Ministry of Education and Youth Policy of the Chuvash Republic
Autonomous institution of the Chuvash Republic
"Tsivilsky Agrarian and Technological College"
Direction - physical and mathematical and information technology
Research:
Location of the roots of a square trinomial
Work completed:
1st year student gr.14 B
specialty "Economics"
Supervisor:
Eshmeykin
Irina Anatolievna,
mathematics teacher
Tsivilsk 2012
1. Introduction.
2. Theoretical part
2.1. The location of the roots of a square trinomial.
2.2. Ten rules for the location of the roots of a square trinomial
3. Practical part
3.1. Examples of problem solving
3.2. The location of the roots relative to one point.
3.3. Location of roots relative to two or more points.
4. Conclusions.
5. Used literature.
6. Applications
Introduction
Relevance: in the tasks of the GIA (part 2) and the USE in mathematics with a detailed answer (part C), there are tasks with parameters that often cause great difficulties for students. Moreover, students often experience psychological problems, they are afraid of such tasks, because at school and technical school they do not solve problems containing parameters much.
Difficulties in solving problems with parameters are due to the fact that the presence of a parameter forces us to solve the problem not according to a template, but to consider different cases, in each of which the solution methods differ significantly from each other.
Many problems with parameters are reduced to studying the location of the roots of a square trinomial relative to a given point or a given interval (segment, interval, ray).
Purpose of the work: to investigate the location of the roots of a square trinomial relative to a given point or a given interval.
Collect material on this topic Consider the rules for the location of the roots of a square trinomial. Solve problems using the rules for the location of the roots of a square trinomial.
Object of study: a square trinomial and the location of its roots.
1. Search - collective.
Practical significance: this material will help students who wish to continue their education at a university in preparing for the exam.
Theoretical part
2.1. Location of the roots of a square trinomial
Many problems with parameters reduce to the study of the location of the roots of a square trinomial relative to a given point or a given interval:
At what values of the parameter the roots (or root) of the quadratic equation are greater (less, no more, no less) of a given number; located between two given numbers; do not belong to the given intervals, etc., etc.
The graph of the quadratic function y \u003d ax² + in + c has the following locations relative to the x-axis.
https://pandia.ru/text/78/376/images/image002_6.jpg" align="right hspace=12" width="196" height="202"> Quadratic equation x²+px+q=0 or not has a solution (a parabola of the form D), or has one or two positive roots (C), or has one or two negative roots (A), or has roots of different signs (B).
Let us analyze the parabola C. In order for the equation to have roots, it is necessary that the discriminant D ≥ 0. Since both roots of the equation must be positive by construction, the abscissa of the vertex of the parabola, which lies between the roots, is positive, xb > 0.
The ordinate of the vertex f(xv) ≤ 0 due to the fact that we required the existence of roots.
If the condition f(0) > 0 is required, then, due to the continuity of the function under study, there is a point x1(0;xb) such that f(x1) = 0. Obviously, this is the smaller root of the equation. So, collecting all the conditions together, we get: The quadratic equation x² + px + q \u003d 0 has two roots, which can be multiples x1, x2>
Arguing in a similar way, we derive the following rules for the location of the roots of a square trinomial.
2.2. Ten rules for the location of the roots of a square trinomial
Rule 1 The quadratic equation ax2 + bx + c = 0 (a ≠ has no solutions then
and only when D< 0.
Rule 2.1. The quadratic equation (1) has two different roots if and only if,
when D > 0.
Rule 2.2. Quadratic equation (1) has two, maybe multiple roots, then and
only when D ≥ 0.
Rule 3.1. Quadratic equation (1) has two roots x1< М и х2 >M then and only
https://pandia.ru/text/78/376/images/image007_15.gif" align="left" width="74 height=42" height="42"> 
only when
Rule 4.1. The quadratic equation x2 + px + q = 0 for a ≠ 0) has two
different roots x1, x2 > M if and only if
where =
Rule 4.2. A quadratic equation has two possible multiple roots
x1, x2 > M if and only if
Rule 4.3. The quadratic equation has two different roots x1, x2 ≥ M then and
only when
https://pandia.ru/text/78/376/images/image018_3.gif" width="162" height="74 src=">
Rule 4.4. A quadratic equation has 2, can be multiple roots
x1, x2 ≥ M if and only if
https://pandia.ru/text/78/376/images/image020_2.gif" width="166" height="74 src=">
Rule 5.1. A quadratic equation has 2 different roots x1, x2< М тогда и
only when
Rule 6.1. < N < M < х2 тогда и
only when
https://pandia.ru/text/78/376/images/image026_1.gif" width="137 height=48" height="48">
Rule 6.2. The quadratic equation has roots x1 = N< М < х2
if and only if
Rule 6.3. The quadratic equation has roots x1< N < M = х2
if and only if
Rule 7.1. The quadratic equation has roots x1< m < x2 < M тогда и только
then when
https://pandia.ru/text/78/376/images/image032_0.gif" width="128 height=48" height="48"> 
Rule 7.2. To quadratic equation has roots N< x1 < M < x2 тогда и только
then when
Rule 8.1. N < x1 < x2 < M (может быть
multiple roots of N< x1 ≤ x2 < M) тогда и только тогда, когда
https://pandia.ru/text/78/376/images/image039_1.gif" width="142" height="98"> 
Rule 8.3. Quadratic equation (1) has different roots N≤ x1< x2 ≤ M (может
be multiple roots of N< x1 ≤ x2 ≤ M) тогда и только тогда, когда
Rule 8.4. Quadratic equation (1) has different roots N ≤ x1< x2 ≤ M (может
be multiple roots N ≤ x1 ≤ x2 ≤ M) if and only if
https://pandia.ru/text/78/376/images/image044_1.gif" width="144" height="98">
Rule 9 The quadratic equation has one root inside the interval (N; M),
and the other is located outside this interval if and only if
f(N) f(M)< 0.
Rule 10 Quadratic equation (1) has a unique solution x1 = x2 > M
(x1 = x2< М) тогда и только тогда, когда
Practical part
3.1. Examples of problem solving.
Example 1. For what values of a the equation x² - 2ax + a² + 2a - 3 = 0
a) has no roots; b) has roots of different signs;
c) has positive roots; d) has two different negative roots?
Solution: a) According to rule 1, there are no solutions when the discriminant D = - 4(2a-3)< 0, откуда а > 3/2.
b) According to rule 3.1 for М = 0 we have f(0)=a² + 2a - 3< 0, откуда а(-3;1).
c) According to the rule 4.2 for М=0
Where .
d) According to the rule 5.1 for М=0
Where a< - 3.
3.2. The location of the roots relative to one point.
Example 2 For what values of the parameter a, the roots of the equation x² + 2(a + 1) x + a² + a + 1 = 0 lie on the ray (-2; + ∞).
Let's make a graphical analysis of the problem. According to the condition of the problem, only the following two cases of the location of the graph of the function f (x) \u003d x² + 2 (a + 1) x + a² + a + 1 relative to the point x \u003d -2 are possible.
xv \u003d - a - 1
Both of these cases are analytically described by the conditions
This implies that 0 ≤ a< .
Example 3 . Find all values of the parameter a for which the roots of the square trinomial x² + x + a are distinct and not greater than a. (Appendix 1)
3.3. Location of roots relative to two or more points.
Example 4. For what values of the parameter m the roots of the equation x² - 2 mx + m² -1= 0 are enclosed between the numbers -2 and 4.
The discriminant of the equation D = 4m² - 4m² + 4 = 4 is a perfect square. Let's find the roots of the equation: x1 = m + 1, x2 = m - 1. These roots satisfy the given condition if
Answer: for m(-1;3).
Example 5 At what values of the parameter a does the equation 2x² + (a-4)x + a + 2 = 0 have different roots that satisfy the inequality │x-1│>2. (Annex 2)
The solution of quadratic equations with parameters can be written as a scheme for studying problems related to the location of the roots of the square trinomial Ax² + Bx + C.
Study of the case A = 0 (if it depends on the parameters).
1. Finding the discriminant D in the case A≠0.
2. If D is the full square of some expression, then finding the roots x1, x2 and subordinating them to the conditions of the problem.
3. If the square root of D is not extracted, then the graphical analysis of the problem.
4. Analytical description of suitable cases for the location of the parabola, for which the following are taken into account:
Ø sign (value) of the coefficient at x²;
Ø sign (value) of the discriminant;
Ø signs (values) of the quadratic function at the points under study;
Ø the location of the top of the parabola relative to the points under study.
4. Combining some inequalities (systems).
5. Solution of the obtained systems.
I found 10 rules for the location of the roots of a square trinomial. Solved problems on the location of the roots relative to one point; location of roots relative to two or more points.
Possession of techniques for solving problems with parameters can be considered a criterion for knowledge of the main sections of mathematics, the level of mathematical and logical thinking, and mathematical culture.
References
1. Mochalov, and inequalities with parameters / , .-
Cheboksary: Chuvash Publishing House. University, 200s.
2. Kozhukhov, methods for solving problems with parameters / // Mathematics at school. - 1998. - No. 6.
3. Weekly educational and methodological supplement to the newspaper "First of September" "Mathematics" No. 18, 2002
Appendix 1
Example 3 . Find all values of the parameter a for which the roots of the square trinomial x² + x + a are distinct and not greater than a.
xv = -1/2
Find the discriminant D = 1 - 4a. given that it is not extracted, let's solve the example graphically.
Let's do a graphical analysis. Since the roots x1, x2 of the function f(x) = x² + x + a are different and x1 ≤ a, x2 ≤ a, its graph can only have the following locations.
Let us describe these graphs analytically.
https://pandia.ru/text/78/376/images/image062_1.gif" width="149" height="48">
We find out for which and the roots of the equation are different, i.e. the discriminant D = a²-16a is positive, and either both are less than -1, or both are greater than 3, or one of them is less than -1, and the other is greater than 3. Graph of the function f( x) \u003d 2x² + (a-4) x + a + 2 in these cases has the following locations:
Analytically, these graphs are described by the conditions
The most powerful tool for solving complex problems with parameters is Vieta's theorem. But here you need to be extremely attentive to the wording.
These two theorems (direct and inverse)
Theorem Vieta
If the equation has roots and ; then the equalities are satisfied.
Features of the theorem:
First
.
The theorem is true only for the equation 
and not true for
In the latter case, you must first divide both parts of the equation by a non-zero coefficient a at x 2, and then apply the Vieta theorem.
Second. To use the results of the theorem, it is necessary to have the fact of the existence of the roots of the equations, i.e. do not forget to impose the condition D>0
Reverse
Vieta's theorem
If there are arbitrary numbers and then they are the roots of the equation
Very important note, facilitating problem solving: the inverse theorem guarantees the existence of roots in the equation, which allows you not to mess with the discriminant. It is automatically non-negative in this case.
| Conditions for roots | Equivalent condition on the coefficients a, b, c, and the discriminant D | |
| Roots exist (and are distinct) | ||
Roots exist and are equal  |  |
|
| Roots exist and |  |
|
| Roots exist and |  |
|
| Roots exist and are different |  |
|
| Roots exist, one root is zero and the other is >0 |  |
one). Set at what values of the parameter the equation
Has no roots.
If the equation has no roots, then it is necessary and sufficient that the discriminant
has different positive roots.
Since there are roots, then if they are both positive, then we use the Vieta formula, then for this equation
⟹
Has various negative roots
Has roots of different sign
Has matching roots
2). At what values of the parameter a both roots of the quadratic equation 
will be positive?
Decision.
Since the given equation is quadratic, then both of its roots (equal or different) will be positive if the discriminant is non-negative, and the sum and product of the roots are positive, that is
As, 
and by Vieta's theorem,
Then we get a system of inequalities
3). Find all parameter values a 
are non-positive.
Since the given equation is quadratic, then . Both of its roots (equal or different) will be negative or equal to zero if the discriminant is non-negative, the sum of the roots is negative or equal to zero, and the product of the roots is non-negative, that is
and by Vieta's theorem
then we get a system of inequalities.
where
4). At what values of the parameter a 
equal to 22.5?
First, we will offer a “solution”, which we have had to meet more than once.
insofar as 
then we get the “Answer” However, with the found value a The original equation has no roots.
In this solution, we encountered one of the “most popular” errors associated with the application of the Vieta theorem:
talk about roots without first finding out whether they exist or not.
So, in this example, first of all, it was necessary to establish that only when the original equation has roots. Only then can one turn to the calculations above.
Answer: Such a does not exist.
5). The roots of the equation are such that 
Define
Decision. According to Vieta's theorem 
Let's square both parts of the first equality Given that a we get or Checking shows that the values satisfy the original equation.
Answer:
6). At what value of the parameter a the sum of the squares of the roots of the equation 
takes the smallest value:
Find the discriminant of this equation. We have Here it is important not to make an erroneous conclusion that the equation has two roots for any a. it really has two roots for any but admissible a, i.e. at at
Using the Vieta theorem, we write
Thus, to obtain an answer, it remains to find the smallest value of the quadratic function 
on the set 
Since at 
and at 
then the function on the specified set takes the smallest value at the point
Tasks for independent solution
one). Find all parameter values a, for which the roots of the quadratic equation
non-negative
2). Calculate the value of the expression , where are the roots of the equation 
3). Find all parameter values a, for which the sum of the squares of the real roots of the equation 
more than 6.
Answer: 
4). At what values of the parameter a the equation ax 2 -4x + a \u003d 0 has:
a) positive roots
b) negative roots
The location of the roots of a quadratic function relative to
given points.
For such problems, the following formulation is typical: for what values of the parameter the roots (only one root) are greater (less, no more, no less) of a given number A; the roots are located between the numbers A and B; the roots do not belong to the interval with ends at points A and B, etc.
When solving problems related to a square trinomial
often we have to deal with the following standard situations (which we will formulate in the form of a “question and answer”.
Question 1. Let a number be given (1) both of its roots and more those. ?
Answer. Coefficients of a square trinomial (7) must meet the conditions
where - abscissa of the top of the parabola.
The validity of what has been said follows from Fig. 1, which separately presents the cases and Note that the two conditions and are still not enough for the roots of and to be larger. 1 dash shows a parabola that satisfies these two conditions, but its roots are smaller. However, if we add to the indicated two conditions that the abscissa of the vertex of the parabola is greater, then the roots will be greater than
Question 2. Let a number be given Under what conditions on the coefficients of a square trinomial (1) its roots and lie on opposite sides of those. ?
Answer. square trinomial coefficients (1) must satisfy the condition
The validity of what has been said follows from Fig. 2, where the cases and are presented separately. Note that the indicated condition guarantees the existence of two different roots and a square trinomial (1).
Question 3. Under what conditions on the coefficients of a square trinomial (1) its roots and are different and only one of them lies in the given interval
Answer. Coefficients of a square trinomial (1) must satisfy the condition 
Question 4. Under what conditions on the coefficients of a square trinomial (1) the set of its roots is not empty and all its roots and lie in the given interval those.
Answer. The coefficients of the square trinomial (1) must satisfy the conditions
To solve such problems, it is useful to work with the table below.
Polynomial roots
.
MOU "Secondary school No. 15"
Michurinsk, Tambov region
Algebra lesson in grade 9
"The location of the roots of a square trinomial depending on the values of the parameter"
Developed
mathematics teacher of the 1st category
Bortnikova M.B.
Michurinsk - science city 2016 year
The lesson is for 2 hours.
Dear Guys! The study of many physical and geometric laws often leads to the solution of problems with parameters. Some universities also include equations, inequalities and their systems in exam tickets, which are often very complex and require a non-standard approach to solving. At school, this one of the most difficult sections of the school course in algebra is considered only in a few elective or subject courses.
In my opinion, the functional-graphical method is a convenient and fast way to solve equations with a parameter.
Lesson Objectives: 1. Expand the idea of quadratic equations 2. Learn to find all values of the parameter, for each of which the solutions of the equation satisfy the given conditions. 3. Develop interest in the subject.
During the classes:
1. What is the parameter
Expression of the form ah 2
+ bx + cin a school algebra course is called a square trinomial with respect toX, where a, b,c are given real numbers, moreover,a=/= 0. The values of the variable x, at which the expression vanishes, are called the roots of a square trinomial. To find the roots of a square trinomial, it is necessary to solve the quadratic equationah 2
+ bx + c =0.
Let's remember the basic equations:ax + b = 0;
ax2 + bx + c = 0.When looking for their roots, the values of the variablesa, b, c,included in the equation are considered fixed and given. The variables themselves are called parameters.
Definition.A parameter is an independent variable, the value of which in the problem is considered to be a given fixed or arbitrary real number, or a number belonging to a predetermined set.
2. Main types and methods for solving problems with parameters
Among the tasks with parameters, the following main types of tasks can be distinguished.
Equations to be solved either for any value of the parameter(s) or for parameter values that belong to a predetermined set. For example. Solve Equations:ax = 1 , (a - 2) x = a 2 – 4.
Equations for which you want to determine the number of solutions depending on the value of the parameter (parameters). For example.
a the equation 4 X 2 – 4 ax + 1 = 0has a single root?
Equations for which, for the desired values of the parameter, the set of solutions satisfies the given conditions in the domain of definition.
For example, find the parameter values for which the roots of the equation (a - 2)
X 2
–
2
ax + a + 3
=
0
positive.
The main ways to solve problems with a parameter: analytical and graphic.
Analytical- this is a method of the so-called direct solution, repeating the standard procedures for finding an answer in problems without a parameter. Let's consider an example of such a task.
Task #1
At what values of the parameter a the equationX 2 – 2 ax + a 2 – 1 = 0 has two different roots belonging to the interval (1; 5)?
Decision
X 2
–
2
ax + a 2
–
1 = 0.
According to the condition of the problem, the equation must have two different roots, and this is possible only under the condition: D > 0.
We have: D = 4a 2
– 2(a 2
– 1) = 4. As you can see, the discriminant does not depend on a, therefore, the equation has two different roots for any values of the parameter a. Let's find the roots of the equation:X 1
=
a + 1,
X 2
=
a – 1
The roots of the equation must belong to the interval (1; 5), i.e.
So, at 2<
a < 4 данное уравнение имеет два различных корня, принадлежащих промежутку (1; 5)
Answer: 2<
a < 4.
Such an approach to solving problems of the type under consideration is possible and rational in cases where the discriminant of the quadratic equation is “good”, i.e. is the exact square of any number or expression, or the roots of the equation can be found by the inverse Vieta theorem. Then, and the roots are not irrational expressions. Otherwise, the solution of problems of this type is associated with rather complicated procedures from a technical point of view. And the solution of irrational inequalities will require new knowledge from you.
Graphic- this is a method in which graphs are used in the coordinate plane (x; y) or (x; a). The visibility and beauty of this method of solution helps to find a quick way to solve the problem. Let's solve problem number 1 graphically.
As you know, the roots of a quadratic equation (square trinomial) are the zeros of the corresponding quadratic function: y =X 2
– 2
Oh +
a 2
– 1. The graph of the function is a parabola, the branches are directed upwards (the first coefficient is equal to 1). The geometric model that meets all the requirements of the problem looks like this.
Now it remains to “fix” the parabola in the desired position with the necessary conditions.
Since the parabola has two points of intersection with the axisX, then D > 0.
The vertex of the parabola lies between the vertical lines.X= 1 and X= 5, hence the abscissa of the vertex of the parabola x about belongs to the interval (1; 5), i.e.
1 < X about< 5.We notice that at(1) > 0, at(5) > 0.
So, passing from the geometric model of the problem to the analytical one, we obtain a system of inequalities.
Answer: 2< a < 4.
As can be seen from the example, a graphical method for solving problems of the type under consideration is possible in the case when the roots are “bad”, i.e. contain a parameter under the radical sign (in this case, the discriminant of the equation is not a perfect square).
In the second solution, we worked with the coefficients of the equation and the range of the functionat =
X 2
– 2
Oh +
a 2
– 1.
This method of solving cannot be called only graphical, because. Here we have to solve a system of inequalities. Rather, this method is combined: functional-graphical. Of these two methods, the latter is not only elegant, but also the most important, since it shows the relationship between all types of a mathematical model: a verbal description of the problem, a geometric model - a graph of a square trinomial, an analytical model - a description of a geometric model by a system of inequalities.
So, we have considered a problem in which the roots of a square trinomial satisfy the given conditions in the domain of definition for the desired values of the parameter.
And what other possible conditions can be satisfied by the roots of a square trinomial for the desired values of the parameter?
Examples of problem solving
3. Investigation of the location of the roots of a square trinomial depending on the desired values of the parameter a.
Task number 2.
At what values of the parametera roots of a quadratic equation
x 2 - 4x - (a - 1) (a - 5) \u003d 0 is more than one?
Decision.
Consider the function: y = x 2 - 4x - (a - 1) (a - 5)
The graph of the function is a parabola. The branches of the parabola are directed upwards.
Let's schematically depict a parabola (a geometric model of the problem).
Now let's move on from the constructed geometric model to the analytical one, i.e. Let us describe this geometric model by a system of conditions adequate to it.
There are points of intersection (or a point of contact) of the parabola with the x-axis, therefore, D≥0, i.e. 16+4(a-1)(a-5)≥0.
We notice that the vertex of the parabola is located in the right half-plane relative to the straight line x=1, i.e. its abscissa is greater than 1, i.e. 2>1 (performed for all values of the parameter a).
Note that y(1)>0, i.e. 1 - 4 - (a - 1) (a - 5)>0
As a result, we arrive at a system of inequalities.
; 
Answer: 2<а<4.
Task number 3.
X 2 + ax - 2 = 0 greater than one?
Decision.
Consider the function: y = -x 2 + ah - 2
The graph of the function is a parabola. The branches of the parabola point downwards. Let us depict the geometric model of the problem under consideration.
U(1)
Let's make a system of inequalities.
, no solutions
Answer. There are no such parameter values.
The conditions of problems No. 2 and No. 3, in which the roots of a square trinomial are greater than a certain number for the desired values of the parameter a, we formulate as follows.
General case #1.
For what values of the parameter a the roots of the square trinomial
f(x) = ax 2 + in + c is greater than some number k, i.e. to<х 1 ≤x 2 .
Let us depict the geometric model of this problem and write down the corresponding system of inequalities.
Table 1. Model - scheme.
Task number 4.
At what values of the parameter a are the roots of the quadratic equation
X 2 +(a+1)x–2a(a–1) = 0 less than one?
Decision.
Consider the function: y = x 2 + (a + 1) x–2a (a–1)
The graph of the function is a parabola. The branches of the parabola are directed upwards. According to the condition of the problem, the roots are less than 1, therefore, the parabola intersects the x-axis (or touches the x-axis to the left of the straight line x=1).
Let's schematically depict a parabola (a geometric model of the problem).
y(1)
Let's move on from the geometric model to the analytical one.
Since there are points of intersection of the parabola with the x-axis, then D≥0.
The vertex of the parabola is located to the left of the straight line x=1, i.e. its abscissa x 0 <1.
Note that y(1)>0, i.e. 1+(a+1)-2a(a-1)>0.
We arrive at a system of inequalities.
; 
Answer: -0.5<а<2.
General case #2.
For what values of the parameter a both roots of the trinomialf(x) = ax 2 + in + c will be less than some number k: x 1 ≤x 2<к.
The geometric model and the corresponding system of inequalities are presented in the table. It is necessary to take into account the fact that there are problems where the first coefficient of the square trinomial depends on the parameter a. And then the branches of the parabola can be directed both up and down, depending on the values of the parameter a. We will take this fact into account when creating a general scheme.
Table number 2.
f(k)
Analytical model
(system of conditions).
Analytical model
(system of conditions).
Task number 5.
At what values of the parameter a 2 -2ax+a=0 belong to the interval (0;3)?
Decision.
Consider the square trinomial y(x) = x 2 -2ax + a.
The graph is a parabola. The branches of the parabola are directed upwards.
The figure shows the geometric model of the problem under consideration.
At
Y(0)
U(3)
0 x 1 x 0 x 1 3 x
From the constructed geometric model, let's move on to the analytical one, i.e. we describe it by a system of inequalities.
There are points of intersection of the parabola with the x-axis (or a point of contact), therefore, D≥0.
The top of the parabola is between the lines x=0 and x=3, i.e. abscissa of the parabola x 0 belongs to the interval (0;3).
Note that y(0)>0 and also y(3)>0.
We come to the system.
; 
Answer: a 
General case #3.
For what values of the parameter a the roots of the square trinomial belong to the interval (k; m), i.e. k<х 1 ≤х 2 < m
Table No. 3. Model - scheme.
f(x)
f(k)
f(m)
k x 1 x 0 x 2 mx
f(x)
0kx 1 x 0 x 2 m
f(k)
f(m)
Analytical model of the problem
Analytical model of the problem
TASK #6.
At what values of the parameter a is only the smaller root of the quadratic equation x 2
+2ax+a=0 belongs to the interval X 
(0;3).
Decision.
2 -2ax + a
The graph is a parabola. The branches of the parabola are directed upwards. Let x 1 smaller root of a square trinomial. According to the condition of the problem x 1 belongs to the interval (0;3). Let us depict a geometric model of the problem that meets the conditions of the problem.
Y(x)
Y(0)
0 x 1 3 x 0 x 2 x
Y(3)
Let's move on to the system of inequalities.
1) Notice that y(0)>0 and y(3)<0. Так как ветви параболы направлены вверх и у(3)<0, то автоматически Д>0. Therefore, this condition does not need to be written into the system of inequalities.
So, we get the following system of inequalities:
Answer: a >1,8.
General case #4.
For what values of the parameter a does the smaller root of the square trinomial belong to the given interval (k; m), i.e. k<х 1 < m<х 2 .
Table No. 4 . Model - scheme.
f(k)kx 1 0 m x 2
f(m)
F(x)
f(m)
kx 1 m x 2 x
f(k)
Analytical model
Analytical model
TASK #7.
At what values of the parameter a only the larger root of the quadratic equation x 2 +4x-(a+1)(a+5)=0 belongs to the interval [-1;0).
Decision.
Consider the square trinomial y(x)=x 2+4x-(a+1)(a+5).
The graph is a parabola. The branches are directed upwards.
Let us depict the geometric model of the problem. Let x 2 is the larger root of the equation. By the condition of the problem, only the larger root belongs to the interval.
y(X)
y(0)
x 1 -1 x 2 0 x
y(-1)
Note that y(0)>0 and y(-1)<0. Кроме этого ветви параболы направлены вверх, значит, при этих условиях Д>0.
Let's create a system of inequalities and solve it.
Answer: 
General case #5.
For what values of the parameter a, the larger root of the square trinomial belongs to the given interval (k; m), i.e. x 1< k<х 2 < m.
Table No. 5. Model - scheme.
f(x)
f(m)
0 x 1 kx 2 m x
f(k)
f(x)
f(k)
x 1 0kx 2 m
f(m)
Analytical model
Analytical model
W ADACHA No. 8.
At what values of the parameter a is the segment [-1; 3] entirely located between the roots of the quadratic equation x 2 -(2a+1)x+a-11=0?
Decision.
Consider the square trinomial y(x)=x 2 - (2a + 1) x + a-11
The graph is a parabola.
The geometric model of this problem is shown in the figure.
Y(x)
X 1 -1 0 3 x 2 x
Y(-1)
Y(3)
Under these conditions, D>0, since the branches of the parabola are directed upwards.
Answer: a 
General case #6.
For what values of the parameter a the roots of the square trinomial are outside the given interval (k; m), i.e. x 1< k < m<х 2 .
x 2 -(2a + 1) x + 4-a \u003d 0 lie on opposite sides of the number from the number 3?
Decision.
Consider the square trinomial y(x)=x 2 - (2a + 1) x + 4-a.
The graph is a parabola, the branches are directed upwards (the first coefficient is 1). Let us depict the geometric model of the problem.
X 1 3 x 2 x
Y(3)
Let's move from a geometric model to an analytical one.
We notice that y(3)<0, а ветви параболы направлены вверх. При этих условиях Д>0 automatically.+in+c is less than some number k: x 1 ≤ x 2 <к
3. For what values of the parameter a roots of the square trinomial ax 2 +in+c belong to the interval (k, t) to<х 1 ≤x 2 <т
4. For what values of the parameter a only the smaller root of the square trinomial ax 2 +in+c belongs to the given interval (k, t), i.e. k<х 1 <т<х 2
1. Depict the geometric model of this problem.
2. Write down the system of conditions to which the solution of this problem is reduced
1. Depict the geometric model of this problem.
2. Write down the system of conditions to which the solution of this problem is reduced
1. Depict the geometric model of this problem.
2. Write down the system of conditions to which the solution of this problem is reduced
The roots of the quadratic equation x 2 -4x-(a-1)(a-5)=0, greater than 1.
Answer: 2<а<4
The roots of the quadratic equation x 2 +(a+1)x-2a(a-1)=0, less than 1.
Answer:
-0,5<а<2
The roots of the quadratic equation x 2 -2ax+a=0, belong to the interval (0;3).
Answer: 1≤a< 9 / 5
Only the smaller root of the equation x 2 -2ax+a=0, belongs to the interval (0;3).
Answer: 1≤a< 9 / 5
1. Depict the geometric model of this problem.2. Write down the system of conditions to which the solution of this problem is reduced
1. Depict the geometric model of this problem.
2. Write down the system of conditions to which the solution of this problem is reduced
1. Depict the geometric model of this problem.
2. Write down the system of conditions to which the solution of this problem is reduced
Only the largest root of the equation x 2 +4x-(a+1)(a+5)=0, belongs to the interval [-1;0).
Answer:(-5;-4]U[-2;-1)
The segment [-1; 3] is entirely between the roots of the quadratic equation x 2 -(2a+1)x+a-11=0.
Answer: -1<а<3
The roots of the quadratic equation x 2 -2 (a + 1) x + 4-a \u003d 0, lie on opposite sides of the number 3.
Answer( 10 / 7 ;∞)
Thanks for the lesson guys!
The square trinomial is the main function of school mathematics - by the way, not the most primitive. The ability to use the resources provided by him to solve problems to a large extent characterizes the level of mathematical thinking of a student of school algebra. In this paper, we substantiate this thesis and give examples of a specific application of the properties of a quadratic function. The stimulating factor is the fact that when solving any problem with parameters, sooner or later it is necessary (and succeeds) to reformulate the problem in terms of a square trinomial and solve it using the properties of this universal function.
Study of the square trinomial
Definition. The square trinomial with respect to x is an expression of the form f(x) = ax 2 + bx + c (1), where a, b, cR, a0.
A square trinomial is an ordinary polynomial of degree 2. The range of questions formulated in terms of a square trinomial is unexpectedly extremely wide. Since the tasks associated with the study of a square trinomial traditionally occupy an honorable and prominent place in the written final school and university entrance exams, it is very important to teach the student (future applicant) an informal (that is, creative) possession of a variety of techniques and methods of such research. In this methodical development, the main statements about the square trinomial (Vieta's theorem, the location of the roots relative to the given points of the numerical axis, the technique of handling the discriminant) are fixed, problems of various types and different levels of complexity are solved. The main ideological conclusion is that in school mathematics there are fragments rich in deep content that are accessible to the student and do not require the use of mathematical analysis and other sections of the so-called “higher mathematics”.
The graph of the trinomial (1) is a parabola; for a 0 - up. The location of the parabola relative to the Ox axis depends on the value of the discriminant D = b 2 - 4ac: for D>0 there are two points of intersection of the parabola with the Ox axis (two different real roots of the trinomial); at D=0 - one point (double real root); at D 0 - above the Ox axis). The standard trick is the following representation of the trinomial (using full square extraction):
f(x) = ax2 + bx + c = 
=
. This representation makes it easy to build a graph through linear transformations of the graph of the function y=x 2 ; parabola vertex coordinates: 
.
The same transformation makes it possible to immediately solve the simplest extremum problem: to find the largest (for a 0) value of function (1); the extreme value is reached at the point and is equal to 
.
One of the main judgments about the square trinomial -
Theorem 1 (Vieta). If x 1, x 2 are the roots of the trinomial (1), then
(Vieta formulas).
With the help of Vieta's theorem, many problems can be solved, in particular, those in which it is required to formulate conditions that determine the signs of the roots. The following two theorems are direct consequences of Vieta's theorem.
Theorem 2. In order for the roots of the square trinomial (1) to be real and have the same signs, it is necessary and sufficient that the following conditions be met:
D \u003d b 2 - 4ac 0, x 1 x 2 \u003d\u003e 0,
both roots are positive for x 1 + x 2 = > 0,
and both roots are negative at x 1 + x 2 =
Theorem 3. In order for the roots of the square trinomial (1) to be real and have different signs, it is necessary and sufficient that the following conditions be met:
D=b 2 - 4ac > 0, x 1 x 2 =
in this case, the positive root has a larger modulus at x 1 + x 2 \u003d\u003e 0,
and the negative root has a larger modulus at x 1 + x 2 =
The theorems and corollaries proved below can (and therefore must) be effectively applied in solving problems with parameters.
Theorem 4. In order for both roots of the square trinomial (1) to be less than the number M, that is, on the real line, the roots lie to the left of the point M, it is necessary and sufficient that the following conditions be satisfied:
, or, by combining the conditions,
(Fig. 1a and 1b).
Proof.
Need. If the trinomial (1) has real roots x 1 and x 2 (maybe the same), x 1 x 2 and x 1, (x 1 - M) (x 2 - M) > 0, x 1 + x 2 0, M > (x 1 + x 2)/2. According to Vieta's formulas 
, therefore , or , etc.
Adequacy
- a contradiction with the condition. If , then (x 1 - M)(x 2 - M)0, x 1 x 2 - (x 1 + x 2)M + M 2 0, whence 
, af(M) 0 - again a contradiction with the condition; only possibility x 1 remains
Theorem 5. In order for one of the roots of the square trinomial (1) to be less than the number M, and the other more than the number M, that is, the point M would lie in the interval between the roots, it is necessary and sufficient that the following conditions be met:
, or, combining conditions, af(M)
(Fig. 2a and 2b).
Proof.
Need. If the trinomial (1) has real roots x 1 and x 2 , x 1 M , then (x 1 - M)(x 2 - M), therefore 
, or af(M)
Adequacy. Let af(M) , or , then (x 1 - M)(x 2 - M)0,
x 1 x 2 - (x 1 + x 2)M + M 2 0, whence 
, af(M)0 - contradiction with the condition; the only possibility remains, which is what needs to be proved. The theorem has been proven.
Theorem 6. In order for both roots of the square trinomial (1) to be greater than the number M, that is, on the real line, the roots lie to the right of the point M, it is necessary and sufficient that the following conditions be met:
, or, by combining the conditions,
(Fig. 3a and 3b).
Proof. Need. If the trinomial (1) has real roots x 1 and x 2 (maybe coinciding), x 1 x 2 and x 1 > M, x 2 > M, then, (x 1 -M)(x 2 -M)> 0, x1 + x2 > 2M; otherwise x 1 x 2 - (x 1 + x 2)M + M 2 > 0, M , therefore , or , etc.
Adequacy. Let be . We argue against the contrary. Suppose that , , then 
- a contradiction with the condition. If , then (x 1 - M)(x 2 - M)0, x 1 x 2 - (x 1 + x 2)M + M 2 0, whence , af(M) 0 - again a contradiction with the condition; only the possibility x 1 > M, x 2 > M remains, which is to be proved. The theorem has been proven.
Corollary 1. In order for both roots of the square trinomial (1) to be greater than the number M, but less than the number N (M
, or, by combining the conditions, 
(Fig. 4a and 4b).
Consequence 2. In order for only the largest root of the square trinomial (1) to belong to the interval (M,N), where M
, or, by combining the conditions,
the smaller root lies outside the segment
(Fig. 5a and 5b).
Corollary 3. In order for only the smaller root of the square trinomial (1) to belong to the interval (M,N), where M
, or, combining the conditions, ;
the larger root lies outside the segment
(Fig. 6a and 6b).
Consequence 4. In order for one of the roots of the square trinomial (1) to be less than M, and the other greater than N (M
, or, by combining the conditions,
(Fig. 7, a and 7, b).
Of course, the analytical and geometric interpretations of the results of Theorems 4-6 and Corollaries 1-4 are equivalent, and the strategic goal is to develop skills for accurate translation from one language to another. It is especially important to demonstrate how “visualization” (“graphical view”) helps to accurately write down the formal conditions necessary and sufficient to fulfill the requirements of the task.
We indicate typical problems that can be solved with the help of proven theorems (more generally, those that can be solved based on the properties of a square trinomial).
Task 1. Find all values of a for which the equations x 2 +ax+1=0 and x 2 +x+a=0 have at least one common root.
Decision. Both equations have exactly the same roots if and only if the coefficients of the corresponding square trinomials are the same (a second-degree polynomial is completely determined by its two roots and the corresponding coefficients of these polynomials are equal), hence we get a=1. However, if only real roots are taken into account, then for a=1 there are none (the discriminant of the corresponding trinomial is negative). For a1, we argue as follows: if x 0 is the root of both equations f(x)=0 and g(x)=0, then x 0 will be the root of the equation f(x)-g(x)=0 (this is only necessary, but not a sufficient condition for the existence of a common root of the two equations f(x)=0 and g(x)=0, since the equation f(x) - g(x)=0 is their consequence); Subtract the second from the first equation and get
(x 2 + ax + 1) - (x 2 + x + a) = 0, x(a-1) - (a-1)=0, whence, since a1, x=1. Thus, if given equations have a common root, then it is equal to 1. Substitute x = 1 into the first equation: 1 + a + 1 = 0, and a = -2.
Answer. a = -2.
Task 2. At what a will the sum of the squares of the roots of the equation x 2 - ax + a - 1 = 0 be the smallest?
Decision. By Vieta's theorem, x 1 + x 2 = a, x 1 x 2 = a - 1. We have:
x 1 2 + x 2 2 = (x 1 + x 2) 2 - 2x 1 x 2 = a 2 - 2(a-1) = a 2 - 2a + 2 = (a-1) 2 + 1 1 and = 1 for a=1.
Answer. a = 1.
Task 3. Are there a such that the roots of the polynomial f(x)=x 2 +2x+a are real, distinct, and both are between -1 and 1?
Decision. In order for both roots x 1 and x 2 of the trinomial f (x) to be enclosed between -1 and 1, it is necessary that the arithmetic mean of these roots be enclosed between -1 and 1: 
; but on Vieta's theorem,
, That's why
Answer. No.
Task 4. For what values of the parameter a are both roots of the quadratic equation x 2 + (2a + 6)x + 4a + 12 = 0 real and both are greater than -1?
Decision. Theorem 6 gives:
,
,
,
.
Answer. .
Task 5. For what values of the parameter a are both roots of the quadratic equation x 2 +4ax+ (1-2a+4a 2) = 0 real and both are less than -1?
Decision. Theorem 4 gives:
,
,
, a>1.
Answer. a > 1.
Task 6. For what values of the parameter a is one root of the quadratic equation f(x) = (a-2)x 2 - 2(a+3)x + 4a = 0 greater than 3, and the other less than 2?
Decision. Note immediately that a2 (otherwise the equation would have only one root). Applicable corollary 4(here M=2, N=3):
,
,
,
2
. Both roots are positive 
(theorem 6), where
and 
;
(theorem 4) - this system of solutions has no; roots have different signs at (a-1)(a+5) theorem 5), i.e. -5 
,
, if 1-p+q=0. The second root is equal to x 2 =1-p; hence, if , then the equation sin 2 t +psint+q=0 (4) has, in addition to those indicated, roots (for p=2, both series of roots coincide).
, if
, then 
, while for p2 there are no roots. If p=2, then q=1, x 2 +2x+1=0, x=-1, t=, and if p=-2, then x=1, t=.
.
.
.4. Location of the roots of a square trinomial depending on the parameter
Often there are problems with parameters in which it is required to determine the location of the roots of a square trinomial on the real axis. Based on the main provisions and notation of the previous paragraph, consider the following cases:
1. Let a square trinomial be given, where 
and dot m on axle Ox. Then both horses 
square trinomial 
will be strictly less m
or 
The geometric illustration is shown in Figures 3.1 and 3.2.
2. Let a square trinomial be given, where and a point m on axle Ox. Inequality 
holds only if and only if the numbers
a and 
have different signs, that is 
(Fig. 4.1 and 4.2.)
3. Let a square trinomial be given, where and the point m on axle Ox. Then both horses 
square trinomial will be strictly larger m if and only if the following conditions are met:
or 
A geometric illustration is shown in Figures 5.1 and 5.2.
4. Let a square trinomial be given, where and the interval (m, M) Then both roots of the square trinomial belong to the indicated interval if and only if the following conditions are satisfied:
or 
The geometric illustration is shown in Figures 6.1 and 6.2.
5. Let a square trinomial be given, where , are its roots and segment 
.
The segment lies in the interval 
if and only if the following conditions are met:
The geometric illustration is shown in Figures 7.1 and 7.2.
Example.Find all parameter valuesa, for each of which both roots of the equation 
more than -2.
Decision. It is specified in the condition of the task. That the equation has two roots, so . The situation under consideration is described by case 3 and is shown in Figure 5.1. and 5.2.
Let's find, 
,
Considering all this, we write the set of two systems:
or 
Solving these two systems, we get .
Answer. For each parameter value a from the gap, both roots of the equation are greater than -2.
Example.At what values of the parameterainequality 
performed for any 
?
Decision. If the set X is the solution of this inequality, then the condition of the problem means that the interval 
must be within the set X, i.e
.
Consider all possible values of the parameter a.
1.If a=0, then the inequality takes the form 
, and its solution will be the interval 
. In this case, the condition is met and a=0 is the solution to the problem.
2.If 
, then the graph of the right side of the inequality is a square trinomial, the branches of which are directed upwards. The solution of the inequality depends on the sign of .
Consider the case when 
. Then, in order for the inequality to hold for all, it is required that the roots of the square trinomial be less than -1, that is:
or 
Solving this system, we get 
.
If a 
, then the parabola lies above the axis Ox, and the solution of the inequality will be any number from the set of real numbers, including the interval . Let's find such a from the condition:
or 
Solving this system, we get 
.
3.If 
, then at 
the solution to the inequality is the interval , which cannot include the interval , and if 
this inequality has no solutions.
Combining all found values a, we get the answer.
Answer. For any parameter value from the interval 
the inequality holds for any .
Example.For what values of the parameter a the set of function values contains the segment 
?
Decision. 1. If 
, then
a) at a = 1 function will take the form y = 2, and the set of its values consists of a single point 2 and does not contain the segment ;
b) when a =-1 function will take the form y = -2
x+2
. Its set of meanings 
contains a segment, so a =-1 is the solution to the problem.
2.If 
, then the branches of the parabola are directed upwards, the function takes the smallest value at the vertex of the parabola 
:
, 
.
The set of function values is an interval 
, which contains the segment 
if the following conditions are met:
.
3. If 
, then the branches of the parabola are directed downwards, the function takes the greatest value at the vertex of the parabola 
. The set of function values is an interval 
, which contains the segment if the following conditions are met: 
Solving this system of inequalities, we obtain 
.
Combining the solutions, we get 
.
Answer. At 
the set of function values contains the segment .
Tasks for independent solution
1. Without calculating the roots of the quadratic equation 
, to find
a) 
, b) 
, in) 
2. Find the set of function values
a) 
, b) 
, in) 
, G) 
3. Solve equations
a) 
, b) 
4. At what values of the parameter a both roots of the equation 
lie on the interval (-5, 4)?
5. At what values of the parameter a the inequality holds for all values x?
6. At what values of the parameter a smallest function value
On the segment 
is -1?
7. At what values of the parameter a the equation 
has roots?
Karpova Irina Viktorovna
PROGRAM AND EDUCATIONAL MATERIALS OF THE ELECTIVE COURSE in mathematics for students of grades 8-9 "Elements of the theory of probability and mathematical statistics"
Explanatory note
At present, the universality of probabilistic-statistical laws is becoming obvious; they have become the basis for describing the scientific picture of the world. Modern physics, chemistry, biology, demography, linguistics, philosophy, the whole complex of socio-economic sciences are developing on a probabilistic-statistical basis.
A child in his life daily encounters probabilistic situations. The range of issues related to understanding the relationship between the concepts of probability and reliability, the problem of choosing the best of several solutions, assessing the degree of risk and chances of success - all this is in the sphere of real interests of the formation and self-development of the individual.
All of the above makes it necessary to familiarize the child with probabilistic-statistical patterns.
Course objective: to acquaint students with some theoretical and probabilistic regularities and statistical methods of data processing.
Course objectives
To acquaint students with the basic conceptual apparatus of probability theory.
Learn to determine the probability of events in the classical test scheme.
To acquaint with the methods of primary processing of statistical data.
Requirements for the level of mastering the course content
As a result of mastering the course program, students should know:
basic concepts of probability theory: test, test outcome, elementary event space, random, certain, impossible events, joint and incompatible events;
conditions of the classical test scheme and determination of the probability of an event in the classical test scheme;
determining the relative frequency of occurrence of the event and the statistical probability;
determination of the variation series and its main numerical characteristics.
During the course, students must acquire skills:
determine all possible outcomes of the test, the compatibility and incompatibility of events;
solve theoretical and probabilistic problems for calculating the probability in the classical test scheme;
calculate the relative frequency of occurrence of an event;
make a statistical distribution of the sample and calculate its numerical characteristics.
The program involves the development of students skills:
use of existing algorithms and, if necessary, their creative processing in the specific conditions of the problem;
independent problem solving;
use in solving problems of generalized schemes containing basic definitions and formulas.
Course scope: the course offered is 20 hours
Thematic planning
Lesson Topics | Number of hours |
|
Basic concepts of probability theory. | ||
Classic test scheme. Determination of probability in the classical test scheme. | ||
The frequency is absolute and relative. | ||
Statistical definition of probability. | ||
General and sample populations. | ||
Statistical distribution of the sample. | ||
Numerical characteristics of the statistical distribution. | ||
Statistical estimation and forecast. | ||
Manual text
Many people love mathematics for its eternal truths: twice two is always four, the sum of even numbers is even, and the area of a rectangle is equal to the product of its adjacent sides. In any problem that you solved in math class, everyone got the same answer - you just had to not make mistakes in the solution.
Real life is not so simple and unambiguous. It is impossible to predict the outcomes of many phenomena in advance, no matter how complete information we have about them. It is impossible, for example, to say for sure which side a tossed coin will land on, when the first snow will fall next year, or how many people in the city will want to make a telephone call within the next hour. Such unpredictable events are called random.
However, the case also has its own laws, which begin to manifest themselves with repeated repetition of random phenomena. If you toss a coin 1000 times, then the "eagle" will fall out about half the time, which cannot be said about two or even ten tosses. Note the word "approximately" - the law does not state that the number of "eagles" will be exactly 500 or fall between 490 and 510. It does not state anything for certain at all, but gives a certain degree of certainty that some random event will occur. . Such regularities are studied by a special branch of mathematics - probability theory.
Probability theory is inextricably linked with our daily life. This provides a remarkable opportunity to establish many probabilistic laws empirically, repeatedly repeating random experiments. The materials for these experiments will most often be an ordinary coin, a dice, a set of dominoes, a roulette wheel, and even a deck of cards. Each of these items, one way or another, is connected with games. The fact is that the case here appears in its purest form, and the first probabilistic problems were associated with assessing the chances of players to win.
Modern probability theory has moved as far from games of chance as geometry from problems of land management, but their props are still the simplest and most reliable source of chance. By practicing with a roulette wheel and a die, you will learn how to calculate the probability of random events in real life situations, which will allow you to assess your chances of success, test hypotheses, and make decisions not only in games and lotteries.
Mathematical statistics is a branch of mathematics that studies methods for collecting, systematizing and processing the results of observations of mass random phenomena in order to identify existing patterns.
In a sense, the problems of mathematical statistics are inverse to the problems of probability theory: dealing only with experimentally obtained values of random variables, statistics aims to put forward and test hypotheses about the distribution of these random variables and evaluate the parameters of their distribution.
1. Random events. How to compare events?
Like any other branch of mathematics, probability theory has its own conceptual apparatus, which is used in formulating definitions, proving theorems, and deriving formulas. Let us consider the concepts that we will use in the further exposition of the theory.
Trial- implementation of a set of conditions.
Outcome of the test (elementary event)– any result that may occur during the test.
Examples.
1) Trial:
Test results:ω 1 - one point has appeared on the upper face of the cube;
ω 2 – two points appeared on the top face of the cube;
ω 3 – three points appeared on the top face of the cube;
ω 4 – four points appeared on the top face of the cube;
ω 5 – five points appeared on the top face of the cube;
ω 6 - six points appeared on the top face of the cube.
In total, 6 test outcomes (or 6 elementary events) are possible.
2) Trial: the student takes the exam.
Test results:ω 1 - the student received a deuce;
ω 2 - the student received a three;
ω 3 - the student received a four;
ω 4 - the student received a five.
In total, 4 test outcomes (or 4 elementary events) are possible.
Comment. The notation ω is the standard notation for an elementary event, in what follows we will use this notation.
We will call the results of this test equally possible if the trial outcomes have the same chance of appearing.
Space of elementary events- the set of all elementary events (test outcomes) that may appear during the test.
In the examples that we considered above, the spaces of elementary events of these tests were actually described.
Comment. The number of points in the space of elementary events (PES), i.e. the number of elementary events will be denoted by the letter n.
Let us consider the main concept, which we will use in what follows.
Definition 1.1.An event is a collection of a certain number of TEC points.
In the future, we will denote events in capital Latin letters: A, B, C.
Definition 1.2.An event that may or may not occur during a test is called a random event.
By buying a lottery ticket, we may or may not win; in the next elections, the ruling party may or may not win; in the lesson you may be called to the board, or they may not be called, etc. These are all examples of random events that, under the same conditions, may or may not occur during a test.
Comment. Any elementary event is also a random event.
Definition 1.3.An event that occurs for any outcome of a trial is called a certain event.
Definition 1.4.An event that cannot occur under any outcome of the test is called an impossible event.
Example.
1) Trial: a dice is thrown.
Event A: an even number of points fell on the top face of the die;
Event B: on the top side of the die, a number of points fell out, a multiple of 3;
Event C: 7 points fell on the top face of the die;
Event D: the number of points less than 7 fell on the top face of the die.
Events BUT and AT may or may not occur during the test, so these are random events.
Event With can never happen, so it is an impossible event.
Event D occurs with any outcome of the test, then this is a reliable event.
We said that random events under the same conditions may or may not occur. At the same time, some random events have more chances to occur (which means they are more probable - closer to reliable), while others have less chances (they are less probable - closer to impossible). Therefore, as a first approximation, it is possible to define the probability as the degree of possibility of the occurrence of an event.
It is clear that more probable events will occur more often than less probable ones. So you can compare probabilities by the frequency with which events occur.
Let's try to place the following events on a special probability scale in order of increasing probability of their occurrence.
Event A: next year the first snow in Khabarovsk will fall on Sunday;
Event B: the sandwich that fell off the table fell butter-side down;
Event C: when throwing a dice, 6 points will fall out;
Event D: when throwing a dice, an even number of points will fall out;
Event E: when throwing a dice, 7 points fell out;
Event F: When a dice is rolled, a number of points less than 7 will come up.
So, at the starting point of our scale, we will place impossible events, since the degree of possibility of their occurrence (probability) is practically equal to 0. Thus, this will be an event E. At the end point of our scale, we place reliable events - F. All other events are random, let's try to arrange them on the scale in order of increasing degree of their occurrence. To do this, we must find out which of them are less likely and which are more likely. Let's start with the event D: When we roll a dice, each of the 6 faces has an equal chance of being on top. An even number of points - on three faces of the cube, on the other three - odd. So exactly half the chance (3 out of 6) that the event D will happen. Therefore, we place the event D in the middle of our scale.
At the event With only one chance in 6 while the event has D- three chances out of 6 (as we found out). So With less likely and will be located on the scale to the left of the event D.
Event BUT even less likely than With, because there are 7 days in weeks and in any of them the first snow can fall with equal probability, so the event has BUT one chance in 7. Event BUT, thus, will be located even more to the left than the event With.
The hardest thing to place on the scale is an event AT. Here it is impossible to accurately calculate the chances, but you can call on life experience to help: a sandwich falls to the floor with butter down much more often (there is even a “sandwich law”), so the event AT much more likely than D, so on the scale we place it to the right than D. Thus, we get the scale:
E A C D B F
impossible random certain
The constructed probability scale is not quite real - it does not have numerical marks, divisions. We are faced with the task of learning how to calculate the degree of possibility of the occurrence (probability) of an event.
