How to solve equations with a parameter graphically. Graphical method for solving equations with parameters

Equations with parameters are considered to be among the most challenging tasks I know school mathematics. It is these tasks that fall from year to year in the list of tasks of type B and C at the unified state USE exam. However, among a large number equations with parameters are those that can be easily solved graphically. Let's consider this method on the example of solving several problems.

Find the sum of integer values of a for which the equation |x 2 – 2x – 3| = a has four roots.

Decision.

To answer the question of the problem, we build on one coordinate plane function graphs

y = |x 2 – 2x – 3| and y = a.

Graph of the first function y = |x 2 – 2x – 3| will be obtained from the graph of the parabola y = x 2 - 2x - 3 by displaying symmetrically about the abscissa axis of the part of the graph that is below the Ox axis. The part of the graph above the x-axis will remain unchanged.

Let's do it step by step. The graph of the function y \u003d x 2 - 2x - 3 is a parabola, the branches of which are directed upwards. To build its graph, we find the coordinates of the vertex. This can be done using the formula x 0 = -b / 2a. Thus, x 0 \u003d 2/2 \u003d 1. To find the coordinate of the top of the parabola along the y-axis, we substitute the obtained value for x 0 into the equation of the function under consideration. We get that y 0 \u003d 1 - 2 - 3 \u003d -4. Hence, the vertex of the parabola has coordinates (1; -4).

Next, you need to find the points of intersection of the branches of the parabola with the coordinate axes. At the points of intersection of the branches of the parabola with the abscissa axis, the value of the function is zero. Therefore, we decide quadratic equation x 2 - 2x - 3 = 0. Its roots will be the desired points. By the Vieta theorem, we have x 1 = -1, x 2 = 3.

At the points of intersection of the branches of the parabola with the y-axis, the value of the argument is zero. Thus, the point y = -3 is the point of intersection of the branches of the parabola with the y-axis. The resulting graph is shown in Figure 1.

To get the graph of the function y = |x 2 - 2x - 3|, we will display the part of the graph, which is below the x-axis, symmetrically about the x-axis. The resulting graph is shown in Figure 2.

The graph of the function y = a is a straight line parallel to the x-axis. It is shown in Figure 3. Using the figure and we find that the graphs have four common points (and the equation has four roots) if a belongs to the interval (0; 4).

Integer values of number a from the received interval: 1; 2; 3. To answer the question of the problem, let's find the sum of these numbers: 1 + 2 + 3 = 6.

Answer: 6.

Find the arithmetic mean of integer values of the number a, for which the equation |x 2 – 4|x| – 1| = a has six roots.

Let's start by plotting the function y = |x 2 – 4|x| – 1|. To do this, we use the equality a 2 = |a| 2 and select full square in a submodule expression written on the right side of the function:

x 2 – 4|x| – 1 = |x| 2 – 4|x| - 1 = (|x| 2 - 4|x| + 4) - 1 - 4 = (|x | - 2) 2 - 5.

Then the original function will look like y = |(|x| – 2) 2 – 5|.

To build a graph of this function, we build successively graphs of functions:

1) y \u003d (x - 2) 2 - 5 - a parabola with a vertex at a point with coordinates (2; -5); (Fig. 1).

2) y = (|x| - 2) 2 - 5 - the part of the parabola constructed in paragraph 1, which is located to the right of the ordinate axis, is symmetrically displayed to the left of the Oy axis; (Fig. 2).

3) y = |(|x| – 2) 2 – 5| - the part of the graph constructed in paragraph 2, which is below the x-axis, is displayed symmetrically with respect to the abscissa axis upwards. (Fig. 3).

Consider the resulting drawings:

The graph of the function y = a is a straight line parallel to the x-axis.

With the help of the figure, we conclude that the function graphs have six common points(the equation has six roots) if a belongs to the interval (1; 5).

This can be seen in the following figure:

Find the arithmetic mean of the integer values of the parameter a:

(2 + 3 + 4)/3 = 3.

Answer: 3.

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In order to fully reveal the possibilities of this method, we will consider the main types of problems.

Sample tasks for developing knowledge and skills in solving problems with parameters using a graphical method (coordinate plane)

Exercise 1.

At what valuesaequation = has two roots?

Decision.

Let's move on to equivalent system:

This system defines a curve on the coordinate plane (;). It is clear that all points of this arc of the parabola (and only they) have coordinates that satisfy the original equation. Therefore, the number of solutions to the equation for each fixed value of the parameter, is equal to the number of points of intersection of the curve with the horizontal line corresponding to this parameter value.

Obviously, for the indicated lines intersect the graph at two points, which is equivalent to the original equation having two roots.

Answer: at.

Task 2.

Find all values of a for which the system has a unique solution.

Decision.

Let's rewrite the original system in this form:

All solutions of this system (view pairs) form the area shown in the figure by hatching. The requirement for the uniqueness of the solution of this system is translated into a graphical language as follows: horizontal lines must have only one common point with the resulting area. It is easy to see that only straight linesand satisfy the stated requirement.

Answer: or.

The two problems just analyzed allow us to give more specific recommendations compared to those given earlier:

try to express the parameter through a variable, i.e. get equalities of the form, then

graph a function on a plane.

Task 3.

At what valuesa does the equation have exactly three roots?

Decision.

We have

The graph of this set is the union of the “corner” and the parabola. Obviously, only the line intersects the resulting union at three points.

Answer: .

Comment: The parameter is usually considered as fixed, but unknown number. Meanwhile, from a formal point of view, a parameter is a variable, moreover, “equal” with others present in the task. With this view of the form parameter, functions are defined not with one, but with two variables.

Task 4.

Find all parameter values, for which the equation has one solution.

Decision.

A fraction is zero if and only if the numerator of the fraction is zero and the denominator is non-zero.

Finding roots square trinomial:

Using the resulting system, it is easy to plot the original equation. It is the presence of "punctures" in this graph that allows for and = to have a unique solution to the equation. This is the determining factor in the decision.

Answer: and.

Task 5.

At what values of the parameter,a the equation has a unique solution.

Decision.

Let us write a system equivalent to the original equation

From here we get

We build a graph and we will draw straight lines perpendicular to the axisa .

The first two inequalities of the system define a set of points shown by hatching, and this set does not include hyperbolas and.

Then a segment and a ray, a segment and a ray, lying respectively on the lines and , are the graph of the original equation. One solution will be if 2< < или < или = .

Answer : 2 < < или < или = .

Task 6.

Find all parameter valuesa , for which the equation

has exactly two various solutions

Decision.

Consider the set of two systems

If a , then.

If a < , then.

From here

Parabolas and a straight line have two common points:BUT (-2; - 2), AT(-1; -1), moreover, AT is the vertex of the first parabola,D - top of the second. So, the graph of the original equation is shown in the figure.

There must be exactly two different solutions. This is done with or.

Answer: or.

Task 7.

Find the set of all numbers, for each of which the equation

has only two distinct roots.

Decision.

Let us rewrite this equation in the form

The roots of the equation, provided that.

Building a graph given equation. AT this case It is convenient to build a graph by assigning the variable the y-axis. Here we “read” the answer with vertical lines, we get that this equation has only two different roots at = -1 or or.

Dotted lines say that.

Answer: at = -1 or or.

Task 8.

For which the set of solutions of the inequality contains a gap.

Decision.

Let's write down the set of two systems, which is equivalent to the original equation:

Since in the solution of the first system neithera segment cannot be included, then we will carry out the necessary studies for the second system.

We have

Denote . Then the second inequality of the system takes the form< - and defines the set shown in the figure on the coordinate plane.

With the help of the figure, we establish that for the resulting set contains all points, the abscissas in which run through all the values of the interval

Then, from here.

Answer : .

Task 9.

Find all non-negative numbers for which there exists singular, satisfying the system

Decision.

We have

The first equation on the coordinate plane defines a family of vertical lines. The straight lines and divide the planes into four regions. Some of them are solutions to the system's inequality. Specifically, which ones can be established by taking a trial point from each area. The area whose point satisfies the inequality is its solution (this technique is associated with the method of intervals when solving inequalities with one variable). We build straight lines

For example, we take a point and substitute it into The coordinates of the point satisfy the inequality.

We get two areas (I) and ( II), but, given that by condition, we take only the area (I). We build straight lines , k .

So, the original system is satisfied by all points (and only them) lying on the rays and highlighted in the drawing by bold lines (i.e., we build points in a given area).

Now we need to find the only one for fixed. Draw parallel lines intersecting the axis. and find where there will be one point of intersection with the line.

We find from the figure that the requirement of uniqueness of the solution is achieved if (for already 2 points),

where is the ordinate of the point of intersection of the lines and,

where is the ordinate of the intersection point of the lines and.

So we get< .

Answer: < .

Task 10.

At what values of the parameter a does the system have solutions?

Decision.

We factorize the left side of the inequality of the system, we have

We build straight lines and We show in the figure by shading the set of points of the plane that satisfies the inequality of the system.

We build a hyperbola = .

Then the abscissas of the distinguished arcs of the hyperbola are solutions of the original system.M , P , N , Q - nodal points. Let's find their abscissas.

For points P , Q we have

It remains to write down the answer: or.

Answer: or.

Task 11.

Find all values for which any solution to the inequality does not exceed two in absolute value ().

Decision .

Let us rewrite this inequality in this form. We construct graphs of equations and =.

Using the “interval method”, we establish that the shaded areas will be the solution to the original inequality.

Now we build the area and see which part of it falls into the shaded area.

Those. now, if for some fixed value, the line at the intersection with the resulting area gives only points whose abscissas satisfy the condition < 2, then is one of the required values of the parameter.

So we see that.

Answer: .

Task 12.

For what values of the parameter does the set of solutions to the inequality contain at most four integer values?

Decision.

Let us transform this inequality to the form This inequality is equivalent to the combination of two systems

Using this set, we depict the solution of the original inequality.

Let's draw straight lines, where. Then the value for which the line intersects the lines at most in four points from the marked set will be the desired one. So we see that either or.

Answer: or or.

Task 13.

At what values of the parametera has a solution system

Decision.

The roots of the square trinomial i.

Then

We build straight lines and

Using the “intervals” method, we find the solution to the system inequality (shaded area).

That part of the circle with the center at the origin and radius 2, which falls into the shaded area and will be the solution of this system. .

Values and find from the system

Values and - from the system.

Answer:

Task 14.

Depending on the parameter valuesa solve inequality > .

Decision.

Let us rewrite this inequality in the form and consider the function, which, expanding the modules, we write as follows:

We build a chart. The graph splits the coordinate plane into two regions. Taking m. (0; 0) and substituting and into the original inequality, we get that 0 > 1, and therefore the original inequality is satisfied in the area lying above the graph.

Directly from the figure we get:

no solutions;

at ;

at.

Answer: no solutions;

at ;

at.

Task 15.

Find all values of the parameter for which the system of inequalities

only one is satisfied.

Decision.

Let's rewrite this system in this form:

Let us construct the area specified by this system.

1) , is the vertex of the parabola.

2) is a straight line passing through the points and.

The requirement for the uniqueness of the solution is translated into a graphical language as follows: horizontal lines with the resulting area must have only one common point. The above requirement is satisfied by the lines and, where is the ordinate of the point of intersection of the parabola and the line.

Let's find the value:

= (not suitable for meaning of the task),

We find the ordinate:

Answer: ,

Task 16.

Find all parameter valuesa, under which the system of inequalities

satisfies only for one x.

Decision .

Let us construct parabolas and show the solution of the last system by shading.

1) , .

2) , .

It can be seen from the figure that the condition of the problem is satisfied for or.

Answer: or.

Task 17.

For what values does the equation have exactly three roots?

Decision.

This equation is equivalent to the set

The population plot is the union of the parabola and angle plots.

The lines intersect the resulting union at three points.

Answer: at.

Task 18.

For what values does the equation have exactly three solutions?

Decision.

Let's transform the left side of this equation. We get a quadratic equation for.

We get the equation

which is equivalent to the aggregate

The union of the graphs of parabolas is the solution of the set.

Find the ordinate of the points of intersection of the parabolas:

We read the necessary information from the figure: this equation has three solutions for or

Answer: at or

Task 19.

Depending on the parameter, determine the number of roots of the equation

Decision .

Consider this equation as a quadratic one with respect to a.

We get the set

We build graphs of the equations of the set and answer the question of the problem.

Answer:: no solutions;

: one solution;

: two solutions;

or: three solutions;

or: four solutions.

Task 20.

How many solutions does the system have

Decision.

It is clear that the number of roots of the second equation of the system is equal to the number of solutions of the system itself.

We have .

Considering this equation as a quadratic one, we obtain the set.

Now referring to the coordinate plane makes the task simple. We find the coordinates of the intersection points by solving the equation

From here

Vertices of parabolas and.

Answer: : four solutions;

: two solutions;

: one solution;

: no solutions.

Task 21.

Find all real values of the parameter for which the equation has only two distinct roots. Write down these roots.

Decision .

Let's find the roots of the square trinomial in brackets:

We depict the set of solutions of this equation in the coordinate plane by plotting graphs provided that

We read the necessary information from the picture. So, this equation has two different roots at (u) and at (u)

Answer: with (and) and

at (and).

Task 2 2 .

Solve the system of inequalities:

Decision.

We build in the plane graphs of a parabola and a straight line.

All points of the shaded area are the solution of the system. Let's divide the constructed area into two parts.

If and, then there are no solutions.

If, then the abscissas of the points of the shaded area will be greater than the abscissas of the points of the straight line, but less than the abscissas (larger root of the equation) of the parabola.

We express through from the equation of a straight line:

Let's find the roots of the equation:

Then.

If, then.

Answer: for and 1 there are no solutions;

at;

at.

Task 23.

Solve the system of inequalities

Decision.

– top of the parabola.

Top of the parabola.

Find the abscissas of the intersection points of the parabolas:

The shaded area is the solution of the system. Let's break it down into two parts.

In the equations of parabolas, we express through:

We write down answer:

if and, then there are no solutions;

if, then< ;

if, then.

Task 24.

At what values, and the equation has no solutions?

Decision.

The equation is equivalent to the system

Let us construct a set of solutions to the system.

Three pieces of a parabola are the solution to this equation.

Let us find under which and exclude it.

So, for there are no solutions;

when there are no solutions;

(note: for the restathere are one or two solutions).

Answer: ; .

Task 25.

For what real values of the parameter there is at least one that satisfies the conditions:

Decision.

Let's graphically solve the inequality in using the "interval method" and build a graph. Let's see what part of the graph falls into the constructed region for solving the inequality, and find the corresponding valuesa.

We build graphs of lines and

They divide the coordinate plane into 4 regions.

Let's graphically solve the last inequality using the "interval method".

The shaded area is its solution. Part of the parabola graph falls into this area. On the interval; (by condition the inequality of the system is strict) exist that satisfy the conditions of the given system.

Answer:

Task 26.

Find all values of the parameter, for each of which the set of solutions to the inequality does not contain any solutions to the inequality.

Decision.

Let us construct a set of solutions to the inequality (“by the method of intervals”). Then we will build a "band" The desired values of the parameterq those for which none of the points of the indicated areas belongs to the "strip"

Answer: or.

Task 27.

For what values of the parameter, the equation has a unique solution.

Decision.

Let's factorize the numerator of the fraction.

This equation is equivalent to the system:

Let's build a population graph in the coordinate plane.

– point of intersection of lines and. A population graph is a union of lines.

We “poke out” the points of the graph with abscissas.

We draw straight lines and see where there is one point of intersection with the graph.

It is obvious that only for or this equation has a unique solution.

Answer: or.

Task 28.

For what real values of the parameter the system of inequalities has no solutions.

Decision.

The set of points in the plane of the shaded area satisfies the given system of inequalities.

We build straight lines. From the figure, we determine that for (- the abscissa of the intersection point of the hyperbola and the line), the lines do not intersect the shaded area.

Answer: at.

Task 29.

At what values of the parametera the system has a unique solution.

Decision.

Let's move on to a system equivalent to the given one.

In the coordinate plane, we plot the graphs of the parabolas and Vertices of the parabolas, respectively, the points and.

Calculate the abscissas of the intersection points of the parabolas by solving the equation

The shaded area is the solution of the system of inequalities. Direct and

has one common point with the shaded area.

Answer: at i.

Task 30.

Solve the inequality:

Decision.

Depending on the parameter, we find the value.

We will solve the inequality using the “interval method”.

Let's build parabolas

: .

Calculate the coordinates of the point of intersection of the parabolas:

The points of the shaded area satisfy this inequality. By drawing a straight line, we divide this area into three parts.

1) If, then there are no solutions.

2) If, then in the equation we express through:

Thus, in the areaI we have.

If, then look:

a) region II .

We express in the equation through.

smaller root,

Bigger root.

So, in the area II we have.

b) area III : .

Answer: when there are no solutions;

at, .

Literature:

Galitsky M. L., Goldman A. M., Zvavich L. I. Collection of problems in algebra for grades 8 - 9: Tutorial for students of schools and classes with in-depth study Mathematics - 2nd ed. – M.: Enlightenment, 1994.

P. I. Gorshtein, V. B. Polonsky, M. S. Yakir. Tasks with parameters. 3rd edition, enlarged and revised. - M .: Ileksa, Kharkov: Gymnasium, 2003.

Faddeev D. K. Algebra 6 - 8. - M .: Education, 1983 (b - ka teacher of mathematics).

A. H. Shakhmeister. Equations and inequalities with parameters. Edited by B. G. Ziva. C - Petersburg. Moscow. 2004.

V. V. Amelkin, V. L. Rabtsevich. Tasks with parameters Minsk "Asar", 2002.

A. H. Shakhmeister. Tasks with parameters in the exam. Moscow University Publishing House, CheRo on the Neva MCNMO.

Otdelkina Olga, 9th grade student

This topic is an integral part of the study school course algebra. The purpose of this work is to study this topic in more depth, to identify the most rational decision leading quickly to an answer. This essay will help other students understand the use of the graphical method for solving equations with parameters, learn about the origin, development of this method.

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Introduction2

Chapter 1

The history of the emergence of equations with parameter 3

Vieta's theorem4

Basic concepts5

Chapter 2. Types of equations with parameters.

Linear Equations6

Quadratic Equations………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………...7

Chapter 3

Analytical method….……………………………………………….......8

Graphic method. History of occurrence……………………………9

Graphical solution algorithm ..…………….....…………….10

Solving an equation with a modulus……………...……………………………….11

Practical part…………………………………………………………………12

Conclusion………………………………………………………………………….19

References…………………………………………………………………20

Introduction.

I chose this topic because it is an integral part of the study of the school algebra course. Cooking this work, I set the goal of a deeper study of this topic, identifying the most rational solution that quickly leads to an answer. My essay will help other students understand the use of the graphical method for solving equations with parameters, learn about the origin, development of this method.

AT modern life study of many physical processes and geometric patterns often leads to solving problems with parameters.

To solve such equations, the graphical method is very effective when it is necessary to establish how many roots the equation has depending on the parameter α.

Problems with parameters are of purely mathematical interest, contribute to intellectual development students, serve as a good material for practicing skills. They have diagnostic value, since they can be used to test knowledge of the main sections of mathematics, the level of mathematical and logical thinking, initial skills research activities and promising opportunities for successful mastery of the course of mathematics in higher educational institutions.

In my abstract, commonly encountered types of equations are considered, and I hope that the knowledge I gained in the process of work will help me when passing school exams, after allequations with parametersrightfully considered one of the most difficult tasks in the course of school mathematics. It is these tasks that fall into the list of tasks on a single state exam USE.

The history of the emergence of equations with a parameter

Problems for equations with a parameter were already encountered in the astronomical treatise "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scholar, Brahmagupta (7th century), expounded general rule solutions of quadratic equations reduced to a single canonical form:

αх 2 + bx = c, α>0

In the equation, the coefficients, except for the parameter, can also be negative.

Quadratic equations in al-Khwarizmi.

Al-Khwarizmi's algebraic treatise gives a classification of linear and quadratic equations with a parameter. The author lists 6 types of equations, expressing them as follows:

1) “Squares are equal to roots”, i.e. αx 2 = bx.

2) "Squares are equal to a number", i.e. αx 2 = c.

3) "The roots are equal to the number", i.e. αx = c.

4) “Squares and numbers are equal to roots”, i.e. αx 2 + c = bx.

5) “Squares and roots are equal to a number”, i.e. αx 2 + bx = c.

6) "Roots and numbers are equal to squares", i.e. bx + c = αx 2 .

The formulas for solving quadratic equations according to al-Khorezmi in Europe were first set forth in the "Book of the Abacus", written in 1202 by the Italian mathematician Leonardo Fibonacci.

Derivation of the formula for solving a quadratic equation with a parameter in general view Vieta has, but Vieta recognized only positive roots. The Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the twelfth century. take into account, in addition to positive, and negative roots. Only in the XVII century. thanks to the works of Girard, Descartes, Newton and others scientists way solving quadratic equations took a modern form.

Vieta's theorem

The theorem expressing the relationship between the parameters, coefficients of a quadratic equation and its roots, bearing the name of Vieta, was formulated by him for the first time in 1591. As follows: “If b + d multiplied by α minus α 2 equals bc, then α equals b and equals d.

To understand Vieta, one should remember that α, like any vowel, meant the unknown (our x), while the vowels b, d are coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means:

If there is

(α + b)x - x 2 \u003d αb,

That is, x 2 - (α -b)x + αb \u003d 0,

then x 1 = α, x 2 = b.

Expressing the relationship between the roots and coefficients of the equations general formulas, written using symbols, Vieta established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from modern look. He didn't acknowledge negative numbers and therefore, when solving equations, he considered only cases where all roots are positive.

Basic concepts

Parameter - an independent variable, the value of which is considered to be a fixed or arbitrary number, or a number belonging to the interval specified by the condition of the problem.

Equation with parameter- mathematicalthe equation, appearance and whose solution depends on the values of one or more parameters.

Decide equation with parameter means for each valuefind x values that satisfy this equation, and also:

1. Investigate for what values of the parameters the equation has roots and how many of them at different meanings parameters.
2. Find all expressions for the roots and indicate for each of them the values of the parameters for which this expression really determines the root of the equation.

Consider the equation α(х+k)= α +c, where α, c, k, x are variables.

The system of admissible values of variables α, c, k, xany system of values of variables is called, in which both the left and right parts of this equation take real values.

Let A be the set of all admissible values of α, K - the set of all admissible values of k, X - the set of all admissible values of x, C - the set of all admissible values of c. If for each of the sets A, K, C, X we choose and fix, respectively, one value α, k, c, and substitute them into the equation, then we obtain an equation for x, i.e. equation with one unknown.

The variables α, k, c, which are considered constant when solving the equation, are called parameters, and the equation itself is called an equation containing parameters.

Parameters are indicated by first letters Latin alphabet: α, b, c, d, …, k , l, m, n, and unknowns - with letters x, y, z.

Two equations containing the same parameters are called equivalent if:

a) they make sense for the same values of the parameters;

b) every solution of the first equation is a solution of the second and vice versa.

Types of equations with parameters

Equations with parameters are: linear and square.

1) Linear equation. General form:

α x = b, where x is unknown;α , b - parameters.

For this equation, the special or control value of the parameter is the one at which the coefficient vanishes in the unknown.

When deciding linear equation with a parameter, cases are considered when the parameter is equal to and different from its special value.

The special value of the parameter α is the valueα = 0.

1.If, a ≠0 , then for any pair of parametersα and b it has a unique solution x = .

2.If, a =0, then the equation takes the form: 0 x = b . In this case, the value b = 0 is special significance parameter b.

2.1. For b ≠ 0 the equation has no solutions.

2.2. For b =0 the equation will take the form: 0 x=0.

The solution to this equation is any real number.

Quadratic equation with a parameter.

General form:

α x 2 + bx + c = 0

where parameter α ≠0, b and c - arbitrary numbers

If α =1, then the equation is called a reduced quadratic equation.

The roots of the quadratic equation are found by the formulas

Expression D = b 2 - 4 α c called the discriminant.

1. If D> 0 - the equation has two different roots.

2. If D< 0 — уравнение не имеет корней.

3. If D = 0 - the equation has two equal roots.

Methods for solving equations with a parameter:

Analytical - way direct solution, repeating the standard procedures for finding the answer in an equation without parameters.
Graphic - depending on the condition of the problem, the position of the graph of the corresponding quadratic function in the coordinate system.

Analytical method

Solution algorithm:

Before proceeding to solve the problem with parameters analytical method, you need to understand the situation for a specific numerical value parameter. For example, take the value of the parameter α =1 and answer the question: is the value of the parameter α =1 the required value for this problem.

Example 1: Decide About X linear equation with parameter m:

According to the meaning of the problem (m-1)(x+3) = 0, that is, m= 1, x = -3.

Multiplying both sides of the equation by (m-1)(x+3), we get the equation

We get

Hence, at m = 2.25.

Now it is necessary to check whether there are no such values of m for which

the x value found is -3.

solving this equation, we get that x is -3 when m = -0.4.

Answer: at m=1, m=2.25.

Graphic method. History of occurrence

The study of general dependencies began in the 14th century. medieval science was scholastic. With such a character, there was no room for the study of quantitative dependencies, it was only about the qualities of objects and their relationships with each other. But among the scholastics, a school arose that asserted that qualities can be more or less intense (the dress of a person who has fallen into the river is wetter than that of someone who has just been caught in the rain)

French scientist Nikolai Oresmus began to represent the intensity of the lengths of the segments. When he arranged these segments perpendicular to some straight line, their ends formed a line, which he called the "line of intensities" or "line of the upper edge" (a graph of the corresponding functional dependence). Oresmus studied even "plane" and "corporeal" qualities, i.e. functions depending on two or three variables.

An important achievement of Oresmes was an attempt to classify the resulting graphs. He singled out three types of qualities: Uniform (with constant intensity), uniformly uneven (with constant speed intensity changes) and uneven-uneven (all others), as well as characteristic properties graphs of such qualities.

To create mathematical apparatus to study the graphs of functions, it took the concept of a variable. This concept was introduced into science by the French philosopher and mathematician René Descartes (1596-1650). It was Descartes who came up with the ideas about the unity of algebra and geometry and about the role variables, Descartes introduced a fixed unit segment and began to consider the relationship of other segments to it.

Thus, the graphs of functions over the entire period of their existence have gone through a series of fundamental transformations that have brought them to the form to which we are accustomed. Each stage or step in the development of graphs of functions is an integral part of the history of modern algebra and geometry.

The graphical method for determining the number of roots of an equation depending on the parameter included in it is more convenient than the analytical one.

Graphical solution algorithm

Function Graph is the set of points whereabscissaare valid values argument, a ordinates- corresponding valuesfunctions.

Algorithm graphic solution equations with parameter:

Find the domain of the equation.
We express α as a function of x.
In the coordinate system we build a graph of the functionα (x) for those values of x that are within the domain of the given equation.
Finding the points of intersection of the lineα =c, with function graph

a(x). If the line α =c crosses the graphα (x), then we determine the abscissas of the intersection points. To do this, it is enough to solve the equation c = α (x) relative to x.

Write down the answer

Solving Equations with Modulus

When solving equations with a modulus containing a parameter graphically, it is necessary to construct graphs of functions and for different values parameter to consider all possible cases.

For example, │х│= a,

Answer: if a < 0, то нет корней, a > 0, then x \u003d a, x \u003d - a, if a \u003d 0, then x \u003d 0.

Problem solving.

Problem 1. How many roots does the equation have| | x | - 2 | = a depending on parameter a?

Decision. In the coordinate system (x; y), we plot the graphs of the functions y = | | x | - 2 | and y= a . Graph of the function y = | | x | - 2 | shown in the figure.

Graph of the function y =α a = 0).

It can be seen from the graph that:

If a = 0, then the line y = a coincides with the Ox axis and has with the graph of the function y = | | x | - 2 | two common points; hence, the original equation has two roots (in this case, the roots can be found: x 1,2 = + 2).
If 0< a < 2, то прямая y = α has with the function graph y = | | x | - 2 | four common points and, therefore, the original equation has four roots.
If a a = 2, then the line y = 2 has three points in common with the graph of the function. Then the original equation has three roots.
If a a > 2, then the line y = a will have two points with the graph of the original function, that is, this equation will have two roots.

Answer: if a < 0, то корней нет;
if a = 0, a > 2, then two roots;
if a = 2, then three roots;
if 0< a < 2, то четыре корня.

Problem 2. How many roots does the equation have| x 2 - 2| x | - 3 | = a depending on parameter a?

Decision. In the coordinate system (x; y), we plot the graphs of the functions y = | x 2 - 2| x | - 3 | and y = a .

Graph of the function y = | x 2 - 2| x | - 3 | shown in the figure. Graph of the function y =α is a line parallel to Ox or coinciding with it (when a = 0).

From the graph you can see:

If a = 0, then the line y = a coincides with the Ox axis and has with the graph of the function y = | x2 - 2| x | - 3 | two common points, as well as a line y = a will have with the function graph y = | x 2 - 2| x | - 3 | two common points a > 4. Hence, for a = 0 and a > 4 the original equation has two roots.
If 0< a< 3, то прямая y = a has with the function graph y = | x 2 - 2| x | - 3 | four common points, as well as a line y= a will have four common points with the graph of the constructed function at a = 4. Hence, at 0< a < 3, a = 4 the original equation has four roots.
If a a = 3, then the line y = a intersects the graph of the function at five points; therefore, the equation has five roots.
If 3< a< 4, прямая y = α intersects the graph of the constructed function at six points; hence, for these values of the parameter, the original equation has six roots.
If a a < 0, уравнение корней не имеет, так как прямая y = α does not intersect the graph of the function y = | x 2 - 2| x | - 3 |.

Answer: if a < 0, то корней нет;
if a = 0, a > 4, then two roots;
if 0< a < 3, a = 4, then four roots;

if a = 3, then five roots;
if 3< a < 4, то шесть корней.

Problem 3. How many roots does the equation have

depending on parameter a?

Decision. We construct in the coordinate system (x; y) the graph of the function

but first let's put it in the form:

The lines x = 1, y = 1 are the asymptotes of the graph of the function. Graph of the function y = | x | + a obtained from the graph of the function y = | x | offset by a units along the Oy axis.

Function Graphs intersect at one point at a > - 1; hence, equation (1) for these values of the parameter has one solution.

For a = - 1, a = - 2 graphs intersect at two points; hence, for these values of the parameter, equation (1) has two roots.
At - 2< a< - 1, a < - 2 графики пересекаются в трех точках; значит, уравнение (1) при этих значениях параметра имеет три решения.

Answer: if a > - 1, then one solution;
if a = - 1, a = - 2, then two solutions;
if - 2< a < - 1, a < - 1, то три решения.

Comment. When solving the equation of the problem, special attention should be paid to the case when a = - 2, since the point (- 1; - 1) does not belong to the graph of the functionbut belongs to the graph of the function y = | x | + a.

Problem 4. How many roots does the equation have

x + 2 = a | x - 1 |

depending on parameter a?

Decision. Note that x = 1 is not a root of this equation, since the equality 3 = a 0 cannot be true for any parameter value a . We divide both sides of the equation by | x - 1 |(| x - 1 |0), then the equation will take the formIn the xOy coordinate system, we plot the function

The graph of this function is shown in the figure. Graph of the function y = a is a straight line parallel to the Ox axis or coinciding with it (for a = 0).

Equations with parameters are rightfully considered one of the most difficult tasks in the course of school mathematics. It is these tasks that fall from year to year in the list of tasks of type B and C at the unified state exam of the Unified State Examination. However, among the large number of equations with parameters, there are those that can be easily solved graphically. Let's consider this method on the example of solving several problems.

Find the sum of integer values of a for which the equation |x 2 – 2x – 3| = a has four roots.

Decision.

To answer the question of the problem, we construct graphs of functions on one coordinate plane

y = |x 2 – 2x – 3| and y = a.

Next, you need to find the points of intersection of the branches of the parabola with the coordinate axes. At the points of intersection of the branches of the parabola with the abscissa axis, the value of the function is zero. Therefore, we solve the quadratic equation x 2 - 2x - 3 \u003d 0. Its roots will be the desired points. By the Vieta theorem, we have x 1 = -1, x 2 = 3.

To get the graph of the function y = |x 2 - 2x - 3|, we will display the part of the graph, which is below the x-axis, symmetrically about the x-axis. The resulting graph is shown in Figure 2.

Integer values of number a from the received interval: 1; 2; 3. To answer the question of the problem, let's find the sum of these numbers: 1 + 2 + 3 = 6.

Answer: 6.

Find the arithmetic mean of integer values of the number a, for which the equation |x 2 – 4|x| – 1| = a has six roots.

Let's start by plotting the function y = |x 2 – 4|x| – 1|. To do this, we use the equality a 2 = |a| 2 and select the full square in the submodule expression written on the right side of the function:

x 2 – 4|x| – 1 = |x| 2 – 4|x| - 1 = (|x| 2 - 4|x| + 4) - 1 - 4 = (|x | - 2) 2 - 5.

Then the original function will look like y = |(|x| – 2) 2 – 5|.

To build a graph of this function, we build successively graphs of functions:

1) y \u003d (x - 2) 2 - 5 - a parabola with a vertex at a point with coordinates (2; -5); (Fig. 1).

2) y = (|x| - 2) 2 - 5 - the part of the parabola constructed in paragraph 1, which is located to the right of the ordinate axis, is symmetrically displayed to the left of the Oy axis; (Fig. 2).

3) y = |(|x| – 2) 2 – 5| - the part of the graph constructed in paragraph 2, which is below the x-axis, is displayed symmetrically with respect to the abscissa axis upwards. (Fig. 3).

Consider the resulting drawings:

The graph of the function y = a is a straight line parallel to the x-axis.

Using the figure, we conclude that the function graphs have six common points (the equation has six roots) if a belongs to the interval (1; 5).

This can be seen in the following figure:

Find the arithmetic mean of the integer values of the parameter a:

(2 + 3 + 4)/3 = 3.

Answer: 3.

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