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».
PAY ATTENTION!!! 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.
*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). 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 symbols for 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 stage
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 more than 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 stage.
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 stage.
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 ∪[ \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 \(2\) roots. What to do in other cases? For example, these:
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\(1) x^2+2x+9>0\) |
\(2) x^2+6x+9≤0\) |
\(3)-x^2-4x-4>0\) |
\(4)-x^2-64<0\) |
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\(D=4-36=-32<0\) |
\(D=-4 \cdot 64<0\) |
If \(D<0\), то квадратный трехчлен имеет постоянный знак, совпадающий со знаком коэффициента \(a\) (тем, что стоит перед \(x^2\)).
That is, the expression:
\(x^2+2x+9\) – positive for any \(x\), because \(a=1>0\)
\(-x^2-64\) - negative for any \(x\), because \(a=-1<0\)
If \(D=0\), then the quadratic 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 equal to zero for \(x=-3\) and positive for all other x's, because \(a=1>0\)
\(-x^2-4x-4\) - equal to zero for \(x=-2\) and negative for all others, because \(a=-1<0\).
How to find x at which the quadratic trinomial is equal to zero? We need to solve the corresponding quadratic equation.
Given this information, let's solve the quadratic inequalities:
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1) \(x^2+2x+9>0\) |
The inequality, one might say, asks us the question: “for which \(x\) is the expression on the left greater than zero?” We have already found out above that for any. In the answer you can write: “for any \(x\)”, but it is better to express the same idea in the language of mathematics. |
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Answer: \(x∈(-∞;∞)\) |
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2) \(x^2+6x+9≤0\) |
Question from inequality: “for which \(x\) is the expression on the left less than or equal to zero?” It cannot be less than zero, but it can be equal to zero. And to find out at what claim this will happen, let’s solve the corresponding quadratic equation. |
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Let's assemble our expression according to \(a^2+2ab+b^2=(a+b)^2\). |
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Now the only thing stopping us is the square. Let's think together - what number squared is equal to zero? Zero! This means that the square of an expression is equal to zero only if the expression itself is equal to zero. |
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\(x+3=0\) |
This number will be the answer. |
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Answer: \(-3\) |
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3)\(-x^2-4x-4>0\) |
When is the expression on the left greater than zero? As was already 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 “empty set” symbol - \(∅\). |
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Answer: \(x∈∅\) |
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4) \(-x^2-64<0\) |
When is the expression on the left less than zero? Always. This means that the inequality holds for any \(x\). |
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Answer: \(x∈(-∞;∞)\) |
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. It 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: $x^2-21.5x-19 > 0$.
