“The history of the emergence of quadratic equations. From the history of quadratic equations

29.09.2019

Research

On the topic

"Methods of solution quadratic equations »

Performed:
group 8 "G" class

Head of work:
Benkovskaya Maria Mikhailovna

Goals and objectives of the project.

1. Show that mathematics, like any other science, has its own unsolved mysteries.
2. Emphasize what makes mathematicians different out-of-the-box thinking. And sometimes the ingenuity and intuition of a good mathematician simply amazes you!
3. Show that the very attempt to solve quadratic equations contributed to the development of new concepts and ideas in mathematics.
4. Learn to work with various sources of information.
5. Continue research work mathematics

Research stages

1. History of the emergence of quadratic equations.

2. Definition of a quadratic equation and its types.

3. Solving quadratic equations using the discriminant formula.

4. Francois Viète and his theorem.

5. Properties of coefficients for quickly finding the roots of a quadratic equation.

6. Practical orientation.

Through equations, theorems

I solved a lot of problems.

(Chaucer, English poet, Middle Ages.)

stage. The history of the emergence of quadratic equations.

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 were able to solve quadratic equations about 2000 BC. The rule for solving these equations, set out in the Babylonian texts, essentially coincides with modern ones, but it is not known how the Babylonians came to find the rule. Almost all cuneiform texts found so far provide only problems with solutions laid out in the form of recipes, with no indication as to how they were found.

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

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

Problems on quadratic equations are found already in the astronomical treatises “Aryabhattiam”, compiled in 499. Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), outlined general rule solving quadratic equations reduced to a unified canonical form:

Al-Khwarizmi's algebraic treatise gives a classification of linear and quadratic equations. The author lists 6 types of equations. For al-Khwarizmi, who did not know negative numbers, the terms of each equation are addends, not subtractables. At the same time, equations that do not have positive solutions are obviously not taken into account; when solving an incomplete quadratic equation, al-Khorezmi, like all scientists until the 17th century, does not take into account the zero solution.

Al-Khwarizmi's treatise is the first book that has come down to us, which systematically sets out the classification of quadratic equations and formulas for their solution.

Formulas for solving quadratic equations modeled after al-Khwarizmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work is distinguished by its completeness and clarity of presentation. The author independently developed some new algebraic methods for solving problems, 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 problems from the “Book of Abacus” were transferred to almost all European textbooks of the 16th - 17th and partly of the 18th centuries.

General rule for solving quadratic equations reduced to a single equation canonical form for all possible combinations of signs coefficients b,c was formulated in Europe only in 1544 by M. Stiefel.

Derivation of the formula for solving a quadratic equation in general view Viet has it, but Viet recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century to take into account not only positive, but also negative roots. Only in the 17th century, thanks to the works of Girrard, Descartes, Newton and other scientists, the method of solving quadratic equations was adopted modern look.

TURNS OUT:

Problems involving quadratic equations were encountered as early as 499.

IN Ancient India public competitions in solving difficult problems were common - OLYMPIADS .

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How Diophantus composed and solved quadratic equations. Hence the equation: (10+x)(10 -x) =96 or: 100 - x2 =96 x2 - 4=0 (1) The solution x = -2 does not exist for Diophantus, since Greek mathematics knew only positive numbers.

Src="https://present5.com/presentation/137369579_55459696/image-4.jpg" alt="Quadratic equations in India. ax2 + bx = c, a>0. (1)"> Квадратные уравнения в Индии. ах2 + bх = с, а>0. (1)!}

Quadratic equations in al-Khorezmi. 1) “Squares are equal roots,” i.e. ax2 + c = bx. 2) “Squares are equal to numbers,” i.e. ax2 = c. 3) “The roots are equal to the number,” i.e. ax = c. 4) “Squares and numbers are equal to roots,” i.e. ax2 + c = bx. 5) “Squares and roots are equal to the number”, i.e. ax2 + bx = c. 6) “Roots and numbers are equal to squares,” i.e. bx + c = ax2.

Quadratic equations in Europe in the 13th and 17th centuries. x2 +bx = c, for all possible combinations of signs of the coefficients b, c was formulated in Europe only in 1544 by M. Stiefel.

About Vieta's theorem. “If B + D times A - A 2 equals BD, then A equals B and equals D.” In the language of modern algebra, the above Vieta formulation means: if (a + b)x - x2 = ab, i.e. x2 - (a + b)x + ab = 0, then x1 = a, x2 = b.

Methods for solving quadratic equations. 1. METHOD: Factoring the left side of the equation. Let's solve the equation x2 + 10 x - 24 = 0. Let's factorize the left side: x2 + 10 x - 24 = x2 + 12 x - 24 = x(x + 12) - 2(x + 12) = (x + 12) (x - 2). Therefore, the equation can be rewritten as follows: (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 becomes zero at x = 2, and also at x = - 12. This means that the number 2 and - 12 are the roots of the equation x2 + 10 x - 24 = 0.

2. METHOD: Full square extraction method. Let's solve the equation x2 + 6 x - 7 = 0. Select on the left side perfect square. To do this, we write the expression x2 + 6 x in the following form: x2 + 6 x = 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 a complete square, you need to add 32, since x2 + 2 x 3 + 32 = (x + 3)2. Now we transform the left side of the equation x2 + 6 x - 7 = 0, adding to it and subtracting 32. We have: x2 + 6 x - 7 = x2 + 2 x 3 + 32 - 7 = (x + 3)2 - 9 - 7 = (x + 3)2 - 16. Thus, given equation can be written as follows: (x + 3)2 - 16 = 0, (x + 3)2 = 16. Therefore, x + 3 - 4 = 0, x1 = 1, or x + 3 = -4, x2 = -7 .

3. METHOD: Solving quadratic equations using the formula. Let's multiply both sides of the equation ax2 + bx + c = 0, a ≠ 0 by 4 a and sequentially we have: 4 a 2 x2 + 4 abx + 4 ac = 0, ((2 ax)2 + 2 ax b + b 2) - b 2 + 4 ac = 0, (2 ax + b)2 = b 2 - 4 ac, 2 ax + b = ± √ b 2 - 4 ac, 2 ax = - b ± √ b 2 - 4 ac,

4. METHOD: Solving equations using Vieta's theorem. As is known, the reduced quadratic equation has the form x2 + px + c = 0. (1) Its roots satisfy Vieta’s theorem, which for a = 1 has the form x 1 x 2 = q, x 1 + x 2 = - p a) x 2 – 3 x + 2 = 0; x 1 = 2 and x 2 = 1, since q = 2 > 0 and p = - 3 0 and p = 8 > 0. b) x 2 + 4 x – 5 = 0; x 1 = - 5 and x 2 = 1, since q= - 5 0; x 2 – 8 x – 9 = 0; x 1 = 9 and x 2 = - 1, since q = - 9

5. METHOD: Solving equations using the “throwing” method. Consider the quadratic equation ax2 + bx + c = 0, where a ≠ 0. Multiplying both sides by a, we obtain the equation a 2 x2 + abx + ac = 0. Let ax = y, whence x = y/a; then we arrive at the equation y2 + by + ac = 0, which is equivalent to the given one. We find its roots y1 and y2 using Vieta’s theorem. We finally get x1 = y1/a and x1 = y2/a.

Example. Let's solve the equation 2 x2 – 11 x + 15 = 0. Solution. Let’s “throw” coefficient 2 to the free term, as a result we get the equation y2 – 11 y + 30 = 0. According to Vieta’s theorem, y1 = 5 y2 = 6 x1 = 5/2 x 2 = 6/2 Answer: 2, 5; 3. x 1 = 2. 5 x 2 = 3.

6. METHOD: Properties of coefficients of a quadratic equation. A. Let the quadratic equation ax2 + bx + c = 0 be given, where a ≠ 0. 1) If a + b + c = 0 (i.e., the sum of the coefficients is zero), then x1 = 1, x2 = c/ A. Proof. Dividing both sides of the equation by a ≠ 0, we obtain 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, whence b = a + c. Thus, x 1 + x 2 = - a + b/a= -1 – c/a, x 1 x 2 = - 1 (- c/a), i.e. x1 = -1 and x2 = c/ a, which is what needed to be proved.

B. If the second coefficient b = 2 k – even number, then the formula for the roots of B. The above equation x2 + px + q = 0 coincides with a general equation in which a = 1, b = p and c = q. Therefore, for the reduced quadratic equation, the root formula is

7. METHOD: Graphic solution of a quadratic equation. If in the equation x2 + px + q = 0 we move the second and third terms to the right side, we get x2 = - px - q. Let's build graphs of the dependence y = x2 and y = - px - q.

Example 1) Let's solve graphically the equation x2 - 3 x - 4 = 0 (Fig. 2). Solution. Let's write the equation in the form x2 = 3 x + 4. Construct a parabola y = x2 and a straight line y = 3 x + 4. The straight line y = 3 x + 4 can be constructed using two points M (0; 4) and N (3; 13) . Answer: x1 = - 1; x2 = 4

8. METHOD: Solving quadratic equations using a compass and ruler. finding the roots of a square compass and ruler (Fig. 5). equations Then, by the secant theorem, we have OB OD = OA OC, whence OC = OB OD/ OA = x1 x2/ 1 = c/a. ax2 + bx + c = 0 using

Src="https://present5.com/presentation/137369579_55459696/image-19.jpg" alt="1) The radius of the circle is greater than the ordinate of the center (AS > SK, or R > a +"> 1) Радиус окружности больше ординаты центра (AS > SK, или R > a + c/2 a), окружность пересекает ось Ох в двух точках (6, а рис.) В(х1; 0) и D(х2; 0), где х1 и х2 - корни квадратного уравнения ах2 + bх + с = 0. 2) Радиус окружности равен ординате центра (AS = SB, или R = a + c/2 a), окружность касается оси Ох (рис. 6, б) в точке В(х1; 0), где х1 - корень квадратного уравнения. 3) Радиус окружности меньше ординаты центра окружность не имеет общих точек с осью абсцисс (рис. 6, в), в этом случае уравнение не имеет решения.!}

9. METHOD: Solving quadratic equations using a nomogram. z 2 + pz + q = 0. The curvilinear scale of the nomogram is constructed 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 obtain the proportion

Examples. 1) For the equation z 2 - 9 z + 8 = 0, the nomogram gives the roots z 1 = 8, 0 and z 2 = 1, 0 (Fig. 12). 2) Using a nomogram, we solve the equation 2 z 2 - 9 z + 2 = 0. Divide the coefficients of this equation by 2, we get the equation z 2 - 4, 5 z + 1 = 0. The nomogram gives the roots z 1 = 4 and z 2 = 0, 5. 3) For the equation z 2 - 25 z + 66 = 0, the coefficients p and q are outside the scale, we perform the substitution z = 5 t, we get the equation t 2 - 5 t + 2, 64 = 0, which we solve using nomograms and get t 1 = 0.6 and t 2 = 4. 4, from which z 1 = 5 t 1 = 3. 0 and z 2 = 5 t 2 = 22. 0.

10. METHOD: Geometric method for solving quadratic equations. Examples. 1) Let’s solve the equation x2 + 10 x = 39. In the original, this problem is formulated as follows: “The square and ten roots are equal to 39” (Fig. 15). For the required side x of the original square we obtain

y2 + 6 y - 16 = 0. The solution is shown in Fig. 16, where y2 + 6 y = 16, or y2 + 6 y + 9 = 16 + 9. Solution. The expressions y2 + 6 y + 9 and 16 + 9 geometrically represent the same square, and the original equation y2 + 6 y - 16 + 9 - 9 = 0 is the same equation. From this we get that y + 3 = ± 5, or y1 = 2, y2 = - 8 (Fig. 16).

Kovalchuk Kirill

The project "Quadratic equations through centuries and countries" introduces students to mathematics scientists, whose discoveries are the basis of scientific and technological progress, develops interest in mathematics as a subject based on familiarity with historical material, broadens students' horizons, and stimulates them cognitive activity and creativity.

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Project work of an 8th grade student of Municipal Educational Institution Secondary School No. 17 in the village of Borisovka Kirill Kovalchuk Supervisor G.V. Mulyukova

Quadratic equations through centuries and countries

Project goal: To introduce students to mathematics scientists, whose discoveries are the basis of scientific and technological progress. Show the significance of the works of scientists for the development of geometry and physics.??????????? Demonstrate application clearly scientific discoveries in life. Develop interest in mathematics as a subject based on familiarity with historical material. Broaden the horizons of students, stimulate their cognitive activity and creativity

The need to solve equations not only of the first degree, but also of the second, in ancient times was caused by the need to solve problems related to finding the areas of land plots, with the development of astronomy and mathematics itself. Quadratic equations could be solved around 2000 BC. e. Babylonians. The rules for solving these equations set out in the Babylonian texts are essentially the same as modern ones, but these texts lack the concept of a negative number and general methods for solving quadratic equations.

. (c. 365 - 300 BC) - ancient Greek mathematician, author of the first theoretical treatises on mathematics that have reached us. Euclid, or Euclid

Euclid Beginnings Where the Nile merges with the sea, In the ancient hot land of the Pyramids The Greek mathematician lived - the Knowledgeable, Wise Euclid. He studied geometry, he taught geometry. He wrote a great work. This book's name is "Beginnings."

Euclid 3rd century BC Euclid solved quadratic equations using a geometric method. Here is one of the problems from the ancient Greek treatise: “There is a city with a border in the form of a square with a side of unknown size, in the center of each side there is a gate. There is a pillar at a distance of 20bu (1bu=1.6m) from the northern gate. If you go from south gate 14b straight ahead, then turn west and go another 1775b, you can see a pillar. The question is: which side of the city border? »

To determine the unknown side of the square, we obtain the quadratic equation x ² +(k+l)x-2kd =0. IN in this case the equation has the form x ² +34x-71000=0, from where x=250bu l x d k

Quadratic equations in India Problems on quadratic equations are also found in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta, set out a general rule for solving quadratic equations reduced to a single canonical form: ax ² +bx=c , a>0 Public competitions in solving difficult problems were common in Ancient India. One of the old Indian books says the following about such competitions: “As the sun eclipses the stars with its brilliance, so learned man eclipse the glory of another in popular assemblies by proposing and solving algebraic problems.”

One of the problems of the famous Indian mathematician of the 12th century Bhaskara A flock of frisky monkeys, having eaten to their hearts' content, had fun. Part eight of them in the square I was having fun in the clearing. And twelve on the vines... They began to jump while hanging... How many monkeys were there, tell me, in this flock?

Solution. () 2 +12 = x, x 2 - 64x +768 = 0, a = 1, b = -64, c = 768, then D = (-64) 2 -4 1 768 = 1024 > 0. X 1, 2 = , x 1 = 48, x 2 = 16. Answer: There were 16 or 48 monkeys. Let's solve it.

The formula for the roots of a quadratic equation has been “rediscovered” several times. One of the first derivations of this formula that has survived to this day belongs to the Indian mathematician Brahmagupta. The Central Asian scientist al-Khwarizmi, in his treatise “Kitab al-jerb wal-mukabala,” obtained this formula by isolating a complete square.

How did al-Khorezmi solve this equation? He wrote: “The rule is this: double the number of roots, x = 2x · 5 in this problem you get five, multiply 5 by this equal to it, it becomes twenty-five, 5 · 5 = 25 add this to thirty-nine, 25 + 39 becomes sixty-four , 64 take the root from this, it will be eight, 8 and subtract from this half the number of roots, i.e. five, 8-5 will remain three - this and 3 Will be the root of the square that you were looking for." What about the second root? The second root was not found, since negative numbers were not known. x 2 +10 x = 39

Quadratic equations in Europe 13-17 centuries. Formulas for solving quadratic equations modeled after al-Khwarizmi in Europe were first set forth in the “Book of Abacus,” written in 1202 by the Italian mathematician Leonardo Fibonacci. This voluminous work, which reflects the influence of mathematics from both Islamic countries and Ancient Greece, is distinguished by both completeness and clarity of presentation. The author independently developed some new algebraic solutions problems 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 problems from the Book of Abacus were used in almost all European textbooks of the 16th and 17th centuries. and partially 18.

Francois Viète - the greatest mathematician of the 16th century

Before F. Vieta, solving a quadratic equation was carried out according to its own rules in the form of very long verbal arguments and descriptions, rather cumbersome actions. They couldn’t even write down the equation itself; this required a rather long and complex verbal description. He coined the term "coefficient". He proposed that the required quantities be denoted by vowels, and the data by consonants. Thanks to Vieta's symbolism, we can write the quadratic equation in the form: ax 2 + bx + c =0. Theorem: 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. Despite the fact that this theorem is called “Vieta’s Theorem,” it was known before him, and he only transformed it into its modern form. Vieta is called the "father of algebra"

Humanity has come a long way from ignorance to knowledge, continuously replacing incomplete and imperfect knowledge with more and more complete and perfect knowledge along the way. Final word

Us living in beginning of XXI century, attracts antiquity. In our ancestors, we notice first of all what they lack from a modern point of view, and usually do not notice what we ourselves lack in comparison with them.

Let us not forget about them...

Thank you for your attention!

Representatives of various civilizations: Ancient Egypt, Ancient Babylon, Ancient Greece, Ancient India, Ancient China, Medieval East, Europe mastered the techniques of solving quadratic equations.

For the first time, the mathematicians of Ancient Egypt were able to solve a quadratic equation. One of the mathematical papyri contains the following problem:

“Find the sides of a field shaped like a rectangle if its area is 12 and its lengths are equal to its width.” “The length of the field is 4,” the papyrus states.

Millennia passed, and negative numbers entered algebra. Solving the equation x²= 16, we get two numbers: 4, –4.

Of course, in the Egyptian problem we would take X = 4, since the length of the field can only be a positive quantity.

Sources that have reached us indicate that ancient scientists had some general techniques for solving problems with unknown quantities. The rule for solving quadratic equations set forth in the Babylonian texts is essentially the same as the modern one, but it is not known how the Babylonians “got this far.” But in almost all papyri and cuneiform texts found, only problems with solutions are given. The authors only occasionally supplied their numerical calculations with skimpy comments such as: “Look!”, “Do this!”, “You found the right one!”

The Greek mathematician Diophantus composed and solved quadratic equations. His Arithmetic does not contain a systematic presentation of algebra, but it contains a systematic series of problems, accompanied by explanations and solved by constructing equations of various degrees.

Problems on composing quadratic equations are found already in the astronomical treatise “Aria-bhatiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta.

Another Indian scientist Brahmagupta (7th century) outlined the general rule for solving quadratic equations of the form ax² + bx = c.

In ancient India, public competitions in solving difficult problems were common. One of the old Indian books about such competitions says the following: “As the sun outshines the stars with its brilliance, so a learned man will outshine the glory of another in public assemblies, proposing and solving algebraic problems.” Problems were often presented in poetic form.

This is one of the problems of the famous Indian mathematician of the 12th century. Bhaskars:

A flock of frisky monkeys

Having eaten to my heart's content, I had fun.

Part eight of them were playing in the clearing in the square.

And twelve on the vines... began to jump, hanging...

How many monkeys were there?

Tell me, in this pack?

Bhaskara's solution shows that he knew that the roots of quadratic equations are two-valued.

The most ancient Chinese mathematical texts that have come down to us date back to the end of the 1st century. BC. In the II century. BC. Mathematics in Nine Books was written. Later, in the 7th century, it was included in the collection “Ten Classical Treatises,” which was studied for many centuries. The treatise "Mathematics in Nine Books" explains how to extract Square root using the formula for the square of the sum of two numbers.

The method was called “tian-yuan” (literally “heavenly element”) - this is how the Chinese designated an unknown quantity.

The first manual for solving problems that became widely known was the work of the Baghdad scientist of the 9th century. Muhammad bin Musa al-Khwarizmi. The word “al-jabr” over time turned into the well-known word “algebra”, and al-Khorezmi’s work itself became the starting point in the development of the science of solving equations. Al-Khwarizmi's algebraic treatise gives a classification of linear and quadratic equations. The author counts six types of equations, expressing them as follows:

-squares equal roots, that is, ah ² = bх;

-squares equal number, that is, ah ² = s;

-the roots are equal to the number, that is, ax = c;

-squares and numbers are equal to roots, that is, ah ²+ с = bх;

-squares and roots are equal to the number, that is, ah ² + bх = с;

-roots and numbers are equal to squares, that is, bx + c = ax ²;

Al-Khwarizmi's treatise is the first book that has come down to us, which systematically sets out the classification of quadratic equations and gives formulas for their solution.

Formulas for solving quadratic equations modeled after al-Khwarizmi in Europe were first set forth in the Book of Abacus, written in 1202 by the Italian mathematician Leonardo Fibonacci. The author independently developed some new algebraic examples solving problems and was the first in Europe to introduce 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 problems from the “Book of Abacus” were included in almost all European textbooks of the 16th-17th centuries. and partly of the 18th century.

General rule for solving quadratic equations reduced to a single canonical form x ² + bх = с, for all possible combinations of signs of the coefficients b and с was formulated in Europe only in 1544 by M. Stiefel.

Vieta has a general derivation of the formula for solving a quadratic equation, but he also recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. In addition to positive and negative roots, they are taken into account. Only in the 17th century, thanks to the works of Girard, Descartes, Newton and other scientists, the method of solving quadratic equations took on its modern form.

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“The history of the emergence of quadratic equations. From the history of quadratic equations

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