A is calculated in accordance with the following rules:

For brevity, notations are used |a|. So, |10| = 10; - 1 / 3 = | 1 / 3 |; | -100| =100, etc.

Every size X corresponds to a fairly accurate value | X|. And that means identity at= |X| sets at like some argument function X.

Schedule this functions presented below.

For x > 0 |x| = x, and for x< 0 |x|= -x; in this regard, the line y = | x| at x> 0 combined with a straight line y = x(bisector of the first coordinate angle), and when X< 0 - с прямой y = -x(bisector of the second coordinate angle).

Separate equations include unknowns under the sign module.

Arbitrary examples of such equations - | X— 1| = 2, |6 — 2X| =3X+ 1, etc.

Solving equations containing an unknown under the modulus sign is based on the fact that if the absolute value of an unknown number x is equal to a positive number a, then this number x itself is equal to either a or -a.

For example:, if | X| = 10, then or X=10, or X = -10.

Let's consider solving individual equations.

Let's analyze the solution to the equation | X- 1| = 2.

Let's expand the module then the difference X- 1 can equal either + 2 or - 2. If x - 1 = 2, then X= 3; if X- 1 = - 2, then X= - 1. We make a substitution and find that both of these values ​​satisfy the equation.

Answer. The above equation has two roots: x 1 = 3, x 2 = - 1.

Let's analyze solution to the equation | 6 — 2X| = 3X+ 1.

After module expansion we get: or 6 - 2 X= 3X+ 1, or 6 - 2 X= - (3X+ 1).

In the first case X= 1, and in the second X= - 7.

Examination. At X= 1 |6 — 2X| = |4| = 4, 3x+ 1 = 4; it follows from the court, X = 1 - root given equations.

At x = - 7 |6 — 2x| = |20| = 20, 3x+ 1= - 20; since 20 ≠ -20, then X= - 7 is not a root of this equation.

Answer. U equation has only one root: X = 1.

Equations of this type can be solve and graphically.

So let's decide For example, graphically equation | X- 1| = 2.

First we will construct function graphics at = |x- 1|. First, let's draw a graph of the function at=X- 1:

That part of it graphic arts, which is located above the axis X We won't change it. For her X- 1 > 0 and therefore | X-1|=X-1.

The part of the graph that is located below the axis X, let's depict symmetrically relative to this axis. Because for this part X - 1 < 0 и соответственно |X - 1|= - (X - 1). The resulting line(solid line) and will function graph y = | X—1|.

This line will intersect with straight at= 2 at two points: M 1 with abscissa -1 and M 2 with abscissa 3. And, accordingly, the equation | X- 1| =2 there will be two roots: X 1 = - 1, X 2 = 3.

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.

It turns out a curious thing: different numbers may 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 these are, whose 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 - 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 singular, which has no 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 break it off: more numbers No. 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 put an arbitrary number $a$ in place of the triple on the right. 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? That's right: because it requires that the modulus be equal to 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 is the expression 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 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 used important property 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 ended - 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”. 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:

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

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 options expansion of the modulus 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:

\[\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.\]

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:

1. Equate each modulus in the equation to zero. We get several 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, at each of which all modules are uniquely revealed;
3. 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:

1. The leftmost one: $x \lt 1$ — 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. 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 online math calculator will help you solve an equation or inequality with moduli. Program for solving equations and inequalities with moduli not only gives the answer to the problem, it leads detailed solution with explanations, i.e. displays the process of obtaining the result.

This program may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.

This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved increases.

|x| or abs(x) - module x

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A little theory.

Equations and inequalities with moduli

In a basic school algebra course, you may encounter the simplest equations and inequalities with moduli. To solve them, you can use a geometric method based on the fact that \(|x-a| \) is the distance on the number line between points x and a: \(|x-a| = \rho (x;\; a) \). For example, to solve the equation \(|x-3|=2\) you need to find points on the number line that are distant from point 3 at a distance of 2. There are two such points: \(x_1=1\) and \(x_2=5\) .

Solving the inequality \(|2x+7|

But the main way to solve equations and inequalities with moduli is associated with the so-called “revelation of the modulus by definition”:
if \(a \geq 0 \), then \(|a|=a \);
if \(a As a rule, an equation (inequality) with moduli is reduced to a set of equations (inequalities) that do not contain the modulus sign.

In addition to the above definition, the following statements are used:
1) If \(c > 0\), then the equation \(|f(x)|=c \) is equivalent to the set of equations: \(\left[\begin(array)(l) f(x)=c \\ f(x)=-c \end(array)\right. \)
2) If \(c > 0 \), then the inequality \(|f(x)| 3) If \(c \geq 0 \), then the inequality \(|f(x)| > c \) is equivalent to a set of inequalities : \(\left[\begin(array)(l) f(x) c \end(array)\right. \)
4) If both sides of the inequality \(f(x) EXAMPLE 1. Solve the equation \(x^2 +2|x-1| -6 = 0\).

If \(x-1 \geq 0\), then \(|x-1| = x-1\) and the given equation takes the form
\(x^2 +2(x-1) -6 = 0 \Rightarrow x^2 +2x -8 = 0 \).
If \(x-1 \(x^2 -2(x-1) -6 = 0 \Rightarrow x^2 -2x -4 = 0 \).
Thus, the given equation should be considered separately in each of the two indicated cases.
1) Let \(x-1 \geq 0 \), i.e. \(x\geq 1\). From the equation \(x^2 +2x -8 = 0\) we find \(x_1=2, \; x_2=-4\). The condition \(x \geq 1 \) is satisfied only by the value \(x_1=2\).
2) Let \(x-1 Answer: \(2; \;\; 1-\sqrt(5) \)

EXAMPLE 2. Solve the equation \(|x^2-6x+7| = \frac(5x-9)(3)\).

First way(module expansion by definition).
Reasoning as in example 1, we come to the conclusion that the given equation needs to be considered separately if two conditions are met: \(x^2-6x+7 \geq 0 \) or \(x^2-6x+7

1) If \(x^2-6x+7 \geq 0 \), then \(|x^2-6x+7| = x^2-6x+7 \) and the given equation takes the form \(x^2 -6x+7 = \frac(5x-9)(3) \Rightarrow 3x^2-23x+30=0 \). Having decided this quadratic equation, we get: \(x_1=6, \; x_2=\frac(5)(3) \).
Let's find out whether the value \(x_1=6\) satisfies the condition \(x^2-6x+7 \geq 0\). To do this, substitute the indicated value into the quadratic inequality. We get: \(6^2-6 \cdot 6+7 \geq 0 \), i.e. \(7 \geq 0 \) is a true inequality. This means that \(x_1=6\) is the root of the given equation.
Let's find out whether the value \(x_2=\frac(5)(3)\) satisfies the condition \(x^2-6x+7 \geq 0\). To do this, substitute the indicated value into the quadratic inequality. We get: \(\left(\frac(5)(3) \right)^2 -\frac(5)(3) \cdot 6 + 7 \geq 0 \), i.e. \(\frac(25)(9) -3 \geq 0 \) is an incorrect inequality. This means that \(x_2=\frac(5)(3)\) is not a root of the given equation.

2) If \(x^2-6x+7 Value \(x_3=3\) satisfies the condition \(x^2-6x+7 Value \(x_4=\frac(4)(3) \) does not satisfy the condition \ (x^2-6x+7 So, the given equation has two roots: \(x=6, \; x=3 \).

Second way. If the equation \(|f(x)| = h(x) \) is given, then with \(h(x) \(\left[\begin(array)(l) x^2-6x+7 = \frac (5x-9)(3) \\ x^2-6x+7 = -\frac(5x-9)(3) \end(array)\right. \)
Both of these equations were solved above (using the first method of solving the given equation), their roots are as follows: \(6,\; \frac(5)(3),\; 3,\; \frac(4)(3)\). Condition \(\frac(5x-9)(3) \geq 0 \) from these four values satisfy only two: 6 and 3. This means that the given equation has two roots: \(x=6, \; x=3\).

Third way(graphic).
1) Let's build a graph of the function \(y = |x^2-6x+7| \). First, let's construct a parabola \(y = x^2-6x+7\). We have \(x^2-6x+7 = (x-3)^2-2 \). The graph of the function \(y = (x-3)^2-2\) can be obtained from the graph of the function \(y = x^2\) by shifting it 3 scale units to the right (along the x-axis) and 2 scale units down ( along the y-axis). The straight line x=3 is the axis of the parabola we are interested in. As control points for more accurate plotting, it is convenient to take point (3; -2) - the vertex of the parabola, point (0; 7) and point (6; 7) symmetrical to it relative to the axis of the parabola.
To now construct a graph of the function \(y = |x^2-6x+7| \), you need to leave unchanged those parts of the constructed parabola that lie not below the x-axis, and mirror that part of the parabola that lies below the x-axis relative to the x axis.
2) Let's build a graph linear function\(y = \frac(5x-9)(3)\). It is convenient to take points (0; –3) and (3; 2) as control points.

It is important that the point x = 1.8 of the intersection of the straight line with the abscissa axis is located to the right of the left point of intersection of the parabola with the abscissa axis - this is the point \(x=3-\sqrt(2) \) (since \(3-\sqrt(2 ) 3) Judging by the drawing, the graphs intersect at two points - A(3; 2) and B(6; 7).Substituting the abscissas of these points x = 3 and x = 6 into the given equation, we are convinced that both In another value, the correct numerical equality is obtained. This means that our hypothesis was confirmed - the equation has two roots: x = 3 and x = 6. Answer: 3; 6.

Comment. Graphic method for all its elegance, it is not very reliable. In the example considered, it worked only because the roots of the equation are integers.

EXAMPLE 3. Solve the equation \(|2x-4|+|x+3| = 8\)

First way
The expression 2x–4 becomes 0 at the point x = 2, and the expression x + 3 becomes 0 at the point x = –3. These two points divide the number line into three intervals: \(x

Consider the first interval: \((-\infty; \; -3) \).
If x Consider the second interval: \([-3; \; 2) \).
If \(-3 \leq x Consider the third interval: \(. Now we expand the internal module for x>2.5. We obtain an equation with one module
|2x-5-1|=x+3;
|2x-6|=x+3.
When expanding the module we get 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 with 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.

MBOU Secondary School No. 17, Ivanovo

« Equations with modulus"
Methodological development

Compiled

math teacher

Lebedeva N.V.

20010

Explanatory note

Chapter 1. Introduction

Section 2. Basic properties Section 3. Geometric interpretation of the concept of modulus of a number Section 4. Graph of the function y = |x| Section 5. Conventions

Chapter 2. Solving equations containing a modulus

Section 1. Equations of the form |F(x)| = m (simplest) Section 2. Equations of the form F(|x|) = m Section 3. Equations of the form |F(x)| = G(x) Section 4. Equations of the form |F(x)| = ± F(x) (most beautiful) Section 5. Equations of the form |F(x)| = |G(x)| Section 6. Examples of solving non-standard equations Section 7. Equations of the form |F(x)| + |G(x)| = 0 Section 8. Equations of the form |a 1 x ± b 1 | ± |a 2 x ± in 2 | ± …|a n x ± in n | = m Section 9. Equations containing several modules

Chapter 3. Examples of solving various equations with modulus.

Section 1. Trigonometric equations Section 2. Exponential equations Section 3. Logarithmic equations Section 4. Irrational equations Section 5. Assignments increased complexity Answers to the exercises Bibliography

Explanatory note.

The concept of absolute value (modulus) of a real number is one of its essential characteristics. This concept is widespread in various sections of physical, mathematical and technical sciences. In the practice of teaching mathematics courses in high school in accordance with the Program of the Ministry of Defense of the Russian Federation, the concept of “absolute value of a number” appears repeatedly: in the 6th grade, the definition of a module is introduced, its geometric meaning; in the 8th grade the concept of absolute error is formed, the solution of the simplest equations and inequalities containing a modulus is considered, the properties of arithmetic are studied square root; in 11th grade the concept is found in the section “Root n-th degree." Teaching experience shows that students often encounter difficulties when solving tasks that require knowledge of this material, and often they skip it without starting to implement it. In the texts exam tasks Similar tasks are also included for the 9th and 11th grade courses. In addition, the requirements that universities place on school graduates differ, namely, more high level than the requirements of the school curriculum. For life in modern society It is very important to develop a mathematical thinking style, which manifests itself in certain mental skills. In the process of solving problems with modules, the ability to use techniques such as generalization and specification, analysis, classification and systematization, and analogy is required. Solving such tasks allows you to test your knowledge of the main sections school course, level logical thinking, initial research skills. this work is devoted to one of the sections - solving equations containing a module. It consists of three chapters. The first chapter introduces basic concepts and the most important theoretical considerations. The second chapter proposes nine main types of equations containing a module, discusses methods for solving them, and examines examples different levels difficulties. The third chapter offers more complex and non-standard equations (trigonometric, exponential, logarithmic and irrational). For each type of equation there are exercises for independent decision(answers and instructions are attached). The main purpose of this work is to provide methodological assistance to teachers in preparing for lessons and in organizing elective courses. The material can also be used as teaching aid for high school students. The tasks proposed in the work are interesting and not always easy to solve, which allows you to learning motivation students to become more aware, test their abilities, and improve the level of preparation of school graduates for entering universities. A differentiated selection of the proposed exercises involves a transition from the reproductive level of mastering the material to the creative one, as well as the opportunity to teach how to apply your knowledge when solving non-standard problems.

Chapter 1. Introduction.

Section 1. Determination of absolute value .

Definition : The absolute value (modulus) of a real number A a non-negative number is called: A or -A. Designation: │ A │ The entry reads as follows: “modulus of the number a” or “absolute value of the number a”

│ a, if a > 0

│a│ = │ 0, if a = 0 (1)

│ - and, if a
Examples: 1) │2,5│ = 2,5 2) │-7│ = 7 3) │1 - √2│ = √2 – 1

Expand expression module:

a) │x - 8│, if x > 12 b) │2x + 3│, if x ≤ -2 │x – 8│= x – 8 │ 2x + 3│= - 2x – 3

Section 2. Basic properties.

Let's consider the basic properties of absolute value. Property #1: Opposite numbers have equal modules, i.e. │а│=│- а│ Let us show that the equality is correct. Let's write down the definition of the number - A : │- a│= (2) Let's compare sets (1) and (2). It is obvious that the definitions absolute values numbers A And - A match up. Hence, │а│=│- а│
When considering the following properties, we will limit ourselves to their formulation, since their proof is given in Property #2: The absolute value of the sum of a finite number of real numbers does not exceed the sum of the absolute values ​​of the terms: │а 1 + а 2 +…+ а n │ ≤│а 1 │+│а 2 │+ … + │а n │ Property #3: The absolute value of the difference between two real numbers does not exceed the sum of their absolute values: │а - в│ ≤│а│+│в│ Property #4: The absolute value of the product of a finite number of real numbers is equal to the product of the absolute values ​​of the factors: │а·в│=│а│·│в│ Property #5: The absolute value of the quotient of real numbers is equal to the quotient of their absolute values:

Section 3. Geometric interpretation of the concept of modulus of a number.

Each real number can be associated with a point on the number line, which will be a geometric image of this real number. Each point on the number line corresponds to its distance from the origin, i.e. the length of the segment from the origin to a given point. This distance is always considered as a non-negative value. Therefore, the length of the corresponding segment will be the geometric interpretation of the absolute value of a given real number

The presented geometric illustration clearly confirms property No. 1, i.e. the moduli of opposite numbers are equal. From here the validity of the equality is easily understood: │х – а│= │а – x│. The solution to the equation │х│= m, where m ≥ 0, namely x 1.2 = ± m, also becomes more obvious. Examples: 1) │х│= 4 x 1.2 = ± 4 2) │х - 3│= 1
x 1.2 = 2; 4

Section 4. Graph of the function y = │х│

The domain of this function is all real numbers.

Section 5. Conventions.

In the future, when considering examples of solving equations, the following will be used symbols: ( - sign of the system [ - sign of the totality When solving a system of equations (inequalities), the intersection of solutions of the equations (inequalities) included in the system is found. When solving a set of equations (inequalities), the union of solutions included in the set of equations (inequalities) is found.

Chapter 2. Solving equations containing a modulus.

In this chapter we will look at algebraic methods for solving equations containing one or more modules.

Section 1. Equations of the form │F (x)│= m

An equation of this type is called the simplest. It has a solution if and only if m ≥ 0. By definition of the modulus, the original equation is equivalent to a set of two equations: │ F(x)│=m
Examples:
№1. Solve the equation: │7х - 2│= 9


Answer: x 1 = - 1; X 2 = 1 4 / 7 №2
│x 2 + 3x + 1│= 1

x 2 + 3x + 2 = 0 x 2 +3x = 0 x 1 = -1; x 2 = -2 x (x + 3) = 0 x 1 = 0; x 2 = -3 Answer: the sum of the roots is - 2.№3
│x 4 -5x 2 + 2│= 2 x 4 – 5x 2 = 0 x 4 – 5x 2 + 4 = 0 x 2 (x 2 – 5) = 0 denote x 2 = m, m ≥ 0 x = 0 ; ±√5 m 2 – 5m + 4 = 0 m = 1; 4 – both values ​​satisfy the condition m ≥ 0 x 2 = 1 x 2 = 4 x = ± 1 x = ± 2 Answer: number of roots of equation 7. Exercises:
№1. Solve the equation and indicate the sum of the roots: │х - 5│= 3 №2 . Solve the equation and indicate the smaller root: │x 2 + x│= 0 №3 . Solve the equation and indicate the larger root: │x 2 – 5x + 4│= 4 №4 .Solve the equation and indicate the whole root: │2x 2 – 7x + 6│= 1 №5 .Solve the equation and indicate the number of roots: │x 4 – 13x 2 + 50│= 14

Section 2. Equations of the form F(│х│) = m

The function argument on the left side is under the modulus sign, and the right side is independent of the variable. Let's consider two ways to solve equations of this type. 1 way: By definition of absolute value, the original equation is equivalent to the combination of two systems. In each of which a condition is imposed on a submodular expression. F(│х│) =m
Since the function F(│x│) is even throughout the entire domain of definition, the roots of the equations F(x) = m and F(- x) = m are pairs of opposite numbers. Therefore, it is enough to solve one of the systems (when considering examples in this way, the solution to one system will be given). Method 2: Application of the method of introducing a new variable. In this case, the notation │x│= a is introduced, where a ≥ 0. This method less voluminous in design.
Examples: №1 . Solve the equation: 3x 2 – 4│x│= - 1 Let’s use the introduction of a new variable. Let us denote │x│= a, where a ≥ 0. We obtain the equation 3a 2 - 4a + 1 = 0 D = 16 – 12 = 4 a 1 = 1 a 2 = 1 / 3 Return to the original variable: │x│=1 and │х│= 1/3. Each equation has two roots. Answer: x 1 = 1; X 2 = - 1; X 3 = 1 / 3 ; X 4 = - 1 / 3 . №2. Solve the equation: 5x 2 + 3│x│- 1 = 1 / 2 │x│ + 3x 2
Let's find the solution to the first system of the population: 4x 2 + 5x – 2 =0 D = 57 x 1 = -5+√57 / 8 x 2 = -5-√57 / 8 Note that x 2 does not satisfy the condition x ≥ 0. Solution the second system will be the number opposite to the value x 1. Answer: x 1 = -5+√57 / 8 ; X 2 = 5-√57 / 8 .№3 . Solve the equation: x 4 – │х│= 0 Let us denote │х│= a, where a ≥ 0. We get the equation a 4 – a = 0 a · (a 3 – 1) = 0 a 1 = 0 a 2 = 1 Return to the original variable: │х│=0 and │х│= 1 x = 0; ± 1 Answer: x 1 = 0; X 2 = 1; X 3 = - 1.
Exercises: №6. Solve the equation: 2│х│ - 4.5 = 5 – 3 / 8 │х│ №7 . Solve the equation, indicate the number of roots in your answer: 3x 2 - 7│x│ + 2 = 0 №8 . Solve the equation, indicate integer solutions in your answer: x 4 + │x│ - 2 = 0

Section 3. Equations of the form │F(x)│ = G(x)

The right-hand side of an equation of this type depends on a variable and, therefore, has a solution if and only if the right-hand side is a function G(x) ≥ 0. The original equation can be solved in two ways: 1 way: Standard, based on the disclosure of a module based on its definition and consists of an equivalent transition to a combination of two systems. │ F(x)│ =G(X)

This method can be rationally used in the case of a complex expression for the function G(x) and a less complex one for the function F(x), since it is assumed that inequalities with the function F(x) will be solved. Method 2: Consists in the transition to an equivalent system in which a condition is imposed on the right side. │ F(x)│= G(x)

This method is more convenient to use if the expression for the function G(x) is less complex than for the function F(x), since the solution to the inequality G(x) ≥ 0 is assumed. In addition, in the case of several modules, it is recommended to use the second option. Examples: №1. Solve the equation: │x + 2│= 6 -2x
(1 way) Answer: x = 1 1 / 3 №2.
│х 2 – 2х - 1│= 2·(x + 1)
(2 way) Answer: The product of roots is 3.
№3. Solve the equation and indicate the sum of the roots in your answer:
│x - 6│= x 2 - 5x + 9

Answer: the sum of the roots is 4.
Exercises: №9. │x + 4│= - 3x №10. Solve the equation, indicate the number of solutions in your answer:│x 2 + x - 1│= 2x – 1 №11 . Solve the equation, indicate the product of the roots in your answer:│x + 3│= x 2 + x – 6

Section 4. Equations of the form │F(x)│= F(x) and │F(x)│= - F(x)

Equations of this type are sometimes called “the most beautiful.” Since the right-hand side of the equations depends on the variable, solutions exist if and only if the right-hand side is non-negative. Therefore, the original equations are equivalent to the inequalities:
│F(x)│= F(x) F(x) ≥ 0 and │F(x)│= - F(x) F(x) Examples: №1 . Solve the equation, indicate the smaller integer root in your answer: │5x - 3│= 5x – 3 5x – 3 ≥ 0 5x ≥ 3 x ≥ 0.6 Answer: x = 1№2. Solve the equation, indicate the length of the interval in your answer: │х 2 - 9│= 9 – x 2 x 2 – 9 ≤ 0 (x – 3) (x + 3) ≤ 0 [- 3; 3] Answer: the length of the gap is 6.№3 . Solve the equation and indicate the number of integer solutions in your answer: │2 + x – x 2 │= 2 + x – x 2 2 + x – x 2 ≥ 0 x 2 – x – 2 ≤ 0 [- 1; 2] Answer: 4 whole solutions.№4 . Solve the equation and indicate the largest root in your answer:
│4 – x -
│= 4 – x –
x 2 – 5x + 5 = 0 D = 5 x 1.2 =
≈ 1,4

Answer: x = 3.

Exercises: №12. Solve the equation, indicate the whole root in your answer: │x 2 + 6x + 8│= x 2 + 6x + 8 №13. Solve the equation, indicate the number of integer solutions in your answer: │13x – x 2 - 36│+ x 2 – 13x + 36 = 0 №14. Solve the equation; in your answer, indicate an integer that is not the root of the equation:

Section 5. Equations of the form │F(x)│= │G(x)│

Since both sides of the equation are non-negative, the solution involves considering two cases: submodular expressions are equal or opposite in sign. Therefore, the original equation is equivalent to the combination of two equations: │ F(x)│= │ G(x)│
Examples: №1. Solve the equation, indicate the whole root in your answer: │x + 3│=│2x - 1│
Answer: whole root x = 4.№2. Solve the equation: │ x – x 2 - 1│=│2x – 3 – x 2 │
Answer: x = 2.№3 . Solve the equation and indicate the product of the roots in your answer:




Root equations 4x 2 + 2x – 1 = 0 x 1.2 = - 1±√5 / 4 Answer: the product of the roots is – 0.25. Exercises: №15 . Solve the equation and indicate the whole solution in your answer: │x 2 – 3x + 2│= │x 2 + 6x - 1│ №16. Solve the equation, indicate the smaller root in your answer:│5x - 3│=│7 - x│ №17 . Solve the equation and indicate the sum of the roots in your answer:

Section 6. Examples of solving non-standard equations

In this section we will look at examples of non-standard equations, when solving which the absolute value of the expression is revealed by definition. Examples:

№1. Solve the equation, indicate the sum of the roots in your answer: x · │x│- 5x – 6 = 0
Answer: the sum of the roots is 1 №2. . Solve the equation, indicate the smaller root in your answer: x 2 - 4x ·
- 5 = 0
Answer: smaller root x = - 5. №3. Solve the equation:

Answer: x = -1. Exercises: №18. Solve the equation and indicate the sum of the roots: x · │3x + 5│= 3x 2 + 4x + 3
№19. Solve the equation: x 2 – 3x =

№20. Solve the equation:

Section 7. Equations of the form │F(x)│+│G(x)│=0

It is easy to notice that on the left side of the equation of this type is the sum of non-negative quantities. Therefore, the original equation has a solution if and only if both terms are equal to zero at the same time. The equation is equivalent to the system of equations: │ F(x)│+│ G(x)│=0
Examples: №1 . Solve the equation:
Answer: x = 2. №2. Solve the equation: Answer: x = 1. Exercises: №21. Solve the equation: №22 . Solve the equation and indicate the sum of the roots in your answer: №23 . Solve the equation and indicate the number of solutions in your answer:

Section 8. Equations of the form │a 1 x + b 1 │±│a 2 x + b 2 │± … │a n x +b n │= m

To solve equations of this type, the interval method is used. If we solve it by sequential expansion of modules, we get n sets of systems, which is very cumbersome and inconvenient. Let's consider the interval method algorithm: 1). Find variable values X, for which each module is equal to zero (zeros of submodular expressions):
2). Mark the found values ​​on a number line, which is divided into intervals (the number of intervals is respectively equal to n+1 ) 3). Determine with what sign each module is revealed at each of the obtained intervals (when making a solution, you can use a number line, marking the signs on it) 4). The original equation is equivalent to the aggregate n+1 systems, in each of which the variable’s membership is indicated X one of the intervals. Examples: №1 . Solve the equation and indicate the largest root in your answer:
1). Let's find the zeros of the submodular expressions: x = 2; x = -3 2). Let's mark the found values ​​on the number line and determine with what sign each module is revealed on the resulting intervals:
x – 2 x – 2 x – 2 - - + - 3 2 x 2x + 6 2x + 6 2x + 6 - + + 3)
- no solutions The equation has two roots. Answer: the largest root x = 2. №2. Solve the equation and provide the whole root in your answer:
1). Let's find the zeros of the submodular expressions: x = 1.5; x = - 1 2). Let's mark the found values ​​on the number line and determine with what sign each module is revealed on the resulting intervals: x + 1 x + 1 x + 1 - + +
-1 1.5 x 2x – 3 2x – 3 2x – 3 - - +
3).
The last system has no solutions, therefore the equation has two roots. When solving the equation, you should pay attention to the “-” sign in front of the second module. Answer: whole root x = 7. №3. Solve the equation, indicate the sum of the roots in your answer: 1). Let's find the zeros of the submodular expressions: x = 5; x = 1; x = - 2 2). Let's mark the found values ​​on the number line and determine with what sign each module is revealed at the resulting intervals: x – 5 x – 5 x – 5 x – 5 - - - +
-2 1 5 x x – 1 x – 1 x – 1 x – 1 - - + + x + 2 x + 2 x + 2 x + 2 - + + +
3).
The equation has two roots x = 0 and 2. Answer: the sum of the roots is 2. №4 . Solve the equation: 1). Let's find the zeros of the submodular expressions: x = 1; x = 2; x = 3. 2). Let us determine with what sign each module is revealed on the resulting intervals. 3).
Let's combine the solutions of the first three systems. Answer: ; x = 5.
Exercises: №24. Solve the equation:
№25. Solve the equation and indicate the sum of the roots in your answer: №26. Solve the equation and indicate the smaller root in your answer: №27. Solve the equation and indicate the larger root in your answer:

Section 9. Equations containing several modules

Equations containing multiple modules assume the presence of absolute values ​​in submodular expressions. The basic principle for solving equations of this type is the sequential disclosure of modules, starting with the “external” one. During the solution, the techniques discussed in sections No. 1, No. 3 are used.

Examples: №1. Solve the equation:
Answer: x = 1; - eleven. №2. Solve the equation:
Answer: x = 0; 4; - 4. №3. Solve the equation and indicate the product of the roots in your answer:
Answer: the product of the roots is – 8. №4. Solve the equation:
Let us denote the equations of the population (1) And (2) and consider the solution to each of them separately for ease of design. Since both equations contain more than one module, it is more convenient to carry out an equivalent transition to sets of systems. (1)

(2)


Answer:
Exercises: №36. Solve the equation, indicate the sum of the roots in your answer: 5 │3x-5│ = 25 x №37. Solve the equation, if there is more than one root, indicate the sum of the roots in your answer: │x + 2│ x – 3x – 10 = 1 №38. Solve the equation: 3 │2x -4│ = 9 │x│ №39. Solve the equation and indicate the number of roots in your answer: 2 │ sin x│ = √2 №40 . Solve the equation and indicate the number of roots in your answer:

Section 3. Logarithmic equations.

Before solving the following equations, it is necessary to review the properties of logarithms and the logarithmic function. Examples: №1. Solve the equation, indicate the product of the roots in your answer: log 2 (x+1) 2 + log 2 │x+1│ = 6 O.D.Z. x+1≠0 x≠ - 1

Case 1: if x ≥ - 1, then log 2 (x+1) 2 + log 2 (x+1) = 6 log 2 (x+1) 3 = log 2 2 6 (x+1) 3 = 2 6 x+1 = 4 x = 3 – satisfies the condition x ≥ - 1 2 case: if x log 2 (x+1) 2 + log 2 (-x-1) = 6 log 2 (x+1) 2 + log 2 (-(x+1)) = 6 log 2 (-(x+1) 3) = log 2 2 6- (x+1) 3 = 2 6- (x+1) = 4 x = - 5 – satisfies condition x - 1
Answer: the product of the roots is – 15.
№2. Solve the equation, indicate the sum of the roots in your answer: lg
O.D.Z.

Answer: the sum of the roots is 0.5.
№3. Solve the equation: log 5
O.D.Z.

Answer: x = 9. №4. Solve the equation: │2 + log 0.2 x│+ 3 = │1 + log 5 x│ O.D.Z. x > 0 Let's use the formula for moving to another base. │2 - log 5 x│+ 3 = │1 + log 5 x│
│2 - log 5 x│- │1 + log 5 x│= - 3 Let's find the zeros of the submodular expressions: x = 25; x = These numbers divide the range of acceptable values ​​into three intervals, so the equation is equivalent to a set of three systems.
Answer: )