Parameters for the location of the roots of a square trinomial. Location of the roots of a square trinomial

Quadratic equations with parameters

(Methodological development for students in grades 9-11)

mathematics teacher of the highest qualification category,

Deputy Director for UVR

Megion 2013

Foreword

https://pandia.ru/text/80/021/images/image002.png" height="22 src=">2. Application of the Vieta theorem

Scientific work mathematical development personality, but also in any other scientific research. Therefore, the solution of problems with parameters and in particular the solution of quadratic equations with parameters is propaedeutics research work of students. At the USE in mathematics (often tasks C5), GIA (tasks of part 2) and at the entrance exams, there are mainly two types of tasks with parameters. First: "For each value of the parameter, find all solutions to some equation or inequality." Second: "Find all the values of the parameter, for each of which for given equation or inequalities, certain conditions are met. Accordingly, the answers in these two types of problems differ in essence. In the answer to the problem of the first type, all possible values of the parameter are listed, and solutions to the equation are written for each of these values. In the answer to the problem of the second type, all parameter values are indicated under which the conditions specified in the problem are met.

As you know, very little attention is paid to solving problems with parameters in school. Therefore, solving problems with parameters always causes great difficulties in students; it is difficult to expect that students whose training did not include “parametric therapy” will be able to successfully cope with such tasks in the tough atmosphere of a competitive exam, therefore, students should prepare specifically for the “encounter with parameters”. Many students perceive the parameter as a "regular" number. Indeed, in some problems the parameter can be considered a constant value, but this constant value takes on unknown values. Therefore, it is necessary to consider the problem for all possible values this constant value. In other problems, it may be convenient to artificially declare one of the unknowns as a parameter.

Tasks with parameters have diagnostic and prognostic value - with the help of tasks with parameters, you can check the knowledge of the main sections school mathematics, the level of mathematical and logical thinking, initial skills of research activities, and most importantly, promising opportunities for successfully mastering the mathematics course of this university.

Analysis USE options in mathematics and entrance exams to various universities shows that most of the proposed tasks with parameters are related to the location of the roots of a square trinomial. Being the main school course mathematics, the quadratic function forms an extensive class of problems with parameters, diverse in form and content, but united common idea– their solution is based on the properties quadratic function. When solving such problems, it is recommended to work with three types of models:

1. verbal model - verbal description tasks;

2. geometric model - a sketch of a graph of a quadratic function;

3. analytical model - a system of inequalities, which describes the geometric model.

The methodological manual contains theorems on the location of the roots of a square trinomial (necessary and sufficient conditions location of the roots of a quadratic function relative to given points), the application of the Vieta theorem to the solution of quadratic equations with parameters. Given detailed solutions 15 tasks with methodological recommendations. Purpose this manual– to help the graduate and teacher of mathematics in preparation for passing the exam and GIA in mathematics, and entrance exam to the university in the form of a test or in the traditional form.

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3. the parabola intersects with the line x = n at a point lying in the upper half-plane for a>0 and at a point lying in the lower half-plane for a<0 (условие a∙f(n) >0).

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Theorem 10. Quadratic equations x2 + p1x + q1 = 0 and x2 + p2x + q2 = 0,

whose discriminants are non-negative, have at least one common root if and only if (q2 – q1)2 = (p2 – p1)(p1q2 – q1p2).

Proof.

Let f1(x) = x2 + p1x + q1, f2(x) = x2 + p2x + q2, and the numbers x1, x2 are the roots of the equation f1(x) = 0. In order for the equations f1(x) = 0 and f2( x) = 0 have at least one common root, it is necessary and sufficient that f1(x)∙f2(x) = 0, i.e., that (x12 + p2x1 + q2)(x22 + p2x2 + q2) = 0 We represent the last equality in the form

(x12 + p1x1 + q1 + (p2 – p1)x1 + q2 – q1) (x22 + p1x2 + q1 + (p2 – p1)x2 + q2 – q1) = 0.

Since x12 + p1x1 + q1 = 0 and x22 + p1x2 + q1 = 0, we get

((p2 – p1)x1 + (q2 – q1))((p2 – p1)x2 + (q2 – q1)) = 0, i.e.

(p2 – p1)2x1x2 + (q2 – q1)(p2 – p1)(x1 + x2) + (q2 – q1)2 = 0.

By the Vieta theorem x1 +x2 = - p1 and x1x2 =q1; Consequently,

(p2 – p1)2q1 – (q2 – q1)(p2 - p1)p1 + (q2 – q1)2 = 0, or

(q2 – q1)2 = (p2 - p1)((q2 – q1)p1 - (p2 - p1)q1) = (p2 – p1)(q2p1 – q1p1 – p2q1 + p1q1) =

(p2 – p1)(q2p1 – p2q1), which was to be proved.

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Quadratic equation ax 2 + bx + c = 0

1) has two real positive roots if and only if the following conditions are met simultaneously:

;

2) has two real negative roots if and only if the conditions are met simultaneously:

;

3) has two real roots of different signs if and only if the following conditions are met simultaneously:

;

4) has two real roots of the same sign if

Remark 1. If the coefficient at X 2 contains a parameter, it is necessary to analyze the case when it vanishes.

Remark 2. If the discriminant quadratic equation is a perfect square, then at first it is more convenient to find explicit expressions for its roots.

Remark 3. If an equation containing several unknowns is quadratic with respect to one of them, then the key to solving the problem is often the study of its discriminant.

We present a scheme for studying problems related to the location of the roots of a square trinomialf(x) = ax2 + bx + c:

1. Study of the case a = o (if the first coefficient depends on the parameters).

2. Finding the discriminant D in the case a≠0.

3.If D - full square some expression, then finding the roots x1, x2 and subordinating the conditions of the problem.

4..png" width="13" height="22 src="> 3. Examples of solving problems for preparing for the GIA and the Unified State Examination in mathematics

Example 1 Solve the equation ( a - 2)x 2 – 2ax + 2a – 3 = 0.

Solution. Consider two cases: a = 2 and a ≠ 2. in the first case, the original equation takes the form - 4 X+ 1 = 0..png" width="255" height="58 src=">

For a \u003d 1 or a \u003d 6, the discriminant is zero and the quadratic equation has one root: , i.e., for a \u003d 1 we get the root , and for a = 6 - the root.

At 1< a < 6 дискриминант положителен и квадратное уравнение имеет два корня: https://pandia.ru/text/80/021/images/image053.png" width="163" height="24 src=">the equation has no roots; for a = 1 the equation has one root X= -1; at the equation has two roots ; at a= 2 the equation has a single root ; at a= 6 the equation has a single root .

Example 2 At what value of the parameter a the equation ( a - 2)X 2 + (4 – 2a)X+ 3 = 0 has a single root?

Solution . If a a= 2, then the equation becomes linear∙ X+ 3 = 0; which has no roots.

If a a≠ 2, then the equation is quadratic and has a single root with zero discriminant D.

D= 0 at a 1 = 2 and a 2 = 5. Meaning a= 2 is excluded, since it contradicts the condition that the original equation is quadratic.

Answer : a = 5.

(a - 1)X 2 + (2a + 3)X + a+ 2 = 0 has roots of the same sign?

Solution. Since, according to the condition of the problem, the considered equation is quadratic, it means that a≠ 1. Obviously, the condition of the problem also implies the existence of roots of the quadratic equation, which means that the discriminant is non-negative

D = (2a + 3)2 – 4(a - 1)(a + 2) = 8a + 17.

Since, by condition, the roots must be of the same sign, then X 1∙X 2 > 0, i.e..png" width="149" height="21 src=">. Subject to conditions D≥ 0 and a≠ 1 we get https://pandia.ru/text/80/021/images/image060.png" width="191" height="52 src=">.

Example 3 Find all values of a for which the equation x2 - 2(a - 1)x + (2a + 1) = 0 has two positive roots.

Solution. From the Vieta theorem, in order for both roots x1 and x2 of this equation to be positive, it is necessary and sufficient that the discriminant of the square trinomial x2 - 2(a - 1)x + (2a + 1) be non-negative, and the product x1 ∙ x2 and the sum x1 + x2 were positive. We get that all a satisfying the system

And only they are the solutions to the problem. This system is equivalent to the system

The solution of which, and hence the problem itself, are all numbers from the interval

Task #3.

At what values of the parameter k the roots of the equation (k-2)x 2 -2kx+2k-3=0

belong to the interval (0;1)?

Solution.

For k≠2, the desired parameter values must satisfy the system of inequalities

Where D= 4k 2 -4(k-2)(2k-3) = -4(k 2 -7k+6), f(0) = 2k-3? F (1) \u003d k-5, x in \u003d k / (k-2).

This system has no solutions.

For k = 2 given equation has the form -4x+1 = 0, its only root

x = ¼, which belongs to the interval (0;1).

Task #4.

At what values of a are both roots of the equation x 2 -2ax + a 2 -a \u003d 0 located on the segment?

The desired values must satisfy the system of inequalities

where D \u003d 4a 2 -4 (a 2 -a) \u003d 4a, f (2) \u003d a 2 -5a + 4, f (6) \u003d a 2 -13a + 36, x in \u003d a.

The only solution of the system is the value, a = 4.

4. Independent work(control - training).

Students work in groups, perform the same option, since the material is very complex and not everyone can do it.

No. 1. At what values of the parameter a do both roots of the equation x 2 -2ax + a 2 - 1 \u003d 0 belong to the interval (-2; 4)?

No. 2. Find all values of k for which there is one root of the equation

(k-5)x 2 -2kx+k-4=0 is less than 1 and the other root is greater than 2.

Number 3. At what values of a is the number 1 between the roots of the square trinomial x 2 + (a + 1) x - a 2?

At the end of the time, the answers are displayed. Self-checking of independent work is carried out.

5. Summary of the lesson. Finish the offer.

"Today in class..."

"I remember..."

"Would like to note …".

The teacher analyzes the entire course of the lesson and its main points, evaluates the activities of each student in the lesson.

6. Homework

(from the collection of tasks for preparing for the GIA in grade 9, author L. V. Kuznetsova)

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)?

Solution

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.

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?

Solution.

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?

Solution.

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?

Solution.

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)?

Solution.

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.

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

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).

Solution.

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 mx 2 x

f(k)

Analytical model

TASK #7.

For what values of the parameter a only larger root quadratic equation x 2 +4x-(a+1)(a+5)=0 belongs to the interval [-1;0).

Solution.

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

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?

Solution.

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=0 lie on different sides numbers from number 3?

Solution.

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!

At what value of the parameter a one root of the equation

greater than 1 and the other less than 1?

Consider the function -

Objective:

The study of all possible features of the location of the roots of a square trinomial relative to a given point and relative to a given segment based on the properties of a quadratic function and graphical interpretations.
Application of the studied properties in solving non-standard problems with a parameter.

Tasks:

To study various methods of solving problems based on the study of the location of the roots of a square trinomial by a graphical method.
Substantiate all possible features of the location of the roots of a square trinomial, develop theoretical recommendations for solving non-standard problems with a parameter.
Master a number of technical and intellectual mathematical skills, learn how to use them in solving problems.