Rational Inequality Calculator. inequalities

And today not everyone can solve rational inequalities. More precisely, not only everyone can decide. Few people can do it.
Klitschko

This lesson is going to be tough. So tough that only the Chosen will reach the end of it. Therefore, before reading, I recommend removing women, cats, pregnant children and ...

Okay, it's actually quite simple. Suppose you have mastered the interval method (if you have not mastered it, I recommend you go back and read it) and learned how to solve inequalities of the form $P\left(x \right) \gt 0$, where $P\left(x \right)$ is some polynomial or product of polynomials.

I believe that it will not be difficult for you to solve, for example, such a game (by the way, try it for a warm-up):

\[\begin(align) & \left(2((x)^(2))+3x+4 \right)\left(4x+25 \right) \gt 0; \\ & x\left(2((x)^(2))-3x-20 \right)\left(x-1 \right)\ge 0; \\ & \left(8x-((x)^(4)) \right)((\left(x-5 \right))^(6))\le 0. \\ \end(align)\]

Now let's complicate the task a little and consider not just polynomials, but the so-called rational fractions of the form:

where $P\left(x \right)$ and $Q\left(x \right)$ are the same polynomials of the form $((a)_(n))((x)^(n))+(( a)_(n-1))((x)^(n-1))+...+((a)_(0))$, or the product of such polynomials.

This will be a rational inequality. The fundamental point is the presence of the variable $x$ in the denominator. For example, here are rational inequalities:

\[\begin(align) & \frac(x-3)(x+7) \lt 0; \\ & \frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0; \\ & \frac(3((x)^(2))+10x+3)(((\left(3-x \right))^(2))\left(4-((x)^( 2)) \right))\ge 0. \\ \end(align)\]

And this is not a rational, but the most common inequality, which is solved by the interval method:

\[\frac(((x)^(2))+6x+9)(5)\ge 0\]

Looking ahead, I’ll say right away: there are at least two ways to solve rational inequalities, but all of them in one way or another are reduced to the method of intervals already known to us. Therefore, before analyzing these methods, let's recall the old facts, otherwise there will be no sense from the new material.

What you already need to know

There are not many important facts. We really only need four.

Abbreviated multiplication formulas

Yes, yes: they will haunt us throughout the school math curriculum. And at the university too. There are quite a few of these formulas, but we only need the following:

\[\begin(align) & ((a)^(2))\pm 2ab+((b)^(2))=((\left(a\pm b \right))^(2)); \\ & ((a)^(2))-((b)^(2))=\left(a-b \right)\left(a+b \right); \\ & ((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b) ^(2))\right); \\ & ((a)^(3))-((b)^(3))=\left(a-b \right)\left(((a)^(2))+ab+((b)^( 2))\right). \\ \end(align)\]

Pay attention to the last two formulas - this is the sum and difference of cubes (and not the cube of the sum or difference!). They are easy to remember if you notice that the sign in the first bracket is the same as the sign in the original expression, and in the second bracket it is opposite to the sign in the original expression.

Linear equations

These are the simplest equations of the form $ax+b=0$, where $a$ and $b$ are ordinary numbers, and $a\ne 0$. This equation is easy to solve:

\[\begin(align) & ax+b=0; \\ &ax=-b; \\ & x=-\frac(b)(a). \\ \end(align)\]

I note that we have the right to divide by the coefficient $a$, because $a\ne 0$. This requirement is quite logical, since with $a=0$ we get this:

First, there is no $x$ variable in this equation. This, generally speaking, should not confuse us (this happens, say, in geometry, and quite often), but still we are no longer a linear equation.

Secondly, the solution of this equation depends solely on the coefficient $b$. If $b$ is also zero, then our equation is $0=0$. This equality is always true; hence $x$ is any number (usually written as $x\in \mathbb(R)$). If the coefficient $b$ is not equal to zero, then the equality $b=0$ is never satisfied, i.e. no answers (written $x\in \varnothing $ and read "solution set is empty").

To avoid all these complexities, we simply assume $a\ne 0$, which does not in any way restrict us from further reflections.

Quadratic equations

Let me remind you that this is called a quadratic equation:

Here on the left is a polynomial of the second degree, and again $a\ne 0$ (otherwise, instead of a quadratic equation, we get a linear one). The following equations are solved through the discriminant:

1. If $D \gt 0$, we get two different roots;
2. If $D=0$, then the root will be one, but of the second multiplicity (what kind of multiplicity it is and how to take it into account - more on that later). Or we can say that the equation has two identical roots;
3. For $D \lt 0$ there are no roots at all, and the sign of the polynomial $a((x)^(2))+bx+c$ for any $x$ coincides with the sign of the coefficient $a$. This, by the way, is a very useful fact, which for some reason is forgotten to be told in algebra classes.

The roots themselves are calculated according to the well-known formula:

\[((x)_(1,2))=\frac(-b\pm \sqrt(D))(2a)\]

Hence, by the way, the restrictions on the discriminant. After all, the square root of a negative number does not exist. As for the roots, many students have a terrible mess in their heads, so I specially recorded a whole lesson: what is a root in algebra and how to calculate it - I highly recommend reading it. :)

Operations with rational fractions

Everything that was written above, you already know if you studied the method of intervals. But what we will analyze now has no analogues in the past - this is a completely new fact.

Definition. A rational fraction is an expression of the form

\[\frac(P\left(x \right))(Q\left(x \right))\]

where $P\left(x \right)$ and $Q\left(x \right)$ are polynomials.

It is obvious that it is easy to obtain an inequality from such a fraction - it is enough just to attribute the sign “greater than” or “less than” to the right. And a little further we will find that solving such problems is a pleasure, everything is very simple there.

Problems begin when there are several such fractions in one expression. They have to be reduced to a common denominator - and it is at this moment that a large number of offensive mistakes are made.

Therefore, in order to successfully solve rational equations, it is necessary to firmly master two skills:

1. Factorization of the polynomial $P\left(x \right)$;
2. Actually, bringing fractions to a common denominator.

How to factorize a polynomial? Very simple. Let we have a polynomial of the form

Let's equate it to zero. We get the $n$-th degree equation:

\[((a)_(n))((x)^(n))+((a)_(n-1))((x)^(n-1))+...+(( a)_(1))x+((a)_(0))=0\]

Let's say we solved this equation and got the roots $((x)_(1)),\ ...,\ ((x)_(n))$ (don't worry: in most cases there will be no more than two of these roots) . In this case, our original polynomial can be rewritten like this:

\[\begin(align) & P\left(x \right)=((a)_(n))((x)^(n))+((a)_(n-1))((x )^(n-1))+...+((a)_(1))x+((a)_(0))= \\ & =((a)_(n))\left(x -((x)_(1)) \right)\cdot \left(x-((x)_(2)) \right)\cdot ...\cdot \left(x-((x)_( n)) \right) \end(align)\]

That's all! Please note: the leading coefficient $((a)_(n))$ has not disappeared anywhere - it will be a separate factor in front of the brackets, and if necessary, it can be inserted into any of these brackets (practice shows that with $((a)_ (n))\ne \pm 1$ there are almost always fractions among the roots).

A task. Simplify the expression:

\[\frac(((x)^(2))+x-20)(x-4)-\frac(2((x)^(2))-5x+3)(2x-3)-\ frac(4-8x-5((x)^(2)))(x+2)\]

Solution. First, let's look at the denominators: they are all linear binomials, and there is nothing to factorize here. So let's factorize the numerators:

\[\begin(align) & ((x)^(2))+x-20=\left(x+5 \right)\left(x-4 \right); \\ & 2((x)^(2))-5x+3=2\left(x-\frac(3)(2) \right)\left(x-1 \right)=\left(2x- 3\right)\left(x-1\right); \\ & 4-8x-5((x)^(2))=-5\left(x+2 \right)\left(x-\frac(2)(5) \right)=\left(x +2 \right)\left(2-5x \right). \\\end(align)\]

Please note: in the second polynomial, the senior coefficient "2", in full accordance with our scheme, first appeared in front of the bracket, and then was included in the first bracket, since a fraction got out there.

The same thing happened in the third polynomial, only there the order of the terms is also confused. However, the coefficient “−5” ended up being included in the second bracket (remember: you can enter a factor in one and only one bracket!), which saved us from the inconvenience associated with fractional roots.

As for the first polynomial, everything is simple there: its roots are sought either in the standard way through the discriminant, or using the Vieta theorem.

Let's go back to the original expression and rewrite it with the numerators decomposed into factors:

\[\begin(matrix) \frac(\left(x+5 \right)\left(x-4 \right))(x-4)-\frac(\left(2x-3 \right)\left( x-1 \right))(2x-3)-\frac(\left(x+2 \right)\left(2-5x \right))(x+2)= \\ =\left(x+5 \right)-\left(x-1 \right)-\left(2-5x \right)= \\ =x+5-x+1-2+5x= \\ =5x+4. \\ \end(matrix)\]

Answer: $5x+4$.

As you can see, nothing complicated. A bit of 7th-8th grade math and that's it. The point of all transformations is to turn a complex and scary expression into something simple and easy to work with.

However, this will not always be the case. So now we will consider a more serious problem.

But first, let's figure out how to bring two fractions to a common denominator. The algorithm is extremely simple:

1. Factorize both denominators;
2. Consider the first denominator and add to it the factors present in the second denominator, but not in the first. The resulting product will be the common denominator;
3. Find out what factors each of the original fractions lacks so that the denominators become equal to the common one.

Perhaps this algorithm will seem to you just a text in which there are “a lot of letters”. So let's take a look at a specific example.

A task. Simplify the expression:

\[\left(\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3) )-8)-\frac(1)(x-2) \right)\cdot \left(\frac(((x)^(2)))(((x)^(2))-4)- \frac(2)(2-x) \right)\]

Solution. Such voluminous tasks are best solved in parts. Let's write out what is in the first bracket:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3))-8 )-\frac(1)(x-2)\]

Unlike the previous problem, here the denominators are not so simple. Let's factorize each of them.

The square trinomial $((x)^(2))+2x+4$ cannot be factorized because the equation $((x)^(2))+2x+4=0$ has no roots (the discriminant is negative). We leave it unchanged.

The second denominator, the cubic polynomial $((x)^(3))-8$, upon closer examination is the difference of cubes and can be easily decomposed using the abbreviated multiplication formulas:

\[((x)^(3))-8=((x)^(3))-((2)^(3))=\left(x-2 \right)\left(((x) ^(2))+2x+4 \right)\]

Nothing else can be factored, since the first bracket contains a linear binomial, and the second one is a construction already familiar to us, which has no real roots.

Finally, the third denominator is a linear binomial that cannot be decomposed. Thus, our equation will take the form:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))-\frac(1)(x-2)\]

It is quite obvious that $\left(x-2 \right)\left(((x)^(2))+2x+4 \right)$ will be the common denominator, and to reduce all fractions to it, you need to multiply the first fraction to $\left(x-2 \right)$, and the last one to $\left(((x)^(2))+2x+4 \right)$. Then it remains only to bring the following:

\[\begin(matrix) \frac(x\cdot \left(x-2 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \ right))+\frac(((x)^(2))+8)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))- \frac(1\cdot \left(((x)^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x +4 \right))= \\ =\frac(x\cdot \left(x-2 \right)+\left(((x)^(2))+8 \right)-\left(((x )^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \\ =\frac (((x)^(2))-2x+((x)^(2))+8-((x)^(2))-2x-4)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))= \\ =\frac(((x)^(2))-4x+4)(\left(x-2 \right)\ left(((x)^(2))+2x+4 \right)). \\ \end(matrix)\]

Pay attention to the second line: when the denominator is already common, i.e. instead of three separate fractions, we wrote one large one, you should not immediately get rid of the brackets. It is better to write an extra line and note that, say, there was a minus before the third fraction - and it will not go anywhere, but will “hang” in the numerator in front of the bracket. This will save you a lot of mistakes.

Well, in the last line it is useful to factorize the numerator. Moreover, this is an exact square, and the abbreviated multiplication formulas again come to our aid. We have:

\[\frac(((x)^(2))-4x+4)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \frac(((\left(x-2 \right))^(2)))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right) )=\frac(x-2)(((x)^(2))+2x+4)\]

Now let's deal with the second bracket in the same way. Here I will simply write a chain of equalities:

\[\begin(matrix) \frac(((x)^(2)))(((x)^(2))-4)-\frac(2)(2-x)=\frac((( x)^(2)))(\left(x-2 \right)\left(x+2 \right))-\frac(2)(2-x)= \\ =\frac(((x) ^(2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2)(x-2)= \\ =\frac(((x)^( 2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2\cdot \left(x+2 \right))(\left(x-2 \right )\cdot \left(x+2 \right))= \\ =\frac(((x)^(2))+2\cdot \left(x+2 \right))(\left(x-2 \right)\left(x+2 \right))=\frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right) ). \\ \end(matrix)\]

We return to the original problem and look at the product:

\[\frac(x-2)(((x)^(2))+2x+4)\cdot \frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right))=\frac(1)(x+2)\]

Answer: \[\frac(1)(x+2)\].

The meaning of this problem is the same as the previous one: to show how much rational expressions can be simplified if you approach their transformation wisely.

And now, when you know all this, let's move on to the main topic of today's lesson - solving fractional rational inequalities. Moreover, after such preparation, the inequalities themselves will click like nuts. :)

The main way to solve rational inequalities

There are at least two approaches to solving rational inequalities. Now we will consider one of them - the one that is generally accepted in the school mathematics course.

But first, let's note an important detail. All inequalities are divided into two types:

1. Strict: $f\left(x \right) \gt 0$ or $f\left(x \right) \lt 0$;
2. Nonstrict: $f\left(x \right)\ge 0$ or $f\left(x \right)\le 0$.

Inequalities of the second type are easily reduced to the first, as well as the equation:

This small "addition" $f\left(x \right)=0$ leads to such an unpleasant thing as filled points - we met them back in the interval method. Otherwise, there are no differences between strict and non-strict inequalities, so let's analyze the universal algorithm:

1. Collect all non-zero elements on one side of the inequality sign. For example, on the left;
2. Bring all fractions to a common denominator (if there are several such fractions), bring similar ones. Then, if possible, factorize into the numerator and denominator. One way or another, we get an inequality of the form $\frac(P\left(x \right))(Q\left(x \right))\vee 0$, where the tick is the inequality sign.
3. Equate the numerator to zero: $P\left(x \right)=0$. We solve this equation and get the roots $((x)_(1))$, $((x)_(2))$, $((x)_(3))$, ... Then we require that the denominator was not equal to zero: $Q\left(x \right)\ne 0$. Of course, in essence, we have to solve the equation $Q\left(x \right)=0$, and we get the roots $x_(1)^(*)$, $x_(2)^(*)$, $x_(3 )^(*)$, ... (in real problems there will hardly be more than three such roots).
4. We mark all these roots (both with and without asterisks) on a single number line, and the roots without stars are painted over, and those with stars are punched out.
5. We place the plus and minus signs, select the intervals that we need. If the inequality has the form $f\left(x \right) \gt 0$, then the answer will be the intervals marked with a "plus". If $f\left(x \right) \lt 0$, then we look at intervals with "minuses".

Practice shows that points 2 and 4 cause the greatest difficulties - competent transformations and the correct arrangement of numbers in ascending order. Well, at the last step, be extremely careful: we always place signs based on the last inequality written before moving on to the equations. This is a universal rule inherited from the interval method.

So, there is a scheme. Let's practice.

A task. Solve the inequality:

\[\frac(x-3)(x+7) \lt 0\]

Solution. We have a strict inequality of the form $f\left(x \right) \lt 0$. Obviously, points 1 and 2 from our scheme have already been completed: all the elements of inequality are collected on the left, nothing needs to be reduced to a common denominator. So let's move on to the third point.

Set the numerator to zero:

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

And the denominator:

\[\begin(align) & x+7=0; \\ & ((x)^(*))=-7. \\ \end(align)\]

In this place, many people get stuck, because in theory you need to write down $x+7\ne 0$, as required by the ODZ (you can’t divide by zero, that’s all). But after all, in the future we will poke out the points that came from the denominator, so you should not complicate your calculations once again - write an equal sign everywhere and don’t worry. No one will deduct points for this. :)

Fourth point. We mark the obtained roots on the number line:

All points are punctured because the inequality is strict

Note: all points are punctured because the original inequality is strict. And here it doesn’t matter anymore: these points came from the numerator or from the denominator.

Well, look at the signs. Take any number $((x)_(0)) \gt 3$. For example, $((x)_(0))=100$ (but you could have just as well taken $((x)_(0))=3.1$ or $((x)_(0)) =1\000\000$). We get:

So, to the right of all the roots we have a positive area. And when passing through each root, the sign changes (this will not always be the case, but more on that later). Therefore, we proceed to the fifth point: we place the signs and choose the right one:

We return to the last inequality, which was before solving the equations. Actually, it coincides with the original one, because we did not perform any transformations in this task.

Since it is necessary to solve an inequality of the form $f\left(x \right) \lt 0$, I shaded the interval $x\in \left(-7;3 \right)$ - it is the only one marked with a minus sign. This is the answer.

Answer: $x\in \left(-7;3 \right)$

That's all! Is it difficult? No, it's not difficult. Indeed, it was an easy task. Now let's complicate the mission a little and consider a more "fancy" inequality. When solving it, I will no longer give such detailed calculations - I will simply outline the key points. In general, we will arrange it the way we would have done it on an independent work or exam. :)

A task. Solve the inequality:

\[\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\ge 0$. All non-zero elements are collected on the left, different denominators no. Let's move on to equations.

Numerator:

\[\begin(align) & \left(7x+1 \right)\left(11x+2 \right)=0 \\ & 7x+1=0\Rightarrow ((x)_(1))=-\ frac(1)(7); \\ & 11x+2=0\Rightarrow ((x)_(2))=-\frac(2)(11). \\ \end(align)\]

Denominator:

\[\begin(align) & 13x-4=0; \\ & 13x=4; \\ & ((x)^(*))=\frac(4)(13). \\ \end(align)\]

I don’t know what kind of pervert made up this problem, but the roots didn’t turn out very well: it will be difficult to arrange them on a number line. And if everything is more or less clear with the root $((x)^(*))=(4)/(13)\;$ (this is the only positive number— it will be on the right), then $((x)_(1))=-(1)/(7)\;$ and $((x)_(2))=-(2)/(11)\ ;$ needs more research: which one is bigger?

You can find this out, for example:

\[((x)_(1))=-\frac(1)(7)=-\frac(2)(14) \gt -\frac(2)(11)=((x)_(2 ))\]

I hope there is no need to explain why the numeric fraction $-(2)/(14)\; \gt -(2)/(11)\;$? If necessary, I recommend remembering how to perform actions with fractions.

And we mark all three roots on the number line:

The points from the numerator are shaded, from the denominator they are cut out

We put up signs. For example, you can take $((x)_(0))=1$ and find out the sign at this point:

\[\begin(align) & f\left(x \right)=\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4); \\ & f\left(1 \right)=\frac(\left(7\cdot 1+1 \right)\left(11\cdot 1+2 \right))(13\cdot 1-4)=\ frac(8\cdot 13)(9) \gt 0. \\\end(align)\]

The last inequality before the equations was $f\left(x \right)\ge 0$, so we are interested in the plus sign.

We got two sets: one is an ordinary segment, and the other is an open ray on the number line.

Answer: $x\in \left[ -\frac(2)(11);-\frac(1)(7) \right]\bigcup \left(\frac(4)(13);+\infty \right )$

An important note about the numbers that we substitute to find out the sign on the rightmost interval. It is not necessary to substitute a number close to the rightmost root. You can take billions or even "plus-infinity" - in this case, the sign of the polynomial in the bracket, numerator or denominator is determined solely by the sign of the leading coefficient.

Let's take another look at the $f\left(x \right)$ function from the last inequality:

It contains three polynomials:

\[\begin(align) & ((P)_(1))\left(x \right)=7x+1; \\ & ((P)_(2))\left(x \right)=11x+2; \\ & Q\left(x\right)=13x-4. \end(align)\]

All of them are linear binomials, and all of them have positive coefficients (numbers 7, 11 and 13). Therefore, when substituting very large numbers, the polynomials themselves will also be positive. :)

This rule may seem overly complicated, but only at first, when we analyze very easy tasks. In serious inequalities, the "plus-infinity" substitution will allow us to figure out the signs much faster than the standard $((x)_(0))=100$.

We will face such challenges very soon. But first, let's look at an alternative way to solve fractional rational inequalities.

Alternative way

This technique was suggested to me by one of my students. I myself have never used it, but practice has shown that it is really more convenient for many students to solve inequalities in this way.

So, the original data is the same. We need to solve a fractional rational inequality:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\]

Let's think: why is the polynomial $Q\left(x \right)$ "worse" than the polynomial $P\left(x \right)$? Why do we have to consider separate groups of roots (with and without an asterisk), think about punched points, etc.? It's simple: a fraction has a domain of definition, according to which the fraction makes sense only when its denominator is different from zero.

Otherwise, there are no differences between the numerator and the denominator: we also equate it to zero, look for the roots, then mark them on the number line. So why not replace the fractional bar (in fact, the division sign) with the usual multiplication, and write all the requirements of the DHS as a separate inequality? For example, like this:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\Rightarrow \left\( \begin(align) & P\left(x \right)\cdot Q \left(x \right) \gt 0, \\ & Q\left(x \right)\ne 0. \\ \end(align) \right.\]

Please note: this approach will allow you to reduce the problem to the method of intervals, but it will not complicate the solution at all. After all, anyway, we will equate the polynomial $Q\left(x \right)$ to zero.

Let's see how it works on real tasks.

A task. Solve the inequality:

\[\frac(x+8)(x-11) \gt 0\]

Solution. So, let's move on to the interval method:

\[\frac(x+8)(x-11) \gt 0\Rightarrow \left\( \begin(align) & \left(x+8 \right)\left(x-11 \right) \gt 0 , \\ & x-11\ne 0. \\ \end(align) \right.\]

The first inequality is solved elementarily. Just set each parenthesis to zero:

\[\begin(align) & x+8=0\Rightarrow ((x)_(1))=-8; \\ & x-11=0\Rightarrow ((x)_(2))=11. \\ \end(align)\]

With the second inequality, everything is also simple:

We mark the points $((x)_(1))$ and $((x)_(2))$ on the real line. All of them are punctured because the inequality is strict:

The right point turned out to be punctured twice. This is fine.

Pay attention to the point $x=11$. It turns out that it is “twice gouged out”: on the one hand, we gouge it out because of the severity of inequality, on the other hand, because of the additional requirement of ODZ.

In any case, it will be just a punctured point. Therefore, we put signs for the inequality $\left(x+8 \right)\left(x-11 \right) \gt 0$ - the last one we saw before we started solving the equations:

We are interested in positive regions, since we are solving an inequality of the form $f\left(x \right) \gt 0$, and we will color them. It remains only to write down the answer.

Answer. $x\in \left(-\infty ;-8 \right)\bigcup \left(11;+\infty \right)$

Using this solution as an example, I would like to warn you against a common mistake among novice students. Namely: never open parentheses in inequalities! On the contrary, try to factor everything - this will simplify the solution and save you a lot of problems.

Now let's try something more difficult.

A task. Solve the inequality:

\[\frac(\left(2x-13 \right)\left(12x-9 \right))(15x+33)\le 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\le 0$, so here you need to carefully monitor the filled points.

Let's move on to the interval method:

\[\left\( \begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)\le 0, \\ & 15x+33\ ne 0. \\ \end(align) \right.\]

Let's move on to the equation:

\[\begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0 \\ & 2x-13=0\Rightarrow ((x )_(1))=6.5; \\ & 12x-9=0\Rightarrow ((x)_(2))=0.75; \\ & 15x+33=0\Rightarrow ((x)_(3))=-2,2. \\ \end(align)\]

We take into account the additional requirement:

We mark all the obtained roots on the number line:

If a point is both punched out and filled in at the same time, it is considered punched out.

Again, two points "overlap" each other - this is normal, it will always be so. It is only important to understand that a point marked both as punched out and filled in is actually a punched out point. Those. "Gouging" is a stronger action than "painting over".

This is absolutely logical, because by puncturing we mark points that affect the sign of the function, but do not themselves participate in the answer. And if at some point the number ceases to suit us (for example, it does not fall into the ODZ), we delete it from consideration until the very end of the task.

In general, stop philosophizing. We arrange the signs and paint over those intervals that are marked with a minus sign:

Answer. $x\in \left(-\infty ;-2,2 \right)\bigcup \left[ 0,75;6,5 \right]$.

And again I wanted to draw your attention to this equation:

\[\left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0\]

Once again: never open parentheses in such equations! You're only making it harder for yourself. Remember: the product is zero when at least one of the factors is zero. Consequently, this equation simply “falls apart” into several smaller ones, which we solved in the previous problem.

Taking into account the multiplicity of roots

From the previous problems, it is easy to see that it is the non-strict inequalities that are the most difficult, because in them you have to keep track of the filled points.

But there is an even greater evil in the world - these are multiple roots in inequalities. Here it is already necessary to follow not some filled points there - here the inequality sign may not suddenly change when passing through these same points.

We have not yet considered anything like this in this lesson (although a similar problem was often encountered in the interval method). So let's introduce a new definition:

Definition. The root of the equation $((\left(x-a \right))^(n))=0$ is equal to $x=a$ and is called the root of the $n$th multiplicity.

Actually, we are not particularly interested in the exact value of the multiplicity. The only important thing is whether this very number $n$ is even or odd. Because:

1. If $x=a$ is a root of even multiplicity, then the sign of the function does not change when passing through it;
2. And vice versa, if $x=a$ is a root of odd multiplicity, then the sign of the function will change.

A special case of a root of odd multiplicity are all the previous problems considered in this lesson: there the multiplicity is equal to one everywhere.

And further. Before we start solving problems, I would like to draw your attention to one subtlety that seems obvious to an experienced student, but drives many beginners into a stupor. Namely:

The multiplicity root $n$ occurs only when the entire expression is raised to this power: $((\left(x-a \right))^(n))$, and not $\left(((x)^( n))-a\right)$.

Once again: the bracket $((\left(x-a \right))^(n))$ gives us the root $x=a$ of multiplicity $n$, but the bracket $\left(((x)^(n)) -a \right)$ or, as often happens, $(a-((x)^(n)))$ gives us a root (or two roots, if $n$ is even) of the first multiplicity, no matter what is equal to $n$.

Compare:

\[((\left(x-3 \right))^(5))=0\Rightarrow x=3\left(5k \right)\]

Everything is clear here: the whole bracket was raised to the fifth power, so at the output we got the root of the fifth degree. And now:

\[\left(((x)^(2))-4 \right)=0\Rightarrow ((x)^(2))=4\Rightarrow x=\pm 2\]

We got two roots, but both of them have the first multiplicity. Or here's another one:

\[\left(((x)^(10))-1024 \right)=0\Rightarrow ((x)^(10))=1024\Rightarrow x=\pm 2\]

And do not be confused by the tenth degree. The main thing is that 10 is an even number, so we have two roots at the output, and both of them again have the first multiplicity.

In general, be careful: multiplicity occurs only when the degree applies to the entire bracket, not just the variable.

A task. Solve the inequality:

\[\frac(((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right))(((\left(x+7 \right))^(5)))\ge 0\]

Solution. Let's try to solve it in an alternative way - through the transition from the particular to the product:

\[\left\( \begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ( (\left(x+7 \right))^(5))\ge 0, \\ & ((\left(x+7 \right))^(5))\ne 0. \\ \end(align )\right.\]

We deal with the first inequality using the interval method:

\[\begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ((\left( x+7 \right))^(5))=0; \\ & ((x)^(2))=0\Rightarrow x=0\left(2k \right); \\ & ((\left(6-x \right))^(3))=0\Rightarrow x=6\left(3k \right); \\ & x+4=0\Rightarrow x=-4; \\ & ((\left(x+7 \right))^(5))=0\Rightarrow x=-7\left(5k \right). \\ \end(align)\]

Additionally, we solve the second inequality. In fact, we have already solved it, but so that the reviewers do not find fault with the solution, it is better to solve it again:

\[((\left(x+7 \right))^(5))\ne 0\Rightarrow x\ne -7\]

Note that there are no multiplicities in the last inequality. Indeed: what difference does it make how many times to cross out the point $x=-7$ on the number line? At least once, at least five times - the result will be the same: a punctured point.

Let's note everything that we got on the number line:

As I said, the $x=-7$ point will eventually be punched out. The multiplicities are arranged based on the solution of the inequality by the interval method.

It remains to place the signs:

Since the point $x=0$ is a root of even multiplicity, the sign does not change when passing through it. The remaining points have an odd multiplicity, and everything is simple with them.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left[ -4;6 \right]$

Pay attention to $x=0$ again. Because of the even multiplicity, an interesting effect arises: everything to the left of it is painted over, to the right - too, and the point itself is completely painted over.

As a consequence, it does not need to be isolated when recording a response. Those. you don't have to write something like $x\in \left[ -4;0 \right]\bigcup \left[ 0;6 \right]$ (although formally such an answer would also be correct). Instead, we immediately write $x\in \left[ -4;6 \right]$.

Such effects are possible only for roots of even multiplicity. And in the next task, we will encounter the reverse "manifestation" of this effect. Ready?

A task. Solve the inequality:

\[\frac(((\left(x-3 \right))^(4))\left(x-4 \right))(((\left(x-1 \right))^(2)) \left(7x-10-((x)^(2)) \right))\ge 0\]

Solution. This time we will follow the standard scheme. Set the numerator to zero:

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

And the denominator:

\[\begin(align) & ((\left(x-1 \right))^(2))\left(7x-10-((x)^(2)) \right)=0; \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(1)^(*)=1\left(2k \right); \\ & 7x-10-((x)^(2))=0\Rightarrow x_(2)^(*)=5;\ x_(3)^(*)=2. \\ \end(align)\]

Since we are solving a non-strict inequality of the form $f\left(x \right)\ge 0$, the roots from the denominator (which have asterisks) will be cut out, and those from the numerator will be painted over.

We arrange the signs and stroke the areas marked with a "plus":

The point $x=3$ is isolated. This is part of the answer

Before writing down the final answer, take a close look at the picture:

1. The point $x=1$ has an even multiplicity, but is itself punctured. Therefore, it will have to be isolated in the answer: you need to write $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)$, and not $x\in \left(-\ infty ;2\right)$.
2. The point $x=3$ also has an even multiplicity and is shaded. The arrangement of signs indicates that the point itself suits us, but a step to the left and right - and we find ourselves in an area that definitely does not suit us. Such points are called isolated and are written as $x\in \left\( 3 \right\)$.

We combine all the obtained pieces into a common set and write down the answer.

Answer: $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;5 \right) $

Definition. Solving the inequality means find the set of all its solutions, or prove that this set is empty.

It would seem: what can be incomprehensible here? Yes, the fact of the matter is that sets can be specified in different ways. Let's rewrite the answer to the last problem:

We literally read what is written. The variable "x" belongs to a certain set, which is obtained by the union (symbol "U") of four separate sets:

• The interval $\left(-\infty ;1 \right)$, which literally means "all numbers less than one, but not one itself";
• The interval is $\left(1;2 \right)$, i.e. "all numbers between 1 and 2, but not the numbers 1 and 2 themselves";
• The set $\left\( 3 \right\)$, consisting of a single number - three;
• The interval $\left[ 4;5 \right)$ containing all numbers between 4 and 5, plus 4 itself, but not 5.

The third point is of interest here. Unlike intervals, which define infinite sets of numbers and only denote only the boundaries of these sets, the set $\left\( 3 \right\)$ defines exactly one number by enumeration.

To understand that we are listing the specific numbers included in the set (and not setting boundaries or anything else), curly braces are used. For example, the notation $\left\( 1;2 \right\)$ means exactly "a set consisting of two numbers: 1 and 2", but not a segment from 1 to 2. In no case do not confuse these concepts.

Multiplicity addition rule

Well, at the end of today's lesson, a little tin from Pavel Berdov. :)

Attentive students have probably already asked themselves the question: what will happen if the same roots are found in the numerator and denominator? So the following rule works:

Multiplicities of identical roots are added. Is always. Even if this root occurs in both the numerator and the denominator.

Sometimes it's better to decide than to talk. Therefore, we solve the following problem:

A task. Solve the inequality:

\[\frac(((x)^(2))+6x+8)(\left(((x)^(2))-16 \right)\left(((x)^(2))+ 9x+14 \right))\ge 0\]

\[\begin(align) & ((x)^(2))+6x+8=0 \\ & ((x)_(1))=-2;\ ((x)_(2))= -four. \\ \end(align)\]

So far, nothing special. Set the denominator to zero:

\[\begin(align) & \left(((x)^(2))-16 \right)\left(((x)^(2))+9x+14 \right)=0 \\ & ( (x)^(2))-16=0\Rightarrow x_(1)^(*)=4;\ x_(2)^(*)=-4; \\ & ((x)^(2))+9x+14=0\Rightarrow x_(3)^(*)=-7;\ x_(4)^(*)=-2. \\ \end(align)\]

Two identical roots are found: $((x)_(1))=-2$ and $x_(4)^(*)=-2$. Both have the first multiplicity. Therefore, we replace them with one root $x_(4)^(*)=-2$, but with a multiplicity of 1+1=2.

In addition, there are also identical roots: $((x)_(2))=-4$ and $x_(2)^(*)=-4$. They are also of the first multiplicity, so only $x_(2)^(*)=-4$ of multiplicity 1+1=2 remains.

Please note: in both cases, we left exactly the “cut out” root, and threw out the “painted over” one from consideration. Because even at the beginning of the lesson, we agreed: if a point is both punched out and painted over at the same time, then we still consider it punched out.

As a result, we have four roots, and all of them turned out to be gouged out:

\[\begin(align) & x_(1)^(*)=4; \\ & x_(2)^(*)=-4\left(2k \right); \\ & x_(3)^(*)=-7; \\ & x_(4)^(*)=-2\left(2k \right). \\ \end(align)\]

We mark them on the number line, taking into account the multiplicity:

We place the signs and paint over the areas of interest to us:

Everything. No isolated points and other perversions. You can write down the answer.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left(4;+\infty \right)$.

multiplication rule

Sometimes an even more unpleasant situation occurs: an equation that has multiple roots is itself raised to a certain power. This changes the multiplicities of all the original roots.

This is rare, so most students do not have experience in solving such problems. And the rule here is:

When an equation is raised to a power $n$, the multiplicity of all its roots also increases by a factor of $n$.

In other words, raising to a power results in multiplying multiplicities by the same power. Let's take this rule as an example:

A task. Solve the inequality:

\[\frac(x((\left(((x)^(2))-6x+9 \right))^(2))((\left(x-4 \right))^(5)) )(((\left(2-x \right))^(3))((\left(x-1 \right))^(2)))\le 0\]

Solution. Set the numerator to zero:

The product is equal to zero when at least one of the factors is equal to zero. Everything is clear with the first multiplier: $x=0$. And here's where the problems start:

\[\begin(align) & ((\left(((x)^(2))-6x+9 \right))^(2))=0; \\ & ((x)^(2))-6x+9=0\left(2k \right); \\ & D=((6)^(3))-4\cdot 9=0 \\ & ((x)_(2))=3\left(2k \right)\left(2k \right) \ \ & ((x)_(2))=3\left(4k \right) \\ \end(align)\]

As you can see, the equation $((x)^(2))-6x+9=0$ has a unique root of the second multiplicity: $x=3$. The whole equation is then squared. Therefore, the multiplicity of the root will be $2\cdot 2=4$, which we finally wrote down.

\[((\left(x-4 \right))^(5))=0\Rightarrow x=4\left(5k \right)\]

No problem with the denominator either:

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

In total, we got five points: two punched out and three filled in. There are no coinciding roots in the numerator and denominator, so we just mark them on the number line:

We arrange the signs taking into account the multiplicities and paint over the intervals of interest to us:

Again one isolated point and one punctured

Because of the roots of even multiplicity, we again received a couple of “non-standard” elements. This is $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)$, not $x\in \left[ 0;2 \right)$, and also an isolated point $ x\in \left\( 3 \right\)$.

Answer. $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;+\infty \right)$

As you can see, everything is not so difficult. The main thing is attentiveness. The last section of this lesson is devoted to transformations - the very ones that we discussed at the very beginning.

Preconversions

The inequalities we will discuss in this section are not complex. However, unlike the previous tasks, here you will have to apply skills from the theory of rational fractions - factorization and reduction to a common denominator.

We discussed this issue in detail at the very beginning of today's lesson. If you are not sure that you understand what it is about, I strongly recommend that you go back and repeat. Because there is no point in cramming the methods for solving inequalities if you "swim" in the conversion of fractions.

In homework, by the way, there will also be many similar tasks. They are placed in a separate subsection. And there you will find very non-trivial examples. But this will be in the homework, but now let's analyze a couple of such inequalities.

A task. Solve the inequality:

\[\frac(x)(x-1)\le \frac(x-2)(x)\]

Solution. Moving everything to the left:

\[\frac(x)(x-1)-\frac(x-2)(x)\le 0\]

We reduce to a common denominator, open the brackets, give like terms in the numerator:

\[\begin(align) & \frac(x\cdot x)(\left(x-1 \right)\cdot x)-\frac(\left(x-2 \right)\left(x-1 \ right))(x\cdot \left(x-1 \right))\le 0; \\ & \frac(((x)^(2))-\left(((x)^(2))-2x-x+2 \right))(x\left(x-1 \right)) \le0; \\ & \frac(((x)^(2))-((x)^(2))+3x-2)(x\left(x-1 \right))\le 0; \\ & \frac(3x-2)(x\left(x-1 \right))\le 0. \\\end(align)\]

Now we have a classical fractional rational inequality, the solution of which is no longer difficult. I propose to solve it by an alternative method - through the method of intervals:

\[\begin(align) & \left(3x-2 \right)\cdot x\cdot \left(x-1 \right)=0; \\ & ((x)_(1))=\frac(2)(3);\ ((x)_(2))=0;\ ((x)_(3))=1. \\ \end(align)\]

Don't forget the constraint that comes from the denominator:

We mark all the numbers and restrictions on the number line:

All roots have first multiplicity. No problem. We just place the signs and paint over the areas we need:

It's all. You can write down the answer.

Answer. $x\in \left(-\infty ;0 \right)\bigcup \left[ (2)/(3)\;;1 \right)$.

Of course, this was a very simple example. So now let's take a closer look at the problem. And by the way, the level of this task is quite consistent with independent and control work on this topic in the 8th grade.

A task. Solve the inequality:

\[\frac(1)(((x)^(2))+8x-9)\ge \frac(1)(3((x)^(2))-5x+2)\]

Solution. Moving everything to the left:

\[\frac(1)(((x)^(2))+8x-9)-\frac(1)(3((x)^(2))-5x+2)\ge 0\]

Before bringing both fractions to a common denominator, we decompose these denominators into factors. Suddenly the same brackets will come out? With the first denominator it's easy:

\[((x)^(2))+8x-9=\left(x-1 \right)\left(x+9 \right)\]

The second one is a little more difficult. Feel free to add a constant multiplier to the bracket where the fraction was found. Remember: the original polynomial had integer coefficients, so it is highly likely that the factorization will also have integer coefficients (in fact, it always will, except when the discriminant is irrational).

\[\begin(align) & 3((x)^(2))-5x+2=3\left(x-1 \right)\left(x-\frac(2)(3) \right)= \\ & =\left(x-1 \right)\left(3x-2 \right) \end(align)\]

As you can see, there is a common bracket: $\left(x-1 \right)$. We return to the inequality and bring both fractions to a common denominator:

\[\begin(align) & \frac(1)(\left(x-1 \right)\left(x+9 \right))-\frac(1)(\left(x-1 \right)\ left(3x-2\right))\ge 0; \\ & \frac(1\cdot \left(3x-2 \right)-1\cdot \left(x+9 \right))(\left(x-1 \right)\left(x+9 \right )\left(3x-2 \right))\ge 0; \\ & \frac(3x-2-x-9)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ & \frac(2x-11)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ \end(align)\]

Set the denominator to zero:

\[\begin(align) & \left(x-1 \right)\left(x+9 \right)\left(3x-2 \right)=0; \\ & x_(1)^(*)=1;\ x_(2)^(*)=-9;\ x_(3)^(*)=\frac(2)(3) \\ \end( align)\]

No multiplicities and no coinciding roots. We mark four numbers on a straight line:

We place the signs:

We write down the answer.

Answer: $x\in \left(-\infty ;-9 \right)\bigcup \left((2)/(3)\;;1 \right)\bigcup \left[ 5,5;+\infty \ right)$.

Everything! Like that, I read up to this line. :)

Solving inequalities online

Before solving inequalities, it is necessary to understand well how equations are solved.

It does not matter whether the inequality is strict () or non-strict (≤, ≥), the first step is to solve the equation by replacing the inequality sign with equality (=).

Explain what it means to solve an inequality?

After studying the equations, the student has the following picture in his head: you need to find such values ​​of the variable for which both parts of the equation take the same values. In other words, find all points where the equality holds. Everything is correct!

When talking about inequalities, they mean finding the intervals (segments) on which the inequality holds. If there are two variables in the inequality, then the solution will no longer be intervals, but some areas on the plane. Guess what will be the solution of the inequality in three variables?

How to solve inequalities?

The method of intervals (aka the method of intervals) is considered to be a universal way to solve inequalities, which consists in determining all the intervals within which the given inequality will be fulfilled.

Without going into the type of inequality, in this case it is not the essence, it is required to solve the corresponding equation and determine its roots, followed by designation of these solutions by numerical axis.

What is the correct way to write the solution to an inequality?

When you have determined the intervals for solving the inequality, you need to correctly write out the solution itself. There is an important nuance - are the boundaries of the intervals included in the solution?

Everything is simple here. If the solution of the equation satisfies the ODZ and the inequality is not strict, then the boundary of the interval is included in the solution of the inequality. Otherwise, no.

Considering each interval, the solution to the inequality can be the interval itself, or a half-interval (when one of its boundaries satisfies the inequality), or a segment - an interval together with its boundaries.

Important point

Do not think that only intervals, half-intervals and segments can be the solution to an inequality. No, individual points can also be included in the solution.

For example, the inequality |x|≤0 has only one solution - point 0.

And the inequality |x|

What is the inequality calculator for?

The inequality calculator gives the correct final answer. In this case, in most cases, an illustration of a numerical axis or plane is given. You can see whether the boundaries of the intervals are included in the solution or not - the points are displayed filled or pierced.

Thanks to online calculator inequalities, you can check whether you have found the roots of the equation correctly, marked them on the real axis and checked the fulfillment of the inequality condition on the intervals (and boundaries)?

If your answer differs from the answer of the calculator, then you definitely need to double-check your solution and identify the mistake made.

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

What "square inequality"? Not a question!) If you take any quadratic equation and change the sign in it "=" (equal) to any inequality icon ( > ≥ < ≤ ≠ ), we get a quadratic inequality. For example:

1. x2 -8x+12 ≥ 0

2. -x 2 +3x > 0

3. x2 ≤ 4

Well, you get the idea...)

I knowingly linked equations and inequalities here. The fact is that the first step in solving any square inequality - solve the equation from which this inequality is made. For this reason - the inability to solve quadratic equations automatically leads to a complete failure in inequalities. Is the hint clear?) If anything, look at how to solve any quadratic equations. Everything is detailed there. And in this lesson we will deal with inequalities.

The inequality ready for solution has the form: left - square trinomial ax 2 +bx+c, on the right - zero. The inequality sign can be absolutely anything. The first two examples are here are ready for a decision. The third example still needs to be prepared.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

For example, the expression \(x>5\) is an inequality.

Types of inequalities:

If \(a\) and \(b\) are numbers or , then the inequality is called numerical. In fact, this is just a comparison of two numbers. These inequalities are subdivided into faithful and unfaithful.

For example:
\(-5<2\) - верное числовое неравенство, ведь \(-5\) действительно меньше \(2\);

\(17+3\geq 115\) is an invalid numerical inequality because \(17+3=20\) and \(20\) is less than \(115\) (not greater than or equal to).

If \(a\) and \(b\) are expressions containing a variable, then we have inequality with variable. Such inequalities are divided into types depending on the content:

\(2x+1\geq4(5-x)\)				Variable only to the first power
\(3x^2-x+5>0\)				There is a variable in the second power (square), but no higher powers (third, fourth, etc.)
\(\log_(4)((x+1))<3\)
\(2^(x)\leq8^(5x-2)\)

... and so on.

What is a solution to an inequality?

If any number is substituted into the inequality instead of a variable, then it will turn into a numeric one.

If the given value for x makes the original inequality true numerical, then it is called solving the inequality. If not, then this value is not a solution. And to solve inequality- you need to find all its solutions (or show that they do not exist).

For example, if we are in the linear inequality \(x+6>10\), we substitute the number \(7\) instead of x, we get the correct numerical inequality: \(13>10\). And if we substitute \(2\), there will be an incorrect numerical inequality \(8>10\). That is, \(7\) is a solution to the original inequality, but \(2\) is not.

However, the inequality \(x+6>10\) has other solutions. Indeed, we will get the correct numerical inequalities when substituting and \(5\), and \(12\), and \(138\) ... And how can we find all possible solutions? To do this, use For our case, we have:

\(x+6>10\) \(|-6\)
\(x>4\)

That is, we can use any number greater than four. Now we need to write down the answer. Solutions to inequalities, as a rule, are written numerically, additionally marking them on the numerical axis with hatching. For our case we have:

Answer: \(x\in(4;+\infty)\)

When does the sign change in an inequality?

There is one big trap in inequalities, which students really “like” to fall into:

When multiplying (or dividing) inequality by a negative number, it is reversed (“greater than” by “less”, “greater than or equal to” by “less than or equal to”, and so on)

Why is this happening? To understand this, let's look at the transformations numerical inequality\(3>1\). It is correct, the triple is really more than one. First, let's try to multiply it by any positive number, for example, two:

\(3>1\) \(|\cdot2\)
\(6>2\)

As you can see, after multiplication, the inequality remains true. And no matter what positive number we multiply, we will always get the correct inequality. And now let's try to multiply by a negative number, for example, minus three:

\(3>1\) \(|\cdot(-3)\)
\(-9>-3\)

It turned out to be an incorrect inequality, because minus nine is less than minus three! That is, in order for the inequality to become true (which means that the transformation of multiplication by a negative was “legal”), you need to flip the comparison sign, like this: \(−9<− 3\).
With division, it will turn out similarly, you can check it yourself.

The rule written above applies to all types of inequalities, and not just to numerical ones.

Example: Solve the inequality \(2(x+1)-1<7+8x\)
Solution:

\(2x+2-1<7+8x\)	Let's move \(8x\) to the left, and \(2\) and \(-1\) to the right, not forgetting to change signs
\(2x-8x<7-2+1\)
\(-6x<6\) \(|:(-6)\)	Divide both sides of the inequality by \(-6\), not forgetting to change from "less" to "greater"
	Let's mark a numerical interval on the axis. Inequality, so the value \(-1\) is “punched out” and we don’t take it in response
	Let's write the answer as an interval

Answer: \(x\in(-1;\infty)\)

Inequalities and DHS

Inequalities, as well as equations, can have restrictions on , that is, on the values ​​of x. Accordingly, those values ​​that are unacceptable according to the ODZ should be excluded from the solution interval.

Example: Solve the inequality \(\sqrt(x+1)<3\)

Solution: It is clear that in order for the left side to be less than \(3\), the root expression must be less than \(9\) (after all, from \(9\) just \(3\)). We get:

\(x+1<9\) \(|-1\)
\(x<8\)

All? Any value of x less than \(8\) will suit us? Not! Because if we take, for example, the value \(-5\) that seems to fit the requirement, it will not be a solution to the original inequality, since it will lead us to calculating the root of a negative number.

\(\sqrt(-5+1)<3\)
\(\sqrt(-4)<3\)

Therefore, we must also take into account the restrictions on the values ​​​​of x - it cannot be such that there is a negative number under the root. Thus, we have the second requirement for x:

\(x+1\geq0\)
\(x\geq-1\)

And for x to be a final solution, it must satisfy both requirements at once: it must be less than \(8\) (to be a solution) and greater than \(-1\) (to be valid in principle). Plotting on the number line, we have the final answer:

Answer: \(\left[-1;8\right)\)

The form ax 2 + bx + 0 0, where (instead of the > sign, there can, of course, be any other inequality sign). We have all the facts of the theory necessary for solving such inequalities, which we will now verify.

Example 1. Solve the inequality:

a) x 2 - 2x - 3 > 0; b) x 2 - 2x - 3< 0;
c) x 2 - 2x - 3 > 0; d) x 2 - 2x - 3< 0.
Solution,

a) Consider the parabola y \u003d x 2 - 2x - 3 shown in fig. 117.

To solve the inequality x 2 - 2x - 3 > 0 - this means answering the question, for which values ​​of x the ordinates of the points of the parabola are positive.

We notice that y > 0, i.e., the graph of the function is located above the x-axis, at x< -1 или при х > 3.

Hence, the solutions of the inequality are all points of the open beam(- 00 , - 1), as well as all points of the open beam (3, +00).

Using the sign U (the sign of the union of sets), the answer can be written as follows: (-00 , - 1) U (3, +00). However, the answer can also be written like this:< - 1; х > 3.

b) Inequality x 2 - 2x - 3< 0, или у < 0, где у = х 2 - 2х - 3, также можно решить с помощью рис. 117: schedule located below the x-axis if -1< х < 3. Поэтому решениями данного неравенства служат все точки интервала (- 1, 3).

c) The inequality x 2 - 2x - 3 > 0 differs from the inequality x 2 - 2x - 3 > 0 in that the answer must also include the roots of the equation x 2 - 2x - 3 = 0, i.e. points x = -1

and x \u003d 3. Thus, the solutions of this non-strict inequality are all points of the beam (-00, - 1], as well as all points of the beam.

Practical mathematicians usually say this: why do we, solving the inequality ax 2 + bx + c > 0, carefully build a parabola graph of a quadratic function

y \u003d ax 2 + bx + c (as was done in example 1)? It is enough to make a schematic sketch of the graph, for which you only need to find roots square trinomial (the point of intersection of the parabola with the x-axis) and determine where the branches of the parabola are directed - up or down. This schematic sketch will give a visual interpretation of the solution of the inequality.

Example 2 Solve the inequality - 2x 2 + 3x + 9< 0.
Solution.

1) Find the roots of the square trinomial - 2x 2 + Zx + 9: x 1 \u003d 3; x 2 \u003d - 1.5.

2) The parabola, which serves as a graph of the function y \u003d -2x 2 + Zx + 9, intersects the x-axis at points 3 and - 1.5, and the branches of the parabola are directed downward, since the older coefficient- negative number - 2. In fig. 118 is a sketch of a graph.

3) Using fig. 118, we conclude:< 0 на тех промежутках оси х, где график расположен ниже оси х, т.е. на открытом луче (-оо, -1,5) или на открытом луче C, +оо).
Answer: x< -1,5; х > 3.

Example 3 Solve the inequality 4x 2 - 4x + 1< 0.
Solution.

1) From the equation 4x 2 - 4x + 1 = 0 we find.

2) The square trinomial has one root; this means that the parabola serving as the graph of a square trinomial does not intersect the x-axis, but touches it at the point. The branches of the parabola are directed upwards (Fig. 119.)

3) Using the geometric model shown in fig. 119, we establish that the specified inequality is satisfied only at the point, since for all other values ​​of x, the ordinates of the graph are positive.
Answer: .
You probably noticed that in fact, in examples 1, 2, 3, a well-defined algorithm solving quadratic inequalities, we will formalize it.

The algorithm for solving the quadratic inequality ax 2 + bx + 0 0 (ax 2 + bx + c< 0)

The first step of this algorithm is to find the roots of a square trinomial. But the roots may not exist, so what to do? Then the algorithm is inapplicable, which means that it is necessary to reason differently. The key to these arguments is given by the following theorems.

In other words, if D< 0, а >0, then the inequality ax 2 + bx + c > 0 is satisfied for all x; on the contrary, the inequality ax 2 + bx + c< 0 не имеет решений.
Proof. schedule functions y \u003d ax 2 + bx + c is a parabola whose branches are directed upwards (since a > 0) and which does not intersect the x axis, since the square trinomial has no roots by condition. The graph is shown in fig. 120. We see that for all x the graph is located above the x axis, which means that for all x the inequality ax 2 + bx + c > 0 is satisfied, which was what was required to be proved.

In other words, if D< 0, а < 0, то неравенство ах 2 + bх + с < 0 выполняется при всех х; напротив, неравенство ах 2 + bх + с >0 has no solutions.

Proof. The graph of the function y \u003d ax 2 + bx + c is a parabola, the branches of which are directed downwards (since a< 0) и которая не пересекает ось х, так как корней у квадратного трехчлена по условию нет. График представлен на рис. 121. Видим, что при всех х график расположен ниже оси х, а это значит, что при всех х выполняется неравенство ах 2 + bх + с < 0, что и требовалось доказать.

Example 4. Solve the inequality:

a) 2x 2 - x + 4 > 0; b) -x 2 + Zx - 8 > 0.

a) Find the discriminant of the square trinomial 2x 2 - x + 4. We have D \u003d (-1) 2 - 4 2 4 \u003d - 31< 0.
The senior coefficient of the trinomial (number 2) is positive.

Hence, by Theorem 1, for all x, the inequality 2x 2 - x + 4 > 0 is satisfied, i.e., the solution to the given inequality is the whole (-00, + 00).

b) Find the discriminant of the square trinomial - x 2 + Zx - 8. We have D \u003d Z2 - 4 (- 1) (- 8) \u003d - 23< 0. Старший коэффициент трехчлена (число - 1) отрицателен. Следовательно, по теореме 2, при всех х выполняется неравенство - х 2 + Зx - 8 < 0. Это значит, что неравенство - х 2 + Зх - 8 0 не выполняется ни при каком значении х, т. е. заданное неравенство не имеет решений.

Answer: a) (-00, + 00); b) there are no solutions.

In the following example, we will get acquainted with another way of reasoning, which is used in solving quadratic inequalities.

Example 5 Solve the inequality 3x 2 - 10x + 3< 0.
Solution. Let's decompose square trinomial 3x 2 - 10x + 3 for multipliers. The roots of the trinomial are the numbers 3 and, therefore, using ax 2 + bx + c \u003d a (x - x 1) (x - x 2), we get Zx 2 - 10x + 3 \u003d 3 (x - 3) (x - )
We note on the number line the roots of the trinomial: 3 and (Fig. 122).

Let x > 3; then x-3>0 and x->0, and hence the product 3(x - 3)(x - ) is positive. Next, let< х < 3; тогда x-3< 0, а х- >0. Therefore, the product 3(x-3)(x-) is negative. Finally, let x<; тогда x-3< 0 и x- < 0. Но в таком случае произведение
3(x -3)(x -) is positive.

Summing up the reasoning, we come to the conclusion: the signs of the square trinomial Zx 2 - 10x + 3 change as shown in Fig. 122. We are interested in for what x the square trinomial takes negative values. From fig. 122 we conclude: the square trinomial 3x 2 - 10x + 3 takes negative values ​​for any value of x from the interval (, 3)
Answer (, 3), or< х < 3.

Comment. The method of reasoning that we applied in Example 5 is usually called the method of intervals (or the method of intervals). It is actively used in mathematics to solve rational inequalities. In 9th grade, we will study the interval method in more detail.

Example 6. At what values ​​of the parameter p is the quadratic equation x 2 - 5x + p 2 \u003d 0:
a) has two different roots;

b) has one root;

c) has no -roots?

Solution. The number of roots of a quadratic equation depends on the sign of its discriminant D. In this case, we find D \u003d 25 - 4p 2.

a) A quadratic equation has two different roots, if D> 0, then the problem is reduced to solving the inequality 25 - 4p 2 > 0. We multiply both parts of this inequality by -1 (remembering to change the inequality sign). We obtain an equivalent inequality 4p 2 - 25< 0. Далее имеем 4 (р - 2,5) (р + 2,5) < 0.

The signs of the expression 4(p - 2.5) (p + 2.5) are shown in fig. 123.

We conclude that the inequality 4(p - 2.5)(p + 2.5)< 0 выполняется для всех значений р из интервала (-2,5; 2,5). Именно при этих значениях параметра р данное квадратное уравнение имеет два различных корня.

b) quadratic equation has one root if D is 0.
As we stated above, D = 0 at p = 2.5 or p = -2.5.

It is for these values ​​of the parameter p that this quadratic equation has only one root.

c) A quadratic equation has no roots if D< 0. Решим неравенство 25 - 4р 2 < 0.

We get 4p 2 - 25 > 0; 4 (p-2.5) (p + 2.5)> 0, whence (see Fig. 123) p< -2,5; р >2.5. For these values ​​of the parameter p, this quadratic equation has no roots.

Answer: a) at p (-2.5, 2.5);

b) at p = 2.5 or p = -2.5;
c) at r< - 2,5 или р > 2,5.

Mordkovich A. G., Algebra. Grade 8: Proc. for general education institutions. - 3rd ed., finalized. - M.: Mnemosyne, 2001. - 223 p.: ill.

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