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Inequalities with modulus. A new look at the solution. Equations and inequalities with modulus

Modulus of numbers 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 the number 6 is 6, and the modulus of the number -6 is also 6.

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

It is designated as follows: |6|, | X|, |A| etc.

(More details in the “Number module” section).

Equations with modulus.

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

Solution.

According to 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.

Let's 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 zero - that means 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 transform it so as to obtain the module in its pure 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 an expression under the modulus on the left side of the equation. Go ahead.
The modulus of a number is a non-negative number - that is, it must be greater than zero 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 answers. Let's check whether they are 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. That is X ≠ 1, X≥ 3/4. Both of these conditions correspond to only one of the two answers received - the number 2. This means that only this is the root of the original equation.

Answer: X = 2.

Inequalities with modulus.

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

Solution.

The module rule states:

|A| = A, If A ≥ 0.

|A| = -A, If A < 0.

The module can have both non-negative and negative numbers. 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 modulus 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 came to the unification 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, our answer is a union of two sets:

3 ≤ X < 7 U -1 < X < 3.

Determine the smallest and highest value. These are -1 and 7. Moreover X greater than -1 but less than 7.
Besides, X≥ 3. This means that 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 - graphically. To do this, you need to draw a horizontal axis (Fig. 1).

Expression | X - 3| < 4 означает, что расстояние от точки X to point 3 is less than four units. We mark the number 3 on the axis and count 4 divisions to the left and to the 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 them 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 simpler even than the graphical method. 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 modulus rule. The non-negative number 4 and the similar negative number -4 are the boundaries for solving the inequality.

4 + 3 < X < 4 + 3

1 < X < 7.

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

Solution.

This example is significantly different from the previous one. The left side is greater than 5 or equal to 5. C geometric point From the point of view, the solution to the inequality is all 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. This means that 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 to the 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]

The example is solved.

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

Solution.

Number X can be a positive number, negative number, 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 modulus sign:

6x 2 - X - 2 ≤ 0.

Now about 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 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 - and this 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 section “Quadratic Equation”. We will immediately name the answer:

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

From the first system of inequalities we obtain 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 at 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 entire 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].

How more people understands, the stronger his desire to understand

Thomas Aquinas

The interval method allows you to solve any equations containing a modulus. The essence of this method is to split the number axis into several sections (intervals), and the axis needs to be split by the zeros of the expressions in the modules. Then, on each of the resulting sections, every submodular expression is either positive or negative. Therefore, each of the modules can be opened either with a minus sign or with a plus sign. After these actions, all that remains is to solve each of the received simple equations on the interval under consideration and combine the received answers.

Let's look at this method using a specific example.

|x + 1| + |2x – 4| – |x + 3| = 2x – 6.

1) Let's find the zeros of the expressions in the modules. To do this, we need to equate them to zero and solve the resulting equations.

x + 1 = 0 2x – 4 = 0 x + 3 = 0

x = -1 2x = 4 x = -3

2) Place the resulting points in the required order on the coordinate line. They will split the entire axis into four sections.

3) Let us determine on each of the resulting sections the signs of the expressions in the modules. To do this, we substitute into them any numbers from the intervals of interest to us. If the result of the calculation is a positive number, then we put “+” in the table, and if the number is negative, then we put “–”. This can be depicted like this:

4) Now we will solve the equation on each of the four intervals, revealing the modules with the signs that are indicated in the table. So, let's look at the first interval:

I interval (-∞; -3). On it, all modules are opened with a “–” sign. We get the following equation:

-(x + 1) – (2x – 4) – (-(x + 3)) = 2x – 6. Let’s present similar terms, first opening the parentheses in the resulting equation:

X – 1 – 2x + 4 + x + 3 = 2x – 6

The received answer is not included in the considered interval, so it is not necessary to write it in the final answer.

II interval [-3; -1). At this interval in the table there are signs “–”, “–”, “+”. This is exactly how we open the modules of the original equation:

-(x + 1) – (2x – 4) – (x + 3) = 2x – 6. Let’s simplify by opening the brackets:

X – 1 – 2x + 4 – x – 3 = 2x – 6. Let us present similar ones in the resulting equation:

x = 6/5. The resulting number does not belong to the interval under consideration, therefore it is not the root of the original equation.

III interval [-1; 2). We expand the modules of the original equation with the signs that appear in the third column in the figure. We get:

(x + 1) – (2x – 4) – (x + 3) = 2x – 6. Let’s get rid of the parentheses and move the terms containing the variable x to the left side of the equation, and those not containing x to the right. Will have:

x + 1 – 2x + 4 – x – 3 = 2x – 6

The number 2 is not included in the interval under consideration.

IV interval) – they will automatically consider this as an incorrect answer. Also, when testing, if a non-strict inequality with modules is given, then look for areas with square brackets among the solutions.

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

Taking into account the area of ​​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 whose submodular function is equal to zero at the point x=3. For smaller values ​​it is negative, for larger values ​​it is positive. Expand the module on the interval x<3.

Finding the discriminant of the equation

and roots

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

Today, friends, there will be no snot or sentimentality. Instead, I will send you, no questions asked, into battle with one of the most formidable opponents in the 8th-9th grade algebra course.

Yes, you understood everything correctly: we are talking about inequalities with modulus. We will look at four basic techniques with which you will learn to solve about 90% of such problems. What about the remaining 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 already need to know

Captain Obviousness seems to hint that to solve inequalities with modulus you need to know two things:

  1. How inequalities are 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 graphical. To begin with - algebraic:

Definition. The modulus of a 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, a modulus is a “number without a minus.” And it is in this duality (in some places you don’t have to do anything with the original number, but in others you will have to remove some kind of minus) that is where the whole difficulty lies for beginning students.

Is there some more geometric definition. It is also useful to know, but we will turn 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 point $a$ be marked on the number line. Then the module $\left| x-a \right|$ is the distance from point $x$ to point $a$ on this line.

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


Graphical module definition

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

Solving inequalities. Interval method

Now let's look at the 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 come down to linear inequalities, as well as to the interval method.

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

  1. Interval method for inequalities (especially watch the video);
  2. Fractional rational inequalities is a very extensive lesson, but after it you won’t have any questions at all.

If you know all this, if the phrase “let’s move from inequality to equation” does not make you have a vague desire to hit yourself against the wall, then you are ready: welcome to hell to the main topic of the lesson. :)

1. Inequalities of the form “Modulus is less than function”

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

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

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

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

All of them can be solved literally in one line according to the following 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 in return we get a double inequality (or, which is the same thing, a system of two inequalities). But this transition takes into account absolutely everything possible problems: if the number under the modulus 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 simpler? Unfortunately, it's not possible. This is the whole point of the module.

However, enough with the 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 classic inequality of the form “the modulus is less”—there’s nothing even 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 that have a “minus” in front of them: it is quite possible that due to your 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 note 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 a little more difficult. First, let’s 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 smaller”, so we get rid of the module using 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 parentheses. But let me remind you once again that our key goal is correctly solve the inequality and get the answer. Later, when you have perfectly mastered everything described in this lesson, you can pervert it yourself as you wish: open brackets, add minuses, etc.

To begin with, we’ll simply 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 the 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 quadratic and can be solved using the interval method (that’s why I say: if you don’t know what this is, it’s better not to take on modules yet). Let's move on 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 look at the second inequality of the system. There you will 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 resulting 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 that after these examples the solution scheme is extremely 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 according to the scheme described above. At some point, it will be necessary to move from double inequality to a system of two independent expressions, each of which can already be solved separately.
  3. Finally, all that remains is to intersect 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 more features. However, there are a couple of serious “buts”. We’ll talk about these “buts” now.

2. Inequalities of the form “Modulus is greater than function”

They look like this:

\[\left| f\right| \gtg\]

Similar to the previous one? It seems. And yet such problems 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 and solve the usual inequality;
  2. Then, in essence, we expand the module with the minus sign, and then multiply both sides of the inequality by −1, while I have the sign.

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

Please note again: this is not a system, but a totality, therefore in the answer the sets are combined rather than intersecting. This is a fundamental difference from the previous point!

In general, many students are completely confused with unions and intersections, so let’s sort this issue out once and for all:

  • "∪" is a union sign. Essentially, this is a stylized letter "U" that came to us from in English and is an abbreviation for “Union”, i.e. "Associations".
  • "∩" is the intersection sign. This crap didn’t come from anywhere, but simply appeared as a counterpoint to “∪”.

To make it even easier to remember, simply draw legs to these signs to make glasses (just don’t now accuse me of 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: the union (totality) includes elements from both sets, therefore it is in no way less than each of them; but the intersection (system) includes only those elements that are simultaneously in both the first set and the second. Therefore, the intersection of sets is never larger than the source sets.

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

Task. Solve the inequality:

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

Solution. We proceed 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 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 combine them:

Union of sets

It is quite obvious that the answer will be $x\in \left(\frac(4)(7);+\infty \right)$

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 move 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 every inequality. Unfortunately, the roots there will not be very good:

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

The second inequality is also a bit wild:

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

Now you need to mark these numbers on two axes - one axis for each inequality. However, points must be marked in in the right order: how larger number, the further we shift the point to the right.

And here a setup awaits us. 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 less), with the numbers $\frac(-3-\sqrt(13))(2) \lt \frac(-1+\sqrt(21))(2)$ there will also be no difficulties (positive number obviously more negative), then with the last couple everything is not so clear. Which is greater: $\frac(-3+\sqrt(21))(2)$ or $\frac(-1+\sqrt(13))(2)$? The placement 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 $\frac(-1+\sqrt(13))(2) \gt \frac(-3+\sqrt(21)) (2)$, the final 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 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 for very tough ones. The only thing " weakness“In this approach, 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”

Now we get to the most interesting part. These are inequalities of the form:

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

Generally speaking, the algorithm that we will talk about now is correct 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 don’t confuse this with taking the root of a 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 this now. Let's solve a couple of problems better:

Task. Solve the inequality:

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

Solution. Let's immediately notice two things:

  1. This is not a strict inequality. Points on the number line will be punctured.
  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, taking advantage of the evenness of the module (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 using 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 modulus sign

Let me remind you for those who are 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 it is $\left(x-3 \right)\left(3x+1 \right)\le 0$.

OK it's all over Now. The problem is 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 everything the same. I won't comment - just look at the sequence of actions.

Square it:

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

Interval 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 interval

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

A small note about the last task. As one of my students accurately noted, both submodular 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 conditionally be called the method of consequences. About it - in a separate lesson. Now let’s move on to the final part of today’s lesson and look at a universal algorithm that always works. Even when all previous approaches were powerless. :)

4. Method of enumeration of options

What if all these techniques don't help? If the inequality cannot be reduced to non-negative tails, if it is impossible to isolate the module, if in general there is pain, sadness, melancholy?

Then the “heavy artillery” of all mathematics comes onto the scene—the brute force method. In relation to inequalities with modulus it looks like this:

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

So how? 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 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 we act ahead.

We write out submodular 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, within which each module is revealed uniquely:

Partitioning the number line by zeros of submodular functions

Let's look at each section separately.

1. Let $x \lt -2$. Then both submodular expressions are negative, and the original inequality will 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 fairly simple limitation. Let's intersect it with the initial 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 be simultaneously less than −2 and greater than 1.5. There are no solutions in this area.

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

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

It is obvious that the chain of calculations has led us to an incorrect inequality. Therefore, the original inequality is also false, and $x=-2$ is not included in the answer.

2. Let now $-2 \lt x \lt 1$. The left module will already open with a “plus”, but the right one will still open 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 set of solutions is empty, 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:

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

Similar 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 opened 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 an interval that 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 usually represent continuous sets on the number line - intervals and segments. Isolated points are much less common. And even less often, it happens that the boundary of the solution (the end of the segment) coincides with the boundary of the range under consideration.

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

Keep this in mind when reviewing your solutions.

Municipal educational institution "Khvastovichskaya" high school»

"The interval method for solving equations and inequalities with multiple modules"

Research paper in mathematics

Performed:

10th grade student

Golysheva Evgenia

Supervisor:

mathematic teacher

Shapenskaya E.N.

Introduction……………………………………………………………………………………… … ….3 Chapter 1. Methods for solving problems with several modules………………… …............4 1.1.Definition of a module. Solution by definition.........4 1.2 Solving equations with multiple modules using the interval method......5 1.3 . Problems with multiple modules. Solution methods……………………………....7 1.4. Method of intervals in problems with modules…………………………………………......9 Chapter 2. Equations and inequalities containing modules………………………….…. 11 2.1 Solving equations with several modules using the interval method..….11 2.2 Solving inequalities with several modules using the interval method.…13 Conclusion…………………………………………………… ………………………...15 Literature……………………………………………………………………………….……….….16

Introduction

The concept of absolute value is one of the most important characteristics of a number, both in the field of real and complex numbers. This concept is widely used not only in various sections school course mathematics, but also in courses of higher mathematics, physics and technical sciences studied at universities. Problems related to absolute values ​​are often found in mathematical Olympiads, university entrance exams and the Unified State Exam.

Subject:"The interval method for solving equations and inequalities with multiple modules by the interval method."

Objective area: mathematics.

Object of study: solving equations and inequalities with modulus.

Subject of study: interval method for solving with several modules.

Purpose of the study: identify the effectiveness of solving equations and inequalities with several modules using the interval method.

Hypothesis: If you use the interval method to solve inequalities and equations with several modules, you can significantly simplify your work.

Working methods: collection of information and its analysis.

Tasks:

    Study the literature on this topic.

    Consider solutions to inequalities and equations with multiple modules.

    Identify the most effective method solutions.

Practical focus of the project:

This work can be used as teaching aid for students and methodological manual for the teacher.

Chapter 1.

1.1.Definition of a module. Solution by definition.

By definition, the modulus, or absolute value, of a non-negative number a coincides with the number itself, and the modulus of a negative number is equal to the opposite number, that is, a:

The modulus of a number is always non-negative. Let's look at examples.

Example 1. Solve the equation |–x| = –3.

There is no need to analyze cases here, because the absolute value of a number is always non-negative, and this means that this equation has no solutions.

Let us write the solution to these simplest equations in general view:

Example 2. Solve the equation |x| = 2 – x.

Solution. At x 0 we have the equation x = 2 – x, i.e. x = 1. Since 1 0, x = 1 is the root of the original equation. In the second case (x

Answer: x = 1.

Example 3. Solve the equation 3|x – 3| + x = –1.

Solution. Here the division into cases is determined by the sign of the expression x – 3. For x – 3 ³ 0 we have 3x – 9 + x = –1 Û x = 2. But 2 – 3 0.

Answer: the equation has no roots.

Example 4. Solve the equation |x – 1| = 1 – x.

Solution. Since 1 – x = – (x – 1), it follows directly from the definition of the modulus that the equation is satisfied by those and only those x for which x – 1 0. This equation has been reduced to an inequality, and the answer is the entire interval (ray).

Answer: x 1.

1.2. Solving equations with modulus using systems.

The examples discussed earlier allow us to formulate rules for eliminating the modulus sign in equations. For equations of the form |f(x)| = g(x) there are two such rules:

1st rule: |f(x)| = g(x) Û (1)
2nd rule: |f(x)| = g(x) Û (2)

Let us explain the notation used here. Curly brackets represent systems, and square brackets represent aggregates.

Solutions to a system of equations are values ​​of a variable that simultaneously satisfy all the equations of the system.

The solutions to a set of equations are all values ​​of a variable, each of which is the root of at least one of the equations in the set.

Two equations are equivalent if any solution of each of them is also a solution of the other, in other words, if the sets of their solutions coincide.

If the equation contains several modules, then you can get rid of them one by one, using the given rules. But usually there is more shortcuts. We'll get to know them later, but now let's look at solving the simplest of these equations:

|f(x)| = |g(x)| Û

This equivalence follows from the obvious fact that if the absolute values ​​of two numbers are equal, then the numbers themselves are either equal or opposite.

Example 1. Solve the equation |x 2 – 7x + 11| = x + 1.
Solution. Let's get rid of the module in two ways described above:

1st way: 2nd way:

As we see, in both cases we have to solve the same two quadratic equations, but in the first case they are accompanied by quadratic inequalities, and in the second – linear. Therefore, the second method for given equation easier. Solving quadratic equations, we find the roots of the first, both roots satisfy the inequality. The discriminant of the second equation is negative, therefore the equation has no roots.

Answer: .
Example 2. Solve the equation |x 2 – x – 6| = |2x 2 + x – 1|.

Solution. We already know that there is no need to consider (as many as 4) variants of the distribution of signs of expressions under modules here: this equation is equivalent to a set of two quadratic equations without any additional inequalities: Which is equivalent to: The first equation of the set of solutions does not have (its discriminant is negative), the second the equation has two roots.

1.3. Problems with multiple modules. Solution methods.

Sequential expansion of modules.

There are two main approaches to solving equations and inequalities that contain multiple modules. We can call them "serial" and "parallel". Now let's get acquainted with the first of them.

Its idea is that first one of the modules is isolated in one part of the equation (or inequality) and is revealed by one of the methods described earlier. Then the same thing is repeated with each of the resulting equations with modules, and so on until we get rid of all the modules.

Example 1. Solve the equation: +

Solution. Let’s isolate the second module and expand it using the first method, that is, simply determining the absolute value:

To the resulting two equations we apply the second method of removing the module:

Finally, we solve the resulting four linear equations and select those roots that satisfy the corresponding inequalities. As a result, only two values ​​remain: x = –1 and .

Answer: -1; .

Parallel expansion of modules.

You can remove all modules in an equation or inequality at once and write down all possible combinations of signs of submodular expressions. If there are n modules in the equation, then there will be 2 n options, because each of the n expressions under the module, when removing the module, can receive one of two signs - plus or minus. In principle, we need to solve all 2 n equations (or inequalities), freed from moduli. But their solutions will also be solutions to the original problem only if they lie in regions where the corresponding equation (inequality) coincides with the original one. These areas are defined by the signs of the expressions under the modules. We have already solved the following inequality, so you can compare different approaches to solving it.

Example 2.+
Solution.

Let's consider 4 possible sets of symbols for expressions under modules.

Only the first and third of these roots satisfy the corresponding inequalities, and therefore the original equation.

Answer: -1; .

Similarly, you can solve any problems with several modules. But, like any universal method, this solution is not always optimal. Below we will see how it can be improved.

1.4. Interval method in problems with modules

Taking a closer look at the conditions that define different variants distribution of signs of submodular expressions in the previous solution, we will see that one of them, 1 – 3x

Imagine that we are solving an equation that includes three modules of linear expressions; for example, |x – a| + |x – b| + |x – c| = m.

The first module is equal to x – a for x ³ a and a – x ​​for x b and x

They form four spaces. On each of them, each of the expressions under the modules retains its sign, therefore, the equation as a whole after expanding the modules has the same form on each interval. So, out of 8 theoretically possible options for opening modules, only 4 turned out to be enough for us!

You can also solve any problem with several modules. Namely, the numerical axis is divided into intervals of constant sign of all expressions under the modules, and then on each of them the equation or inequality into which the given problem turns into on this interval is solved. In particular, if all expressions under the modules are rational, then it is enough to mark their roots on the axis, as well as the points where they are not defined, that is, the roots of their denominators. The marked points define the required intervals of constant sign. We act in exactly the same way when solving rational inequalities using the interval method. And the method we described for solving problems with modules has the same name.

Example 1. Solve the equation.

Solution. Let's find the zeros of the function, from where. We solve the problem on each interval:

So, this equation has no solutions.

Example 2. Solve the equation.

Solution. Let's find the zeros of the function. We solve the problem on each interval:

1) (no solutions);

Example 3. Solve the equation.

Solution. Expressions under the absolute value sign vanish at . Accordingly, we need to consider three cases:

2) - root of the equation;

3) is the root of this equation.

Chapter 2. Equations and inequalities containing modules.

2.1 Solving equations with several modules using the interval method.

Example 1.

Solve the equation:

|x+2| = |x-1|+x-3

-(x+2) = -(x-1) + x-3

X-2=-x+1+x-3

x=2 – does not satisfy

condition x

no solutions

2. If -2≤х

x+2 = -(x-1)+x-3

satisfies

condition -2

3. If x≥1, then

Answer: x=6

Example 2.

Solve the equation:

1) Find zeros of submodular expressions

The zeros of submodular expressions split the number line into several intervals. We arrange the signs of submodular expressions on these intervals.

At each interval we open the modules and solve the resulting equation. After finding the root, we check that it belongs to the interval on which we are this moment we are working.

1. :

- fits.

2. :

– doesn’t fit.

3. :

– fits.

4. :

– doesn’t fit. Answer:

2.2 Solving inequalities with several modules using the interval method.

Example 1.

Solve the inequality:

|x-1| + |x-3| 4


-(x-1) - (x-3) 4

2. If 1≤x

x-1– (x-3) 4

24 is not correct

no solutions

3. If x≥3, then

Answer: xЄ (-∞;0) U (4;+∞)

Example 2.

Let's solve the inequality

Solution. The points and (the roots of the expressions under the module) divide the entire numerical axis into three intervals, at each of which the modules should be expanded.

1) When , and the inequality has the form , that is . In this case the answer is .

2) When , the inequality has the form , that is . This inequality is true for any values ​​of the variable, and, taking into account the fact that we solve it on the set, we get the answer in the second case.

3) When , the inequality is transformed to , and the solution in this case is . Common decision inequalities --- Union three responses received.

Thus, to solve equations and inequalities containing several modules, it is convenient to use the interval method. To do this, you need to find the zeros of all submodular functions and designate them on the ODZ of equations and inequalities.

Conclusion

IN Lately In mathematics, methods are widely used to simplify the solution of problems, in particular the interval method, which can significantly speed up calculations. Therefore, the study of the interval method for solving equations and inequalities with several modules is relevant.

In the process of working on the topic “Solving equations and inequalities containing an unknown under the modulus sign using the interval method,” I: studied the literature on this issue, got acquainted with the algebraic and graphical approach to solving equations and inequalities containing an unknown under the modulus sign, and came to the conclusion:

    In some cases, when solving equations with a modulus, it is possible to solve the equations according to the rules, and sometimes it is more convenient to use the interval method.

    When solving equations and inequalities containing a modulus, the interval method is more visual and comparatively simpler.

During writing research work I have uncovered many problems that can be solved using the interval method. Most important task is the solution of equations and inequalities with several modules.

In the course of my work on solving inequalities and equations with several modules using the interval method, I found that the speed of solving problems doubled. This allows you to significantly speed up the work process and reduce time costs. Thus, my hypothesis “if you use the interval method to solve inequalities and equations with several modules, you can significantly simplify your work” was confirmed. While working on the research, I gained experience in solving equations and inequalities with multiple modules. I think that the knowledge I have acquired will allow me to avoid mistakes when making decisions.

Literature

    http://padabum.com

  1. http://yukhym.com

    http://www.tutoronline.ru

    http://fizmat.by

    http://diffur.kemsu.ru

    http://solverbook.com

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    Sevryukov P.F., Smolyakov A.N. Equations and inequalities with moduli and methods for solving them. M.: Publishing house Enlightenment 2005. - 112 p.

    Sadovnichy Yu.V. Unified State Exam. Workshop on mathematics. Solving equations and inequalities. Conversion algebraic expressions. M.: Legion Publishing House 2015 - 128 p.

    Shevkin A.V. Quadratic inequalities. Interval method. M.: LLC " Russian word– educational book”, 2003. – 32 p.