Non-standard methods for solving equations. Coursework: Non-standard methods for solving equations and inequalities

"Non-standard methods for solving equations"

Kubanova Olga Nikolaevna, teacher of mathematics,

MBOU "Plesetsk secondary school"

« The process of solving the equation -

there is simply the act of reducing it to a simpler form.

But in some forms it is not easy to read.

Its solution is similar to the translation

unfamiliar phrase into a language we understand"

To solve most of the equations encountered in the exams, it is enough to master the school mathematics course, but at the same time it is necessary to be able to solve them not only with the help of standard techniques intended for well-defined types of equations, but also with those “non-standard” methods that I want to talk about.

The essence of these methods is to implement a "different view" of the problem, which allows, without going beyond the school curriculum, to significantly simplify the solution of some problems, that is, we will apply well-known statements, but in situations where they are used relatively rarely.

Along with the main task of teaching mathematics - ensuring a strong and conscious mastery of the system of mathematical knowledge and skills by students, non-standard methods provide for the formation of a sustainable interest in the subject, the identification and development mathematical ability in children, as well as improving the quality of teaching mathematics.

I will focus on a method where the properties of the functions included in the equation are used to solve equations.

Investigation of the domain of definitions and the scope of functions:

Note that and and

Therefore equality is impossible.

Answer: no roots.

Function monotonicity properties:

This equation can be solved in a standard way, or it can be simpler. The left side of the equation is an increasing function, and the right side is a decreasing function. Therefore, this equation cannot have more than one root. The number 1 is the root of the equation, which can be checked by substitution.

Raising to the fifth power seems futile. Let , then . Consider the functions: and . These functions are mutually inverse, increases, then it is equivalent to the equation .

There is only one root, because on the left is an increasing function, on the right is a decreasing function.

Using the "non-negativity" of functions:

.

All terms on the left side are non-negative, therefore, equality is possible only if each of the terms is equal to zero.

These two equalities contradict each other. The system has no solutions.

Answer: There are no solutions.

To use these methods for solving equations, it is necessary to have a good knowledge of the theoretical material. Using these methods, time is saved, which allows you to solve more tasks. And this is very important when writing tests and passing the exam.

Function properties:

T-1:

Using function superpositions:

T -2:

"Non-negativity" of functions.

Function properties:

The domain of definition and the domain of the square root.

Function monotonicity properties:

T-1: Let y \u003d f (x) is a function that increases on the interval L, and y \u003d g (x) is a function that decreases on the same interval L. Then the equation f(x)=g(x) has at most one root on the interval L.

Using function superpositions:

T -2: If the functions f (x) and g (x) are mutually inverse and the function f (x) increases, then the equation f (x) \u003d g (x) and the equation f (x) \u003d x are equivalent.

"Non-negativity" of functions.

Function properties:

The domain of definition and the domain of the square root.

Function monotonicity properties:

T-1: Let y \u003d f (x) is a function that increases on the interval L, and y \u003d g (x) is a function that decreases on the same interval L. Then the equation f(x)=g(x) has at most one root on the interval L.

Using function superpositions:

T -2: If the functions f (x) and g (x) are mutually inverse and the function f (x) increases, then the equation f (x) \u003d g (x) and the equation f (x) \u003d x are equivalent.

"Non-negativity" of functions.

Non-standard ways to solve quadratic equations

9th grade student

Work manager:

Firsova Daria Evgenievna

mathematic teacher

It is often more useful for a student of algebra to solve the same problem in three different ways than to solve three or four problems. By solving one problem in different ways, you can find out by comparison which one is shorter and more efficient. This is how experience is made.

W.U. Sawyer (English mathematician of the 20th century)

Objective

Learn all the existing ways to solve a quadratic equation. Learn how to use these.

Tasks

• Understand what is called a quadratic equation.
• Find out what types of quadratic equations exist.
• Find information about how to solve a quadratic equation and study it.

Relevance of the topic: People have been studying quadratic equations since ancient times. I wanted to know the history of the development of quadratic equations.

School textbooks do not provide complete information about quadratic equations and how to solve them.

An object: Quadratic equations.

Thing: Methods for solving these equations.

Research methods: analytical.

Hypothesis - if I can realize the goal and tasks set by me while studying this topic, then I will accordingly go to the implementation of pre-profile training in the field of mathematical education.

Research methods:

• Work with educational and popular science literature.
• Observation, comparison, analysis.
• Problem solving.

Expected results: In the course of studying this work, I will really be able to assess my intellectual potential and, accordingly, in the future decide on the profile of training, create a project product on the topic under study in the form of a computer presentation, studying this issue will allow me to compensate for the lack of knowledge on the designated topic.

I consider my work promising, since in the future both students can use this material to improve mathematical literacy, and teachers in optional classes

Quadratic Equations in Ancient Babylon

The need to solve equations not only of the first, but also of the second degree, even in ancient times, was caused by the need to solve problems related to finding the areas of land and earthworks of a military nature., as well as with the development of astronomy and mathematics itself. The Babylonians knew how to solve quadratic equations around 2000 BC. Using modern algebraic notation, we can say that in their cuneiform texts, in addition to incomplete ones, there are such, for example, complete quadratic equations:

The rule for solving these equations, set forth in the Babylonian texts, coincides with the modern one, but it is not known how the Babylonians came to this rule. Almost all the cuneiform texts found so far give only problems with solutions set out in the form of recipes, with no indication of how they were found. Despite the high level of development of algebra in Babylonia, the concept of a negative number and general methods for solving quadratic equations are absent in cuneiform texts.

How Diophantus compiled and solved

quadratic equations

THE EQUATION:

"Find two numbers knowing that their sum is 20 and their product is 96"

Diophantus argues as follows: from the condition of the problem it follows that the desired numbers not are equal, because if they were equal, then their product would not be 96, but 100. Thus, one of them will be more than half of their sum, i.e. 10+X , the other is smaller, i.e. 10-X .

Difference between them 2 X

From here X=2 . One of the desired numbers is 12, the other is 8. Solution X = -2 for Diophantus does not exist, since Greek mathematics knew only positive numbers.

Quadratic equations in India

Problems on quadratic equations are also found in the astronomical treatise "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta, outlined a general rule for solving quadratic equations reduced to a single canonical form: ax ² +bx=c, a0

One of the problems of the famous Indian mathematician of the XII century Bhaskara

Frisky flock of monkeys

Eating well, having fun.

Them squared part eight

Having fun in the meadow.

And twelve in vines ...

They began to jump hanging ...

How many monkeys were

You tell me, in this flock?.

The equation corresponding to the problem is:

Baskara writes under the guise:

Complement the left side to a square,

Quadratic Equations in Ancient Asia

X 2 +10 x = 39

Here is how the Central Asian scientist al-Khwarizmi solved this equation:

He wrote: "The rule is this:

bifurcate number of roots, x=2x ·5

get five in this problem, 5

multiply by this equal to him, there will be twenty-five, 5 5=25

add that to thirty nine, 25+39

will be sixty four 64

take the root out of it, there will be eight, 8

and subtract from this half the number of roots, i.e. five, 8-5

will remain 3

this will be the root of the square you were looking for."

What about the second root? The second root was not found, since negative numbers were not known.

Quadratic equations in Europe XIII-XVII centuries.

The general rule for solving quadratic equations reduced to a single canonical form x2 + in + c \u003d 0 was formulated in Europe only in 1544 Mr. Stiefel.

Formulas for solving quadratic equations in Europe were first stated in 1202 by an Italian mathematician

Leonard Fibonacci.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. Only in the 17th century thanks to the work Descartes, Newton and other scientists method for solving quadratic equations takes modern look

About Vieta's theorem

The theorem expressing the relationship between the 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 times A-A is equal to BD, then A is equal to B and equal to D” .

To understand Vieta, it should be remembered that A, 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 the given quadratic equation x 2 +px+q=0 has real roots, then their sum is equal to -p, and the product is q, i.e x 1 + x 2 =-p, x 1 x 2 = q

(the sum of the roots of the given quadratic equation is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term).

• Factoring the left side of the equation
• Vieta's theorem
• Applying Quadratic Coefficient Properties
• The solution of quadratic equations by the method of "transferring" the highest coefficient
• Full square selection method
• Graphical way to solve quadratic equations
• Solving quadratic equations with a compass and straightedge
• Solving quadratic equations using a nomogram
• Geometric way of solving quadratic equations

Factorization method

give a quadratic equation general view to the view:

A(x) B(x)=0,

where A(x) and B(x) are polynomials with respect to x.

Target:

Ways:

• Taking the common factor out of brackets;
• Using abbreviated multiplication formulas;
• grouping method.

Example:

: X 2 + 10x - 24 = 0

Let's factorize the left side of the equation:

X 2 + 10x - 24 = x 2 + 12x - 2x - 24 \u003d x (x + 12) - 2 (x + 12) \u003d \u003d (x + 12) (x - 2);

(x + 12) (x - 2) = 0;

x + 12 = 0 or x - 2 = 0;

X 1 = -12 x 2 = 2 ;

The numbers - 12 and 2 are the roots of this equation.

Answer: x 1 = -12; X 2 = 2.

Solving equations using Vieta's theorem

x 1 and X 2 are the roots of the equation

for example :

X 2 + 3X – 10 = 0

X 1 ·X 2 = - 10, so the roots have different

signs

X 1 + X 2 = - 3, means greater in modulus

root - negative

By selection we find the roots: X 1 = – 5, Х 2 = 2

Properties of the coefficients of a quadratic equation

Let the quadratic equation ax be given 2 + bx + c = 0

If a a + b + c = 0 (i.e. the sum of the coefficients

equation is zero), then X 1 = 1 , X 2 = c/a

If a a - b + c = 0 , or b = a + c , then X 1 = – 1 , X 2 = – s/a .

Example :

137x 2 + 20x – 157 = 0.

a = 137, b = 20, c = -157.

a + b + c = 137 + 20 – 157 =0.

x 1 = 1,

Answer: 1;

Solving equations using the "transfer" method

Roots of quadratic equations ax 2 + bx + c = 0 and y 2 + by + ac = 0 related by the ratio : x = y/a .

Consider the quadratic equation ax ² + bx + c = 0 , where a ≠ 0. Multiplying both sides by a , we get the equation a²x² + abx + ac = 0. Let be ah = y , where X = u/a; then we come to the equation y² + bu + ac = 0 , which is equivalent to the given one. its roots at 1 and at 2 find with the help of Vieta's theorem. Finally we get X 1 =y 1 /a and X 2 =y 2 /a .

Solve the equation: 2x 2 - 11x +15 = 0.

Let's transfer the coefficient 2 to the free term

at 2 - 11y + 30 = 0. D0, according to the theorem, the inverse of Vieta's theorem, we get the roots: 5; 6, then we return to the roots of the original equation: 2.5; 3.

Full square selection method

X 2 + 6x - 7 = 0

Let's select a full square on the left side. To do this, we write the expression X 2 + 6x in the following form:

X 2 + 6x = x 2 + 2 x 3

In the resulting expression, the first term is the square of the number X, and the second is the double product X on the 3 , so to get a full square, you need to add 3 2 , as

X 2 + 2 x 3 + 3 2 = (x + 3) 2

We now transform the left side of the equation X 2 + 6x - 7 = 0, adding to it and subtracting 3 2 , we have:

X 2 + 6x - 7 = x 2 + 2 x 3 + 3 2 – 3 2 – 7 =

= (x + 3) 2 – 9 – 7 = (x + 3) 2 – 16

Thus, this equation can be written as follows:

(x + 3) 2 –16 = 0 , i.e. (x + 3) 2 = 16 .

Hence, x + 3 - 4 = 0 or x + 3 + 4 = 0

X 1 = 1 X 2 = -7

Answer: -7; one.

Graphical way to solve a quadratic equation

Without using formulas, a quadratic equation can be solved graphically

way. Let's solve the equation

To do this, we will build two graphs:

The abscissas of the intersection points of the graphs will be the roots of the equation.

If the graphs intersect at two points, then the equation has two roots.

If the graphs intersect at one point, then the equation has one root.

If the graphs do not intersect, then the equation has no roots.

Answer:

Solving quadratic equations with

compasses and rulers

1. Let's choose a coordinate system.

2. Let's build points S (-b/ 2 a; a+c/ 2 a) is the center of the circle and BUT( 0; 1 ) .

3. Draw a circle with a radius SA .

Abscissas points of intersection of the circle with the x-axis are roots given quadratic equation.

x 1

x 2

Solving quadratic equations using a nomogram

This is an old and undeservedly forgotten way to solve quadratic equations, placed on p. 83 "Four-digit mathematical tables" Bradis V.M.

For the equation

nomogram gives roots

Table XXII. Nomogram for Equation Solving

This nomogram allows, without solving the quadratic equation, to determine the roots of the equation by its coefficients.

Geometric way of solving quadratic equations

An example that has become famous is from the "Algebra" of al-Khwarizmi: X 2 + 10x = 39. In the original, this problem is formulated as follows: "The square and ten roots are equal to 39."

S=x 2 + 10 x + 25 (X 2 + 10 x = 39 )

S= 39 + 25 = 64 , whence follows,

what is the side of the square ABCD ,

those. line segment AB = 8 .

x = 8 - 2,5 - 2,5 = 3

Based on the survey, it was found that:

• The most difficult were the following methods:

Factoring the left side of the equation,

Full square selection method.

• Rational solution methods:

Solution of quadratic equations by formula;

Solving equations using Vieta's theorem

• Has no practical application

Geometric way of solving quadratic equations.

• Never heard of methods before:

Application of the properties of the coefficients of a quadratic equation;

With the help of a nomogram;

Solving quadratic equations with a compass and straightedge;

The method of "transfer" (this method aroused interest among the students).

Conclusion

• these decision methods deserve attention, since they are not all reflected in school mathematics textbooks;

• mastering these techniques will help students save time and solve equations efficiently;

• the need for a quick solution is due to the use of a test system of entrance exams;

THANK YOU BEHIND ATTENTION!

The theory of equations in the school course of algebra occupies a leading place. More time is devoted to their study than to any other topic of the school mathematics course. This is due to the fact that most of life's tasks come down to solving various kinds equations.

In the algebra textbook for grade 8, we get acquainted with several types of quadratic equations, and work out their solution using formulas. Are there other methods for solving quadratic equations? How complex are these methods and can they be used in practice?

The research work "Non-standard ways of solving quadratic equations" is devoted to these issues.

Relevance of this topic is that in the lessons of algebra, geometry, physics, we very often meet with the solution of quadratic equations. Therefore, each 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 the 9th grade when passing exams.

The purpose of the research work: identify ways to solve quadratic equations, find out if any quadratic equation can be solved by these methods and highlight the features and disadvantages of these methods.

Research tasks: analyze literature sources to identify ways to solve quadratic equations, show different ways to solve quadratic equations; identify the most convenient ways to solve quadratic equations; learn how to solve quadratic equations in various ways.

Research methods: literature analysis, sociological survey, observation, comparison and generalization of results .

View presentation content
"Non-standard ways to solve quadratic equations"

Municipal state educational institution

"Bogucharskaya middle comprehensive school№1"

Research work on the topic: "Non-standard ways to solve quadratic equations "

Pryadkova Ekaterina Sergeevna

Head: Alabina Galina Yurievna

Reveal ways to solve quadratic equations

Find out whether it is possible to solve any quadratic equation by these methods and highlight the features and disadvantages of these methods

• Analyze literature sources to identify ways to solve quadratic equations
• Show different ways to solve quadratic equations
• Identify the most convenient ways to solve quadratic equations
• Learn to solve quadratic equations in various ways

Ways to solve quadratic equations

Factoring the Left Side

According to the formula

Main

Using the Vieta theorem (direct and inverse)

Graphical way

According to the properties of the coefficients

The method of "transfer"

Additional

Using a compass and ruler

With the help of a nomogram

Geometric way

Coefficient properties

Properties:

Transfer method

Multiplying both sides of the equation by a, we get

Let be

, where

Then we get an equation with a new variable

Its roots are 1 and at 2 . Finally

Using a compass and ruler

The radius of the circle is greater than the ordinate of the center

, the circle intersects the x-axis at two points, where the roots of the original equation.

The radius of the circle is equal to the ordinate of the center

, the circle intersects the Ox axis at one point where is the root of the original equation.

The radius of the circle is less than the ordinate of the center

, the circle has no common points with the x-axis. In this case, the original equation has no roots.

With the help of a nomogram

X 2 -9x+8=0

X 1 =8; X 2 =1

Geometric way

Consider how the ancient Greeks solved the equation

The solution is shown in the figure, where

or

Expressions

and 16 + 9

geometrically represent the same square with a side of 5.

So

Data processing

Selection method

full square

Decomposition of the left side

multiplier equations

Answer: -4,5; 1.

Data processing

Using

Vieta formulas

According to the formula

has two different

by root sign

greater in modulus

root is negative

Answer: -4,5; 1.

Data processing

By property of coefficients

The method of "transfer"

We transfer the coefficient a \u003d 2 to the free term and get the equation:

As

from which, according to the Vieta formulas

The roots of the original equation will be

Answer: -4,5; 1.

Data processing

Graphic method

Using a compass and ruler

We write the equation in the form

Determine the coordinates of the center of the circle by the formulas:

Let us construct graphs of functions in one coordinate system

Let's draw a circle of radius SA, where BUT (0;1).

Answer: -4,5; 1.

Data processing

Geometric way

With the help of a nomogram

Let's represent the equation in the form:

Let's represent the equation in the form:

Area of ​​the resulting square:

As

The nomogram gives a positive root

Thus, we got the equation:

negative root

Answer: -4,5; 1.

Positive aspects and disadvantages

Positive sides

Factoring the left side of the equation

disadvantages

It makes it possible to immediately see the roots of the equation.

Full square selection method

It is necessary to correctly divide the terms for

groupings.

According to the formula

For the minimum number of actions, you can find the roots of the equations

It is necessary to correctly find all the terms to select the full square.

Can be applied to all quadratic equations.

Using Vieta formulas

You need to learn the formulas.

Enough easy way, makes it possible to immediately see the roots of the equation.

Only whole roots are easily found.

The name of the method for solving quadratic equations

Positive sides

disadvantages

The method of "transfer"

For the minimum number of actions, you can find the roots of the equation, it is used in conjunction with the method of Vieta's theorem.

According to the properties of the coefficients

Graphical way

It is easy to find only whole roots.

Doesn't require much effort

visual way

Fits only some equations

Using a compass and ruler

There may be inaccuracies in scheduling

visual way

With the help of a nomogram

Intuitive, easy to use.

There may be inaccuracies

Geometric way

Visual way.

Not always at hand there is a nomogram.

Similar to the way to select a full square

In order to solve any quadratic equation well, it is necessary

know:

 the formula for finding the discriminant;

 the formula for finding the roots of a quadratic equation;

 Algorithms for solving equations of this type.

be able to:

 solve incomplete quadratic equations;

 solve complete quadratic equations;

 solve the given quadratic equations;

 find errors in the solved equations and correct them;

 Check.

Municipal competition of research and creative works schoolchildren

"Step into Science"

Section of MATHEMATICS

Subject: Non-standard methods for solving irrational

equations.

Nuzhdina Maria, MAOU secondary school №2

Grade 10, Karymskoye village

Scientific adviser: Vasilyeva Elena Valerievna,

mathematic teacher

MAOU secondary school No. 2, Karymskoye village

settlement Karymskoe, 2013

Abstract…………………………………………………………………….3

Study plan…………………………………………………………………………………………………………..4-5

Description of work:

§one. Basic techniques for solving irrational equations………………6-9

§2. Solving irrational equations by the method of replacing the unknown ... 10-14

§3. Irrational equations reduced to the modulus ………….15-17

§4. Factorization…………………………………………………..18-19

§5. Equations of the form ………………………………………20-22

§6. Geometric mean theorem in irrational equations

; ……………………………23-24

4) References…………………………………………………….....25

Annotation.

The theme of our research work: "Non-standard methods for solving irrational equations."

When performing the work, it was necessary to: compare different methods of solution; move from common methods to private, and vice versa; argue and prove the statements made; study and summarize information collected from various sources. In this regard, one can single out following methods research activities: empirical; logical and theoretical (research); step by step; reproductive and heuristic;

As a result of the work carried out, the following results and conclusions:

There are many tricks for solving irrational equations;

Not all irrational equations are solved using standard tricks;

We have studied the frequent substitutions by which complex irrational equations are reduced to the simplest;

We considered non-standard methods for solving irrational equations

Topic: "Non-standard methods for solving irrational equations"

Nuzhdina M.P., Trans-Baikal Territory, Karymskoye settlement, MAOU secondary school No. 2, grade 10.

Research plan.

Object area in which we have been researching is algebra. An object research- solution of equations. Among the many equations, we considered irrational equations - thing our research.

In the school course of algebra, only standard methods and techniques for solving (raised to a power and simple replacement techniques) are considered. But in the process of research, it turned out 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 the study of such methods of solution is a necessary and interesting work.

In the process of research, it turned out that there are a great many irrational equations and it is problematic to group them according to types and methods.

aim research is the study and systematization of methods for solving irrational equations.

Hypothesis: If you know non-standard methods for solving irrational equations, then this will improve the quality of the performance of some Olympiad and test tasks of the Unified State Examination.

To achieve the set goals and test the hypothesis, it is necessary to solve the following tasks:

Characterize the types of irrational equations.

Establish links between types and methods of solution.

Assess the value of checking and finding ODZ.

Consider non-standard cases when solving irrational equations (geometric mean theorem, monotonicity properties of functions).

In the course of the study, many textbooks by such authors as M.I.Skanavi, I.F.Sharygin, O.Yu.Cherkasov, A.N.Rurukin, I.T. methodical journal "Mathematics at School".

Topic: "Non-standard methods for solving irrational equations"

Nuzhdina M.P., Trans-Baikal Territory, Karymskoye settlement, MAOU secondary school No. 2, grade 10.

Description of work.

§1 Basic techniques for solving irrational equations

The equation y(x)=0 is irrational if the function y(x) contains roots from an unknown value x or expressions that depend 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 (ODV), but there are other methods, some of which will be discussed in the work.

The main technique for solving irrational equations is considered to be seclusion in one part of the radical equation, followed by raising both parts of the equation to the appropriate degree. If there are several such radicals, then the equation must be raised to the original power repeatedly, by the way, while there is no need to take care that the expression under the sign of the solitary radical would be 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 solution, the roots must be checked and extraneous ones are discarded, in this case the check is an element of the solution and is necessary even in cases where extra roots did not appear, but the course of the solution was such that they could appear. On the other hand, sometimes it is easier to make 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 the 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 even-degree radicals, it is sufficient to solve the system of inequalities.

3x -2x 2 +5 ≥0 (ODZ equation conditions)

4x 2 -26x +40 ≥0

Solving this system of inequalities, we get:

x € Where x = 2.5.

x € (-∞ ; 2.5] ᴗ )