Quadratic inequalities, examples, solutions. Solving quadratic inequalities by the interval method

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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
Decision:
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.
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

2. Solve the inequality: $5x-6

Decision:
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 $x_2=3$.

Let's build a graph of a quadratic equation, the abscissa axis intersects at points 2 and 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 Answer: $x 3$.

3. Solve the inequality: $2^2+2x+1≥0$.

Decision:
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.

4. Solve the inequality: $x^2+x-2
Decision:
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 $x

Problems for solving quadratic inequalities

Solve inequalities:
a) $x^2-11x+30 b) $2x+15≥x^2$.
c) $3x^2+4x+3 d) $4x^2-5x+2>0$.

Square inequality - "FROM and TO".In this article, we will consider the solution of quadratic inequalities, which is called to the subtleties. I recommend studying the material of the article carefully without missing anything. You won’t be able to master the article right away, I recommend doing it in several approaches, there is a lot of information.

Content:

Introduction. Important!

Introduction. Important!

A quadratic inequality is an inequality of the form:

If you take a quadratic equation and replace the equal sign with any of the above, you get a quadratic inequality. To solve an inequality means to answer the question for what values ​​of x the given inequality will be true. Examples:

10 x 2 – 6 x+12 ≤ 0

2 x 2 + 5 x –500 > 0

– 15 x 2 – 2 x+13 > 0

8 x 2 – 15 x+45≠ 0

The quadratic inequality can be specified implicitly, for example:

10 x 2 – 6 x+14 x 2 –5 x +2≤ 56

2 x 2 > 36

8 x 2 <–15 x 2 – 2 x+13

0> – 15 x 2 – 2 x+13

In this case, it is necessary to perform algebraic transformations and bring it to the standard form (1).

* The coefficients can be both fractional and irrational, but such examples are rare in the school curriculum, and they are not found at all in the USE assignments. But do not be afraid if, for example, you meet:

This is also a quadratic inequality.

First, consider a simple solution algorithm that does not require an understanding of what a quadratic function is and how its graph looks on the coordinate plane relative to the coordinate axes. If you are able to remember information firmly and for a long time, while regularly reinforcing it with practice, then the algorithm will help you. Also, if you, as they say, need to solve such an inequality “at once”, then the algorithm will help you. By following it, you will easily implement the solution.

If you study at school, then I strongly recommend that you start studying the article from the second part, which tells the whole meaning of the solution (see below from paragraph -). If there is an understanding of the essence, then it will not be necessary not to learn, not to memorize the specified algorithm, you can easily quickly solve any quadratic inequality.

Of course, one should immediately begin the explanation with the graph of the quadratic function and explain the meaning itself, but I decided to “build” the article in this way.

Another theoretical moment! Look at the formula for factoring a square trinomial into factors:

where x 1 and x 2 are the roots of the quadratic equation ax 2+ bx+c=0

*In order to solve the quadratic inequality, it will be necessary to factorize the square trinomial.

The algorithm presented below is also called the interval method. It is suitable for solving inequalities of the form f(x)>0, f(x)<0 , f(x)≥0 andf(x)≤0 . Please note that there can be more than two multipliers, for example:

(x–10)(x+5)(x–1)(x+104)(x+6)(x–1)<0

Solution algorithm. interval method. Examples.

Given the inequality ax 2 + bx+ c > 0 (any sign).

1. Write a quadratic equation ax 2 + bx+ c = 0 and we solve it. We get x 1 and x 2 are the roots of the quadratic equation.

2. Substitute in formula (2) coefficient a and roots. :

a(x – x 1 )(x – x 2)>0

3. Determine the intervals on the number line (the roots of the equation divide the number axis into intervals):

4. We determine the "signs" on the intervals (+ or -) by substituting an arbitrary value of "x" from each received interval into the expression:

a(x – x 1 )(x – x2)

and celebrate them.

5. It remains only to write out the intervals of interest to us, they are marked:

- sign "+" if the inequality was ">0" or "≥0".

- sign "-", if the inequality was "<0» или «≤0».

NOTE!!! The signs themselves in the inequality can be:

strict is ">", "<» и нестрогими – это «≥», «≤».

How does this affect the outcome of the decision?

With strict inequality signs, the boundaries of the interval are NOT INCLUDED in the solution, while in the answer the interval itself is written as ( x 1 ; x 2 ) are round brackets.

For non-strict inequality signs, the boundaries of the interval ENTER the solution, and the answer is written as [ x 1 ; x 2 ] – square brackets.

*This applies not only to square inequalities. The square bracket means that the interval boundary itself is included in the solution.

You will see this in the examples. Let's take a look at a few to remove all questions about this. In theory, the algorithm may seem somewhat complicated, in fact, everything is simple.

EXAMPLE 1: Decide x 2 – 60 x+500 ≤ 0

We solve a quadratic equation x 2 –60 x+500=0

D = b 2 –4 ac = (–60) 2 –4∙1∙500 = 3600–2000 = 1600

Finding roots:

We substitute the coefficient a

x 2 –60 x+500 = (x-50)(x-10)

We write the inequality in the form (х–50)(х–10) ≤ 0

The roots of the equation divide the number line into intervals. Let's show them on the number line:

We got three intervals (–∞;10), (10;50) and (50;+∞).

We determine the “signs” on the intervals, do this by substituting arbitrary values ​​​​of each received interval into the expression (x–50) (x–10) and look at the correspondence of the obtained “sign” to the sign in the inequality (х–50)(х–10) ≤ 0:

at x=2 (x–50)(x–10) = 384 > 0 is wrong

at x=20 (x–50)(x–10) = –300 < 0 верно

at x=60 (x–50)(x–10) = 500 > 0 is false

The solution will be the interval.

For all values ​​of x from this interval, the inequality will be true.

*Please note that we have included square brackets.

For x = 10 and x = 50, the inequality will also be true, that is, the boundaries are included in the solution.

Answer: x∊

Again:

- The boundaries of the interval ARE INCLUDED in the solution of the inequality when the condition contains the sign ≤ or ≥ (non-strict inequality). At the same time, it is customary to display the obtained roots in the sketch with a HASHED circle.

- The boundaries of the interval are NOT INCLUDED in the solution of the inequality when the condition contains the sign< или >(strict inequality). At the same time, it is customary to display the root in the sketch with an UNSHATCHED circle.

EXAMPLE 2: Solve x 2 + 4 x–21 > 0

We solve a quadratic equation x 2 + 4 x–21 = 0

D = b 2 –4 ac = 4 2 –4∙1∙(–21) =16+84 = 100

Finding roots:

We substitute the coefficient a and roots into formula (2), we get:

x 2 + 4 x–21 = (x–3)(x+7)

We write the inequality in the form (х–3)(х+7) > 0.

The roots of the equation divide the number line into intervals. Let's mark them on the number line:

*The inequality is not strict, so the notation of the roots is NOT shaded. We got three intervals (–∞;–7), (–7;3) and (3;+∞).

We determine the “signs” on the intervals, we do this by substituting arbitrary values ​​​​of these intervals into the expression (x–3) (x + 7) and look at the correspondence to the inequality (х–3)(х+7)> 0:

at x= -10 (-10-3)(-10 +7) = 39 > 0 true

at x \u003d 0 (0–3) (0 + 7) \u003d -21< 0 неверно

at x=10 (10–3)(10 +7) = 119 > 0 true

The solution will be two intervals (–∞;–7) and (3;+∞). For all values ​​of x from these intervals, the inequality will be true.

*Please note that we have included parentheses. For x = 3 and x = -7, the inequality will be wrong - the boundaries are not included in the solution.

Answer: x∊(–∞;–7) U (3;+∞)

EXAMPLE 3: Solve – x 2 –9 x–20 > 0

We solve a quadratic equation – x 2 –9 x–20 = 0.

a = –1 b = –9 c = –20

D = b 2 –4 ac = (–9) 2 –4∙(–1)∙ (–20) =81–80 = 1.

Finding roots:

We substitute the coefficient a and roots into formula (2), we get:

– x 2 –9 x–20 =–(x–(–5))(x–(–4))= –(x+5)(x+4)

We write the inequality in the form –(x+5)(x+4) > 0.

The roots of the equation divide the number line into intervals. Note on the number line:

*The inequality is strict, so the symbols for the roots are not shaded. We got three intervals (–∞;–5), (–5; –4) and (–4;+∞).

We determine the "signs" on the intervals, we do this by substituting into the expression –(x+5)(x+4) arbitrary values ​​of these intervals and look at the correspondence to the inequality –(x+5)(x+4)>0:

at x= -10 - (-10+5)(-10 +4) = -30< 0 неверно

at x= -4.5 - (-4.5+5)(-4.5+4) = 0.25 > 0 true

at x \u003d 0 - (0 + 5) (0 + 4) \u003d -20< 0 неверно

The solution will be the interval (-5; -4). For all values ​​\u200b\u200bof "x" belonging to it, the inequality will be true.

*Please note that boundaries are not included in the solution. For x = -5 and x = -4, the inequality will not be true.

COMMENT!

When solving a quadratic equation, we may get one root or there will be no roots at all, then when using this method blindly, it may be difficult to determine the solution.

Small summary! The method is good and convenient to use, especially if you are familiar with the quadratic function and know the properties of its graph. If not, then please read it, proceed to the next section.

Using a graph of a quadratic function. Recommend!

Quadratic is a function of the form:

Its graph is a parabola, the branches of the parabola are directed up or down:

The graph can be located as follows: it can cross the x-axis at two points, it can touch it at one point (top), it can not cross. More on this later.

Now let's look at this approach with an example. The whole decision process consists of three stages. Let's solve the inequality x 2 +2 x –8 >0.

First stage

Solve the equation x 2 +2 x–8=0.

D = b 2 –4 ac = 2 2 –4∙1∙(–8) = 4+32 = 36

Finding roots:

We got x 1 \u003d 2 and x 2 \u003d - 4.

Second phase

Building a parabola y=x 2 +2 x–8 by points:

Points - 4 and 2 are the points of intersection of the parabola and the x-axis. Everything is simple! What did they do? We have solved the quadratic equation x 2 +2 x–8=0. Check out his post like this:

0 = x2+2x-8

Zero for us is the value of "y". When y = 0, we get the abscissas of the points of intersection of the parabola with the x-axis. We can say that the zero value of "y" is the x-axis.

Now look at what values ​​of x the expression x 2 +2 x – 8 greater (or less) than zero? According to the parabola graph, this is not difficult to determine, as they say, everything is in plain sight:

1. At x< – 4 ветвь параболы лежит выше оси ох. То есть при указанных х трёхчлен x 2 +2 x –8 will be positive.

2. At -4< х < 2 график ниже оси ох. При этих х трёхчлен x 2 +2 x –8 will be negative.

3. For x > 2, the branch of the parabola lies above the x-axis. For the given x, the trinomial x 2 +2 x –8 will be positive.

Third stage

From the parabola, we can immediately see for which x the expression x 2 +2 x–8 greater than zero, equal to zero, less than zero. This is the essence of the third stage of the solution, namely to see and determine the positive and negative areas in the figure. We compare the result with the original inequality and write down the answer. In our example, it is necessary to determine all values ​​of x for which the expression x 2 +2 x–8 Above zero. We did this in the second step.

It remains to write down the answer.

Answer: x∊(–∞;–4) U (2;∞).

To summarize: having calculated the roots of the equation in the first step, we can mark the obtained points on the x-axis (these are the points of intersection of the parabola with the x-axis). Next, we schematically build a parabola and we can already see the solution. Why sketchy? We do not need a mathematically accurate schedule. Yes, and imagine, for example, if the roots turn out to be 10 and 1500, try to build an accurate graph on a sheet in a cell with such a range of values. The question arises! Well, we got the roots, well, we marked them on the x-axis, and sketch the location of the parabola itself - with the branches up or down? Everything is simple here! The coefficient at x 2 will tell you:

- if it is greater than zero, then the branches of the parabola are directed upwards.

- if less than zero, then the branches of the parabola are directed downwards.

In our example, it is equal to one, that is, it is positive.

*Note! If there is a non-strict sign in the inequality, that is, ≤ or ≥, then the roots on the number line should be shaded, this conditionally indicates that the boundary of the interval itself is included in the solution of the inequality. In this case, the roots are not shaded (punched out), since our inequality is strict (there is a “>” sign). What does the answer, in this case, put round brackets, not square brackets (boundaries are not included in the solution).

Written a lot, someone confused, probably. But if you solve at least 5 inequalities using parabolas, then there will be no limit to your admiration. Everything is simple!

So, briefly:

1. We write down the inequality, we bring it to the standard one.

2. We write down the quadratic equation and solve it.

3. Draw the x-axis, mark the obtained roots, schematically draw a parabola, branches up if the coefficient at x 2 is positive, or branches down if it is negative.

4. We determine visually positive or negative areas and write down the answer according to the original inequality.

Consider examples.

EXAMPLE 1: Decide x 2 –15 x+50 > 0

First stage.

We solve a quadratic equation x 2 –15 x+50=0

D = b 2 –4 ac = (–15) 2 –4∙1∙50 = 225–200 = 25

Finding roots:

Second phase.

We build an axis oh. Let's mark the obtained roots. Since our inequality is strict, we will not shade them. We schematically build a parabola, it is located with branches up, since the coefficient at x 2 is positive:

Third stage.

We define visually positive and negative areas, we marked them here different colors for clarity, you can not do this.

We write down the answer.

Answer: x∊(–∞;5) U (10;∞).

*The sign U denotes a union solution. Figuratively speaking, the solution is “this” AND “this” interval.

EXAMPLE 2: Solve – x 2 + x+20 ≤ 0

First stage.

We solve a quadratic equation – x 2 + x+20=0

D = b 2 –4 ac = 1 2 –4∙(–1)∙20 = 1+80 = 81

Finding roots:

Second phase.

We build an axis oh. Let's mark the obtained roots. Since our inequality is not strict, we shade the notation of the roots. We schematically build a parabola, it is located with branches down, since the coefficient at x 2 is negative (it is equal to -1):

Third stage.

We define visually positive and negative areas. Compare with the original inequality (our sign ≤ 0). The inequality will be true for x ≤ - 4 and x ≥ 5.

We write down the answer.

Answer: x∊(–∞;–4] U ∪[ \frac(2)(3);∞)\)

Quadratic inequalities with negative and zero discriminant

The algorithm above works when the discriminant is greater than zero, that is, it has a \(2\) root. What to do in other cases? For example, these:

\(1)x^2+2x+9>0\)	\(2) x^2+6x+9≤0\)	\(3)-x^2-4x-4>0\)	\(4) -x^2-64<0\)
\(D=4-36=-32<0\)			\(D=-4 \cdot 64<0\)

If \(D<0\), то квадратный трехчлен имеет постоянный знак, совпадающий со знаком коэффициента \(a\) (тем, что стоит перед \(x^2\)).

That is, the expression:
\(x^2+2x+9\) is positive for any \(x\), because \(a=1>0\)
\(-x^2-64\) - negative for any \(x\), because \(a=-1<0\)

If \(D=0\), then the square trinomial for one value \(x\) is equal to zero, and for all others it has a constant sign, which coincides with the sign of the coefficient \(a\).

That is, the expression:
\(x^2+6x+9\) is zero for \(x=-3\) and positive for all other x's, because \(a=1>0\)
\(-x^2-4x-4\) is equal to zero for \(x=-2\) and negative for all others, because \(a=-1<0\).

How to find x, at which the square trinomial is equal to zero? You need to solve the corresponding quadratic equation.

With this information, let's solve quadratic inequalities:

1) \(x^2+2x+9>0\) \(D=4-36=-32<0\)		The inequality, one might say, asks us the question: "for what \(x\) is the expression on the left greater than zero?". Above, we have already found out that for any. In the answer, you can write like this: “for any \ (x \)”, but it’s better to express the same idea in the language of mathematics.
Answer: \(x∈(-∞;∞)\)
2) \(x^2+6x+9≤0\) \(D=36-36=0\)		Question from inequality: “for what \(x\) expression on the left is less than or equal to zero?” It cannot be less than zero, but it is equal to zero - completely. And in order to find out under what claim this will happen, we will solve the corresponding quadratic equation.
		Let's build our expression on \(a^2+2ab+b^2=(a+b)^2\).
		Now we are only hindered by a square. Let's think together - what number in the square is zero? Zero! So the square of an expression is zero only if the expression itself is zero.
\(x+3=0\) \(x=-3\)		This number will be the answer.
Answer: \(-3\)
3)\(-x^2-4x-4>0\) \(D=16-16=0\)		When is the expression on the left greater than zero? As it was said above, the expression on the left is either negative or equal to zero, it cannot be positive. So the answer is never. Let's write "never" in the language of mathematics, using the symbol "empty set" - \(∅\).
Answer: \(x∈∅\)
4) \(-x^2-64<0\) \(D=-4 \cdot 64<0\)		When is the expression on the left less than zero? Always. This means that the inequality holds for any \(x\).
Answer: \(x∈(-∞;∞)\)

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

What "square inequality"? Not a question!) If you take any quadratic equation and change the sign in it "=" (equal) to any inequality icon ( > ≥ < ≤ ≠ ), we get a quadratic inequality. For example:

1. x2 -8x+12 ≥ 0

2. -x 2 +3x > 0

3. x2 ≤ 4

Well, you get the idea...)

I knowingly linked equations and inequalities here. The fact is that the first step in solving any square inequality - solve the equation from which this inequality is made. For this reason - the inability to solve quadratic equations automatically leads to a complete failure in inequalities. Is the hint clear?) If anything, look at how to solve any quadratic equations. Everything is detailed there. And in this lesson we will deal with inequalities.

The inequality ready for solution has the form: left - square trinomial ax 2 +bx+c, on the right - zero. The inequality sign can be absolutely anything. The first two examples are here are ready for a decision. The third example still needs to be prepared.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Definition of quadratic inequality

Remark 1

Square inequality is called because. variable is squared. Also called quadratic inequalities inequalities of the second degree.

Example 1

Example.

$7x^2-18x+3 0$, $11z^2+8 \le 0$ are 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 $c$), but in the inequality $11z^2+8 \le 0$ there is no term with coefficient $b$. Such inequalities are also square inequalities, but they are also called incomplete quadratic inequalities. It only 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;
• selection of the square of the binomial.

Graphical way

Remark 2

A graphical way to solve square inequalities $ax^2+bx+c > 0$ (or with $ sign

These intervals are solution of the quadratic inequality.

Spacing method

Remark 3

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

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

Selection of the square of the binomial

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

Inequalities that reduce to square

Remark 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 square ones.

Remark 5

The simplest way to reduce inequalities to square ones can be 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 the terms of the inequality $7x > 6-3x^2$ from the right side to the left side, a quadratic inequality of the form $3x^2+7x-6 > 0$ is obtained.

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, one often uses their reduction to quadratic inequalities. In this case, it is necessary to transfer all the terms to the left side and convert the resulting expression to the form of a square trinomial.

Example 2

Example.

Square the inequality $7 \cdot (x+0,5) \cdot x > (3+4x)^2-10x^2+10$.

Decision.

We transfer 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 the abbreviated multiplication formulas and expanding the brackets, 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: $x^2-21.5x-19 > 0$.