Otdelkina Olga, 9th grade student
This topic is an integral part of the study of the school course of 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. Preparing 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.
V modern life the study of many physical processes and geometric regularities often leads to the solution of 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.
Tasks with parameters are of purely mathematical interest, contribute to the intellectual development of students, serve good material for skill development. 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 challenging tasks in high school mathematics. It is these tasks that fall into the list of tasks at the unified 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 other scientists, the method of solving quadratic equations has taken on a modern look.
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 letter, 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 recognize 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 denoted by the first letters of the Latin alphabet: α, b, c, d, …, k , l, m, n, and unknowns - by 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 only decision x = .
2.If, a =0, then the equation takes the form: 0 x = b . In this case, the value b = 0 is a special parameter value 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 - a direct solution method that repeats 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 of the 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)
The French scientist Nicholas Oresme began to depict 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 "planar" 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 the characteristic properties of graphs of such qualities.
To create a mathematical apparatus for studying function graphs, 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 of variables, Descartes introduced a fixed unit segment and began to consider the relations of other segments to it.
Thus, the graphs of functions have gone through a series of fundamental transformations throughout their existence, leading 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 argument values, 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?
Solution. 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 = 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 > 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?
Solution. 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 = 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 < 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?
Solution. 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?
Solution. 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).
1. Determination of personal motivation of students. To continue education, for self-development and intellectual growth, it is necessary to study diligently and consciously and take care of your health. 2. Access to the concept of "parameter". Parameter - a value that characterizes the main properties of a change in a system or phenomenon. ( Dictionary)
In equations (inequalities), the coefficients for unknowns or free terms given not by specific numerical values, but indicated by letters, are called parameters. Example: To solve a problem with a parameter means, for each value of the parameter, to find x values that satisfy the condition of this problem.
X y x y a > 0 a 0, (2 roots) 0 a 0, (2 roots)"> 0 a 0, (2 roots)"> 0 a 0, (2 roots)" title="(!LANG:x y x y a > 0 a 0, (2 roots)"> title="x y x y a > 0 a 0, (2 roots)"> !}
x uuuuuuuuuuh
2. when the equation takes the form, and has a root x \u003d 0. 3. when we find the roots of the equation according to the formula Answer: when there are no roots; with one root x = 0. with two roots 1. the left side of the equation is non-negative for any value of the unknown x,. no solutions. x y 0 y = a "LOOK!" Method 1 (analytical) Method 2 (graphic)
At what values of the parameter a does the equation have one solution? Let's write the equation in the form: x Let's plot the graphs of the functions: Answer: a = 3 and a moving straight line y = a. a
For what values of the parameter a does the equation have no solutions? x y Let's build a graph According to the figure, we see at and a straight line y \u003d a. there are no solutions. a Answer:
(Graphic method for solving problems with a parameter) A problem with a parameter can be considered as a function f (x; a) =0 1. We build a graphic image 2. We intersect the resulting graph with straight lines parallel to the x-axis 3. “Read” the necessary information Solution scheme: !!!
3 Answer: 1 root "title="(!LANG: Specify the number of roots of the equation f(x)= a for all values of parameter a. 1 35-2 1 x a -5 3 1 root, a3 Answer: 1 root" class="link_thumb"> 15 !} Indicate the number of roots of the equation f (x) \u003d a for all values of the parameter a x a root, a3 Answer: 1 root for a 3 2 roots for a \u003d -5, a \u003d 3 3 roots for 1 3 Answer: 1 root "> 3 Answer: 1 root with a 3 2 roots with a \u003d -5, a \u003d 3 3 roots with 1 3 Answer: 1 root "title="(!LANG: Specify the number of roots of the equation f(x)= a for all values of parameter a. 1 35-2 1 x a -5 3 1 root, a3 Answer: 1 root"> title="Indicate the number of roots of the equation f(x)= a for all values of the parameter a. 1 35-2 1 x a -5 3 1 root, a3 Answer: 1 root">!}
X y y For what values of the parameter a does the equation have two roots? x y x
1) When a \u003d 3, the top right angle; Find the sum of integer values of the parameter a for which the equation has three roots. The original equation is equivalent to the set B Expressing the parameter a, we get: It can be seen from the figure that the equation has three roots in 3 cases x a a 1 = 3 a 2 = ? and 3 = ? Then a = = 5. Answer. 8. 2) For x 4, a 2 = 5 a 3 a 3 4, a 2 \u003d 5 a 3 a 3 "\u003e 
TO tasks with parameter include, for example, the search for a solution to linear and quadratic equations in a general form, the study of the equation for the number of roots available, depending on the value of the parameter.
Without giving detailed definitions, consider the following equations as examples:
y = kx, where x, y are variables, k is a parameter;
y = kx + b, where x, y are variables, k and b are parameters;
ax 2 + bx + c = 0, where x are variables, a, b and c are parameters.
To solve an equation (inequality, system) with a parameter means, as a rule, to solve an infinite set of equations (inequalities, systems).
Tasks with a parameter can be conditionally divided into two types:
a) the condition says: solve the equation (inequality, system) - this means, for all values of the parameter, find all solutions. If at least one case remains unexplored, such a solution cannot be considered satisfactory.
b) required to specify possible values parameter under which the equation (inequality, system) has certain properties. For example, has one solution, has no solutions, has solutions, belonging to the interval etc. In such tasks, it is necessary to clearly indicate at what value of the parameter the required condition is satisfied.
The parameter, being an unknown fixed number, has, as it were, a special duality. First of all, it must be taken into account that the alleged fame suggests that the parameter must be perceived as a number. Secondly, the freedom to handle a parameter is limited by its unknown. So, for example, the operations of dividing by an expression in which there is a parameter or extracting a root of an even degree from a similar expression require preliminary research. Therefore, care must be taken in handling the parameter.
For example, to compare two numbers -6a and 3a, three cases need to be considered:
1) -6a will be greater than 3a if a is a negative number;
2) -6a = 3a in the case when a = 0;
3) -6a will be less than 3a if a is a positive number 0.
The decision will be the answer.
Let the equation kx = b be given. This equation is shorthand for an infinite set of equations in one variable.
When solving such equations, there may be cases:
1. Let k be any non-zero real number and b any number from R, then x = b/k.
2. Let k = 0 and b ≠ 0, the original equation will take the form 0 · x = b. Obviously, this equation has no solutions.
3. Let k and b be numbers, zero, then we have the equality 0 · x = 0. Its solution is any real number.
The algorithm for solving this type of equations:
1. Determine the "control" values of the parameter.
2. Solve the original equation for x with the values of the parameter that were determined in the first paragraph.
3. Solve the original equation for x with parameter values that differ from those selected in the first paragraph.
4. You can write down the answer in the following form:
1) when ... (parameter value), the equation has roots ...;
2) when ... (parameter value), there are no roots in the equation.
Example 1
Solve the equation with the parameter |6 – x| = a.
Solution.
It is easy to see that here a ≥ 0.
By the rule of modulo 6 – x = ±a, we express x:
Answer: x = 6 ± a, where a ≥ 0.
Example 2
Solve the equation a(x - 1) + 2(x - 1) = 0 with respect to the variable x.
Solution.
Let's open the brackets: ax - a + 2x - 2 \u003d 0
Let's write the equation in standard form: x(a + 2) = a + 2.
If the expression a + 2 is not zero, i.e. if a ≠ -2, we have the solution x = (a + 2) / (a + 2), i.e. x = 1.
If a + 2 is equal to zero, i.e. a \u003d -2, then we have the correct equality 0 x \u003d 0, therefore x is any real number.
Answer: x \u003d 1 for a ≠ -2 and x € R for a \u003d -2.
Example 3
Solve the equation x/a + 1 = a + x with respect to the variable x.
Solution.
If a \u003d 0, then we transform the equation to the form a + x \u003d a 2 + ax or (a - 1) x \u003d -a (a - 1). The last equation for a = 1 has the form 0 · x = 0, therefore, x is any number.
If a ≠ 1, then the last equation will take the form x = -a.
This solution can be illustrated on the coordinate line (Fig. 1)
Answer: there are no solutions for a = 0; x - any number at a = 1; x \u003d -a with a ≠ 0 and a ≠ 1.
Graphic method
Consider another way to solve equations with a parameter - graphical. This method is used quite often.
Example 4
How many roots, depending on the parameter a, does the equation ||x| – 2| = a?
Solution.
To solve by a graphical method, we construct graphs of functions y = ||x| – 2| and y = a (Fig. 2).
The drawing clearly shows the possible cases of the location of the line y = a and the number of roots in each of them.
Answer: the equation will have no roots if a< 0; два корня будет в случае, если a >2 and a = 0; the equation will have three roots in the case a = 2; four roots - at 0< a < 2.
Example 5
For which a the equation 2|x| + |x – 1| = a has a single root?
Solution.
Let's draw graphs of functions y = 2|x| + |x – 1| and y = a. For y = 2|x| + |x - 1|, expanding the modules by the gap method, we get:
(-3x + 1, at x< 0,
y = (x + 1, for 0 ≤ x ≤ 1,
(3x – 1, for x > 1.
On the figure 3 it is clearly seen that the equation will have a unique root only when a = 1.
Answer: a = 1.
Example 6
Determine the number of solutions of the equation |x + 1| + |x + 2| = a depending on the parameter a?
Solution.
Graph of the function y = |x + 1| + |x + 2| will be a broken line. Its vertices will be located at the points (-2; 1) and (-1; 1) (picture 4).
Answer: if the parameter a is less than one, then the equation will have no roots; if a = 1, then the solution of the equation is an infinite set of numbers from the segment [-2; -one]; if the values of the parameter a are greater than one, then the equation will have two roots.
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