Rational inequalities with modulus. Solving inequalities with modulus

Today, friends, there will be no snot and sentiment. Instead, I will send you into battle with one of the most formidable opponents in the 8th-9th grade algebra course without further 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 to solve about 90% of these problems. What about the other 10%? Well, we'll talk about them in a separate lesson. :)

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

What you already need to know

Captain Evidence, as it were, hints that in order to solve inequalities with a modulus, you need to know two things:

1. How are inequalities resolved?
2. What is a module.

Let's start with the second point.

Module Definition

Everything is simple here. There are two definitions: algebraic and graphic. Let's start with the algebra:

Definition. The module of the number $x$ is either the 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, the modulus is “a number without a minus”. And it is in this duality (somewhere you don’t need to do anything with the original number, but somewhere you have to remove some minus there) and all the 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 real 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:

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

Solution of 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 are reduced to linear inequalities, as well as to the method of intervals.

I have two big tutorials on this topic (by the way, very, VERY useful - I recommend studying):

1. The interval method for inequalities (especially watch the video);
2. Fractional-rational inequalities is a very voluminous lesson, but after it you will not have any questions left at all.

If you know all this, if the phrase "let's move from inequality to 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 frequently encountered tasks with modules. It is required to solve an inequality of the form:

\[\left| f\right| \ltg\]

Anything can act as functions $f$ and $g$, but usually they are polynomials. Examples of such inequalities:

\[\begin(align) & \left| 2x+3\right| \ltx+7; \\ & \left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0; \\ & \left| ((x)^(2))-2\left| x \right|-3 \right| \lt 2. \\\end(align)\]

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 thing, 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: is it not easier? Unfortunately, you can't. This is the whole point of the module.

But enough of the philosophizing. Let's solve a couple of problems:

A task. Solve the inequality:

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

Solution. So, we have a classical inequality of the form “the module is less than” - 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 brackets that are preceded by a “minus”: it is quite possible that because of the 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 has been reduced to two elementary inequalities. We note their solutions on parallel real lines:

Intersection of many

The intersection of these sets will be the answer.

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

A task. Solve the inequality:

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

Solution. This task is a little more difficult. To begin with, we isolate the module by moving the second term to the right:

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

Obviously, we again have 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'm a bit of a pervert with all these brackets. But once again I remind you that our key goal is correctly solve the inequality and get the answer. Later, when you have perfectly mastered everything that is described in this lesson, you can pervert yourself as you like: open brackets, add minuses, etc.

And for starters, we 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 open all the brackets in the double inequality:

Let's move on 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 interval method (that's why I say: if you don't know what it is, it's better not to take on modules yet). 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 turned out to be an incomplete quadratic equation, which is solved elementarily. 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 obtained numbers on two parallel lines (separate for the first inequality and separate 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:

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

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

2. Inequalities of the form "Module is greater than function"

They look like this:

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

Similar to the previous one? Looks like. 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 consider two cases:

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

In this case, the options are combined with a square bracket, i.e. We have a combination of two requirements.

Pay attention again: before us is not a system, but an aggregate, therefore in the answer, the sets are combined, not intersected. This is a fundamental difference from the previous paragraph!

In general, many students have a lot of confusion with unions and intersections, so let's look into this issue once and for all:

• "∪" is a concatenation sign. 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 the intersection sign. This crap didn't come from anywhere, but 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 accuse me of promoting drug addiction and alcoholism now: if you are seriously studying this lesson, then you are already a drug addict):

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

So it became clearer? That is great. Let's move on to practice.

A task. Solve the inequality:

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

Solution. We act according to the scheme:

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

We solve each population inequality:

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

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

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

We mark each resulting set on the number line, and then combine them:

Union of sets

Obviously the answer is $x\in \left(\frac(4)(7);+\infty \right)$

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

A task. Solve the inequality:

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

Solution. Well? No, it's all the same. We pass from an inequality with a 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 bit of 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 we 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 we are waiting for a setup. If everything is clear with the numbers $\frac(-3-\sqrt(21))(2) \lt \frac(-1-\sqrt(13))(2)$ (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 smaller), with the numbers $\frac(-3-\sqrt(13))(2) \lt \frac(-1+\sqrt(21))(2)$ there will also be no difficulty (a positive number obviously more negative), but with the last couple, everything is not so simple. Which is larger: $\frac(-3+\sqrt(21))(2)$ 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 isolated the root, 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 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 arranged like this:

Case of ugly roots

Let me remind you that we are solving a set, so the answer will be the union, and not the 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 both for simple tasks and for very hard ones. The only “weak spot” in this approach is that you need to correctly compare irrational numbers (and believe me: these are not only roots). But a separate (and very serious lesson) will be devoted to questions of comparison. And we move on.

3. Inequalities with non-negative "tails"

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

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

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

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 do not confuse this with taking the root of the square:

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

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

A task. Solve the inequality:

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

Solution. We immediately notice two things:

1. This is a non-strict inequality. Points on the number line will be punched out.
2. Both sides of the inequality are obviously 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 interval method. Let's move from inequality to 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 points are shaded because the original inequality is not strict!

Getting rid of the module sign

Let me remind you for the especially stubborn: we take the signs from the last inequality, which was written down before moving on to the equation. And we paint over the areas required in the same inequality. In our case, this is $\left(x-3 \right)\left(3x+1 \right)\le 0$.

Well that's all. Problem solved.

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

A task. Solve the inequality:

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

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

Let's square it:

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

Spacing method:

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

There is only one root on the number line:

The answer is a whole range

Answer: $x\in \left[ -1.5;+\infty \right)$.

A small note about 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 already 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. And now let's move on to the final part of today's lesson and consider a universal algorithm that always works. Even when all previous approaches were powerless. :)

4. Method of enumeration of options

What if all these tricks don't work? If the inequality does not reduce to non-negative tails, if it is impossible to isolate the module, if at all pain-sadness-longing?

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

1. Write out all submodule expressions and equate them to zero;
2. Solve the resulting equations and mark the found roots on one number line;
3. The straight line will be divided into several sections, within which each module has a fixed sign and therefore unambiguously expands;
4. Solve the inequality on each such section (you can separately consider the boundary roots obtained in paragraph 2 - for reliability). Combine the results - this will be the answer. :)

Well, how? Weak? Easily! Only for a long time. Let's see in practice:

A task. Solve the inequality:

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

Solution. This crap doesn't boil down to inequalities like $\left| f\right| \lt g$, $\left| f\right| \gt g$ or $\left| f\right| \lt\left| g \right|$, so let's go 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 that divide the number line into three sections, inside which each module is revealed uniquely:

Splitting the number line by zeros of submodular functions

Let's consider each section separately.

1. Let $x \lt -2$. Then both submodule expressions are negative, and the original inequality is 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 fairly simple constraint. Let's intersect 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 greater than 1.5. There are no solutions in this area.

1.1. Let's separately consider the boundary case: $x=-2$. Let's just substitute this number into the original inequality and check: does it hold?

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

Obviously, the chain of calculations has led us to the wrong inequality. Therefore, the original inequality is also false, 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:

\[\begin(align) & x+2 \lt -\left(x-1 \right)+x-1.5 \\ & x+2 \lt -x+1+x-1.5 \\& x \lt -2.5 \\\end(align)\]

Again we intersect 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 both less than −2.5 and greater than −2.

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

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

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

3. The last piece of the 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 have found the interval, which will be the answer.

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

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

Solutions of inequalities with modules are usually continuous sets on the number line - intervals and segments. Isolated points are much rarer. And even more rarely, it happens that the boundaries of the solution (the end of the segment) coincide with the boundary of the range under consideration.

Therefore, if the boundaries (those very “special cases”) are not included in the answer, then the areas to the left-right of these boundaries will almost certainly not be included in the answer either. And vice versa: the border entered in response, which means that some areas around it will also be responses.

Keep this in mind when you check your solutions.

Maths is a symbol of the wisdom of science,

an example of scientific rigor and simplicity,

the standard of perfection and beauty in science.

Russian philosopher, professor A.V. Voloshinov

Modulo inequalities

The most difficult problems to solve in school mathematics are the inequalities, containing variables under the module sign. To successfully solve such inequalities, it is necessary to know the properties of the module well and have the skills to use them.

Basic concepts and properties

Modulus (absolute value) of a real number denoted and is defined as follows:

The simple properties of the module include the following relationships:

AND .

Note, that the last two properties hold for any even degree.

Also, if , where , then and

More complex module properties, which can be effectively used in solving equations and inequalities with modules, are formulated by means of the following theorems:

Theorem 1.For any analytic functions And the inequality.

Theorem 2. Equality is equivalent to the inequality.

Theorem 3. Equality is equivalent to the inequality.

The most common inequalities in school mathematics, containing unknown variables under the modulo sign, are inequalities of the form and where some positive constant.

Theorem 4. Inequality is equivalent to a double inequality, and the solution to the inequalityreduces to solving the set of inequalities And .

This theorem is a particular case of Theorems 6 and 7.

More complex inequalities, containing the module are inequalities of the form, And .

Methods for solving such inequalities can be formulated using the following three theorems.

Theorem 5. Inequality is equivalent to the combination of two systems of inequalities

AND (1)

Proof. Since then

This implies the validity of (1).

Theorem 6. Inequality is equivalent to the system of inequalities

Proof. Because , then from the inequality follows that . Under this condition, the inequalityand in this case the second system of inequalities (1) turns out to be inconsistent.

The theorem has been proven.

Theorem 7. Inequality is equivalent to the combination of one inequality and two systems of inequalities

AND (3)

Proof. Since , then the inequality always executed, if .

Let be , then the inequalitywill be tantamount to inequality, from which the set of two inequalities follows And .

The theorem has been proven.

Consider typical examples of solving problems on the topic “Inequalities, containing variables under the module sign.

Solving inequalities with modulus

The simplest method for solving inequalities with modulus is the method, based on module expansion. This method is generic, however, in the general case, its application can lead to very cumbersome calculations. Therefore, students should also know other (more efficient) methods and techniques for solving such inequalities. In particular, need to have the skills to apply theorems, given in this article.

Example 1Solve the inequality

. (4)

Solution.Inequality (4) will be solved by the "classical" method - the moduli expansion method. To this end, we break the numerical axis dots and intervals and consider three cases.

1. If , then , , , and inequality (4) takes the form or .

Since the case is considered here, , is a solution to inequality (4).

2. If , then from inequality (4) we obtain or . Since the intersection of intervals And is empty, then there are no solutions to inequality (4) on the considered interval.

3. If , then inequality (4) takes the form or . It's obvious that is also a solution to inequality (4).

Answer: , .

Example 2 Solve the inequality.

Solution. Let's assume that . Because , then the given inequality takes the form or . Because , then and hence follows or .

However , therefore or .

Example 3 Solve the inequality

. (5)

Solution. Because , then inequality (5) is equivalent to the inequalities or . From here, according to Theorem 4, we have a set of inequalities And .

Answer: , .

Example 4Solve the inequality

. (6)

Solution. Let's denote . Then from inequality (6) we obtain the inequalities , , or .

From here, using the interval method, we get . Because , then here we have a system of inequalities

The solution to the first inequality of system (7) is the union of two intervals And , and the solution of the second inequality is the double inequality. This implies , that the solution to the system of inequalities (7) is the union of two intervals And .

Answer: ,

Example 5Solve the inequality

. (8)

Solution. We transform inequality (8) as follows:

Or .

Applying the interval method, we obtain a solution to inequality (8).

Answer: .

Note. If we put and in the condition of Theorem 5, then we obtain .

Example 6 Solve the inequality

. (9)

Solution. From inequality (9) it follows. We transform inequality (9) as follows:

Or

Since , then or .

Answer: .

Example 7Solve the inequality

. (10)

Solution. Since and , then or .

In this connection and inequality (10) takes the form

Or

. (11)

It follows from this that or . Since , then inequality (11) also implies or .

Answer: .

Note. If we apply Theorem 1 to the left side of inequality (10), then we get . From here and from inequality (10) it follows, that or . Because , then inequality (10) takes the form or .

Example 8 Solve the inequality

. (12)

Solution. Since then and inequality (12) implies or . However , therefore or . From here we get or .

Answer: .

Example 9 Solve the inequality

. (13)

Solution. According to Theorem 7, the solutions to inequality (13) are or .

Let now. In this case and inequality (13) takes the form or .

If we combine intervals And , then we obtain a solution to inequality (13) of the form.

Example 10 Solve the inequality

. (14)

Solution. Let us rewrite inequality (14) in an equivalent form: . If we apply Theorem 1 to the left side of this inequality, then we obtain the inequality .

From here and from Theorem 1 it follows, that inequality (14) is satisfied for any values.

Answer: any number.

Example 11. Solve the inequality

. (15)

Solution. Applying Theorem 1 to the left side of inequality (15), we get . From here and from inequality (15) follows the equation, which looks like.

According to Theorem 3, the equation is equivalent to the inequality. From here we get.

Example 12.Solve the inequality

. (16)

Solution. From inequality (16), according to Theorem 4, we obtain the system of inequalities

When solving the inequalitywe use Theorem 6 and obtain the system of inequalitiesfrom which follows.

Consider the inequality. According to Theorem 7, we obtain a set of inequalities And . The second population inequality holds for any real.

Consequently , the solution of inequality (16) are.

Example 13Solve the inequality

. (17)

Solution. According to Theorem 1, we can write

(18)

Taking into account inequality (17), we conclude that both inequalities (18) turn into equalities, i.e. there is a system of equations

By Theorem 3, this system of equations is equivalent to the system of inequalities

or

Example 14Solve the inequality

. (19)

Solution. Since , then . Let us multiply both parts of inequality (19) by the expression , which for any values ​​takes only positive values. Then we obtain an inequality that is equivalent to inequality (19), of the form

From here we get or , where . Since and then the solutions to inequality (19) are And .

Answer: , .

For a deeper study of methods for solving inequalities with a module, it is advisable to refer to tutorials, listed in the list of recommended readings.

1. Collection of tasks in mathematics for applicants to technical universities / Ed. M.I. Scanavi. - M .: World and Education, 2013. - 608 p.

2. Suprun V.P. Mathematics for high school students: methods for solving and proving inequalities. – M.: Lenand / URSS, 2018. - 264 p.

3. Suprun V.P. Mathematics for high school students: non-standard methods for solving problems. - M .: KD "Librocom" / URSS, 2017. - 296 p.

Do you have any questions?

To get the help of a tutor - register.

site, with full or partial copying of the material, a link to the source is required.

Methods (rules) for revealing inequalities with modules consist in the sequential disclosure of modules, while using intervals of constant sign of submodule functions. In the final version, several inequalities are obtained from which they find intervals or intervals that satisfy the condition of the problem.

Let's move on to solving examples that are common in practice.

Linear inequalities with modules

By linear we mean equations in which the variable enters the equation linearly.

Example 1. Find a solution to an inequality

Solution:
It follows from the condition of the problem that the modules turn into zero at x=-1 and x=-2. These points divide the numerical axis into intervals

In each of these intervals, we solve the given inequality. To do this, first of all, we draw up graphic drawings of the areas of constant sign of submodular functions. They are depicted as areas with signs of each of the functions.

or intervals with signs of all functions.

On the first interval, open the modules

We multiply both parts by minus one, while the sign in the inequality will change to the opposite. If it is difficult for you to get used to this rule, then you can move each of the parts beyond the sign to get rid of the minus. In the end, you will receive

The intersection of the set x>-3 with the area on which the equations were solved will be the interval (-3;-2) . For those who find it easier to look for solutions graphically, you can draw the intersection of these areas

General intersection of areas will be the solution. With strict unevenness, the edges are not included. If nonstrict is checked by substitution.

On the second interval, we get

The section will be the interval (-2; -5/3). Graphically, the solution will look like

On the third interval, we get

This condition does not give solutions on the required area.

Since the two solutions found (-3;-2) and (-2;-5/3) border the point x=-2 , we check it too.

Thus the point x=-2 is the solution. The general solution with this in mind will look like (-3;5/3).

Example 2. Find a solution to the inequality
|x-2|-|x-3|>=|x-4|

Solution:
The zeros of the submodule functions will be the points x=2, x=3, x=4 . When the values ​​of the arguments are less than these points, the submodule functions are negative, and when the values ​​are large, they are positive.

The points divide the real axis into four intervals. We open the modules according to the intervals of constancy of sign and solve the inequalities.

1) On the first interval, all submodular functions are negative, therefore, when expanding the modules, we change the sign to the opposite.

The intersection of the found x values ​​with the considered interval will be the set of points

2) In the interval between the points x=2 and x=3, the first submodule function is positive, the second and third are negative. Expanding the modules, we get

an inequality that, in intersection with the interval on which we are solving, gives one solution - x=3.

3) In the interval between the points x=3 and x=4, the first and second submodule functions are positive, and the third one is negative. Based on this, we get

This condition shows that the whole interval will satisfy the inequality with modules.

4) For values ​​x>4, all functions are sign-positive. When expanding modules, we do not change their sign.

The found condition at the intersection with the interval gives the following set of solutions

Since the inequality is solved on all intervals, it remains to find the common value of all found x values. The solution is two intervals

This example is solved.

Example 3. Find a solution to the inequality
||x-1|-5|>3-2x

Solution:
We have an inequality with a module from a module. Such inequalities are revealed as modules are nested, starting with those that are placed deeper.

The submodule function x-1 is converted to zero at the point x=1 . For smaller values ​​beyond 1 it is negative and positive for x>1 . Based on this, we open the inner module and consider the inequality on each of the intervals.

First consider the interval from minus infinity to one

The submodule function is zero at the point x=-4 . For smaller values ​​it is positive, for larger values ​​it is negative. Expand the module for x<-4:

At the intersection with the area on which we consider, we obtain a set of solutions

The next step is to expand the module on the interval (-4; 1)

Taking into account the expansion area of ​​the module, we obtain the interval of solutions

REMEMBER: if you get two intervals bordering a common point in such irregularities with modules, then, as a rule, this is also a solution.

To do this, you just need to check.

In this case, we substitute the point x=-4.

So x=-4 is the solution.
Expand the inner module for x>1

Submodule function is negative for x<6.
Expanding the module, we get

This condition in the section with the interval (1;6) gives an empty set of solutions.

For x>6 we get the inequality

Also solving we got an empty set.
Given all of the above, the only solution to the inequality with modules will be the following interval.

Inequalities with modules containing quadratic equations

Example 4. Find a solution to the inequality
|x^2+3x|>=2-x^2

Solution:
The submodule function vanishes at the points x=0, x=-3. By simple substitution minus one

we set that it is less than zero on the interval (-3; 0) and positive beyond it.
Expand the module in areas where the submodule function is positive

It remains to determine the areas where the square function is positive. To do this, we determine the roots of the quadratic equation

For convenience, we substitute the point x=0, which belongs to the interval (-2;1/2). The function is negative in this interval, so the solution will be the following sets x

Here, brackets indicate the edges of the areas with solutions; this was done deliberately, taking into account the following rule.

REMEMBER: If the inequality with modules, or a simple inequality is strict, then the edges of the found areas are not solutions, but if the inequalities are not strict (), then the edges are solutions (indicated by square brackets).

This rule is used by many teachers: if a strict inequality is given, and you write a square bracket ([,]) in the solution during calculations, they will automatically consider this to be an incorrect answer. Also, when testing, if a non-strict inequality with modules is specified, then among the solutions, look for areas with square brackets.

On the interval (-3; 0), expanding the module, we change the sign of the function to the opposite

Taking into account the scope of the inequality disclosure, the solution will have the form

Together with the previous area, this will give two half-intervals

Example 5. Find a solution to the inequality
9x^2-|x-3|>=9x-2

Solution:
A non-strict inequality is given, the submodule function of which is equal to zero at the point x=3. At smaller values ​​it is negative, at larger values ​​it is positive. We expand the module on the interval x<3.

Finding the discriminant of the equation

and roots

Substituting the zero point, we find out that on the interval [-1/9; 1] the quadratic function is negative, therefore the interval is a solution. Next, open the module for x>3

modulo number this number itself is called if it is non-negative, or the same number with the opposite sign if it is negative.

For example, the modulus of 6 is 6, and the modulus of -6 is also 6.

That is, the modulus of a number is understood as an absolute value, the absolute value of this number without taking into account its sign.

Denoted as follows: |6|, | X|, |but| etc.

(For more details, see the "Module of Number" section).

Modulo Equations.

Example 1 . solve the equation|10 X - 5| = 15.

Solution.

In accordance with the rule, the equation is equivalent to the combination of two equations:

10X - 5 = 15
10X - 5 = -15

We decide:

10X = 15 + 5 = 20
10X = -15 + 5 = -10

X = 20: 10
X = -10: 10

X = 2
X = -1

Answer: X 1 = 2, X 2 = -1.

Example 2 . solve the equation|2 X + 1| = X + 2.

Solution.

Since the modulus is a non-negative number, then X+ 2 ≥ 0. Accordingly:

X ≥ -2.

We make two equations:

2X + 1 = X + 2
2X + 1 = -(X + 2)

We decide:

2X + 1 = X + 2
2X + 1 = -X - 2

2X - X = 2 - 1
2X + X = -2 - 1

X = 1
X = -1

Both numbers are greater than -2. So both are roots of the equation.

Answer: X 1 = -1, X 2 = 1.

Example 3 . solve the equation

|X + 3| - 1
————— = 4
X - 1

Solution.

The equation makes sense if the denominator is not equal to zero - so if X≠ 1. Let's take this condition into account. Our first action is simple - we don’t just get rid of the fraction, but we transform it in such a way as to get the module in its purest form:

|X+ 3| - 1 = 4 ( X - 1),

|X + 3| - 1 = 4X - 4,

|X + 3| = 4X - 4 + 1,

|X + 3| = 4X - 3.

Now we have only the expression under the modulus on the left side of the equation. Move on.
The modulus of a number is a non-negative number - that is, it must be greater than or equal to zero. Accordingly, we solve the inequality:

4X - 3 ≥ 0

4X ≥ 3

X ≥ 3/4

Thus, we have a second condition: the root of the equation must be at least 3/4.

In accordance with the rule, we compose a set of two equations and solve them:

X + 3 = 4X - 3
X + 3 = -(4X - 3)

X + 3 = 4X - 3
X + 3 = -4X + 3

X - 4X = -3 - 3
X + 4X = 3 - 3

X = 2
X = 0

We received two responses. Let's check if they are the roots of the original equation.

We had two conditions: the root of the equation cannot be equal to 1, and it must be at least 3/4. I.e X ≠ 1, X≥ 3/4. Both of these conditions correspond to only one of the two answers received - the number 2. Hence, only it is the root of the original equation.

Answer: X = 2.

Inequalities with the modulus.

Example 1 . Solve the inequality| X - 3| < 4

Solution.

The module rule says:

|but| = but, if but ≥ 0.

|but| = -but, if but < 0.

The modulus can have both a non-negative and a negative number. So we have to consider both cases: X- 3 ≥ 0 and X - 3 < 0.

1) When X- 3 ≥ 0 our original inequality remains as it is, only without the modulo sign:
X - 3 < 4.

2) When X - 3 < 0 в исходном неравенстве надо поставить знак минус перед всем подмодульным выражением:

-(X - 3) < 4.

Opening the brackets, we get:

-X + 3 < 4.

Thus, from these two conditions, we have come to the union of two systems of inequalities:

X - 3 ≥ 0
X - 3 < 4

X - 3 < 0
-X + 3 < 4

Let's solve them:

X ≥ 3
X < 7

X < 3
X > -1

So, in our answer we have the union of two sets:

3 ≤ X < 7 U -1 < X < 3.

Determine the smallest and largest values. These are -1 and 7. At the same time X greater than -1 but less than 7.
Besides, X≥ 3. Hence, the solution to the inequality is the entire set of numbers from -1 to 7, excluding these extreme numbers.

Answer: -1 < X < 7.

Or: X ∈ (-1; 7).

Add-ons.

1) There is a simpler and shorter way to solve our inequality - graphical. To do this, draw a horizontal axis (Fig. 1).

Expression | X - 3| < 4 означает, что расстояние от точки X to point 3 less than four units. We mark the number 3 on the axis and count 4 divisions to the left and right of it. On the left we will come to point -1, on the right - to point 7. Thus, the points X we just saw without calculating them.

Moreover, according to the inequality condition, -1 and 7 themselves are not included in the set of solutions. Thus, we get the answer:

1 < X < 7.

2) But there is another solution that is even simpler than the graphical way. To do this, our inequality must be presented in the following form:

4 < X - 3 < 4.

After all, this is how it is according to the rule of the module. The non-negative number 4 and the similar negative number -4 are the boundaries of the solution to the inequality.

4 + 3 < X < 4 + 3

1 < X < 7.

Example 2 . Solve the inequality| X - 2| ≥ 5

Solution.

This example differs significantly from the previous one. The left side is greater than 5 or equal to 5. From a geometric point of view, the solution to the inequality is all the numbers that are at a distance of 5 units or more from point 2 (Fig. 2). The graph shows that these are all numbers that are less than or equal to -3 and greater than or equal to 7. So, we have already received the answer.

Answer: -3 ≥ X ≥ 7.

Along the way, we solve the same inequality by rearranging the free term to the left and right with the opposite sign:

5 ≥ X - 2 ≥ 5

5 + 2 ≥ X ≥ 5 + 2

The answer is the same: -3 ≥ X ≥ 7.

Or: X ∈ [-3; 7]

Example solved.

Example 3 . Solve the inequality 6 X 2 - | X| - 2 ≤ 0

Solution.

Number X can be positive, negative or zero. Therefore, we need to take into account all three circumstances. As you know, they are taken into account in two inequalities: X≥ 0 and X < 0. При X≥ 0, we simply rewrite our original inequality as is, only without the modulo sign:

6x 2 - X - 2 ≤ 0.

Now for the second case: if X < 0. Модулем отрицательного числа является это же число с противоположным знаком. То есть пишем число под модулем с обратным знаком и опять же освобождаемся от знака модуля:

6X 2 - (-X) - 2 ≤ 0.

Expanding the brackets:

6X 2 + X - 2 ≤ 0.

Thus, we have received two systems of equations:

6X 2 - X - 2 ≤ 0
X ≥ 0

6X 2 + X - 2 ≤ 0
X < 0

We need to solve inequalities in systems - which means we need to find the roots of two quadratic equations. To do this, we equate the left-hand sides of the inequalities to zero.

Let's start with the first one:

6X 2 - X - 2 = 0.

How to solve a quadratic equation - see the "Quadric Equation" section. We will immediately name the answer:

X 1 \u003d -1/2, x 2 \u003d 2/3.

From the first system of inequalities, we get that the solution to the original inequality is the entire set of numbers from -1/2 to 2/3. We write the union of solutions for X ≥ 0:
[-1/2; 2/3].

Now let's solve the second quadratic equation:

6X 2 + X - 2 = 0.

Its roots:

X 1 = -2/3, X 2 = 1/2.

Conclusion: when X < 0 корнями исходного неравенства являются также все числа от -2/3 до 1/2.

Let's combine the two answers and get the final answer: the solution is the whole set of numbers from -2/3 to 2/3, including these extreme numbers.

Answer: -2/3 ≤ X ≤ 2/3.

Or: X ∈ [-2/3; 2/3].