One of the most difficult topics for students is solving equations containing a variable under the modulus sign. Let's see for a start what is it connected with? Why, for example, quadratic equations most children click like nuts, but with such a far from the most complex concept as a module has so many problems?

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. But what if a module is encountered in the equation? We will try to clearly describe the necessary plan of action in the case when the equation contains an unknown under the modulus sign. We give several examples for each case.

But first, let's remember module definition. So, the modulus of the number a the number itself is called if a non-negative and -a if the 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 to coordinate. So, the module or the absolute value of a number is the distance from this point to the origin of the numerical axis. The distance is always given 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. Any number can be in the module, but the result of applying the module is always a positive number.

Now let's move on 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 definition of the modulus.

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. An equation of the form |f(x)| = b, where b > 0. To solve this equation, it is necessary to get rid of the modulus. We do it like this: f(x) = b or f(x) = -b. Now it is necessary to solve separately each of the obtained equations. 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 -eight< 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 side is greater than or equal to zero, i.e. g(x) ≥ 0. Then we 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. Combine O.D.Z. and the solution, we get:

The root x \u003d 11/7 does not fit according to O.D.Z., it is less than 2, and x \u003d 3 satisfies this condition.

Answer: x = 3

2) |x – 1| \u003d 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 \u003d 1 - x 2 or x - 1 \u003d - (1 - x 2)

x 2 + x - 2 = 0 x 2 - x = 0

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

3. Combine solution and O.D.Z.:

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

Answer: x = 0, x = 1.

4. An 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 (change of variable). This solution method is easiest to explain with a specific example. So, let a quadratic equation with a modulus be given:

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

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

t 2 - 6t + 5 \u003d 0. Solving this equation, we get that t \u003d 1 or t \u003d 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 property of the module x 2 = |x| 2 , so

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

t 2 + t - 2 \u003d 0. Solving this equation, we get, t \u003d -2 or t \u003d 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's express the module 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 -one< 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. There are no roots.

Answer: x = -3, x = 1.

There is also a universal method for solving equations with a modulus. This is the spacing method. But we will consider it further.

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A module is one of those things that everyone seems to have heard about, but in reality no one really understands. Therefore, today there will be a big lesson devoted to solving equations with modules.

I'll tell you right away: the lesson will be simple. In general, modules are generally a relatively simple topic. “Yes, of course, it’s easy! It makes my brain explode!" - many students will say, but all these brain breaks are due to the fact that most people have not knowledge in their heads, but some kind of crap. And the purpose of this lesson is to turn crap into knowledge. :)

A bit of theory

So let's go. Let's start with the most important: what is a module? Let me remind you that the modulus of a number is simply the same number, but taken without the minus sign. That is, for example, $\left| -5 \right|=5$. Or $\left| -129.5\right|=129.5$.

Is it that simple? Yes, simple. What then is the modulus of a positive number? Here it is even simpler: the modulus of a positive number is equal to this number itself: $\left| 5\right|=5$; $\left| 129.5 \right|=129.5$ etc.

It turns out a curious thing: different numbers can have the same module. For example: $\left| -5 \right|=\left| 5\right|=5$; $\left| -129.5 \right|=\left| 129.5 \right|=129.5$. It is easy to see what kind of numbers are these, in which the modules are the same: these numbers are opposite. Thus, we note for ourselves that the modules of opposite numbers are equal:

\[\left| -a \right|=\left| a\right|\]

Another important fact: modulus is never negative. Whatever number we take - even positive, even negative - its modulus always turns out to be positive (or in extreme cases, zero). That is why the modulus is often called the absolute value of a number.

In addition, if we combine the definition of the modulus for a positive and negative number, then we get a global definition of the modulus for all numbers. Namely: the modulus of a number is equal to this number itself, if the number is positive (or zero), or equal to the opposite number, if the number is negative. You can write this as a formula:

There is also a module of zero, but it is always equal to zero. Also, zero is the only number that doesn't have an opposite.

Thus, if we consider the function $y=\left| x \right|$ and try to draw its graph, you will get such a “daw”:

Modulus graph and equation solution example

From this picture you can immediately see that $\left| -m \right|=\left| m \right|$, and the module plot never falls below the x-axis. But that's not all: the red line marks the straight line $y=a$, which, with positive $a$, gives us two roots at once: $((x)_(1))$ and $((x)_(2)) $, but we'll talk about that later. :)

In addition to a purely algebraic definition, there is a geometric one. Let's say there are two points on the number line: $((x)_(1))$ and $((x)_(2))$. In this case, the expression $\left| ((x)_(1))-((x)_(2)) \right|$ is just the distance between the specified points. Or, if you like, the length of the segment connecting these points:

Modulus is the distance between points on the number line

It also follows from this definition that the modulus is always non-negative. But enough definitions and theory - let's move on to real equations. :)

Basic Formula

Okay, we've figured out the definition. But it didn't get any easier. How to solve equations containing this very module?

Calm, just calm. Let's start with the simplest things. Consider something like this:

\[\left| x\right|=3\]

So the modulo$x$ is 3. What can $x$ be equal to? Well, judging by the definition, $x=3$ will suit us just fine. Really:

\[\left| 3\right|=3\]

Are there other numbers? Cap seems to hint that there is. For example, $x=-3$ — $\left| -3 \right|=3$, i.e. the required equality is satisfied.

So maybe if we search, think, we will find more numbers? But break off: there are no more numbers. Equation $\left| x \right|=3$ has only two roots: $x=3$ and $x=-3$.

Now let's complicate the task a little. Let, instead of the variable $x$, the function $f\left(x \right)$ hang under the modulus sign, and on the right, instead of the triple, we put an arbitrary number $a$. We get the equation:

\[\left| f\left(x \right) \right|=a\]

Well, how do you decide? Let me remind you: $f\left(x \right)$ is an arbitrary function, $a$ is any number. Those. any at all! For example:

\[\left| 2x+1 \right|=5\]

\[\left| 10x-5 \right|=-65\]

Let's look at the second equation. You can immediately say about him: he has no roots. Why? That's right: because it requires the modulus to be equal to a negative number, which never happens, since we already know that the modulus is always a positive number or, in extreme cases, zero.

But with the first equation, everything is more fun. There are two options: either there is a positive expression under the module sign, and then $\left| 2x+1 \right|=2x+1$, or this expression is still negative, in which case $\left| 2x+1 \right|=-\left(2x+1 \right)=-2x-1$. In the first case, our equation will be rewritten as:

\[\left| 2x+1 \right|=5\Rightarrow 2x+1=5\]

And suddenly it turns out that the submodule expression $2x+1$ is indeed positive - it is equal to the number 5. That is, we can safely solve this equation - the resulting root will be a piece of the answer:

Those who are especially incredulous can try to substitute the found root into the original equation and make sure that there really will be a positive number under the modulus.

Now let's look at the case of a negative submodule expression:

\[\left\( \begin(align)& \left| 2x+1 \right|=5 \\& 2x+1 \lt 0 \\\end(align) \right.\Rightarrow -2x-1=5 \Rightarrow 2x+1=-5\]

Oops! Again, everything is clear: we assumed that $2x+1 \lt 0$, and as a result we got that $2x+1=-5$ - indeed, this expression is less than zero. We solve the resulting equation, while already knowing for sure that the found root will suit us:

In total, we again received two answers: $x=2$ and $x=3$. Yes, the amount of calculations turned out to be a little more than in the very simple equation $\left| x \right|=3$, but fundamentally nothing has changed. So maybe there is some kind of universal algorithm?

Yes, such an algorithm exists. And now we will analyze it.

Getting rid of the module sign

Let us be given the equation $\left| f\left(x \right) \right|=a$, and $a\ge 0$ (otherwise, as we already know, there are no roots). Then you can get rid of the modulo sign according to the following rule:

\[\left| f\left(x \right) \right|=a\Rightarrow f\left(x \right)=\pm a\]

Thus, our equation with the modulus splits into two, but without the modulus. That's the whole technology! Let's try to solve a couple of equations. Let's start with this

\[\left| 5x+4 \right|=10\Rightarrow 5x+4=\pm 10\]

We will separately consider when there is a ten with a plus on the right, and separately when it is with a minus. We have:

\[\begin(align)& 5x+4=10\Rightarrow 5x=6\Rightarrow x=\frac(6)(5)=1,2; \\& 5x+4=-10\Rightarrow 5x=-14\Rightarrow x=-\frac(14)(5)=-2.8. \\\end(align)\]

That's all! We got two roots: $x=1.2$ and $x=-2.8$. The whole solution took literally two lines.

Ok, no question, let's look at something a little more serious:

\[\left| 7-5x \right|=13\]

Again, open the module with a plus and a minus:

\[\begin(align)& 7-5x=13\Rightarrow -5x=6\Rightarrow x=-\frac(6)(5)=-1,2; \\& 7-5x=-13\Rightarrow -5x=-20\Rightarrow x=4. \\\end(align)\]

Again a couple of lines - and the answer is ready! As I said, there is nothing complicated in modules. You just need to remember a few rules. Therefore, we go further and proceed with really more difficult tasks.

Variable right side case

Now consider this equation:

\[\left| 3x-2 \right|=2x\]

This equation is fundamentally different from all the previous ones. How? And the fact that the expression $2x$ is to the right of the equal sign - and we cannot know in advance whether it is positive or negative.

How to be in that case? First, we must understand once and for all that if the right side of the equation is negative, then the equation will have no roots- we already know that the modulus cannot be equal to a negative number.

And secondly, if the right part is still positive (or equal to zero), then you can proceed in exactly the same way as before: just open the module separately with the plus sign and separately with the minus sign.

Thus, we formulate a rule for arbitrary functions $f\left(x \right)$ and $g\left(x \right)$ :

\[\left| f\left(x \right) \right|=g\left(x \right)\Rightarrow \left\( \begin(align)& f\left(x \right)=\pm g\left(x \right ), \\& g\left(x \right)\ge 0. \\\end(align) \right.\]

With regard to our equation, we get:

\[\left| 3x-2 \right|=2x\Rightarrow \left\( \begin(align)& 3x-2=\pm 2x, \\& 2x\ge 0. \\\end(align) \right.\]

Well, we can handle the $2x\ge 0$ requirement somehow. In the end, we can stupidly substitute the roots that we get from the first equation and check whether the inequality holds or not.

So let's solve the equation itself:

\[\begin(align)& 3x-2=2\Rightarrow 3x=4\Rightarrow x=\frac(4)(3); \\& 3x-2=-2\Rightarrow 3x=0\Rightarrow x=0. \\\end(align)\]

Well, which of these two roots satisfies the requirement $2x\ge 0$? Yes, both! Therefore, the answer will be two numbers: $x=(4)/(3)\;$ and $x=0$. That's the solution. :)

I suspect that one of the students has already begun to get bored? Well, consider an even more complex equation:

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

Although it looks evil, in fact it is all the same equation of the form "modulus equals function":

\[\left| f\left(x \right) \right|=g\left(x \right)\]

And it is solved in the same way:

\[\left| ((x)^(3))-3((x)^(2))+x \right|=x-((x)^(3))\Rightarrow \left\( \begin(align)& ( (x)^(3))-3((x)^(2))+x=\pm \left(x-((x)^(3)) \right), \\& x-((x )^(3))\ge 0. \\\end(align) \right.\]

We will deal with inequality later - it is somehow too vicious (actually simple, but we will not solve it). For now, let's take a look at the resulting equations. Consider the first case - this is when the module is expanded with a plus sign:

\[((x)^(3))-3((x)^(2))+x=x-((x)^(3))\]

Well, here it’s a no brainer that you need to collect everything on the left, bring similar ones and see what happens. And this is what happens:

\[\begin(align)& ((x)^(3))-3((x)^(2))+x=x-((x)^(3)); \\& 2((x)^(3))-3((x)^(2))=0; \\\end(align)\]

Putting the common factor $((x)^(2))$ out of the bracket, we get a very simple equation:

\[((x)^(2))\left(2x-3 \right)=0\Rightarrow \left[ \begin(align)& ((x)^(2))=0 \\& 2x-3 =0 \\\end(align) \right.\]

\[((x)_(1))=0;\quad ((x)_(2))=\frac(3)(2)=1.5.\]

Here we used an important property of the product, for the sake of which we factored the original polynomial: the product is equal to zero when at least one of the factors is equal to zero.

Now, in the same way, we will deal with the second equation, which is obtained by expanding the module with a minus sign:

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

Again, the same thing: the product is zero when at least one of the factors is zero. We have:

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

Well, we got three roots: $x=0$, $x=1.5$ and $x=(2)/(3)\;$. Well, what will go into the final answer from this set? To do this, remember that we have an additional inequality constraint:

How to take into account this requirement? Let's just substitute the found roots and check whether the inequality holds for these $x$ or not. We have:

\[\begin(align)& x=0\Rightarrow x-((x)^(3))=0-0=0\ge 0; \\& x=1,5\Rightarrow x-((x)^(3))=1,5-((1,5)^(3)) \lt 0; \\& x=\frac(2)(3)\Rightarrow x-((x)^(3))=\frac(2)(3)-\frac(8)(27)=\frac(10) (27)\ge 0; \\\end(align)\]

Thus, the root $x=1.5$ does not suit us. And only two roots will go in response:

\[((x)_(1))=0;\quad ((x)_(2))=\frac(2)(3).\]

As you can see, even in this case there was nothing difficult - equations with modules are always solved according to the algorithm. You just need to have a good understanding of polynomials and inequalities. Therefore, we move on to more complex tasks - there will already be not one, but two modules.

Equations with two modules

So far, we have studied only the simplest equations - there was one module and something else. We sent this “something else” to another part of the inequality, away from the module, so that in the end everything would be reduced to an equation like $\left| f\left(x \right) \right|=g\left(x \right)$ or even simpler $\left| f\left(x \right) \right|=a$.

But kindergarten is over - it's time to consider something more serious. Let's start with equations like this:

\[\left| f\left(x \right) \right|=\left| g\left(x \right) \right|\]

This is an equation of the form "the modulus is equal to the modulus". A fundamentally important point is the absence of other terms and factors: only one module on the left, one more module on the right - and nothing more.

One would now think that such equations are more difficult to solve than what we have studied so far. But no: these equations are solved even easier. Here is the formula:

\[\left| f\left(x \right) \right|=\left| g\left(x \right) \right|\Rightarrow f\left(x \right)=\pm g\left(x \right)\]

Everything! We simply equate submodule expressions by prefixing one of them with a plus or minus sign. And then we solve the resulting two equations - and the roots are ready! No additional restrictions, no inequalities, etc. Everything is very simple.

Let's try to solve this problem:

\[\left| 2x+3 \right|=\left| 2x-7 \right|\]

Elementary Watson! Opening the modules:

\[\left| 2x+3 \right|=\left| 2x-7 \right|\Rightarrow 2x+3=\pm \left(2x-7 \right)\]

Let's consider each case separately:

\[\begin(align)& 2x+3=2x-7\Rightarrow 3=-7\Rightarrow \emptyset ; \\& 2x+3=-\left(2x-7 \right)\Rightarrow 2x+3=-2x+7. \\\end(align)\]

The first equation has no roots. Because when is $3=-7$? For what values ​​of $x$? “What the fuck is $x$? Are you stoned? There is no $x$ at all,” you say. And you will be right. We have obtained an equality that does not depend on the variable $x$, and at the same time the equality itself is incorrect. That's why there are no roots.

With the second equation, everything is a little more interesting, but also very, very simple:

As you can see, everything was decided literally in a couple of lines - we didn’t expect anything else from a linear equation. :)

As a result, the final answer is: $x=1$.

Well, how? Complicated? Of course not. Let's try something else:

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

Again we have an equation like $\left| f\left(x \right) \right|=\left| g\left(x \right) \right|$. Therefore, we immediately rewrite it, revealing the module sign:

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

Perhaps someone will now ask: “Hey, what kind of nonsense? Why is plus-minus on the right side and not on the left side? Calm down, I'll explain everything. Indeed, in a good way, we should have rewritten our equation as follows:

Then you need to open the brackets, move all the terms in one direction from the equal sign (since the equation, obviously, will be square in both cases), and then find the roots. But you must admit: when “plus-minus” is in front of three terms (especially when one of these terms is a square expression), it somehow looks more complicated than the situation when “plus-minus” is only in front of two terms.

But nothing prevents us from rewriting the original equation as follows:

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

What happened? Yes, nothing special: just swapped the left and right sides. A trifle, which in the end will simplify our lives a little. :)

In general, we solve this equation, considering options with a plus and a minus:

\[\begin(align)& ((x)^(2))-3x+2=x-1\Rightarrow ((x)^(2))-4x+3=0; \\& ((x)^(2))-3x+2=-\left(x-1 \right)\Rightarrow ((x)^(2))-2x+1=0. \\\end(align)\]

The first equation has roots $x=3$ and $x=1$. The second is generally an exact square:

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

Therefore, it has a single root: $x=1$. But we have already received this root earlier. Thus, only two numbers will go into the final answer:

\[((x)_(1))=3;\quad ((x)_(2))=1.\]

Mission Complete! You can take it from the shelf and eat a pie. There are 2 of them, your average. :)

Important note. The presence of the same roots for different versions of the expansion of the module means that the original polynomials are decomposed into factors, and among these factors there will necessarily be a common one. Really:

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

One of the module properties: $\left| a\cdot b \right|=\left| a \right|\cdot \left| b \right|$ (that is, the modulus of the product is equal to the product of the moduli), so the original equation can be rewritten as

\[\left| x-1 \right|=\left| x-1 \right|\cdot \left| x-2 \right|\]

As you can see, we really have a common factor. Now, if you collect all the modules on one side, then you can take this multiplier out of the bracket:

\[\begin(align)& \left| x-1 \right|=\left| x-1 \right|\cdot \left| x-2 \right|; \\&\left| x-1 \right|-\left| x-1 \right|\cdot \left| x-2 \right|=0; \\&\left| x-1 \right|\cdot \left(1-\left| x-2 \right| \right)=0. \\\end(align)\]

Well, now we recall that the product is equal to zero when at least one of the factors is equal to zero:

\[\left[ \begin(align)& \left| x-1 \right|=0, \\& \left| x-2 \right|=1. \\\end(align) \right.\]

Thus, the original equation with two modules has been reduced to the two simplest equations that we talked about at the very beginning of the lesson. Such equations can be solved in just a couple of lines. :)

This remark may seem unnecessarily complicated and inapplicable in practice. However, in reality, you may encounter much more complex tasks than those that we are analyzing today. In them, modules can be combined with polynomials, arithmetic roots, logarithms, etc. And in such situations, the ability to lower the overall degree of the equation by putting something out of the bracket can be very, very handy. :)

Now I would like to analyze another equation, which at first glance may seem crazy. Many students “stick” on it - even those who believe that they have a good understanding of the modules.

However, this equation is even easier to solve than what we considered earlier. And if you understand why, you will get another trick for quickly solving equations with modules.

So the equation is:

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

No, this is not a typo: it is a plus between the modules. And we need to find for which $x$ the sum of two modules is equal to zero. :)

What is the problem? And the problem is that each module is a positive number, or in extreme cases, zero. What happens when you add two positive numbers? Obviously, again a positive number:

\[\begin(align)& 5+7=12 \gt 0; \\& 0.004+0.0001=0.0041 \gt 0; \\& 5+0=5 \gt 0. \\\end(align)\]

The last line may give you an idea: the only case where the sum of the moduli is zero is if each modulus is equal to zero:

\[\left| x-((x)^(3)) \right|+\left| ((x)^(2))+x-2 \right|=0\Rightarrow \left\( \begin(align)& \left| x-((x)^(3)) \right|=0, \\& \left|((x)^(2))+x-2 \right|=0. \\\end(align) \right.\]

When is the modulus equal to zero? Only in one case - when the submodule expression is equal to zero:

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

Thus, we have three points at which the first modulus is set to zero: 0, 1, and −1; as well as two points at which the second module is zeroed: −2 and 1. However, we need both modules to be zeroed at the same time, so among the numbers found, we need to choose those that are included in both sets. Obviously, there is only one such number: $x=1$ - this will be the final answer.

splitting method

Well, we have already covered a bunch of tasks and learned a lot of tricks. Do you think that's it? But no! Now we will consider the final technique - and at the same time the most important. We will talk about splitting equations with a modulus. What will be discussed? Let's go back a little and consider some simple equation. For example, this:

\[\left| 3x-5\right|=5-3x\]

In principle, we already know how to solve such an equation, because it is a standard $\left| f\left(x \right) \right|=g\left(x \right)$. But let's try to look at this equation from a slightly different angle. More precisely, consider the expression under the module sign. Let me remind you that the modulus of any number can be equal to the number itself, or it can be opposite to this number:

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

Actually, this ambiguity is the whole problem: since the number under the modulus changes (it depends on the variable), it is not clear to us whether it is positive or negative.

But what if we initially require that this number be positive? For example, let's demand that $3x-5 \gt 0$ - in this case, we are guaranteed to get a positive number under the modulus sign, and we can completely get rid of this modulus:

Thus, our equation will turn into a linear one, which is easily solved:

True, all these considerations make sense only under the condition $3x-5 \gt 0$ - we ourselves introduced this requirement in order to unambiguously reveal the module. So let's substitute the found $x=\frac(5)(3)$ into this condition and check:

It turns out that for the specified value of $x$, our requirement is not met, because expression turned out to be equal to zero, and we need it to be strictly greater than zero. Sad. :(

But that's okay! After all, there is another option $3x-5 \lt 0$. Moreover: there is also the case $3x-5=0$ - this must also be considered, otherwise the solution will be incomplete. So, consider the $3x-5 \lt 0$ case:

It is obvious that the module will open with a minus sign. But then a strange situation arises: the same expression will stick out both on the left and on the right in the original equation:

I wonder for what such $x$ the expression $5-3x$ will be equal to the expression $5-3x$? From such equations, even the Captain would obviously choke on saliva, but we know that this equation is an identity, i.e. it is true for any value of the variable!

And this means that any $x$ will suit us. However, we have a limitation:

In other words, the answer will not be a single number, but a whole interval:

Finally, there is one more case left to consider: $3x-5=0$. Everything is simple here: there will be zero under the modulus, and the modulus of zero is also equal to zero (this directly follows from the definition):

But then the original equation $\left| 3x-5 \right|=5-3x$ will be rewritten like this:

We have already obtained this root above when we considered the case $3x-5 \gt 0$. Moreover, this root is a solution to the equation $3x-5=0$ - this is the restriction that we ourselves introduced to nullify the modulus. :)

Thus, in addition to the interval, we will also be satisfied with the number lying at the very end of this interval:

Combining Roots in Equations with Modulus

Total final answer: $x\in \left(-\infty ;\frac(5)(3) \right]$. It's not very common to see such crap in the answer to a rather simple (essentially linear) equation with modulus Well, get used to it: the complexity of the module lies in the fact that the answers in such equations can be completely unpredictable.

Much more important is something else: we have just dismantled a universal algorithm for solving an equation with a modulus! And this algorithm consists of the following steps:

1. Equate each modulus in the equation to zero. Let's get some equations;
2. Solve all these equations and mark the roots on the number line. As a result, the straight line will be divided into several intervals, on each of which all modules are uniquely expanded;
3. Solve the original equation for each interval and combine the answers.

That's all! There remains only one question: what to do with the roots themselves, obtained at the 1st step? Let's say we have two roots: $x=1$ and $x=5$. They will break the number line into 3 pieces:

Splitting a number line into intervals using points

So what are the intervals? It is clear that there are three of them:

1. Leftmost: $x \lt 1$ - the unit itself is not included in the interval;
2. Central: $1\le x \lt 5$ - here one is included in the interval, but five is not included;
3. The rightmost one: $x\ge 5$ — the five is included only here!

I think you already understand the pattern. Each interval includes the left end and does not include the right end.

At first glance, such a record may seem uncomfortable, illogical, and generally some kind of crazy. But believe me: after a little practice, you will find that this is the most reliable approach and at the same time does not interfere with unambiguously revealing modules. It is better to use such a scheme than to think every time: give the left / right end to the current interval or “throw” it to the next one.

In this article, we will analyze in detail the absolute value of a number. We will give various definitions of the modulus of a number, introduce notation and give graphic illustrations. In this case, we consider various examples of finding the modulus of a number by definition. After that, we list and justify the main properties of the module. At the end of the article, we will talk about how the modulus of a complex number is determined and found.

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Modulus of number - definition, notation and examples

First we introduce modulus designation. The module of the number a will be written as , that is, to the left and to the right of the number we will put vertical lines that form the sign of the module. Let's give a couple of examples. For example, modulo -7 can be written as ; module 4,125 is written as , and module is written as .

The following definition of the module refers to, and therefore, to, and to integers, and to rational and irrational numbers, as to the constituent parts of the set of real numbers. We will talk about the modulus of a complex number in.

Definition.

Modulus of a is either the number a itself, if a is a positive number, or the number −a, the opposite of the number a, if a is a negative number, or 0, if a=0 .

The voiced definition of the modulus of a number is often written in the following form , this notation means that if a>0 , if a=0 , and if a<0 .

The record can be represented in a more compact form . This notation means that if (a is greater than or equal to 0 ), and if a<0 .

There is also a record . Here, the case when a=0 should be explained separately. In this case, we have , but −0=0 , since zero is considered a number that is opposite to itself.

Let's bring examples of finding the modulus of a number with a given definition. For example, let's find modules of numbers 15 and . Let's start with finding . Since the number 15 is positive, its modulus is, by definition, equal to this number itself, that is, . What is the modulus of a number? Since is a negative number, then its modulus is equal to the number opposite to the number, that is, the number . Thus, .

In conclusion of this paragraph, we give one conclusion, which is very convenient to apply in practice when finding the modulus of a number. From the definition of the modulus of a number it follows that the modulus of a number is equal to the number under the sign of the modulus, regardless of its sign, and from the examples discussed above, this is very clearly visible. The voiced statement explains why the modulus of a number is also called the absolute value of the number. So the modulus of a number and the absolute value of a number are one and the same.

Modulus of a number as a distance

Geometrically, the modulus of a number can be interpreted as distance. Let's bring determination of the modulus of a number in terms of distance.

Definition.

Modulus of a is the distance from the origin on the coordinate line to the point corresponding to the number a.

This definition is consistent with the definition of the modulus of a number given in the first paragraph. Let's explain this point. The distance from the origin to the point corresponding to a positive number is equal to this number. Zero corresponds to the origin, so the distance from the origin to the point with coordinate 0 is zero (no single segment and no segment that makes up any fraction of the unit segment needs to be postponed in order to get from point O to the point with coordinate 0). The distance from the origin to a point with a negative coordinate is equal to the number opposite to the coordinate of the given point, since it is equal to the distance from the origin to the point whose coordinate is the opposite number.

For example, the modulus of the number 9 is 9, since the distance from the origin to the point with coordinate 9 is nine. Let's take another example. The point with coordinate −3.25 is at a distance of 3.25 from point O, so .

The sounded definition of the modulus of a number is a special case of defining the modulus of the difference of two numbers.

Definition.

Difference modulus of two numbers a and b is equal to the distance between the points of the coordinate line with coordinates a and b .

That is, if points on the coordinate line A(a) and B(b) are given, then the distance from point A to point B is equal to the modulus of the difference between the numbers a and b. If we take point O (reference point) as point B, then we will get the definition of the modulus of the number given at the beginning of this paragraph.

Determining the modulus of a number through the arithmetic square root

Sometimes found determination of the modulus through the arithmetic square root.

For example, let's calculate the modules of the numbers −30 and based on this definition. We have . Similarly, we calculate the modulus of two-thirds: .

The definition of the modulus of a number in terms of the arithmetic square root is also consistent with the definition given in the first paragraph of this article. Let's show it. Let a be a positive number, and let −a be negative. Then and , if a=0 , then .

Module properties

The module has a number of characteristic results - module properties. Now we will give the main and most commonly used of them. When substantiating these properties, we will rely on the definition of the modulus of a number in terms of distance.

Let's start with the most obvious module property − modulus of a number cannot be a negative number. In literal form, this property has the form for any number a . This property is very easy to justify: the modulus of a number is the distance, and the distance cannot be expressed as a negative number.

Let's move on to the next property of the module. The modulus of a number is equal to zero if and only if this number is zero. The modulus of zero is zero by definition. Zero corresponds to the origin, no other point on the coordinate line corresponds to zero, since each real number is associated with a single point on the coordinate line. For the same reason, any number other than zero corresponds to a point other than the origin. And the distance from the origin to any point other than the point O is not equal to zero, since the distance between two points is equal to zero if and only if these points coincide. The above reasoning proves that only the modulus of zero is equal to zero.

Move on. Opposite numbers have equal modules, that is, for any number a . Indeed, two points on the coordinate line, whose coordinates are opposite numbers, are at the same distance from the origin, which means that the modules of opposite numbers are equal.

The next module property is: the modulus of the product of two numbers is equal to the product of the modules of these numbers, i.e, . By definition, the modulus of the product of numbers a and b is either a b if , or −(a b) if . It follows from the rules of multiplication of real numbers that the product of moduli of numbers a and b is equal to either a b , , or −(a b) , if , which proves the considered property.

The modulus of the quotient of dividing a by b is equal to the quotient of dividing the modulus of a by the modulus of b, i.e, . Let us justify this property of the module. Since the quotient is equal to the product, then . By virtue of the previous property, we have . It remains only to use the equality , which is valid due to the definition of the modulus of the number.

The following module property is written as an inequality: , a , b and c are arbitrary real numbers. The written inequality is nothing more than triangle inequality. To make this clear, let's take the points A(a) , B(b) , C(c) on the coordinate line, and consider the degenerate triangle ABC, whose vertices lie on the same line. By definition, the modulus of the difference is equal to the length of the segment AB, - the length of the segment AC, and - the length of the segment CB. Since the length of any side of a triangle does not exceed the sum of the lengths of the other two sides, the inequality , therefore, the inequality also holds.

The inequality just proved is much more common in the form . The written inequality is usually considered as a separate property of the module with the formulation: “ The modulus of the sum of two numbers does not exceed the sum of the moduli of these numbers". But the inequality directly follows from the inequality , if we put −b instead of b in it, and take c=0 .

Complex number modulus

Let's give determination of the modulus of a complex number. Let us be given complex number, written in algebraic form , where x and y are some real numbers, representing, respectively, the real and imaginary parts of a given complex number z, and is an imaginary unit.

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The absolute value of a number a is the distance from the origin to the point BUT(a).

To understand this definition, we substitute instead of a variable a any number, for example 3 and try to read it again:

The absolute value of a number 3 is the distance from the origin to the point BUT(3 ).

It becomes clear that the module is nothing more than the usual distance. Let's try to see the distance from the origin to point A( 3 )

The distance from the origin of coordinates to point A( 3 ) is equal to 3 (three units or three steps).

The modulus 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:

Reads like: "The modulus of 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 new point B. point A we have already used in the first example.

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

The distance from one point to another cannot be negative. Therefore, the modulus of any negative number, being a distance, will also not be negative. The module of the number -3 will be the number 3. The distance from the origin to the point B(-3) is also equal to three units:

Reads like: "The modulus of a number minus three is three"

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

"The modulus of zero is zero"

We 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 one, to 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 2 has a plus sign, but we don’t see it, because plus, as we said earlier, is traditionally not written.

More examples of opposite numbers:

Opposite numbers have equal modules. For example, let's find modules for −2 and 2

The figure shows that the distance from the origin to the points A(−2) and B(2) equal to two steps.

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