The method of intervals is considered to be a universal method for solving inequalities. This is the easiest way to use it to solve quadratic inequalities with one variable. In this material, we will consider all aspects of using the interval method to solve quadratic inequalities. To facilitate the assimilation of the material, we will consider a large number of examples of varying degrees of complexity.
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Algorithm for applying the interval method
Consider an algorithm for applying the interval method in an adapted version, which is suitable for solving quadratic inequalities. It is with this version of the interval method that students are introduced to algebra lessons. Let's not complicate the task and we.
Let's move on to the algorithm itself.
We have a square trinomial a x 2 + b x + c from the left side of the square inequality. We find zeros from this trinomial.
Draw a coordinate line in a coordinate system. We mark the roots on it. For convenience, we can introduce different ways of designating points for strict and non-strict inequalities. Let's agree that we will mark the coordinates with "empty" points when solving a strict inequality, and with ordinary points - a non-strict one. By marking the points, we get several gaps on the coordinate axis.
If at the first step we found zeros, then we determine the signs of the values of the trinomial for each of the obtained intervals. If we did not receive zeros, then we perform this action for the entire number line. We mark the gaps with the signs "+" or "-".
Additionally, we will introduce shading in those cases when we solve inequalities with signs > or ≥ and< или ≤ . В первом случае штриховка будет наноситься над промежутками, отмеченными « + », во втором над участками, отмеченными « - ».
By marking the signs of the values of the trinomial and by hatching over the segments, we obtain a geometric image of a certain numerical set, which is actually a solution to the inequality. We just need to write down the answer.
Let us dwell in more detail on the third step of the algorithm, which involves determining the sign of the gap. There are several ways to define signs. Let's consider them in order, starting with the most accurate, although not the fastest. This method involves calculating the values of the trinomial at several points of the obtained intervals.
Example 1
For example, take the trinomial x 2 + 4 · x − 5 .
The roots of this trinomial 1 and - 5 divide the coordinate axis into three intervals (− ∞ , − 5) , (− 5 , 1) and (1 , + ∞) .
Let's start with the interval (1 , + ∞) . In order to simplify the task for ourselves, let's take x \u003d 2. We get 2 2 + 4 2 − 5 = 7 .
7 is a positive number. This means that the values of this square trinomial on the interval (1 , + ∞) are positive and it can be denoted by the “+” sign.
To determine the sign of the interval (− 5 , 1) we take x = 0 . We have 0 2 + 4 0 − 5 = − 5 . We put a "-" sign above the interval.
For the interval (− ∞ , − 5) we take x = − 6 , we get (− 6) 2 + 4 · (− 6) − 5 = 7 . We mark this interval with a “+” sign.
It is much faster to determine the signs, taking into account the following facts.
With a positive discriminant, a square trinomial with two roots gives an alternation of signs of its values on the intervals into which the numerical axis is divided by the roots of this trinomial. This means that we do not have to define signs for each of the intervals. It is enough to carry out calculations for one and put down signs for the rest, taking into account the principle of alternation.
If desired, you can do without calculations altogether, drawing conclusions about the signs from the value of the leading coefficient. If a > 0 , then we get a sequence of characters + , − , + , and if a< 0 – то − , + , − .
For square trinomials with one root, when the discriminant is zero, we get two gaps on the coordinate axis with the same signs. This means that we determine the sign for one of the intervals and set the same for the second.
Here we also apply the method of determining the sign based on the value of the coefficient a: if a > 0 , then it will be + , + , and if a< 0 , то − , − .
If the square trinomial has no roots, then the signs of its values for the entire coordinate line coincide with both the sign of the leading coefficient a and the sign of the free term c.
For example, if we take a square trinomial - 4 x 2 - 7, it has no roots (its discriminant is negative). The coefficient at x 2 is a negative number - 4, and the free term - 7 is also negative. This means that on the interval (− ∞ , + ∞) its values are negative.
Consider examples of solving quadratic inequalities using the algorithm discussed above.
Example 2
Solve the inequality 8 · x 2 − 4 · x − 1 ≥ 0 .
Solution
We use the interval method to solve the inequality. To do this, we find the roots of the square trinomial 8 · x 2 − 4 · x − 1 . Due to the fact that the coefficient at x is even, it will be more convenient for us to calculate not the discriminant, but the fourth part of the discriminant: D " = (− 2) 2 − 8 (− 1) = 12.
The discriminant is greater than zero. This allows us to find two roots of the square trinomial: x 1 = 2 - 12 9 , x 1 = 1 - 3 4 and x 2 = 2 + 12 8 , x 2 = 1 + 3 4 . Note these values on the number line. Since the equation is not strict, we use ordinary points on the graph.
Now, using the interval method, we determine the signs of the three intervals obtained. The coefficient at x 2 is equal to 8, that is, it is positive, therefore, the sequence of signs will be + , − , + .
Since we are solving the inequality with the sign ≥ , we draw hatching over the gaps with plus signs: 
Let's write down analytically the numerical set according to the obtained graphic image. We can do this in two ways:
Answer:(- ∞ ; 1 - 3 4 ] ∪ [ 1 + 3 4 , + ∞) or x ≤ 1 - 3 4 , x ≥ 1 + 3 4 .
Example 3
Solve the quadratic inequality - 1 7 x 2 + 2 x - 7< 0 методом интервалов.
Solution
First, let's find the roots of the square trinomial from the left side of the inequality:
D " \u003d 1 2 - - 1 7 - 7 \u003d 0 x 0 \u003d - 1 - 1 7 x 0 \u003d 7
This is a strict inequality, so we use an “empty” point on the graph. With coordinate 7 .
Now we need to determine the signs on the obtained intervals (− ∞ , 7) and (7 , + ∞) . Since the discriminant of the square trinomial is equal to zero, and the leading coefficient is negative, we put down the signs − , − :
Since we are solving a signed inequality< , то изображаем штриховку над интервалами со знаками минус:
In this case, the solutions are both intervals (− ∞ , 7) , (7 , + ∞) .
Answer:(− ∞ , 7) ∪ (7 , + ∞) or in other notation x ≠ 7 .
Example 4
Does the quadratic inequality x 2 + x + 7< 0 решения?
Solution
Let's find the roots of the square trinomial from the left side of the inequality. To do this, we find the discriminant: D = 1 2 − 4 1 7 = 1 − 28 = − 27 . The discriminant is less than zero, so there are no real roots.
The graphic image will look like a number line without points marked on it.
Let us determine the sign of the values of the square trinomial. At D< 0 он совпадает со знаком коэффициента при x 2 , то есть, со знаком числа 1 , оно положительное, следовательно, имеем знак + :
In this case, we could apply hatching over the gaps with the “-” sign. But we do not have such gaps. So the drawing looks like this:
As a result of calculations, we got an empty set. This means that this quadratic inequality has no solutions.
Answer: No.
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It has been necessary to compare values and quantities in solving practical problems since ancient times. At the same time, such words 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 the counting of objects, the measurement and comparison of quantities. For example, the 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 of the triangle lies opposite the larger angle. Archimedes, while calculating the circumference of a circle, found 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-firsts of the diameter.
Symbolically write relationships between numbers and quantities using the > and b signs. Entries in which two numbers are connected by one of the signs: > (greater than), You also met with numerical inequalities in elementary grades. You know that inequalities may or may not be true. For example, \(\frac(1)(2) > \frac(1)(3) \) is a valid numerical inequality, 0.23 > 0.235 is an invalid numerical inequality.
Inequalities that include 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. Problems of solving inequalities in practice are posed and solved no less frequently than problems of solving equations. For example, many economic problems are reduced 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 to prove or disprove the existence of a certain object, for example, the root of an equation.
Numerical inequalities
You can compare integers and decimals. 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.
Comparison of 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 Theorem. If a > b and b > c, then a > c.
Theorem. If the same number is added to both sides of the inequality, then the sign of the inequality does not change.
Consequence. Any term can be transferred 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 parts of the inequality are divided by the same positive number, then the sign of the inequality does not change. If both parts 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 you solve the problems of evaluating and comparing expression values.
When solving various problems, it is often necessary to add or multiply term by term the left and right parts of inequalities. It is sometimes said that inequalities are added or multiplied. For example, if a tourist walked more than 20 km on the first day, and more than 25 km on the second day, then it can be argued 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 it can be argued that the area of this rectangle is less than 65 cm2.
In considering these examples, the following theorems on addition and multiplication of inequalities:
Theorem. When adding inequalities of the same sign, we get an inequality of the same sign: if a > b and c > d, then a + c > b + d.
Theorem. When multiplying inequalities of the same sign, for which the 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 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 than b.
Inequalities containing the sign \(\geq \) or the sign \(\leq \) 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 in order to solve a number of applied problems, you have to draw up a mathematical model in the form of an equation or a system of equations. Further, you will learn that the mathematical models for solving many problems are inequalities with unknowns. We will introduce the concept of solving an inequality and show how to check whether a given number is a solution to a particular inequality.
Inequalities of the form
\(ax > b, \quad ax where a and b are given numbers and x is unknown, is called linear inequalities with one unknown.
Definition. The solution of an inequality with one unknown is the value of the unknown for which this inequality turns into a true numerical inequality. To solve an inequality means to find all its solutions or establish that there are none.
You solved the equations by reducing them to the simplest equations. Similarly, when solving inequalities, one tends to reduce them with the help of properties to the form of the simplest inequalities.
Solution of 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 \(a \neq 0 \) are called second-degree inequalities with one variable.
Solving the inequality
\(ax^2+bx+c >0 \) or \(ax^2+bx+c \) can be thought of as finding gaps where 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 if the trinomial has roots;
2) if the trinomial has roots, then mark them on the x axis and schematically draw a parabola through the marked points, the branches of which are directed upwards at a > 0 or downwards at a 0 or at the bottom at a 3) find gaps 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 Solution of inequalities by the method of intervals
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 the function into intervals \((-\infty; -2), \; (-2; 3), \; (3; 5) \) and \( (5; +\infty)\)
Let us find out what are the signs of this function 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 considered intervals 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 not equal numbers. 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 the 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 not equal numbers
Considered method solving inequalities is called the method of intervals.
Let us give examples of solving inequalities by the interval method.
Solve the inequality:
\(x(0.5-x)(x+4) Obviously, the zeros of the function f(x) = x(0.5-x)(x+4) are the points \frac(1)(2) , \; x=-4 \)We plot the zeros of the function on the real axis and calculate the sign on each interval:
We select those intervals on which the function is less than or equal to zero and write down the answer.
Answer:
\(x \in \left(-\infty; \; 1 \right) \cup \left[ 4; \; +\infty \right) \)
Lesson and presentation on the topic: "Square inequalities, examples of solutions"
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Guys, we already know how to solve quadratic equations. Now let's learn how to solve quadratic inequalities.
Square inequality An inequality like this is called:
$ax^2+bx+c>0$.
The inequality sign can be any, the coefficients a, b, c are any numbers ($a≠0$).
All the rules that we defined for linear inequalities work here as well. Repeat these rules yourself!
Let's introduce another important rule:
If the trinomial $ax^2+bx+c$ has a negative discriminant, then if we substitute any value of x, the sign of the trinomial will be the same as the sign of y of the coefficient a.
Examples of solving quadratic inequality
can be solved by plotting graphs or plotting intervals. Let's see examples of solutions to inequalities.Examples.
1. Solve the inequality: $x^2-2x-8
Solution:
Find the roots of the equation $x^2-2x-8=0$.
$x_1=4$ and $x_2=-2$.
Let's plot a quadratic equation. The abscissa axis intersects at points 4 and -2. 2. Solve the inequality: $5x-6 Solution: Let's build a graph of a quadratic equation, the abscissa axis intersects at points 2 and 3. 3. Solve the inequality: $2^2+2x+1≥0$. Solution: 4. Solve the inequality: $x^2+x-2 This article contains material covering the topic " solution of square inequalities". First, it is shown what quadratic inequalities with one variable are, their general form is given. And then it is analyzed in detail how to solve quadratic inequalities. The main approaches to the solution are shown: a graphical method, a method of intervals, and by highlighting the square of the binomial on the left side of the inequality. Solutions of typical examples are given. Page navigation. Naturally, before talking about solving quadratic inequalities, one must clearly understand what a quadratic inequality is. In other words, you need to be able to distinguish square inequalities from inequalities of other types by the type of record. Definition.
Square 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 - inequality of the second degree. 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 that quadratic inequalities are called quadratic inequalities. This is not entirely correct: the definition of "quadratic" refers to functions given 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 square 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 the quadratic inequality, the coefficient a at x 2 is considered non-zero. 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 square 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 and b and c are zero. Now you can be puzzled by the question of how to solve quadratic inequalities. Basically, three main methods are used to solve: Let us make a reservation right away that the method of solving quadratic inequalities, which we are starting to consider, is not called graphical in algebra school textbooks. However, in essence, this is what he is. Moreover, the first acquaintance with graphical way of solving inequalities usually begins when the question arises of how to solve quadratic inequalities. Graphical way to solve quadratic inequalities a x 2 +b x+c<0
(≤, >, ≥) is to analyze 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 of 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. To solve square inequalities with one variable, in addition to the graphical method, the interval method is quite convenient, which in itself is very versatile, and is suitable for solving various inequalities, not just square ones. Its theoretical side lies outside the algebra course of grades 8, 9, 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 the solution of square inequalities a x 2 +b x + c<0
(≤, >, ≥), consists in determining the signs that have the values of the square 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). The gaps with minus signs make up the solutions of 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 the intervals and consider ready-made solutions for typical examples with the illustrations given by referring to the material of the article solving quadratic inequalities by the interval method. In addition to the graphical method and the interval method, there are other approaches that allow solving quadratic inequalities. And we come to one of them, which is based on squaring a binomial on the left side of the quadratic inequality. The principle of this method of solving square inequalities is to perform equivalent transformations of the inequality , allowing one to pass to the solution of an equivalent inequality of the form (x−p) 2 And how is the transition to the inequality (x−p) 2 In practice, very often one has to deal with inequalities that can be reduced with the help of 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 can be reduced to square ones. Sometimes, in order to pass to a quadratic inequality, it is enough to rearrange the terms in this inequality or transfer 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 side, then we get a quadratic inequality in the form specified above 3 x 2 −2 x+5≤0 . Another example: rearranging the inequality 5+0.6 x 2 −x on the left side<0
слагаемые по убыванию степени переменной, придем к равносильному квадратному неравенству в привычной форме 0,6·x 2 −x+5<0
. At school, in algebra lessons, when they learn to solve quadratic inequalities, they simultaneously deal with solution of rational inequalities, reducing to square. Their solution involves the transfer of all terms to the left side with the subsequent transformation of the expression formed there to the form a x 2 +b x + c by executing . Consider an example. Example. Find a set of solutions to the inequality 3 (x−1) (x+1)<(x−2) 2 +x 2 +5
.irrational inequality Bibliography. Average level To figure out how to solve quadratic equations, we need to figure out what a quadratic function is and what properties it has. Surely you 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 every day in everyday life you encounter it. Have you noticed how a thrown ball flies in physical education? "In an arc"? The most correct answer would be "in a parabola"! And along what trajectory does the jet move in the fountain? Yes, also in a parabola! And how does a bullet or projectile 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 the ball be thrown to provide the greatest range? Or where would the projectile end up if fired at a certain angle? etc. So, let's figure it out. For example, . What are equal here, and? Well, of course, and! What if, i.e. less than zero? Well, of course, we are “sad”, which means that the branches will be directed downwards! Let's look at the chart. This figure shows a graph of a function. Since, i.e. less than zero, the branches of the parabola point downwards. 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): no restrictions are 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 considered functions (u) have shifted so that their vertices are now at the point with coordinates, that is, at the intersection of the axes and, this did not affect 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 function graph touches the axis at a point. So 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. So 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 to do is to find the roots of the equation. This will be very useful to us. When solving such inequalities, we will need the ability to determine where the quadratic function is greater, less, or equal to zero. That is: If the inequalities are not strict (u), then the roots (the coordinates of the intersections of the parabola with the axis) are included in the desired numerical interval, with strict inequalities they are excluded. This is all quite formalized, but do not despair and be afraid! Now let's look at examples, and everything will fall into place. When solving quadratic inequalities, we will adhere to the above algorithm, and we will inevitably succeed! Got it? Then fasten ahead! Example: Well, did it work? If you have any difficulties, then understand the solutions. Solution: Let's write out the intervals corresponding to the sign " ", since the inequality sign is " ". The inequality is not strict, so the roots are included in the intervals: We write the corresponding quadratic equation: Find the roots of this quadratic equation: We schematically mark the obtained roots on the axis and arrange the signs: Let's write out the intervals corresponding to the sign " ", since the inequality sign is " ". The inequality is strict, so the roots are not included in the intervals: We write the corresponding quadratic equation: Find the roots of this quadratic equation: this equation has one root We schematically mark the obtained roots on the axis and arrange the signs: Let's write out the intervals corresponding to the sign " ", since the inequality sign is " ". For any function takes non-negative values. Since the inequality is not strict, the answer is Let's write the corresponding quadratic equation: Find the roots of this quadratic equation: Schematically draw a graph of a parabola and place the signs: Let's write out 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: Before talking about the topic of "square inequalities", let's remember what a quadratic function is and what its graph is. In other words, this second degree polynomial. The graph of a quadratic function is a parabola (remember what that is?). Its branches are directed upwards if "a) the function takes only positive values for all, and in the second () - only negative: 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, similarly to the previous case, for " . So, after all, we have recently learned to determine where the 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 strict, they are not. If there is only one root, it's okay, there will be the same sign 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: quadratic function is a function of the form: The graph of a quadratic function is a parabola. Its branches are directed upwards if, and downwards if: Types of square inequalities: All quadratic inequalities are reduced to the following four types: Solution algorithm:
Our square trinomial takes on values less than zero where the graph of the function is located below the x-axis.
Looking at the graph of the function, we get the answer: $x^2-2x-8 Answer: $-2
Let's transform the inequality: $-x^2+5x-6 Divide the inequality by minus one. Let's not forget to change the sign: $x^2-5x+6>0$.
Let's find the roots of the trinomial: $x_1=2$ and $x_2=3$.
Our square trinomial takes on values greater than zero where the graph of the function is located above the x-axis. Looking at the graph of the function, we get the answer: $5x-6
Let's find the roots of our trinomial, for this we calculate the discriminant: $D=2^2-4*2=-4 The discriminant is less than zero. Let's use the rule that we introduced at the beginning. The sign of the inequality will be the same as the sign of the square coefficient. In our case, the coefficient is positive, which means that our equation will be positive for any value of x.
Answer: For all x, the inequality is greater than zero.
Solution:
Let's find the roots of the trinomial and place them on the coordinate line: $x_1=-2$ and $x_2=1$. 
If $x>1$ and $x If $x>-2$ and $x Answer: $x>-2$ and $xProblems for solving quadratic inequalities
Solve inequalities:
a) $x^2-11x+30 b) $2x+15≥x^2$.
c) $3x^2+4x+3 d) $4x^2-5x+2>0$. What is a quadratic inequality?
.How to solve quadratic inequalities?
Graphically
interval method
By isolating the square of the binomial
, ≥), where p and q are some numbers.
, ≥) and how to solve it, the material of the article explains the solution of quadratic inequalities by highlighting the square of the binomial. There are also examples of solving quadratic inequalities in this way and the necessary graphic illustrations are given.
Quadratic inequalities
is equivalent to the quadratic inequality x 2 −6 x−9<0
, а logarithmic inequality 
– inequality x 2 +x−2≥0 .Square inequalities. Comprehensive Guide (2019)
quadratic function
Square inequality
Algorithm
Example:
1) Let's write the quadratic equation corresponding to the inequality (simply change the inequality sign to the equal sign "=").
2) 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 on the axis the signs corresponding to the sign of the quadratic function: where the parabola is above the axis, we put "", and where it is lower - "".
5) We 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 included.
SQUARE INEQUALITIES. AVERAGE LEVEL
Quadratic function.
A quadratic function is a function of the form
SQUARE INEQUALITIES. BRIEFLY ABOUT THE MAIN
Algorithm
Example:
1) Let's write the quadratic equation corresponding to the inequality (simply change the inequality sign to the equal sign "").
2) 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 on the axis the signs corresponding to the sign of the quadratic function: where the parabola is above the axis, we put "", and where it is lower - "".
5) We write out the interval (s) corresponding to (s) "" or "", depending on the inequality sign. If the inequality is not strict, the roots are included in the interval; if the inequality is strict, they are not included.
