Equations under the modulus sign. Modulus of a number (absolute value of a number), definitions, examples, properties. Collection and use of personal information

The 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 dedicated to solving equations with modules.

I’ll say right away: the lesson will not be difficult. And in general, modules are a relatively simple topic. “Yes, of course, it’s not complicated! It blows my mind!” - many students will say, but all these brain breaks occur due to the fact that most people do not have knowledge in their heads, but some kind of crap. And the goal of this lesson is to turn crap into knowledge. :)

A little theory

So, let's go. Let's start with the most important thing: 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 absolute value of a positive number? It’s even simpler here: the modulus of a positive number is equal to this number itself: $\left| 5 \right|=5$; $\left| 129.5 \right|=$129.5, etc.

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

Another important fact: modulus is never negative. Whatever number we take - be it positive or negative - its modulus always turns out to be positive (or, in extreme cases, zero). This 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, we obtain a global definition of the modulus for all numbers. Namely: the modulus of a number is equal to the 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 modulus of zero, but it is always equal to zero. In addition, zero is the only number that does not have an opposite.

Thus, if we consider the function $y=\left| x \right|$ and try to draw its graph, you will get something like this:

Modulus graph and example of solving the equation

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

In addition to the 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 simply the distance between the specified points. Or, if you prefer, the length of the segment connecting these points:

Modulus is the distance between points on a number line

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

Basic formula

Okay, we've sorted out the definition. But that didn’t make it 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 modulus of $x$ is 3. What could $x$ be equal to? Well, judging by the definition, we are quite happy with $x=3$. Really:

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

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

So maybe if we search and think, we will find more numbers? But let's face it: 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 the function $f\left(x \right)$ hang out under the modulus sign instead of the variable $x$, and on the right instead of the triple we put an arbitrary number $a$. We get the equation:

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

So how can we solve this? Let me remind you: $f\left(x \right)$ is an arbitrary function, $a$ is any number. Those. Anything at all! For example:

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

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

Let's pay attention to the second equation. You can immediately say about him: he has no roots. Why? Everything is correct: because it requires that the modulus 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 modulus sign, and then $\left| 2x+1 \right|=2x+1$, or this expression is still negative, and then $\left| 2x+1 \right|=-\left(2x+1 \right)=-2x-1$. In the first case, our equation will be rewritten as follows:

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

And suddenly it turns out that the submodular expression $2x+1$ is really 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 particularly distrustful can try to substitute the found root into the original equation and make sure that there really is a positive number under the modulus.

Now let's look at the case of a negative submodular 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 larger than in the very simple equation $\left| x \right|=3$, but nothing fundamentally 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 modulus 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 modulus sign using the following rule:

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

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

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

Let’s consider separately when there is a ten plus on the right, and separately when there is 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 entire solution took literally two lines.

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

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

Again we open the module with plus and minus:

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

A couple of lines again - and the answer is ready! As I said, there is nothing complicated about modules. You just need to remember a few rules. Therefore, we move on and begin with truly more complex tasks.

The case of a right-hand side variable

Now consider this equation:

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

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

What to do in this case? First, we must understand once and for all that if the right side of the equation turns out to be negative, then the equation will have no roots- we already know that the module cannot be equal to a negative number.

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

In relation 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 will somehow cope with the requirement $2x\ge 0$. 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 some of the students are already starting to get bored? Well, let's look at an even more complex equation:

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

Although it looks evil, in fact it is still the same equation of the form “modulus equals function”:

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

And it is solved in exactly 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 evil (in fact, it is simple, but we will not solve it). For now, it’s better to deal with the resulting equations. Let's 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, it’s a no brainer that you need to collect everything from the left, bring similar ones and see what happens. And this is what happens:

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

We take the common factor $((x)^(2))$ out of brackets and 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 took advantage of 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 let’s deal with the second equation in exactly the same way, 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 equal to zero when at least one of the factors is equal to 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, which of this set will go into the final answer? To do this, remember that we have an additional constraint in the form of inequality:

How to take this requirement into account? 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 in response there will be only two roots:

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

As you can see, even in this case there was nothing complicated - equations with modules are always solved using an 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

Until now, 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 of the form $\left| f\left(x \right) \right|=g\left(x \right)$ 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 “modulus equals modulus”. The 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.

Someone will now think that such equations are more difficult to solve than what we have studied so far. But no: these equations are even easier to solve. Here's the formula:

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

All! We simply equate submodular expressions by putting a plus or minus sign in front of one of them. 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! Expanding 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 $3=-7$? At what values of $x$? “What the hell is $x$? Are you stoned? There’s no $x$ there at all,” you say. And you'll 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 solved literally in a couple of lines - we didn’t expect anything else from a linear equation. :)

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

So how? Difficult? Of course not. Let's try something else:

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

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

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

Perhaps someone will now ask: “Hey, what nonsense? Why does “plus-minus” appear on the right-hand expression and not on the left?” Calm down, I’ll explain everything now. Indeed, in a good way we should have rewritten our equation as follows:

Then you need to open the brackets, move all the terms to one side of 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” appears before three terms (especially when one of these terms is a quadratic expression), it somehow looks more complicated than the situation when “plus-minus” appears before only two terms.

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

What happened? Nothing special: they just swapped the left and right sides. A little thing that will ultimately make our life a little easier. :)

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 only one root: $x=1$. But we have already obtained 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 a pie from the shelf and eat it. There are 2 of them, yours is the middle one. :)

Important Note. The presence of identical roots for different variants of expansion of the module means that the original polynomials are factorized, and among these factors there will definitely 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: $\left| a\cdot b \right|=\left| a \right|\cdot \left| b \right|$ (i.e. the modulus of the product is equal to the product of the moduli), so the original equation can be rewritten as follows:

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

As you can see, we really have a common factor. Now, if you collect all the modules on one side, you can take this factor 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 remember 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 literally in a couple of lines. :)

This remark may seem unnecessarily complex and inapplicable in practice. However, in reality, you may encounter much more complex problems than those we are looking at 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 taking something out of brackets can be very, very useful. :)

Now I would like to look at another equation, which at first glance may seem crazy. Many students get stuck on it, even those who think they have a good understanding of the modules.

However, this equation is even easier to solve than what we looked at earlier. And if you understand why, you'll get another trick for quickly solving equations with moduli.

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 at what $x$ the sum of two modules is equal to zero. :)

What's the problem anyway? But the problem is that each module is a positive number, or, in extreme cases, zero. What happens if you add two positive numbers? Obviously a positive number again:

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

The last line might give you an idea: the only time the sum of the modules is zero is if each module is zero:

And when is the module equal to zero? Only in one case - when the submodular 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 module is reset to zero: 0, 1 and −1; as well as two points at which the second module is reset to zero: −2 and 1. However, we need both modules to be reset to zero at the same time, so among the found numbers 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.

Cleavage method

Well, we've already covered a bunch of problems and learned a lot of techniques. Do you think that's all? But no! Now we will look at the final technique - and at the same time the most important. We will talk about splitting equations with modulus. What will we even talk about? Let's go back a little and look at 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 construction of the form $\left| f\left(x \right) \right|=g\left(x \right)$. But let's try to look at this equation from a slightly different angle. More precisely, consider the expression under the modulus 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 you initially require that this number be positive? For example, we require 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 very modulus:

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

True, all these thoughts make sense only under the condition $3x-5 \gt 0$ - we ourselves introduced this requirement in order to unambiguously reveal the module. Therefore, 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 the expression turned out to be equal to zero, and we need it to be strictly greater than zero. Sad. :(

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

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

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

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: under the modulus there will be zero, and the modulus of zero is also equal to zero (this follows directly from the definition):

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

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

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 modulo equations

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 fairly simple (essentially linear) equation with modulus , really? Well, get used to it: the difficulty of the module is that the answers in such equations can turn out to be completely unpredictable.

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

Equate each modulus in the equation to zero. We get several equations;
Solve all these equations and mark the roots on the number line. As a result, the straight line will be divided into several intervals, at each of which all modules are uniquely revealed;
Solve the original equation for each interval and combine your answers.

That's all! There is only one question left: what to do with the roots obtained in step 1? Let's say we have two roots: $x=1$ and $x=5$. They will split the number line into 3 pieces:

Splitting the number line into intervals using points

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

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

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

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

This concludes the lesson. Download problems to solve on your own, practice, compare with the answers - and see you in the next lesson, which will be devoted to inequalities with moduli. :)

We don't choose mathematics her profession, and she chooses us.

Russian mathematician Yu.I. Manin

Equations with modulus

The most difficult problems to solve in school mathematics are equations containing variables under the modulus sign. To successfully solve such equations, you need to know the definition and basic properties of the module. Naturally, students must have the skills to solve equations of this type.

Basic concepts and properties

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

The simple properties of a module include the following relationships:

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

Moreover, if, where, then and

More complex module properties, which can be effectively used when solving equations with moduli, are formulated through the following theorems:

Theorem 1.For any analytical functions And inequality is true

Theorem 2. Equality is equivalent to inequality.

Theorem 3. Equality tantamount to inequality.

Let's look at typical examples of solving problems on the topic “Equations, containing variables under the modulus sign."

Solving equations with modulus

The most common method in school mathematics for solving equations with a modulus is the method, based on module expansion. This method is universal, however, in the general case, its use can lead to very cumbersome calculations. In this regard, students should know other, more effective methods and techniques for solving such equations. In particular, it is necessary to have skills in applying theorems, given in this article.

Example 1. Solve the equation. (1)

Solution. We will solve Equation (1) using the “classical” method – the method of revealing modules. To do this, let's split the number axis dots and into intervals and consider three cases.

1. If , then , , , and equation (1) takes the form . It follows from this. However, here , therefore the value found is not the root of equation (1).

2. If, then from equation (1) we obtain or .

Since then root of equation (1).

3. If, then equation (1) takes the form or . Let's note that.

Answer: , .

When solving subsequent equations with a module, we will actively use the properties of modules in order to increase the efficiency of solving such equations.

Example 2. Solve the equation.

Solution. Since and then from the equation it follows. In this regard, , , and the equation takes the form. From here we get. However , therefore the original equation has no roots.

Answer: no roots.

Example 3. Solve the equation.

Solution. Since, then. If , then and the equation takes the form.

From here we get .

Example 4. Solve the equation.

Solution.Let us rewrite the equation in equivalent form. (2)

The resulting equation belongs to equations of type .

Taking into account Theorem 2, it can be argued that equation (2) is equivalent to the inequality . From here we get .

Answer: .

Example 5. Solve the equation.

Solution. This equation has the form. That's why , according to Theorem 3, here we have inequality or .

Example 6. Solve the equation.

Solution. Let's assume that. Because , then the given equation takes the form of a quadratic equation, (3)

Where . Since equation (3) has a single positive root and , then . From here we get two roots of the original equation: And .

Example 7. Solve the equation. (4)

Solution. Since the equationis equivalent to the combination of two equations: And , then when solving equation (4) it is necessary to consider two cases.

1. If , then or .

From here we get , and .

2. If , then or .

Since, then.

Answer: , , , .

Example 8.Solve the equation . (5)

Solution. Since and , then . From here and from equation (5) it follows that and , i.e. here we have a system of equations

However, this system of equations is inconsistent.

Answer: no roots.

Example 9. Solve the equation. (6)

Solution. If we denote , then and from equation (6) we obtain

Or . (7)

Since equation (7) has the form , this equation is equivalent to the inequality . From here we get . Since , then or .

Answer: .

Example 10.Solve the equation. (8)

Solution.According to Theorem 1, we can write

(9)

Taking into account equation (8), we conclude that both inequalities (9) turn into equalities, i.e. there is a system of equations

However, according to Theorem 3, the above system of equations is equivalent to the system of inequalities

(10)

Solving the system of inequalities (10) we obtain . Since the system of inequalities (10) is equivalent to equation (8), the original equation has a single root.

Answer: .

Example 11. Solve the equation. (11)

Solution. Let and , then the equality follows from equation (11).

It follows that and . Thus, here we have a system of inequalities

The solution to this system of inequalities is And .

Answer: , .

Example 12.Solve the equation. (12)

Solution. Equation (12) will be solved by the method of sequential expansion of modules. To do this, let's consider several cases.

1. If , then .

1.1. If , then and , .

1.2. If, then. However , therefore, in this case, equation (12) has no roots.

2. If , then .

2.1. If , then and , .

2.2. If , then and .

Answer: , , , , .

Example 13.Solve the equation. (13)

Solution. Since the left side of equation (13) is non-negative, then . In this regard, and equation (13)

takes the form or .

It is known that the equation is equivalent to the combination of two equations And , solving which we get, . Because , then equation (13) has one root.

Answer: .

Example 14. Solve system of equations (14)

Solution. Since and , then and . Consequently, from the system of equations (14) we obtain four systems of equations:

The roots of the above systems of equations are the roots of the system of equations (14).

Answer: ,, , , , , , .

Example 15. Solve system of equations (15)

Solution. Since, then. In this regard, from the system of equations (15) we obtain two systems of equations

The roots of the first system of equations are and , and from the second system of equations we obtain and .

Answer: , , , .

Example 16. Solve system of equations (16)

Solution. From the first equation of system (16) it follows that .

Since then . Let's consider the second equation of the system. Because the, That , and the equation takes the form, , or .

If you substitute the valueinto the first equation of system (16), then , or .

Answer: , .

For a deeper study of problem solving methods, related to solving equations, containing variables under the modulus sign, You can recommend tutorials from the list of recommended literature.

1. Collection of problems in mathematics for applicants to colleges / Ed. M.I. Scanavi. – M.: Peace and Education, 2013. – 608 p.

2. Suprun V.P. Mathematics for high school students: tasks of increased complexity. – M.: CD “Librocom” / URSS, 2017. – 200 p.

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

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The modulus of a number is easy to find, and the theory behind it is important when solving problems.

The properties and rules of disclosure used in solving exercises and exams will be useful for schoolchildren and students. Earn money using your knowledge on https://teachs.ru!

What is a module in mathematics

The modulus of a number describes the distance on a number line from zero to a point, without taking into account the direction in which the point lies from zero. Mathematical notation : |x|.

In other words, it is the absolute value of a number. The definition proves that the value is never negative.

Module properties

It is important to remember the following properties:

Modulus of a complex number

The absolute value of a complex number is the length of a directed segment drawn from the beginning of the complex plane to the point (a, b).

This directed segment is also a vector representing a complex number a+bi, so the absolute value of a complex number is the same as the magnitude (or length) of the vector representing a+ bi.

How to solve equations with modulus

An equation with a modulus is an equality that contains an absolute value expression. If for a real number it represents its distance from the origin on the number line, then inequalities with modulus are the type of inequalities that consist of absolute values.

Equations like |x| =a

Equation |x| = a has two answers x = a and x = –a, because both options are on the coordinate line at a distance a from 0.

Equality with an absolute value has no solution if the value is negative.

If |x|< a представляет собой расстояние чисел от начала координат, это значит, что нужно искать все числа, чье расстояние от начала координат меньше a.

Equations like |x| = |y|

When there are absolute values on both sides of the equations, we need to consider both possibilities for acceptable definitions—positive and negative expressions.

For example, for the equality |x − a| = |x + b| there are two options: (x − a) = − (x + b) or (x − a) = (x + b).

Equations like |x| = y

Equations of this type contain the absolute value of an expression with a variable to the left of zero and another unknown to the right. The variable y can be either greater than or less than zero.

To obtain an answer to such an equality, you need to solve a system of several equations, in which you need to make sure that y is a non-negative quantity:

Solving inequalities with modulus

To better understand how to expand a module in different types of equalities and inequalities, you need to analyze the examples.

Equations of the form |x| =a

Example 1(algebra 6th grade). Solve: |x| + 2 = 4.

Solution.

Such equations are solved in the same way as equalities without absolute values. This means that by moving the unknowns to the left and the constants to the right, the expression does not change.

After moving the constant to the right, we get: |x| = 2.

Since the unknowns are related to the absolute value, this equation has two answers: 2 And −2 .

Answer: 2 And −2 .

Example 2(7th grade algebra). Solve the inequality |x + 2| ≥ 1.

Solution.

The first thing to do is to find the points where the absolute value will change. To do this, the expression is equated to 0 . Received: x = –2.

It means that –2 – turning point.

Let's divide the interval into 2 parts:

for x + 2 ≥ 0

[−1; + ∞).

for x+2< 0

The common answer for these two inequalities is the interval (−∞; –3].

Final decision – combining the answers of the individual parts:

x∈ (–∞; –3] ∪ [–1; + ∞).

Answer: x∈ (–∞; –3] ∪ [–1; + ∞) .

Equations of the form |x| = |y|

Example 1(8th grade algebra). Solve the equation with two modules: 2 * |x – 1| + 3 = 9 – |x – 1|.

Solution:

Answer: x 1 = 3; x 2 = − 1.

Example 2(8th grade algebra). Solve inequality:

Solution:

Equations of the form |x| = y

Example 1(algebra 10th grade). Find x:

Solution:

It is very important to check the right-hand side, otherwise you may write erroneous roots in your answer. It is clear from the system that it does not lie in the gap.

Answer: x = 0.

Sum module

Modulus of difference

Absolute value of the difference between two numbers x and y is equal to the distance between points with coordinates X And Y on the coordinate line.

Example 1.

Example 2.

Modulus of a negative number

To find the absolute value of a number that is less than zero, you need to find out how far it is from zero. Since the distance is always positive (it is impossible to take "negative" steps, they are just steps in the other direction), the result is always positive. That is,

Simply put, the absolute value of a negative number has the opposite meaning.

Zero module

Known property:

This is why absolute value cannot be said to be a positive number: zero is neither negative nor positive.

Squared module

The squared modulus is always equal to the squared expression:

Examples of graphs with a module

Often in tests and exams there are tasks that can only be solved by analyzing the graphs. Let's consider such tasks.

Example 1.

Given a function f(x) = |x|. It is necessary to construct a graph from – 3 to 3 with a step of 1.

Solution:

Explanation: The figure shows that the graph is symmetrical about the Y axis.

Example 2. It is necessary to draw and compare graphs of the functions f(x) = |x–2| and g(x) = |x|–2.

Solution:

Explanation: A constant inside an absolute value moves the entire graph to the right if its value is negative, and to the left if its value is positive. But a constant outside will move the graph up if the value is positive and down if it is negative (like - 2 in function g(x)).

Vertex coordinate x(the point at which two lines connect, the vertex of the graph) is the number by which the graph is shifted to the left or right. A coordinate y– this is the value by which the graph moves up or down.

You can build such graphs using online plotting applications. With their help, you can clearly see how constants affect functions.

Interval method in problems with modulus

The interval method is one of the best ways to find the answer in problems with a modulus, especially if there are several modulus in the expression.

To use the method you need to do the following:

Equate each expression to zero.
Find the values of the variables.
Plot the points obtained in step 2 on the number line.
Determine the sign of the expressions (negative or positive value) on the intervals and draw a – or + symbol, respectively. The easiest way to determine the sign is using the substitution method (substituting any value from the interval).
Solve inequalities with the given signs.

Example 1. Solve using the interval method.

Solution:

The term (module) literally translated from Latin means “measure”. This concept was introduced into mathematics by the English scientist R. Cotes. And the German mathematician K. Weierstrass introduced the modulus sign - a symbol that denotes this concept when writing.

In contact with

For the first time this concept is studied in mathematics in the 6th grade curriculum of secondary school. According to one definition, the modulus is the absolute value of a real number. In other words, to find out the modulus of a real number, you need to discard its sign.

Graphically absolute value A denoted as |a|.

The main distinguishing feature of this concept is that it is always a non-negative quantity.

Numbers that differ from each other only in sign are called opposite numbers. If a value is positive, then its opposite is negative, and zero is its opposite.

Geometric meaning

If we consider the concept of a module from the perspective of geometry, then it will denote the distance that is measured in unit segments from the origin of coordinates to a given point. This definition fully reveals the geometric meaning of the term being studied.

Graphically this can be expressed as follows: |a| = OA.

Properties of absolute value

Below we will consider all the mathematical properties of this concept and ways of writing it in the form of literal expressions:

Features of solving equations with modulus

If we talk about solving mathematical equations and inequalities that contain module, then we must remember that to solve them you will need to open this sign.

For example, if the sign of an absolute value contains some mathematical expression, then before opening the module, it is necessary to take into account the current mathematical definitions.

|A + 5| = A + 5, if, A is greater than or equal to zero.

5-A, if, A value is less than zero.

In some cases, the sign can be revealed unambiguously for any value of the variable.

Let's look at another example. Let's construct a coordinate line on which we mark all numerical values whose absolute value will be 5.

First you need to draw a coordinate line, mark the origin of coordinates on it and set the size of a unit segment. In addition, the straight line must have a direction. Now on this line it is necessary to apply markings that will be equal to the size of a unit segment.

Thus, we can see that on this coordinate line there will be two points of interest to us with values 5 and -5.

Instructions

If a module is represented as a continuous function, then the value of its argument can be either positive or negative: |x| = x, x ≥ 0; |x| = - x, x

The modulus is zero, and the modulus of any positive number is . If the argument is negative, then after opening the brackets its sign changes from minus to plus. Based on this, the conclusion follows that the modules of opposites are equal: |-x| = |x| = x.

The modulus of a complex number is found by the formula: |a| = √b ² + c ², and |a + b| ≤ |a| + |b|. If the argument contains a positive number as a multiplier, then it can be taken out of the bracket sign, for example: |4*b| = 4*|b|.

If the argument is presented as a complex number, then for convenience of calculations the order of the terms of the expression enclosed in rectangular brackets is allowed: |2-3| = |3-2| = 3-2 = 1 because (2-3) is less than zero.

The argument raised to a power is simultaneously under the sign of a root of the same order - it is solved using: √a² = |a| = ±a.

If you have a task in which the condition for expanding the module brackets is not specified, then there is no need to get rid of them - this will be the end result. And if you need to open them, then you must indicate the ± sign. For example, you need to find the value of the expression √(2 * (4-b))². His solution looks like this: √(2 * (4-b))² = |2 * (4-b)| = 2 * |4-b|. Since the sign of expression 4-b is unknown, it must be left in parentheses. If you add an additional condition, for example, |4-b| >

The modulus of zero is equal to zero, and the modulus of any positive number is equal to itself. If the argument is negative, then after opening the brackets its sign changes from minus to plus. Based on this, the conclusion follows that the modules of opposite numbers are equal: |-x| = |x| = x.

The modulus of a complex number is found by the formula: |a| = √b ² + c ², and |a + b| ≤ |a| + |b|. If the argument contains a positive integer as a factor, then it can be taken out of the bracket sign, for example: |4*b| = 4*|b|.

The modulus cannot be negative, so any negative number is converted to positive: |-x| = x, |-2| = 2, |-1/7| = 1/7, |-2.5| = 2.5.

If the argument is presented in the form of a complex number, then for the convenience of calculations it is allowed to change the order of the terms of the expression enclosed in rectangular brackets: |2-3| = |3-2| = 3-2 = 1 because (2-3) is less than zero.