Methods for solving non-standard problems. Non-standard methods for solving problems
“Solving quadratic inequalities” - Solve an inequality. What are function zeros? Solution quadratic inequalities. How to find the zeros of a function? Objective of the lesson: What depends on the sign of the first coefficient of a quadratic function? How does the sign of the discriminant affect the solution to the quadratic inequality?
"Solving Inequalities 2" - Review the properties of numerical inequalities. Mental arithmetic is an exercise for the mind. Cultivating interest in mathematics. Stages of graphical solution of equations. Markers, crayons different colors, rulers, computers. Solving inequalities of the first degree with one variable (graphical solution). Equipment. Updating knowledge.
“Non-standard lessons” - Refusal from a template in organizing a lesson, from routine and formalism in conducting. The influence of non-standard lesson forms on the educational process. Maximum involvement of class students in active activities during the lesson. MO plan: Preparatory period, actual lesson analysis. Using assessment as a formative (and not just an effective) tool.
“Properties of inequalities” - Properties of inequalities. What is inequality? Oral work. What properties of inequalities do you know? Addition and multiplication of numerical inequalities. Prove the inequality. Solve the inequality. Definition of inequality. What properties did you use to solve the inequality? Solving inequalities. Inequalities.
“Irrational equations and inequalities” - Irrational inequalities. Irrational equations and inequalities. 3. Introduction of auxiliary variables. Solution methods. 5. Narrowing the search area for the roots of the equation by finding the ODZ. Irrational equations Methods of solution. 1. Exponentiation. 6. Graphic method. Irrational equations and inequalities with a parameter.
“Equations and Inequalities” - System Solution graphically. 2. Find the sum of numbers that satisfy the inequality. Find the domain of definition of the function. "Solving equations and inequalities." Task formulations. Inequalities in CIMs. Find the product x*y, where (x;y) is the solution of the system. Y=x+2. x2 – 2x – 3 =0 Let’s represent it as x2 –3 = 2x.
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Introduction
Mathematics education received in school is an essential component general education and general culture modern man. Almost everything that surrounds modern man is all somehow connected with mathematics. A latest achievements in physics, technology and information technology leave no doubt that in the future the state of affairs will remain the same. Therefore, solving many practical problems comes down to solving various types equations.
Equations occupy a leading place in the school algebra course. More time is devoted to their study than to any other topic. school course mathematics. The strength of the theory of equations is that it not only has theoretical significance for the knowledge of natural laws, but also serves specific practical purposes.
Relevance of the topic is that in algebra, geometry, and physics lessons we very often encounter solving quadratic equations. Most problems about spatial forms and quantitative relationships real world comes down to solving various types of equations. By mastering ways to solve them, people find answers to various questions from science and technology (transport, Agriculture, industry, communications, etc.). Therefore, every student should be able to correctly and rationally solve quadratic equations; this can also be useful to me when solving more complex problems, including in grade 9, as well as grades 10 and 11, and when passing exams.
Target: Explore standard and non-standard ways to solve quadratic equations
Tasks
- State the most known methods solving equations
- Explain non-standard ways to solve equations
- Draw a conclusion
Object of study: quadratic equations
Subject of study: ways to solve quadratic equations
Research methods:
- Theoretical: study of literature on the research topic;
- Analysis: information obtained from studying the literature; results obtained by solving quadratic equations in various ways.
- Comparison of methods for the rationality of their use in solving quadratic equations.
Chapter 1. Quadratic equations and standard solutions
1.1.Definition quadratic equation
Quadratic equation called an equation of the form ax 2 + bx + c= 0, where X- variable , a, b And With- some numbers, and A≠ 0.
Numbers a, b And With - coefficients of a quadratic equation. Number A called the first coefficient, the number b- second coefficient and number c- a free member.
Complete quadratic equation is a quadratic equation in which all three terms are present, i.e. coefficients in and с are different from zero.
Incomplete quadratic equation is an equation in which at least one of the coefficients in or, c is equal to zero.
Definition 3. Root of a quadratic equation Oh 2 + bX + With= 0 is any value of the variable x for which the quadratic trinomial Oh 2 + bX+ With goes to zero.
Definition 4. Solving a quadratic equation means finding all of it
roots or establish that there are no roots.
Example: - 7 x+ 3 =0
In each of the equations of the form a + bx + c= 0, where A≠ 0, highest degree of variable x- square. Hence the name: quadratic equation.
A quadratic equation in which the coefficient at X 2 equals 1, called given quadratic equation.
Example
X 2 - 11x+ 30=0, X 2 -8x= 0.
1.2.Standard methods for solving quadratic equations
Solving quadratic equations by squaring the binomial
Solving a quadratic equation in which both the coefficients of the unknowns and the free term are nonzero. This method of solving a quadratic equation is called squaring the binomial.
Factoring the Left Side of the Equation.
Let's solve the equation x 2 + 10x - 24 = 0. Let's factorize the left side:
x 2 + 10x - 24 = x 2 + 12x - 2x - 24 = x(x + 12) - 2(x + 12) = (x + 12)(x - 2).
Therefore, the equation can be rewritten as follows: (x + 12)(x - 2) = 0
A product of factors is zero if at least one of its factors is zero.
Answer: -12; 2.
Solving a quadratic equation using the formula.
Discriminant of a quadratic equationax 2 + bx + c= 0 expression b 2 - 4ac = D - by the sign of which one judges whether this equation has real roots.
Possible cases depending on the value of D:
- If D>0, then the equation has two roots.
- If D= 0, then the equation has one root: x =
- If D< 0, then the equation has no roots.
Solving equations using Vieta's theorem.
Theorem: The sum of the roots of the given quadratic equation is equal to the second coefficient taken from opposite sign, and the product of the roots is equal to the free term.
The quadratic equation given is:
x 2 + bx + c= 0.
Let's denote the second coefficient by the letter p, and the free term by the letter q:
x 2 + px + q= 0, then
x 1 + x 2 = - p; x 1 x 2 = q
Chapter 2. Non-standard methods for solving quadratic equations
2.1. Solving using the properties of the coefficients of a quadratic equation
Properties of coefficients of a quadratic equation is a way to solve quadratic equations that will help you quickly and verbally find the roots of the equation:
ax 2 + bx + c= 0
- Ifa+ b+c= 0, thenx 1 = 1, x 2 =
Example. Consider the equation x 2 + 3x - 4 = 0.
a+ b + c = 0, then x 1 = 1, x 2 =
1+3+(-4) = 0, then x 1 = 1, x 2 = = - 4
Let's check the obtained roots by finding the discriminant:
D= b 2- 4ac= 3 2 - 4·1·(-4) = 9+16= 25
x 1 = = = = = - 4
Therefore, if +b +c= 0, then x 1 = 1, x 2 =
- Ifb = a + c , Thatx 1 = -1, x 2 =
x 2 + 4X+1 = 0, a=3, b=4, c=1
If b=a + c, then x 1 = -1, x 2 = , then 4 = 3 + 1
Roots of the equation: x 1 = -1, x 2 =
So the roots of this equation are -1 and. Let's check this by finding the discriminant:
D= b 2- 4ac= 4 2 - 4 3 1 = 16 - 12 = 4
x 1 = = = = = - 1
Hence, b=a + c, then x 1 = -1, x 2 =
2.2.Method of “transfer”
With this method the coefficient A multiplied by the free term, as if “thrown” to it, which is why it is called transfer method. This method is used when the roots of the equation can be easily found using Vieta's theorem and, most importantly, when the discriminant is an exact square.
If A± b+c≠0, then the transfer technique is used:
3x 2 +4x+ 1=0; 3+4+1 ≠ 0
Using the “transfer” method we get:
X 2 + 4x+3= 0
Thus, using Vieta’s theorem, we obtain the roots of the equation:
x 1 = - 3, x 2 = -1.
However, the roots of the equation must be divided by 3 (the number that was “thrown over”):
This means we get the roots: x 1 = -1, x 2 = .
Answer: ; - 1
2.3. Solution using the regularity of coefficients
- If the equationax 2 + bx + c= 0, coefficientb= (a 2 +1), and coefficientc = a, then its roots are x 1 = - a, x 2 =
ax 2 +(a 2 + 1)∙ x + a= 0
Example. Consider Equation 3 x 2 +10x+3 = 0.
Thus, the roots of the equation are: x 1 = -3 , x 2 =
D= b 2- 4ac= 10 2 - 4 3 3 = 100 - 36 = 64
x 1 = = = = = - 3
x 2 = = = = = ; Therefore x 1 = - a, x 2 =
- If the equationax 2 - bx + c= 0, coefficientb= (a 2 +1), and coefficientc = a, then its roots are x 1 = a, x 2 =
Thus, the equation to be solved should have the form
ax 2 -(a 2 + 1)∙ x+ a= 0
Example. Consider Equation 3 x 2 - 10x+3 = 0.
, x 2 =
Let's check this solution using the discriminant:
D= b 2- 4ac= 10 2 - 4 3 3 = 100 - 36 = 64
a, x 2 =
- If the equationax 2 + bx - c= 0, coefficientb= (a 2 -1), and coefficientc = a, then its roots are x 1 = - a, x 2 =
Thus, the equation to be solved should have the form
ax 2 +(and 2 - 1)∙ x - a= 0
Example. Consider Equation 3 x 2 + 8x - 3 = 0..
Thus, the roots of the equation are: x 1 = - 3, x 2 =
Let's check this solution using the discriminant:
D= b 2- 4ac= 8 2 + 4 3 3 = 64 + 36 = 100
x 1 = = = = = - 3
x 2 = = = = = ;Hence, x 1 = - a, x 2 =
- If the equationax 2 - bx - c= 0, coefficientb= (a 2 -1), and coefficientc = a, then its roots are x 1 = a, x 2 =
Thus, the equation to be solved should have the form
ax 2 -(and 2 - 1)∙ x - a= 0
Example. Consider Equation 3 x 2 - 8x - 3 = 0..
Thus, the roots of the equation are: x 1 = 3 , x 2 = -
Let's check this solution using the discriminant:
D= b 2- 4ac= 8 2 + 4 3 3 = 64 + 36 = 100
x 2 = = = = = 3; Therefore x 1 = a, x 2 = -
2.4. Solution using a compass and ruler
I propose the following method for finding the roots of a quadratic equation ah 2 +bx + c = 0 using a compass and ruler (Fig. 6).
Let us assume that the desired circle intersects the axis
abscissa in points B(x 1; 0) And D(x 2 ; 0), Where x 1 And x 2- roots of the equation ah 2 +bx + c = 0, and passes through the points
A(0; 1) And C(0;c/ a) on the ordinate axis. Then, by the secant theorem, we have O.B. . O.D. = O.A. . O.C., where O.C. = = =
The center of the circle is at the point of intersection of the perpendiculars SF And S.K., restored in the middles of the chords A.C. And BD, That's why
1) construct points S (center of the circle) and A(0; 1) ;
2) draw a circle with radius S.A.;
3) abscissa of the points of intersection of this circle with the axis Oh are the roots of the original quadratic equation.
In this case, three cases are possible.
1) The radius of the circle is greater than the ordinate of the center (AS > S.K., orR > a + c/2 a) , the circle intersects the Ox axis at two points (Fig. 7a) B(x 1; 0) And D(x 2; 0), Where x 1 And x 2- roots of quadratic equation ah 2 +bx + c = 0.
2) The radius of the circle is equal to the ordinate of the center (AS = S.B., orR = a + c/2 a) , the circle touches the Ox axis (Fig. 8b) at the point B(x 1; 0), where x 1 is the root of the quadratic equation.
3) The radius of the circle is less than the ordinate of the center AS< S, R<
the circle has no common points with the abscissa axis (Fig. 7c), in this case the equation has no solution.
A)AS>SB, R>
b) AS=SB, R=
V) AS
Two solutions x 1 Andx 2 One solution x 1 There is no decision
Example.
Let's solve the equation x 2 - 2x - 3 = 0(Fig. 8).
Solution. Let's determine the coordinates of the center point of the circle using the formulas:
x = - = - = 1,
y = = = -1
Let's draw a circle of radius SA, where A (0; 1).
Answer: x 1 = - 1; x 2 = 3.
2.5. Geometric method for solving quadratic equations.
In ancient times, when geometry was more developed than algebra, quadratic equations were solved not algebraically, but geometrically. I will give a famous example from the “Algebra” of al-Khorezmi.
Examples.
1) Let's solve the equation x 2 + 10x = 39.
In the original, this problem is formulated as follows: “A square and ten roots are equal to 39” (Fig. 9).
Solution. Consider a square with side x, rectangles are constructed on its sides so that the other side of each of them is 2.5, therefore, the area of each is 2.5x. The resulting figure is then supplemented to a new square ABCD, building four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25.
Square S square ABCD can be represented as the sum of areas:
original square x 2, four rectangles (4. 2.5x = 10x) and four attached squares (6,25. 4 = 25) , i.e. S = x 2 + 10x + 25. Replacing
x 2 + 10x number 39 , we get that S = 39 + 25 = 64 , which means that the side of the square ABCD, i.e. line segment AB = 8. For the required side X of the original square we get:
x = 8 - 2 - 2 = 3
2) But, for example, how the ancient Greeks solved the equation y 2 + 6y - 16 = 0.
Solution presented in Fig. 10. where
y 2 + 6y = 16, or y 2 + 6y + 9 = 16 + 9.
Solution. Expressions y 2 + 6y + 9 And 16 + 9 geometrically represent
the same square, and the original equation y 2 + 6y - 16 + 9 - 9 = 0- the same equation. From where we get that y + 3 = ± 5, or y 1 = 2, y 2 = - 8(rice. .
Fig.10
3) Solve the geometric equation y 2 - 6y - 16 = 0.
Transforming the equation, we get
y 2 - 6y = 16.
In Fig. 11 we find “images” of the expression y 2 - 6y, those. from the area of a square with side y, subtract the area of a square with side equal to 3 . This means that if to the expression y 2 - 6y add 9 , then we get the area of a square with side y - 3. Replacing the expression y 2 - 6y its equal number 16,
we get: (y - 3) 2 = 16 + 9, those. y - 3 = ± √25, or y - 3 = ± 5, where y 1 = 8 And y 2 = - 2.
Conclusion
In the course of carrying out my research work, I believe that I coped with the set goal and objectives, I was able to generalize and systematize the material studied on the above mentioned topic.
It should be noted that each method of solving quadratic equations is unique in its own way. Some solutions help save time, which is important when solving tasks on tests and exams. When working on the topic, I set the task to find out which methods are standard and which are non-standard.
So, standard methods(used more often when solving quadratic equations):
- Solving by squaring the binomial
- Factoring the left-hand side
- Solving quadratic equations using the formula
- Solution using Vieta's theorem
- Graphical solution of equations
- Properties of coefficients of a quadratic equation
- Solution by transferring coefficients
- Solution using the coefficient pattern
- Solving quadratic equations using compass and ruler.
- Study of the equation on intervals of the real axis
- Geometric method
It should be noted that each method has its own characteristics and limits of application.
Solving equations using Vieta's theorem
A fairly easy way, it makes it possible to immediately see the roots of the equation, while only whole roots are easily found.
Solving equations using the transfer method
In a minimum number of steps, you can find the roots of an equation, used in conjunction with the Vieta theorem method, and it is also easy to find only whole roots.
Properties of coefficients of a quadratic equation
An accessible method for finding the roots of a quadratic equation verbally, but only suitable for some equations
Graphical solution of a quadratic equation
A visual way to solve a quadratic equation, but errors may occur when drawing graphs
Solving quadratic equations using compass and ruler
A visual way to solve a quadratic equation, but errors may also occur
Geometric method for solving quadratic equations
A visual method, similar to the method of selecting a complete square
Solving equations in different ways, I came to the conclusion that knowing a set of methods for solving quadratic equations, you can solve any equation proposed during the learning process.
At the same time, it should be noted that one of the more rational ways to solve quadratic equations is the method of “transferring” the coefficient. However, the most universal method can be considered the standard method of solving equations using a formula, because this method allows you to solve any quadratic equation, although sometimes in a longer time. Also, such solution methods as the “transfer” method, the property of coefficients and Vieta’s theorem help to save time, which is very important when solving tasks in exams and tests.
I think that my work will be of interest to students in grades 9-11, as well as those who want to learn how to solve rational quadratic equations and prepare well for final exams. It will also be of interest to mathematics teachers, due to the consideration of the history of quadratic equations and the systematization of methods for solving them.
Bibliography
- Glaser, G.I. History of mathematics at school / G.I. Glazer.-M.: Enlightenment, 1982- 340 p.
- Gusev, V.A. Mathematics. Reference materials/ V.A. Gusev, A.G. Mordkovich - M.: Education, 1988, 372 p.
- Kovaleva G. I., Konkina E. V. “Functional method for solving equations and inequalities”, 2014
- Kulagin E. D. “300 competitive problems in mathematics”, 2013
- Potapov M.K. “Equations and inequalities. Non-standard solution methods" M. "Drofa", 2012
- .Barvenov S. A “Methods for solving algebraic equations”, M. “Aversev”, 2006
- Suprun V.P. “Non-standard methods for solving problems in mathematics” - Minsk “Polymya”, 2010
- Shabunin M.I. “Mathematics manual for applicants to universities,” 2005.
- Bashmakov M.I. Algebra: textbook. for 8th grade. general education institutions. - M.: Education, 2004. - 287 p.
- Shatalova S. Lesson - workshop on the topic “Quadratic equations”. - 2004.
Municipal competition of research and creative works of schoolchildren
"Step into Science"
MATHEMATICS section
Subject: Non-standard methods for solving irrational problems
equations.
Nuzhdina Maria, MAOU Secondary School No. 2
10th grade, Karymskoe village
Scientific supervisor: Vasilyeva Elena Valerievna,
mathematic teacher
MAOU secondary school No. 2, Karymskoe village
Karymskoe village, 2013
Abstract……………………………………………………………………………….3
Research plan……………………………………………………......4-5
Description of work:
§1. Basic techniques for solving irrational equations………………6-9
§2. Solving irrational equations by replacing the unknown…10-14
§3. Irrational equations reducible to modulus………….15-17
§4. Factorization…………………………………………...…..18-19
§5. Equations of the form………………………………………………………20-22
§6. Geometric mean theorem in irrational equations
; ……………………………23-24
4) List of references…………………………………………………….....25
Annotation.
The topic of our research work: “Non-standard techniques for solving irrational equations.”
When performing the work it was necessary to: compare different solution methods; move from general methods to specific ones, and vice versa; argue and prove the assertions made; study and synthesize information collected from various sources. In this regard, the following research methods can be distinguished: empirical; logical and theoretical (research); step by step; reproductive and heuristic;
As a result of the work carried out, the following were obtained results and conclusions:
There are many techniques for solving irrational equations;
Not all irrational equations can be solved using standard techniques;
We have studied frequently occurring substitutions with the help of which complex irrational equations are reduced to the simplest ones;
We looked at non-standard techniques for solving irrational equations
Topic: “Non-standard techniques for solving irrational equations”
Nuzhdina M.P., Trans-Baikal Territory, Karymskoye village, MAOU Secondary School No. 2, 10th grade.
Research plan.
Object area The area in which we conducted our research is algebra. An object research- solving equations. Among the many equations, we considered irrational equations - item our research.
In the school algebra course, only standard methods and solving techniques (raised to a power and simple substitution techniques) are considered. But during the research it became clear that there are irrational equations for which standard techniques and methods are not enough to solve. Such equations are solved using other, more rational methods.
Therefore, we believe that studying such solution techniques is necessary and interesting work.
During the research, it became clear that there are a great many irrational equations and grouping them by types and methods is problematic.
Purpose research is the study and systematization of methods for solving irrational equations.
Hypothesis: If you know non-standard methods for solving irrational equations, this will improve the quality of performing some Olympiad and Unified State Exam test tasks.
To achieve the goals and test the hypothesis, it is necessary to solve the following tasks:
Describe the types of irrational equations.
Establish connections between types and methods of solution.
Assess the importance of checking and finding DL.
Consider non-standard cases when solving irrational equations (geometric mean theorem, properties of monotonicity of functions).
During the research process, many textbooks were studied by such authors as M.I. Skanavi, I.F. Sharygina, O.Yu. Cherkasov, A.N. Rurukin, I.T. Borodulya, as well as articles from scientific-theoretical and methodological journal "Mathematics at school".
Topic: “Non-standard techniques for solving irrational equations”
Nuzhdina M.P., Trans-Baikal Territory, Karymskoye village, MAOU Secondary School No. 2, 10th grade.
Description of work.
§1 Basic techniques for solving irrational equations
The equation y(x)=0 is irrational if the function y(x) contains roots of an unknown quantity x or expressions depending on x.
Many irrational equations can be solved based only on the concepts of the root and the range of permissible values of the equation (ADV), but there are other methods, some of which will be discussed in the work.
The main technique for solving irrational equations is to isolate the radical in one part of the equation, then raise both parts of the equation to the appropriate power. If there are several such radicals, then the equation must be raised to its original power repeatedly; by the way, there is no need to care that the expression under the sign of the solitary radical is non-negative.
However, when raised to an even power, extraneous roots may appear, that is, roots that are not a solution to the original equation.
Therefore, when using such a decision method, the roots must be checked and extraneous ones discarded; in this case, verification is an element of the decision and is necessary even in cases where unnecessary roots did not appear, but the course of the decision was such that they could appear. On the other hand, sometimes it is easier to do a check than to prove that it is necessary.
Let's look at a few examples:
Answer: no roots
– extraneous root
In these examples, we looked at standard methods for solving irrational equations (raising both sides to a power and checking the roots).
However, many irrational equations can be solved by
based only on the concepts of the root and ODZ of the equation.
Since the equation includes only radicals of even degrees, it is enough to solve the system of inequalities.
3x -2x 2 +5 ≥0 (conditions of the ODZ equation)
4х 2 -26х +40 ≥0
Solving this system of inequalities we obtain:
x € Where x = 2.5.
x € (-∞; 2.5] ᴗ )
