Of the form ax 2 + bx + 0 0, where (instead of the > sign there can, of course, be any other inequality sign). We have all the theoretical facts necessary to solve such inequalities, as we will now see.
Example 1. Solve inequality:
a) x 2 - 2x - 3 >0; b) x 2 - 2x - 3< 0;
c) x 2 - 2x - 3 > 0; d) x 2 - 2x - 3< 0.
Solution,
a) Consider the parabola y = x 2 - 2x - 3, shown in Fig. 117.
Solving the inequality x 2 - 2x - 3 > 0 means answering the question at what values of x the ordinates of the points of the parabola are positive.
We note that y > 0, i.e. the graph of the function is located above the x axis, at x< -1 или при х > 3.
This means that the solutions to the inequality are all points of the open beam(- 00 , - 1), as well as all points of the open beam (3, +00).
Using the sign U (the sign for combining sets), the answer can be written as follows: (-00, - 1) U (3, +00). However, the answer can be written like this: x< - 1; х > 3.
b) Inequality x 2 - 2x - 3< 0, или у < 0, где у = х 2 - 2х - 3, также можно решить с помощью рис. 117: schedule located below the x-axis if -1< х < 3. Поэтому решениями данного неравенства служат все точки интервала (- 1, 3).
c) The inequality x 2 - 2x - 3 > 0 differs from the inequality x 2 - 2x - 3 > 0 in that the answer must also include the roots of the equation x 2 - 2x - 3 = 0, i.e. points x = -1
and x = 3. Thus, the solutions to this non-strict inequality are all points of the ray (-00, - 1], as well as all points of the ray.
Practical mathematicians usually say this: why should we, when solving the inequality ax 2 + bx + c > 0, carefully construct a parabola graph of a quadratic function
y = ax 2 + bx + c (as was done in example 1)? It is enough to make a schematic sketch of the graph, for which you just need to find roots quadratic trinomial (the point of intersection of the parabola with the x-axis) and determine whether the branches of the parabola are directed up or down. This schematic sketch will give a visual interpretation of the solution to the inequality.
Example 2. Solve the inequality - 2x 2 + 3x + 9< 0.
Solution.
1) Find the roots of the square trinomial - 2x 2 + 3x + 9: x 1 = 3; x 2 = - 1.5.
2) The parabola, which serves as a graph of the function y = -2x 2 + 3x + 9, intersects the x axis at points 3 and - 1.5, and the branches of the parabola are directed downward, since the highest coefficient- negative number - 2. In Fig. 118 shows a sketch of the graph.
3) Using fig. 118, we conclude:< 0 на тех промежутках оси х, где график расположен ниже оси х, т.е. на открытом луче (-оо, -1,5) или на открытом луче C, +оо).
Answer: x< -1,5; х > 3.
Example 3. Solve the inequality 4x 2 - 4x + 1< 0.
Solution.
1) From the equation 4x 2 - 4x + 1 = 0 we find .
2) A square trinomial has one root; this means that the parabola serving as the graph of a quadratic trinomial does not intersect the x-axis, but touches it at point . The branches of the parabola are directed upward (Fig. 119.)
3) Using the geometric model presented in Fig. 119, we establish that the given inequality is satisfied only at the point, since for all other values of x the ordinates of the graph are positive.
Answer: .
You probably noticed that in fact, in examples 1, 2, 3, a very specific algorithm solution of quadratic inequalities, let's formalize it.
Algorithm for solving the quadratic inequality ax 2 + bx + 0 0 (ax 2 + bx + c< 0)
The first step of this algorithm is to find the roots of a quadratic trinomial. But the roots may not exist, so what can we do? Then the algorithm is not applicable, which means we need to think differently. The key to these arguments is given by the following theorems.
In other words, if D< 0, а >0, then the inequality ax 2 + bx + c > 0 holds for all x; on the contrary, the inequality ax 2 + bx + c< 0 не имеет решений.
Proof.
Schedule functions y = ax 2 + bx + c is a parabola whose branches are directed upward (since a > 0) and which does not intersect the x axis, since the quadratic trinomial has no roots by condition. The graph is shown in Fig. 120. We see that for all x the graph is located above the x axis, which means that for all x the inequality ax 2 + bx + c > 0 holds, which is what needed to be proven.
In other words, if D< 0, а < 0, то неравенство ах 2 + bх + с < 0 выполняется при всех х; напротив, неравенство ах 2 + bх + с >0 has no solutions.
Proof. The graph of the function y = ax 2 + bx +c is a parabola, the branches of which are directed downward (since a< 0) и которая не пересекает ось х, так как корней у квадратного трехчлена по условию нет. График представлен на рис. 121. Видим, что при всех х график расположен ниже оси х, а это значит, что при всех х выполняется неравенство ах 2 + bх + с < 0, что и требовалось доказать.
Example 4. Solve inequality:
a) 2x 2 - x + 4 >0; b) -x 2 + 3x - 8 >0.
a) Find the discriminant of the square trinomial 2x 2 - x + 4. We have D = (-1) 2 - 4 2 4 = - 31< 0.
The leading coefficient of the trinomial (number 2) is positive.
This means, according to Theorem 1, for all x the inequality 2x 2 - x + 4 > 0 holds, that is, the solution to the given inequality is the whole (-00, + 00).
b) Find the discriminant of the square trinomial - x 2 + 3x - 8. We have D = 32 - 4 (- 1) (- 8) = - 23< 0. Старший коэффициент трехчлена (число - 1) отрицателен. Следовательно, по теореме 2, при всех х выполняется неравенство - х 2 + Зx - 8 < 0. Это значит, что неравенство - х 2 + Зх - 8 0 не выполняется ни при каком значении х, т. е. заданное неравенство не имеет решений.
Answer: a) (-00, + 00); b) no solutions.
In the following example, we will introduce another method of reasoning that is used to solve quadratic inequalities.
Example 5. Solve the inequality 3x 2 - 10x + 3< 0.
Solution. Let's decompose quadratic trinomial 3x 2 - 10x + 3 for multipliers. The roots of the trinomial are the numbers 3 and , so using ax 2 + bx + c = a (x - x 1)(x - x 2), we get 3x 2 - 10x + 3 = 3(x - 3) (x - )
Let us mark the roots of the trinomial on the number line: 3 and (Fig. 122).
Let x > 3; then x-3>0 and x->0, and therefore the product 3(x - 3)(x - ) is positive. Next, let< х < 3; тогда x-3< 0, а х- >0. Therefore, the product 3(x-3)(x-) is negative. Finally, let x<; тогда x-3< 0 и x- < 0. Но в таком случае произведение
3(x -3)(x -) is positive.
Summarizing the reasoning, we come to the conclusion: the signs of the square trinomial 3x 2 - 10x + 3 change as shown in Fig. 122. We are interested in at what x the square trinomial takes negative values. From Fig. 122 we conclude: the square trinomial 3x 2 - 10x + 3 takes negative values for any value of x from the interval (, 3)
Answer (, 3), or< х < 3.
Comment. The reasoning method we used in Example 5 is usually called the method of intervals (or the method of intervals). It is actively used in mathematics to solve rational inequalities In 9th grade we will study the interval method in more detail.
Example 6. At what values of the parameter p is the quadratic equation x 2 - 5x + p 2 = 0:
a) has two different roots;
b) has one root;
c) has no roots?
Solution. Number of roots quadratic equation depends on the sign of its discriminant D. In this case we find D = 25 - 4p 2.
a) The quadratic equation has two different roots, if D>0, then the problem is reduced to solving the inequality 25 - 4р 2 > 0. Let's multiply both sides of this inequality by -1 (not forgetting to change the sign of the inequality). We obtain the equivalent inequality 4p 2 - 25< 0. Далее имеем 4 (р - 2,5) (р + 2,5) < 0.
The signs of the expression 4(p - 2.5) (p + 2.5) are shown in Fig. 123.
We conclude that inequality 4(p - 2.5)(p + 2.5)< 0 выполняется для всех значений р из интервала (-2,5; 2,5). Именно при этих значениях параметра р данное квадратное уравнение имеет два различных корня.
b) quadratic equation has one root if D - 0.
As we established above, D = 0 at p = 2.5 or p = -2.5.
It is for these values of the parameter p that this quadratic equation has only one root.
c) A quadratic equation has no roots if D< 0. Решим неравенство 25 - 4р 2 < 0.
We get 4p 2 - 25 > 0; 4 (p-2.5)(p + 2.5)>0, whence (see Fig. 123) p< -2,5; р >2.5. For these values of the parameter p, this quadratic equation has no roots.
Answer: a) at p (-2.5, 2.5);
b) at p = 2.5 or = -2.5;
c) at p< - 2,5 или р > 2,5.
Mordkovich A. G., Algebra. 8th grade: Textbook. for general education institutions. - 3rd ed., revised. - M.: Mnemosyne, 2001. - 223 p.: ill.
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Interval method– a simple way to solve fractional rational inequalities. This is the name for inequalities containing rational (or fractional-rational) expressions that depend on a variable.
1. Consider, for example, the following inequality
The interval method allows you to solve it in a couple of minutes.
On the left side of this inequality – fractional rational function. Rational because it does not contain roots, sines, or logarithms - only rational expressions. On the right is zero.
The interval method is based on the following property of a fractional rational function.
A fractional rational function can change sign only at those points at which it is equal to zero or does not exist.
Let us recall how a quadratic trinomial is factored, that is, an expression of the form .
Where and are the roots of the quadratic equation.
We draw an axis and place the points at which the numerator and denominator go to zero.
The zeros of the denominator and are punctured points, since at these points the function on the left side of the inequality is not defined (you cannot divide by zero). The zeros of the numerator and - are shaded, since the inequality is not strict. When and our inequality is satisfied, since both its sides are equal to zero.
These points break the axis into intervals.
Let us determine the sign of the fractional rational function on the left side of our inequality on each of these intervals. We remember that a fractional rational function can change sign only at those points at which it is equal to zero or does not exist. This means that at each of the intervals between the points where the numerator or denominator goes to zero, the sign of the expression on the left side of the inequality will be constant - either “plus” or “minus”.
And therefore, to determine the sign of the function on each such interval, we take any point belonging to this interval. The one that is convenient for us.
. Take, for example, and check the sign of the expression on the left side of the inequality. Each of the "brackets" is negative. The left side has a sign.
Next interval: . Let's check the sign at . We find that the left side has changed its sign to .
Let's take it. When the expression is positive - therefore, it is positive over the entire interval from to.
When the left side of the inequality is negative.
And finally, class="tex" alt="x>7"> . Подставим и проверим знак выражения в левой части неравенства. Каждая "скобочка" положительна. Следовательно, левая часть имеет знак .!}
We have found at what intervals the expression is positive. All that remains is to write down the answer:
Answer: .
Please note: the signs alternate between intervals. This happened because when passing through each point, exactly one of the linear factors changed sign, while the rest kept it unchanged.
We see that the interval method is very simple. To solve the fractional-rational inequality using the interval method, we reduce it to the form:
Or class="tex" alt="\genfrac())()(0)(\displaystyle P\left(x \right))(\displaystyle Q\left(x \right)) > 0"> !}, or , or .
(on the left side is a fractional rational function, on the right side is zero).
Then we mark on the number line the points at which the numerator or denominator goes to zero.
These points divide the entire number line into intervals, on each of which the fractional-rational function retains its sign.
All that remains is to find out its sign at each interval.
We do this by checking the sign of the expression at any point belonging to a given interval. After that, we write down the answer. That's it.
But the question arises: do the signs always alternate? No, not always! You must be careful and not place signs mechanically and thoughtlessly.
2. Let's consider another inequality.
Class="tex" alt="\genfrac())()(0)(\displaystyle \left(x-2 \right)^2)(\displaystyle \left(x-1 \right) \left(x-3 \right))>0"> !}
Place the points on the axis again. The dots and are punctured because they are zeros of the denominator. The point is also cut out, since the inequality is strict.
When the numerator is positive, both factors in the denominator are negative. This can be easily checked by taking any number from a given interval, for example, . The left side has the sign:
When the numerator is positive; The first factor in the denominator is positive, the second factor is negative. The left side has the sign:
The situation is the same! The numerator is positive, the first factor in the denominator is positive, the second is negative. The left side has the sign:
Finally, with class="tex" alt="x>3">
все множители положительны, и левая часть имеет знак :!} 
Answer: .
Why was the alternation of signs disrupted? Because when passing through a point the multiplier is “responsible” for it didn't change sign. Consequently, the entire left side of our inequality did not change sign.
Conclusion: if the linear multiplier is an even power (for example, squared), then when passing through a point the sign of the expression on the left side does not change. In the case of an odd degree, the sign, of course, changes.
3. Let's consider a more complex case. It differs from the previous one in that the inequality is not strict:
The left side is the same as in previous task. The picture of signs will be the same:
Maybe the answer will be the same? No! A solution is added This happens because at both the left and right sides of the inequality are equal to zero - therefore, this point is a solution.
Answer: .
This situation often occurs in problems on the Unified State Examination in mathematics. This is where applicants fall into a trap and lose points. Be careful!
4. What to do if the numerator or denominator cannot be factored into linear factors? Consider this inequality:
A square trinomial cannot be factorized: the discriminant is negative, there are no roots. But this is good! This means that the sign of the expression for all is the same, and specifically, positive. You can read more about this in the article on properties of quadratic functions.
And now we can divide both sides of our inequality by a value that is positive for all. Let us arrive at an equivalent inequality:
Which is easily solved using the interval method.
Please note that we divided both sides of the inequality by a value that we knew for sure was positive. Of course, in general, you should not multiply or divide an inequality by variable value, whose sign is unknown.
5 . Let's consider another inequality, seemingly quite simple:
I just want to multiply it by . But we are already smart, and we won’t do this. After all, it can be both positive and negative. And we know that if both sides of the inequality are multiplied by a negative value, the sign of the inequality changes.
We will do it differently - we will collect everything in one part and bring it to a common denominator. The right side will remain zero:
Class="tex" alt="\genfrac())()()(0)(\displaystyle x-2)(\displaystyle x)>0"> !}
And after that - apply interval method.
Inequalities are called linear the left and right sides of which are linear functions with respect to the unknown quantity. These include, for example, inequalities:
2x-1-x+3; 7x0;
5 >4 - 6x 9- x< x + 5 .
1) Strict inequalities: ax +b>0 or ax+b<0
2) Non-strict inequalities: ax +b≤0 or ax+b≫ 0
Let's analyze this task. One of the sides of the parallelogram is 7 cm. What must be the length of the other side so that the perimeter of the parallelogram is greater than 44 cm?
Let the required side be X cm. In this case, the perimeter of the parallelogram will be represented by (14 + 2x) cm. The inequality 14 + 2x > 44 is a mathematical model of the problem of the perimeter of a parallelogram. If we replace the variable in this inequality X on, for example, the number 16, then we obtain the correct numerical inequality 14 + 32 > 44. In this case, they say that the number 16 is a solution to the inequality 14 + 2x > 44.
Solving the inequality name the value of a variable that turns it into a true numerical inequality.
Therefore, each of the numbers is 15.1; 20;73 act as a solution to the inequality 14 + 2x > 44, but the number 10, for example, is not its solution.
Solve inequality means to establish all its solutions or to prove that there are no solutions.
The formulation of the solution to the inequality is similar to the formulation of the root of the equation. And yet it is not customary to designate the “root of inequality.”
The properties of numerical equalities helped us solve equations. Similarly, the properties of numerical inequalities will help solve inequalities.
When solving an equation, we replace it with another, simpler equation, but equivalent to the given one. The answer to inequalities is found in a similar way. When changing an equation to an equivalent equation, they use the theorem about transferring terms from one side of the equation to the opposite and about multiplying both sides of the equation by the same non-zero number. When solving an inequality, there is a significant difference between it and an equation, which lies in the fact that any solution to an equation can be verified simply by substitution into the original equation. In inequalities, this method is absent, since it is not possible to substitute countless solutions into the original inequality. Therefore, there is an important concept, these arrows<=>is a sign of equivalent, or equivalent, transformations. The transformation is called equivalent, or equivalent, if they do not change the set of solutions.
Similar rules for solving inequalities.
If we move any term from one part of the inequality to another, replacing its sign with the opposite one, we obtain an inequality equivalent to this one.
If both sides of the inequality are multiplied (divided) by the same positive number, we obtain an inequality equivalent to this one.
If both sides of the inequality are multiplied (divided) by the same negative number, replacing the inequality sign with the opposite one, we obtain an inequality equivalent to the given one.
Using these rules Let us calculate the following inequalities.
1) Let's analyze the inequality 2x - 5 > 9.
This linear inequality, we will find its solution and discuss the basic concepts.
2x - 5 > 9<=>2x>14(5 was moved to the left side with the opposite sign), then we divided everything by 2 and we have x > 7. Let us plot the set of solutions on the axis x
We have obtained a positively directed beam. We note the set of solutions either in the form of inequality x > 7, or in the form of the interval x(7; ∞). What is a particular solution to this inequality? For example, x = 10 is a particular solution to this inequality, x = 12- this is also a particular solution to this inequality.
There are many partial solutions, but our task is to find all the solutions. And there are usually countless solutions.
Let's sort it out example 2:
2) Solve inequality 4a - 11 > a + 13.
Let's solve it: A move it to one side 11 move it to the other side, we get 3a< 24, и в результате после деления обеих частей на 3 the inequality has the form a<8 .
4a - 11 > a + 13<=>3a< 24 <=>a< 8 .
We will also display the set a< 8 , but already on the axis A.
We either write the answer in the form of inequality a< 8, либо A(-∞;8), 8 does not turn on.
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For example, the inequality is the expression \(x>5\).
Types of inequalities:
If \(a\) and \(b\) are numbers or , then the inequality is called numerical. It's actually just comparing two numbers. Such inequalities are divided into faithful And unfaithful.
For example:
\(-5<2\) - верное числовое неравенство, ведь \(-5\) действительно меньше \(2\);
\(17+3\geq 115\) is an incorrect numerical inequality, since \(17+3=20\), and \(20\) is less than \(115\) (and not greater than or equal to).
If \(a\) and \(b\) are expressions containing a variable, then we have inequality with variable. Such inequalities are divided into types depending on the content:
|
\(2x+1\geq4(5-x)\) |
Variable only to the first power |
|||
|
\(3x^2-x+5>0\) |
There is a variable in the second power (square), but there are no higher powers (third, fourth, etc.) |
|||
|
\(\log_(4)((x+1))<3\) |
||||
|
\(2^(x)\leq8^(5x-2)\) |
What is the solution to an inequality?
If you substitute a number instead of a variable into an inequality, it will turn into a numeric one.
If a given value for x turns the original inequality into a true numerical one, then it is called solution to inequality. If not, then this value is not a solution. And so that solve inequality– you need to find all its solutions (or show that there are none).
For example, if we substitute the number \(7\) into the linear inequality \(x+6>10\), we get the correct numerical inequality: \(13>10\). And if we substitute \(2\), there will be an incorrect numerical inequality \(8>10\). That is, \(7\) is a solution to the original inequality, but \(2\) is not.
However, the inequality \(x+6>10\) has other solutions. Indeed, we will get the correct numerical inequalities when substituting \(5\), and \(12\), and \(138\)... And how can we find all possible solutions? For this they use For our case we have:
\(x+6>10\) \(|-6\)
\(x>4\)
That is, any number greater than four will suit us. Now you need to write down the answer. Solutions to inequalities are usually written numerically, additionally marking them on the number axis with shading. For our case we have:
Answer:
\(x\in(4;+\infty)\)
When does the sign of an inequality change?
There is one big trap in inequalities that students really “love” to fall into:
When multiplying (or dividing) an inequality by a negative number, it is reversed (“more” by “less”, “more or equal” by “less than or equal”, and so on)
Why is this happening? To understand this, let's look at the transformations of the numerical inequality \(3>1\). It is correct, three is indeed greater than one. First, let's try to multiply it by any positive number, for example, two:
\(3>1\) \(|\cdot2\)
\(6>2\)
As we can see, after multiplication the inequality remains true. And no matter what positive number we multiply by, we will always get the correct inequality. Now let’s try to multiply by a negative number, for example, minus three:
\(3>1\) \(|\cdot(-3)\)
\(-9>-3\)
The result is an incorrect inequality, because minus nine is less than minus three! That is, in order for the inequality to become true (and therefore, the transformation of multiplication by negative was “legal”), you need to reverse the comparison sign, like this: \(−9<− 3\).
With division it will work out the same way, you can check it yourself.
The rule written above applies to all types of inequalities, not just numerical ones.
Example: Solve the inequality \(2(x+1)-1<7+8x\)Solution:
|
\(2x+2-1<7+8x\) |
Let's move \(8x\) to the left, and \(2\) and \(-1\) to the right, not forgetting to change the signs |
|
\(2x-8x<7-2+1\) |
|
|
\(-6x<6\) \(|:(-6)\) |
Let's divide both sides of the inequality by \(-6\), not forgetting to change from “less” to “more” |
|
Let's mark a numerical interval on the axis. Inequality, therefore we “prick out” the value \(-1\) itself and do not take it as an answer |
|
|
Let's write the answer as an interval |
Answer: \(x\in(-1;\infty)\)
Inequalities and disability
Inequalities, just like equations, can have restrictions on , that is, on the values of x. Accordingly, those values that are unacceptable according to the DZ should be excluded from the range of solutions.
Example: Solve the inequality \(\sqrt(x+1)<3\)
Solution: It is clear that in order for the left side to be less than \(3\), the radical expression must be less than \(9\) (after all, from \(9\) just \(3\)). We get:
\(x+1<9\) \(|-1\)
\(x<8\)
All? Any value of x smaller than \(8\) will suit us? No! Because if we take, for example, the value \(-5\) that seems to fit the requirement, it will not be a solution to the original inequality, since it will lead us to calculating the root of a negative number.
\(\sqrt(-5+1)<3\)
\(\sqrt(-4)<3\)
Therefore, we must also take into account the restrictions on the value of X - it cannot be such that there is a negative number under the root. Thus, we have the second requirement for x:
\(x+1\geq0\)
\(x\geq-1\)
And for x to be the final solution, it must satisfy both requirements at once: it must be less than \(8\) (to be a solution) and greater than \(-1\) (to be admissible in principle). Plotting it on the number line, we have the final answer:
Answer: \(\left[-1;8\right)\)