Solving complex inequalities with modulus. Equations and inequalities with modulus
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Today, friends, there will be no snot or sentimentality. Instead, I will send you, no questions asked, into battle with one of the most formidable opponents in the 8th-9th grade algebra course.
Yes, you understood everything correctly: we are talking about inequalities with modulus. We will look at four basic techniques with which you will learn to solve about 90% of such problems. What about the remaining 10%? Well, we'll talk about them in a separate lesson. :)
However, before analyzing any of the techniques, I would like to remind you of two facts that you already need to know. Otherwise, you risk not understanding the material of today’s lesson at all.
What you already need to know
Captain Obviousness seems to hint that to solve inequalities with modulus you need to know two things:
- How inequalities are resolved;
- What is a module?
Let's start with the second point.
Module Definition
Everything is simple here. There are two definitions: algebraic and graphical. To begin with - algebraic:
Definition. The modulus of a number $x$ is either the number itself, if it is non-negative, or the number opposite to it, if the original $x$ is still negative.
It is written like this:
\[\left| x \right|=\left\( \begin(align) & x,\ x\ge 0, \\ & -x,\ x \lt 0. \\\end(align) \right.\]
Speaking in simple language, the modulus is “a number without a minus”. And it is in this duality (in some places you don’t have to do anything with the original number, but in others you have to remove some kind of minus) that is where the whole difficulty lies for beginning students.
Is there some more geometric definition. It is also useful to know, but we will turn to it only in complex and some special cases, where the geometric approach is more convenient than the algebraic one (spoiler: not today).
Definition. Let point $a$ be marked on the number line. Then the module $\left| x-a \right|$ is the distance from point $x$ to point $a$ on this line.
If you draw a picture, you will get something like this:
Graphical module definition One way or another, from the definition of a module its key property immediately follows: the modulus of a number is always a non-negative quantity. This fact will be a red thread running through our entire narrative today.
Solving inequalities. Interval method
Now let's look at the inequalities. There are a great many of them, but our task now is to be able to solve at least the simplest of them. Those that come down to linear inequalities, as well as to the interval method.
I have two big lessons on this topic (by the way, very, VERY useful - I recommend studying them):
- Interval method for inequalities (especially watch the video);
- Fractional rational inequalities is a very extensive lesson, but after it you won’t have any questions at all.
If you know all this, if the phrase “let’s move from inequality to equation” does not make you have a vague desire to hit yourself against the wall, then you are ready: welcome to hell to the main topic of the lesson. :)
1. Inequalities of the form “Modulus is less than function”
This is one of the most common problems with modules. It is required to solve an inequality of the form:
\[\left| f\right| \ltg\]
The functions $f$ and $g$ can be anything, but usually they are polynomials. Examples of such inequalities:
\[\begin(align) & \left| 2x+3 \right| \lt x+7; \\ & \left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0; \\ & \left| ((x)^(2))-2\left| x \right|-3 \right| \lt 2. \\\end(align)\]
All of them can be solved literally in one line according to the following scheme:
\[\left| f\right| \lt g\Rightarrow -g \lt f \lt g\quad \left(\Rightarrow \left\( \begin(align) & f \lt g, \\ & f \gt -g \\\end(align) \right.\right)\]
It is easy to see that we get rid of the module, but in return we get a double inequality (or, which is the same thing, a system of two inequalities). But this transition takes into account absolutely everything possible problems: if the number under the modulus is positive, the method works; if negative, it still works; and even with the most inadequate function in place of $f$ or $g$, the method will still work.
Naturally, the question arises: couldn’t it be simpler? Unfortunately, it's not possible. This is the whole point of the module.
However, enough with the philosophizing. Let's solve a couple of problems:
Task. Solve the inequality:
\[\left| 2x+3 \right| \lt x+7\]
Solution. So, we have before us a classic inequality of the form “the modulus is less” - there’s even nothing to transform. We work according to the algorithm:
\[\begin(align) & \left| f\right| \lt g\Rightarrow -g \lt f \lt g; \\ & \left| 2x+3 \right| \lt x+7\Rightarrow -\left(x+7 \right) \lt 2x+3 \lt x+7 \\\end(align)\]
Do not rush to open the parentheses preceded by a “minus”: it is quite possible that due to your haste you will make an offensive mistake.
\[-x-7 \lt 2x+3 \lt x+7\]
\[\left\( \begin(align) & -x-7 \lt 2x+3 \\ & 2x+3 \lt x+7 \\ \end(align) \right.\]
\[\left\( \begin(align) & -3x \lt 10 \\ & x \lt 4 \\ \end(align) \right.\]
\[\left\( \begin(align) & x \gt -\frac(10)(3) \\ & x \lt 4 \\ \end(align) \right.\]
The problem was reduced to two elementary inequalities. Let us note their solutions on parallel number lines:
Intersection of many
The intersection of these sets will be the answer.
Answer: $x\in \left(-\frac(10)(3);4 \right)$
Task. Solve the inequality:
\[\left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0\]
Solution. This task is a little more difficult. First, let’s isolate the module by moving the second term to the right:
\[\left| ((x)^(2))+2x-3 \right| \lt -3\left(x+1 \right)\]
Obviously, we again have an inequality of the form “the module is smaller”, so we get rid of the module using the already known algorithm:
\[-\left(-3\left(x+1 \right) \right) \lt ((x)^(2))+2x-3 \lt -3\left(x+1 \right)\]
Now attention: someone will say that I'm a bit of a pervert with all these parentheses. But let me remind you once again that our key goal is correctly solve the inequality and get the answer. Later, when you have perfectly mastered everything described in this lesson, you can pervert it yourself as you wish: open parentheses, add minuses, etc.
Let's start by just getting rid of double minus left:
\[-\left(-3\left(x+1 \right) \right)=\left(-1 \right)\cdot \left(-3 \right)\cdot \left(x+1 \right) =3\left(x+1 \right)\]
Now let's open all the brackets in the double inequality:
Let's move on to the double inequality. This time the calculations will be more serious:
\[\left\( \begin(align) & ((x)^(2))+2x-3 \lt -3x-3 \\ & 3x+3 \lt ((x)^(2))+2x -3 \\ \end(align) \right.\]
\[\left\( \begin(align) & ((x)^(2))+5x \lt 0 \\ & ((x)^(2))-x-6 \gt 0 \\ \end( align)\right.\]
Both inequalities are quadratic and can be solved using the interval method (that’s why I say: if you don’t know what this is, it’s better not to take on modules yet). Let's move on to the equation in the first inequality:
\[\begin(align) & ((x)^(2))+5x=0; \\ & x\left(x+5 \right)=0; \\ & ((x)_(1))=0;((x)_(2))=-5. \\\end(align)\]
As you can see, the output was incomplete quadratic equation, which can be solved in an elementary way. Now let's look at the second inequality of the system. There you will have to apply Vieta’s theorem:
\[\begin(align) & ((x)^(2))-x-6=0; \\ & \left(x-3 \right)\left(x+2 \right)=0; \\& ((x)_(1))=3;((x)_(2))=-2. \\\end(align)\]
We mark the resulting numbers on two parallel lines (separate for the first inequality and separate for the second):
Again, since we are solving a system of inequalities, we are interested in the intersection of the shaded sets: $x\in \left(-5;-2 \right)$. This is the answer.
Answer: $x\in \left(-5;-2 \right)$
I think that after these examples the solution scheme is extremely clear:
- Isolate the module by moving all other terms to the opposite side of the inequality. Thus we get an inequality of the form $\left| f\right| \ltg$.
- Solve this inequality by getting rid of the module according to the scheme described above. At some point, it will be necessary to move from double inequality to a system of two independent expressions, each of which can already be solved separately.
- Finally, all that remains is to intersect the solutions of these two independent expressions - and that’s it, we will get the final answer.
A similar algorithm exists for inequalities next type, when the module is larger than the function. However, there are a couple of serious “buts”. We’ll talk about these “buts” now.
2. Inequalities of the form “Modulus is greater than function”
They look like this:
\[\left| f\right| \gtg\]
Similar to the previous one? It seems. And yet such problems are solved in a completely different way. Formally, the scheme is as follows:
\[\left| f\right| \gt g\Rightarrow \left[ \begin(align) & f \gt g, \\ & f \lt -g \\\end(align) \right.\]
In other words, we consider two cases:
- First, we simply ignore the module and solve the usual inequality;
- Then, in essence, we expand the module with the minus sign, and then multiply both sides of the inequality by −1, while I have the sign.
In this case, the options are combined with a square bracket, i.e. We have before us a combination of two requirements.
Please note again: this is not a system, but a totality, therefore in the answer the sets are combined rather than intersecting. This is a fundamental difference from the previous point!
In general, many students are completely confused with unions and intersections, so let’s sort this issue out once and for all:
- "∪" is a union sign. Essentially, this is a stylized letter "U" that came to us from in English and is an abbreviation for “Union”, i.e. "Associations".
- "∩" is the intersection sign. This crap didn’t come from anywhere, but simply appeared as a counterpoint to “∪”.
To make it even easier to remember, just draw legs to these signs to make glasses (just don’t now accuse me of promoting drug addiction and alcoholism: if you are seriously studying this lesson, then you are already a drug addict):
Difference between intersection and union of sets Translated into Russian, this means the following: the union (totality) includes elements from both sets, therefore it is in no way less than each of them; but the intersection (system) includes only those elements that are simultaneously in both the first set and the second. Therefore, the intersection of sets is never larger than the source sets.
So it became clearer? That is great. Let's move on to practice.
Task. Solve the inequality:
\[\left| 3x+1 \right| \gt 5-4x\]
Solution. We proceed according to the scheme:
\[\left| 3x+1 \right| \gt 5-4x\Rightarrow \left[ \begin(align) & 3x+1 \gt 5-4x \\ & 3x+1 \lt -\left(5-4x \right) \\\end(align) \ right.\]
We solve each inequality in the population:
\[\left[ \begin(align) & 3x+4x \gt 5-1 \\ & 3x-4x \lt -5-1 \\ \end(align) \right.\]
\[\left[ \begin(align) & 7x \gt 4 \\ & -x \lt -6 \\ \end(align) \right.\]
\[\left[ \begin(align) & x \gt 4/7\ \\ & x \gt 6 \\ \end(align) \right.\]
We mark each resulting set on the number line, and then combine them:
Union of sets
It is quite obvious that the answer will be $x\in \left(\frac(4)(7);+\infty \right)$
Answer: $x\in \left(\frac(4)(7);+\infty \right)$
Task. Solve the inequality:
\[\left| ((x)^(2))+2x-3 \right| \gt x\]
Solution. Well? Nothing - everything is the same. We move from an inequality with a modulus to a set of two inequalities:
\[\left| ((x)^(2))+2x-3 \right| \gt x\Rightarrow \left[ \begin(align) & ((x)^(2))+2x-3 \gt x \\ & ((x)^(2))+2x-3 \lt -x \\\end(align) \right.\]
We solve every inequality. Unfortunately, the roots there will not be very good:
\[\begin(align) & ((x)^(2))+2x-3 \gt x; \\ & ((x)^(2))+x-3 \gt 0; \\&D=1+12=13; \\ & x=\frac(-1\pm \sqrt(13))(2). \\\end(align)\]
The second inequality is also a bit wild:
\[\begin(align) & ((x)^(2))+2x-3 \lt -x; \\ & ((x)^(2))+3x-3 \lt 0; \\&D=9+12=21; \\ & x=\frac(-3\pm \sqrt(21))(2). \\\end(align)\]
Now you need to mark these numbers on two axes - one axis for each inequality. However, points must be marked in in the right order: the larger the number, the further the point moves to the right.
And here a setup awaits us. If everything is clear with the numbers $\frac(-3-\sqrt(21))(2) \lt \frac(-1-\sqrt(13))(2)$ (the terms in the numerator of the first fraction are less than the terms in the numerator of the second , so the sum is also less), with the numbers $\frac(-3-\sqrt(13))(2) \lt \frac(-1+\sqrt(21))(2)$ there will also be no difficulties (positive number obviously more negative), then with the last couple everything is not so clear. Which is greater: $\frac(-3+\sqrt(21))(2)$ or $\frac(-1+\sqrt(13))(2)$? The placement of points on the number lines and, in fact, the answer will depend on the answer to this question.
So let's compare:
\[\begin(matrix) \frac(-1+\sqrt(13))(2)\vee \frac(-3+\sqrt(21))(2) \\ -1+\sqrt(13)\ vee -3+\sqrt(21) \\ 2+\sqrt(13)\vee \sqrt(21) \\\end(matrix)\]
We isolated the root, got non-negative numbers on both sides of the inequality, so we have the right to square both sides:
\[\begin(matrix) ((\left(2+\sqrt(13) \right))^(2))\vee ((\left(\sqrt(21) \right))^(2)) \ \ 4+4\sqrt(13)+13\vee 21 \\ 4\sqrt(13)\vee 3 \\\end(matrix)\]
I think it’s a no brainer that $4\sqrt(13) \gt 3$, so $\frac(-1+\sqrt(13))(2) \gt \frac(-3+\sqrt(21)) (2)$, the final points on the axes will be placed like this:
A case of ugly roots
Let me remind you that we are solving a collection, so the answer will be a union, not an intersection of shaded sets.
Answer: $x\in \left(-\infty ;\frac(-3+\sqrt(21))(2) \right)\bigcup \left(\frac(-1+\sqrt(13))(2 );+\infty \right)$
As you can see, our scheme works great for both simple tasks, and for very tough ones. The only thing " weakness“In this approach, you need to competently compare irrational numbers (and believe me: these are not only roots). But a separate (and very serious) lesson will be devoted to comparison issues. And we move on.
3. Inequalities with non-negative “tails”
Now we get to the most interesting part. These are inequalities of the form:
\[\left| f\right| \gt\left| g\right|\]
Generally speaking, the algorithm that we will talk about now is correct only for the module. It works in all inequalities where there are guaranteed non-negative expressions on the left and right:
What to do with these tasks? Just remember:
In inequalities with non-negative “tails”, both sides can be raised to any natural power. There will be no additional restrictions.
First of all, we will be interested in squaring - it burns modules and roots:
\[\begin(align) & ((\left(\left| f \right| \right))^(2))=((f)^(2)); \\ & ((\left(\sqrt(f) \right))^(2))=f. \\\end(align)\]
Just don’t confuse this with taking the root of a square:
\[\sqrt(((f)^(2)))=\left| f \right|\ne f\]
Countless mistakes were made when a student forgot to install a module! But that's a completely different story (it's like irrational equations), so we won’t go into this now. Let's solve a couple of problems better:
Task. Solve the inequality:
\[\left| x+2 \right|\ge \left| 1-2x \right|\]
Solution. Let's immediately notice two things:
- This is not a strict inequality. Points on the number line will be punctured.
- Both sides of the inequality are obviously non-negative (this is a property of the module: $\left| f\left(x \right) \right|\ge 0$).
Therefore, we can square both sides of the inequality to get rid of the modulus and solve the problem using the usual interval method:
\[\begin(align) & ((\left(\left| x+2 \right| \right))^(2))\ge ((\left(\left| 1-2x \right| \right) )^(2)); \\ & ((\left(x+2 \right))^(2))\ge ((\left(2x-1 \right))^(2)). \\\end(align)\]
At the last step, I cheated a little: I changed the sequence of terms, taking advantage of the evenness of the module (in fact, I multiplied the expression $1-2x$ by −1).
\[\begin(align) & ((\left(2x-1 \right))^(2))-((\left(x+2 \right))^(2))\le 0; \\ & \left(\left(2x-1 \right)-\left(x+2 \right) \right)\cdot \left(\left(2x-1 \right)+\left(x+2 \ right)\right)\le 0; \\ & \left(2x-1-x-2 \right)\cdot \left(2x-1+x+2 \right)\le 0; \\ & \left(x-3 \right)\cdot \left(3x+1 \right)\le 0. \\\end(align)\]
We solve using the interval method. Let's move from inequality to equation:
\[\begin(align) & \left(x-3 \right)\left(3x+1 \right)=0; \\ & ((x)_(1))=3;((x)_(2))=-\frac(1)(3). \\\end(align)\]
We mark the found roots on the number line. Once again: all points are shaded because the original inequality is not strict!
Getting rid of the modulus sign
Let me remind you for those who are especially stubborn: we take the signs from the last inequality, which was written down before moving on to the equation. And we paint over the areas required in the same inequality. In our case it is $\left(x-3 \right)\left(3x+1 \right)\le 0$.
OK it's all over Now. The problem is solved.
Answer: $x\in \left[ -\frac(1)(3);3 \right]$.
Task. Solve the inequality:
\[\left| ((x)^(2))+x+1 \right|\le \left| ((x)^(2))+3x+4 \right|\]
Solution. We do everything the same. I won't comment - just look at the sequence of actions.
Square it:
\[\begin(align) & ((\left(\left| ((x)^(2))+x+1 \right| \right))^(2))\le ((\left(\left | ((x)^(2))+3x+4 \right| \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))\le ((\left(((x)^(2))+3x+4 \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))-((\left(((x)^(2))+3x+4 \ right))^(2))\le 0; \\ & \left(((x)^(2))+x+1-((x)^(2))-3x-4 \right)\times \\ & \times \left(((x) ^(2))+x+1+((x)^(2))+3x+4 \right)\le 0; \\ & \left(-2x-3 \right)\left(2((x)^(2))+4x+5 \right)\le 0. \\\end(align)\]
Interval method:
\[\begin(align) & \left(-2x-3 \right)\left(2((x)^(2))+4x+5 \right)=0 \\ & -2x-3=0\ Rightarrow x=-1.5; \\ & 2((x)^(2))+4x+5=0\Rightarrow D=16-40 \lt 0\Rightarrow \varnothing . \\\end(align)\]
There is only one root on the number line:
The answer is a whole interval
Answer: $x\in \left[ -1.5;+\infty \right)$.
A small note about the last task. As one of my students accurately noted, both submodular expressions in this inequality are obviously positive, so the modulus sign can be omitted without harm to health.
But this is a completely different level of thinking and a different approach - it can conditionally be called the method of consequences. About it - in a separate lesson. Now let’s move on to the final part of today’s lesson and look at a universal algorithm that always works. Even when all previous approaches were powerless. :)
4. Method of enumeration of options
What if all these techniques don't help? If the inequality cannot be reduced to non-negative tails, if it is impossible to isolate the module, if in general there is pain, sadness, melancholy?
Then the “heavy artillery” of all mathematics comes onto the scene—the brute force method. In relation to inequalities with modulus it looks like this:
- Write out all submodular expressions and set them equal to zero;
- Solve the resulting equations and mark the roots found on one number line;
- The straight line will be divided into several sections, within which each module has a fixed sign and therefore is uniquely revealed;
- Solve the inequality on each such section (you can separately consider the roots-boundaries obtained in step 2 - for reliability). Combine the results - this will be the answer. :)
So how? Weak? Easily! Only for a long time. Let's see in practice:
Task. Solve the inequality:
\[\left| x+2 \right| \lt \left| x-1 \right|+x-\frac(3)(2)\]
Solution. This crap doesn't boil down to inequalities like $\left| f\right| \lt g$, $\left| f\right| \gt g$ or $\left| f\right| \lt \left| g \right|$, so we act ahead.
We write out submodular expressions, equate them to zero and find the roots:
\[\begin(align) & x+2=0\Rightarrow x=-2; \\ & x-1=0\Rightarrow x=1. \\\end(align)\]
In total, we have two roots that divide the number line into three sections, within which each module is revealed uniquely:
Partitioning the number line by zeros of submodular functions
Let's look at each section separately.
1. Let $x \lt -2$. Then both submodular expressions are negative, and the original inequality will be rewritten as follows:
\[\begin(align) & -\left(x+2 \right) \lt -\left(x-1 \right)+x-1.5 \\ & -x-2 \lt -x+1+ x-1.5 \\ & x \gt 1.5 \\\end(align)\]
We got a fairly simple limitation. Let's intersect it with the initial assumption that $x \lt -2$:
\[\left\( \begin(align) & x \lt -2 \\ & x \gt 1.5 \\\end(align) \right.\Rightarrow x\in \varnothing \]
Obviously, the variable $x$ cannot simultaneously be less than −2 and greater than 1.5. There are no solutions in this area.
1.1. Let us separately consider the borderline case: $x=-2$. Let's just substitute this number into the original inequality and check: is it true?
\[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1.5 \right|)_(x=-2) ) \\ & 0 \lt \left| -3\right|-2-1.5; \\ & 0 \lt 3-3.5; \\ & 0 \lt -0.5\Rightarrow \varnothing . \\\end(align)\]
It is obvious that the chain of calculations has led us to an incorrect inequality. Therefore, the original inequality is also false, and $x=-2$ is not included in the answer.
2. Let now $-2 \lt x \lt 1$. The left module will already open with a “plus”, but the right one will still open with a “minus”. We have:
\[\begin(align) & x+2 \lt -\left(x-1 \right)+x-1.5 \\ & x+2 \lt -x+1+x-1.5 \\& x \lt -2.5 \\\end(align)\]
Again we intersect with the original requirement:
\[\left\( \begin(align) & x \lt -2.5 \\ & -2 \lt x \lt 1 \\\end(align) \right.\Rightarrow x\in \varnothing \]
And again, the set of solutions is empty, since there are no numbers that are both less than −2.5 and greater than −2.
2.1. And again special case: $x=1$. We substitute into the original inequality:
\[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1.5 \right|)_(x=1)) \\ & \left| 3\right| \lt \left| 0 \right|+1-1.5; \\ & 3 \lt -0.5; \\ & 3 \lt -0.5\Rightarrow \varnothing . \\\end(align)\]
Similar to the previous “special case”, the number $x=1$ is clearly not included in the answer.
3. The last piece of the line: $x \gt 1$. Here all modules are opened with a plus sign:
\[\begin(align) & x+2 \lt x-1+x-1.5 \\ & x+2 \lt x-1+x-1.5 \\ & x \gt 4.5 \\ \end(align)\]
And again we intersect the found set with the original constraint:
\[\left\( \begin(align) & x \gt 4.5 \\ & x \gt 1 \\\end(align) \right.\Rightarrow x\in \left(4.5;+\infty \right)\]
Finally! We have found an interval that will be the answer.
Answer: $x\in \left(4,5;+\infty \right)$
Finally, one remark that may save you from stupid mistakes when solving real problems:
Solutions to inequalities with moduli usually represent continuous sets on the number line - intervals and segments. Isolated points are much less common. And even less often, it happens that the boundary of the solution (the end of the segment) coincides with the boundary of the range under consideration.
Consequently, if boundaries (the same “special cases”) are not included in the answer, then the areas to the left and right of these boundaries will almost certainly not be included in the answer. And vice versa: the border entered into the answer, which means that some areas around it will also be answers.
Keep this in mind when reviewing your solutions.
Mathematics is a symbol of the wisdom of science,
a model of scientific rigor and simplicity,
the standard of excellence and beauty in science.
Russian philosopher, professor A.V. Voloshinov
Inequalities with modulus
The most difficult problems to solve in school mathematics are inequalities, containing variables under the modulus sign. To successfully solve such inequalities, you must have a good knowledge of the properties of the module and have the skills to use them.
Basic concepts and properties
Module ( absolute value) real number denoted by and is defined as follows:
The simple properties of a module include the following relationships:
AND .
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 and inequalities with moduli, are formulated through the following theorems:
Theorem 1.For any analytical functions And inequality is true.
Theorem 2. Equality tantamount to inequality.
Theorem 3. Equality tantamount to inequality.
The most common inequalities in school mathematics, containing unknown variables under the modulus sign, are inequalities of the form and where some positive constant.
Theorem 4. Inequality equivalent to double inequality, and the solution to the inequalityreduces to solving a set of inequalities And .
This theorem is a special case of Theorems 6 and 7.
More complex inequalities, containing a module are inequalities of the form, And .
Methods for solving such inequalities can be formulated using the following three theorems.
Theorem 5. Inequality is equivalent to the combination of two systems of inequalities
I (1)
Proof. Since then
This implies the validity of (1).
Theorem 6. Inequality is equivalent to the system of inequalities
Proof. Because , then from inequality follows that . Under this condition, the inequalityand in this case the second system of inequalities (1) will turn out to be inconsistent.
The theorem is proven.
Theorem 7. Inequality is equivalent to the combination of one inequality and two systems of inequalities
I (3)
Proof. Since , then the inequality always executed, If .
Let , then inequalitywill be equivalent to inequality, from which follows a set of two inequalities And .
The theorem is proven.
Let's consider typical examples solving problems on the topic “Inequalities, containing variables under the modulus sign."
Solving inequalities with modulus
Most simple method solving inequalities with 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. Therefore, students should know other (more effective) methods and techniques for solving such inequalities. In particular, it is necessary to have skills in applying theorems, given in this article.
Example 1.Solve inequality
. (4)
Solution.We will solve inequality (4) using the “classical” method – the method of revealing modules. For this purpose, we divide the number axis dots and into intervals and consider three cases.
1. If , then , , , and inequality (4) takes the form or .
Since the case is considered here, it is a solution to inequality (4).
2. If, then from inequality (4) we obtain or . Since the intersection of intervals And is empty, then on the considered solution interval there is no inequality (4).
3. If, then inequality (4) takes the form or . It's obvious that is also a solution to inequality (4).
Answer: , .
Example 2. Solve inequality.
Solution. Let's assume that. Because , then the given inequality takes the form or . Since then and from here it follows or .
However, therefore or.
Example 3. Solve inequality
. (5)
Solution. Because , then inequality (5) is equivalent to the inequalities or . From here, according to Theorem 4, we have a set of inequalities And .
Answer: , .
Example 4.Solve inequality
. (6)
Solution. Let's denote . Then from inequality (6) we obtain the inequalities , , or .
From here, using the interval method, we get . Because , then here we have a system of inequalities
The solution to the first inequality of system (7) is the union of two intervals And , and the solution to the second inequality is the double inequality. This implies , that the solution to the system of inequalities (7) is the union of two intervals And .
Answer: ,
Example 5.Solve inequality
. (8)
Solution. Let us transform inequality (8) as follows:
Or .
Using the interval method, we obtain a solution to inequality (8).
Answer: .
Note. If we put and in the conditions of Theorem 5, we obtain .
Example 6. Solve inequality
. (9)
Solution. From inequality (9) it follows. Let us transform inequality (9) as follows:
Or
Since , then or .
Answer: .
Example 7.Solve inequality
. (10)
Solution. Since and , then or .
In this regard and inequality (10) takes the form
Or
. (11)
It follows that or . Since , then inequality (11) also implies or .
Answer: .
Note. If we apply Theorem 1 to the left side of inequality (10), then we get . From this and inequality (10) it follows, what or . Because , then inequality (10) takes the form or .
Example 8. Solve inequality
. (12)
Solution. Since then and from inequality (12) it follows or . However, therefore or. From here we get or .
Answer: .
Example 9. Solve inequality
. (13)
Solution. According to Theorem 7, the solution to inequality (13) is or .
Let it be now. In this case and inequality (13) takes the form or .
If you combine the intervals And , then we obtain a solution to inequality (13) of the form.
Example 10. Solve inequality
. (14)
Solution. Let us rewrite inequality (14) in an equivalent form: . If we apply Theorem 1 to the left side of this inequality, we obtain the inequality .
From here and from Theorem 1 it follows, that inequality (14) is satisfied for any values.
Answer: any number.
Example 11. Solve inequality
. (15)
Solution. Applying Theorem 1 to the left side of inequality (15), we get . This and inequality (15) yield the equation, which has the form.
According to Theorem 3, the equation tantamount to inequality. From here we get.
Example 12.Solve inequality
. (16)
Solution. From inequality (16), according to Theorem 4, we obtain a system of inequalities
When solving the inequalityLet us use Theorem 6 and obtain a system of inequalitiesfrom which it follows.
Consider the inequality. According to Theorem 7, we obtain a set of inequalities And . The second population inequality is valid for any real.
Hence , the solution to inequality (16) is.
Example 13.Solve inequality
. (17)
Solution. According to Theorem 1, we can write
(18)
Taking into account inequality (17), we conclude that both inequalities (18) turn into equalities, i.e. there is a system of equations
By Theorem 3 this system equations is equivalent to the system of inequalities
or
Example 14.Solve inequality
. (19)
Solution. Since, then. Let us multiply both sides of inequality (19) by the expression , which takes only positive values for any values. Then we obtain an inequality that is equivalent to inequality (19), of the form
From here we get or , where . Since and then the solution to inequality (19) is And .
Answer: , .
For a more in-depth study of methods for solving inequalities with a modulus, we recommend turning to textbooks, given in 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: methods of solving and proving inequalities. – M.: Lenand / URSS, 2018. – 264 p.
3. Suprun V.P. Mathematics for high school students: non-standard methods problem solving. – M.: CD “Librocom” / URSS, 2017. – 296 p.
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There are several ways to solve inequalities containing a modulus. Let's look at some of them.
1) Solving the inequality using the geometric property of the module.
Let me remind you what the geometric property of a modulus is: the modulus of a number x is the distance from the origin to the point with coordinate x.
When solving inequalities using this method, two cases may arise:
1. |x| ≤ b, 
And the inequality with modulus obviously reduces to a system of two inequalities. Here the sign can be strict, in which case the dots in the picture will be “punctured”.
2. |x| ≥ b, then the solution picture looks like this:
And the inequality with modulus obviously reduces to a combination of two inequalities. Here the sign can be strict, in which case the dots in the picture will be “punctured”.
Example 1.
Solve the inequality |4 – |x|| ≥ 3.
Solution.
This inequality is equivalent to the following set:
U [-1;1] U
Example 2.
Solve the inequality ||x+2| – 3| ≤ 2.
Solution.
This inequality is equivalent to the following system.
(|x + 2| – 3 ≥ -2
(|x + 2| – 3 ≤ 2,
(|x + 2| ≥ 1
(|x + 2| ≤ 5.
Let us solve separately the first inequality of the system. It is equivalent to the following set:
U[-1; 3].
2) Solving inequalities using the definition of the modulus.
Let me remind you first module definition.
|a| = a if a ≥ 0 and |a| = -a if a< 0.
For example, |34| = 34, |-21| = -(-21) = 21.
Example 1.
Solve the inequality 3|x – 1| ≤ x+3.
Solution.
Using the module definition we get two systems:
(x – 1 ≥ 0
(3(x – 1) ≤ x + 3
(x – 1< 0
(-3(x – 1) ≤ x + 3.
Solving the first and second systems separately, we obtain:
(x ≥ 1
(x ≤ 3,
(x< 1
(x ≥ 0.
The solution to the original inequality will be all solutions of the first system and all solutions of the second system.
Answer: x € .
3) Solving inequalities by squaring.
Example 1.
Solve the inequality |x 2 – 1|< | x 2 – x + 1|.
Solution.
Let's square both sides of the inequality. Let me note that it is possible to square both sides of the inequality only if they are both positive. IN in this case We have modules on both the left and right, so we can do this.
(|x 2 – 1|) 2< (|x 2 – x + 1|) 2 .
Now let's use the following property modulus: (|x|) 2 = x 2 .
(x 2 – 1) 2< (x 2 – x + 1) 2 ,
(x 2 – 1) 2 – (x 2 – x + 1) 2< 0.
(x 2 – 1 – x 2 + x – 1)(x 2 – 1 + x 2 – x + 1)< 0,
(x – 2)(2x 2 – x)< 0,
x(x – 2)(2x – 1)< 0.
We solve using the interval method.
Answer: x € (-∞; 0) U (1/2; 2)
4) Solving inequalities by changing variables.
Example.
Solve the inequality (2x + 3) 2 – |2x + 3| ≤ 30.
Solution.
Note that (2x + 3) 2 = (|2x + 3|) 2 . Then we get the inequality
(|2x + 3|) 2 – |2x + 3| ≤ 30.
Let's make the change y = |2x + 3|.
Let's rewrite our inequality taking into account the replacement.
y 2 – y ≤ 30,
y 2 – y – 30 ≤ 0.
Let's decompose quadratic trinomial, standing on the left, into factors.
y1 = (1 + 11) / 2,
y2 = (1 – 11) / 2,
(y – 6)(y + 5) ≤ 0.
Let's solve using the interval method and get:
Let's go back to the replacement:
5 ≤ |2x + 3| ≤ 6.
This double inequality is equivalent to the system of inequalities:
(|2x + 3| ≤ 6
(|2x + 3| ≥ -5.
Let's solve each of the inequalities separately.
The first is equivalent to the system
(2x + 3 ≤ 6
(2x + 3 ≥ -6.
Let's solve it.
(x ≤ 1.5
(x ≥ -4.5.
The second inequality obviously holds for all x, since the modulus is, by definition, a positive number. Since the solution to the system is all x that simultaneously satisfy both the first and second inequality of the system, then the solution to the original system will be the solution to its first double inequality (after all, the second is true for all x).
Answer: x € [-4.5; 1.5].
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