop philosophizing. Let's solve a couple of problems:

Task. Solve the inequality:

\ [\ left | 2x + 3 \ right | \ lt x + 7 \]

Solution. So, we have before us a classical inequality of the form "the modulus is less" - there is even nothing to transform. We work according to the algorithm:

\ [\ begin (align) & \ left | f \ right | \ lt g \ Rightarrow -g \ lt f \ lt g; \\ & \ left | 2x + 3 \ right | \ lt x + 7 \ Rightarrow - \ left (x + 7 \ right) \ lt 2x + 3 \ lt x + 7 \\\ end (align) \]

Do not rush to open the parentheses in front of which there is a minus: it is quite possible that out of haste you will make an offensive mistake.

\ [- x-7 \ lt 2x + 3 \ lt x + 7 \]

\ [\ left \ (\ begin (align) & -x-7 \ lt 2x + 3 \\ & 2x + 3 \ lt x + 7 \\ \ end (align) \ right. \]

\ [\ left \ (\ begin (align) & -3x \ lt 10 \\ & x \ lt 4 \\ \ end (align) \ right. \]

\ [\ left \ (\ begin (align) & x \ gt - \ frac (10) (3) \\ & x \ lt 4 \\ \ end (align) \ right. \]

The problem was reduced to two elementary inequalities. Let us mark their solutions on parallel number lines:
Intersection of many
The intersection of these sets will be the answer.

Answer: $ x \ in \ left (- \ frac (10) (3); 4 \ right) $

Task. Solve the inequality:

\ [\ left | ((x) ^ (2)) + 2x-3 \ right | +3 \ left (x + 1 \ right) \ lt 0 \]

Solution. This task is already a little more difficult. To begin with, let's seclude the module by moving the second term to the right:

\ [\ left | ((x) ^ (2)) + 2x-3 \ right | \ lt -3 \ left (x + 1 \ right) \]

Obviously, we are again faced with an inequality of the form "the module is less", so we get rid of the module according to the already known algorithm:

\ [- \ left (-3 \ left (x + 1 \ right) \ right) \ lt ((x) ^ (2)) + 2x-3 \ lt -3 \ left (x + 1 \ right) \]

Now attention: someone will say that I am a bit of a pervert with all these brackets. But let me remind you once again that our key goal is competently solve inequality and get an answer... Later, when you have perfectly mastered everything that is described in this lesson, you yourself can be perverted as you like: open parentheses, add minuses, etc.

For a start, we'll just get rid of the double minus on the left:

\ [- \ left (-3 \ left (x + 1 \ right) \ right) = \ left (-1 \ right) \ cdot \ left (-3 \ right) \ cdot \ left (x + 1 \ right) = 3 \ left (x + 1 \ right) \]

Now let's expand all the parentheses in the double inequality:

We pass to double inequality. This time the calculations will be more serious:

\ [\ left \ (\ begin (align) & ((x) ^ (2)) + 2x-3 \ lt -3x-3 \\ & 3x + 3 \ lt ((x) ^ (2)) + 2x -3 \\ \ end (align) \ right. \]

\ [\ left \ (\ begin (align) & ((x) ^ (2)) + 5x \ lt 0 \\ & ((x) ^ (2)) - x-6 \ gt 0 \\ \ end ( align) \ right. \]

Both inequalities are square and are solved by the method of intervals (that's why I say: if you don't know what it is, it's better not to take up modules for now). We pass to the equation in the first inequality:

\ [\ begin (align) & ((x) ^ (2)) + 5x = 0; \\ & x \ left (x + 5 \ right) = 0; \\ & ((x) _ (1)) = 0; ((x) _ (2)) = - 5. \\\ end (align) \]

As you can see, the output is an incomplete quadratic equation, which can be solved in an elementary way. Now let's deal with the second inequality of the system. There you have to apply Vieta's theorem:

\ [\ begin (align) & ((x) ^ (2)) - x-6 = 0; \\ & \ left (x-3 \ right) \ left (x + 2 \ right) = 0; \\ & ((x) _ (1)) = 3; ((x) _ (2)) = - 2. \\\ end (align) \]

We mark the numbers obtained on two parallel lines (one for the first inequality and a separate one for the second):
Again, since we are solving a system of inequalities, we are interested in the intersection of the shaded sets: $ x \ in \ left (-5; -2 \ right) $. This is the answer.
Answer: $ x \ in \ left (-5; -2 \ right) $

I think after these examples the solution scheme is very clear:

Solve the module by transferring all other terms to the opposite side of the inequality. Thus, we get an inequality of the form $ \ left | f \ right | \ lt g $.
Solve this inequality by getting rid of the module as described above. At some point, it will be necessary to go from double inequality to a system of two independent expressions, each of which can already be solved separately.
Finally, it remains only to intersect the solutions of these two independent expressions - and that's it, we will get the final answer.

A similar algorithm also exists for inequalities of the following type, when the modulus is greater than the function. However, there are a couple of serious "buts" there. We will now talk about these "but".

2. Inequalities of the form "A module is more than a function"

They look like this:

\ [\ left | f \ right | \ gt g \]

Similar to the previous one? It seems. And nevertheless, such tasks are solved in a completely different way. Formally, the scheme is as follows:

\ [\ left | f \ right | \ gt g \ Rightarrow \ left [\ begin (align) & f \ gt g, \\ & f \ lt -g \\\ end (align) \ right. \]

In other words, we are considering two cases:

First, we simply ignore the module - solve the usual inequality;
Then, in fact, we expand the module with a minus sign, and then we multiply both sides of the inequality by −1, with me the sign.

In this case, the options are combined with a square bracket, i.e. before us is a combination of two requirements.

Note again: we have before us not a system, but an aggregate, therefore in the answer, the sets combine, but do not intersect... This is a fundamental difference from the previous point!

In general, many students have a complete confusion with unions and intersections, so let's figure this out once and for all:

"∪" is the sign of union. In fact, this is a stylized letter "U", which came to us from the English language and is an abbreviation for "Union", i.e. "Associations".
"∩" is an intersection sign. This crap didn't come out of nowhere, it just appeared as an opposition to "∪".

To make it even easier to remember, just add legs to these signs to make glasses (just don't blame me now for promoting drug addiction and alcoholism: if you are seriously studying this lesson, then you are already a drug addict):

Difference between intersection and union of sets

Translated into Russian, this means the following: a union (set) includes elements from both sets, therefore, no less than each of them; but the intersection (system) includes only those elements that are simultaneously in the first set and in the second. Therefore, the intersection of sets is never larger than the source sets.

So it became clearer? That is great. Let's get down to practice.

Task. Solve the inequality:

\ [\ left | 3x + 1 \ right | \ gt 5-4x \]

Solution. We act according to the scheme:

\ [\ left | 3x + 1 \ right | \ gt 5-4x \ Rightarrow \ left [\ begin (align) & 3x + 1 \ gt 5-4x \\ & 3x + 1 \ lt - \ left (5-4x \ right) \\\ end (align) \ right. \]

Solve each inequality in the population:

\ [\ left [\ begin (align) & 3x + 4x \ gt 5-1 \\ & 3x-4x \ lt -5-1 \\ \ end (align) \ right. \]

\ [\ left [\ begin (align) & 7x \ gt 4 \\ & -x \ lt -6 \\ \ end (align) \ right. \]

\ [\ left [\ begin (align) & x \ gt 4/7 \ \\ & x \ gt 6 \\ \ end (align) \ right. \]

We mark each resulting set on the number line, and then we combine them:
Union of sets
Obviously, the answer is $ x \ in \ left (\ frac (4) (7); + \ infty \ right) $

Answer: $ x \ in \ left (\ frac (4) (7); + \ infty \ right) $

Task. Solve the inequality:

\ [\ left | ((x) ^ (2)) + 2x-3 \ right | \ gt x \]

Solution. Well? Nothing - everything is the same. We pass from an inequality with modulus to a set of two inequalities:

\ [\ left | ((x) ^ (2)) + 2x-3 \ right | \ gt x \ Rightarrow \ left [\ begin (align) & ((x) ^ (2)) + 2x-3 \ gt x \\ & ((x) ^ (2)) + 2x-3 \ lt -x \\\ end (align) \ right. \]

We solve each inequality. Unfortunately, the roots will not be very good there:

\ [\ begin (align) & ((x) ^ (2)) + 2x-3 \ gt x; \\ & ((x) ^ (2)) + x-3 \ gt 0; \\ & D = 1 + 12 = 13; \\ & x = \ frac (-1 \ pm \ sqrt (13)) (2). \\\ end (align) \]

In the second inequality, there is also a little game:

\ [\ begin (align) & ((x) ^ (2)) + 2x-3 \ lt -x; \\ & ((x) ^ (2)) + 3x-3 \ lt 0; \\ & D = 9 + 12 = 21; \\ & x = \ frac (-3 \ pm \ sqrt (21)) (2). \\\ end (align) \]

Now you need to mark these numbers on two axes - one axis for each inequality. However, you need to mark the points in the correct order: the larger the number, the further the point shifts to the right.

And here a setup awaits us. If the numbers $ \ frac (-3- \ sqrt (21)) (2) \ lt \ frac (-1- \ sqrt (13)) (2) $ are clear (the terms in the numerator of the first fraction are less than the terms in the numerator of the second , so the sum is also less), with the numbers $ \ frac (-3- \ sqrt (13)) (2) \ lt \ frac (-1+ \ sqrt (21)) (2) $ there will be no difficulties either (positive number obviously more negative), then with the last couple everything is not so simple. Which is more: $ \ frac (-3+ \ sqrt (21)) (2) $ or $ \ frac (-1+ \ sqrt (13)) (2) $? The arrangement of points on the number lines and, in fact, the answer will depend on the answer to this question.

So let's compare:

\ [\ begin (matrix) \ frac (-1+ \ sqrt (13)) (2) \ vee \ frac (-3+ \ sqrt (21)) (2) \\ -1+ \ sqrt (13) \ vee -3+ \ sqrt (21) \\ 2+ \ sqrt (13) \ vee \ sqrt (21) \\\ end (matrix) \]

We have removed the root, we got non-negative numbers on both sides of the inequality, so we have the right to square both sides:

\ [\ begin (matrix) ((\ left (2+ \ sqrt (13) \ right)) ^ (2)) \ vee ((\ left (\ sqrt (21) \ right)) ^ (2)) \ \ 4 + 4 \ sqrt (13) +13 \ vee 21 \\ 4 \ sqrt (13) \ vee 3 \\\ end (matrix) \]

I think it's a no brainer here that $ 4 \ sqrt (13) \ gt 3 $, so $ \ frac (-1+ \ sqrt (13)) (2) \ gt \ frac (-3+ \ sqrt (21)) (2) $, finally the points on the axes will be placed like this:
A case of ugly roots
Let me remind you that we are solving a collection, so the answer will be a union, not an intersection of the shaded sets.

Answer: $ x \ in \ left (- \ infty; \ frac (-3+ \ sqrt (21)) (2) \ right) \ bigcup \ left (\ frac (-1+ \ sqrt (13)) (2 ); + \ infty \ right) $

As you can see, our scheme works great for both simple tasks and very hard ones. The only "weak point" in this approach is that you need to competently compare irrational numbers (and believe me: these are not only roots). But a separate (and very serious lesson) will be devoted to comparison issues. And we move on.

3. Inequalities with non-negative "tails"

So we got to the most interesting thing. These are inequalities of the form:

\ [\ left | f \ right | \ gt \ left | g \ right | \]

Generally speaking, the algorithm we are going to talk about now is only valid for a module. It works in all inequalities where the left and right are guaranteed non-negative expressions:

What to do with these tasks? Just remember:

In inequalities with non-negative "tails", both sides can be raised to any natural power. There will be no additional restrictions.

First of all, we will be interested in squaring - it burns modules and roots:

\ [\ begin (align) & ((\ left (\ left | f \ right | \ right)) ^ (2)) = ((f) ^ (2)); \\ & ((\ left (\ sqrt (f) \ right)) ^ (2)) = f. \\\ end (align) \]

Just don't confuse this with square root extraction:

\ [\ sqrt (((f) ^ (2))) = \ left | f \ right | \ ne f \]

Countless mistakes were made at the moment when the student forgot to install the module! But this is a completely different story (these are, as it were, irrational equations), so we will not delve into this now. Let's better solve a couple of problems:

Task. Solve the inequality:

\ [\ left | x + 2 \ right | \ ge \ left | 1-2x \ right | \]

Solution. Let's immediately notice two things:

This is a loose inequality. The points on the number line will be gouged out.

Both sides of the inequality are certainly non-negative (this is a property of the module: $ \ left | f \ left (x \ right) \ right | \ ge 0 $).

Therefore, we can square both sides of the inequality to get rid of the modulus and solve the problem using the usual interval method:

\ [\ begin (align) & ((\ left (\ left | x + 2 \ right | \ right)) ^ (2)) \ ge ((\ left (\ left | 1-2x \ right | \ right) ) ^ (2)); \\ & ((\ left (x + 2 \ right)) ^ (2)) \ ge ((\ left (2x-1 \ right)) ^ (2)). \\\ end (align) \]

At the last step, I cheated a little: I changed the sequence of terms using the parity of the modulus (in fact, I multiplied the expression $ 1-2x $ by −1).

\ [\ begin (align) & ((\ left (2x-1 \ right)) ^ (2)) - ((\ left (x + 2 \ right)) ^ (2)) \ le 0; \\ & \ left (\ left (2x-1 \ right) - \ left (x + 2 \ right) \ right) \ cdot \ left (\ left (2x-1 \ right) + \ left (x + 2 \ right) \ right) \ le 0; \\ & \ left (2x-1-x-2 \ right) \ cdot \ left (2x-1 + x + 2 \ right) \ le 0; \\ & \ left (x-3 \ right) \ cdot \ left (3x + 1 \ right) \ le 0. \\\ end (align) \]

We solve by the method of intervals. We pass from inequality to the equation:

\ [\ begin (align) & \ left (x-3 \ right) \ left (3x + 1 \ right) = 0; \\ & ((x) _ (1)) = 3; ((x) _ (2)) = - \ frac (1) (3). \\\ end (align) \]

We mark the found roots on the number line. Once again: all dots are filled, since the original inequality is not strict!
Getting rid of the modulus sign
Let me remind you for especially stubborn ones: we take the signs from the last inequality, which was written down before proceeding to the equation. And paint over the areas required in the same inequality. In our case, this is $ \ left (x-3 \ right) \ left (3x + 1 \ right) \ le 0 $.

So that is all. The problem has been solved.

Answer: $ x \ in \ left [- \ frac (1) (3); 3 \ right] $.

Task. Solve the inequality:

\ [\ left | ((x) ^ (2)) + x + 1 \ right | \ le \ left | ((x) ^ (2)) + 3x + 4 \ right | \]

Solution. We do all the same. I will not comment - just look at the sequence of actions.

Squaring:

\ [\ begin (align) & ((\ left (\ left | ((x) ^ (2)) + x + 1 \ right | \ right)) ^ (2)) \ le ((\ left (\ left | ((x) ^ (2)) + 3x + 4 \ right | \ right)) ^ (2)); \\ & ((\ left (((x) ^ (2)) + x + 1 \ right)) ^ (2)) \ le ((\ left (((x) ^ (2)) + 3x + 4 \ right)) ^ (2)); \\ & ((\ left (((x) ^ (2)) + x + 1 \ right)) ^ (2)) - ((\ left (((x) ^ (2)) + 3x + 4 \ right)) ^ (2)) \ le 0; \\ & \ left (((x) ^ (2)) + x + 1 - ((x) ^ (2)) - 3x-4 \ right) \ times \\ & \ times \ left (((x) ^ (2)) + x + 1 + ((x) ^ (2)) + 3x + 4 \ right) \ le 0; \\ & \ left (-2x-3 \ right) \ left (2 ((x) ^ (2)) + 4x + 5 \ right) \ le 0. \\\ end (align) \]

Spacing method:

\ [\ begin (align) & \ left (-2x-3 \ right) \ left (2 ((x) ^ (2)) + 4x + 5 \ right) = 0 \\ & -2x-3 = 0 \ Rightarrow x = -1.5; \\ & 2 ((x) ^ (2)) + 4x + 5 = 0 \ Rightarrow D = 16-40 \ lt 0 \ Rightarrow \ varnothing. \\\ end (align) \]

Just one root on the number line:
The answer is a whole interval
Answer: $ x \ in \ left [-1,5; + \ infty \ right) $.

A quick note on the last task. As one of my students accurately noted, both submodule expressions in this inequality are obviously positive, so the modulus sign can be omitted without harm to health.

But this is a completely different level of thinking and a different approach - it can be conditionally called the method of consequences. About him - in a separate lesson. Now let's move on to the final part of today's lesson and consider a universal algorithm that always works. Even when all the previous approaches were powerless. :)

4. Method of enumerating options

But what if all these techniques don't work? If inequality does not reduce to non-negative tails, if the module cannot be secluded, if at all pain-sadness-longing?

Then the "heavy artillery" of all mathematics enters the scene - the brute force method. With regard to inequalities with the modulus, it looks like this:

Write out all submodule expressions and set them to zero;
Solve the obtained equations and mark the found roots on one number line;
The straight line will be split into several sections, inside which each module has a fixed sign and therefore unambiguously unfolds;
Solve the inequality at each such site (you can separately consider the roots-boundaries obtained in paragraph 2 - for reliability). Combine the results - this will be the answer. :)

How is it? Weak? Easily! Only for a long time. Let's see in practice:

Task. Solve the inequality:

\ [\ left | x + 2 \ right | \ lt \ left | x-1 \ right | + x- \ frac (3) (2) \]

Solution. This crap is not reduced to inequalities like $ \ left | f \ right | \ lt g $, $ \ left | f \ right | \ gt g $ or $ \ left | f \ right | \ lt \ left | g \ right | $, so we go straight ahead.

We write out submodule expressions, equate them to zero and find the roots:

\ [\ begin (align) & x + 2 = 0 \ Rightarrow x = -2; \\ & x-1 = 0 \ Rightarrow x = 1. \\\ end (align) \]

In total, we have two roots, which divide the number line into three sections, within which each module is unambiguously revealed:
Partition of a numeric line by zeros of submodular functions
Let's consider each site separately.

1. Let $ x \ lt -2 $. Then both submodule expressions are negative, and the original inequality can be rewritten as follows:

\ [\ begin (align) & - \ left (x + 2 \ right) \ lt - \ left (x-1 \ right) + x-1,5 \\ & -x-2 \ lt -x + 1 + x-1,5 \\ & x \ gt 1,5 \\\ end (align) \]

We got a pretty simple limitation. Let's cross it with the original assumption that $ x \ lt -2 $:

\ [\ left \ (\ begin (align) & x \ lt -2 \\ & x \ gt 1,5 \\\ end (align) \ right. \ Rightarrow x \ in \ varnothing \]

Obviously, the variable $ x $ cannot simultaneously be less than −2, but more than 1.5. There are no decisions on this site.

1.1. Let us consider separately the borderline case: $ x = -2 $. We just substitute this number into the original inequality and check: is it true?

\ [\ begin (align) & ((\ left. \ left | x + 2 \ right | \ lt \ left | x-1 \ right | + x-1,5 \ right |) _ (x = -2) ) \\ & 0 \ lt \ left | -3 \ right | -2-1.5; \\ & 0 \ lt 3-3.5; \\ & 0 \ lt -0.5 \ Rightarrow \ varnothing. \\\ end (align) \]

Obviously, the chain of calculations led us to the wrong inequality. Therefore, the original inequality is also wrong, and $ x = -2 $ is not included in the answer.

2. Now let $ -2 \ lt x \ lt 1 $. The left module will already open with a "plus", but the right one is still with a "minus". We have:

\ [\ begin (align) & x + 2 \ lt - \ left (x-1 \ right) + x-1,5 \\ & x + 2 \ lt -x + 1 + x-1,5 \\ & x \ lt -2.5 \\\ end (align) \]

We cross again with the original requirement:

\ [\ left \ (\ begin (align) & x \ lt -2.5 \\ & -2 \ lt x \ lt 1 \\\ end (align) \ right. \ Rightarrow x \ in \ varnothing \]

And again, the empty set of solutions, since there are no numbers that are simultaneously less than −2.5, but more than −2.

2.1. And again a special case: $ x = 1 $. We substitute in the original inequality:

\ [\ begin (align) & ((\ left. \ left | x + 2 \ right | \ lt \ left | x-1 \ right | + x-1,5 \ right |) _ (x = 1)) \\ & \ left | 3 \ right | \ lt \ left | 0 \ right | + 1-1.5; \\ & 3 \ lt -0.5; \\ & 3 \ lt -0.5 \ Rightarrow \ varnothing. \\\ end (align) \]

Similar to the previous "special case", the number $ x = 1 $ is clearly not included in the answer.

3. The last piece of the straight line: $ x \ gt 1 $. Here all modules are expanded with a plus sign:

\ [\ begin (align) & x + 2 \ lt x-1 + x-1.5 \\ & x + 2 \ lt x-1 + x-1.5 \\ & x \ gt 4.5 \\ \ end (align) \]

And again we intersect the found set with the original constraint:

\ [\ left \ (\ begin (align) & x \ gt 4,5 \\ & x \ gt 1 \\\ end (align) \ right. \ Rightarrow x \ in \ left (4,5; + \ infty \ right) \]

Finally! We found the interval, which will be the answer.

Answer: $ x \ in \ left (4,5; + \ infty \ right) $

Finally, one remark that may save you from stupid mistakes when solving real problems:

Solutions to inequalities with moduli are usually solid sets on the number line - intervals and segments. Isolated points are much less common. And even less often it happens that the boundaries of the solution (the end of the segment) coincides with the boundary of the considered range.

Consequently, if the boundaries (those "special cases") are not included in the answer, then almost certainly the areas to the left and right of these boundaries will not be included in the answer. And on the contrary: the border entered into the answer, which means that some areas around it will also be the answers.

Keep this in mind when testing your solutions.

Solving inequalities online

Before solving inequalities, it is necessary to understand well how the equations are solved.

It does not matter whether the inequality is strict () or non-strict (≤, ≥), the first step is to solve the equation, replacing the inequality sign with equality (=).

Let us explain what it means to solve inequality?

After studying the equations in the student's head, the following picture develops: you need to find such values of the variable for which both sides of the equation take on the same values. In other words, find all the points where equality holds. That's right!

When we talk about inequalities, we mean finding the intervals (segments) on which the inequality holds. If there are two variables in the inequality, then the solution will no longer be intervals, but some areas on the plane. Guess what will be the solution to the inequality in three variables?

How to deal with inequalities?

A universal method for solving inequalities is considered the method of intervals (aka the method of intervals), which consists in determining all the intervals within the boundaries of which a given inequality will be satisfied.

Without going into the type of inequality, in this case it is not the essence, it is required to solve the corresponding equation and determine its roots, followed by the designation of these solutions on the number axis.

How to write down the solution to an inequality correctly?

When you have determined the intervals of solutions to an inequality, you need to correctly write out the solution itself. There is an important nuance - are the boundaries of the intervals included in the solution?

Everything is simple here. If the solution of the equation satisfies the GDV and the inequality is not strict, then the boundary of the interval is included in the solution of the inequality. Otherwise, no.

Considering each interval, the solution to the inequality can be the interval itself, or a half-interval (when one of its boundaries satisfies the inequality), or a segment - an interval together with its boundaries.

An important point

Do not think that only intervals, half-intervals, and line segments can be a solution to an inequality. No, the solution may include individual points.

For example, the inequality | x | ≤0 has only one solution - this is the point 0.

And the inequality | x |

What is the inequality calculator for?

The inequality calculator gives the correct final answer. In this case, in most cases, an illustration of a numerical axis or plane is given. It can be seen whether the boundaries of the intervals are included in the solution or not - the points are displayed filled or punctured.

Thanks to the online inequality calculator, you can check if you found the roots of the equation correctly, marked them on the number axis and checked the inequality condition on the intervals (and boundaries)?

If your answer differs from the answer of the calculator, then you definitely need to double-check your decision and identify the mistake.

In the article we will consider solution of inequalities... We will tell you in an accessible way about how to construct a solution to inequalities, with clear examples!

Before considering the solution of inequalities using examples, let's understand the basic concepts.

General information about inequalities

Inequality is called an expression in which functions are connected by relation signs>,. Inequalities are both numerical and alphabetic.
Inequalities with two signs of the relationship are called double, with three - triple, etc. For example:
a (x)> b (x),
a (x) a (x) b (x),
a (x) b (x).
a (x) Inequalities containing the sign> or or are not strict.
Solving inequality is any value of the change at which this inequality is true.
"Solve inequality"means that it is necessary to find many of all its solutions. There are various methods for solving inequalities... For solutions to inequality use the number line, which is infinite. For example, solution of inequality x> 3 is an interval from 3 to +, and the number 3 is not included in this interval, therefore a point on a straight line is denoted by an empty circle, since the inequality is strict.
+
The answer will be: x (3; +).
The value x = 3 is not included in the solution set, so the parenthesis is round. The infinity sign is always surrounded by a parenthesis. The sign means "belonging".
Consider how to solve inequalities using another signed example:
x 2
-+
The value x = 2 is included in the set of solutions, therefore the bracket is square and a point on the line is denoted by a filled circle.
The answer will be: x)