How to determine whether a quadratic inequality has a solution. Systems of linear and quadratic inequalities

Consider an example where there is a negative coefficient in front of “x 2” in the quadratic inequality.

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.

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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 of solving quadratic inequalities, which we are now considering, is not called graphical in school algebra textbooks. 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, very often one has to deal with inequalities that can be reduced 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.

Definition of quadratic inequality

Note 1

The inequality is called quadratic because the variable is squared. Quadratic inequalities are also called inequalities of the second degree.

Example 1

Example.

$7x^2-18x+3 0$, $11z^2+8 \le 0$ – quadratic inequalities.

As can be seen from the example, not all elements of the inequality of the form $ax^2+bx+c > 0$ are present.

For example, in the inequality $\frac(5)(11) y^2+\sqrt(11) y>0$ there is no free term (term $с$), and in the inequality $11z^2+8 \le 0$ there is no term with coefficient $b$. Such inequalities are also quadratic, but they are also called incomplete quadratic inequalities. This just means that the coefficients $b$ or $c$ are equal to zero.

Methods for solving quadratic inequalities

When solving quadratic inequalities, the following basic methods are used:

• graphic;
• interval method;
• isolating the square of a binomial.

Graphic method

Note 2

Graphical method for solving quadratic inequalities $ax^2+bx+c > 0$ (or with the $ sign

These intervals are solving the quadratic inequality.

Interval method

Note 3

Interval method for solving quadratic inequalities of the form $ax^2+bx+c > 0$ (the inequality sign can also be $

Solutions to quadratic inequalities with the sign $""$ - positive intervals, with the signs $"≤"$ and $"≥"$ - negative and positive intervals (respectively), including points that correspond to the zeros of the trinomial.

Isolating the square of a binomial

The method for solving a quadratic inequality by isolating the square of the binomial is to pass to an equivalent inequality of the form $(x-n)^2 > m$ (or with the sign $

Inequalities that reduce to quadratic

Note 4

Often, when solving inequalities, they need to be reduced to quadratic inequalities of the form $ax^2+bx+c > 0$ (the inequality sign can also be $ inequalities that reduce to quadratic ones.

Note 5

The simplest way to reduce inequalities to quadratic ones is to rearrange the terms in the original inequality or transfer them, for example, from the right side to the left.

For example, when transferring all terms of the inequality $7x > 6-3x^2$ from the right side to the left, we obtain a quadratic inequality of the form $3x^2+7x-6 > 0$.

If we rearrange the terms on the left side of the inequality $1.5y-2+5.3x^2 \ge 0$ in descending order of the degree of the variable $y$, then this will lead to an equivalent quadratic inequality of the form $5.3x^2+1.5y-2 \ge 0$.

When solving rational inequalities, they are often reduced to quadratic inequalities. In this case, it is necessary to transfer all terms to the left side and transform the resulting expression to the form of a quadratic trinomial.

Example 2

Example.

Reduce the inequality $7 \cdot (x+0.5) \cdot x > (3+4x)^2-10x^2+10$ to a quadratic one.

Solution.

Let's move all the terms to the left side of the inequality:

$7 \cdot (x+0.5) \cdot x-(3+4x)^2+10x^2-10 > 0$.

Using abbreviated multiplication formulas and opening parentheses, we simplify the expression on the left side of the inequality:

$7x^2+3.5x-9-24x-16x^2+10x^2-10 > 0$;

$x^2-21.5x-19 > 0$.

Answer: $x^2-21.5x-19 > 0$.

A quadratic inequality is an inequality in which a variable is squared ( x 2 (\displaystyle x^(2))) and has two roots. The graph of such an inequality is a parabola and intersects the X-axis at two points. Solving the inequality implies finding such values x (\displaystyle x), for which the inequality is true. The roots of inequalities can be written in algebraic form and also displayed on the number line or coordinate plane.

Steps

Part 1

Write the inequality in standard form. The standard form of the quadratic inequality is the following trinomial: a x 2 + b x + c< 0 {\displaystyle ax^{2}+bx+c<0} , Where a (\displaystyle a), b (\displaystyle b), c (\displaystyle c)– coefficients, and a ≠ 0 (\displaystyle a\neq 0).

Find two monomials that, when multiplied, will yield the first term of the inequality. To solve an inequality, you need to decompose it into two binomials (binomials), when multiplied, you get the original inequality, written in standard form. A binomial is an expression with two monomials. Remember that binomials are multiplied according to a certain rule. First, find two monomials, each of which is the first monomial of the corresponding binomial.

Find two numbers that, when multiplied, yield the third term of the inequality written in standard form. In this case, the sum of such numbers must be equal to the coefficient of the second term of the inequality. Most likely, here the numbers need to be looked for by trial and error so that they satisfy the two described conditions at once. Pay attention to the sign (“plus” or “minus”) that appears before the third term of the inequality.

Part 2

Finding the roots of inequality

1. Determine whether both binomials have the same signs. If the product of binomials is greater than zero, then both binomials will be either negative (less than 0) or positive (greater than 0), because minus by minus gives a plus, and plus by plus also gives a plus.

   Determine whether both binomials have different (opposite) signs. If the product of binomials is less than zero, then one binomial will be negative (less than 0), and the second will be positive (greater than 0), because minus times plus gives minus.

   Write down variations of two inequalities to find the roots of the original inequality. To do this, turn each binomial into an inequality, taking into account the fact that both binomials have the same or different signs.

   Solve the two inequalities of the first option. x (\displaystyle x)

   • For example, two inequalities of the first option: x+7< 0 {\displaystyle x+7<0} AND x − 3 > 0 (\displaystyle x-3>0)
   • Thus, the first pair of roots of the original inequality: x< − 7 {\displaystyle x<-7} And x > 3 (\displaystyle x>3)

2. Check the validity of the first pair of roots. To do this, find the values x (\displaystyle x)

   Solve the two inequalities of the second option. To do this, isolate the variable x (\displaystyle x) in every inequality. Remember that if you multiply or divide both sides of an inequality by a negative number, the sign of the inequality is reversed.

   • For example, two inequalities of the second option: x + 7 > 0 (\displaystyle x+7>0) AND x − 3< 0 {\displaystyle x-3<0}
   • Thus, the second pair of roots of the original inequality: x > − 7 (\displaystyle x>-7) And x< 3 {\displaystyle x<3}

3. Check the validity of the second pair of roots. To do this, find the values x (\displaystyle x), satisfying both found roots. If such values ​​exist, the roots are valid; otherwise the roots can be neglected.

Part 3

Showing roots of inequalities on the number line

Draw a number line. Do it as required (in the task or by the teacher). If there are no specific requirements, under the number line write the numbers corresponding to the previously found roots (values x (\displaystyle x)). You can also write several numbers that are greater or less than the found values; This will make it easier for you to work with the number line.

Draw circles on the number line to represent the values ​​found. x (\displaystyle x) . Draw circles directly above the numbers. If the variable is less than ( < {\displaystyle <} ) or more ( > (\displaystyle >)) of the found value, the circle is not filled in. If the variable is less than or equal to ( ≤ (\displaystyle \leq )) or greater than or equal to ( ≥ (\displaystyle \geq )) the value found, the circle is filled in because the solution set includes this value.

On the number line, shade the region that defines the solution set. If x (\displaystyle x) greater than the number found, shade the area to the right of it, because the solution set includes all values ​​that are greater than the found number. If x (\displaystyle x) is less than the found number, shade the area to the left of it because the solution set includes all values ​​that are less than the found number. If the solution set lies between two numbers, shade the area between the numbers.

Part 4

Displaying the roots of inequalities on the coordinate plane

On the coordinate plane, plot the intersection points with the X axis. The found roots are the “x” coordinates of the points of intersection of the graph with the X axis.

Find the axis of symmetry. The axis of symmetry is a straight line that passes through the vertex of the parabola and divides it into two mirror-symmetrical branches. To find the axis of symmetry, use the formula x = − b 2 a (\displaystyle x=(\frac (-b)(2a))), Where a (\displaystyle a) And b (\displaystyle b) are the coefficients in the original quadratic inequality.

In this section we have collected information about quadratic inequalities and the main approaches to solving them. Let's consolidate the material with an analysis of examples.

What is a quadratic inequality

Let's see how to distinguish between different types of inequalities based on the type of recording and identify quadratic ones among them.

Definition 1

Quadratic inequality is an inequality that has the form a x 2 + b x + c< 0 , where a, b and c– some numbers, and a not equal to zero. x is a variable, and in place of the sign < Any other inequality sign can appear.

The second name for quadratic equations is the name “inequalities of the second degree”. The presence of the second name can be explained as follows. On the left side of the inequality there is a polynomial of the second degree - a square trinomial. Applying the term “quadratic inequalities” to quadratic inequalities is incorrect, since functions that are given by equations of the form are quadratic y = a x 2 + b x + c.

Here is an example of a quadratic inequality:

Example 1

Let's take 5 x 2 − 3 x + 1 > 0. In this case a = 5, b = − 3 and c = 1.

Or this inequality:

Example 2

− 2 , 2 z 2 − 0 , 5 z − 11 ≤ 0, where a = − 2, 2, b = − 0, 5 and c = − 11.

Let's show some examples of quadratic inequalities:

Example 3

Particular attention should be paid to the fact that the coefficient at x 2 is considered not equal to zero. This is explained by the fact that otherwise we get a linear inequality of the form b x + c > 0, since a quadratic variable when multiplied by zero will itself become equal to zero. At the same time, the coefficients b And c can be equal to zero both together and separately.

Example 4

An example of such inequality x 2 − 5 ≥ 0.

Methods for solving quadratic inequalities

There are three main methods:

Definition 2

• graphic;
• interval method;
• by selecting the square of the binomial on the left side.

Graphical method

The method involves constructing and analyzing a graph of a quadratic function y = a x 2 + b x + c for quadratic inequalities a x 2 + b x + c< 0 (≤ , >, ≥) . The solution to the quadratic inequality is the intervals or intervals at which the specified function takes on positive and negative values.

Interval method

You can solve a quadratic inequality in one variable using the interval method. The method is applicable to solving any type of inequalities, not only quadratic ones. The essence of the method is to determine the signs of the intervals into which the coordinate axis is divided by the zeros of the trinomial a x 2 + b x + c if available.

For inequality a x 2 + b x + c< 0 solutions are intervals with a minus sign, for the inequality a x 2 + b x + c > 0, spaces with a plus sign. If we are dealing with loose inequalities, then the solution becomes an interval that includes points that correspond to the zeros of the trinomial.

Isolating the square of a binomial

The principle of isolating the square of a binomial on the left side of a quadratic inequality is to perform equivalent transformations that allow us to proceed to solving an equivalent inequality of the form (x − p) 2< q (≤ , >, ≥) , where p And q- some numbers.

Quadratic inequalities can be obtained using equivalent transformations from inequalities of other types. This can be done in different ways. For example, by rearranging terms in a given inequality or transferring terms from one part to another.

Let's give an example. Consider the equivalent transformation of the inequality 5 ≤ 2 x − 3 x 2. If we move all the terms from the right side to the left, we get a quadratic inequality of the form 3 x 2 − 2 x + 5 ≤ 0.

Example 5

It is necessary to find a set of solutions to the inequality 3 (x − 1) (x + 1)< (x − 2) 2 + x 2 + 5 .

Solution

To solve the problem we use abbreviated multiplication formulas. To do this, we collect all the terms on the left side of the inequality, open the brackets and present similar terms:

3 · (x − 1) · (x + 1) − (x − 2) 2 − x 2 − 5< 0 , 3 · (x 2 − 1) − (x 2 − 4 · x + 4) − x 2 − 5 < 0 , 3 · x 2 − 3 − x 2 + 4 · x − 4 − x 2 − 5 < 0 , x 2 + 4 · x − 12 < 0 .

We have obtained an equivalent quadratic inequality, which can be solved graphically by determining the discriminant and the intercept points.

D ’ = 2 2 − 1 · (− 12) = 16 , x 1 = − 6 , x 2 = 2

By plotting the graph, we can see that the solution set is the interval (− 6, 2).

Answer: (− 6 , 2) .

Examples of inequalities that often reduce to quadratic inequalities include irrational and logarithmic inequalities. So, for example, the inequality 2 x 2 + 5< x 2 + 6 · x + 14

is equivalent to the quadratic inequality x 2 − 6 x − 9< 0 , and the logarithmic inequality log 3 (x 2 + x + 7) ≥ 2 – the inequality x 2 + x − 2 ≥ 0.

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