Teacher highest category: Minaichenko N.S., gymnasium No. 24, Sevastopol
Lesson in 8th grade: "Square trinomial and its roots"
Lesson type : lesson of new knowledge.
The purpose of the lesson:
organize student activities to consolidate and develop knowledge about the decomposition of a quadratic trinomial into linear factors and the reduction of fractions;
develop skills in applying knowledge of all methods of factorization: bracketing, using abbreviated multiplication formulas and grouping methods in order to prepare for successful completion algebra exam;
create conditions for development cognitive interest to the subject, formation logical thinking and self-control when using factorization.
Equipment: multimedia projector, screen, presentation: “Roots of the square trinomial”, crossword puzzle, test, handouts.
Basic Concepts . Factoring a quadratic trinomial.
Independent activity of students. Application of the theorem on the factorization of a quadratic trinomial in solving problems.
Lesson Plan
Problem solving.Answers to student questions
IV. Primary test of knowledge acquisition. Reflection
Teacher's message.
Student message
V. Homework
Writing on the board
Methodological comment:
This topic is fundamental in the section " Identity transformations algebraic expressions" Therefore, it is important that students automatically be able not only to see factorization formulas in examples, but also to apply them in other tasks: such as solving equations, transforming expressions, proving identities.
This topic focuses on factoring a quadratic trinomial:
ax+ bx + c = a(x – x)(x – x
),
where x and x – roots quadratic equation ax + bx + c = 0.
This allows you to expand the student’s field of vision, teach him to think in non-standard situation, using the material being studied, i.e. using the formula for factoring a quadratic trinomial:
ability to reduce algebraic fractions;
ability to simplify algebraic expressions;
ability to solve equations;
ability to prove identities.
Main lesson content:
a) 3x + 5x – 2;
b) –x + 16x – 15;
c) x – 12x + 24;
d) –5x + 6x – 1.
№2. Reduce the fraction:
№3. Simplify the expression:
№4. Solve the equation:
b) 
During the classes:
I. Stage of updating knowledge.
Motivation for learning activities.
a) from history:
b) crossword:
Warm-up-train the mind – crossword puzzle:
Horizontally:
1) The root of the second degree is called…. (square)
2) Values of the variable at which the equation becomes a true equality (roots)
3) An equality containing an unknown is called... (equation)
4) Indian scientist, which outlined general rule solving quadratic equations (Brahmagupta)
5) The coefficients of the quadratic equation are... (numbers)
6) Ancient Greek scientist who invented a geometric method for solving equations (Euclid)
7) Theorem relating coefficients and roots of a quadratic equation (Vieta)
8) “discriminant”, determining the roots of a quadratic equation – this is... (discriminant)
Additionally:
If D>0, how many roots? (two)
If D=0, how many roots? (one)
If D<0, сколько корней? (нет действительных корней)
Horizontally and vertically the topic of the lesson: “ Square trinomial»
b) motivation:
This topic is fundamental in the section “Identical transformations of algebraic expressions.” Therefore, it is important that you automatically be able not only to see factorization formulas in examples, but also to apply them in other tasks: such as reducing fractions, solving equations, transforming expressions, proving identities.
Today we will focus on factoring the quadratic trinomial:
II. Learning new material.
Topic: Square trinomial and its roots.
The general theory of polynomials of many variables goes far beyond the scope of the school course. Therefore, we will limit ourselves to studying polynomials of one real variable, and only in the simplest cases. Let us consider polynomials of one variable, reduced to standard form.
Root of a polynomial is the value of a variable at which the value of the polynomial is equal to zero. This means that to find the roots of a polynomial, you need to equate it to zero, i.e. solve the equation.
Root of a polynomial of the first degree 
easy to find 
. Examination: 
.
The roots of a quadratic trinomial can be found by solving the equation: 
.
Using the formula for the roots of a quadratic equation we find:
; 
If 
And 
-roots of a square trinomial 
, Where 
≠ 0,
That .
Proof:
Let us perform the following transformations of the quadratic trinomial:
=
=
=
=
=
=
=
=
Since the discriminant 
, we get:
=
=
Let us apply the difference of squares formula in brackets and get:
=
=
,
because ; . The theorem has been proven.
The resulting formula is called the formulafactoring a quadratic trinomial.
III. Formation of skills and abilities.
№1. Factor the quadratic trinomial:
a) 3x + 5x – 2;Solution:
Answer: 3x+5x–2=3(x+2)(x-)=(x+2)(3x-1)
On the desk:
b) –5x + 6x – 1;
Additionally:
c) x – 12x + 24;
d) –x + 16x – 15.
№2. Reduce the fraction:
A)
№4. Solve the equation:
b)
IV. Primary test of knowledge acquisition.
A) Test.
Option 1.
1. Find the roots of the quadratic trinomial:2x 2 -9x-5
Answer:
2. Which polynomial must be substituted for the ellipsis in order for the equality to be true:
b) Mutual verification of options (answers and evaluation parameters are illustrated).
c) Reflection.
V. Homework.
The study of many physical and geometric patterns often leads to solving problems with parameters. Some universities also include equations, inequalities and their systems in exam papers, which are often very complex and require a non-standard approach to solution. At school, this one of the most difficult sections of the school algebra course 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.
As is known, in relation to equations with parameters there are two formulations of the problem.
- Solve the equation (for each parameter value, find all solutions to the equation).
- Find all values of the parameter for each of which the solutions to the equation satisfy the given conditions.
In this paper, a problem of the second type is considered and studied in relation to the roots of a square trinomial, the finding of which is reduced to solving a quadratic equation.
The author hopes that this work will help teachers when developing lessons and preparing students for the Unified State Exam.
1. What is a parameter
Expression of the form ah 2 + bx + c in the school algebra course they call the quadratic trinomial with respect to X, Where a, b, c are given real numbers, and, a=/= 0. The values of the variable x at which the expression becomes zero are called the roots of the square trinomial. To find the roots of a quadratic trinomial, you need to solve the quadratic equation ah 2 + bх + c = 0.
Let's remember the basic equations from the school algebra course ax + b = 0;
aх2 + bх + c = 0. When searching for their roots, the values of the variables a, b, c, included in the equation are considered fixed and given. The variables themselves are called parameters. Since there is no definition of the parameter in school textbooks, I propose to take the following simplest version as a basis.
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. Basic types and methods for solving problems with parameters
Among tasks with parameters, the following main types of tasks can be distinguished.
- Equations that must be solved either for any value of a parameter(s) or for parameter values belonging to a pre-specified set. For example. Solve equations: ax = 1, (a – 2)x = a 2 – 4.
- Equations for which it is necessary to determine the number of solutions depending on the value of the parameter (parameters). For example. At what parameter values a the equation 4X 2 – 4ax + 1 = 0 has a single root?
- Equations for which, for the required parameter values, the set of solutions satisfies the specified conditions in the domain of definition.
For example, find the parameter values at which the roots of the equation ( a – 2)X 2
–
2ax + a + 3 =
0
positive.
The main ways to solve problems with a parameter: analytical and graphical.
Analytical- This is a method of the so-called direct solution, repeating standard procedures for finding the answer in problems without a parameter. Let's look at an example of such a task.
Task No. 1
At what values of the parameter a does the equation X 2 – 2ax + a 2 – 1 = 0 has two different roots belonging to the interval (1; 5)?
Solution
X 2 –
2ax + a 2 –
1 = 0.
According to the conditions of the problem, the equation must have two different roots, and this is only possible under the condition: D > 0.
We have: D = 4 a 2 – 2(A 2 – 1) = 4. As we 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.
This 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 using the inverse theorem of Vieta. Then, the roots do not represent irrational expressions. Otherwise, solving problems of this type involves quite complex procedures from a technical point of view. And solving irrational inequalities requires new knowledge from the student.
Graphic- this is a method in which graphs are used in the coordinate plane (x; y) or (x; a). The clarity and beauty of this method of solution helps to find a quick way to solve the problem. Let's solve problem No. 1 graphically.
As you know from an algebra course, the roots of a quadratic equation (quadratic trinomial) are the zeros of the corresponding quadratic function: U = X 2
– 2Oh + A 2 – 1. The graph of the function is a parabola, the branches are directed upward (the first coefficient is 1). A geometric model that meets all the requirements of the problem looks like this.
Now all that remains is to “fix” the parabola in the desired position using the necessary conditions.
- Since a parabola has two points of intersection with the axis X, then D > 0.
- The vertex of the parabola is between the vertical lines X= 1 and X= 5, therefore the abscissa of the vertex of the parabola x o belongs to the interval (1; 5), i.e.
1 <X O< 5. - We notice that at(1) > 0, at(5) > 0.
So, moving 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 method, we worked with the coefficients of the equation and the range of the function at = X 2 – 2Oh + A 2
– 1.
This method of solution cannot be called only graphical, because here we have to solve a system of inequalities. Rather, this method is combined: functional and graphic. Of these two methods, the latter is not only elegant, but also the most important, since it shows the relationship between all types of mathematical models: a verbal description of the problem, a geometric model - a graph of a quadratic 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 quadratic trinomial satisfy given conditions in the domain of definition for the desired parameter values.
What other possible conditions can the roots of a quadratic trinomial satisfy for the desired parameter values?
Finding the roots of a quadratic trinomial
Goals: introduce the concept of a quadratic trinomial and its roots; develop the ability to find the roots of a quadratic trinomial.
During the classes
I. Organizational moment.
II. Oral work.
Which of the numbers: –2; -1; 1; 2 – are the roots of the equations?
a) 8 X+ 16 = 0; V) X 2 + 3X – 4 = 0;
b) 5 X 2 – 5 = 0; G) X 3 – 3X – 2 = 0.
III. Explanation of new material.
Explanation of new material should be carried out according to the following scheme:
1) Introduce the concept of the root of a polynomial.
2) Introduce the concept of a quadratic trinomial and its roots.
3) Analyze the question of the possible number of roots of a square trinomial.
The question of isolating the square of a binomial from a square trinomial is best discussed in the next lesson.
At each stage of explaining new material, it is necessary to offer students an oral task to test their understanding of the main points of the theory.
Task 1. Which of the numbers: –1; 1; ; 0 – are the roots of the polynomial X 4 + 2X 2 – 3?
Assignment 2. Which of the following polynomials are quadratic trinomials?
1) 2X 2 + 5X – 1; 6) X 2 – X – ;
2) 2X – ; 7) 3 – 4X + X 2 ;
3) 4X 2 + 2X + X 3 ; 8) X + 4X 2 ;
4) 3X 2 – ; 9) 
+ 3X – 6;
5) 5X 2 – 3X; 10) 7X 2 .
Which quadratic trinomials have root 0?
Task 3. Can a square trinomial have three roots? Why? How many roots does a square trinomial have? X 2 + X – 5?
IV. Formation of skills and abilities.
Exercises:
1. № 55, № 56, № 58.
2. No. 59 (a, c, d), No. 60 (a, c).
In this task you do not need to look for the roots of quadratic trinomials. It is enough to find their discriminant and answer the question posed.
a) 5 X 2 – 8X + 3 = 0;
D 1 = 16 – 15 = 1;
D 1 0, which means that this quadratic trinomial has two roots.
b) 9 X 2 + 6X + 1 = 0;
D 1 = 9 – 9 = 0;
D 1 = 0, which means the square trinomial has one root.
at 7 X 2 + 6X – 2 = 0;
7X 2 – 6X + 2 = 0;
D 1 = 9 – 14 = –5;
If there is time left, you can do No. 63.
Solution
Let ax 2 + bx + c is a given quadratic trinomial. Because the a+ b +
+ c= 0, then one of the roots of this trinomial is equal to 1. By Vieta’s theorem, the second root is equal to . According to the condition, With = 4A, so the second root of this quadratic trinomial is equal to 
.
ANSWER: 1 and 4.
V. Lesson summary.
Frequently asked questions:
– What is the root of a polynomial?
– Which polynomial is called a quadratic trinomial?
– How to find the roots of a quadratic trinomial?
– What is the discriminant of a quadratic trinomial?
– How many roots can a square trinomial have? What does this depend on?
Homework: No. 57, No. 59 (b, d, f), No. 60 (b, d), No. 62.
Lesson topic: "Square trinomial and its roots."
The purpose of the lesson: to introduce students to the concept of a square trinomial and its roots, to improve their skills in solving tasks for isolating the square of a binomial from a square trinomial.
Lesson includes four main stages:
Knowledge control
Explanation of new material
Reproductive consolidation.
Training reinforcement.
Reflection.
Stage 1. Knowledge control.
The teacher conducts a mathematical dictation “as a carbon copy” based on the material from the previous cycle. For dictation, cards of two colors are used: blue for 1 option, red for 2 options.
From the given analytical models of functions, select only quadratic ones.
Option 1. y=ax+4, y=45-4x, y=x²+4x-5, y=x³+x²-1.
Option 2. y=8x-b, y=13+2x, y= -x²+4x, y=-x³+4x²-1.
Sketch quadratic functions. Is it possible to uniquely determine the position of a quadratic function on the coordinate plane. Try to justify your answer.
Solve quadratic equations.
Option 1. a) x² +11x-12=0
B) x² +11x =0
Option 2. a) x² -9x+20=0
B) x² -9 x =0
4. Without solving the equation, find out whether it has roots.
Option 1. A) x² + x +12=0
Option 2. A) x² + x - 12=0
The teacher checks the answers received from the first two pairs. Incorrect answers received are discussed with the whole class.
| Option 1. | Option 2. |
| 1. y=x²+4x-5 | 1. y= -x²+4x |
| 2. The branches are up, but the position cannot be determined unambiguously because there is not enough data. | branches down, but it is impossible to unambiguously determine the position because there is not enough data. |
| 3. a) –12; 1 b) –11;0 | 3. a) 4;5 b) 9;0 |
| 4. D0, there are two roots |
Stage 2. Let's create a cluster. What associations do you have when considering the quadratic trinomial?
Creating a cluster.
Possible answers:
the quadratic trinomial is used to consider square. functions;
you can find the zeros of the square. functions
Using the discriminant value, estimate the number of roots.
Describe real processes, etc.
Explanation of new material.
Paragraph 2. clause 3 pp. 19-22.
Expressions are considered, and the definition of a quadratic trinomial and the root of a polynomial is given (during the discussion of previously discussed expressions)
The definition of the root of a polynomial is formulated.
The definition of a quadratic trinomial is formulated.
Examples of solving a trinomial are analyzed:
Find the roots of a quadratic trinomial.
Let us isolate the square binomial from the square trinomial.
3x²-36x+140=0.
A diagram of the approximate basis of the action is drawn up.
Algorithm for separating a binomial from a square trinomial.
1. Determine the numerical value of the leading square coefficient trinomial.
2. Perform identical and 2. Transform the expression,
equivalent transformations using formulas
(put the common factor out of brackets; the square of the sum and the difference.
convert the expression in parentheses
building it up to the formula for the square of the sum
or differences)
a²+2ab+b²= (a+c)² a²-2ab+b²= (a-c)²
Stage 3. Solving typical tasks from the textbook (No. 60 a, c; 61 a, 64 a, c) They are done at the board and commented on.
Stage 4. Independent work on 2 options (No. 60a, b; 65 a, b). Students check the sample solutions on the board.
Homework: P.3 (learn the theory, No. 56, 61g, 64g)
Reflection. The teacher gives the task: evaluate your progress at each stage of the lesson using a drawing and turn it in to the teacher. (the task is completed on separate sheets, a sample is provided).
Sample: 
Using the order of the elements in the picture, determine at what stage of the lesson your ignorance prevailed. Highlight this stage in red.
The practice of mathematics exams shows that problems with parameters are the most difficult, both logically and technically, and therefore the ability to solve them largely determines the successful passing of an exam at any level.
In problems with parameters, along with unknown quantities, there appear quantities whose numerical values, although not specifically indicated, are considered known and specified on a certain numerical set. In this case, the parameters included in the condition significantly influence the logical and technical course of the solution and the form of the answer. Such problems can be found in the book “514 Problems with Parameters.” In the literature on elementary mathematics there are many textbooks, problem books, and methodological manuals that contain problems with parameters. But most of them cover a narrow range of issues, placing the main emphasis on the recipe, rather than on the logic of solving problems. In addition, the most successful of the books have long become a bibliographic rarity. At the end of the work there is a list of books, articles from which helped to compile a classification of statements on the topic of the work. The most significant is the manual by A. Kh. Schachmeister. Equations and inequalities with parameters.
The main goal of this work is to fill some substantive gaps in the basic algebra course and establish the facts of using the properties of a quadratic function, which can significantly simplify the solution of problems related to the location of the roots of a quadratic equation with respect to certain characteristic points.
Job objectives:
Establish possible cases of the location of the roots of a square trinomial on the number line;
Identify algorithms that allow solving quadratic equations with a parameter based on the location of the roots of a quadratic trinomial on the number line;
Learn to solve problems of higher complexity than the required level; master a number of technical and intellectual mathematical skills at the level of their free use; improve mathematical culture as part of the school mathematics course.
Object of study: location of the roots of a square trinomial on a coordinate line.
Subject of research: quadratic equations with a parameter.
Methods of research. The main methods of studying problems with a parameter: analytical, graphical and combined (functional - graphical). Analytical is a method of so-called direct solution, repeating standard procedures for finding the answer in problems without a parameter. Graphic is a method that uses graphs in the coordinate plane (x; y). The clarity of the graphical method helps to find a quick way to solve a problem. Of these two methods, the latter is not only elegant, but also the most important, since it shows the relationship between all types of mathematical models: a verbal description of the problem, a geometric model - a graph of a quadratic trinomial, an analytical model - a description of a geometric model by a system of inequalities compiled on the basis mathematical statements identified from the graph of a quadratic function.
In many cases, solving quadratic equations with a parameter leads to cumbersome transformations. Hypothesis: using the properties of a quadratic function will significantly simplify the solution, reducing it to solving rational inequalities.
Main part. The location of the roots of a quadratic trinomial on the coordinate line
Let's consider some statements related to the location of the roots of the square trinomial f(x)=ax2+bx+c on the number line relative to points m and n such that m
x1 and x2 are the roots of the quadratic trinomial,
D=b2-4ac- discriminant of a square trinomial, D≥0.
m, n, m1, m2, n1, n2 - given numbers.
All arguments are considered for a>0, the case for a
Statement one
In order for the number m to be located between the roots of the square trinomial (x1
Proof.
provided x1
Geometric interpretation
Let x1 and x2 be the roots of the equation. For a > 0 f(x)
Problem 1. For what values of k does the equation x2-(2k+1)x + 3k-4=0 have two roots, one of which is less than 2 and the other is greater than 2?
Solution. f(x)=x2-(2k+1)x + 3k-4; x1
For k>-2, the equation x2-(2k+1)x + 3k-4=0 has two roots, one of which is less than 2 and the other is greater than 2.
Answer: k>-2.
Problem 2. For what values of k does the equation kx2+(3k-2)x + k-3=0 have roots of different signs?
This problem can be formulated as follows: for what values of k does the number 0 lie between the roots of this equation.
Solution (1 way) f(x)= kx2+(3k-2)x + k-3; x1
Method 2 (using Vieta's theorem). If a quadratic equation has roots (D>0) and c/a
Problem 3. For what values of k does the equation (k2-2)x2+(k2+k-1)x – k3+k2=0 have two roots, one of which is less than k and the other is greater than k?
f(x)=(k2-2)x2+(k2+k-1)x – k3+k2; x1 Substituting the values of k from the found set, we make sure that for these values of k D>0.
Statement two(a)
In order for the roots of a quadratic trinomial to be less number m(x1
Proof: x1-m>0, x2-m 0; m2-mx1-mx2+x1x2>0; m2-(x1+x2)m+x1x2
Problem 4. At what values of the parameter are the roots of the equation x2-(3k+1)x+2k2+4k-6=0 less than -1?
D≥0; (3k+1)2-4(2k2+4k-6) ≥0; (k-5)2≥0; k- any; x0-3/2; k0. 1+(3k+1)+(2k2+4k-6)>0. 2(k+4)(k-1/2)>0. k1/2
Statement two (b)
In order for the roots of a quadratic trinomial to be more number m(m
D ≥0; x0>m; af(m)>0.
If the condition m m. Since m does not belong to the interval (x1; x2), then f(m) > O for a > 0 and f(m)
Conversely, let the system of inequalities be satisfied. The condition D > 0 implies the existence of roots x1 and x2 (x1 m.
It remains to show that x1 > m. If D = 0, then x1 = x2 > m. If D > 0, then f(x0) = -D/4a and af(x0) 0, therefore, at points x0 and m the function takes values of opposite signs and x1 belongs to the interval (m; x0).
Problem 5. For what values of the parameter m are the roots of the equation x2-(3m+1)x+2m2+4m-6=0 a) greater than 1? b) less than -1?
Solution a) D≥0; D≥0; (3m+1)2-4(2m2+4m-6) ≥0; x0>m; x0>1; ½(3m+1)>1; f(m)>0. f(1)>0. 1-(3m+1)+(2m2+4m-6)>0.
(m-5)2≥0; m - any m>1/3; m>1/3;
(2km-3)(m+2)>0. m3/2. Answer: m>3/2.
b) D≥0; (3m+1)2-4(2m2+4m-6)≥0; (m-5)2 ≥0; m - any x0-3/2; m0. 1+(3m+1)+(2m2+4m-6)>0. 2(m+4)(m-1/2)>0. m1/2.
Problem 6. At what values of the parameter are the roots of the equation kx2-(2k +1)x+3 k -1=0 greater than 1?
Solution. Obviously, the problem is equivalent to the following: for what values of the parameter m are the roots of a quadratic trinomial greater than 1?
D≥0; D≥0 (2k+1)2-4k (3k-1) ≥0; 8k2-8k-1≤0; x0>m; x0>1 (2k+1)/ (2k) >1; 2k+1 > 2k; af(m)>0. af(1)>0. k(k-(2k+1)+(3k-1)) >0. 2k2-2k>0.
Having solved this system, we find that
Statement three
In order for the roots of a square trinomial to be greater than the number m and less than n (m
D ≥0; m 0 af(n)>0.
Note character traits graphic arts.
1) The equation has roots, which means D > 0.
2) The axis of symmetry is located between the lines x = m and x = n, which means m
3) At points x = m and x = n, the graph is located above the OX axis, therefore f(m) > 0 and f(n) > 0 (at m
The conditions listed above (1; 2; 3) are necessary and sufficient for the desired parameter values.
Problem 7. For what m x2-2mx+m2-2m+5=0 do the numbers not exceed 4 in absolute value?
Solution. The condition of the problem can be formulated as follows: for what m does the relation -4
We find the values of m from the system
D > 0; m2 - (m2 – 2m + 5) ≥ 0;
4 ≤ x0 ≤ 4; -4 ≤ m≤ 4; f(-4)≥ 0; 16 + 8m+ m2 – 2m + 5 ≥ 0; f(4)≥0; 16-8m + m2-2m + 5 ≥0; whose solution is the segment . Answer: m.
Problem 8. For what values of m are the roots of the quadratic trinomial
(2m - 2)x2 + (m+1)x + 1 is greater than -1, but less than 0?
Solution. The values of m can be found from the system
D≥0; (m+1)2-4(2m-2) ≥ 0;
(2m - 2)/(-1) > 0 (2m -2)(2m -2 -m -1 +1) > 0;
(2m-2)f(0)>0; (2m-2)>0;
Answer: m > 2.
Statement four(s)
In order for the smaller root of the square trinomial to belong to the interval (m;n), and the larger one not to belong to (m
D ≥0; af(m)>0 af(n)
The graph of a quadratic trinomial intersects the OX axis exactly once on the interval (m; n). This means that at the points x=m and x=n the square trinomial takes on values of different signs.
Problem 10. For what values of the parameter a does only the smaller root of the quadratic equation x2+2ax+a=0 belong to the interval X(0;3).
Solution. Consider the quadratic trinomial y(x) = x2-2ax+a. The graph is a parabola. The branches of the parabola are directed upward. Let x1 be the lesser root of the square trinomial. According to the conditions of the problem, x1 belongs to the interval (0;3). Let us depict a geometric model of the problem that meets the conditions of the problem.
Let's move on to the system of inequalities.
1) We note that y(0)>0 and y(3) 0. Therefore, this condition does not need to be written into the system of inequalities.
So, we obtain the following system of inequalities:
Answer: a>1.8.
Statement four (b)
In order for the larger root of the square trinomial to belong to the interval (m; n), and the smaller one not to belong to (x1
D ≥0; af(m) 0.
Statement four (combined)
Comment. Let the problem be formulated as follows: for what values of the parameter does one root of the equation belong to the interval (b;m), and the other does not? To solve this problem, there is no need to distinguish between two subcases; we find the answer from the inequality f(m) f(n)
D ≥0; f(m) f(n)
Problem 11. For what m does only one root of the equation x2-mх+6=0 satisfy condition 2
Solution. Based on statement 4(b), we find the value of m from the condition f(2)f(5) (10 – 2m)(31 – 5m) m2 - 24 = 0, i.e. for m = ±2√6, For m = -2√6 x = - √6, which does not belong to the interval (2; 5), with m = 2√6 x =√6, belonging to the interval (2; 5).
Answer: m (2√6) U (5; 31/5).
Statement five
In order for the roots of a quadratic trinomial to satisfy the relation (x1
D ≥0; af(m)Problem 12. Find all values of m for which the inequality x2+2(m-3)x + m2-6m
Solution. By condition, the interval (0; 2) must be contained in the set of solutions to the inequality x2 + 2(m - 3)x + m2 – 6m Based on Statement 5, we find the values of m from the system of inequalities f(0) ≤ 0;m2-6m ≤ 0; m f(2) ≤ 0. 4 + 4(m-3) + m2-6m ≤ 0. m [-2;4], whence m.
Answer: m.
Statement six
In order for the smaller root of the square trinomial to belong to the interval (m1; m2), and the larger root to belong to the interval (n1; n2) (m2
D ≥0; af(m1)>0; af(m2)This statement is a combination of statements 4a and 4b. The first two inequalities guarantee that x1(m1, n1), and the last two inequalities guarantee that x2(m2, n2),
Problem 13. At what m is one of the roots of the equation x2 - (2m + l)x + m2 + m- 2 = 0 located between the numbers 1 and 3, and the second - between the numbers 4 and 6?
Solution. 1 way. Considering that a = 1, the values of m can be found from the system f(1) > 0; 1 -2m- 1+m2 + m-2 >0; m2-m-2>0 m (-∞;-1) U (2;+∞) f(3)
4(4) 0; 36-12m-6 + m2 + m-2 0 m (-∞;4)U (7;+∞), whence m(2; 4).
Answer: m(2; 4).
Thus, we have established statements related to the location of the roots of the square trinomial f(x)=ax2+bx+ on the number line with respect to certain points.
Conclusion
In the course of my work, I mastered a number of technical and mathematical skills at the level of free use and improved my mathematical culture as part of the school mathematics course.
As a result of the work, the set goal was achieved: the properties of the quadratic function were established, which make it possible to significantly simplify the solution of problems related to the location of the roots of a quadratic equation relative to certain characteristic points. Possible cases of the location of the roots of a square trinomial on the number line are established. Algorithms have been identified that allow solving quadratic equations with a parameter based on the location of the roots of a square trinomial on the number line; tasks of higher complexity than the required level were solved. The work presents a solution to only 12 problems due to the limited number of pages of the work. Of course, the problems discussed in the work can be solved in other ways: using formulas for the roots of a quadratic equation, using the property of roots (Vieta’s theorem).
In fact, a significant number of problems were solved. Therefore, it was decided to create a collection of problems on the topic of the design and research work “Solver of problems on the application of the properties of a square trinomial related to the location of its roots on the coordinate line.” In addition, the result of the work (the product of design and research work) is a computer presentation that can be used in classes of the elective subject “Solving problems with parameters.”
