The principle of solving inequalities by the interval method. Interval method: solving the simplest strict inequalities

The interval method is universal method solves inequalities, in particular, it allows you to solve quadratic inequalities in one variable. In this article we will cover in detail all the nuances of solving quadratic inequalities using the interval method. First, we present the algorithm, after which we will analyze in detail ready-made solutions typical examples.

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Algorithm

The first acquaintance with the interval method usually occurs in algebra lessons, when they learn to solve quadratic inequalities. In this case, the interval method algorithm is given in a form adapted specifically to solving quadratic inequalities. Paying tribute to simplicity, we will also give it in this form, and general algorithm You can look at the interval method using the link at the very beginning of this article.

So, algorithm for solving quadratic inequalities using the interval method is:

• Finding zeros quadratic trinomial a·x 2 +b·x+c from the left side of the quadratic inequality.
• We draw it and, if there are roots, mark them on it. Moreover, if we solve a strict inequality, then we mark them with empty (punctured) points, and if we solve a non-strict inequality, then with ordinary points. They break coordinate axis at intervals.
• We determine which signs have the values ​​of the trinomial on each interval (if zeros were found in the first step) or on the entire number line (if there are no zeros), we will tell you how to do this below. And we put + or − above these intervals in accordance with certain signs.
• If we decide quadratic inequality with the > or ≥ sign, then we apply shading over the intervals with + signs, but if we solve the inequality with the sign< или ≤, то наносим штриховку над промежутками со знаком −. В результате получаем , которое и является искомым решением неравенства.
• We write down the answer.

As promised, we explain the third step of the announced algorithm. There are several basic approaches that allow you to find signs on intervals. We will study them using examples, and start with the most reliable, but not the most fast way, which consists in calculating the values ​​of the trinomial at individual points of the intervals.

Let's take the trinomial x 2 +4·x−5, its roots are the numbers −5 and 1, they divide the number line into three intervals (−∞, −5), (−5, 1) and (1, +∞).

Let us determine the sign of the trinomial x 2 +4·x−5 on the interval (1, +∞) . To do this, we calculate the value of this trinomial for a certain value of x from this interval. It is advisable to take the value of the variable so that the calculations are simple. In our case, for example, we can take x=2 (with this number it is easier to carry out calculations than, for example, with 1.3, 74 or). We substitute it into the trinomial instead of the variable x, as a result we get 2 2 +4 2−5=7. 7 is a positive number, which means that any value of the quadratic trinomial on the interval (1, +∞) will be positive. This is how we defined the + sign.

To consolidate the skills, we will determine the signs on the remaining two spaces. Let's start with the sign on the interval (−5, 1) . From this interval it is best to take x=0 and calculate the value of the quadratic trinomial for this value of the variable, we have 0 2 +4·0−5=−5. Since −5 – a negative number, then on this interval all values ​​of the trinomial will be negative, therefore, we have defined the minus sign.

It remains to find out the sign on the interval (−∞, −5) . Let's take x=−6, substitute it for x, we get (−6) 2 +4·(−6)−5=7, therefore, the required sign will be plus.

But the following facts allow you to place signs faster:

• When a square trinomial has two roots (with a positive discriminant), then the signs of its values ​​on the intervals into which these roots divide the number line alternate (as in the previous example). That is, it is enough to determine the sign on one of the three intervals, and place the signs over the remaining intervals, alternating them. As a result, one of two sequences of characters is possible: +, −, + or −, +, −. Moreover, you can generally do without calculating the value of the quadratic trinomial at the point of the interval, and draw conclusions about the signs based on the value of the leading coefficient a: if a>0, then we have a sequence of signs +, −, +, and if a<0 – то −, +, −.
• If the square trinomial has one root (when the discriminant is zero), then this root splits the number line into two intervals, and the signs above them will be the same. That is, it is enough to determine a sign above one of them, and above the other - put the same one. This will result in either +, +, or −, −. A conclusion based on the signs can also be made based on the value of the coefficient a: if a>0, then it will be +, +, and if a<0 , то −, −.
• When a square trinomial has no roots, then the signs of its values ​​on the entire number line coincide with both the sign of the leading coefficient a and the sign of the free term c. For example, consider the square trinomial −4 x 2 −7, it has no roots (its discriminant is negative), and on the interval (−∞, +∞) its values ​​are negative, since the coefficient of x 2 is a negative number −4, and the free term −7 is also negative.

Now all the steps of the algorithm have been analyzed and it remains to consider examples of solving quadratic inequalities using it.

Examples with solutions

Let's move on to practice. Let's solve several quadratic inequalities using the interval method and touch on the main characteristic cases.

Example.

Solve the inequality 8 x 2 −4 x−1≥0 .

Solution.

Let's solve this quadratic inequality using the interval method. In the first step, it involves searching for the roots of the quadratic trinomial 8 x 2 −4 x −1 . The coefficient of x is even, so it is more convenient to calculate not the discriminant, but its fourth part: D"=(−2) 2 −8·(−1)=12. Since it is greater than zero, we find two roots And .

Now we mark them on the coordinate line. It is easy to see that x 1

Next, using the interval method, we determine the signs on each of the three resulting intervals. This is most convenient and fastest to do based on the value of the coefficient at x 2, it is equal to 8, that is, positive, therefore, the sequence of signs will be +, −, +:

Since we are solving an inequality with a ≥ sign, we draw shading over the intervals with plus signs:

Based on the resulting image of a numerical set, it is not difficult to describe it analytically: or so . This is how we solved the original quadratic inequality.

Answer:

or .

Example.

Solve quadratic inequality interval method.

Solution.

Find the roots of the quadratic trinomial on the left side of the inequality:

Since we are solving a strict inequality, we depict a punctured point with coordinate 7 on the coordinate line:

Now we determine the signs on the two resulting intervals (−∞, 7) and (7, +∞). This is easy to do, given that the discriminant of a quadratic trinomial is equal to zeros and the leading coefficient is negative. We have signs −, −:

Since we are solving an inequality with a sign<, то изображаем штриховку над интервалами со знаками минус:

It is clearly seen that both intervals (−∞, 7) , (7, +∞) are solutions.

Answer:

(−∞, 7)∪(7, +∞) or in another notation x≠7 .

Example.

Does the quadratic inequality x 2 +x+7<0 решения?

Solution.

To answer the question posed, we will solve this quadratic inequality, and since we are analyzing the method of intervals, we will use it. As usual, we start by finding the roots of the square trinomial on the left side. We find the discriminant: D=1 2 −4·1·7=1−28=−27, it is less than zero, which means there are no real roots.

Therefore, we simply draw a coordinate line without marking any points on it:

Now we determine the sign of the values ​​of the quadratic trinomial. At D<0 он совпадает со знаком коэффициента при x 2 , то есть, со знаком числа 1 , оно положительное, следовательно, имеем знак +:

We solve signed inequality<, поэтому штриховку следует изобразить над промежутками со знаком −, но таковых нет, и в силу этого штриховку не наносим, а чертеж сохраняет свой вид.

As a result, we have an empty set, which means that the original quadratic inequality has no solutions.

Answer:

Bibliography.

• Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• 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. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• 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.
• Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. In 2 hours. Part 1. Textbook for students of general education institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

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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 brackets 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 the 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, it 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:

1. We move everything to the left side, leaving only zero on the right;
2. We find ODZ;
3. We plot all the roots of the inequality on the axis;
4. 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;
5. 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 what points each of the factors becomes zero, 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 circle the point with a square 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:

1. 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:

1. If $D \gt 0$, we get two different roots;
2. If $D=0$, then the root will be the same, 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;
3. For $D \lt 0$ there are no roots at all, and the sign of the polynomial $a((x)^(2))+bx+c$ for any $x$ coincides with the sign of the coefficient $a$. This, by the way, is a very useful fact, which for some reason 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:

1. Factoring the polynomial $P\left(x \right)$;
2. 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:

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

1. Strict: $f\left(x \right) \gt 0$ or $f\left(x \right) \lt 0$;
2. 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 got acquainted 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:

1. Collect all non-zero elements on one side of the inequality sign. For example, on the left;
2. 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.
3. 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).
4. We mark all these roots (both with and without asterisks) on a single number line, and the roots without stars are painted over, and those with stars are punctured.
5. We place the “plus” and “minus” signs, select the intervals that we need. If the inequality has the form $f\left(x \right) \gt 0$, then the answer will be the intervals marked with a “plus”. If $f\left(x \right) \lt 0$, then we look at 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 scheme 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’ll format it the 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 are analyzing 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:

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

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:

1. 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 $x\in \left(-\ infty ;2 \right)$.
2. The point $x=3$ also has an even multiplicity and is shaded. The arrangement of signs indicates that the point itself suits us, but a step 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 tasks, here you will have to apply skills from the theory of rational fractions - factorization and reduction to a common denominator.

We discussed this issue in detail at the very beginning of today's lesson. If you'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 bring 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. So 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: $x\in \left(-\infty ;-9 \right)\bigcup \left((2)/(3)\;;1 \right)\bigcup \left[ 5.5;+\infty \ right)$.

It has been necessary to compare quantities and quantities when solving practical problems since ancient times. At the same time, words such as more and less, higher and lower, lighter and heavier, quieter and louder, cheaper and more expensive, etc. appeared, denoting the results of comparing homogeneous quantities.

The concepts of more and less arose in connection with counting objects, measuring and comparing quantities. For example, mathematicians of Ancient Greece knew that the side of any triangle is less than the sum of the other two sides and that the larger side lies opposite the larger angle in a triangle. Archimedes, while calculating the circumference, established that the perimeter of any circle is equal to three times the diameter with an excess that is less than a seventh of the diameter, but more than ten seventy times the diameter.

Symbolically write relationships between numbers and quantities using the signs > and b. Records in which two numbers are connected by one of the signs: > (greater than), You also encountered numerical inequalities in the lower grades. You know that inequalities can be true, or they can be false. For example, \(\frac(1)(2) > \frac(1)(3)\) is a correct numerical inequality, 0.23 > 0.235 is an incorrect numerical inequality.

Inequalities involving unknowns may be true for some values ​​of the unknowns and false for others. For example, the inequality 2x+1>5 is true for x = 3, but false for x = -3. For an inequality with one unknown, you can set the task: solve the inequality. In practice, problems of solving inequalities are posed and solved no less often than problems of solving equations. For example, many economic problems come down to the study and solution of systems of linear inequalities. In many branches of mathematics, inequalities are more common than equations.

Some inequalities serve as the only auxiliary means of proving or disproving the existence of a certain object, for example, the root of an equation.

Numerical inequalities

You can compare whole numbers and decimal fractions. Know the rules for comparing ordinary fractions with the same denominators but different numerators; with the same numerators but different denominators. Here you will learn how to compare any two numbers by finding the sign of their difference.

Comparing numbers is widely used in practice. For example, an economist compares planned indicators with actual ones, a doctor compares a patient’s temperature with normal, a turner compares the dimensions of a machined part with a standard. In all such cases, some numbers are compared. As a result of comparing numbers, numerical inequalities arise.

Definition. The number a is greater than the number b if the difference a-b is positive. The number a is less than the number b if the difference a-b is negative.

If a is greater than b, then they write: a > b; if a is less than b, then they write: a Thus, the inequality a > b means that the difference a - b is positive, i.e. a - b > 0. Inequality a For any two numbers a and b from the following three relations a > b, a = b, a To compare the numbers a and b means to find out which of the signs >, = or Theorem. If a > b and b > c, then a > c.

Theorem. If you add the same number to both sides of the inequality, the sign of the inequality will not change.
Consequence. Any term can be moved from one part of the inequality to another by changing the sign of this term to the opposite.

Theorem. If both sides of the inequality are multiplied by the same positive number, then the sign of the inequality does not change. If both sides of the inequality are multiplied by the same negative number, then the sign of the inequality will change to the opposite.
Consequence. If both sides of the inequality are divided by the same positive number, then the sign of the inequality will not change. If both sides of the inequality are divided by the same negative number, then the sign of the inequality will change to the opposite.

You know that numerical equalities can be added and multiplied term by term. Next, you will learn how to perform similar actions with inequalities. The ability to add and multiply inequalities term by term is often used in practice. These actions help solve problems of evaluating and comparing the meanings of expressions.

When solving various problems, it is often necessary to add or multiply the left and right sides of inequalities term by term. At the same time, it is sometimes said that inequalities add up or multiply. For example, if a tourist walked more than 20 km on the first day, and more than 25 km on the second, then we can say that in two days he walked more than 45 km. Similarly, if the length of a rectangle is less than 13 cm and the width is less than 5 cm, then we can say that the area of ​​this rectangle is less than 65 cm2.

When considering these examples, the following were used: theorems on addition and multiplication of inequalities:

Theorem. When adding inequalities of the same sign, an inequality of the same sign is obtained: if a > b and c > d, then a + c > b + d.

Theorem. When multiplying inequalities of the same sign, whose left and right sides are positive, an inequality of the same sign is obtained: if a > b, c > d and a, b, c, d are positive numbers, then ac > bd.

Inequalities with the sign > (greater than) and 1/2, 3/4 b, c Along with the signs of strict inequalities > and In the same way, the inequality \(a \geq b \) means that the number a is greater than or equal to b, i.e. .and not less b.

Inequalities containing the \(\geq \) sign or the \(\leq \) sign are called non-strict. For example, \(18 \geq 12 , \; 11 \leq 12 \) are not strict inequalities.

All properties of strict inequalities are also valid for non-strict inequalities. Moreover, if for strict inequalities the signs > were considered opposite and you know that to solve a number of applied problems you have to create a mathematical model in the form of an equation or a system of equations. Next, you will learn that mathematical models for solving many problems are inequalities with unknowns. The concept of solving an inequality will be introduced and how to test whether a given number is a solution to a particular inequality will be shown.

Inequalities of the form
\(ax > b, \quad ax in which a and b are given numbers, and x is an unknown, are called linear inequalities with one unknown.

Definition. The solution to an inequality with one unknown is the value of the unknown at which this inequality becomes a true numerical inequality. Solving an inequality means finding all its solutions or establishing that there are none.

You solved the equations by reducing them to the simplest equations. Similarly, when solving inequalities, one tries to reduce them, using properties, to the form of simple inequalities.

Solving second degree inequalities with one variable

Inequalities of the form
\(ax^2+bx+c >0 \) and \(ax^2+bx+c where x is a variable, a, b and c are some numbers and \(a \neq 0 \), called inequalities of the second degree with one variable.

Solution to inequality
\(ax^2+bx+c >0 \) or \(ax^2+bx+c can be considered as finding intervals in which the function \(y= ax^2+bx+c \) takes positive or negative values To do this, it is enough to analyze how the graph of the function \(y= ax^2+bx+c\) is located in the coordinate plane: where the branches of the parabola are directed - up or down, whether the parabola intersects the x axis and if it does, then at what points.

Algorithm for solving second degree inequalities with one variable:
1) find the discriminant of the square trinomial \(ax^2+bx+c\) and find out whether the trinomial has roots;
2) if the trinomial has roots, then mark them on the x-axis and through the marked points draw a schematic parabola, the branches of which are directed upward for a > 0 or downward for a 0 or at the bottom for a 3) find intervals on the x-axis for which the points parabolas are located above the x-axis (if they solve the inequality \(ax^2+bx+c >0\)) or below the x-axis (if they solve the inequality
\(ax^2+bx+c Solving inequalities using the interval method

Consider the function
f(x) = (x + 2)(x - 3)(x - 5)

The domain of this function is the set of all numbers. The zeros of the function are the numbers -2, 3, 5. They divide the domain of definition of the function into the intervals \((-\infty; -2), \; (-2; 3), \; (3; 5) \) and \( (5; +\infty)\)

Let us find out what the signs of this function are in each of the indicated intervals.

The expression (x + 2)(x - 3)(x - 5) is the product of three factors. The sign of each of these factors in the intervals under consideration is indicated in the table:

In general, let the function be given by the formula
f(x) = (x-x 1)(x-x 2) ... (x-x n),
where x is a variable, and x 1, x 2, ..., x n are numbers that are not equal to each other. The numbers x 1 , x 2 , ..., x n are the zeros of the function. In each of the intervals into which the domain of definition is divided by zeros of the function, the sign of the function is preserved, and when passing through zero its sign changes.

This property is used to solve inequalities of the form
(x-x 1)(x-x 2) ... (x-x n) > 0,
(x-x 1)(x-x 2) ... (x-x n) where x 1, x 2, ..., x n are numbers not equal to each other

Considered method solving inequalities is called the interval method.

Let us give examples of solving inequalities using the interval method.

Solve inequality:

\(x(0.5-x)(x+4) Obviously, the zeros of the function f(x) = x(0.5-x)(x+4) are the points \(x=0, \; x= \frac(1)(2) , \; x=-4 \)

We plot the zeros of the function on the number axis and calculate the sign on each interval:

We select those intervals at which the function is less than or equal to zero and write down the answer.

Answer:
\(x \in \left(-\infty; \; 1 \right) \cup \left[ 4; \; +\infty \right) \)

Interval method is a special algorithm designed to solve complex inequalities of the form f(x) > 0. The algorithm consists of 5 steps:

1. Solve the equation f(x) = 0. Thus, instead of an inequality, we get an equation that is much simpler to solve;
2. Mark all obtained roots on the coordinate line. Thus, the straight line will be divided into several intervals;
3. Find the multiplicity of the roots. If the roots are of even multiplicity, then draw a loop above the root. (A root is considered a multiple if there are an even number of identical solutions)
4. Find out the sign (plus or minus) of the function f(x) on the rightmost interval. To do this, it is enough to substitute into f(x) any number that will be to the right of all the marked roots;
5. Mark the signs at the remaining intervals, alternating them.

After this, all that remains is to write down the intervals that interest us. They are marked with a “+” sign if the inequality was of the form f(x) > 0, or with a “−” sign if the inequality was of the form f(x)< 0.

In the case of non-strict inequalities (≤ , ≥), it is necessary to include in the intervals points that are a solution to the equation f(x) = 0;

Example 1:

Solve inequality:

(x - 2)(x + 7)< 0

We work using the interval method.

Step 1: replace the inequality with an equation and solve it:

(x - 2)(x + 7) = 0

The product is zero if and only if at least one of the factors is zero:

x - 2 = 0 => x = 2

x + 7 = 0 => x = -7

We got two roots.

Step 2: We mark these roots on the coordinate line. We have:

Step 3: we find the sign of the function on the rightmost interval (to the right of the marked point x = 2). To do this, you need to take any number that is greater than the number x = 2. For example, let's take x = 3 (but no one forbids taking x = 4, x = 10 and even x = 10,000).

f(x) = (x - 2)(x + 7)

f(3)=(3 - 2)(3 + 7) = 1*10 = 10

We get that f(3) = 10 > 0 (10 is a positive number), so we put a plus sign in the rightmost interval.

Step 4: you need to note the signs on the remaining intervals. We remember that when passing through each root the sign must change. For example, to the right of the root x = 2 there is a plus (we made sure of this in the previous step), so there must be a minus to the left. This minus extends to the entire interval (−7; 2), so there is a minus to the right of the root x = −7. Therefore, to the left of the root x = −7 there is a plus. It remains to mark these signs on the coordinate axis.

Let's return to the original inequality, which had the form:

(x - 2)(x + 7)< 0

So the function must be less than zero. This means that we are interested in the minus sign, which appears only on one interval: (−7; 2). This will be the answer.

Example 2:

Solve inequality:

(9x 2 - 6x + 1)(x - 2) ≥ 0

Solution:

First you need to find the roots of the equation

(9x 2 - 6x + 1)(x - 2) = 0

Let's collapse the first bracket and get:

(3x - 1) 2 (x - 2) = 0

x - 2 = 0; (3x - 1) 2 = 0

Solving these equations we get:

Let's plot the points on the number line:

Because x 2 and x 3 are multiple roots, then there will be one point on the line and above it “ a loop”.

Let's take any number less than the leftmost point and substitute it into the original inequality. Let's take the number -1.

Don’t forget to include the solution to the equation (found X), because our inequality is not strict.

Answer: ()U)