Ministry of Education of the Russian Federation

Municipal educational institution

"Secondary school No. 22"

Quadratic and higher order equations

Completed:

Pupils of 8 "B" class

Kuznetsov Evgeniy and Rudi Alexey

Supervisor:

Zenina Alevtina Dmitrievna

mathematics teacher

Introduction

1.1 Equations in Ancient Babylon

1.2 Arab equations

1.3 Equations in India

Chapter 2. Theory of quadratic equations and higher order equations

2.1 Basic concepts

2.2 Formulas for even coefficient at x

2.3 Vieta's theorem

2.4 Quadratic equations of a particular nature

2.5 Vieta’s theorem for polynomials (equations) of higher degrees

2.6 Equations reducible to quadratic (biquadratic)

2.7 Study of biquadratic equations

2.8 Cordano formulas

2.9 Symmetric equations of the third degree

2.10 Reciprocal equations

2.11 Horner circuit

Conclusion

List of used literature

Appendix 1

Appendix 2

Appendix 3

Introduction

Equations occupy a leading place in the school algebra course. More time is devoted to their study than to any other topic. Indeed, equations not only have important theoretical significance, but also serve purely practical purposes. The overwhelming number of problems about spatial forms and quantitative relationships in the real world comes down to solving various types of equations. By mastering ways to solve them, we find answers to various questions from science and technology (transport, agriculture, industry, communications, etc.).

In this essay I would like to display formulas and methods for solving various equations. For this purpose, equations are given that are not studied in the school curriculum. These are mainly equations of a particular nature and equations of higher degrees. To expand on this topic, proofs of these formulas are given.

Objectives of our essay:

Improve equation solving skills

Develop new ways to solve equations

Learn some new ways and formulas to solve these equations.

The object of study is elementary algebra. The object of study is equations. The choice of this topic was based on the fact that equations are included both in the primary curriculum and in each subsequent grade of secondary schools, lyceums, and colleges. Many geometric problems, problems in physics, chemistry and biology are solved using equations. The equations were solved twenty-five centuries ago. They are still being created today - both for use in the educational process, and for competitive exams at universities, for olympiads of the highest level.

Chapter 1. History of quadratic and higher order equations

1.1 Equations in Ancient Babylon

Algebra arose in connection with solving various problems using equations. Typically, problems require finding one or more unknowns, while knowing the results of some actions performed on the desired and given quantities. Such problems come down to solving one or a system of several equations, to finding the required ones using algebraic operations on given quantities. Algebra studies the general properties of operations on quantities.

Some algebraic techniques for solving linear and quadratic equations were known 4000 years ago 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 plots and land works of a military nature, as well as with the development of astronomy and mathematics itself. As mentioned earlier, quadratic equations were able to be solved around 2000 BC by the Babylonians. Using modern algebraic notation, we can say that both incomplete and complete quadratic equations occur in their cuneiform texts.

The rule for solving these equations, set out in Babylonian texts, essentially coincides with modern ones, but it is not known how the Babylonians arrived at this 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 the high level of development of algebra in Babylon, the cuneiform texts lack the concept of a negative number and general methods for solving a quadratic equation.

1.2 Arab equations

Some methods for solving both quadratic and higher-order equations were developed by the Arabs. Thus, the famous Arab mathematician Al-Khorezmi in his book “Al-Jabar” described many ways to solve various equations. Their peculiarity was that Al-Khorezmi used complex radicals to find the roots (solutions) of equations. The need to solve such equations was needed in questions about the division of inheritance.

1.3 Equations in India

Quadratic equations were also solved in India. Problems on quadratic equations are found already in the astronomical treatise “Aryabhattiam”, compiled in 499 by the Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (7th century), set out a general rule for solving quadratic equations reduced to a single conic form:

aх² + bx= c, where a > 0

In this equation, the coefficients, except a, can be negative. Brahmagupta's rule is essentially the same as ours.

In ancient India, public competitions in solving difficult problems were common. One of the old Indian books says the following about such competitions: “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.

Various equations, both quadratic and equations of higher degrees, were solved by our distant ancestors. These equations were solved in very different and distant countries. The need for equations was great. The equations were used in construction, in military affairs, and in everyday situations.

Chapter 2. Quadratic equations and higher order equations

2.1 Basic concepts

A quadratic equation is an equation of the form

where coefficients a, b, c are any real numbers, and a ≠ 0.

A quadratic equation is called reduced if its leading coefficient is 1.

Example :

x 2 + 2x + 6 = 0.

A quadratic equation is called unreduced if the leading coefficient is different from 1.

Example :

2x 2 + 8x + 3 = 0.

A complete quadratic equation is a quadratic equation in which all three terms are present, in other words, it is an equation in which the coefficients b and c are non-zero.

Example :

3x 2 + 4x + 2 = 0.

An incomplete quadratic equation is a quadratic equation in which at least one coefficient b, c is equal to zero.

Thus, there are three types of incomplete quadratic equations:

1) ax² = 0 (has two coinciding roots x = 0).

2) ax² + bx = 0 (has two roots x 1 = 0 and x 2 = -)

Example :

x 1 = 0, x 2 = -5.

Answer: x 1 =0, x 2 = -5.

If -<0 - уравнение не имеет корней.

Example :

Answer: The equation has no roots.

If –> 0, then x 1,2 = ±

Example :

Answer: x 1.2 =±

Any quadratic equation can be solved using the discriminant (b² - 4ac). Usually the expression b² - 4ac is denoted by the letter D and is called the discriminant of the quadratic equation ax² + bx + c = 0 (or the discriminant of the quadratic three term ax² + bx + c)

Example :

x 2 +14x – 23 = 0

D = b 2 – 4ac = 144 + 92 = 256

x 2 =

Answer: x 1 = 1, x 2 = - 15.

Depending on the discriminant, the equation may or may not have a solution.

1) If D< 0, то не имеет решения.

2) If D = 0, then the equation has two coinciding solutions x 1,2 =

3) If D > 0, then it has two solutions found according to the formula:

x 1.2 =

2.2 Formulas for even coefficient at x

We are accustomed to the fact that the roots of a quadratic equation

ax² + bx + c = 0 are found by the formula

x 1.2 =

But mathematicians will never pass up the opportunity to make their calculations easier. They found that this formula can be simplified in the case where the coefficient b is b = 2k, in particular if b is an even number.

In fact, let the coefficient b of the quadratic equation ax² + bx + c = 0 be b = 2k. Substituting the number 2k instead of b into our formula, we get:

So, the roots of the quadratic equation ax² + 2kx + c = 0 can be calculated using the formula:

x 1.2 =

Example :

5x 2 - 2x + 1 = 0

The advantage of this formula is that it is not the number b that is squared, but its half; it is not 4ac that is subtracted from this square, but simply ac, and, finally, that the denominator contains not 2a, but simply a.

If the quadratic equation is given, then our formula will look like this:

Example :

x 2 – 4x + 3 = 0

Answer: x 1 = 3, x 2 = 1.

2.3 Vieta's theorem

A very interesting property of the roots of a quadratic equation was discovered by the French mathematician Francois Viète. This property was called Vieta's theorem:

So that the numbers x 1 and x 2 are the roots of the equation:

ax² + bx + c = 0

it is necessary and sufficient to fulfill the equality

x 1 + x 2 = -b/a and x 1 x 2 = c/a

Vieta's theorem allows us to judge the signs and absolute value of a quadratic equation

x² + bx + c = 0

1. If b>0, c>0 then both roots are negative.

2. If b<0, c>0 then both roots are positive.

3. If b>0, c<0 то уравнение имеет корни разных знаков, причём отрицательный корень по абсолютной величине больше положительного.

4. If b<0, c<0 то уравнение имеет корни разных знаков, причём отрицательный корень по абсолютной величине меньше положительного.

2.4 Quadratic equations of a particular nature

1) If a + b + c = 0 in the equation ax² + bx + c = 0, then

x 1 = 1, and x 2 = .

Proof :

In the equation ax² + bx + c = 0, its roots

x 1.2 = (1).

Let us represent b from the equality a + b + c = 0

Let's substitute this expression into formula (1):

If we consider the two roots of the equation separately, we get:

1) x 1 =

2) x 2 =

It follows: x 1 = 1, and x 2 =.

1. Example :

2x² - 3x + 1 = 0

a = 2, b = -3, c = 1.

a + b + c = 0, therefore

2. Example :

418x² - 1254x + 836 = 0

This example is very difficult to solve using a discriminant, but knowing the above formula it can be easily solved.

a = 418, b = -1254, c = 836.

x 1 = 1 x 2 = 2

2) If a - b + c = 0, in the equation ax² + bx + c = 0, then:

x 1 =-1, and x 2 =-.

Proof :

Consider the equation ax² + bx + c = 0, it follows that:

x 1.2 = (2).

Let us represent b from the equality a - b + c = 0

b = a + c, substitute into formula (2):

We get two expressions:

1) x 1 =

2) x 2 =

This formula is similar to the previous one, but it is also important because... Examples of this type are common.

1) Example :

2x² + 3x + 1 = 0

a = 2, b = 3, c = 1.

a - b + c = 0, therefore

2)Example :

Answer: x 1 = -1; x 2 = -

3) Method “ transfers ”

The roots of the quadratic equations y² + by + ac = 0 and ax² + bx + c = 0 are related by the following relations:

x 1 = and x 2 =

Proof :

a) Consider the equation ax² + bx + c = 0

x 1.2 = =

b) Consider the equation y² + by + ac = 0

y 1,2 =

Note that the discriminants of both solutions are equal; let us compare the roots of these two equations. They differ from each other by a leading factor, the roots of the first equation are less than the roots of the second by a. Using Vieta's theorem and the above rule, it is not difficult to solve various equations.

Example :

We have an arbitrary quadratic equation

10x² - 11x + 3 = 0

Let's transform this equation according to the given rule

y² - 11y + 30 = 0

We obtain the reduced quadratic equation, which can be solved quite easily using Vieta’s theorem.

Let y 1 and y 2 be the roots of the equation y² - 11y + 30 = 0

y 1 y 2 = 30 y 1 = 6

y 1 + y 2 = 11 y 2 = 5

Knowing that the roots of these equations differ from each other by a, then

x 1 = 6/10 = 0.6

x 2 = 5/10 = 0.5

In some cases, it is convenient to first solve not the given equation ax² + bx + c = 0, but the reduced y² + by + ac = 0, which is obtained from the given “transfer” coefficient a, and then divide the found roots by a to find the original equation.

2.5 Vieta formula for polynomials (equations) of higher degrees

The formulas derived by Viète for quadratic equations are also true for polynomials of higher degrees.

Let the polynomial

P(x) = a 0 x n + a 1 x n -1 + … +a n

Has n different roots x 1, x 2..., x n.

In this case, it has a factorization of the form:

a 0 x n + a 1 x n-1 +…+ a n = a 0 (x – x 1)(x – x 2)…(x – x n)

Let's divide both sides of this equality by a 0 ≠ 0 and open the brackets in the first part. We get the equality:

x n + ()x n -1 + … + () = x n – (x 1 + x 2 + … + x n) x n -1 + (x 1 x 2 + x 2 x 3 + … + x n -1 x n)x n - 2 + … +(-1) n x 1 x 2 … x n

But two polynomials are identically equal if and only if the coefficients of the same powers are equal. It follows that the equality

x 1 + x 2 + … + x n = -

x 1 x 2 + x 2 x 3 + … + x n -1 x n =

x 1 x 2 … x n = (-1) n

For example, for polynomials of third degree

a 0 x³ + a 1 x² + a 2 x + a 3

We have identities

x 1 + x 2 + x 3 = -

x 1 x 2 + x 1 x 3 + x 2 x 3 =

x 1 x 2 x 3 = -

As for quadratic equations, this formula is called Vieta's formulas. The left-hand sides of these formulas are symmetric polynomials from the roots x 1, x 2 ..., x n of this equation, and the right-hand sides are expressed through the coefficient of the polynomial.

2.6 Equations reducible to quadratic (biquadratic)

Equations of the fourth degree are reduced to quadratic equations:

ax 4 + bx 2 + c = 0,

called biquadratic, and a ≠ 0.

It is enough to put x 2 = y in this equation, therefore,

ay² + by + c = 0

let's find the roots of the resulting quadratic equation

y 1,2 =

To immediately find the roots x 1, x 2, x 3, x 4, replace y with x and get

x² =

x 1,2,3,4 = .

If a fourth degree equation has x 1, then it also has a root x 2 = -x 1,

If has x 3, then x 4 = - x 3. The sum of the roots of such an equation is zero.

Example :

2x 4 - 9x² + 4 = 0

Let's substitute the equation into the formula for the roots of biquadratic equations:

x 1,2,3,4 = ,

knowing that x 1 = -x 2, and x 3 = -x 4, then:

x 3.4 =

Answer: x 1.2 = ±2; x 1.2 =

2.7 Study of biquadratic equations

Let's take the biquadratic equation

ax 4 + bx 2 + c = 0,

where a, b, c are real numbers, and a > 0. By introducing the auxiliary unknown y = x², we examine the roots of this equation and enter the results into the table (see Appendix No. 1)

2.8 Cardano formula

If we use modern symbolism, the derivation of the Cardano formula can look like this:

x =

This formula determines the roots of a general third-degree equation:

ax 3 + 3bx 2 + 3cx + d = 0.

This formula is very cumbersome and complex (it contains several complex radicals). It will not always apply, because... very difficult to fill out.

2.9 Symmetric equations of the third degree

Symmetric equations of the third degree are equations of the form

ax³ + bx² +bx + a = 0 ( 1 )

ax³ + bx² - bx – a = 0 ( 2 )

where a and b are given numbers, with a¹0.

Let us show how the equation ( 1 ).

ax³ + bx² + bx + a = a(x³ + 1) + bx(x + 1) = a(x + 1) (x² - x + 1) + bx(x + 1) = (x + 1) (ax² +(b – a)x + a).

We find that the equation ( 1 ) is equivalent to the equation

(x + 1) (ax² +(b – a)x + a) = 0.

This means that its roots will be the roots of the equation

ax² +(b – a)x + a = 0

and number x = -1

the equation ( 2 )

ax³ + bx² - bx - a = a(x³ - 1) + bx(x - 1) = a(x - 1) (x² + x + 1) + bx(x - 1) = (x - 1) (ax 2 + ax + a + bx) = (x - 1) (ax² +(b + a)x + a).

1) Example :

2x³ + 3x² - 3x – 2 = 0

It is clear that x 1 = 1, and

x 2 and x 3 roots of the equation 2x² + 5x + 2 = 0,

Let's find them through the discriminant:

x 1.2 =

x 2 = -, x 3 = -2

2) Example :

5x³ + 21x² + 21x + 5 = 0

It is clear that x 1 = -1, and

x 2 and x 3 roots of the equation 5x² + 26x + 5 = 0,

Let's find them through the discriminant:

x 1.2 =

x 2 = -5, x 3 = -0.2.

2.10 Reciprocal equations

Reciprocal equation – algebraic equation

a 0 x n + a 1 x n – 1 + … + a n – 1 x + a n =0,

in which a k = a n – k, where k = 0, 1, 2 …n, and a ≠ 0.

The problem of finding the roots of a reciprocal equation is reduced to the problem of finding solutions to an algebraic equation of a lower degree. The term reciprocal equations was introduced by L. Euler.

Fourth degree equation of the form:

ax 4 + bx 3 + cx 2 + bmx + am² = 0, (a ≠ 0).

Reducing this equation to the form

a (x² + m²/x²) + b(x + m/x) + c = 0, and y = x + m/x and y² - 2m = x² + m²/x²,

from where the equation is reduced to quadratic

ay² + by + (c-2am) = 0.

3x 4 + 5x 3 – 14x 2 – 10x + 12 = 0

Dividing it by x 2 gives the equivalent equation

3x 2 + 5x – 14 – 5 ×, or

Where and

3(y 2 - 4) + 5y – 14 = 0, whence

y 1 = y 2 = -2, therefore

And where from

Answer: x 1.2 = x 3.4 = .

A special case of reciprocal equations are symmetric equations. We talked about symmetric equations of the third degree earlier, but there are symmetric equations of the fourth degree.

Symmetric equations of the fourth degree.

1) If m = 1, then this is a symmetric equation of the first kind, having the form

ax 4 + bx 3 + cx 2 + bx + a = 0 and solved by a new substitution

2) If m = -1, then this is a symmetric equation of the second kind, having the form

ax 4 + bx 3 + cx 2 - bx + a = 0 and solved by a new substitution

2.11 Horner scheme

To divide polynomials, the “division by angle” rule, or Horner’s scheme, is used . For this purpose, polynomials are arranged in descending degrees X and find the leading term of the quotient Q(x) from the condition that when multiplied by the leading term of the divisor D(x), the leading term of the dividend P(x) is obtained. The found term of the quotient is multiplied, then by the divisor and subtracted from the dividend. The leading term of the quotient is determined from the condition that, when multiplied by the leading term of the divisor, it gives the leading term of the difference polynomial, etc. The process continues until the degree of the difference is less than the degree of the divisor (see Appendix No. 2).

In the case of equations R = 0, this algorithm is replaced by Horner's scheme.

Example :

x 3 + 4x 2 + x – 6 = 0

Find the divisors of the free term ±1; ± 2; ± 3; ± 6.

Let's denote the left side of the equation by f(x). Obviously, f(1) = 0, x1 = 1. Divide f(x) by x – 1. (see Appendix No. 3)

x 3 + 4x 2 + x – 6 = (x – 1) (x 2 + 5x + 6)

We denote the last factor by Q(x). We solve the equation Q(x) = 0.

x 2.3 =

Answer : 1; -2; -3.

In this chapter, we have given some formulas for solving various equations. Most of these formulas for solving partial equations. These properties are very convenient because it is much easier to solve equations using a separate formula for this equation, rather than using the general principle. We have provided a proof and several examples for each method.

Conclusion

The first chapter examined the history of the emergence of quadratic equations and higher order equations. Various equations were solved more than 25 centuries ago. Many methods for solving such equations were created in Babylon, India. There has been and will continue to be a need for equations.

The second chapter provides various ways to solve (find roots) quadratic equations and higher order equations. Basically, these are methods for solving equations of a particular nature, that is, for each group of equations united by some common properties or type, a special rule is given that applies only to this group of equations. This method (selecting your own formula for each equation) is much easier than finding roots through a discriminant.

In this abstract, all goals have been achieved and the main tasks have been completed, new, previously unknown formulas have been proven and learned. We worked through many examples of examples before including them in the abstract, so we already have an idea of how to solve some equations. Each solution will be useful to us in further studies. This essay helped to classify old knowledge and learn new ones.

References

1. Vilenkin N.Ya. “Algebra for 8th grade”, M., 1995.

2. Galitsky M.L. “Collection of problems in algebra”, M. 2002.

3. Daan-Dalmedico D. “Paths and labyrinths”, M., 1986.

4. Zvavich L.I. “Algebra 8th grade”, M., 2002.

5. Kushnir I.A. “Equations”, Kyiv 1996.

6. Savin Yu.P. “Encyclopedic Dictionary of a Young Mathematician”, M., 1985.

7. Mordkovich A.G. “Algebra 8th grade”, M., 2003.

8. Khudobin A.I. “Collection of problems in algebra”, M., 1973.

9. Sharygin I.F. “Optional course in algebra”, M., 1989.

Appendix 1

Study of biquadratic equations

C	b		Conclusions
C	b		On the roots of the auxiliary equation ay² +by+c=0	About the roots of this equation a(x²)² +bx² +c=0
C< 0	b- any real number		y< 0 ; y > 0 1 2	x = ±Öy
C > 0	b<0	D > 0		x = ±Öy
		D=0	y > 0	x = ±Öy
		D< 0	No roots	No roots
	b ≥ 0			No roots
			No roots	No roots
			y > 0 ; y< 0 1 2	x = ±Öy
C=0	b > 0		y = 0	x = 0
	b = 0		y = 0	x = 0
	b< 0		y = 0	x = 0

Appendix 2

Dividing a polynomial into a polynomial using a corner

A 0	a 1	a 2	...	a n	c
+
b 0 c	b 1 c	…	b n-1 c
B 0	b 1	b 2	…	b n	= R (remainder)

Appendix 3

Horner scheme

Root
	1	4	1	-6	1
x 1 = 1
demolishing	5	6	0
	1	1×1 +4 = 5	5×1 + 1 = 6	6×1 – 6 = 0
root
x 1 = 1

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 plots and with excavation work 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 years before our faith. Using modern algebraic notation, we can say that in their cuneiform texts, in addition to incomplete ones, there are, for example, complete quadratic equations: The rule for solving these equations, set out in the Babylonian texts, coincides with the modern one, but it is not known how the Babylonians got there rules. Almost all cuneiform texts found so far present only problems with solutions laid out in the form of recipes, with no indication as to how they were found. Despite the high level of development of algebra in Babylonia, the cuneiform texts lack the concept of a negative number and general methods for solving quadratic equations.

How Diophantus composed and solved quadratic equations “Find two numbers, knowing that their sum is 20 and their product is 96.” Diophantus reasons as follows: from the conditions of the problem it follows that the required numbers are not equal, because if they were equal, then their product would not be 96, but 100. Thus, one of them would be more than half of their sum, i.e. 10+X, the other is less, i.e. 10-X. The difference between them is 2X Hence X=2. One of the required numbers is 12, the other is 8. The solution X = -2 does not exist for Diophantus, since Greek mathematics knew only positive numbers. EQUATION: or:

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, outlined the general rule for solving quadratic equations reduced to a single canonical form: ax ² +bx=c, a>0 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? The equation corresponding to the problem is: Baskara writes under the form: Completed the left side to a square, 0 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? The equation corresponding to the problem: Baskara writes under the form: Completed the left side to a square,">

Quadratic equations in Ancient Asia This is how the Central Asian scientist al-Khwarizmi solved this equation: He wrote: “The rule is: double the number of roots, x = 2x 5, you get five in this problem, multiply 5 by this equal to it, it will be twenty-five, 5 ·5=25 add this to thirty-nine, there will be 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 3 this will be the root of the square, which you I was looking." What about the second root? The second root was not found, since negative numbers were not known. x x = 39

Quadratic equations in Europe XIII-XVII centuries. The general rule for solving quadratic equations reduced to a single canonical form x2+inx+c=0 was formulated in Europe only in 1544 by Stiefel. Formulas for solving quadratic equations in Europe were first stated in 1202 by the Italian mathematician Leonard Fibonacci. The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète recognized only positive roots. Only in the 17th century. thanks to the works of 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 Vieta, was formulated by him for the first time in 1591 as follows: “If B + D multiplied by A-A is equal to BD, then A is equal to B and equals D." To understand Vieta, one should remember that A, like any vowel letter, meant the unknown (our x), while the vowels B, D are coefficients for the unknown. In the language of modern algebra, the above Vieta formulation 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 equal to q, that is, x 1 + x 2 = -p, x 1 x 2 = q (the sum of the roots of the above 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).

The factorization method brings a general quadratic equation to the form: A(x)·B(x)=0, where A(x) and B(x) are polynomials with respect to x. Goal: Taking the common factor out of brackets; Using abbreviated multiplication formulas; Grouping method. Methods: Example:

Roots of a quadratic equation: If D>0, If D 0, If D"> 0, If D"> 0, If D" title="Roots of a quadratic equation: If D>0, If D"> title="Roots of a quadratic equation: If D>0, If D"> !}

X 1 and x 2 – roots of the equation Solving equations using Vieta’s theorem X 2 + 3X – 10 = 0 X 1 · X 2 = – 10, which means the roots have different signs X 1 + X 2 = – 3, which means the root has a larger modulus - negative By selection we find the roots: X 1 = – 5, X 2 = 2 For example:

0, by the theorem inverse to Vieta’s theorem, we obtain the roots: 5;6, then we return to the roots of the original equation: 2.5; 3. Answer: 2.5; 3. Solution of the equation" title="Solve the equation: 2x 2 - 11x +15 = 0. Let’s transfer the coefficient 2 to the free term of 2 - 11y +30= 0. D>0, according to the theorem inverse to Vieta’s theorem, we get the roots: 5;6, then we return to the roots of the original equation: 2.5; 3. Answer: 2.5; 3. Solution of the equation." class="link_thumb"> 14 !} Solve the equation: 2x x +15 = 0. Let’s transfer the coefficient 2 to the free term y y +30= 0. D>0, according to the theorem inverse to Vieta’s theorem, we get the roots: 5;6, then we return to the roots of the original equation: 2, 5; 3. Answer: 2.5; 3. Solving equations using the “throw” method 0, by the theorem inverse to Vieta’s theorem, we obtain the roots: 5;6, then we return to the roots of the original equation: 2.5; 3. Answer: 2.5; 3. Solution of the equation "> 0, according to the theorem inverse to Vieta's theorem, we obtain the roots: 5;6, then we return to the roots of the original equation: 2.5; 3. Answer: 2.5; 3. Solution of the equations using the "transfer" method." > 0, by the theorem inverse to Vieta’s theorem, we obtain the roots: 5;6, then we return to the roots of the original equation: 2.5; 3. Answer: 2.5; 3. Solution of the equation" title="Solve the equation: 2x 2 - 11x +15 = 0. Let’s transfer the coefficient 2 to the free term of 2 - 11y +30= 0. D>0, according to the theorem inverse to Vieta’s theorem, we get the roots: 5;6, then we return to the roots of the original equation: 2.5; 3. Answer: 2.5; 3. Solution of the equation."> title="Solve the equation: 2x 2 - 11x +15 = 0. Let's transfer the coefficient 2 to the free term y 2 - 11y +30= 0. D>0, according to the theorem inverse to Vieta's theorem, we get the roots: 5;6, then we return to the roots of the original equations: 2.5; 3. Answer: 2.5; 3. Solution of the equation"> !}

If in a quadratic equation a+b+c=0, then one of the roots is equal to 1, and the second by Vieta’s theorem is equal to the second by Vieta’s theorem is equal to If in a quadratic equation a+c=b, then one of the roots is equal to (-1), and the second according to Vieta’s theorem is equal to Example: Properties of the coefficients of the quadratic equation 137x x – 157 = 0. a = 137, b = 20, c = a + b+ c = – 157 =0. x 1 = 1, Answer: 1; 137x x – 157 = 0. a = 137, b = 20, c = a + b+ c = – 157 =0. x 1 = 1, Answer: 1;

Graphical method for solving a quadratic equation Without using formulas, a quadratic equation can be solved graphically. Let's solve the equation. To do this, we'll build two graphs: X Y X 01 Y012 Answer: The abscissas of the points of intersection 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. 1)y=x2 2)y=x+1

Solving quadratic equations using a nomogram This is an old and undeservedly forgotten method of solving quadratic equations, placed on p. 83 “Four-digit mathematical tables” Bradis V.M. Table XXII. Nomogram for solving an equation This nomogram allows, without solving a quadratic equation, to determine the roots of the equation from its coefficients. For the equation, the nomogram gives the roots

Geometric method for solving quadratic equations In ancient times, when geometry was more developed than algebra, quadratic equations were solved not algebraically, but geometrically. But, for example, how the ancient Greeks solved the equation: or Expressions and geometrically represent the same square, and the original equation is the same equation. Where do we get what, or

Conclusion These solution methods deserve attention, since they are not all reflected in school mathematics textbooks; mastering these techniques will help students save time and solve equations effectively; the need for a quick solution is due to the use of a test system for entrance exams;

Research work

On the topic

"Methods of solving quadratic equations"

Completed:
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 that mathematicians are distinguished by non-standard 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 in 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 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.

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

The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète 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 took on its modern form.

TURNS OUT:

Problems involving quadratic equations were encountered as early as 499.

In ancient India, public competitions in solving difficult problems were common - OLYMPIADS .

Ministry of Education and Science of the Republic of Tatarstan

Municipal budgetary educational institution

"Usad Secondary School

Vysokogorsky municipal district of the Republic of Tatarstan"

Research work:

"Story emergencesquare equations»

Completed by: Andreeva Ekaterina,

8B grade student

Scientific supervisor:

Pozharskaya Tatyana Leonidovna,

math teacher

Introduction

Who wants to limit themselves to the present?

without knowledge of the past,

he will never understand him.

G.V. Leibniz

Equations occupy a leading place in the school mathematics course, but none of the types of equations have found such wide application as quadratic equations.

People were able to solve equations of the second degree or quadratic equations back in Ancient Babylon in the 2nd millennium BC. Problems leading to quadratic equations are discussed in many ancient mathematical manuscripts and treatises. And nowadays, many problems in algebra, geometry, and physics are also solved using quadratic equations. By solving them, people find answers to various questions of science and technology.

Target This study is to study the history of the emergence of quadratic equations.

To achieve this goal, it is necessary to solve the following tasks:

Study scientific literature on the topic.
Trace the history of the emergence of quadratic equations.

Object of study: quadratic equations.

Subject of research: history of the emergence of quadratic equations.

Relevance of the topic :

People have been solving quadratic equations since ancient times. I wanted to know the history of quadratic equations.
There is no information in school textbooks about the history of quadratic equations.

Research methods:

Working with educational and popular science literature.
Observation, comparison, analysis.

The scientific value of the work, in my opinion, lies in the fact that this material may be of interest to schoolchildren who are interested in mathematics, and to teachers in extracurricular classes.

Quadratic equations in Ancient Babylon.

In Ancient Babylon, the need to solve equations not only of the first, but also of the second degree was caused by the need to solve problems associated with finding the areas of land and with excavation work of a military nature, as well as with the development of astronomy and mathematics itself.

Using 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 = 14.5

The rule for solving these equations, set out in the Babylonian texts, essentially coincides with the modern one, but it is not known how the Babylonians arrived at this 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 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.

An example taken from one of the clay tablets from this period.

“The area of the sum of two squares is 1000. The side of one of the squares is the side of the other square reduced by 10. What are the sides of the squares?”

This leads to equations whose solution reduces to solving a quadratic equation with a positive root.

In reality, the solution in the cuneiform text is limited, as in all Eastern problems, to a simple listing of the steps of calculation required to solve the quadratic equation:

“Square 10; this gives 100; subtract 100 from 1000; this gives 900" etc

How Diophantus composed and solved quadratic equations

Diophantus presents one of the most difficult mysteries in the history of science. He was one of the most original ancient Greek mathematicians, Diophantus of Alexandria, whose works were of great importance for algebra and number theory. Neither the year of birth nor the date of death of Diophantus has yet been clarified. The period of time when Diophantus could have lived is half a millennium! It is believed that he lived in the 3rd century AD. But the place of residence of Diophantus is well known - this is the famous Alexandria, the center of scientific thought of the Hellenistic world.

Of the works of Diophantus, the most important is Arithmetic, of which 13 books only 6 have survived to this day.

Diophantus' 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.

When composing equations, Diophantus skillfully selects unknowns to simplify the solution.

Here, for example, is one of his tasks.

Task: “Find two numbers, knowing that their sum is 20 and their product is 96”

Diophantus reasons as follows: from the conditions of the problem it follows that the required numbers are not equal, since if they were equal, then their product would not be equal to 96, but to 100. Thus, one of them will be more than half of their sum, i.e. . 10 + x, the other is less, i.e. 10's. The difference between them 2x.

Hence the equation:

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

100 - x 2 = 96

x 2 - 4 = 0 (1)

From here x = 2. One of the required numbers is equal to 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 required numbers as the unknown, then we will come to a solution to the equation

y(20 - y) = 96,

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

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

Quadratic equations from Diophantus' arithmetic:

12x 2 +x = 1
630x 2 +73x=6.

Even in ancient times, India was famous for its knowledge in the field of astronomy, grammar and other sciences.

Indian scientists have achieved the greatest success in the field mathematicians. They were the founders of arithmetic and algebra, in the development of which they went further than the Greeks.

Problems on quadratic equations are found already in the astronomical treatise “Aryabhattiam”, compiled in 499. Indian mathematician and astronomer Aryabhatta. Another Indian scientist, Brahmagupta (VII century), outlined the general rule for solving quadratic equations reduced to a single canonical form: ax 2 + bx = c, a> 0.

Brahmagupta's rule is essentially the same as ours.
Public competitions were common in ancient India
in solving difficult problems. One of the old Indian books says the following about such competitions: “Just 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.

There are part eight of them squared,

I was having fun in the clearing.

And twelve along the vines...

They started jumping, hanging...

How many monkeys were there?

Tell me, in this pack?

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

The equation corresponding to the problem

Bhaskara writes in the form x 2 - 64x = -768 and, to complete the left side of this equation to a square, add 32 2 to both sides, then obtaining:

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

x 1 =16, x 2 =48.

Quadratic equations in China (1st millennium BC).

The first Chinese written monuments that have reached us date back to the Shang era (XVIII-XII centuries BC). And already on the fortune telling bones of the 14th century. BC BC, found in Henan, the designations of numbers have been preserved. But the real flourishing of science began after the 12th century. BC e. China was conquered by the Zhou nomads. During these years, Chinese mathematics and astronomy emerged and reached amazing heights. The first accurate calendars and mathematics textbooks appeared. Unfortunately, the “extermination of books” by Emperor Qin Shi Huang (Shi Huangdi) did not allow the early books to reach us, but they most likely formed the basis for subsequent works.

“Mathematics in Nine Books” is the first mathematical work from a number of classics in ancient China, a remarkable monument of ancient China during the Early Han Dynasty (206 BC - 7 AD). This essay contains varied and rich mathematical material, including quadratic equations.

Chinese challenge: “There is a reservoir with a side of 10 cm. In its center there is a reed that protrudes above the water by 1 hour. If you pull the reed towards the shore, it will just touch it. The question is: what is the depth of the water and what is the length of the reeds?

(x+1) 2 =x 2 +5 2,

x 2 +2x+1= x 2 +25,

Answer: 12chi; 13pm

Quadratic equations by al-Khwarizmi

“I have compiled a short book on the calculus of algebra and almukabala, containing simple and complex questions of arithmetic, because this is necessary for people.” Al-Khorezmi Mohammed ben Musa.

Al-Khorezmi (Uzbekistan) is best known for his “Book of Completion and Opposition” (“Al-kitab al-mukhtasar fi hisab al-jabr wa-l-mukabala”), from the name of which the word “algebra” is derived. This 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.

In the theoretical part of his treatise, al-Khorezmi gives a Classification of equations of the 1st and 2nd degrees and identifies six of their types:

1) “Squares are equal to roots,” i.e. ax 2 = bx. (example:)

2) “Squares are equal to numbers,” i.e. ax 2 = s. (example:)

3) “The roots are equal to the number,” i.e. ax = c. (example:)

4) “Squares and numbers are equal to roots,” i.e. ax 2 + c = bx. (example:)

5) “Squares and roots are equal to the number,” i.e. ax 2 + bx = c.

6) “Roots and numbers are equal to squares,” i.e. bx + c == ax 2. (example:)

For al-Khwarizmi, who avoided the use of negative numbers, the terms of each of these equations are addends and not subtractables. 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 using the techniques of al-jabr and al-mukabal. His decision, of course, does not completely coincide with ours. Not to mention 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 until the 17th century, does not take into account the zero solution, probably because in specific practical it doesn't matter in tasks. 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 give an example.

“The square and the number 21 are equal to 10 roots. Find the root"(implying the root of the equation x 2 + 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, what remains is 4. Take the root from 4, you get 2. Subtract 2 from 5, you get 3, this will be the desired root . Or add 2 to 5, which gives 7, this is also a root.”

Al-Khwarizmi's famous equation: "A square and ten roots equal 39." x 2 + 10x= 39 (IX century). In his treatise he writes: “The rule is this: double the number of roots, you get five in this problem. Add that to thirty-nine, it becomes sixty-four. Take the root of this, it becomes eight, and subtract half the number of roots from this, i.e. five, that leaves three: this will be the root of the square you were looking for.”

Quadratic equations in Europe in the 12th-17th centuries.

Forms for solving quadratic equations following the model of Al-Khwarizmi in Europe were first set forth in the “Book of the Abacus,” written in 1202. Italian mathematician Leonard Fibonacci. The author independently developed some new algebraic examples of solving problems and was the first in Europe to approach the introduction of negative numbers.

This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many problems from this book were used in almost all European textbooks of the 14th-17th centuries. The general rule for solving quadratic equations reduced to the form x 2 + bх = с for all possible combinations of signs and coefficients b, c was formulated in Europe in 1544 by M. Stiefel.

The derivation of the formula for solving a quadratic equation in general form is available from Viète, but Viète recognized only positive roots. Italian mathematicians Tartaglia, Cardano, Bombelli were among the first in the 16th century. In addition to positive ones, negative roots are also 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 takes on a modern form.

Conclusion.

Quadratic equations are the foundation on which the majestic edifice of algebra rests. Various equations, both quadratic and equations of higher degrees, were solved by our distant ancestors. These equations were solved in very different and distant countries. The need for equations was great. The equations were used in construction, in military affairs, and in everyday situations.

Nowadays, the ability to solve quadratic equations is necessary for everyone. The ability to quickly, rationally and correctly solve quadratic equations makes it easier to complete many topics in a mathematics course. Quadratic equations are solved not only in mathematics lessons, but also in physics, chemistry, and computer science lessons. Most practical problems in the real world also come down to solving quadratic equations.

Literature

Bashmakova I. G. Diophantus and Diophantine equations. M.: Nauka, 1972.
Berezkina E.I. Mathematics of ancient China - M.: Nauka, 1980
Pichurin L.F. Behind the pages of an algebra textbook: Book. for students

7-9 grades school average - M.: Education, 1990

Glazer G.I. History of mathematics in school VII - VIII grades. Manual for teachers. - M.: Education, 1982.

Representatives of various civilizations: Ancient Egypt, Ancient Babylon, Ancient Greece, Ancient India, Ancient China, the Medieval East, Europe mastered techniques for 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. Mathematics in Nine Books explains how to find the 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 of 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.