Modulus 4 x plus 5 equals 1. How to solve equations with modulus. Collection and use of personal information

Number modulus a is the distance from the origin to the point A(a) .

To understand this definition, let’s substitute for the variable a any number, for example 3, and read it again:

The modulus of the number 3 is the distance from the origin to the point A(3 ).

That is, the module is nothing more than an ordinary distance. Let's try to see the distance from the origin to the point A(3)

Distance from origin to point A(3) is 3 (three units or three steps).

The module of a number is indicated by two vertical lines, for example:

The modulus of the number 3 is denoted as follows: |3|

The modulus of the number 4 is denoted as follows: |4|

The modulus of the number 5 is denoted as follows: |5|

We looked for the modulus of the number 3 and found out that it is equal to 3. So we write it down:

|3| = 3

Reads like "The modulus of the number three is three"

Now let's try to find the modulus of the number −3. Again, we return to the definition and substitute the number −3 into it. Only instead of a dot A use a new point B. Full stop A we already used in the first example.

The modulus of the number −3 is the distance from the origin to the point B(−3 ).

The distance from one point to another cannot be negative. The modulus is also a distance, so it also cannot be negative.

The modulus of the number −3 is 3. Distance from origin to point B(−3) is equal to three units:

|−3| = 3

Reads like “The modulus of minus three is three.”

The modulus of the number 0 is equal to 0, since the point with coordinate 0 coincides with the origin. That is, the distance from the origin to the point O(0) equals zero:

|0| = 0

"The modulus of zero is zero"

Let's draw conclusions:

• The modulus of a number cannot be negative;
• For a positive number and zero, the modulus is equal to the number itself, and for a negative number – the opposite number;
• Opposite numbers have equal modules.

Opposite numbers

Numbers that differ only in signs are called opposite.

For example, the numbers −2 and 2 are opposites. They differ only in signs. The number −2 has a minus sign, and the number 2 has a plus sign, but we don’t see it, since plus, as mentioned earlier, is not written down.

More examples of opposite numbers:

−1 and 1

−3 and 3

−5 and 5

−9 and 9

Opposite numbers have equal modules. For example, let’s find the moduli of the numbers −3 and 3

|−3| and |3|

3 = 3

The figure shows that the distance from the origin to the points A(−3) and B(3) is equally equal to two steps.

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One of the most difficult topics for students is solving equations containing a variable under the modulus sign. Let's first figure out what this is connected with? Why, for example, do most children crack quadratic equations like nuts, but have so many problems with such a far from complex concept as a module?

In my opinion, all these difficulties are associated with the lack of clearly formulated rules for solving equations with a modulus. So, when solving a quadratic equation, the student knows for sure that he needs to first apply the discriminant formula, and then the formulas for the roots of the quadratic equation. What to do if a modulus is found in the equation? We will try to clearly describe the necessary action plan for the case when the equation contains an unknown under the modulus sign. We will give several examples for each case.

But first, let's remember module definition. So, modulo the number a this number itself is called if a non-negative and -a, if number a less than zero. You can write it like this:

|a| = a if a ≥ 0 and |a| = -a if a< 0

Speaking about the geometric meaning of the module, it should be remembered that each real number corresponds to a certain point on the number axis - its coordinate. So, the module or absolute value of a number is the distance from this point to the origin of the numerical axis. The distance is always specified as a positive number. Thus, the modulus of any negative number is a positive number. By the way, even at this stage, many students begin to get confused. The module can contain any number, but the result of using the module is always a positive number.

Now let's move directly to solving the equations.

1. Consider an equation of the form |x| = c, where c is a real number. This equation can be solved using the modulus definition.

We divide all real numbers into three groups: those that are greater than zero, those that are less than zero, and the third group is the number 0. We write the solution in the form of a diagram:

(±c, if c > 0

If |x| = c, then x = (0, if c = 0

(no roots if with< 0

1) |x| = 5, because 5 > 0, then x = ±5;

2) |x| = -5, because -5< 0, то уравнение не имеет корней;

3) |x| = 0, then x = 0.

2. Equation of the form |f(x)| = b, where b > 0. To solve this equation it is necessary to get rid of the module. We do it this way: f(x) = b or f(x) = -b. Now you need to solve each of the resulting equations separately. If in the original equation b< 0, решений не будет.

1) |x + 2| = 4, because 4 > 0, then

x + 2 = 4 or x + 2 = -4

2) |x 2 – 5| = 11, because 11 > 0, then

x 2 – 5 = 11 or x 2 – 5 = -11

x 2 = 16 x 2 = -6

x = ± 4 no roots

3) |x 2 – 5x| = -8, because -8< 0, то уравнение не имеет корней.

3. An equation of the form |f(x)| = g(x). According to the meaning of the module, such an equation will have solutions if its right-hand side is greater than or equal to zero, i.e. g(x) ≥ 0. Then we will have:

f(x) = g(x) or f(x) = -g(x).

1) |2x – 1| = 5x – 10. This equation will have roots if 5x – 10 ≥ 0. This is where the solution of such equations begins.

1. O.D.Z. 5x – 10 ≥ 0

2. Solution:

2x – 1 = 5x – 10 or 2x – 1 = -(5x – 10)

3. We combine O.D.Z. and the solution, we get:

The root x = 11/7 does not fit the O.D.Z., it is less than 2, but x = 3 satisfies this condition.

Answer: x = 3

2) |x – 1| = 1 – x 2 .

1. O.D.Z. 1 – x 2 ≥ 0. Let’s solve this inequality using the interval method:

(1 – x)(1 + x) ≥ 0

2. Solution:

x – 1 = 1 – x 2 or x – 1 = -(1 – x 2)

x 2 + x – 2 = 0 x 2 – x = 0

x = -2 or x = 1 x = 0 or x = 1

3. We combine the solution and O.D.Z.:

Only roots x = 1 and x = 0 are suitable.

Answer: x = 0, x = 1.

4. Equation of the form |f(x)| = |g(x)|. Such an equation is equivalent to the following two equations f(x) = g(x) or f(x) = -g(x).

1) |x 2 – 5x + 7| = |2x – 5|. This equation is equivalent to the following two:

x 2 – 5x + 7 = 2x – 5 or x 2 – 5x +7 = -2x + 5

x 2 – 7x + 12 = 0 x 2 – 3x + 2 = 0

x = 3 or x = 4 x = 2 or x = 1

Answer: x = 1, x = 2, x = 3, x = 4.

5. Equations solved by the substitution method (variable replacement). This solution method is easiest to explain with a specific example. So, let us be given a quadratic equation with modulus:

x 2 – 6|x| + 5 = 0. By the modulus property x 2 = |x| 2, so the equation can be rewritten as follows:

|x| 2 – 6|x| + 5 = 0. Let's make the replacement |x| = t ≥ 0, then we will have:

t 2 – 6t + 5 = 0. Solving this equation, we find that t = 1 or t = 5. Let’s return to the replacement:

|x| = 1 or |x| = 5

x = ±1 x = ±5

Answer: x = -5, x = -1, x = 1, x = 5.

Let's look at another example:

x 2 + |x| – 2 = 0. By the modulus property x 2 = |x| 2, therefore

|x| 2 + |x| – 2 = 0. Let’s make the replacement |x| = t ≥ 0, then:

t 2 + t – 2 = 0. Solving this equation, we get t = -2 or t = 1. Let’s return to the replacement:

|x| = -2 or |x| = 1

No roots x = ± 1

Answer: x = -1, x = 1.

6. Another type of equations is equations with a “complex” modulus. Such equations include equations that have “modules within a module.” Equations of this type can be solved using the properties of the module.

1) |3 – |x|| = 4. We will act in the same way as in equations of the second type. Because 4 > 0, then we get two equations:

3 – |x| = 4 or 3 – |x| = -4.

Now let us express the modulus x in each equation, then |x| = -1 or |x| = 7.

We solve each of the resulting equations. There are no roots in the first equation, because -1< 0, а во втором x = ±7.

Answer x = -7, x = 7.

2) |3 + |x + 1|| = 5. We solve this equation in a similar way:

3 + |x + 1| = 5 or 3 + |x + 1| = -5

|x + 1| = 2 |x + 1| = -8

x + 1 = 2 or x + 1 = -2. No roots.

Answer: x = -3, x = 1.

There is also a universal method for solving equations with a modulus. This is the interval method. But we will look at it later.

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Solving equations and inequalities with modulus often causes difficulties. However, if you understand well what it is the absolute value of a number, And how to correctly expand expressions containing a modulus sign, then the presence in the equation expression under the modulus sign, ceases to be an obstacle to its solution.

A little theory. Each number has two characteristics: the absolute value of the number and its sign.

For example, the number +5, or simply 5, has a “+” sign and an absolute value of 5.

The number -5 has a "-" sign and an absolute value of 5.

The absolute values ​​of the numbers 5 and -5 are 5.

The absolute value of a number x is called the modulus of the number and is denoted by |x|.

As we see, the modulus of a number is equal to the number itself if this number is greater than or equal to zero, and to this number with the opposite sign if this number is negative.

The same applies to any expressions that appear under the modulus sign.

The module expansion rule looks like this:

|f(x)|= f(x) if f(x) ≥ 0, and

|f(x)|= - f(x), if f(x)< 0

For example |x-3|=x-3, if x-3≥0 and |x-3|=-(x-3)=3-x, if x-3<0.

To solve an equation containing an expression under the modulus sign, you must first expand a module according to the module expansion rule.

Then our equation or inequality becomes into two different equations existing on two different numerical intervals.

One equation exists on a numerical interval on which the expression under the modulus sign is non-negative.

And the second equation exists on the interval on which the expression under the modulus sign is negative.

Let's look at a simple example.

Let's solve the equation:

|x-3|=-x 2 +4x-3

1. Let's open the module.

|x-3|=x-3, if x-3≥0, i.e. if x≥3

|x-3|=-(x-3)=3-x if x-3<0, т.е. если х<3

2. We received two numerical intervals: x≥3 and x<3.

Let us consider into which equations the original equation is transformed on each interval:

A) For x≥3 |x-3|=x-3, and our wounding has the form:

Attention! This equation exists only on the interval x≥3!

Let's open the brackets and present similar terms:

and solve this equation.

This equation has roots:

x 1 =0, x 2 =3

Attention! since the equation x-3=-x 2 +4x-3 exists only on the interval x≥3, we are only interested in those roots that belong to this interval. This condition is satisfied only by x 2 =3.

B) At x<0 |x-3|=-(x-3) = 3-x, и наше уравнение приобретает вид:

Attention! This equation exists only on the interval x<3!

Let's open the brackets and present similar terms. We get the equation:

x 1 =2, x 2 =3

Attention! since the equation 3-x=-x 2 +4x-3 exists only on the interval x<3, нас интересуют только те корни, которые принадлежат этому промежутку. Этому условию удовлетворяет только х 1 =2.

So: from the first interval we take only the root x=3, from the second - the root x=2.

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 precisely 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.

There is also a 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 narrative 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 reduce 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 all 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 even nothing to transform. We work according to the algorithm:

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

Do not rush to open the parentheses preceded by a “minus”: 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 parentheses, 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 by 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 is greater than the function. 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. 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 simply appeared as a counterpoint to “∪”.

To make it even easier to remember, just 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, you need to mark the points in the correct order: the larger the number, the further the point moves 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 set, 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 and very tough problems. The only “weak point” 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 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 simultaneously be 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.

Among examples per module Often there are equations where you need to find module roots in a module, that is, an equation of the form
||a*x-b|-c|=k*x+m .
If k=0, that is, the right side is equal to a constant (m), then it’s easier to look for a solution equations with modules graphically. Below is the method opening of double modules using examples common in practice. Understand the algorithm for calculating equations with modules well, so that you don’t have problems on quizzes, tests, and just to know.

Example 1. Solve the equation modulo |3|x|-5|=-2x-2.
Solution: Always start opening equations from the internal module
|x|=0 <->x=0.
At the point x=0, the equation with modulus is divided by 2.
At x< 0 подмодульная функция отрицательная, поэтому при раскрытии знак меняем на противоположный
|-3x-5|=-2x-2.
For x>0 or equal, expanding the module we get
|3x-5|=-2x-2 .
Let's solve the equation for negative variables (x< 0) . Оно разлагается на две системы уравнений. Первое уравнение получаем из условия, что функция после знака равенства неотрицательна. Второе - раскрывая модуль в одной системе принимаем, что подмодульная функция положительная, в иной отрицательная - меняем знак правой или левой части (зависит от методики преподавания).

From the first equation we get that the solution should not exceed (-1), i.e.

This limitation entirely belongs to the area in which we are solving. Let's move variables and constants to opposite sides of equality in the first and second systems

and find a solution


Both values ​​belong to the interval that is being considered, that is, they are roots.
Consider an equation with moduli for positive variables
|3x-5|=-2x-2.
Expanding the module we get two systems of equations

From the first equation, which is common to the two systems, we obtain the familiar condition

which, in intersection with the set on which we are looking for a solution, gives an empty set (there are no points of intersection). So the only roots of a module with a module are the values
x=-3; x=-1.4.

Example 2. Solve the equation with modulus ||x-1|-2|=3x-4.
Solution: Let's start by opening the internal module
|x-1|=0 <=>x=1.
A submodular function changes sign at one. For smaller values ​​it is negative, for larger values ​​it is positive. In accordance with this, when expanding the internal module, we obtain two equations with the module
x |-(x-1)-2|=3x-4;
x>=1 -> |x-1-2|=3x-4.
Be sure to check the right side of the modulus equation; it must be greater than zero.
3x-4>=0 -> x>=4/3.
This means that there is no need to solve the first equation, since it was written for x< 1, что не соответствует найденному условию. Раскроем модуль во втором уравнении
|x-3|=3x-4 ->
x-3=3x-4 or x-3=4-3x;
4-3=3x-x or x+3x=4+3;
2x=1 or 4x=7;
x=1/2 or x=7/4.
We received two values, the first of which is rejected because it does not belong to the required interval. Finally, the equation has one solution x=7/4.

Example 3. Solve the equation with modulus ||2x-5|-1|=x+3.
Solution: Let's open the internal module
|2x-5|=0 <=>x=5/2=2.5.
The point x=2.5 splits the number line into two intervals. Respectively, submodular function changes sign when passing through 2.5. Let us write down the condition for the solution on the right side of the equation with modulus.
x+3>=0 -> x>=-3.
So the solution can be values ​​no less than (-3) . Let's expand the module for the negative value of the internal module
|-(2x-5)-1|=x+3;
|-2x+4|=x+3.
This module will also give 2 equations when expanded
-2x+4=x+3 or 2x-4=x+3;
2x+x=4-3 or 2x-x=3+4;
3x=1; x=1/3 or x=7 .
We reject the value x=7, since we were looking for a solution in the interval [-3;2.5]. Now we open the internal module for x>2.5. We get an equation with one module
|2x-5-1|=x+3;
|2x-6|=x+3.
When expanding the module, we obtain the following linear equations
-2x+6=x+3 or 2x-6=x+3;
2x+x=6-3 or 2x-x=3+6;
3x=3; x=1 or x=9 .
The first value x=1 does not satisfy the condition x>2.5. So on this interval we have one root of the equation with modulus x=9, and there are two in total (x=1/3). By substitution you can check the correctness of the calculations performed
Answer: x=1/3; x=9.

Example 4. Find solutions to the double module ||3x-1|-5|=2x-3.
Solution: Let's expand the internal module of the equation
|3x-1|=0 <=>x=1/3.
The point x=2.5 divides the number line into two intervals and the given equation into two cases. We write down the condition for the solution based on the form of the equation on the right side
2x-3>=0 -> x>=3/2=1.5.
It follows that we are interested in values ​​>=1.5. Thus modular equation consider on two intervals
,
|-(3x-1)-5|=2x-3;
|-3x-4|=2x-3.
The resulting module, when expanded, is divided into 2 equations
-3x-4=2x-3 or 3x+4=2x-3;
2x+3x=-4+3 or 3x-2x=-3-4;
5x=-1; x=-1/5 or x=-7 .
Both values ​​do not fall into the interval, that is, they are not solutions to the equation with moduli. Next, we will expand the module for x>2.5. We get the following equation
|3x-1-5|=2x-3;
|3x-6|=2x-3.
Expanding the module, we get 2 linear equations
3x-6=2x-3 or –(3x-6)=2x-3;
3x-2x=-3+6 or 2x+3x=6+3;
x=3 or 5x=9; x=9/5=1.8.
The second value found does not correspond to the condition x>2.5, we reject it.
Finally we have one root of the equation with moduli x=3.
Performing a check
||3*3-1|-5|=2*3-3 3=3 .
The root of the equation with the modulus was calculated correctly.
Answer: x=1/3; x=9.