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Quadratic inequalities. Comprehensive Guide (2019)

To figure out how to solve quadratic equations, we need to understand what it is quadratic function, 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 every day in Everyday life you encounter her. 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 out? 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.

Quadratic inequality – “FROM and TO”.In this article we will look at the solution of quadratic inequalities, as they say, down to the subtleties. I recommend studying the material in 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. Solving an inequality means answering the question for what values ​​of x this 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 standard form (1).

*Coefficients can be fractional and irrational, but such examples are rare in the school curriculum, and are not found at all in Unified State Examination tasks. But don’t be alarmed if, for example, you come across:

This is also a quadratic inequality.

First, let's 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, and regularly reinforce 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 are studying 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 point -). If you understand the essence, then there will be no need to learn or memorize the specified algorithm; you can easily quickly solve any quadratic inequality.

Of course, I should have immediately started the explanation with the graph of the quadratic function and an explanation of the meaning itself, but I decided to “construct” the article this way.

Another theoretical point! Look at the formula for factoring a quadratic trinomial:

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

*In order to solve a quadratic inequality, it will be necessary to factor the quadratic 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 inequality ax 2 + bx+ c > 0 (any sign).

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

2. Substitute the coefficient into formula (2) a and roots. :

a(x – x 1 )(x – x 2)>0

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

4. Determine the “signs” on the intervals (+ or –) by substituting an arbitrary “x” value from each resulting interval into the expression:

a(x – x 1 )(x – x2)

and celebrate them.

5. All that remains is to write down the intervals that interest us, they are marked:

- with a “+” sign if the inequality contained “>0” or “≥0”.

- sign “–” if the inequality included “<0» или «≤0».

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

strict – this 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 in the form ( x 1 ; x 2 ) – round brackets.

For weak inequality signs, the boundaries of the interval are included in the solution, and the answer is written in the form [ x 1 ; x 2 ] – square brackets.

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

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

EXAMPLE 1: Solve x 2 – 60 x+500 ≤ 0

Solving a quadratic equation x 2 –60 x+500=0

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

Finding the roots:

Substitute the coefficient a

x 2 –60 x+500 = (x–50)(x–10)

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

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

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

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

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

at x=20 (x–50)(x–10) = –300 < 0 верно

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

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.

At 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). In this case, it is customary to display the resulting roots in a 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). In this case, it is customary to display the root in the sketch as an UNHASHED circle.

EXAMPLE 2: Solve x 2 + 4 x–21 > 0

Solving a quadratic equation x 2 + 4 x–21 = 0

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

Finding the roots:

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 (x–3)(x+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 designations of the roots are NOT shaded. We obtained 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 for compliance with the inequality (x–3)(x+7)> 0:

at x= –10 (–10–3)(–10 +7) = 39 > 0 correct

at x= 0 (0–3)(0 +7) = –21< 0 неверно

at x=10 (10–3)(10 +7) = 119 > 0 correct

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

*Note that we have included parentheses. At x = 3 and x = –7 the inequality will be incorrect - the boundaries are not included in the solution.

Answer: x∊(–∞;–7) U (3;+∞)

EXAMPLE 3: Solve – x 2 –9 x–20 > 0

Solving 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 the roots:

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. Let's mark 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 define “signs” on 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 correct

at x= 0 – (0+5)(0 +4) = –20< 0 неверно

The solution will be the interval (–5,–4). For all values ​​of “x” belonging to it, the inequality will be true.

*Please note that boundaries are not part of the solution. For x = –5 and x = –4 the inequality will not be true.

COMMENT!

When solving a quadratic equation, we may end up with one root or no roots at all, then when using this method blindly, difficulties may arise in determining the solution.

A 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, please take a look and move on to the next section.

Using the graph of a quadratic function. I recommend!

Quadratic is a function of the form:

Its graph is a parabola, the branches of the parabola are directed upward or downward:

The graph can be positioned as follows: it can intersect the x-axis at two points, it can touch it at one point (vertex), or it can not intersect. More on this later.

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

First stage

Solving the equation x 2 +2 x–8=0.

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

Finding the roots:

We got x 1 = 2 and x 2 = – 4.

Second phase

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

Points 4 and 2 are the intersection points of the parabola and the x axis. It's simple! What did you do? We solved the quadratic equation x 2 +2 x–8=0. Check out his post like this:

0 = x 2+2x – 8

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

Now look at what values ​​of x the expression x 2 +2 x – 8 greater (or less) than zero? This is not difficult to determine from the parabola graph; as they say, everything is in 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 specified x, the trinomial x 2 +2 x –8 will be positive.

Third stage

From the parabola we can immediately see at what 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 identify the positive and negative areas in the drawing. We compare the result obtained 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 stage.

All that remains is to write down the answer.

Answer: x∊(–∞;–4) U (2;∞).

Let's summarize: having calculated the roots of the equation in the first step, we can mark the resulting points on the x-axis (these are the points of intersection of the parabola with the x-axis). Next, we schematically construct a parabola and we can already see the solution. Why schematic? We don't need a mathematically precise schedule. And imagine, for example, if the roots turn out to be 10 and 1500, try to build an accurate graph on a sheet of paper with such a range of values. The question arises! Well, we got the roots, well, we marked them on the o-axis, but should we sketch the location of the parabola itself - with its branches up or down? Everything is simple here! The coefficient for x 2 will tell you:

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

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

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

*Note! If the inequality contains a non-strict sign, 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 (punctured out), since our inequality is strict (there is a “>” sign). Moreover, in this case, the answer uses parentheses rather than square ones (borders are not included in the solution).

A lot has been written, I probably confused someone. But if you solve at least 5 inequalities using parabolas, then your admiration will know no bounds. It's simple!

So, briefly:

1. We write down the inequality and reduce it to the standard one.

2. Write down a quadratic equation and solve it.

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

4. Visually identify positive or negative areas and write down the answer to the original inequality.

Let's look at examples.

EXAMPLE 1: Solve x 2 –15 x+50 > 0

First stage.

Solving a quadratic equation x 2 –15 x+50=0

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

Finding the roots:

Second phase.

We are building the axis o. Let's mark the resulting roots. Since our inequality is strict, we will not shade them. We schematically construct a parabola, it is located with its branches up, since the coefficient of x 2 is positive:

Third stage.

We define visually positive and negative areas, here we marked them in different colors for clarity, you don’t have to do this.

We write down the answer.

Answer: x∊(–∞;5) U (10;∞).

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

EXAMPLE 2: Solve – x 2 + x+20 ≤ 0

First stage.

Solving a quadratic equation – x 2 + x+20=0

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

Finding the roots:

Second phase.

We are building the axis o. Let's mark the resulting roots. Since our inequality is not strict, we shade the designations of the roots. We schematically construct a parabola, it is located with the branches down, since the coefficient of x 2 is negative (it is equal to –1):

Third stage.

We visually identify positive and negative areas. We compare it with the original inequality (our sign is ≤ 0). The inequality will be true for x ≤ – 4 and x ≥ 5.

We write down the answer.

Answer: x∊(–∞;–4] U; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 pp.: ill. - ISBN 978-5-09 -019243-9.