Games

10 ways to solve square. Methods for solving quadratic equations. History of the development of quadratic equations

https://pandia.ru/text/78/082/images/image002_237.gif" height="952"> MOU "Sergievskaya Secondary School"

Completed by: Sizikov Stanislav

Teacher:

With. Sergievka, 2007

1. Introduction. Quadratic Equations in Ancient Babylon……………….3

2. Quadratic equations in Diaphant…………..………………………….4

3. Quadratic equations in India …………………………………………………………………………………………………………………………………………………………………………………………

4. Quadratic equations in al-Khorezmi …………………………………..6

5. Quadratic equations in Europe XIII - XYII…………………………...7

6. About the Vieta theorem …………………………………………………………..9

7. Ten ways to solve quadratic equations……………………..10

8. Conclusion …………………………………………………………………20

9. References ……………………………………………………...21

Introduction

Quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational equations. We all know how to solve quadratic equations, starting from grade 8. But how did the history of solving quadratic equations originate and develop?

Quadratic Equations in Ancient Babylon

The need to solve equations not only of the first, but also of the second degree, back in antiquity, was caused by the need to solve problems related to finding the areas of land; earthworks of a military nature, as well as with the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians. Using modern algebraic notation, we can say that in their cuneiform texts, in addition to incomplete ones, there are, for example, complete quadratic equations: x2 + x = , : x2 - x = 14https://pandia.ru/text/78/082 /images/image005_150.gif" width="16" height="41 src=">)2 + 12 = x; Bhaskara writes under the guise

x2- 64X = - 768

and, to complete the left side of this equation to the square, he adds 322 to both sides, getting then: x2- 64x + 322 = - 768 + 1024;

(X- 32)2 = 256; X - 32 = ± 16, xt = 16, hg= 48.

Quadratic equations in al - Khorezmi

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

1) "Squares are equal to roots", i.e. ax2 = in.

2) “Squares are equal to number”, i.e. ah2= With.

3) "The roots are equal to the number", i.e. ah = s.

4) “Squares and numbers are equal to roots”, i.e. ah2+ c = in.

5) “Squares and roots are equal to number”, i.e. ah2+ in = s.

6) "Roots and numbers are equal to squares", i.e. in+ c \u003d ax2. For al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends, not subtractions. In this case, equations that do not have positive solutions are obviously not taken into account. The author sets out methods for solving these equations. His decision, of course, does not completely coincide with ours. Not to mention the fact that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type, al-Khwarizmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because in specific practical tasks, it does not matter. When solving complete quadratic equations, al-Khwarizmi sets out the rules for solving them using particular numerical examples, and then their geometric proofs.

Let's take an example.

Problem 14. “The square and the number 21 are equal to 10 roots. Find the root "(meaning the root of the equation x2+ 21 = 10X).

The author's solution goes something like this: divide the number of roots in half, you get 5, multiply 5 by itself, subtract 21 from the product, 4 remains. Take the root of 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root. Or add 2 to 5, which will give 7, this is also a root.

The treatise of al-Khwarizmi is the first book that has come down to us, in which the classification of quadratic equations is systematically presented and formulas for their solution are given.

Quadratic equations in EuropeXIII- XVIIcenturies

The formulas for solving quadratic equations on the model of al-Khwarizmi in Europe were first set forth in the Book of the Abacus (published in Rome in the middle of the last century, the Fibonacci Book of the Abacus contains 459 pages), written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics from both the countries of Islam and Ancient Greece, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic examples of problem solving and the first in Europe approached the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from the Book of the Abacus passed into almost all European textbooks of the 16th-17th centuries. and partly XVIII.

General rule for solving quadratic equations reduced to a single canonical form x2+ in = s, for all possible combinations of signs of the coefficients in, with was formulated in Europe only in 1544. M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. The Italian mathematicians Tartaglia, Cardaco, Bombelli were among the first in the 16th 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 takes on a modern form.

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 AT+ D, multiplied by BUT minus A2, equals BD, then BUT equals AT and equal D».

To understand Vieta, one must remember that BUT, like any
vowel, meant for him unknown (our X), vowels
AT,D- coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means: if

(a+ c) x - x 2 = ab, x2 - (a+ b) x + ab = 0, x1 = a, x2 = b.

Expressing the relationship between the roots and coefficients of equations by general formulas written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from its modern form. He did not recognize negative numbers and therefore, when solving equations, he considered only cases where all roots are positive.

Ten Ways to Solve Quadratic Equations

In the school course of mathematics, the formulas of the roots of quadratic equations are studied, with the help of which you can solve any quadratic equations. However, there are other ways to solve quadratic equations that allow you to solve many equations very quickly and rationally. There are ten ways to solve quadratic equations. Let's consider each of them.

1. Factorization of the left side of the equation

Let's solve the equation x2+ 10X- 24 = 0. Let's factorize the left side of the equation:

x2 + 10x - 24 = x2 + 12x - 2x - 24 =

X(x + x + 12) = (x + 12)(x - 2).

Therefore, the equation can be rewritten as:

( X + 12)(x - 2) = 0.

Since the product is zero, at least one of its factors is zero. Therefore, the left side of the equation vanishes when x = 2, as well as X= - 12. This means that the numbers 2 and - 12 are the roots of the equation x2 + 10x - 24 = 0.

2. Full square selection method

Let's explain this method with an example.

Let's solve the equation x2 + 6x - 7 = 0. Select a full square on the left side. To do this, we write the expression x2 + 6x in the following form:

x2 + 6x = x2 + 2*x*3.

In the resulting expression, the first term is the square of the number x, and the second is the double product of x by 3. Therefore, to get the full square, you need to add 32, since

x2 + 2 x 3 + 32 = (x + 3)2.

We now transform the left side of the equation

x2 + 6x - 7 = 0,

adding to it and subtracting 32. We have:

x2 + 6x - 7 = x2 + 2 X 3 +– 7 = (X- \u003d (x - Z) 2 - 16 .

Thus, this equation can be written as follows:

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

Consequently, X+ 3 \u003d 4 x1 \u003d 1, or x + 3 \u003d - 4, x2 \u003d - 7.

3. Solution of quadratic equations by the formula

Multiply both sides of the equation

ah2+ in+ c = 0, a ≠ 0, on 4a and successively we have:

4a2 x2 + 4abx+ 4ac = 0,

((2ax)2 + 2 axb + b2 ) - b2 + 4ac= 0,

(2ax +b)2 = in2- 4ac,

2ax+ b= ± https://pandia.ru/text/78/082/images/image006_128.gif" width="71" height="27">, x1,2 =

In the case of a positive discriminant, i.e., with v2 - 4ac > 0, equation ah2+ in + s= 0 has two different roots.

If the discriminant is zero, i.e. v2 - 4ac = 0, then the equation ah2+ in+ With= 0 has a single root, x = - https://pandia.ru/text/78/082/images/image009_95.gif" width="14" height="62"> Its roots satisfy the Vieta theorem, which, when a= 1 has the form

x1 x2 = q,

x1 + x2 = - R.

From this we can draw the following conclusions (by the coefficients R and q root signs can be predicted).

a) If a free member q reduced equation (1)
positive (q> 0), then the equation has two identical
by the sign of the root and it depends on the second coefficient R
If a R> 0, then both roots are negative if R< 0, then both
roots are positive.

For example,

x2- 3X + 2 = 0; x1= 2 and x2 = 1, because q = 2 > 0 u p = - 3 < 0;

x2 + 8x + 7 = 0; x 1 \u003d - 7 and x2 \u003d - 1, since q= 7 > 0 and R = 8 > 0.

b) If a free member q reduced equation (1)
negative (q < 0), then the equation has two roots of different sign, and the larger root in absolute value will be positive if R< 0, or negative if p > 0.

For example,

x2 + 4x - 5 = 0; x1 \u003d - 5 and x2 \u003d 1, since q = - 5 < 0 и R= 4 > 0;

x2 - 8x - 9 = 0; x1 = 9 and x2= - 1 because q = - 9 < и R= - 8 < 0.

5. Solution of equations by the method of "transfer"

Consider the quadratic equation ax2 + in+ c = 0, where a ≠ 0. Multiplying both its parts by a, we get the equation a2x2 +abx+ ace= 0.

Let ah = y where X=; then we come to the equation

y2+ by+ ac = 0,

equivalent to this one. its roots y1 and y2 find with the help of Vieta's theorem. Finally we get x1= https://pandia.ru/text/78/082/images/image012_77.gif" width="24" height="43">.

With this method, the coefficient a is multiplied by the free term, as if “thrown” to it, which is why it is called transfer method. This method is used when it is easy to find the roots of an equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.

1. Solve the equation 2x2 - 11x + 15 = 0.

Solution. Let's "transfer" the coefficient 2 to the free term, as a result we get the equation

y2 - 11 at+ 30 = 0.

According to the Vieta theorem, y1 = 5, y2 = 6, hence x1 = https://pandia.ru/text/78/082/images/image014_69.gif" width="16 height=41" height="41">, t e.

x1 = 2.5 x2 = 3.

Answer: 2,5; 3.

6. Properties of the coefficients of the squareequations

A. Let a quadratic equation be given

ax2 + in + c= 0, where a ≠ 0.

1. If a + in + with= 0 (i.e., the sum of the coefficients of the equation is equal to zero), then x1 = 1, x2 = .

2. If a - b + c= 0, orb = a + c, then x1 = - 1, X 2 = - https://pandia.ru/text/78/082/images/image016_58.gif" width="44 height=41" height="41">.

Answer: 1; 184">

The following cases are possible:

A straight line and a parabola can intersect at two points, the abscissas of the intersection points are the roots of a quadratic equation;

A straight line and a parabola can touch (only one common point), that is, the equation has one solution;

The straight line and the parabola do not have common points, that is, the quadratic equation has no roots.

Examples.

1. Let's graphically solve the equation x2 - 3x - 4 = 0 (Fig. 2).

Solution. We write the equation in the form x2 = 3x + 4.

Let's build a parabola y = x2 and direct y= 3x + 4. Direct at= 3x + 4 can be constructed from two points M(0; 4) and N(3; 13). A line and a parabola intersect at two points A to B with abscissa x1= - 1 and x2 = 4.

Answer: x1= - 1, x, = 4.

8. Solving quadratic equations with a compass and straightedge

The graphical way to solve quadratic equations using a parabola is inconvenient. If you build a parabola point by point, then it takes a lot of time, and the degree of accuracy of the results obtained is low.

We propose the following method for finding the roots of a quadratic equation

ah2+ in+ With= 0

using a compass and ruler (Fig.).

Let us assume that the desired circle intersects the abscissa axis at the points B(x1; 0) and D(x2 ; 0), where x1 and x2- roots of the equation ax2 + in+With=0,
and passes through points A(0; 1) and C(0; ) on the y-axis..gif" width="197" height="123">

So: 1) build points https://pandia.ru/text/78/082/images/image023_40.gif" width="171" height="45"> the circle intersects the OX axis at point B(x1;0), and D(x1 ; 0), where x1 and x2 - roots of the quadratic equation ax2+bx+c = 0.

2) The radius of the circle is equal to the ordinate of the center , the circle touches the x-axis at the point B(x1; 0), where xx is the root of the quadratic equation.

3) The radius of the circle is less than the ordinate of the center left">

https://pandia.ru/text/78/082/images/image029_34.gif" width="612" height="372">40" height="14">

https://pandia.ru/text/78/082/images/image031_28.gif" width="612" height="432 src=">

Whence after substitutions and

simplifications, the equation z2+pz+q=0 follows, and the letter z means the label of any point of the curvilinear scale.

10. Geometric method for solving quadratic equations

In ancient times, when geometry was more developed than algebra, quadratic equations were solved not algebraically, but geometrically. Let us give an example that has become famous from Algebra by al-Khwarizmi.

And four attached squares i.e. S=x2+10x+25. Replacing x2+10x with 39, we get S = 39 + 25 = 64, which means that the side of the square ABCD, i.e. segment AB= 8. For the desired side X the original square we get

Conclusion

We all know how to solve quadratic equations, from school to graduation. But in the school course of mathematics, the formulas of the roots of quadratic equations are studied, with the help of which any quadratic equations can be solved. However, having studied this issue more deeply, I was convinced that there are other ways to solve quadratic equations that allow you to solve many equations very quickly and rationally.

Maybe mathematics is somewhere out there in other dimensions, not visible to the eye - everything is written down and we just get all the new facts from the hole with the worlds? ... God knows; but it turns out that if physicists, chemists, economists or archaeologists need a new model of the structure of the world, this model can always be taken from the shelf where mathematicians put it three hundred years ago, or assembled from parts lying on the same shelf. Perhaps these parts will have to be twisted, adjusted to each other, polished, quickly machined a couple of new theorem bushings; but the theory of the result will not only describe the actual situation that has arisen, but also predict the consequences! ...

A strange thing is this game of the mind, which is always right ...

Literature

1. Alimov SHA., Ilyin VA. et al. Algebra, 6-8. Trial textbook for 6-8 grades of high school. - M., Education, 1981.

2.Bradis math tables for high school. Ed. 57th. - M., Education, 1990. S. 83.

3. Zlotsky - tasks in teaching mathematics. The book for the teacher. - M., Education, 1992.

4.M., Mathematics (supplement to the newspaper "First of September"), Nos. 21/96, 10/97, 24/97, 18/98, 21/98.

5. Okunev functions, equations and inequalities. A guide for the teacher. - M., Education, 1972.

6. Solomnik B. C., Sweet questions and problems in mathematics. Ed. 4th, add. - M., Higher School, 1973.

7.M., Mathematics (supplement to the newspaper "First of September"), No. 40, 2000.

Review

for the work of a student of the 11th grade of the MOU "Sergievskaya secondary

comprehensive school"

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

s.Kopyevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

1.2 How Diophantus compiled and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations in al-Khwarizmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1 .1 Square equationsstrife in ancient Babylon

The need to solve equations not only of the first, but also of the second degree 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 the development of astronomy and mathematics itself. Quadratic equations were able to solve about 2000 BC. e. Babylonians.

Applying modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X 2 + X = ѕ; X 2 - X = 14,5

The rule for solving these equations, stated in the Babylonian texts, coincides essentially 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 stated in the form of recipes, with no indication of how they were found.

Despite the high level of development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.

1.2 How Diophantus compiled and solved quadratic equations.

Diophantus' Arithmetic does not contain a systematic exposition of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by drawing up equations of various degrees.

When compiling equations, Diophantus skillfully chooses unknowns to simplify the solution.

Here, for example, is one of his tasks.

Task 11."Find two numbers knowing that their sum is 20 and their product is 96"

Diophantus argues as follows: it follows from the condition of the problem that the desired numbers are not equal, since 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's. The difference between them 2x.

Hence the equation:

(10 + x)(10 - x) = 96

100's 2 = 96

X 2 - 4 = 0 (1)

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

If we solve this problem by choosing one of the desired numbers as the unknown, then we will come to the solution of the equation

y(20 - y) = 96,

at 2 - 20y + 96 = 0. (2)

It is clear that Diophantus simplifies the solution by choosing the half-difference of the desired numbers as the unknown; he manages to reduce the problem to solving an incomplete quadratic equation (1).

1.3 Quadratic equations in India

Problems for quadratic equations are already found in the astronomical tract "Aryabhattam", compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined the general rule for solving quadratic equations reduced to a single canonical form:

Oh 2 + bx = c, a > 0. (1)

In equation (1), the coefficients, except for a, can also be negative. Brahmagupta's rule essentially coincides with ours.

In ancient India, public competitions in solving difficult problems were common. In one of the old Indian books, the following is said about such competitions: “As the sun outshines the stars with its brilliance, so a learned person will outshine the glory of another in public meetings, proposing and solving algebraic problems.” Tasks were often dressed in poetic form.

Here is one of the problems of the famous Indian mathematician of the XII century. Bhaskara.

Task 13.

“A frisky flock of monkeys And twelve in vines ...

Having eaten power, had fun. They began to jump, hanging ...

Part eight of them in a square How many monkeys were there,

Having fun in the meadow. You tell me, in this flock?

Bhaskara's solution indicates that he knew about the two-valuedness of the roots of quadratic equations (Fig. 3).

The equation corresponding to problem 13 is:

(x/8) 2 + 12 = x

Bhaskara writes under the guise of:

X 2 - 64x = -768

and, to complete the left side of this equation to a square, he adds to both sides 32 2 , getting then:

X 2 - 64x + 32 2 = -768 + 1024,

(x - 32) 2 = 256,

x - 32 = ± 16,

X 1 = 16, X 2 = 48.

1.4 Square equationsal-Khorezmi

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

1) "Squares are equal to roots", i.e. Oh 2 + with =bX.

2) "Squares are equal to number", i.e. Oh 2 = s.

3) "The roots are equal to the number", i.e. ah = s.

4) "Squares and numbers are equal to roots", i.e. Oh 2 + with =bX.

5) "Squares and roots are equal to the number", i.e. Oh 2 + bx= s.

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

For al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends, not subtractions. In this case, equations that do not have positive solutions are obviously not taken into account. The author outlines the methods for solving these equations, using the methods of al-jabr and al-muqabala. His decisions, of course, do not completely coincide with ours. Not to mention the fact that it is purely rhetorical, it should be noted, for example, that when solving an incomplete quadratic equation of the first type

al-Khorezmi, like all mathematicians before the 17th century, does not take into account the zero solution, probably because it does not matter in specific practical problems. When solving complete quadratic equations, al-Khorezmi sets out the rules for solving, and then geometric proofs, using particular numerical examples.

Task 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming the root of the equation x 2 + 21 = 10x).

Treatise al - Khorezmi is the first book that has come down to us, in which the classification of quadratic equations is systematically stated and formulas for their solution are given.

1.5 Quadratic equations in EuropeXIII - XVIIcenturies

Formulas for solving quadratic equations on the model of al - Khorezmi in Europe were first set forth in the "Book of the Abacus", written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics, both the countries of Islam and Ancient Greece, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic examples of problem solving and was the first in Europe to approach the introduction of negative numbers. His book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from the "Book of the Abacus" passed into almost all European textbooks of the 16th - 17th centuries. and partly XVIII.

The general rule for solving quadratic equations reduced to a single canonical form:

X 2 + bx= with,

for all possible combinations of signs of the coefficients b, With was formulated in Europe only in 1544 by M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but Vieta recognized only positive roots. The Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. Take into account, in addition to positive, and negative roots. Only in the XVII century. Thanks to the work of Girard, Descartes, Newton and other scientists, the way to solve quadratic equations takes on a modern look.

1.6 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 multiplied by A - A 2 , equals BD, then A equals AT and equal D».

To understand Vieta, one must remember that BUT, like any vowel, meant for him the unknown (our X), the vowels AT,D- coefficients for the unknown. In the language of modern algebra, Vieta's formulation above means: if

(a +b)x - x 2 = ab,

X 2 - (a +b)x + ab = 0,

X 1 = a, X 2 = b.

Expressing the relationship between the roots and coefficients of equations by general formulas written using symbols, Viet established uniformity in the methods of solving equations. At the same time, the symbolism of Vieta is still far from its modern look. He did not recognize negative numbers, and therefore, when solving equations, he considered only cases where all roots are positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (grade 8) until graduation.

In the school course of mathematics, the formulas of the roots of quadratic equations are studied, with the help of which you can solve any quadratic equations. At the same time, there are other ways to solve quadratic equations, which allow you to solve many equations very quickly and rationally. There are ten ways to solve quadratic equations. In my work, I analyzed each of them in detail.

1. METHOD : Factorization of 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 \u003d x (x + 12) - 2 (x + 12) \u003d (x + 12) (x - 2).

Therefore, the equation can be rewritten as:

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

Since the product is zero, then at least one of its factors is zero. Therefore, the left side of the equation vanishes at x = 2, as well as at x = - 12. This means that the number 2 and - 12 are the roots of the equation X 2 + 10x - 24 = 0.

2. METHOD : Full square selection method.

Let's solve the equation 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 of x by 3. Therefore, to get the full square, you need to add 3 2, since

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, (x + 3) 2 = 16.

Consequently, x + 3 - 4 = 0, x 1 = 1, or x + 3 = -4, x 2 = -7.

3. METHOD :Solution of quadratic equations by formula.

Multiply both sides of the equation

Oh 2 + bx + c = 0, huh? 0

on 4a and successively we have:

4a 2 X 2 + 4abx + 4ac = 0,

((2ah) 2 + 2ax *b + b 2 ) - b 2 + 4 ac = 0,

(2ax+b) 2 = b 2 - 4ac,

2ax + b = ± vb 2 - 4ac,

2ax = - b ± v b 2 - 4ac,

Examples.

a) Let's solve the equation: 4x 2 + 7x + 3 = 0.

a = 4,b= 7, c = 3,D = b 2 - 4 ac = 7 2 - 4 * 4 * 3 = 49 - 48 = 1,

D > 0, two different roots;

Thus, in the case of a positive discriminant, i.e. at

b 2 - 4 ac >0 , the equation Oh 2 + bx + c = 0 has two different roots.

b) Let's solve the equation: 4x 2 - 4x + 1 = 0,

a = 4,b= - 4, c = 1,D = b 2 - 4 ac = (-4) 2 - 4 * 4 * 1= 16 - 16 = 0,

D = 0, one root;

So, if the discriminant is zero, i.e. b 2 - 4 ac = 0 , then the equation

Oh 2 + bx + c = 0 has a single root

in) Let's solve the equation: 2x 2 + 3x + 4 = 0,

a = 2,b= 3, c = 4,D = b 2 - 4 ac = 3 2 - 4 * 2 * 4 = 9 - 32 = - 13 , D < 0.

This equation has no roots.

So, if the discriminant is negative, i.e. b 2 - 4 ac < 0 ,

the equation Oh 2 + bx + c = 0 has no roots.

Formula (1) of the roots of the quadratic equation Oh 2 + bx + c = 0 allows you to find the roots any quadratic equation (if any), including reduced and incomplete. Formula (1) is expressed verbally as follows: the roots of a quadratic equation are equal to a fraction whose numerator is equal to the second coefficient, taken with the opposite sign, plus minus the square root of the square of this coefficient without quadruple the product of the first coefficient by the free term, and the denominator is twice the first coefficient.

4. METHOD: Solution of equations using Vieta's theorem.

As is known, the given quadratic equation has the form

X 2 + px + c = 0. (1)

Its roots satisfy the Vieta theorem, which, when a =1 has the form

x 1 x 2 = q,

x 1 + x 2 = - p

From this we can draw the following conclusions (the signs of the roots can be predicted from the coefficients p and q).

a) If the summary term q of the reduced equation (1) is positive ( q > 0 ), then the equation has two roots of the same sign and this is the envy of the second coefficient p. If a R< 0 , then both roots are negative if R< 0 , then both roots are positive.

For example,

x 2 - 3 x + 2 = 0; x 1 = 2 and x 2 = 1, because q = 2 > 0 and p = - 3 < 0;

x 2 + 8 x + 7 = 0; x 1 = - 7 and x 2 = - 1, because q = 7 > 0 and p= 8 > 0.

b) If a free member q of the reduced equation (1) is negative ( q < 0 ), then the equation has two roots of different sign, and the larger root in absolute value will be positive if p < 0 , or negative if p > 0 .

For example,

x 2 + 4 x - 5 = 0; x 1 = - 5 and x 2 = 1, because q= - 5 < 0 and p = 4 > 0;

x 2 - 8 x - 9 = 0; x 1 = 9 and x 2 = - 1, because q = - 9 < 0 and p = - 8 < 0.

5. METHOD: Solving equations using the "transfer" method.

Consider the quadratic equation

Oh 2 + bx + c = 0, where a? 0.

Multiplying both its parts by a, we obtain the equation

a 2 X 2 + abx + ac = 0.

Let ah = y, where x = y/a; then we come to the equation

at 2 + by+ ac = 0,

equivalent to this one. its roots at 1 and at 2 can be found using Vieta's theorem.

Finally we get

X 1 = y 1 /a and X 1 = y 2 /a.

With this method, the coefficient a is multiplied by the free term, as if “thrown” to it, therefore it is called transfer method. This method is used when it is easy to find the roots of an equation using Vieta's theorem and, most importantly, when the discriminant is an exact square.

Example.

Let's solve the equation 2x 2 - 11x + 15 = 0.

Solution. Let's "transfer" the coefficient 2 to the free term, as a result we get the equation

at 2 - 11y + 30 = 0.

According to Vieta's theorem

at 1 = 5 X 1 = 5/2 x 1 = 2,5

at 2 = 6 x 2 = 6/2 x 2 = 3.

Answer: 2.5; 3.

6. METHOD: Properties of the coefficients of a quadratic equation.

BUT. Let the quadratic equation

Oh 2 + bx + c = 0, where a? 0.

1) If, a+b+ c = 0 (i.e. the sum of the coefficients is zero), then x 1 = 1,

X 2 = s/a.

Proof. Divide both sides of the equation by a? 0, we get the reduced quadratic equation

x 2 + b/ a * x + c/ a = 0.

According to Vieta's theorem

x 1 + x 2 = - b/ a,

x 1 x 2 = 1* c/ a.

By condition a -b + c = 0, where b= a + c. In this way,

x 1 + x 2 = - a+ b / a \u003d -1 - c / a,

x 1 x 2 = - 1* (-c/a),

those. X 1 = -1 and X 2 = c/ a, which we needed to prove.

Examples.

1) Solve the equation 345x 2 - 137x - 208 = 0.

Solution. Because a +b+ c = 0 (345 - 137 - 208 = 0), then

X 1 = 1, X 2 = c/ a = -208/345.

Answer: 1; -208/345.

2) Solve the equation 132x 2 - 247x + 115 = 0.

Solution. Because a +b+ c = 0 (132 - 247 + 115 = 0), then

X 1 = 1, X 2 = c/ a = 115/132.

Answer: 1; 115/132.

B. If the second coefficient b = 2 k is an even number, then the formula of the roots

Example.

Let's solve the equation 3x2 -- 14x + 16 = 0.

Solution. We have: a = 3,b= -- 14, c = 16,k = -- 7 ;

D = k 2 - ac = (- 7) 2 - 3 * 16 = 49 - 48 = 1, D > 0, two different roots;

Answer: 2; 8/3

AT. Reduced Equation

X 2 +px+q= 0

coincides with the general equation, in which a = 1, b= p and c =q. Therefore, for the reduced quadratic equation, the formula for the roots

takes the form:

Formula (3) is especially convenient to use when R-- even number.

Example. Let's solve the equation X 2 - 14x - 15 = 0.

Solution. We have: X 1,2 =7±

Answer: x 1 = 15; X 2 = -1.

7. METHOD: Graphical solution of a quadratic equation.

If in the equation

X 2 + px + q = 0

move the second and third terms to the right side, we get

X 2 = - px - q.

Let's build dependence graphs y \u003d x 2 and y \u003d - px - q.

The graph of the first dependence is a parabola passing through the origin. Graph of the second dependency -

straight line (Fig. 1). The following cases are possible:

A straight line and a parabola can intersect at two points, the abscissas of the intersection points are the roots of a quadratic equation;

The line and the parabola can touch (only one common point), i.e. the equation has one solution;

The straight line and the parabola do not have common points, i.e. a quadratic equation has no roots.

Examples.

1) Let's solve the equation graphically X 2 - 3x - 4 = 0(Fig. 2).

Solution. We write the equation in the form X 2 = 3x + 4.

Let's build a parabola y = x 2 and direct y = 3x + 4. direct

y = 3x + 4 can be built from two points M (0; 4) and

N (3; 13) . A line and a parabola intersect at two points

BUT and AT with abscissa X 1 = - 1 and X 2 = 4 . Answer: X 1 = - 1;

X 2 = 4.

2) Let's solve the equation graphically (Fig. 3) X 2 - 2x + 1 = 0.

Solution. We write the equation in the form X 2 = 2x - 1.

Let's build a parabola y = x 2 and direct y = 2x - 1.

direct y = 2x - 1 build on two points M (0; - 1)

and N(1/2; 0) . Line and parabola intersect at a point BUT With

abscissa x = 1. Answer:x = 1.

3) Let's solve the equation graphically X 2 - 2x + 5 = 0(Fig. 4).

Solution. We write the equation in the form X 2 = 5x - 5. Let's build a parabola y = x 2 and direct y = 2x - 5. direct y = 2x - 5 construct by two points M(0; - 5) and N(2.5; 0). The straight line and the parabola have no intersection points, i.e. This equation has no roots.

Answer. The equation X 2 - 2x + 5 = 0 has no roots.

8. METHOD: Solving quadratic equations with a compass and rulers.

The graphical way to solve quadratic equations using a parabola is inconvenient. If you build a parabola point by point, then it takes a lot of time, and with all this, the degree of accuracy of the results obtained is low.

I propose the following method for finding the roots of a quadratic equation Oh 2 + bx + c = 0 using a compass and ruler (Fig. 5).

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 Oh 2 + bx + c = 0, and passes through the points

A(0; 1) and C(0;c/ a) on the y-axis. Then, by the secant theorem, we have OB * OD = OA * OC, where OC = OB * OD/ OA= x 1 X 2 / 1 = c/ a.

The center of the circle is at the point of intersection of the perpendiculars SF and SK, restored at the midpoints of the chords AC and BD, that's why

1) construct points (the center of the circle) and A(0; 1) ;

2) draw a circle with a radius SA;

3) the abscissas 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 > SK, or R > a + c/2 a) , the circle intersects the x-axis at two points (Fig. 6,a) B(x 1 ; 0) and D(X 2 ; 0) , where X 1 and X 2 - roots of the quadratic equation Oh 2 + bx + c = 0.

2) The radius of the circle is equal to the ordinate of the center (AS = SB, orR = a + c/2 a) , the circle touches the Ox axis (Fig. 6,b) 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, the circle has no common points with the abscissa axis (Fig. 6, c), in this case the equation has no solution.

Example.

Let's solve the equation X 2 - 2x - 3 = 0 (Fig. 7).

Solution. Determine the coordinates of the point of the center of the circle by the formulas:

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

Answer: X 1 = - 1; X 2 = 3.

9. METHOD: Solving quadratic equations with nomograms.

This is an old and undeservedly forgotten method for solving quadratic equations, placed on p. 83 (see Bradis V.M. Four-valued mathematical tables. - M., Enlightenment, 1990).

Table XXII. Nomogram for Equation Solving z 2 + pz + q = 0 . This nomogram allows, without solving the quadratic equation, to determine the roots of the equation by its coefficients.

The curvilinear scale of the nomogram is built according to the formulas (Fig. 11):

Assuming OS = p,ED = q, OE = a(all in cm), from the similarity of triangles SAN and CDF we get the proportion

whence, after substitutions and simplifications, the equation follows

z 2 + pz + q = 0,

and the letter z means the label of any point on the curved scale.

Examples.

1) For the equation z 2 - 9 z + 8 = 0 nomogram gives roots

z 1 = 8,0 and z 2 = 1,0 (Fig. 12).

2) We solve the equation using the nomogram

2 z 2 - 9 z + 2 = 0.

We divide the coefficients of this equation by 2, we get the equation

z 2 - 4,5 z + 1 = 0.

Nomogram gives roots z 1 = 4 and z 2 = 0,5.

3) For the equation

z 2 - 25 z + 66 = 0

coefficients p and q are out of scale, we will perform the substitution z = 5 t, we get the equation

t 2 - 5 t + 2,64 = 0,

which we solve by means of a nomogram and get t 1 = 0,6 and t 2 = 4,4, where z 1 = 5 t 1 = 3,0 and z 2 = 5 t 2 = 22,0.

10. METHOD: Geometric way of solving square equations.

In ancient times, when geometry was more developed than algebra, quadratic equations were solved not algebraically, but geometrically. I will give an example that has become famous from the "Algebra" of al-Khwarizmi.

Examples.

1) Solve the equation X 2 + 10x = 39.

In the original, this problem is formulated as follows: “The square and ten roots are equal to 39” (Fig. 15).

Solution. Consider a square with side x, rectangles are built on its sides so that the other side of each of them is 2.5, therefore, the area of \u200b\u200beach is 2.5x. The resulting figure is then supplemented to a new square ABCD, completing 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 the areas: the 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 , whence it follows that the side of the square ABCD, i.e. line segment AB = 8. For the desired side X the original square we get

2) But, for example, how the ancient Greeks solved the equation at 2 + 6y - 16 = 0.

Solution shown in fig. 16, where

at 2 + 6y = 16, or at 2 + 6y + 9 = 16 + 9.

Solution. Expressions at 2 + 6y + 9 and 16 + 9 geometrically represent the same square, and the original equation at 2 + 6y - 16 + 9 - 9 = 0 is the same equation. From where we get that y + 3 = ± 5, or at 1 = 2, y 2 = - 8 (Fig. 16).

3) Solve geometric equation at 2 - 6y - 16 = 0.

Transforming the equation, we get

at 2 - 6y = 16.

On fig. 17 find the "images" of the expression at 2 - 6y, those. from the area of a square with side y subtract twice the area of a square with side equal to 3 . So, if the expression at 2 - 6y add 9 , then we get the area of a square with a side at - 3 . Replacing the expression at 2 - 6y its equal number 16,

we get: (y - 3) 2 = 16 + 9, those. y - 3 = ± v25, or y - 3 = ± 5, where at 1 = 8 and at 2 = - 2.

Conclusion

Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities.

At the same time, the value of quadratic equations lies not only in the elegance and brevity of solving problems, although this is very significant. No less important is the fact that as a result of the use of quadratic equations in solving problems, new details are often discovered, interesting generalizations can be made and refinements made, which are prompted by an analysis of the obtained formulas and relations.

I would also like to note that the topic presented in this work is still little studied at all, they just do not deal with it, therefore it is fraught with a lot of hidden and unknown, which provides an excellent opportunity for further work on it.

Here I settled on the question of solving quadratic equations, and what,

if there are other ways to solve them?! Again, find beautiful patterns, some facts, clarifications, make generalizations, discover everything new and new. But these are questions for future works.

Summing up, we can conclude: quadratic equations play a huge role in the development of mathematics. We all know how to solve quadratic equations from school (grade 8) until graduation. This knowledge can be useful to us throughout life.

Since these methods for solving quadratic equations are easy to use, they should certainly be of interest to students who are fond of mathematics. My work makes it possible to take a different look at the problems that mathematics sets before us.

Literature:

1. Alimov Sh.A., Ilyin V.A. et al. Algebra, 6-8. Trial textbook for 6-8 grade high school. - M., Education, 1981.

2. Bradis V.M. Four-digit mathematical tables for high school. Ed. 57th. - M., Education, 1990. S. 83.

3. Kruzhepov A.K., Rubanov A.T. Problem book on algebra and elementary functions. Textbook for secondary specialized educational institutions. - M., higher school, 1969.

4. Okunev A.K. Quadratic functions, equations and inequalities. A guide for the teacher. - M., Education, 1972.

5. Presman A.A. Solving a quadratic equation with a compass and straightedge. - M., Kvant, No. 4/72. S. 34.

6. Solomnik V.S., Milov P.I. Collection of questions and tasks in mathematics. Ed. - 4th, add. - M., Higher School, 1973.

7. Khudobin A.I. Collection of problems in algebra and elementary functions. A guide for the teacher. Ed. 2nd. - M., Education, 1970.

Shapovalova L.A. (station Egorlykskaya, MBOU ESOSH No. 11)

1. Mordkovich A.G. Algebra.8 class. Textbook for educational institutions / A.G. Mordkovich. No. 8622 / 0790 - M.: Mnemozina, 2013. No. 8622 / 0790 - 260 p.

2. Mordkovich A.G. Algebra.8 class. Taskbook for educational institutions / A.G. Mordkovich. No. 8622 / 0790 - M.: Mnemozina, 2013. No. 8622 / 0790 - 270 p.

3. Glazer G.I. History of mathematics in school No. 8622 / 0790 / G.I. Glaser. No. 8622 / 0790 - M .: Education, 1982. No. 8622 / 0790 - 340 p.

4. Gusev V.A. Maths. Reference materials / V.A. Gusev, A.G. Mordkovich. No. 8622 / 0790 - M .: Prosveshchenie, 1988. No. 8622 / 0790 - 372 p.

5. Bradis V.M. Four-digit mathematical tables for secondary school / V.M. Bradis. No. 8622 / 0790 - M .: Education, 1990. No. 8622 / 0790 - 83 p.

6. Vieta's theorem. No. 8622 / 0790 - Access mode: http://phizmat.org.ua/2009-10-27-13-31-30/817-stihi-o-francua-vieta/ Vieta's theorem (remote access resources (Internet)) . 01/20/2016.

7. Quadratic equations. No. 8622 / 0790 - Access mode: http://revolution.allbest.ru/pedagogics/00249255_0.html (remote access resources (Internet)). 01/20/2016.

The theory of equations occupies a leading place in algebra and mathematics in general. Its significance lies not only in its theoretical significance for the knowledge of natural laws, but also serves practical purposes. Most of life's problems come down to solving various types of equations, and more often these are equations of a quadratic form.

The school curriculum considers only 3 ways to solve them. In preparation for the upcoming exams, I became interested in other ways of these equations. Therefore, I chose the topic "10 Ways to Solve Quadratic Equations".

The relevance of this topic lies in the fact 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, which is also useful in solving more complex problems, including when passing exams.

The purpose of the work: to study various ways of solving quadratic equations, to learn how to solve quadratic equations.

Consider standard and non-standard methods for solving quadratic equations;

Identify the most convenient ways to solve quadratic equations;

Learn to solve quadratic equations in various ways.

Object of study: quadratic equations.

Subject of study: ways to solve quadratic equations.

Research methods:

Theoretical: study of literature on the research topic, study of thematic Internet resources;

Analysis of the received information;

Comparison of methods for solving quadratic equations for convenience and rationality.

Methods for solving quadratic equations

A quadratic equation is an equation of the form ax 2 + bx + c \u003d 0, where x is a variable, a, b and c are some numbers, while a? 0. The root of such an equation is the value of the variable that turns the square trinomial to zero, that is, the value that turns the quadratic equation into an identity. The coefficients of the quadratic equation have their own names: the coefficient a is called the first or senior, the coefficient b is called the second or the coefficient at x, c is called the free member of this equation.

A complete quadratic equation is one whose coefficients are all non-zero (a, b, c - 0).

A reduced quadratic equation is called, in which the leading coefficient is equal to one. Such an equation can be obtained by dividing the entire expression by the leading coefficient a: x 2 + px + q \u003d 0, p \u003d b / a, q \u003d c / a.

Incomplete quadratic equations are of three types:

1) ax 2 + c = 0, where c is 0;

2) ax 2 + bx = 0, where b - 0;

In the framework of this work, we will consider methods for solving only complete quadratic equations.

Solving quadratic equations by the general formula

To solve quadratic equations, the method of finding roots through the discriminant is used. The following formula is used to find the discriminant: D = b 2 - 4ac. After finding D, we use the formula to find the roots of the equation

It is worth noting that if:

D > 0 - the equation has two roots;

D \u003d 0 - the equation has one root;

D< 0 - уравнение не имеет корней.

An example of solving the equation in this way is shown in fig. 1(1.1).

Rice. 1. Practical part

Factoring the Left Side

To demonstrate the method, we solve the equation x 2 + 10x - 24 = 0.

Let's factorize the left side:

x 2 + 10x - 24 = x + 12x - 2x - 24 = = x(x + 12) - 2(x + 12) = (x + 12)(x - 2).

Therefore, the equation can be rewritten as:

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

Since the product is zero, then at least one of its factors is zero. Therefore, the left side of the equation vanishes at x = 2, and also at x = -12.

An example of solving the equation in this way is shown in fig. 1(1.2).

The selection of the full square is such an identity transformation in which the given trinomial is represented as (a ± b) 2 the sum or difference of the square of the binomial and some numeric or literal expression.

Let's solve the equation x 2 + 14x + 40 = 0.

Let us decompose the polynomial into factors using the full square method.

To apply the first formula, you need to get the expression

x2 + 14x + 49 = 0.

Therefore, we add and subtract the number 9 from the polynomial x 2 + 14x + 40 to select the full square

x 2 + 14x + 40 + 9 - 9 = 0

(x + 14x + 40 + 9) - 9 = 0

(x + 14x + 49) - 9 = 0

(x + 7) 2 - 9 = 0

Let's apply the formula "difference of squares" a2 - b2 = (a - b) (a + b)

(x + 7) 2 - 32 = 0

(x + 7 - 3)(x + 7 + 3) = 0

(x + 4)(x + 10) = 0

x + 4 = 0x + 10 = 0

x1 = - 4x2 = - 10

Answer: -4; - ten.

An example of solving the equation in this way is shown in fig. 1(1.3).

Solving equations using Vieta's theorem

To solve the complete quadratic equation according to the Vieta theorem, you need to divide the entire equation by the coefficient a. For the equation x 2 + px + q = 0, if x1 and x2 are its roots, the formulas are valid:

An example of solving the equation in this way is shown in fig. 1(1.4).

Solving Equations Using Properties of Coefficients

If the following condition is satisfied: a + c = b, then x1 = - 1; x2 = - s/a.

4x2 + 3x - 1 = 04 - 1 = 3

x1 = - 1x2 = - 1/4

If the following condition is met:

a + b + c = 0, then x1 = 1; x2 = s/a.

5x2 + 2x - 7 = 05 + 2 -7 = 0

An example of the impossibility of solving the equation in this way is shown in fig. 1(1.5).

Solving equations using the "transfer" method

The so-called "transfer" method makes it possible to reduce the solution of non-reduced and non-transformable equations to the form of reduced ones with integer coefficients by dividing them by the leading coefficient of equations to the solution of equations reduced with integer coefficients. It is as follows: multiply the equation ax 2 + bx + c = 0 by a.

We get: a 2 x2 + abx + aс = 0. Let's introduce a new variable y = ax. We get y 2 +by+ac = 0. The roots of this equation are y1 and y2. Therefore, x1 = y1/a; x2 = y2/a.

An example of solving the equation in this way is shown in fig. 1(1.6).

Let's solve the equation x 2 - 4x - 12 = 0.

Let's represent it as x 2 - 4x = 12.

On fig. 2 "depicts" the expression x - 4x, i.e. the area of a square with side x is subtracted twice from the area of a square with side 2. So x 2 - 4x + 4 is the area of a square with side x - 2.

After replacing x 2 - 4x = 12, we get

(x - 2)2 = 12 + 4

x - 2 = 4x - 2 = - 4

Answer: x1 = 6, x1 = - 2.

An example of solving the equation in this way is shown in fig. 1(1.7).

In the equation x 2 + px + q = 0, we move the second and third terms to the right side of the equation. We get: x 2 \u003d - px - q. Let's build graphs of functions

y = x 2 (parabola);

y = - qx - p (straight line).

It should be noted that:

If a line and a parabola can intersect at two points, the abscissas of the points of intersection are the roots of a quadratic equation;

If the line touches the parabola (only one common point), then the equation has one root;

If the line and the parabola do not have common points, i.e. a quadratic equation has no roots.

Solving an equation with a compass and straightedge

Let's solve the equation ax 2 + bx + c = 0:

1) construct points on the coordinate plane:

A(- b/2a; (a + c)/2a) is the center of the circle and B(0; 1)

2) Draw a circle r = AB

3) The abscissas of the points of intersection with the Ox axis are the roots of the original equation

It should be noted that:

If the radius of the circle is greater than the ordinate of the center (AB > AC, or R > (a + c) / 2a), the circle.

Crosses the x-axis at two points K(x1; 0) and N(x2; 0), where x1 and x2 are the roots of the quadratic equation x2 + bx + c = 0.

If the radius of the circle is equal to the ordinate of the center (AB \u003d AC, or R \u003d (a + c) / 2a), the circle touches the abscissa axis at the point C (x; 0), where x1 is the root of the quadratic equation.

If the radius of the circle is less than the ordinate of the center (AB< AС, или R < (a + c)/2a), окружность не имеет общих точек с осью абсцисс, в этом случае уравнение не имеет решения.

An example of solving the equation in this way is shown in fig. 1(1.9).

This is an old and now forgotten way of solving quadratic equations.

The nomogram gives the values of the positive roots of the equation z 2 + pz + q \u003d 0. If the equation has roots of different signs, then, having found a positive root from the nomogram, a negative one is found by subtracting the positive from - p.

Rice. 6. Type of monogram for solving the equation z 2 + pz + q = 0

In the case when both roots are negative, they take z = - t and find two positive roots t1 from the nomogram; t 2 equations t 2 + - pt + z = 0 and then z1 = - t1; z 2 \u003d - t2.

If the coefficients p and q are out of scale, perform the substitution z = kt and solve the equation using the nomogram

where k is taken in such a way that the inequalities

The form of the monogram for solving the equation z 2 + pz + q = 0 can be found in fig. 6.

"Pros" and "cons" of various solutions

The name of the method for solving quadratic equations
Solving quadratic equations by formula	Can be applied to all quadratic equations.	You need to learn the formulas.
Factoring the left side of the equation	It makes it possible to immediately see the roots of the equation.	It is necessary to correctly calculate the terms for grouping.
Full square selection method	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.
Solving equations using Vieta's theorem	A fairly easy way, makes it possible to immediately see the roots of the equation.	only whole roots are easily found.
Properties of the coefficients of a quadratic equation	Doesn't require much effort	Fits only some equations
Solution of equations by transfer method	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.	it is easy to find only whole roots.
Geometric way of solving quadratic equations	Visual way.	similar to the way to select a full square
Graphical solution of a quadratic equation	visual way	There may be inaccuracies in scheduling
Solving quadratic equations with a compass and straightedge	visual way	May not be accurate
Solving quadratic equations using a nomogram	Intuitive, easy to use.	Not always at hand there is a nomogram.

Conclusion

In the course of this research work, I managed to generalize and systematize the studied material on the chosen topic, to study various ways of solving quadratic equations, to learn how to solve quadratic equations in 10 ways. It should be noted that not all of them are convenient for solving, but each of them is interesting in its own way. From my point of view, the methods studied at school will be the most rational for use: 1.1. (according to the formula); 1.4. (according to the Vieta theorem); as well as method 1.5. (using the properties of the coefficients).

Summing up, we can conclude: quadratic equations play a huge role in mathematics. This knowledge can be useful to us not only at school and at the university, but also throughout our lives.

Bibliographic link

Ulevsky S.A. TEN WAYS OF SOLVING QUADRATIC EQUATIONS // Start in science. - 2016. - No. 1. - P. 75-79;
URL: http://science-start.ru/ru/article/view?id=15 (date of access: 12/30/2019).

slide 1

slide 2

Course objectives: Acquaintance with new methods for solving quadratic equations Deepening knowledge on the topic "Quadric Equations" Development of mathematical, intellectual abilities, research skills Creating conditions for self-realization of the individual

slide 3

Course objectives: To introduce students to new ways of solving quadratic equations To reinforce the ability to solve equations using known methods To introduce theorems that allow solving equations in non-standard ways To continue the formation of general educational skills, mathematical culture To promote the formation of interest in research activities To create conditions for students to realize and develop interest in the subject of mathematics Prepare students for the right choice of profile direction

slide 4

Contents of the program Topic 1. Introduction. 1 hour. Definition of a quadratic equation. Full and incomplete sq. equations. Methods for their solution. Questioning. Topic 2. Solution of sq. equations. Factoring method Full square selection method Solution sq. equations by formulas Solution square. equations by transfer method Solution sq. equations using t. Vieta Solution sq. equations using the coefficient Solution sq. equations in a graphical way Solution sq. equations using a compass and ruler Solution sq. equations in a geometric way Solution sq. equations using "nomograms"

slide 5

A bit of history ... Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. Quadratic Equations in Ancient Babylon. Quadratic equations in India. Quadratic equations in al-Khorezmi. Quadratic equations in Europe XIII - XVII centuries.

slide 6

Slide 7

Slide 8

Slide 9

slide 10

The famous French scientist Francois Viet (1540-1603) was a lawyer by profession. He devoted his free time to astronomy. Astronomy classes required knowledge of trigonometry and algebra. Viet took up these sciences and soon came to the conclusion that it was necessary to improve them, which he worked on for a number of years. Thanks to his work, algebra becomes the general science of algebraic equations based on literal calculus. Therefore, it became possible to express the properties of equations and their roots by general formulas.

slide 11

When doing the work, the following were noticed: The methods that I will use: Vieta's theorem Properties of the coefficients The "transfer" method Factorization of the left side into factors Graphical method The methods are interesting, but they take a lot of time and are not always convenient. Graphical method With the help of a nomogram Rulers and compasses Selection of a full square I bow to the scientists who discovered these methods and gave science an impetus for development in the topic “Solving quadratic equations”

slide 12

Factorization of the left side of the equation Let's solve the equation x2 + 10x - 24=0. Factoring the left side: x2 + 10x - 24= x2 + 12x -2x - 24= x(x + 12) - 2(x + 12)= (x + 12)(x - 2). (x + 12)(x - 2)=0 x + 12=0 or x - 2=0 x= -12 x= 2 Answer: x1= -12, x2 = 2. Solve equations: x2 - x=0 x2 + 2x=0 x2 - 81=0 x2 + 4x + 3=0 x2 + 2x - 3=0

slide 13

Full square selection method Solve the equation x2 + 6x - 7=0 x2 + 6x - 7=x2 + 2x3 + 32 - 32 - 7=(x-3)2 - 9- 7= (x-3)2 - 16 (x -3)2 -16=0 (x-3)2 =16 x-3=4 or x-3=-4 x=1 x=-7 Answer: x1=1, x2=-7. Solve equations: x2 - 8x+15=0 x2 +12x +20=0 x2 + 4x + 3=0 x2 + 2x - 2=0 x2 - 6x + 8=0

slide 14

Solution of quadratic equations according to the formula Basic formulas: If b is odd, then D= b2-4ac and x 1.2=, (if D> 0) If b is even, then D1= and x1.2=, (if D >0) Solve the equations: 2x2 - 5x + 2=0 6x2 + 5x +1=0 4x2 - 5x + 2=0 2x2 - 6x + 4=0 x2 - 18x +17=0 =

slide 15

Solution of equations by the transfer method Let's solve the equation ax2 +bx+c=0. Multiply both sides of the equation by a, we get a2 x2 +abx+ac=0. Let ax = y, whence x = y/a. Then U2 +buy+ac=0. Its roots are y1 and y2. Finally x1 = y1/a, x1 = y2/a. Let's solve the equation 2x2 -11x + 15=0. Let's transfer coefficient 2 to the free term: Y2 -11y+30=0. According to the Vieta theorem, y1 =5 and y2 =6. x1 = 5/2 and x2 = 6/2 x1 = 2.5 and x2 = 3 Answer: x1 = 2.5, x2 = 3 Solve the equation: 2x2 -9x +9=0 10x2 -11x + 3=0 3x2 + 11x +6=0 6x2 +5x - 6=0 3x2 +1x - 4=0

slide 16

Solving equations using Vieta's theorem Let's solve the equation x2 +10x-24=0. Since x1 * x2 \u003d -24 x1 + x2 \u003d -10, then 24 \u003d 2 * 12, but -10 \u003d -12 + 2, then x1 \u003d -12 x2 \u003d 2 Answer: x1 \u003d 2, x2 \u003d -12. Solve equations: x2 - 7x - 30 =0 x2 +2x - 15=0 x2 - 7x + 6=0 3x2 - 5x + 2=0 5x2 + 4x - 9=0

slide 17

Properties of coefficients of a quadratic equation If a+b+c=0, then x2 = 1, x2 = c/a 7= 0 Let's solve the equation 2x2 + 3x +1= 0 1 + 6 - 7 = 0, so x1=1, x2 = -7/1=-7. 2 - 3+1=0, so x1= - 1, x2 = -1/2 Answer: x1=1, x2 = -7. Answer: x1=-1, x2=-1/2. Solve equations: 5x2 - 7x +2 =0 Solve equations: 5x2 - 7x -12 =0 11x2 +25x - 36=0 11x2 +25x +14=0 345x2 -137x -208=0 3x2 +5x +2=0 3x2 + 5x - 8=0 5x2 + 4x - 1=0 5x2 + 4x - 9=0 x2 + 4x +3=0

Kopyevskaya rural secondary school

10 Ways to Solve Quadratic Equations

Head: Patrikeeva Galina Anatolyevna,

mathematic teacher

s.Kopyevo, 2007

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

1.2 How Diophantus compiled and solved quadratic equations

1.3 Quadratic equations in India

1.4 Quadratic equations in al-Khwarizmi

1.5 Quadratic equations in Europe XIII - XVII centuries

1.6 About Vieta's theorem

2. Methods for solving quadratic equations

Conclusion

Literature

1. History of the development of quadratic equations

1.1 Quadratic equations in ancient Babylon

Applying modern algebraic notation, we can say that in their cuneiform texts there are, in addition to incomplete ones, such, for example, complete quadratic equations:

X 2 + X = ¾; X 2 - X = 14,5

Despite the high level of development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.

1.2 How Diophantus compiled and solved quadratic equations.

When compiling equations, Diophantus skillfully chooses unknowns to simplify the solution.

Here, for example, is one of his tasks.

Task 11."Find two numbers knowing that their sum is 20 and their product is 96"

Hence the equation:

(10 + x)(10 - x) = 96

100 - x 2 = 96

x 2 - 4 = 0 (1)

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

If we solve this problem by choosing one of the desired numbers as the unknown, then we will come to the solution of the equation

y(20 - y) = 96,

y 2 - 20y + 96 = 0. (2)

1.3 Quadratic equations in India

ah 2+bx = c, a > 0. (1)

In equation (1), the coefficients, except for a, can also be negative. Brahmagupta's rule essentially coincides with ours.

Here is one of the problems of the famous Indian mathematician of the XII century. Bhaskara.

Task 13.

“A frisky flock of monkeys And twelve in vines ...

Having eaten power, had fun. They began to jump, hanging ...

Part eight of them in a square How many monkeys were there,

Having fun in the meadow. You tell me, in this flock?

Bhaskara's solution indicates that he knew about the two-valuedness of the roots of quadratic equations (Fig. 3).

The equation corresponding to problem 13 is:

(x/8) 2 + 12 = x

Bhaskara writes under the guise of:

x 2 - 64x = -768

and, to complete the left side of this equation to a square, he adds to both sides 32 2 , getting then:

x 2 - 64x + 32 2 = -768 + 1024,

(x - 32) 2 = 256,

x - 32 = ± 16,

x 1 = 16, x 2 = 48.

1.4 Quadratic equations in al-Khorezmi

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

1) "Squares are equal to roots", i.e. ax 2 + c =bX.

2) "Squares are equal to number", i.e. ax 2 = s.

3) "The roots are equal to the number", i.e. ah = s.

4) "Squares and numbers are equal to roots", i.e. ax 2 + c =bX.

5) "Squares and roots are equal to the number", i.e. ah 2+bx= s.

6) "Roots and numbers are equal to squares", i.e.bx+ c \u003d ax 2.

Task 14.“The square and the number 21 are equal to 10 roots. Find the root" (assuming the root of the equation x 2 + 21 = 10x).

Treatise al - Khorezmi is the first book that has come down to us, in which the classification of quadratic equations is systematically stated and formulas for their solution are given.

1.5 Quadratic equations in EuropeXIII - XVIIcenturies

The general rule for solving quadratic equations reduced to a single canonical form:

x 2+bx= with,

for all possible combinations of signs of the coefficients b, With was formulated in Europe only in 1544 by M. Stiefel.

1.6 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 multiplied by A - A 2 , equals BD, then A equals AT and equal D».

(a +b)x - x 2 =ab,

x 2 - (a +b)x + ab = 0,

x 1 = a, x 2 =b.

Expressing the relationship between the roots and coefficients of equations by general formulas written using symbols, Viet established uniformity in the methods of solving equations. However, the symbolism of Vieta is still far from its modern form. He did not recognize negative numbers, and therefore, when solving equations, he considered only cases where all roots are positive.

2. Methods for solving quadratic equations

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Quadratic equations are widely used in solving trigonometric, exponential, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve quadratic equations from school (grade 8) until graduation.