How to solve a system using the interval method. Solving rational inequalities using the interval method

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You just need to understand this method and know it like the back of your hand! If only because it is used to solve rational inequalities and because, knowing this method properly, solving these inequalities is surprisingly simple. A little later I’ll tell you a couple of secrets on how to save time solving these inequalities. Well, are you intrigued? Then let's go!

The essence of the method is to factor the inequality into factors (repeat the topic) and determine the ODZ and the sign of the factors; now I’ll explain everything. Let's take the simplest example: .

There is no need to write the range of acceptable values () here, since there is no division by the variable, and there are no radicals (roots) observed here. Everything here is already factorized for us. But don’t relax, this is all to remind you of the basics and understand the essence!

Let's say you don't know the interval method, how would you solve this inequality? Approach logically and build on what you already know. Firstly, the left side will be greater than zero if both expressions in parentheses are either greater than zero or less than zero, because “plus” for “plus” gives “plus” and “minus” for “minus” gives “plus”, right? And if the signs of the expressions in brackets are different, then in the end the left side will be less than zero. What do we need to find out those values at which the expressions in brackets will be negative or positive?

We need to solve an equation, it is exactly the same as an inequality, only instead of a sign there will be a sign, the roots of this equation will allow us to determine those boundary values, when departing from which the factors will be greater or less than zero.

And now the intervals themselves. What is an interval? This is a certain interval of the number line, that is, all possible numbers contained between two numbers - the ends of the interval. It’s not so easy to imagine these intervals in your head, so it’s common to draw intervals, I’ll teach you now.

We draw an axis; the entire number series from and to is located on it. Points are plotted on the axis, the very so-called zeros of the function, the values at which the expression equals zero. These points are "pinned out" which means that they are not among those values at which the inequality is true. In this case, they are punctured because sign in the inequality and not, that is, strictly greater than and not greater than or equal to.

I want to say that it is not necessary to mark zero, it is here without circles, but just for understanding and orientation along the axis. Okay, we’ve drawn the axis, put the dots (more precisely, circles), what next, how will this help me in solving? - you ask. Now just take the value for x from the intervals in order and substitute them into your inequality and see what sign the multiplication results in.

In short, we just take for example, substitute it here, it will work out, which means that the inequality will be valid over the entire interval (over the entire interval) from to, from which we took it. In other words, if x is from to, then the inequality is true.

We do the same with the interval from to, take or, for example, substitute in, determine the sign, the sign will be “minus”. And we do the same with the last, third interval from to, where the sign turns out to be “plus”. There’s such a lot of text, but not enough clarity, right?

Take another look at inequality.

Now we also apply the signs that will be obtained as a result on the same axis. In my example, a broken line denotes the positive and negative sections of the axis.

Look at the inequality - at the drawing, again at the inequality - and again at the drawing, is anything clear? Now try to say on what intervals X, the inequality will be true. That's right, from to the inequality will also be true from to, but on the interval from to the inequality is zero and this interval is of little interest to us, because we have a sign in the inequality.

Well, now that you’ve figured it out, the only thing left to do is write down the answer! In response, we write those intervals for which the left side is greater than zero, which reads as X belongs to the interval from minus infinity to minus one and from two to plus infinity. It is worth clarifying that the parentheses mean that the values by which the interval is limited are not solutions to the inequality, that is, they are not included in the answer, but only indicate that up to, for example, is not a solution.

Now an example in which you will not only have to draw the interval:

What do you think needs to be done before putting points on the axis? Yeah, factor it into factors:

We draw intervals and place signs, notice that we have punctured dots because the sign is strictly less than zero:

It's time to tell you one secret that I promised at the beginning of this topic! What if I told you that you don’t have to substitute the values from each interval to determine the sign, but you can determine the sign in one of the intervals, and simply alternate the signs in the rest!

Thus, we saved a little time on putting down signs - I think this gained time on the Unified State Exam will not hurt!

We write the answer:

Now consider an example of a fractional-rational inequality - an inequality, both parts of which are rational expressions (see).

What can you say about this inequality? And you look at it as a fractional-rational equation, what do we do first? We immediately see that there are no roots, which means it’s definitely rational, but then it’s a fraction, and even with an unknown in the denominator!

That's right, we need ODZ!

So, let's go further, here all the factors except one have a variable of the first degree, but there is a factor where x has a second degree. Usually, our sign changed after passing through one of the points at which the left side of the inequality takes on a zero value, for which we determined what x should be equal to in each factor. But here, it’s always positive, because any number squared > zero and a positive term.

Do you think this will affect the meaning of inequality? That's right - it won't affect! We can safely divide the inequality into both parts and thereby remove this factor so that it is not an eyesore.

The time has come to draw the intervals; to do this, you need to determine those boundary values, when departing from which the multipliers will be greater and less than zero. But pay attention that there is a sign here, which means that we will not pick out the point at which the left side of the inequality takes on a zero value, it is included in the number of solutions, we have only one such point, this is the point where x is equal to one. Shall we color the point where the denominator is negative? - Of course not!

The denominator must not be zero, so the interval will look like this:

Using this diagram, you can easily write the answer, I’ll just say that now you have a new type of bracket at your disposal - square! Here's a bracket [ says that the value is included in the solution interval, i.e. is part of the answer, this bracket corresponds to a filled (not pinned) point on the axis.

So, did you get the same answer?

We factor it into factors and move everything to one side; after all, we only need to leave zero on the right to compare with it:

I draw your attention to the fact that in the last transformation, in order to obtain in the numerator as well as in the denominator, I multiply both sides of the inequality by. Remember that when both sides of an inequality are multiplied by, the sign of the inequality changes to the opposite!!!

We write ODZ:

Otherwise, the denominator will go to zero, and, as you remember, you cannot divide by zero!

Agree, the resulting inequality is tempting to reduce the numerator and denominator! This cannot be done; you may lose some of the decisions or ODZ!

Now try to put the points on the axis yourself. I will only note that when plotting points, you need to pay attention to the fact that a point with a value, which, based on the sign, would seem to be plotted on the axis as shaded, will not be shaded, it will be gouged out! Why do you ask? And remember the ODZ, you’re not going to divide by zero like that?

Remember, ODZ comes first! If all the inequalities and equal signs say one thing, and the ODZ says another, trust the ODZ, great and powerful!

Well, you built the intervals, I'm sure you took my hint about alternation and you got it like this (see picture below) Now cross it out and don't make that mistake again! What error? - you ask.

The fact is that in this inequality the factor was repeated twice (remember how you tried to reduce it?). So, if some factor is repeated in the inequality an even number of times, then when passing through a point on the axis that turns this factor to zero (in this case, a point), the sign will not change; if it is odd, then the sign changes!

The following axis with intervals and signs will be correct:

And, please note that the sign we are interested in is not the one that was at the beginning (when we first saw the inequality, the sign was there), after the transformations, the sign changed to, which means we are interested in intervals with a sign.

Answer:

I will also say that there are situations when there are roots of inequality that do not fall into any interval, in response they are written in curly brackets, like this, for example: . You can read more about such situations in the article average level.

Let's summarize how to solve inequalities using the interval method:
We move everything to the left side, leaving only zero on the right;
We find ODZ;
We take an arbitrary one from one of the intervals and determine the sign in the interval to which the root belongs, alternate the signs, paying attention to the roots that are repeated several times in the inequality; whether the sign changes when passing through them depends on the evenness or oddness of the number of times they are repeated or not;
In response, we write intervals, observing the punctured and non-punctured points (see ODZ), placing the necessary types of brackets between them.

And finally, our favorite section, “do it yourself”!

Examples:

Answers:

INTERVAL METHOD. AVERAGE LEVEL

Linear function

A function of the form is called linear. Let's take a function as an example. It is positive at and negative at. The point is the zero of the function (). Let's show the signs of this function on the number axis:

We say that “the function changes sign when passing through the point”.

It can be seen that the signs of the function correspond to the position of the function graph: if the graph is above the axis, the sign is “ ”, if below it is “ ”.

If we generalize the resulting rule to an arbitrary linear function, we obtain the following algorithm:

Finding the zero of the function;
We mark it on the number axis;
We determine the sign of the function on opposite sides of zero.

Quadratic function

I hope you remember how to solve quadratic inequalities? If not, read the topic. Let me remind you of the general form of a quadratic function: .

Now let's remember what signs the quadratic function takes. Its graph is a parabola, and the function takes the sign " " for those in which the parabola is above the axis, and " " - if the parabola is below the axis:

If a function has zeros (values at which), the parabola intersects the axis at two points - the roots of the corresponding quadratic equation. Thus, the axis is divided into three intervals, and the signs of the function alternately change when passing through each root.

Is it possible to somehow determine the signs without drawing a parabola every time?

Recall that a square trinomial can be factorized:

For example: .

Let's mark the roots on the axis:

We remember that the sign of a function can only change when passing through the root. Let's use this fact: for each of the three intervals into which the axis is divided by roots, it is enough to determine the sign of the function at only one arbitrarily chosen point: at the remaining points of the interval the sign will be the same.

In our example: at both expressions in brackets are positive (substitute, for example:). We put a “ ” sign on the axis:

Well, when (substitute, for example), both brackets are negative, which means the product is positive:

That's what it is interval method: knowing the signs of the factors on each interval, we determine the sign of the entire product.

Let's also consider cases when the function has no zeros, or only one.

If they are not there, then there are no roots. This means that there will be no “passing through the root”. This means that the function takes only one sign on the entire number line. It can be easily determined by substituting it into a function.

If there is only one root, the parabola touches the axis, so the sign of the function does not change when passing through the root. What rule can we come up with for such situations?

If you factor such a function, you get two identical factors:

And any squared expression is non-negative! Therefore, the sign of the function does not change. In such cases, we will highlight the root, when passing through which the sign does not change, by circling it with a square:

We will call such a root a multiple.

Interval method in inequalities

Now any quadratic inequality can be solved without drawing a parabola. It is enough just to place the signs of the quadratic function on the axis and select intervals depending on the sign of the inequality. For example:

Let's measure the roots on the axis and place the signs:

We need the part of the axis with the " " sign; since the inequality is not strict, the roots themselves are also included in the solution:

Now consider a rational inequality - an inequality, both sides of which are rational expressions (see).

Example:

All factors except one are “linear” here, that is, they contain a variable only to the first power. We need such linear factors to apply the interval method - the sign changes when passing through their roots. But the multiplier has no roots at all. This means that it is always positive (check this for yourself), and therefore does not affect the sign of the entire inequality. This means that we can divide the left and right sides of the inequality by it, and thus get rid of it:

Now everything is the same as it was with quadratic inequalities: we determine at which points each of the factors vanishes, mark these points on the axis and arrange the signs. I would like to draw your attention to a very important fact:

Answer: . Example: .

To apply the interval method, one of the parts of the inequality must have. Therefore, let's move the right side to the left:

The numerator and denominator have the same factor, but don’t rush to reduce it! After all, then we may forget to prick out this point. It is better to mark this root as a multiple, that is, when passing through it, the sign will not change:

Answer: .

And one more very illustrative example:

Again, we don't cancel the same factors of the numerator and denominator, because if we do, we'll have to specifically remember to puncture the dot.

: repeated times;
: times;
: times (in the numerator and one in the denominator).

In the case of an even number, we do the same as before: we draw a square around the point and do not change the sign when passing through the root. But in the case of an odd number, this rule does not apply: the sign will still change when passing through the root. Therefore, we do not do anything additional with such a root, as if it were not a multiple. The above rules apply to all even and odd powers.

What should we write in the answer?

If the alternation of signs is violated, you need to be very careful, because if the inequality is not strict, the answer should include all shaded points. But some of them often stand apart, that is, they are not included in the shaded area. In this case, we add them to the answer as isolated points (in curly braces):

Examples (decide for yourself):

Answers:

If among the factors it is simple, it is a root, because it can be represented as.
.

INTERVAL METHOD. BRIEFLY ABOUT THE MAIN THINGS

The interval method is used to solve rational inequalities. It consists in determining the sign of the product from the signs of the factors on various intervals.

Algorithm for solving rational inequalities using the interval method.

We move everything to the left side, leaving only zero on the right;
We find ODZ;
We plot all the roots of the inequality on the axis;
We take an arbitrary one from one of the intervals and determine the sign in the interval to which the root belongs, alternate the signs, paying attention to the roots that are repeated several times in the inequality; whether the sign changes when passing through them depends on the evenness or oddness of the number of times they are repeated or not;
In response, we write intervals, observing the punctured and non-punctured points (see ODZ), placing the necessary types of brackets between them.

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

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If you don't like our tasks, find others. Just don't stop at theory.

“Understood” and “I can solve” are completely different skills. You need both.

Find problems and solve them!

But today rational inequalities cannot solve everything. More precisely, not only everyone can decide. Few people can do this.
Klitschko

This lesson will be tough. So tough that only the Chosen will reach the end. Therefore, before starting reading, I recommend removing women, cats, pregnant children and... from screens.

Come on, it's actually simple. Let’s say you have mastered the interval method (if you haven’t mastered it, I recommend going back and reading 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 won’t be difficult for you to solve, for example, something like this (by the way, try it as 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 problem a little and consider not just polynomials, but 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, these 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 inequality, but the most common inequality, which can be 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, one way or another, come down to the method of intervals already known to us. Therefore, before we analyze these methods, let's remember the old facts, otherwise there will be no sense from the new material.

What you already need to know

There are never too many important facts. We really only need four.

Abbreviated multiplication formulas

Yes, yes: they will haunt us throughout the school mathematics 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 - these are 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 coincides with the sign in the original expression, and in the second one 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 can be solved simply:

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

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

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

Secondly, the solution to this equation depends solely on the coefficient $b$. If $b$ is also zero, then our equation has the form $0=0$. This equality is always true; this means $x$ is any number (usually written like this: $x\in \mathbb(R)$). If the coefficient $b$ is not equal to zero, then the equality $b=0$ is never satisfied, i.e. there are no answers (write $x\in \varnothing $ and read “the solution set is empty”).

To avoid all these difficulties, we simply assume $a\ne 0$, which does not at all limit us in further thinking.

Quadratic equations

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

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

If $D \gt 0$, we get two different roots;
If $D=0$, then there will be one root, but of the second multiplicity (what kind of multiplicity is this and how to take it into account - more on that later). Or we can say that the equation has two identical roots;
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 they forget to talk about in algebra lessons.

The roots themselves are calculated using 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. Many students have a terrible mess in their heads about roots, so I specially wrote down a whole lesson: what is a root in algebra and how to calculate it - I highly recommend reading it. :)

Operations with rational fractions

You already know everything that was written above if you have studied the interval method. 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.

Obviously, it’s easy to get an inequality from such a fraction—you just need to add the “greater than” or “less than” sign to the right. And a little further we will discover that solving such problems is a pleasure, everything is very simple.

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

Therefore, to successfully solve rational equations, you need to firmly grasp two skills:

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

How to factor a polynomial? Very simple. Let us have a polynomial of the form

We equate it to zero. We obtain an equation of $n$th degree:

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

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

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

That's all! Please note: the leading coefficient $((a)_(n))$ has not disappeared anywhere - it will be a separate multiplier 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).

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 factor here. So let's factor 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 leading coefficient “2”, in full accordance with our scheme, first appeared in front of the bracket, and then was included in the first bracket, since the fraction appeared there.

The same thing happened in the third polynomial, only there the order of the terms is also reversed. However, the coefficient “−5” ended up being included in the second bracket (remember: you can enter the 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: its roots are sought either standardly through the discriminant or using Vieta’s theorem.

Let's return to the original expression and rewrite it with the numerators factored:

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

Answer: $5x+4$.

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

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

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

Factor both denominators;
Consider the first denominator and add to it factors that are present in the second denominator, but not in the first. The resulting product will be the common denominator;
Find out what factors each of the original fractions is missing so that the denominators become equal to the common.

This algorithm may seem to you like just text with “a lot of letters.” Therefore, let’s look at everything using a specific example.

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. It is better to solve such large-scale problems in parts. Let's write down what's 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 factor each of them.

The square trinomial $((x)^(2))+2x+4$ cannot be factorized, since 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 careful examination is the difference of cubes and is easily expanded 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 factorized, since in the first bracket there is a linear binomial, and in the second there is a construction that is already familiar to us, which has no real roots.

Finally, the third denominator is a linear binomial that cannot be expanded. 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 the common denominator will be precisely $\left(x-2 \right)\left(((x)^(2))+2x+4 \right)$, and to reduce all fractions to it it is necessary to multiply the first fraction on $\left(x-2 \right)$, and the last one - on $\left(((x)^(2))+2x+4 \right)$. Then all that remains is to give similar ones:

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

Pay attention to the second line: when the denominator is already common, i.e. Instead of three separate fractions, we wrote one big one; you shouldn’t get rid of the parentheses right away. It’s better to write an extra line and note that, say, there was a minus before the third fraction - and it won’t go anywhere, but will “hang” in the numerator in front of the bracket. This will save you from a lot of mistakes.

Well, in the last line it’s useful to factor the numerator. Moreover, this is an exact square, and 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 exactly the same way. Here I’ll just write a chain of equalities:

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

Let's 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 task is the same as the previous one: to show how rational expressions can be simplified if you approach their transformation wisely.

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

The main way to solve rational inequalities

There are at least two approaches to solving rational inequalities. Now we will look at 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:

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

Inequalities of the second type can easily be 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 became familiar with them in the interval method. Otherwise, there are no differences between strict and non-strict inequalities, so let's look at the universal algorithm:

Collect all non-zero elements on one side of the inequality sign. For example, on the left;

Reduce all fractions to a common denominator (if there are several such fractions), bring similar ones. Then, if possible, factor the numerator and denominator. One way or another, we will get an inequality of the form $\frac(P\left(x \right))(Q\left(x \right))\vee 0$, where the “tick” is the inequality sign.

We 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).

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 punctured.

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 the intervals with “minuses”.

Practice shows that the greatest difficulties are caused by points 2 and 4 - 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 very 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.

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 diagram have already been fulfilled: all the elements of inequality are collected on the left, there is no need to bring anything to a common denominator. Therefore, let's move straight to the third point.

We equate the numerator to zero:

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

And the denominator:

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

This is where many people get stuck, because in theory you need to write $x+7\ne 0$, as required by the ODZ (you can’t divide by zero, that’s all). But in the future we will be pricking out the points that came from the denominator, so there is no need to complicate your calculations again - write an equal sign everywhere and don’t worry. Nobody will deduct points for this. :)

Fourth point. We mark the resulting roots on the number line:
All points are pinned out, since the inequality is strict
Note: all points are pinned out, since the original inequality is strict. And here it doesn’t matter whether these points came from the numerator or the denominator.

Well, let's look at the signs. Let's take any number $((x)_(0)) \gt 3$. For example, $((x)_(0))=100$ (but with the same success one could take $((x)_(0))=3.1$ or $((x)_(0)) =1\ 000\ 000$). We get:

So, to the right of all the roots we have a positive region. And when passing through each root, the sign changes (this will not always be the case, but more on that later). Therefore, let’s move on to the fifth point: arrange the signs and select the one you need:

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

Since we need 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. True, the task was easy. Now let’s complicate the mission a little and consider a more “sophisticated” inequality. When solving it, I will no longer give such detailed calculations - I will simply outline the key points. In general, we will format it the same way we would format it during independent work or an exam. :)

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, there are no different denominators. Let's move on to the 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 created this problem, but the roots didn’t turn out very well: it would be difficult to place them on the number line. And if with the root $((x)^(*))=(4)/(13)\;$ everything is more or less clear (this is the only positive number - it will be on the right), then $((x)_(1 ))=-(1)/(7)\;$ and $((x)_(2))=-(2)/(11)\;$ require additional research: which one is larger?

You can find this out, for example, like this:

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

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

And we mark all three roots on the number line:
The dots from the numerator are filled in, the dots from the denominator are punctured
We are putting 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 absolutely not necessary to substitute the number closest 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 look again at the function $f\left(x \right)$ from the last inequality:

Its notation 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 their leading coefficients (numbers 7, 11 and 13) are positive. Consequently, 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 problems. In serious inequalities, substituting “plus-infinity” will allow us to figure out the signs much faster than the standard $((x)_(0))=100$.

We will be faced with 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 many students really find it more convenient to solve inequalities this way.

So, the initial data is the same. We need to solve the 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 punctured points, etc.? It's simple: a fraction has a domain of definition, according to which the fraction makes sense only when its denominator is non-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 line (in fact, the division sign) with ordinary multiplication, and write down all the requirements of the ODZ in the form of 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 reduce the problem to the interval method, but will not complicate the solution at all. After all, we will still equate the polynomial $Q\left(x \right)$ to zero.

Let's see how this works on real problems.

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 can be solved in an elementary way. We simply equate each bracket to zero:

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

The second inequality is also simple:

Mark the points $((x)_(1))$ and $((x)_(2))$ on the number line. All of them are knocked out, since the inequality is strict:
The right point was gouged out twice. This is fine.
Pay attention to the point $x=11$. It turns out that it is “double-punctured”: on the one hand, we prick it out because of the severity of inequality, on the other hand, because of the additional requirement of DL.

In any case, it will just be a punctured point. Therefore, we arrange the 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 $f\left(x \right) \gt 0$ - we will shade them. All that remains is 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 beginning students. Namely: never open parentheses in inequalities! On the contrary, try to factor everything - this will simplify the solution and save you from many problems.

Now let's try something more complicated.

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 pay close attention to the shaded 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 go 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 resulting roots on the number line:
If a point is both punctured and filled in, it is considered to be punctured
Again, two points “overlap” each other - this is normal, it will always be like this. It is only important to understand that a point marked as both punctured and painted over is actually a punctured point. Those. “pricking” is a stronger action than “painting.”

This is absolutely logical, because by pinching 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 no longer suits us (for example, it does not fall into the ODZ), we cross it out from consideration until the very end of the task.

In general, stop philosophizing. We place 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 the brackets in such equations! You will only make things more difficult for yourself. Remember: the product is equal to zero when at least one of the factors is equal to 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 shaded points.

But there is an even greater evil in the world - these are multiple roots in inequalities. Here you no longer have to keep track of some shaded dots - here the inequality sign may not suddenly change when passing through these same dots.

We have not yet considered anything like this in this lesson (although a similar problem was often encountered in the interval method). Therefore, we 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 thing that matters is whether this same number $n$ is even or odd. Because:

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

All previous problems discussed in this lesson are a special case of a root of odd multiplicity: everywhere the multiplicity is equal to one.

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 root of multiplicity $n$ arises only in the case 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, regardless of what equals $n$.

Compare:

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

Everything is clear here: the entire bracket was raised to the fifth power, so at the output we got the root of the fifth power. 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 first multiplicity. Or here's another one:

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

And don't let the tenth degree bother you. The main thing is that 10 is an even number, so at the output we have two roots, and both of them again have the first multiple.

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

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 quotient 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.\]

Let's 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:

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

Please note: there are no multiplicities in the last inequality. In fact: what difference does it make how many times you 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 mark everything we got on the number line:

As I said, the point $x=-7$ will eventually be punctured. The multiplicities are arranged based on solving the inequality using the interval method.

All that remains is 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. $x\in \left(-\infty ;-7 \right)\bigcup \left[ -4;6 \right]$

Once again, pay attention to $x=0$. Due to the even multiplicity, an interesting effect arises: everything to the left of it is painted over, everything to the right is also painted over, and the point itself is completely painted over.

As a result, it does not need to be isolated when recording the answer. Those. there is no need 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 with roots of even multiplicity. And in the next problem we will encounter the reverse “manifestation” of this effect. Ready?

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. We equate 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 taken out, and those from the numerator will be shaded.

We place signs and shade the areas marked with a “plus”:
Point $x=3$ is isolated. This is part of the answer
Before writing down the final answer, let's take a close look at the picture:

The point $x=1$ has an even multiplicity, but is itself punctured. Consequently, 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 $x\in \left(-\ infty ;2 \right)$.

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 left or right - and we find ourselves in an area that definitely does not suit us. Such points are called isolated and are written in the form $x\in \left$ 3 \right$$.

We combine all the received 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 inequality means find the set of all its solutions, or prove that this set is empty.

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

We literally read what is written. The variable “x” belongs to a certain set, which is obtained by combining (the “U” sign) four separate sets:

Interval $\left(-\infty ;1 \right)$, which literally means “all numbers smaller than one, but not the unit itself”;
Interval $\left(1;2 \right)$, i.e. “all numbers in the range from 1 to 2, but not the numbers 1 and 2 themselves”;
The set $\left$ 3 \right$$, consisting of one single number - three;
The interval $\left[ 4;5 \right)$ containing all numbers in the range from 4 to 5, as well as the four itself, but not the five.

The third point is of interest here. Unlike intervals, which define infinite sets of numbers and only indicate the boundaries of these sets, the set $\left$ 3 \right$$ specifies strictly one number by enumeration.

To understand that we are listing 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. Do not confuse these concepts under any circumstances.

Rule for adding multiples

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

Attentive students have probably already wondered: what will happen if the numerator and denominator have the same roots? So, the following rule works:

The multiplicities of identical roots are added. 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:

Task. Solve the inequality:

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

\[\begin(align) & ((x)^(2))+6x+8=0 \\ & ((x)_(1))=-2;\ ((x)_(2))= -4. \\ \end(align)\]

Nothing special yet. We equate 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 were discovered: $((x)_(1))=-2$ 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 will remain.

Please note: in both cases, we left exactly the “punctured” root, and excluded the “painted” one from consideration. Because at the beginning of the lesson we agreed: if a point is both punctured and painted over, then we still consider it to be punctured.

As a result, we have four roots, and all of them were cut 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 signs and paint over the areas of interest to us:

All. No isolated points or other perversions. You can write down the answer.

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

Rule for multiplying multiples

Sometimes an even more unpleasant situation occurs: an equation that has multiple roots is itself raised to some power. In this case, the multiplicities of all original roots change.

This is rare, so most students have no experience solving such problems. And the rule here is:

When an equation is raised to the $n$ power, the multiplicities of all its roots also increase by $n$ times.

In other words, raising to a power leads to multiplying the multiples by the same power. Let's look at this rule using an example:

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. We equate the numerator to zero:

The product is zero when at least one of the factors is zero. Everything is clear with the first factor: $x=0$. But then the problems begin:

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

As we see, the equation $((x)^(2))-6x+9=0$ has a single root of the second multiplicity: $x=3$. This entire equation is then squared. Therefore, the multiplicity of the root will be $2\cdot 2=4$, which is what we eventually wrote down.

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

There are no problems 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 dots: two punctured and three painted. There are no coinciding roots in the numerator and denominator, so we simply mark them on the number line:

We arrange the signs taking into account multiplicities and paint over the intervals that interest us:
Again one isolated point and one punctured
Due to the roots of even multiplicity, we again got a couple of “non-standard” elements. This is $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)$, and 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 complicated. The main thing is attentiveness. The last section of this lesson is devoted to transformations - the same ones that we discussed at the very beginning.

Pre-conversions

The inequalities that we will examine in this section cannot be called complex. However, unlike previous problems, 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're not sure you understand what I'm talking about, I highly recommend going back and repeating it. Because there is no point in cramming methods for solving inequalities if you “float” in converting 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 homework, and now let's look at a couple of such inequalities.

Task. Solve the inequality:

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

Solution. Move everything to the left:

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

We reduce to a common denominator, open the brackets, and bring similar terms in the numerator:

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

Now we have before us a classical fractional-rational inequality, the solution of which is no longer difficult. I propose to solve it using 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 simply place signs and paint over the areas we need:

This is 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. Therefore, now let’s look at the problem more seriously. And by the way, the level of this task is quite consistent with independent and test work on this topic in 8th grade.

Task. Solve the inequality:

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

Solution. Move 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, let's factorize these denominators. What if the same brackets come out? With the first denominator it is 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 factor into the bracket where the fraction appears. Remember: the original polynomial had integer coefficients, so there is a good chance that the factorization will have integer coefficients (in fact, it always will, unless 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:

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

We equate 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 multiples or coinciding roots. We mark four numbers on the line:

We are placing signs:

We write down the answer.

Answer: $x\in \left(-\infty ;-9 \right)\bigcup \left((2)/(3)\;;1 \right)\bigcup \left[ 5.5;+\infty \ right)$.

Interval method(or as it is sometimes called the interval method) is a universal method for solving inequalities. It is suitable for solving a variety of inequalities, but is most convenient in solving rational inequalities with one variable. Therefore, in the school algebra course, the method of intervals is closely tied specifically to rational inequalities, and practically no attention is paid to solving other inequalities with its help.

In this article we will analyze the interval method in detail and touch on all the intricacies of solving inequalities with one variable using it. Let's start by presenting an algorithm for solving inequalities using the interval method. Next, we will explain what theoretical aspects it is based on and analyze the steps of the algorithm, in particular, we will dwell in detail on the determination of signs on intervals. After this, we will move on to practice and show solutions to several typical examples. And in conclusion, we will consider the interval method in general form (that is, without reference to rational inequalities), in other words, the generalized interval method.

Page navigation.

Algorithm

Acquaintance with the interval method in school begins with solving inequalities of the form f(x)<0 (знак неравенства может быть и другим ≤, >or ≥), where f(x) is either , represented as a product linear binomials with 1 for variable x and/or square trinomials with a leading coefficient of 1 and with a negative discriminant and their degrees, or the ratio of such polynomials. For clarity, we give examples of such inequalities: (x−5)·(x+5)≤0 , (x+3)·(x 2 −x+1)·(x+2) 3 ≥0, .

To make further conversation substantive, let’s immediately write down an algorithm for solving inequalities of the above type using the interval method, and then we’ll figure out what, how and why. So, using the interval method:

First, the zeros of the numerator and zeros of the denominator are found. To do this, the numerator and denominator of the expression on the left side of the inequality are equal to zero, and the resulting equations are solved.
After this, the points corresponding to the found zeros are marked with dashes. A schematic drawing is enough, in which it is not necessary to observe the scale, the main thing is to adhere to the location of the points relative to each other: the point with the smaller coordinate is located to the left of the point with the larger coordinate. After this, it becomes clear how they should be depicted: regular or punctured (with an empty center). When solving a strict inequality (with sign< или >) all points are depicted as punctured. When solving a non-strict inequality (with a sign ≤ or ≥), the points corresponding to the zeros of the denominator are punctured, and the remaining points marked with dashes are ordinary. These points divide the coordinate line into several numerical intervals.
Next, the signs of the expression f(x) are determined from the left side of the inequality being solved on each interval (we will describe in detail how this is done in one of the following paragraphs), and + or − are placed above them in accordance with the signs defined on them.
Finally, when solving the signed inequality< или ≤ изображается штриховка над промежутками, отмеченными знаком −, а при решении неравенства со знаком >or ≥ - over spaces marked with a + sign. The result is , which is the desired solution to the inequality.

Note that the above algorithm is consistent with the description of the interval method in school textbooks.

What is the method based on?

The approach underlying the interval method takes place due to the following property of a continuous function: if on the interval (a, b) the function f is continuous and does not vanish, then it retains a constant sign on this interval (we would add that a similar property this is also true for the number rays (−∞, a) and (a, +∞) ). And this property, in turn, follows from the Bolzano-Cauchy theorem (its consideration is beyond the scope of the school curriculum), the formulation and proof of which, if necessary, can be found, for example, in the book.

For expressions f(x) having the form indicated in the previous paragraph, the constancy of the sign on intervals can be justified in another way, starting from the properties of numerical inequalities and taking into account the rules for multiplying and dividing numbers with the same signs and different signs.

As an example, consider the inequality. The zeros of its numerator and denominator divide the number line into three intervals (−∞, −1), (−1, 5) and (5, +∞). Let us show that on the interval (−∞, −1) the expression on the left side of the inequality has a constant sign (we can take another interval, the reasoning will be similar). Let's take any number t from this interval. It will obviously satisfy the inequality t<−1 , и так как −1<5 , то по свойству транзитивности, оно же будет удовлетворять и неравенству t<5 . Из этих неравенств в силу свойств числовых неравенств следует, что t+1<0 и t−5<0. То есть, t+1 и t−5 – отрицательные числа, не зависимо от того, какое конкретно число t мы возьмем из промежутка (−∞, −1) . Тогда позволяет констатировать, что значение выражения будет положительным, откуда следует, что значение выражения будет положительным при любом значении x из промежутка (−∞, −1) . Итак, на указанном промежутке выражение имеет постоянный знак, причем, это знак +.

So we smoothly approached the issue of determining signs on intervals, but we will not skip over the first step of the interval method, which involves finding the zeros of the numerator and denominator.

How to find the zeros of the numerator and denominator?

Finding the zeros of the numerator and denominator of a fraction of the type indicated in the first paragraph usually does not pose any problems. For this, the expressions from the numerator and denominator are set equal to zero, and the resulting equations are solved. The principle of solving equations of this type is described in detail in the article solving equations by factorization method. Here we will just limit ourselves to an example.

Consider the fraction and find the zeros of its numerator and denominator. Let's start with the zeros of the numerator. We equate the numerator to zero, we obtain the equation x·(x−0.6)=0, from which we proceed to the set of two equations x=0 and x−0.6=0, from where we find two roots 0 and 0.6. These are the desired zeros of the numerator. Now we find the zeros of the denominator. Let's make an equation x 7 ·(x 2 +2·x+7) 2 ·(x+5) 3 =0, it is equivalent to a set of three equations x 7 =0, (x 2 +2 x+7) 2 =0, (x+5) 3 =0, and then x=0, x 2 +2 x+7=0 , x+5=0 . The root of the first of these equations is obvious, it is 0, the second equation has no roots, since its discriminant is negative, and the root of the third equation is −5. So, we found the zeros of the denominator, there were two of them: 0 and −5. Note that 0 turned out to be both a zero in the numerator and a zero in the denominator.

To find the zeros of the numerator and denominator in the general case, when the left side of the inequality is a fraction, but not necessarily rational, the numerator and denominator are also equated to zero, and the corresponding equations are solved.

How to determine signs at intervals?

The most reliable way to determine the sign of the expression on the left side of the inequality on each interval is to calculate the value of this expression at any one point in each interval. In this case, the desired sign on the interval coincides with the sign of the value of the expression at any point in this interval. Let's explain this with an example.

Let's take inequality . The expression on its left side has no zeros in the numerator, and the zero in the denominator is the number −3. It divides the number line into two intervals (−∞, −3) and (−3, +∞). Let's determine the signs on them. To do this, take one point from these intervals and calculate the values of the expression in them. Let us immediately note that it is advisable to take such points so that it is easy to carry out calculations. For example, from the first interval (−∞, −3) we can take −4. For x=−4 we have , received a value with a minus sign (negative), therefore, there will be a minus sign on this interval. We move on to determining the sign on the second interval (−3, +∞). It is convenient to take 0 from it (if 0 is included in the interval, then it is advisable to always take it, since at x=0 the calculations are the simplest). At x=0 we have . This value has a plus sign (positive), so there will be a plus sign on this interval.

There is another approach to determining signs, which consists of finding the sign at one of the intervals and maintaining it or changing it when moving to the adjacent interval through zero. You must adhere to the following rule. When passing through the zero of the numerator, but not the denominator, or through the zero of the denominator, but not the numerator, the sign changes if the degree of the expression giving this zero is odd, and does not change if it is even. And when passing through a point that is both the zero of the numerator and the zero of the denominator, the sign changes if the sum of the powers of the expressions giving this zero is odd, and does not change if it is even.

By the way, if the expression on the right side of the inequality has the form indicated at the beginning of the first paragraph of this article, then there will be a plus sign in the rightmost gap.

To make everything clear, let's look at an example.

Let there be inequality before us , and we solve it using the interval method. To do this, we find the zeros of the numerator 2, 3, 4 and the zeros of the denominator 1, 3, 4, mark them on the coordinate line first with dashes

then we replace the zeros of the denominator with images of punctured dots

and since we are solving a non-strict inequality, we replace the remaining dashes with ordinary dots

And then comes the moment of identifying signs at intervals. As we noticed before this example, on the rightmost interval (4, +∞) there will be a + sign:

Let's determine the remaining signs, while moving from gap to gap from right to left. Moving on to the next interval (3, 4), we pass through the point with coordinate 4. This is the zero of both the numerator and the denominator, these zeros give the expressions (x−4) 2 and x−4, the sum of their powers is 2+1=3, and this is an odd number, which means that when passing through this point you need to change the sign. Therefore, on the interval (3, 4) there will be a minus sign:

We go further to the interval (2, 3), while passing through the point with coordinate 3. This is also the zero of both the numerator and the denominator, it is given by the expressions (x−3) 3 and (x−3) 5, the sum of their powers is 3+5=8, and this is an even number, therefore, the sign will remain unchanged:

We move further to the interval (1, 2). The path to it is blocked by a point with coordinate 2. This is the zero of the numerator, it is given by the expression x−2, its degree is 1, that is, it is odd, therefore, when passing through this point, the sign will change:

Finally, it remains to determine the sign on the last interval (−∞, 1) . To get to it, we need to overcome the point with coordinate 1. This is the zero of the denominator, it is given by the expression (x−1) 4, its degree is 4, that is, it is even, therefore, the sign will not change when passing through this point. So we have identified all the signs, and the drawing takes on the following form:

It is clear that the use of the considered method is especially justified when calculating the value of an expression involves a large amount of work. For example, calculate the value of the expression at any point in the interval .

Examples of solving inequalities using the interval method

Now you can put together all the information presented, sufficient to solve inequalities using the interval method, and analyze the solutions of several examples.

Example.

Solve the inequality .

Solution.

Let us solve this inequality using the interval method. Obviously, the zeros of the numerator are 1 and −5, and the zeros of the denominator are 1. We mark them on the number line, with the points with coordinates and 1 pricked out as zeros of the denominator, and the remaining zero of the numerator −5 is represented by an ordinary point, since we are solving a non-strict inequality:

Now we put signs on the intervals, adhering to the rule of maintaining or changing the sign when passing through zeros. There will be a + sign above the rightmost gap (this can be checked by calculating the value of the expression on the left side of the inequality at some point in this gap, for example, at x=3). When passing through the sign we change, when passing through 1 we leave it the same, and when passing through −5 we again leave the sign unchanged:

Since we are solving the inequality with the ≤ sign, it remains to draw shading over the intervals marked with the sign − and write down the answer from the resulting image.

So, the solution we are looking for is: .

To be fair, let us draw attention to the fact that in the overwhelming majority of cases, when solving rational inequalities, they first have to be transformed to the required form in order to make it possible to solve them using the method of intervals. We will discuss in detail how to carry out such transformations in the article. solving rational inequalities, and now we will give an example illustrating one important point regarding square trinomials in the recording of inequalities.

Example.

Find the solution to the inequality .

Solution.

At first glance at this inequality, it seems that its form is suitable for applying the interval method. But it doesn’t hurt to check whether the discriminants of the quadratic trinomials in his notation are really negative. Let's figure them out to ease our conscience. For the trinomial x 2 +3 x+3 we have D=3 2 −4 1 3=−3<0 , а для трехчлена x 2 +2·x−8 получаем D’=1 2 −1·(−8)=9>0 . This means that transformations are required to give this inequality the desired form. In this case, it is enough to represent the trinomial x 2 +2 x−8 as (x+4) (x−2) , and then solve the inequality using the method of intervals .

Generalized interval method

The generalized interval method allows you to solve inequalities of the form f(x)<0 (≤, >, ≥), where f(x) is arbitrary with one variable x. Let's write it down algorithm for solving inequalities using the generalized interval method:

First you need f and the zeros of this function.

The boundary points, including individual points, of the domain of definition are marked on the number line. For example, if the domain of a function is the set (−5, 1]∪(3)∪ (we do not define the sign on the interval (−6, 4), since it is not part of the domain of definition of the function.) To do this, take one point from each interval, for example, 16 , 8 , 6 and −8, and calculate the value of the function f in them:

If you have questions about how it was found out what the calculated values of the function are, positive or negative, then study the material in the article comparison of numbers.

We place the just defined signs, and apply shading over the spaces with a minus sign:

In the answer we write the union of two intervals with the sign −, we have (−∞, −6]∪(7, 12). Note that −6 is included in the answer (the corresponding point is solid, not punctured). The fact is that this not the zero of the function (which, when solving a strict inequality, we would not include in the answer), but the boundary point of the domain of definition (it is colored, not black), and the value of the function at this point is negative (as evidenced by the minus sign). over the corresponding interval), that is, it satisfies the inequality. But 4 does not need to be included in the answer (as well as the entire interval ∪(7, 12) .

Bibliography.

Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., erased. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
Kudryavtsev L. D. Course of mathematical analysis (in two volumes): Textbook for university and college students. – M.: Higher. school, 1981, vol. 1. – 687 p., ill.

Interval method– a simple way to solve fractional rational inequalities. This is the name for inequalities containing rational (or fractional-rational) expressions that depend on a variable.

1. Consider, for example, the following inequality

The interval method allows you to solve it in a couple of minutes.

On the left side of this inequality is a fractional rational function. Rational because it does not contain roots, sines, or logarithms - only rational expressions. On the right is zero.

The interval method is based on the following property of a fractional rational function.

A fractional rational function can change sign only at those points at which it is equal to zero or does not exist.

Let us recall how a quadratic trinomial is factored, that is, an expression of the form .

Where and are the roots of the quadratic equation.

We draw an axis and place the points at which the numerator and denominator go to zero.

The zeros of the denominator and are punctured points, since at these points the function on the left side of the inequality is not defined (you cannot divide by zero). The zeros of the numerator and - are shaded, since the inequality is not strict. When and our inequality is satisfied, since both its sides are equal to zero.

These points break the axis into intervals.

Let us determine the sign of the fractional rational function on the left side of our inequality on each of these intervals. We remember that a fractional rational function can change sign only at those points at which it is equal to zero or does not exist.

This means that at each of the intervals between the points where the numerator or denominator goes to zero, the sign of the expression on the left side of the inequality will be constant - either “plus” or “minus”.
And therefore, to determine the sign of the function on each such interval, we take any point belonging to this interval. The one that is convenient for us.

. Take, for example, and check the sign of the expression on the left side of the inequality. Each of the "brackets" is negative. The left side has a sign.

Next interval: . Let's check the sign at . We find that the left side has changed its sign to .

Let's take it. When the expression is positive - therefore, it is positive over the entire interval from to.

When the left side of the inequality is negative."> . Подставим и проверим знак выражения в левой части неравенства. Каждая "скобочка" положительна. Следовательно, левая часть имеет знак .!}

And finally, class="tex" alt="x>7

We have found at what intervals the expression is positive. All that remains is to write down the answer:

Answer: . Please note: the signs alternate between intervals. This happened because.

when passing through each point, exactly one of the linear factors changed sign, while the rest kept it unchanged

We see that the interval method is very simple. To solve the fractional-rational inequality using the interval method, we reduce it to the form: Or"> !} class="tex" alt="\genfrac())()(0)(\displaystyle P\left(x \right))(\displaystyle Q\left(x \right)) > 0

, or or .

(on the left side is a fractional rational function, on the right side is zero).
Then we mark on the number line the points at which the numerator or denominator goes to zero.
These points divide the entire number line into intervals, on each of which the fractional-rational function retains its sign.
All that remains is to find out its sign at each interval.

We do this by checking the sign of the expression at any point belonging to a given interval. After that, we write down the answer. That's all.

2. But the question arises: do the signs always alternate? No not always! You must be careful and not place signs mechanically and thoughtlessly.

Class="tex" alt="\genfrac())()(0)(\displaystyle \left(x-2 \right)^2)(\displaystyle \left(x-1 \right) \left(x-3 \right))>0"> !}

Place the points on the axis again. The dots and are punctured because they are zeros of the denominator. The point is also cut out, since the inequality is strict.

When the numerator is positive, both factors in the denominator are negative. This can be easily checked by taking any number from a given interval, for example, . The left side has the sign:

When the numerator is positive; The first factor in the denominator is positive, the second factor is negative. The left side has the sign:

The situation is the same! The numerator is positive, the first factor in the denominator is positive, the second is negative. The left side has the sign:

Finally, with class="tex" alt="x>3"> все множители положительны, и левая часть имеет знак :!}

We have found at what intervals the expression is positive. All that remains is to write down the answer:

Why was the alternation of signs disrupted? Because when passing through a point the multiplier is “responsible” for it didn't change sign. Consequently, the entire left side of our inequality did not change sign.

Conclusion: if the linear multiplier is an even power (for example, squared), then when passing through a point the sign of the expression on the left side does not change. In the case of an odd degree, the sign, of course, changes.

3. Let's consider a more complex case. It differs from the previous one in that the inequality is not strict:

The left side is the same as in the previous problem. The picture of signs will be the same:

Maybe the answer will be the same? No! A solution is added This happens because at both the left and right sides of the inequality are equal to zero - therefore, this point is a solution.

We have found at what intervals the expression is positive. All that remains is to write down the answer:

This situation often occurs in problems on the Unified State Examination in mathematics. This is where applicants fall into a trap and lose points. Be careful!

4. What to do if the numerator or denominator cannot be factored into linear factors? Consider this inequality:

A square trinomial cannot be factorized: the discriminant is negative, there are no roots. But this is good! This means that the sign of the expression for all is the same, and specifically, positive. You can read more about this in the article on properties of quadratic functions.

And now we can divide both sides of our inequality by a value that is positive for all. Let us arrive at an equivalent inequality:

Which is easily solved using the interval method.

Please note that we divided both sides of the inequality by a value that we knew for sure was positive. Of course, in general, you should not multiply or divide an inequality by a variable whose sign is unknown.

5 . Let's consider another inequality, seemingly quite simple:

I just want to multiply it by . But we are already smart, and we won’t do this. After all, it can be both positive and negative. And we know that if both sides of the inequality are multiplied by a negative value, the sign of the inequality changes.

We will do it differently - we will collect everything in one part and bring it to a common denominator. The right side will remain zero:

Class="tex" alt="\genfrac())()()(0)(\displaystyle x-2)(\displaystyle x)>0"> !}

And after that - apply interval method.

How to solve inequalities using the interval method (algorithm with examples)

Example . (assignment from the OGE) Solve the inequality using the interval method $(x-7)^2< \sqrt{11}(x-7)$
Solution:

Answer : $(7;7+\sqrt(11))$

Example . Solve the inequality using the interval method $≥0$
Solution:

$\frac((4-x)^3 (x+6)(6-x)^4)((x+7.5))$$≥0$		Here, at first glance, everything seems normal, and the inequality is initially brought to the desired form. But this is not so - after all, in the first and third brackets of the numerator, the x appears with a minus sign. We transform the brackets, taking into account the fact that the fourth degree is even (i.e., it will remove the minus sign), and the third is odd (i.e., it will not remove). $(4-x)^3=(-x+4)^3=(-(x-4))^3=-(x-4)^3$ $(6-x)^4=(-x+6)^4=(-(x-6))^4=(x-6)^4$ Like this. Now we return the brackets “in place” already transformed.
$\frac(-(x-4)^3 (x+6)(x-6)^4)((x+7.5))$$≥0$		Now all the parentheses look as they should (the unsigned name comes first and then the number). But a minus appeared in front of the numerator. We remove it by multiplying the inequality by $-1$, not forgetting to reverse the comparison sign
$\frac((x-4)^3 (x+6)(x-6)^4)((x+7.5))$$≤0$		Ready. Now the inequality looks as it should. You can use the interval method.
$x=4;$ $x=-6;$ $x=6;$ $x=-7.5$		Let's place points on the axis, signs and paint over the necessary intervals.
		In the interval from $4$ to $6$, the sign does not need to be changed, because the bracket $(x-6)$ is to an even power (see point 4 of the algorithm). The flag will be a reminder that six is also a solution to inequality. Let's write down the answer.

Answer : $(-∞;7,5]∪[-6,4]∪\left\(6\right$\)

Example.(Assignment from the OGE) Solve the inequality using the interval method $x^2 (-x^2-64)≤64(-x^2-64)$
Solution:

$x^2 (-x^2-64)≤64(-x^2-64)$		There are identical ones on the left and right - this is clearly not a coincidence. The first desire is to divide by $-x^2-64$, but this is a mistake, because there is a chance of losing the root. Instead, move $64(-x^2-64)$ to the left
$x^2 (-x^2-64)-64(-x^2-64)≤0$
$(-x^2-64)(x^2-64)≤0$		Let's take out the minus in the first bracket and factor the second
$-(x^2+64)(x-8)(x+8)≤0$		Note that $x^2$ is either equal to zero or greater than zero. This means that $x^2+64$ is uniquely positive for any value of x, that is, this expression does not affect the sign of the left side in any way. Therefore, we can safely divide both sides of the inequality by this expression. Let's also divide the inequality by $-1$ to get rid of the minus.
$(x-8)(x+8)≥0$		Now you can use the interval method
$x=8;$ $x=-8$		Let's write down the answer

Answer : \((-∞;-8]∪}

\(\frac((4-x)^3 (x+6)(6-x)^4)((x+7.5))\)\(≥0\)		Here, at first glance, everything seems normal, and the inequality is initially brought to the desired form. But this is not so - after all, in the first and third brackets of the numerator, the x appears with a minus sign. We transform the brackets, taking into account the fact that the fourth degree is even (i.e., it will remove the minus sign), and the third is odd (i.e., it will not remove). \((4-x)^3=(-x+4)^3=(-(x-4))^3=-(x-4)^3\) \((6-x)^4=(-x+6)^4=(-(x-6))^4=(x-6)^4\) Like this. Now we return the brackets “in place” already transformed.
\(\frac(-(x-4)^3 (x+6)(x-6)^4)((x+7.5))\)\(≥0\)		Now all the parentheses look as they should (the unsigned name comes first and then the number). But a minus appeared in front of the numerator. We remove it by multiplying the inequality by \(-1\), not forgetting to reverse the comparison sign
\(\frac((x-4)^3 (x+6)(x-6)^4)((x+7.5))\)\(≤0\)		Ready. Now the inequality looks as it should. You can use the interval method.
\(x=4;\) \(x=-6;\) \(x=6;\) \(x=-7.5\)		Let's place points on the axis, signs and paint over the necessary intervals.
		In the interval from \(4\) to \(6\), the sign does not need to be changed, because the bracket \((x-6)\) is to an even power (see point 4 of the algorithm). The flag will be a reminder that six is also a solution to inequality. Let's write down the answer.

\(x^2 (-x^2-64)≤64(-x^2-64)\)		There are identical ones on the left and right - this is clearly not a coincidence. The first desire is to divide by \(-x^2-64\), but this is a mistake, because there is a chance of losing the root. Instead, move \(64(-x^2-64)\) to the left
\(x^2 (-x^2-64)-64(-x^2-64)≤0\)
\((-x^2-64)(x^2-64)≤0\)		Let's take out the minus in the first bracket and factor the second
\(-(x^2+64)(x-8)(x+8)≤0\)		Note that \(x^2\) is either equal to zero or greater than zero. This means that \(x^2+64\) is uniquely positive for any value of x, that is, this expression does not affect the sign of the left side in any way. Therefore, we can safely divide both sides of the inequality by this expression. Let's also divide the inequality by \(-1\) to get rid of the minus.
\((x-8)(x+8)≥0\)		Now you can use the interval method
\(x=8;\) \(x=-8\)		Let's write down the answer