Inequalities are quadratic graphically. Solving quadratic inequalities using the interval method

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One of the most convenient methods for solving quadratic inequalities is the graphical method. In this article we will look at how quadratic inequalities are solved graphically. First, let's discuss what the essence of this method is. Next, we will present the algorithm and consider examples of solving quadratic inequalities graphically.

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The essence of the graphic method At all graphical method for solving inequalities with one variable is used not only to solve quadratic inequalities, but also other types of inequalities. The essence of the graphical method for solving inequalities

• next: consider the functions y=f(x) and y=g(x), which correspond to the left and right sides of the inequality, build their graphs in one rectangular coordinate system and find out at what intervals the graph of one of them is lower or higher than the other. Those intervals where
• the graph of function f above the graph of function g are solutions to the inequality f(x)>g(x) ;
• the graph of the function f not lower than the graph of the function g are solutions to the inequality f(x)≥g(x) ;
• the graph of f below the graph of g are solutions to the inequality f(x)

the graph of a function f not higher than the graph of a function g are solutions to the inequality f(x)≤g(x) .

We will also say that the abscissas of the intersection points of the graphs of the functions f and g are solutions to the equation f(x)=g(x) .<0 (≤, >, ≥).

Let's transfer these results to our case - to solve the quadratic inequality a x 2 +b x+c We introduce two functions: the first y=a x 2 +b x+c (with f(x)=a x 2 +b x+c) corresponding to the left side of the quadratic inequality, the second y=0 (with g (x)=0 ) corresponds to the right side of the inequality. Schedule quadratic function f is a parabola and the graph constant function

g – straight line coinciding with the abscissa axis Ox.

Next, according to the graphical method of solving inequalities, it is necessary to analyze at what intervals the graph of one function is located above or below another, which will allow us to write down the desired solution to the quadratic inequality. In our case, we need to analyze the position of the parabola relative to the Ox axis.

In this drawing we see a parabola, the branches of which are directed upward, and which intersects the Ox axis at two points, the abscissa of which are x 1 and x 2. This drawing corresponds to the option when the coefficient a is positive (it is responsible for the upward direction of the parabola branches), and when the value is positive discriminant of a quadratic trinomial a x 2 +b x+c (in this case, the trinomial has two roots, which we denoted as x 1 and x 2, and we assumed that x 1 0 , D=b 2 −4·a·c=(−1) 2 −4·1·(−6)=25>0, x 1 =−2 , x 2 =3 .

For clarity, let’s depict in red the parts of the parabola located above the x-axis, and in blue – those located below the x-axis.


Now let's find out which intervals correspond to these parts. The following drawing will help you identify them (in the future we will make similar selections in the form of rectangles mentally):


So on the abscissa axis two intervals (−∞, x 1) and (x 2 , +∞) were highlighted in red, on them the parabola is above the Ox axis, they constitute a solution to the quadratic inequality a x 2 +b x+c>0 , and the interval (x 1 , x 2) is highlighted in blue, there is a parabola below the Ox axis, it represents the solution to the inequality a x 2 +b x+c<0 . Решениями нестрогих квадратных неравенств a·x 2 +b·x+c≥0 и a·x 2 +b·x+c≤0 будут те же промежутки, но в них следует включить числа x 1 и x 2 , отвечающие равенству a·x 2 +b·x+c=0 .

And now briefly: for a>0 and D=b 2 −4 a c>0 (or D"=D/4>0 for an even coefficient b)

• the solution to the quadratic inequality a x 2 +b x+c>0 is (−∞, x 1)∪(x 2 , +∞) or in another notation x x2;
• the solution to the quadratic inequality a x 2 +b x+c≥0 is (−∞, x 1 ]∪ or in another notation x 1 ≤x≤x 2 ,

where x 1 and x 2 are the roots of the quadratic trinomial a x 2 +b x+c, and x 1


Here we see a parabola, the branches of which are directed upward, and which touches the abscissa axis, that is, it has one common point with it; we denote the abscissa of this point as x 0. The presented case corresponds to a>0 (the branches are directed upward) and D=0 (the square trinomial has one root x 0). For example, you can take the quadratic function y=x 2 −4·x+4, here a=1>0, D=(−4) 2 −4·1·4=0 and x 0 =2.

The drawing clearly shows that the parabola is located above the Ox axis everywhere except the point of contact, that is, on the intervals (−∞, x 0), (x 0, ∞). For clarity, let’s highlight areas in the drawing by analogy with the previous paragraph.


We draw conclusions: for a>0 and D=0

• the solution to the quadratic inequality a·x 2 +b·x+c>0 is (−∞, x 0)∪(x 0, +∞) or in another notation x≠x 0;
• the solution to the quadratic inequality a·x 2 +b·x+c≥0 is (−∞, +∞) or in another notation x∈R ;
• quadratic inequality a x 2 +b x+c<0 не имеет решений (нет интервалов, на которых парабола расположена ниже оси Ox );
• the quadratic inequality a x 2 +b x+c≤0 has a unique solution x=x 0 (it is given by the point of tangency),

where x 0 is the root of the square trinomial a x 2 + b x + c.


In this case, the branches of the parabola are directed upward, and it does not have common points with the abscissa axis. Here we have the conditions a>0 (branches are directed upward) and D<0 (квадратный трехчлен не имеет действительных корней). Для примера можно построить график функции y=2·x 2 +1 , здесь a=2>0 , D=0 2 −4·2·1=−8<0 .

Obviously, the parabola is located above the Ox axis throughout its entire length (there are no intervals at which it is below the Ox axis, there is no point of tangency).


Thus, for a>0 and D<0 решением квадратных неравенств a·x 2 +b·x+c>0 and a x 2 +b x+c≥0 is the set of all real numbers, and the inequalities a x 2 +b x+c<0 и a·x 2 +b·x+c≤0 не имеют решений.

And there remain three options for the location of the parabola with branches directed downward, not upward, relative to the Ox axis. In principle, they need not be considered, since multiplying both sides of the inequality by −1 allows us to go to an equivalent inequality with a positive coefficient for x 2. But it still doesn’t hurt to get an idea about these cases. The reasoning here is similar, so we will write down only the main results.

Solution algorithm

The result of all previous calculations is algorithm for solving quadratic inequalities graphically:

A schematic drawing is made on the coordinate plane, which depicts the Ox axis (it is not necessary to depict the Oy axis) and a sketch of a parabola corresponding to the quadratic function y=a·x 2 +b·x+c. To draw a sketch of a parabola, it is enough to clarify two points:

• Firstly, by the value of the coefficient a it is determined where its branches are directed (for a>0 - upward, for a<0 – вниз).
• And secondly, by the value of the discriminant of the square trinomial a x 2 + b x + c it is determined whether the parabola intersects the abscissa axis at two points (for D>0), touches it at one point (for D=0), or has no common points with the Ox axis (at D<0 ). Для удобства на чертеже указываются координаты точек пересечения или координата точки касания (при наличии этих точек), а сами точки изображаются выколотыми при решении строгих неравенств, или обычными при решении нестрогих неравенств.

• When the drawing is ready, use it in the second step of the algorithm

  • when solving the quadratic inequality a·x 2 +b·x+c>0, the intervals are determined at which the parabola is located above the abscissa;
  • when solving the inequality a·x 2 +b·x+c≥0, the intervals at which the parabola is located above the abscissa axis are determined and the abscissas of the intersection points (or the abscissa of the tangent point) are added to them;
  • when solving the inequality a x 2 +b x+c<0 находятся промежутки, на которых парабола ниже оси Ox ;
  • finally, when solving a quadratic inequality of the form a·x 2 +b·x+c≤0, intervals are found in which the parabola is below the Ox axis and the abscissa of the intersection points (or the abscissa of the tangent point) is added to them;

  they constitute the desired solution to the quadratic inequality, and if there are no such intervals and no points of tangency, then the original quadratic inequality has no solutions.

All that remains is to solve a few quadratic inequalities using this algorithm.

Examples with solutions

Example.

Solve the inequality .

Solution.

We need to solve a quadratic inequality, let's use the algorithm from the previous paragraph. In the first step we need to draw a sketch of the graph of the quadratic function . The coefficient of x 2 is equal to 2, it is positive, therefore, the branches of the parabola are directed upward. Let’s also find out whether the parabola has common points with the x-axis; to do this, we’ll calculate the discriminant of the quadratic trinomial . We have . The discriminant turned out to be greater than zero, therefore the trinomial has two real roots: And , that is, x 1 =−3 and x 2 =1/3.

From this it is clear that the parabola intersects the Ox axis at two points with abscissas −3 and 1/3. We will depict these points in the drawing as ordinary points, since we are solving a non-strict inequality. Based on the clarified data, we obtain the following drawing (it fits the first template from the first paragraph of the article):

Let's move on to the second step of the algorithm. Since we are solving a non-strict quadratic inequality with the sign ≤, we need to determine the intervals at which the parabola is located below the abscissa and add to them the abscissas of the intersection points.

From the drawing it is clear that the parabola is below the x-axis on the interval (−3, 1/3) and to it we add the abscissas of the intersection points, that is, the numbers −3 and 1/3. As a result, we arrive at the numerical interval [−3, 1/3] . This is the solution we are looking for. It can be written as a double inequality −3≤x≤1/3.

Answer:

[−3, 1/3] or −3≤x≤1/3 .

Example.

Find the solution to the quadratic inequality −x 2 +16 x−63<0 .

Solution.

As usual, we start with a drawing. The numerical coefficient for the square of the variable is negative, −1, therefore, the branches of the parabola are directed downward. Let's calculate the discriminant, or better yet, its fourth part: D"=8 2 −(−1)·(−63)=64−63=1. Its value is positive, let's calculate the roots of the square trinomial: And , x 1 =7 and x 2 =9. So the parabola intersects the Ox axis at two points with abscissas 7 and 9 (the original inequality is strict, so we will depict these points with an empty center). Now we can make a schematic drawing:

Since we are solving a strict quadratic inequality with a sign<, то нас интересуют промежутки, на которых парабола расположена ниже оси абсцисс:

The drawing shows that the solutions to the original quadratic inequality are two intervals (−∞, 7) , (9, +∞) .

Answer:

(−∞, 7)∪(9, +∞) or in another notation x<7 , x>9 .

When solving quadratic inequalities, when the discriminant of a quadratic trinomial on its left side is zero, you need to be careful about including or excluding the abscissa of the tangent point from the answer. This depends on the sign of the inequality: if the inequality is strict, then it is not a solution to the inequality, but if it is not strict, then it is.

Example.

Does the quadratic inequality 10 x 2 −14 x+4.9≤0 have at least one solution?

Solution.

Let's plot the function y=10 x 2 −14 x+4.9. Its branches are directed upward, since the coefficient of x 2 is positive, and it touches the abscissa axis at the point with the abscissa 0.7, since D"=(−7) 2 −10 4.9=0, whence or 0.7 in the form of a decimal fraction. Schematically it looks like this:

Since we are solving a quadratic inequality with the ≤ sign, its solution will be the intervals on which the parabola is below the Ox axis, as well as the abscissa of the tangent point. From the drawing it is clear that there is not a single gap where the parabola would be below the Ox axis, so its solution will be only the abscissa of the tangent point, that is, 0.7.

Answer:

this inequality has a unique solution 0.7.

Example.

Solve the quadratic inequality –x 2 +8 x−16<0 .

Solution.

We follow the algorithm for solving quadratic inequalities and start by constructing a graph. The branches of the parabola are directed downward, since the coefficient of x 2 is negative, −1. Let us find the discriminant of the square trinomial –x 2 +8 x−16, we have D’=4 2 −(−1)·(−16)=16−16=0 and then x 0 =−4/(−1) , x 0 =4 . So, the parabola touches the Ox axis at the abscissa point 4. Let's make the drawing:

We look at the sign of the original inequality, it is there<. Согласно алгоритму, решение неравенства в этом случае составляют все промежутки, на которых парабола расположена строго ниже оси абсцисс.

In our case, these are open rays (−∞, 4) , (4, +∞) . Separately, we note that 4 - the abscissa of the point of contact - is not a solution, since at the point of contact the parabola is not lower than the Ox axis.

Answer:

(−∞, 4)∪(4, +∞) or in another notation x≠4 .

Pay special attention to cases where the discriminant of the quadratic trinomial on the left side of the quadratic inequality is less than zero. There is no need to rush here and say that the inequality has no solutions (we are used to making such a conclusion for quadratic equations with a negative discriminant). The point is that the quadratic inequality for D<0 может иметь решение, которым является множество всех действительных чисел.

Example.

Find the solution to the quadratic inequality 3 x 2 +1>0.

Solution.

As usual, we start with a drawing. The coefficient a is 3, it is positive, therefore, the branches of the parabola are directed upward. We calculate the discriminant: D=0 2 −4·3·1=−12 . Since the discriminant is negative, the parabola has no common points with the Ox axis. The information obtained is sufficient for a schematic graph:

We solve a strict quadratic inequality with a > sign. Its solution will be all intervals in which the parabola is above the Ox axis. In our case, the parabola is above the x-axis along its entire length, so the desired solution will be the set of all real numbers.

Ox , and also to them you need to add the abscissa of the points of intersection or the abscissa of the point of tangency. But the drawing clearly shows that there are no such gaps (since the parabola is everywhere below the abscissa axis), just as there are no points of intersection, and there are no points of tangency. Therefore, the original quadratic inequality has no solutions.

Answer:

no solutions or in another entry ∅.

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.

This article contains material covering the topic “ solving quadratic inequalities" First, it is shown what quadratic inequalities with one variable are and their general form is given. And then we look in detail at how to solve quadratic inequalities. The main approaches to the solution are shown: the graphical method, the method of intervals and by selecting the square of the binomial on the left side of the inequality. Solutions to typical examples are given.

One of the most convenient methods for solving quadratic inequalities is the graphical method. In this article we will look at how quadratic inequalities are solved graphically. First, let's discuss what the essence of this method is. Next, we will present the algorithm and consider examples of solving quadratic inequalities graphically.

What is a quadratic inequality?

Naturally, before talking about solving quadratic inequalities, we must clearly understand what a quadratic inequality is. In other words, you need to be able to distinguish quadratic inequalities from other types of inequalities by the type of recording.

Definition.

Quadratic inequality is an inequality of the form a x 2 +b x+c<0 (вместо знака >there can be any other inequality sign ≤, >, ≥), where a, b and c are some numbers, and a≠0, and x is a variable (the variable can be denoted by any other letter).

Let's immediately give another name for quadratic inequalities - second degree inequalities. This name is explained by the fact that on the left side of the inequalities a x 2 +b x+c<0 находится второй степени - квадратный трехчлен. Термин «неравенства второй степени» используется в учебниках алгебры Ю. Н. Макарычева, а Мордкович А. Г. придерживается названия «квадратные неравенства».

You can also sometimes hear quadratic inequalities called quadratic inequalities. This is not entirely correct: the definition of “quadratic” refers to functions defined by equations of the form y=a·x 2 +b·x+c. So, there are quadratic inequalities and quadratic functions, but not quadratic inequalities.

Let's show some examples of quadratic inequalities: 5 x 2 −3 x+1>0, here a=5, b=−3 and c=1; −2.2·z 2 −0.5·z−11≤0, the coefficients of this quadratic inequality are a=−2.2, b=−0.5 and c=−11; , in this case .

Note that in the definition of a quadratic inequality, the coefficient a of x 2 is considered to be nonzero. This is understandable; the equality of the coefficient a to zero will actually “remove” the square, and we will be dealing with a linear inequality of the form b x+c>0 without the square of the variable. But the coefficients b and c can be equal to zero, both separately and simultaneously. Here are examples of such quadratic inequalities: x 2 −5≥0, here the coefficient b for the variable x is equal to zero; −3 x 2 −0.6 x<0 , здесь c=0 ; наконец, в квадратном неравенстве вида 5·z 2 >0 both b and c are zero.

How to solve quadratic inequalities?

Now you can be puzzled by the question of how to solve quadratic inequalities. Basically, three main methods are used to solve:

• graphical method (or, as in A.G. Mordkovich, functional-graphic),
• interval method,
• and solving quadratic inequalities by isolating the square of the binomial on the left side.

Graphically

Let us immediately make a reservation that the method for solving quadratic inequalities, which we are now considering, school textbooks algebra is not called graphical. However, in essence this is what he is. Moreover, the first acquaintance with graphical method for solving inequalities usually begins when the question arises of how to solve quadratic inequalities.

Graphical method for solving quadratic inequalities a x 2 +b x+c<0 (≤, >, ≥) consists of analyzing the graph of the quadratic function y=a·x 2 +b·x+c to find the intervals in which the specified function takes negative, positive, non-positive or non-negative values. These intervals constitute the solutions to the quadratic inequalities a x 2 +b x+c<0 , a·x 2 +b·x+c>0, a x 2 +b x+c≤0 and a x 2 +b x+c≥0, respectively.

Interval method

To solve quadratic inequalities with one variable, in addition to the graphical method, the interval method is quite convenient, which in itself is very universal and is suitable for solving various inequalities, not just quadratic ones. Its theoretical side lies beyond the limits of the 8th and 9th grade algebra course, when they learn to solve quadratic inequalities. Therefore, here we will not go into the theoretical justification of the interval method, but will focus on how quadratic inequalities are solved with its help.

The essence of the interval method in relation to solving quadratic inequalities a x 2 +b x+c<0 (≤, >, ≥), consists in determining the signs that have the values ​​of the quadratic trinomial a·x 2 +b·x+c on the intervals into which the coordinate axis is divided by the zeros of this trinomial (if any). Intervals with minus signs constitute solutions to the quadratic inequality a x 2 +b x+c<0 , со знаками плюс – неравенства a·x 2 +b·x+c>0, and when solving non-strict inequalities, points corresponding to the zeros of the trinomial are added to the indicated intervals.

You can get acquainted with all the details of this method, its algorithm, the rules for placing signs on intervals and consider ready-made solutions to typical examples with the illustrations provided by referring to the material in the article solving quadratic inequalities using the interval method.

By squaring the binomial

In addition to the graphical method and the interval method, there are other approaches that allow you to solve quadratic inequalities. And we come to one of them, which is based on squared binomial on the left side of the quadratic inequality.

The principle of this method of solving quadratic inequalities is to perform equivalent transformations of the inequality, allowing one to proceed to solving an equivalent inequality of the form (x−p) 2 , ≥), where p and q are some numbers.

And how does the transition to inequality (x−p) 2 take place? , ≥) and how to solve it, the article explains the solution of quadratic inequalities by isolating the square of the binomial. There are also examples of solving quadratic inequalities using this method and the necessary graphic illustrations.

Inequalities that reduce to quadratic

In practice, one very often encounters inequalities given using equivalent transformations to quadratic inequalities of the form a x 2 +b x+c<0 (знаки, естественно, могут быть и другими). Их можно назвать неравенствами, сводящимися к квадратным неравенствам.

Let's start with examples of the simplest inequalities that reduce to quadratic inequalities. Sometimes, in order to move to a quadratic inequality, it is enough to rearrange the terms in this inequality or move them from one part to another. For example, if we transfer all the terms from the right side of the inequality 5≤2·x−3·x 2 to the left, we obtain a quadratic inequality in the form specified above 3·x 2 −2·x+5≤0. Another example: rearranging the left side of the inequality 5+0.6 x 2 −x<0 слагаемые по убыванию степени переменной, придем к равносильному квадратному неравенству в привычной форме 0,6·x 2 −x+5<0 .

At school, during algebra lessons, when they learn to solve quadratic inequalities, they also deal with solving rational inequalities, reducing to square ones. Their solution involves transferring all terms to the left side and then transforming the expression formed there to the form a·x 2 +b·x+c by executing . Let's look at an example.

Example.

Find many solutions to the inequality 3·(x−1)·(x+1)<(x−2) 2 +x 2 +5 .irrational inequality is equivalent to the quadratic inequality x 2 −6 x−9<0 , а logarithmic inequality – inequality x 2 +x−2≥0.

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.

Average level

Quadratic inequalities. Comprehensive Guide (2019)

To figure out how to solve quadratic equations, we need to understand what a quadratic function is and what properties it has.

You've probably wondered why a quadratic function is needed at all? Where is its graph (parabola) applicable? Yes, you just have to look around and you will notice that you come across it every day in everyday life. Have you noticed how a thrown ball flies in physical education? "Along the arc"? The most correct answer would be “parabola”! And along what trajectory does the jet move in the fountain? Yes, also in a parabola! How does a bullet or shell fly? That's right, also in a parabola! Thus, knowing the properties of a quadratic function, it will be possible to solve many practical problems. For example, at what angle should a ball be thrown to ensure the greatest distance? Or, where will the projectile end up if you launch it at a certain angle? etc.

Quadratic function

So, let's figure it out.

Eg, . What are the equals here, and? Well, of course!

What if, i.e. less than zero? Well, of course, we are “sad,” which means the branches will be directed downward! Let's look at the graph.

This figure shows the graph of a function. Since, i.e. less than zero, the branches of the parabola are directed downward. In addition, you probably already noticed that the branches of this parabola intersect the axis, which means that the equation has 2 roots, and the function takes both positive and negative values!

At the very beginning, when we gave the definition of a quadratic function, it was said that and are some numbers. Can they be equal to zero? Well, of course they can! I’ll even reveal an even bigger secret (which is not a secret at all, but it’s worth mentioning): there are no restrictions imposed on these numbers (and) at all!

Well, let's see what happens to the graphs if and are equal to zero.

As you can see, the graphs of the functions (and) under consideration have shifted so that their vertices are now at the point with coordinates, that is, at the intersection of the axes and, this has no effect on the direction of the branches. Thus, we can conclude that they are responsible for the “movement” of the parabola graph along the coordinate system.

The graph of a function touches the axis at a point. This means that the equation has one root. Thus, the function takes values ​​greater than or equal to zero.

We follow the same logic with the graph of the function. It touches the x-axis at a point. This means that the equation has one root. Thus, the function takes values ​​less than or equal to zero, that is.

Thus, to determine the sign of an expression, the first thing you need to do is find the roots of the equation. This will be very useful to us.

Quadratic inequality

When solving such inequalities, we will need the ability to determine where a quadratic function is greater, less, or equal to zero. That is:

• if we have an inequality of the form, then in fact the task comes down to determining the numerical interval of values ​​for which the parabola lies above the axis.
• if we have an inequality of the form, then in fact the task comes down to determining the numerical interval of x values ​​for which the parabola lies below the axis.

If the inequalities are not strict, then the roots (the coordinates of the intersection of the parabola with the axis) are included in the desired numerical interval; in the case of strict inequalities, they are excluded.

This is all quite formalized, but don’t despair or be scared! Now let's look at the examples, and everything will fall into place.

When solving quadratic inequalities, we will adhere to the given algorithm, and inevitable success awaits us!

Algorithm	Example:
1) Let us write down the corresponding inequality quadratic equation(just change the inequality sign to the equal sign “=”).
2) Let's find the roots of this equation.
3) Mark the roots on the axis and schematically show the orientation of the branches of the parabola (“up” or “down”)
4) Let’s place signs on the axis corresponding to the sign of the quadratic function: where the parabola is above the axis, we put “ ”, and where below - “ “.
5) Write out the interval(s) corresponding to “ ” or “ ”, depending on the inequality sign. If the inequality is not strict, the roots are included in the interval; if it is strict, they are not.

Got it? Then go ahead and pin it!

Example:

Well, did it work? If you have any difficulties, look for solutions.

Solution:

Let's write down the intervals corresponding to the sign " ", since the inequality sign is " ". The inequality is not strict, so the roots are included in the intervals:

Let's write the corresponding quadratic equation:

Let's find the roots of this quadratic equation:

Let us schematically mark the obtained roots on the axis and arrange the signs:

Let's write down the intervals corresponding to the sign " ", since the inequality sign is " ". The inequality is strict, so the roots are not included in the intervals:

Let's write the corresponding quadratic equation:

Let's find the roots of this quadratic equation:

this equation has one root

Let us schematically mark the obtained roots on the axis and arrange the signs:

Let's write down the intervals corresponding to the sign " ", since the inequality sign is " ". For any, the function takes non-negative values. Since the inequality is not strict, the answer will be.

Let's write the corresponding quadratic equation:

Let's find the roots of this quadratic equation:

Let's schematically draw a graph of a parabola and arrange the signs:

Let's write down the intervals corresponding to the sign " ", since the inequality sign is " ". For any, the function takes positive values, therefore, the solution to the inequality will be the interval:

SQUARE INEQUALITIES. AVERAGE LEVEL

Quadratic function.

Before talking about the topic “quadratic inequalities,” let us remember what a quadratic function is and what its graph is.

In other words, this polynomial of the second degree.

The graph of a quadratic function is a parabola (remember what that is?). Its branches are directed upward if "a) the function takes only positive values ​​for all, and in the second () - only negative ones:

In the case when the equation () has exactly one root (for example, if the discriminant is zero), this means that the graph touches the axis:

Then, similar to the previous case, for " .

So, we recently learned how to determine where a quadratic function is greater than zero and where it is less:

If the quadratic inequality is not strict, then the roots are included in the numerical interval; if it is strict, they are not.

If there is only one root, it’s okay, the same sign will be everywhere. If there are no roots, everything depends only on the coefficient: if "25((x)^(2))-30x+9

Answers:

2) 25((x)^(2))-30x+9>

There are no roots, so the entire expression on the left side takes the sign of the coefficient before:

• If you want to find a numerical interval on which the quadratic trinomial is greater than zero, then this is the numerical interval where the parabola lies above the axis.
• If you want to find a numerical interval on which the quadratic trinomial is less than zero, then this is the numerical interval where the parabola lies below the axis.

SQUARE INEQUALITIES. BRIEFLY ABOUT THE MAIN THINGS

Quadratic function is a function of the form: ,

The graph of a quadratic function is a parabola. Its branches are directed upward if, and downward if:

Types of quadratic inequalities:

All quadratic inequalities are reduced to the following four types:

Solution algorithm:

Algorithm	Example:
1) Let's write the quadratic equation corresponding to the inequality (simply change the inequality sign to the equal sign " ").
2) Let's find the roots of this equation.
3) Mark the roots on the axis and schematically show the orientation of the branches of the parabola (“up” or “down”)
4) Let’s place signs on the axis corresponding to the sign of the quadratic function: where the parabola is above the axis, we put “ ”, and where below - “ “.
5) Write down the interval(s) corresponding to “ ” or “ ”, depending on the inequality sign. If the inequality is not strict, the roots are included in the interval; if it is strict, they are not.

The concept of mathematical inequality arose in ancient times. This happened when primitive man There was a need to compare their quantity and size when counting and handling various objects. Since ancient times, Archimedes, Euclid and other famous scientists: mathematicians, astronomers, designers and philosophers used inequalities in their reasoning.

But they, as a rule, used verbal terminology in their works. For the first time, modern signs to denote the concepts of “more” and “less” in the form in which every schoolchild knows them today were invented and put into practice in England. The mathematician Thomas Harriot provided such a service to his descendants. And this happened about four centuries ago.

There are many types of inequalities known. Among them are simple ones, containing one, two or more variables, quadratic, fractional, complex ratios, and even those represented by a system of expressions. The best way to understand how to solve inequalities is to use various examples.

Don't miss the train

To begin with, let’s imagine that a resident rural areas hurries to railway station, which is located at a distance of 20 km from his village. In order not to miss the train leaving at 11 o'clock, he must leave the house on time. At what time should this be done if its speed is 5 km/h? The solution to this practical problem comes down to fulfilling the conditions of the expression: 5 (11 - X) ≥ 20, where X is the departure time.

This is understandable, because the distance that a villager needs to cover to the station is equal to the speed of movement multiplied by the number of hours on the road. Come formerly man maybe, but there’s no way he can be late. Knowing how to solve inequalities and applying your skills in practice, you will end up with X ≤ 7, which is the answer. This means that the villager should go to the railway station at seven in the morning or a little earlier.

Numerical intervals on a coordinate line

Now let's find out how to map the described relations onto the The inequality obtained above is not strict. It means that the variable can take values ​​less than 7, or it can be equal to this number. Let's give other examples. To do this, carefully consider the four figures presented below.

On the first of them you can see a graphical representation of the interval [-7; 7]. It consists of a set of numbers placed on a coordinate line and located between -7 and 7, including the boundaries. In this case, the points on the graph are depicted as filled circles, and the interval is recorded using

The second figure is a graphical representation of the strict inequality. In this case, the borderline numbers -7 and 7, shown by punctured (not filled in) dots, are not included in the specified set. And the interval itself is written in parentheses as follows: (-7; 7).

That is, having figured out how to solve inequalities of this type and received a similar answer, we can conclude that it consists of numbers that are between the boundaries in question, except -7 and 7. The next two cases must be evaluated in a similar way. The third figure shows images of the intervals (-∞; -7] U)