10 methods for solving a square equation. Methods for solving square equations

In the school course of mathematics are studied formulas of the roots of square equations, with which any square equations can be solved. However, there are other ways to solve square equations that allow many equations very quickly and rationally. There are ten ways to solve square equations. In detail in my work, I disassembled each of them.

1. Method : Decomposition of the left part of the factory equation.

Resolving equation

x 2 + 10x - 24 \u003d 0.

Spatulate the left side of the factors:

x 2 + 10x - 24 \u003d x 2 + 12x - 2x - 24 \u003d x (x + 12) - 2 (x + 12) \u003d (x + 12) (x - 2).

Consequently, the equation can be rewritten so:

(x + 12) (x - 2) \u003d 0

Since the product is zero, at least one of its factor is zero. Therefore, the left part of the equation is drawn by zero x \u003d 2.as well as x \u003d - 12. This means that the number 2 and - 12 are roots equations x 2 + 10x - 24 \u003d 0.

2. Method : Method of allocation of a full square.

Resolving equation x 2 + 6x - 7 \u003d 0.

We highlight the full square in the left side.

To do this, write the expression x 2 + 6x in the following form:

x 2 + 6x \u003d x 2 + 2 x 3.

In the resulting expression, the first term is the square of the number X, and the second - a double product X by 3. On this to get a full square, you need to add 3 2, since

x 2 +. 2 x 3 + 3 2 \u003d (x + 3) 2.

We now transform the left part of the equation

x 2 + 6x - 7 \u003d 0,

adding it and subtracting 3 2. We have:

x 2 + 6x - 7 \u003dx 2 +. 2 x 3 + 3 2 - 3 2 - 7 \u003d (x + 3) 2 - 9 - 7 \u003d (x + 3) 2 - 16.

Thus, this equation can be written as:

(x + 3) 2 - 16 \u003d 0, (x + 3) 2 \u003d 16.

Hence, x + 3 - 4 \u003d 0, x 1 \u003d 1, or x + 3 \u003d -4, x 2 \u003d -7.

3. Method : Solution of square equations by the formula.

Multiply both parts of the equation

ah 2 +.b.x + c \u003d 0, and ≠ 0

on 4a and consistently we have:

4a 2 x 2 + 4ab.x + 4as \u003d 0,

((2AH) 2 + 2Ahb. + b. 2 ) - b. 2 + 4 aC = 0,

(2AX + B) 2 \u003d B 2 - 4AC,

2AX + B \u003d ± √ B 2 - 4AC,

2AX \u003d - B ± √ B 2 - 4AC,

Examples.

but) Resolving equation: 4x 2 + 7x + 3 \u003d 0.

a \u003d 4,b. \u003d 7, C \u003d 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. for

b. 2 - 4 aC >0 , the equation ah 2 +.b.x + c \u003d 0 It has two different roots.

b) Resolving equation: 4x 2 - 4x + 1 \u003d 0,

a \u003d 4,b. \u003d - 4, C \u003d 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 equation

ah 2 +.b.x + c \u003d 0 has the only root

in) Resolving equation: 2x 2 + 3x + 4 \u003d 0,

a \u003d 2,b. \u003d 3, C \u003d 4,D. = b. 2 - 4 aC = 3 2 - 4 2 4 = 9 - 32 = - 13 , D. < 0.

This equation has no root.

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

the equation ah 2 +.b.x + c \u003d 0 It does not have roots.

Formula (1) of the roots of the square equation ah 2 +.b.x + c \u003d 0 Allows you to find roots anyone Square equation (if any), including the above and incomplete. Valid formula (1) is expressed as: the roots of the square equation are equal to the fraction, the numerator of which is equal to the second coefficient taken with the opposite sign, plus minus the root square from the square of this coefficient without the still-standing product of the first coefficient to the free member, and the denominator has a double-coefficient.

4. Method: Solving equations using the Vieta Theorem.

As you know, the reduced square equation has the form

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

His roots satisfy the Vieta theorem, which a \u003d 1. Has appearance

x. 1 x. 2 = q.,

x. 1 + x. 2 = - p.

From here you can draw the following conclusions (according to the coefficients p and q you can predict the signs of the roots).

a) if a consolidated member q. The given equation (1) is positive ( q. > 0 ), the equation has two identical root signs and is the envy of the second coefficient p.. If a r< 0 then both root are negative if r< 0 , both root are positive.

For example,

x. 2 – 3 x. + 2 = 0; x. 1 = 2 and x. 2 = 1, as q. = 2 > 0 and p. = - 3 < 0;

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

b) if a free member q. The given equation (1) is negative ( q. < 0 ), the equation has two different on the root sign, and the root larger in the module 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, as q.= - 5 < 0 and p. = 4 > 0;

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

5. Method: Solving equations by the method of "transit".

Consider a square equation

ah 2 +.b.x + c \u003d 0,where A ≠ 0.

Multiplying both parts by a, we get the equation

a 2 x 2 + ab.x + ac \u003d 0.

Let be ah \u003d uFrom! x \u003d y / a; then come to the equation

in 2 +.by + ac \u003d 0,

equivalent to this. His roots in 1.and w. 2 We will find with the help of the Vieta theorem.

Finally get

x 1 \u003d y 1 / aand x 1 \u003d y 2 / a.

With this method coefficient but multiplied by a free member, as if "moves" to him, so it is called howling "transit". This method is used when you can easily find the roots of the equation using the Vieta theorem and, most importantly, when the discriminant is an accurate square.

Example.

Resolving equation 2x 2 - 11x + 15 \u003d 0.

Decision. "We will transfer" a coefficient 2 to a free member, as a result we obtain the equation

in 2 - 11th + 30 \u003d 0.

According to Viet theorem

in 1 \u003d 5 x 1 \u003d 5/2x. 1 = 2,5

U 2 \u003d 6x. 2 = 6/2 x. 2 = 3.

Answer: 2.5; 3.

6. Method: Properties of the coefficients of the square equation.

BUT. Let the square equation be given

ah 2 +.b.x + c \u003d 0,where A ≠ 0.

1) if, a +b. + C \u003d 0 (i.e., the sum of the coefficients is zero), then x 1 \u003d 1,

x 2 \u003d s / a.

Evidence. We divide both parts of the equation on a ≠ 0, we obtain the reduced square equation

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

According to Viet theorem

x. 1 + x. 2 = - b./ a.,

x. 1 x. 2 = 1 c./ a..

By condition but -b. + C \u003d 0,from b. \u003d a + s.In this way,

x 1 + x 2 \u003d -but + b / a \u003d -1 - C / A,

x 1 x 2 \u003d - 1 (- C / A),

those. x 1 \u003d -1 and x 2 \u003dc./ a.that m required to prove.

Examples.

1) solve equation 345x 2 - 137x - 208 \u003d 0.

Decision.As a +.b. + C \u003d 0 (345 - 137 - 208 \u003d 0),that

x 1 \u003d 1, x 2 \u003dc./ a. = -208/345.

Answer: 1; -208/345.

2) solutions equation 132x 2 - 247x + 115 \u003d 0.

Decision.As a +.b. + C \u003d 0 (132 - 247 + 115 \u003d 0), that

x 1 \u003d 1, x 2 \u003dc./ a. = 115/132.

Answer: 1; 115/132.

B. If the second coefficient b. = 2 k.- even number, then the root formula

Example.

Resolving equation 3x2 - 14x + 16 \u003d 0.

Decision. We have: a \u003d 3,b. \u003d - 14, C \u003d 16,k. = - 7 ;

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

Ministry of Education and Science of the Russian Federation

Bryansk region Zhukovsky district

MOU RJANITSKA Middle School

RESEARCH

Solutions

Pavlikov Dmitry, 9 cl.

Leader: Prikhodko Yuri
Vladimirovich,

mathematic teacher.

Bryansk, 2009

I.. The history of the development of square equations ……………………….2

1. Square equations in the ancient Babylon .............................2

2. As accounted for and solved diofant square equations ............ ... 2

3. Square equations in India .......................................... ... 3

4. Square equations in alcohol .................................... 4

5. Square equations in Europe XIII - XVII centuries .................. .......... 5

6. About the Vieta Theorem ............................................................... 6

II.. Methods for solving square equations ……………………….7

Method ........................................................................... 7.

Method ........................................................................ .... 9

Method ........................................................................ ... 10

Method ........................................................................ ... 12

Method ........................................................................ ... 13

Method ........................................................................ ... 15

Method ........................................................................ ... 16

III. Conclusion…………………………………………………..............18

Literature……………………………………………………………….19

The history of the development of square equations.

1. Square equations in ancient Babylon.

The need to solve equations not only the first, but also a second degree in antiquity was caused by the need to solve the tasks related to the location of land areas and with earthworks of a military nature, as well as with the development of astronomy and mathematics itself. Square equations were able to solve about 2000 years before. e. Babylonian.

By applying a modern algebraic record, we can say that in their clinox texts there are, except for incomplete, and such, for example, full square equations:

X. 2 + X. = ¾; X. 2 - X. = 14,5

The rule of solving these equations set forth in the Babylonian texts coincides essentially with modern, but it is not known how Babylonians reached this rule. Almost all clinbow texts found until now, only tasks with decisions set forth in the form of recipes, without indication as to how they were found.

Despite the high level of development of algebra in Babylon, there are no concept of negative number and general methods for solving square equations in clinox texts.

2. As accounted for and solved diophant square equations.

In the "arithmetic" of Diophanta there is no systematic presentation of the algebra, but it contains a systematic number of tasks accompanied by explanations and solved with the preparation of equations of different degrees.

When drawing up the Diofant equations to simplify the solution skillfully chooses unknown.

Here, for example, one of his tasks.

Task 11. "Find two numbers, knowing that their sum is 20, and the work is 96"

Diofant argues as follows: From the condition of the problem, it follows that the desired numbers are not equal, since if they were equal, then their work would not be 96, and 100. Thus, one of them will be more than half of their sum, i.e. . 10 + H.The other is less, i.e. 10 - H.. The difference between them 2x.

Hence the equation:

(10 + x) (10 - x) \u003d 96

100 - H. 2 = 96

h. 2 - 4 = 0 (1)

From here x \u003d 2.. One of the desired numbers is 12 , Other 8 . Decision x \u003d -2. It does not exist for diophanta, as Greek mathematics knew only positive numbers.

If we decide this task, choosing one of the desired numbers as an unknown, we will come to solve the equation

y (20 - y) \u003d 96,

w. 2 - 20U + 96 \u003d 0. (2)

It is clear that, choosing as an unknown game of the desired numbers, Diofant simplifies the decision; He can reduce the task of solving an incomplete square equation (1).

3. Square equations in India.

The tasks per square equations are already found in the astronomical tract "Ariabhatti", compiled in 499. Indian mathematician and astronomer Ariabhatta. Another Indian scientist, brahmagupta (VII century), outlined the general rule of solving the square equations given to a single canonical form:

oh 2 + b.x \u003d C, and 0. (1)

In equation (1) coefficients except butmay be negative. The brahmagupta rule essentially coincides with our.

In ancient India, public competitions were distributed in solving difficult tasks. In one of the old Indian books, the following competitions are said about such competitions: "As the sun is glittering with its own stars, so the scientist is overshadowing the falsities of another in the national assembly, offering and solving algebraic tasks." The tasks are often enjoyed in a poetic shape.

Here is one of the tasks of the famous Indian mathematics XII century. Bhaskara.

Task 13.

"Stating monkeys and twelve on Lianam ...

The power of the facing, having fun. Began to jump, hanging ...

They are in the square part of the eighth how many monkeys were,

In the glade was amused. Do you tell me, in this stack? "

The decision of Bhaskara testifies to the fact that he knew about the doubleness of the roots of square equations (Fig. 3).

The corresponding task 13 Equation:

(x./8) 2 + 12 = x.

Bhaskara writes under the guise of:

h. 2 - 64x \u003d -768

and to supplement the left part of this equation to the square adds to both parts 32 2 , getting then:

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

(x - 32) 2 = 256,

x - 32 \u003d ± 16,

h. 1 \u003d 16, x 2 = 48.

4. Square equations in al - Khorezmi.

In the algebraic treatise al - Khorezmi gives the classification of linear and square equations. The author includes 6 species of equations, expressing them as follows:

1) "Squares are roots", i.e. Oh 2 + C \u003d.b.x.

2) "Squares are equal to the number", i.e. Oh 2 \u003d s.

3) "The roots are equal to the number", i.e. ah \u003d s.

4) "Squares and numbers are equal to roots", i.e. Oh 2 + C \u003d.b.x.

5) "Squares and roots are equal to the number", i.e. Oh 2 + bX. \u003d s.

6) "Roots and numbers are equal to squares", i.e.bX. + C \u003d ah 2 .

For al-Khorezmi, avoiding the use of negative numbers, the members of each of these equations are the components, and not subtracted. At the same time, it is not obviously taken into account the equations that have no positive solutions. The author sets out ways to solve these equations, using the techniques of al - Jabr and Al - Mukabala. His decisions, of course, does not coincide with our. Already not to mention that it is purely rhetorical, it should be noted, for example, in solving an incomplete square equation of the first type of al - chores, like all mathematics until the XVII century, it takes into account the zero solution, probably because in concrete practical It does not matter tasks. When solving complete square al-chores equations on private numeric examples, it sets out the rules of decision, and then geometrical evidence.

Let us give an example:

Task 14. "Square and number 21 are equal to 10 roots. Find the root »

(Means the root of the equation x 2 + 21 \u003d 10x).

The decision of the author reads something like this: we divide the number of roots, you will get 5, you will multiply on yourself, from the work of one 21, will remain 4. Removing the root out of 4, you will receive 2. ONDE 2 OT5, you will receive 3, it will be the desired root. Or add 2 to 5, which will give 7, it also has a root.

The Al-Khorezmi treatise is the first, which came to us the book in which the classification of square equations systematically set out and the formulas are given.

5. Square equations in EuropeXIII. - XVII explosive

The formulas for solving square equations for the al-Khorezmi in Europe were first set out in the "Book of Abaka", written in 1202 by the Italian mathematician Leonardo Fibonacci. This thorough work, which reflects the influence of mathematics, both countries of Islam and ancient Greece, is distinguished by both completeness, and clarity of presentation. The author developed independently some new algebraic examples of solving problems and the first in Europe approached the introduction of negative numbers. His book promoted the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many challenges from the "Abaka book" passed almost all European textbooks XVI - XVII centuries. and partially XVIII.

The general rule of solving the square equations given to the same canonical form:

h. 2 + bX. \u003d C,

for all sorts of combinations of coefficient signs b., fromit was formulated in Europe only in 1544 M. Stiffel.

The output of the formula of the solution of the square equation in general is available in Vieta, but Viet recognized only positive roots. Italian mathematicians Tartalia, Kardano, Bombelly among the first in the XVI century. Given, in addition to positive, and negative roots. Only in the XVII century. Due to the labor of Girard, Descartes, Newton and other scientists, the method of solving square equations takes a modern appearance.

6. About Vieta Theorem.

The theorem expressing the relationship between the coefficients of the square equation and its roots, which is the name of the Vieta, was formulated for the first time in 1591 as follows: "If B. + D.multiplied by A. - A. 2 well BD.T. A. equally IN And equal D.».

To understand Vieta, you should remember that BUTlike every vowel letter meant he has an unknown (our h.), vowels IN,D. - The coefficients at the unknown. In the language of modern algebra above, the wording of the Vieta means: if there is

(A +.b.) x - x 2 = aB,

h. 2 - (A +b.) x + ab. = 0,

h. 1 \u003d a, x 2 = b..

Expressing the relationship between the roots and coefficients of the equations with common formulas recorded using symbols, the visiet has set uniformity in the methods of solving equations. However, the symbolism of Viet is still far from the current species. He did not recognize the negative numbers and for this, when solving the equations, considered only cases when all the roots are positive.

So: Square equations are a foundation on which the majestic building of the algebra is resting. Square equations are widely used in solving trigonometric, indicative, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve square equations from school bench (grade 8), before the end of the university.

1. Method : Decomposition of the left part of the factory equation.

Resolving equation h. 2 + 10x - 24 \u003d 0. Spatulate the left side of the factors:

h. 2 + 10x - 24 \u003d x 2 + 12x - 2x - 24 \u003d x (x + 12) - 2 (x + 12) \u003d (x + 12) (x - 2).

Consequently, the equation can be rewritten so:

(x + 12) (x - 2) \u003d 0

2. Method : Method of allocation of a full square.

Resolving equation h. 2 + 6x - 7 \u003d 0. We highlight the full square in the left side.

To do this, write the expression x 2 + 6x in the following form:

h. 2 + 6x \u003d x 2 + 2 H. 3.

In the resulting expression, the first term is the square of the number X, and the second - a double product X by 3. On this to get a full square, you need to add 3 2, since

x 2 +. 2 H. 3 + 3 2 \u003d (x + 3) 2 .

We now transform the left part of the equation

h. 2 + 6x - 7 \u003d 0,

adding it and subtracting 3 2. We have:

h. 2 + 6x - 7 \u003dx 2 +. 2 H. 3 + 3 2 - 3 2 - 7 \u003d (x + 3) 2 - 9 - 7 \u003d (x + 3) 2 - 16.

Thus, this equation can be written as:

(x + 3) 2 - 16 \u003d 0, (x + 3) 2 = 16.

Hence, x + 3 - 4 \u003d 0, x 1 \u003d 1, or x + 3 \u003d -4, x 2 = -7.

3. Method : Solution of square equations by the formula.

Multiply both parts of the equation

oh 2 + b.x + c \u003d 0, and ≠ 0

on 4a and consistently we have:

4A. 2 h. 2 + 4A.b.x + 4as \u003d 0,

((2ah) 2 + 2akh b. + b. 2 ) - b. 2 + 4 aC = 0,

(2AX + B) 2 \u003d B. 2 - 4ac,

2AX + B \u003d ± √ b 2 - 4ac,

2AX \u003d - B ± √ b 2 - 4ac,

Examples.

but) Resolving equation: 4x 2 + 7x + 3 \u003d 0.

a \u003d 4,b. \u003d 7, C \u003d 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. for

b. 2 - 4 aC 0 , the equation Oh 2 + b.x + c \u003d 0 It has two different roots.

b) Resolving equation: 4x 2 - 4x + 1 \u003d 0,

a \u003d 4,b. \u003d - 4, C \u003d 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 equation

oh 2 + b.x + c \u003d 0 has the only root

in) Resolving equation: 2x 2 + 3x + 4 \u003d 0,

a \u003d 2,b. \u003d 3, C \u003d 4,D. = b. 2 - 4 aC = 3 2 - 4 2 4 = 9 - 32 = - 13 , D.

This equation has no root.

So, if the discriminant is negative, i.e. b. 2 - 4 aC , the equation

oh 2 + b.x + c \u003d 0 It does not have roots.

Formula (1) of the roots of the square equation oh 2 + b.x + c \u003d 0 Allows you to find roots anyone Square equation (if any), including the above and incomplete. Valid formula (1) is expressed as: the roots of the square equation are equal to the fraction, the numerator of which is equal to the second coefficient taken with the opposite sign, plus minus the root square from the square of this coefficient without the still-standing product of the first coefficient to the free member, and the denominator has a double-coefficient.

4. Method: Solving equations using the Vieta Theorem.

As you know, the reduced square equation has the form

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

His roots satisfy the Vieta theorem, which a \u003d 1. Has appearance

x. 1 x. 2 = q.,

x. 1 + x. 2 = - p.

From here you can draw the following conclusions (according to the coefficients p and q you can predict the signs of the roots).

a) if a consolidated member q. The given equation (1) is positive ( q. 0 ), the equation has two identical root signs and is the envy of the second coefficient p.. If a p, both root are negative if p, both root are positive.

For example,

x. 2 – 3 x. + 2 = 0; x. 1 = 2 and x. 2 = 1, as q. = 2 0 and p. = - 3

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

b) if a free member q. The given equation (1) is negative ( q.), the equation has two different on the root sign, and the root larger in the module will be positive if p. or negative if p. 0 .

For example,

x. 2 + 4 x. – 5 = 0; x. 1 = - 5 and x. 2 = 1, as q.\u003d - 5 and p. = 4 0;

x. 2 – 8 x. – 9 = 0; x. 1 = 9 and x. 2 = - 1, as q. \u003d - 9 and p. = - 8

5. Method: Solving equations by the method of "transit".

Consider a square equation

oh 2 + b.x + c \u003d 0,where A ≠ 0.

Multiplying both parts by a, we get the equation

but 2 h. 2 + A.b.x + ac \u003d 0.

Let be ah \u003d uFrom! x \u003d y / a; then come to the equation

w. 2 + by + ac \u003d 0,

equivalent to this. His roots w. 1 and w. 2 We will find with the help of the Vieta theorem.

Finally get h. 1 \u003d W. 1 /butand h. 1 \u003d W. 2 /but. With this method coefficient but multiplied by a free member, as if "moves" to him, so it is called howling "transit". This method is used when you can easily find the roots of the equation using the Vieta theorem and, most importantly, when the discriminant is an accurate square.

Example.

Resolving equation 2x 2 - 11x + 15 \u003d 0.

Decision. "We will transfer" a coefficient 2 to a free member, as a result we obtain the equation

w. 2 - 11th + 30 \u003d 0.

According to Viet theorem

w.1 \u003d 5 x 1 = 5/2 x. 1 = 2,5

w. 2 = 6 x. 2 = 6/2 x. 2 = 3.

Answer: 2.5; 3.

6. Method: Properties of the coefficients of the square equation.

BUT. Let the square equation be given oh 2 + b.x + c \u003d 0,where A ≠ 0.

1) if, a +b. + C \u003d 0 (i.e., the sum of the coefficients is zero), then x 1 = 1,

h. 2 \u003d s / a.

Evidence. We divide both parts of the equation on a ≠ 0, we obtain the reduced square equation

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

According to Viet theorem

x. 1 + x. 2 = - b./ a.,

x. 1 x. 2 = 1 c./ a..

By condition but -b. + C \u003d 0,from b. \u003d a + s.In this way,

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

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

those. h. 1 = -1 and h. 2 = c./ a.that m required to prove.

Examples.

Resolving equation 345x 2 - 137x - 208 \u003d 0.

Decision.As a +.b. + C \u003d 0 (345 - 137 - 208 \u003d 0),that

h. 1 \u003d 1, x 2 = c./ a. = -208/345.

Answer: 1; -208/345.

2) solutions equation 132x 2 - 247x + 115 \u003d 0.

Decision.As a +.b. + C \u003d 0 (132 - 247 + 115 \u003d 0), that

h. 1 \u003d 1, x 2 = c./ a. = 115/132.

Answer: 1; 115/132.

B. If the second coefficient b. = 2 k. - even number, then the root formula

Example.

Resolving equation 3x2 - 14x + 16 \u003d 0.

Decision. We have: a \u003d 3,b. \u003d - 14, C \u003d 16,k. = - 7 ;

D. = k. 2 – aC = (- 7) 2 – 3 16 = 49 – 48 = 1, D. 0, two different roots;

Answer: 2; 8/3

IN. The reduced equation

h. 2 + RH +.q.= 0

coincides with the general view equation in which a \u003d 1., b. \u003d R. and c \u003d.q.. Therefore, for the reduced square equation of the formula roots

takes the View:

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

Example. Resolving equation h. 2 - 14x - 15 \u003d 0.

Decision.We have: h. 1,2 \u003d 7 ±

Answer: H. 1 \u003d 15; H. 2 = -1.

7. Method: Graphic solution of the square equation.

E. if in equation

h. 2 + px. + q. = 0

transfer the second and third members on the right side, then we get

h. 2 = - px. - q..

We construct graphs of dependence y \u003d x 2 and y \u003d - px - q.

The first dependence schedule is parabola, passing through the origin of the coordinates. Graph of the second dependence -

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

Direct and Parabola can intersect in two points,

the abscissions of the intersection points are the roots of the quad ratio;

Direct and Parabola can touch (only one common point), i.e. The equation has one solution;

Direct and Parabola do not have common points, i.e. The square equation does not have roots.

Examples.

1) Grandically equation h. 2 - 3x - 4 \u003d 0 (Fig. 2).

Decision. We write an equation in the form H. 2 \u003d 3x + 4.

Let's build a parabola y \u003d x. 2 And straight y \u003d 3x + 4. Straight

y \u003d 3x + 4 can be built on two points M (0; 4) and

N. (3; 13) . Direct and Parabolas intersect at two points

BUT and IN With abscissions h. 1 = - 1 and h. 2 = 4 . Answer : H. 1 = - 1;

h. 2 = 4.

2) Resist graphically equation (Fig. 3) h. 2 - 2x + 1 \u003d 0.

Decision. We write an equation in the form h. 2 \u003d 2x - 1.

Let's build a parabola y \u003d x. 2 and straight y \u003d 2x - 1.

Straight y \u003d 2x - 1 Build on two points M (0; - 1)

and N.(1/2; 0) . Direct and Parabola intersect at the point BUT from

abscissa x \u003d 1.. Answer: x \u003d 1.

3) Grandically equation h. 2 - 2x + 5 \u003d 0(Fig. 4).

Decision. We write an equation in the form h. 2 \u003d 5x - 5. Let's build a parabola y \u003d x. 2 And straight y \u003d 2x - 5. Straight y \u003d 2x - 5 We construct along two points M (0; - 5) and N (2.5; 0). Direct and Parabola do not have intersection points, i.e. This equation has no root.

Answer. The equation h. 2 - 2x + 5 \u003d 0 No roots have.

8. Method: Solution of square equations with a circulation and

line.

The graphic method of solving square equations with a parabola is inconvenient. If you build a parabola at points, it takes a lot of time, and the degree of accuracy of the resulting results is small.

I propose the following method of finding the roots of the square equation oh 2 + b.x + c \u003d 0 With the help of a circulation and ruler (Fig. 5).

Suppose that the desired circle crosses the axis

abscissa at points In (x. 1 ; 0) and D. (H. 2 ; 0), Where h. 1 and h. 2 - Roots of the equation oh 2 + b.x + c \u003d 0and passes through points

A (0; 1)and C (0;c./ a.) On the axis of the ordinate. Then, by the theorem on the sequencing we have OB. OD. = Oa. OC.From! OC. = OB. OD./ Oa.\u003d H. 1 h. 2 / 1 = c./ a..

The center of the circle is located at the point of intersection of perpendiculars Sf. and SKrestored in the middle of the chord AC and BD., so

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

2) We will conduct a circle with a radius SA;

3) abscissa points of intersection of this circle with axis Oh are roots of the original square equation.

It is possible three cases.

1) Radius of the circle more ordinate center (As SK, orR. a. + c./2 a.) , the circle crosses the axis oh at two points (Fig. 6, a) In (x. 1 ; 0) and D.(H. 2 ; 0) where h. 1 and h. 2 - roots of the square equation oh 2 + b.x + c \u003d 0.

2) Radius of the circle is equal to the ordinate center (As = SB., orR. = a. + c./2 a.) , the circle concerns the axis oh (Fig. 6, b) at the point In (x. 1 ; 0) where x 1 is the root of the square equation.

3) Radius of the circle less ordinate center

the circle does not have common points with the abscissa axis (Fig. 6, B), in this case the equation has no solution.

Example.

Resolving equation h. 2 - 2x - 3 \u003d 0 (Fig. 7).

Decision.We define the coordinates of the center of the center of the circumference by formulas:

We carry out the SA radius circle, where A (0; 1).

Answer: h. 1 \u003d - 1; H. 2 = 3.

9. Method: Solution of square equations using

nomograms.

This is an old and undeservedly forgotten way to solve square equations,

posted on S.83 (see Bradis V.M. Four-digit Mathematical Tables. - M., Enlightenment, 1990).

Table XXII. Nomogram for solving the equation z. 2 + pz. + q. = 0 . This nomogram allows, without solving a square equation, according to its coefficient

there to determine the roots of the equation.

Curvoline scale nomogram built

according to formulas (Fig.11):

Believed OS \u003d P,ED = q., O \u003d a (all in see), from

like triangles San and CDF. Receive

proportion

where after substitutions and simplifications follow the equation

z. 2 + pz. + q. = 0,

moreover, the letter z. means a label of any point of the curvilinear scale.

Examples.

1) For equation z. 2 - 9 z. + 8 = 0 The nomogram gives roots z. 1 = 8,0 and z. 2 = 1,0 (Fig.12).

2) Resolving the nomogram equation

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.

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

3) For equation

z. 2 - 25 z. + 66 = 0

the P and Q coefficients go beyond the scale of the scale, perform the substitution z. = 5 t.,

we get the equation

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

which we solve the nomogram and get t. 1 = 0,6 and t. 2 = 4,4, from z. 1 = 5 t. 1 = 3,0 and z. 2 = 5 t. 2 = 22,0.

10. Method: Geometric way of solving square

equations.

In antiquity, when the geometry was more developed than algebra, the square equations were not solved algebraically, but geometrically. I will give the famous example from the algebra algebra al - Khorezmi.

Examples.

1) solve equation h. 2 + 10x \u003d 39.

In the original, this task is formulated as follows: "Square and ten roots are 39" (Fig.15).

Decision. Consider the square from the side X, the rectangles are built on its parties so that the other side of each of them is 2.5, therefore, each area is 2.5x. The resulting figure is then complemented to a new ABCD square, completing four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25.

Area S. Square Abcd. can be represented as a sum of the square: the original square H. 2 , four rectangles (4 2.5x \u003d 10x) and four attached squares (6,25 4 = 25) . S. = h. 2 + 10x + 25.Replacing

h. 2 + 10x Number 39 , I get that S. = 39 + 25 = 64 where it follows that the side of the square Abcd.. section AV \u003d 8.. For the desired side h. Initial square get

2) But, for example, as ancient Greeks solved the equation w. 2 + 6th - 16 \u003d 0.

Decisionpresented in Fig. 16, where

w. 2 + 6th \u003d 16, or 2 + 6U + 9 \u003d 16 + 9.

Decision. Expressions w. 2 + 6U + 9 and 16 + 9 Geometrically represent

same square, and the initial equation w. 2 + 6th - 16 + 9 - 9 \u003d 0 - the same equation. Where and get that y + 3 \u003d ± 5, or w. 1 \u003d 2, 2 = - 8 (Fig.16).

3) solve geometrically equation w. 2 - 6th - 16 \u003d 0.

Converting the equation, get

w. 2 - 6th \u003d 16.

In fig. 17 Find "Images" of expressions w. 2 - 6th, those. From the square of the square of the side, the square of the square of the side of the side is subtracted 3 . So if to express w. 2 - 6U. add 9 , then we get the square square with the side y - 3.. Replacing expression w. 2 - 6U. equal to it number 16,

we get: (y - 3) 2 = 16 + 9, those. y - 3 \u003d ± √25, or y - 3 \u003d ± 5, where w. 1 = 8 and W. 2 = - 2.

Conclusion

Square equations are widely used in solving trigonometric, indicative, logarithmic, irrational and transcendental equations and inequalities.

However, the value of square equations is not only in the grace and shortness of solving problems, although it is very significant. Equally important is that, as a result of the use of square equations, new parts are not rarely detected when solving problems, new parts are detected, it is possible to make interesting generalizations and make clarifications that are prompted by the analysis of the obtained formulas and ratios.

I would like to note the fact that there is still little studied theme in this work at all, just do not do it, so it is a lot of hidden and unknown, which gives an excellent opportunity for further work on it.

Here we stopped on the question of solving square equations, and what if there are other ways to solve them?! Again finding beautiful patterns, some facts, clarifications, make generalizations, open all new and new. But these are questions already following work.

Summing up, we can conclude: Square equations play a huge role in the development of mathematics. We all know how to solve square equations from school bench (grade 8), before the end of the university. These knowledge can come in handy throughout life.

Since these methods for solving square equations are easy to use, they will certainly be interested in fond of mathematics of students. Our job makes it possible to look different about the tasks that mathematics poses.

Literature:

1. Alimov Sh.A., Ilyin V.A. and others. Algebra, 6-8. Trial tutorial for 6-8 class high school. - M., Enlightenment, 1981.

2. Bradis V.M. Four-digit mathematical tables for high school.

Ed. 57th. - M., Enlightenment, 1990. P. 83.

3. Krozhapov A.K., Rubanov A.T. Problem on algebra and elementary functions. Tutorial for secondary special educational institutions. - M., Higher School, 1969.

4. Okunev A.K. Quadratic functions, equations and inequalities. Manual for teacher. - M., Enlightenment, 1972.

5. Presman A.A. Solving a square equation with a circulation and a ruler. - M., Kvant, No. 4/72. P. 34.

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

7. Khudobin A.I. Collection of tasks on algebra and elementary functions. Manual for teacher. Ed. 2nd. - M., Enlightenment, 1970.

Application for management

research work

Leader: Prikhodko Yury Vladimirovich (Mathematics teacher)

Estimated topic: "10 ways to solve square equations"

Consultants:

Prikhodko Yuri Vladimirovich (Mathematics teacher);

Eroshenkov Dmitry Aleksandrovich (Teacher of Informatics)

Educational area of \u200b\u200bknowledge, educational subject, within the framework of the project mathematics

Educational disciplines close to the topic of the project: mathematics

Training class: Grade 9.

The composition of the research group: Kursin Dmitry, Pavlikov Dmitry

View of the project on the dominant activity of the student: study of rational ways to solve square equations

Type of duration project: long-term

Type of education: elective course

Necessary equipment: scientific and popular literature related to the consideration of various ways to solve square equations

Estimated Project Product: creation of educational and methodological material on the use of rational ways to solve square equations

https://pandia.ru/text/78/082/images/image002_237.gif "height \u003d" 952 "\u003e MOU" Sergievsky Secondary School "

Performed: Sizikov Stanislav

Teacher:

from. Sergievka, 2007

1. Introduction. Square equations in ancient Babylon .................. .3

2. Square equations at the diaphlast ............ .. .............................. .4

3. Square equations in India ................................................ 5

4. Square equations in Al - Khorezmi ........................................................

5. Square equations in Europe XIII - XYII .............................. ... 7

6. About the Vieta Theorem .................................................................. ..9

7. Ten ways to solve square equations ........................ ..10

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

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

Introduction

Quadratic equations

Square equations are a foundation on which the majestic building of the algebra is resting. Square equations are widely used in solving trigonometric, indicative, logarithmic, irrational equations. We all know how to solve square equations, starting from the 8th grade. But what caused the history of the solutions of square equations?

Square equations in ancient Babylon

The need to solve equations not only the first, but also a second degree in antiquity was caused by the need to solve problems associated with finding the area of \u200b\u200bland plots; Earthworks of a military nature, as well as with the development of astronomy and mathematics itself. Square equations were able to solve about 2000 years before. e. Babylonian. Applying a modern algebraic record, we can say that in their clinox texts there are, except for incomplete, and such, for example, full square equations: x2 + x \u003d,: x2 - x \u003d 14https: //pandia.ru/text/78/082 /images/image005_150.gif "width \u003d" 16 "height \u003d" 41 src \u003d "\u003e) 2 + 12 \u003d x; bhaskara writes under the guise

x2- 64h. = - 768

and to supplement the left part of this equation to the square, adds to both parts 322, then obtaining: x2- 64x + 322 \u003d - 768 + 1024;

(H.- 32)2 = 256; x -32 \u003d ± 16, xt. = 16, xg= 48.

Square Equations in Al - Khorezmi

Al-Khorezmi algebraic treatise provides a classification of linear and square equations. The author includes 6 species of equal, expressing them as follows:

1) "Squares are equal to roots", i.e. ah2 \u003d wt.

2) "Squares are equal to the number", i.e. ah2= from.

3) "The roots are equal to the number", i.e. ah \u003d s.

4) "Squares and numbers are equal to roots", i.e. ah2+ c \u003d wk.

5) "Squares and roots are equal to the number", i.e. ah2+ in \u003d s.

6) "Roots and numbers are equal to squares", i.e. vK+ c \u003d ah2. For al-Khorezmi, avoiding the use of negative numbers, members of each of these equations are components, and not subtracted. At the same time, it is not obviously taken into account the equations that have no positive solutions. The author sets out ways to solve these equations. His decision, of course, does not coincide with our. Already not to mention that it is purely rhetorical, it should be noted, for example, that when solving an incomplete square equation of the first type of al-Khorezmi, like all mathematics until the XVII century, does not take into account the zero solution, probably because in concrete practical practical It does not matter tasks. When solving full square equations, al-chores on private numeric examples sets out the rules for the solution, and then their geometrical evidence.

Let us give an example.

Task 14. "Square and number 21 are equal to 10 roots. Find the root "(Measures the root of the equation x2 +. 21 = 10x).

The decision of the author reads something like this: we divide the number of roots, you will get 5, multiply 5 by itself, from the work of one 21, will remain 4. Removing the root out of 4, you will get 2. Taken 2 from 5, you will receive 3, this will be the desired root. Or add 2 to 5, which will give 7, it also has a root.

Treatise by al-Khorezmi is the first book that has reached us, in which the classification of square equations is systematically presented and their formulas are given.

Square equations in EuropeXIII.- XVII explosive

Formulas for the solution of square equations for the al-Khorezmi in Europe were first set forth in the "Abaka book" (published in Rome in the middle of the last century "Abaca Book" Fibonacci contains 459 pages), written in 1202 by the Italian mathematician Leonardo Fibonacci. This thorough work, which reflects the influence of mathematics of both the countries of Islam and ancient Greece, is also distinguished by both the completeness and clarity of the presentation. The author has developed independently some new algebraic examples of solving problems and the first inEurope approached the introduction of negative numbers. His book promoted the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many challenges from the "Book of Abaka" passed almost all European textbooks of the XVI-XVII centuries. and partially XVIII.

General rule of solving square equations given to a single canonical form x2+ q \u003d s, For all sorts of combinations of coefficient signs b, S.it was formulated in Europe only in 1544. M. Stiffel.

The output of the formula of the solution of the square equation in general is available in Vieta, but Viet recognized only positive roots. Italian mathematicians Tartalia, Kardako, Bombelly among the first in the XVI century. Given, in addition to positive, and negative roots. Only in the XVII century. Thanks to the works of Girard, Decartes, Newton and other scientists, the method of solving square equations takes a modern appearance.

About Viet Theorem

The theorem expressing the relationship between the coefficients of the square equation and its roots, which is the name of the Vieta, was formulated for the first time in 1591 as follows: "If IN+ D., multiplicated by BUTminus A2,equally BD., that BUTequally INand equal D.».

To understand Vieta, you should remember that BUT,like everyone
the vowel letter meant that he was unknown (our x)vowels
IN,D. - The coefficients at the unknown. In the language of modern algebra above, the wording of the Vieta means: if there is

(but+ c) x - x2 = aB, x2 - (A +. b.) x. + aB = 0, x1 \u003d a, x2 \u003d c.

Expressing the relationship between the roots and coefficients of the equations with common formulas recorded using symbols, the visiet has set uniformity in the methods of solving equations. However, the symbolism of Viet is still far from the current species. He did not recognize negative numbers and therefore, when solving the equations, he considered only cases when all the roots are positive

Ten ways to solve square equations

1. Decomposition of the left part of the factory equation

Resolving equation x2+ 10h. - 24 \u003d 0. Spread the left part of the factors' equation:

x2 + 10x - 24 \u003d x2 + 12x - 2x - 24 \u003d

X (x + x + 12) \u003d (x + 12) (x - 2).

Consequently, the equation can be rewritten so:

(h. + 12) (x - 2) \u003d 0.

Since the work is zero, at least one of his factor is zero. Therefore, the left part of the equation appeals to zero x \u003d2, as well as h.\u003d - 12. This means that the numbers 2 and - 12 are the roots of the x2 + 10x equation - 24 \u003d 0.

2. Method of allocation of a full square

Let us explain this method on the example.

I solve equation x2 + 6x - 7 \u003d 0. We allocate the full square on the left side. To do this, write the expression x2 + 6x in the following form:

x2 + 6x \u003d x2 + 2 * x * 3.

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

x2 + 2 x 3 + 32 \u003d (x + 3) 2.

We now transform the left part of the equation

x2 + 6x - 7 \u003d 0,

adding it and subtracting 32. We have:

x2 + 6x - 7 \u003d x2 + 2 h. 3 +– 7 = (H.- \u003d (x - z) 2 - 16 .

Therefore, this yard can be written as follows:

(x + \u003d 0, i.e. (x + 3) 2 \u003d 16.

Hence, h.+ 3 \u003d 4 x1 \u003d 1, or x + 3 \u003d - 4, x2 \u003d - 7.

3. Solution of square equations on the formula

Multiply both parts of the equation

ah2+ vK+ c \u003d.0, a ≠0, by 4A.and consistently we have:

4A2 x2 + 4aBX. + 4As \u003d.0,

((2AH) 2 + 2 aXB. + b.2 ) - b.2 + 4As.= 0,

(2ah +.b.) 2 \u003d B2- 4As,

2akh+ b. \u003d ± https://pandia.ru/text/78/082/images/image006_128.gif "width \u003d" 71 "height \u003d" 27 "\u003e, x1,2 \u003d

In the case of a positive discriminant, i.e. b2 - 4As\u003e0, equation ah2+ vK + S.\u003d 0 has two different roots.

If the discriminant is zero, i.e. b2 - 4As \u003d0, then equation ah2+ vK+ from\u003d 0 has the only root, x \u003d - https://pandia.ru/text/78/082/images/image009_95.gif "width \u003d" 14 "height \u003d" 62 "\u003e its roots satisfy the theorem of the Vieta, which but\u003d 1 has the view

x1 x2 \u003d q.,

x1 + x2 \u003d - r.

From here you can draw the following conclusions (by coefficients rand q. You can predict root signs).

a) if a free member q. The given equation (1)
positive (q. \u003e 0), the equation has two identical
on the root sign and it depends on the second coefficient r
If a r\u003e 0, both root are negative if r< 0, then both
the root is positive.

For example,

x2- 3h. + 2 = 0; x1\u003d 2 and x2 \u003d 1, since q. = 2 > 0 u. p. = - 3 < 0;

x2 + 8x + 7 \u003d 0; x 1 \u003d - 7 and x2 \u003d - 1, since q. \u003d 7\u003e 0 and r = 8 > 0.

b) if a free member q. The given equation (1)
negative (q. < 0), the equation has two different on the root sign, and the root larger in the module will be positive if r< 0, or negative if P\u003e0.

For example,

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

x2 - 8x - 9 \u003d 0; x1 \u003d9 I. x2\u003d - 1, since q. = - 9 < и r= - 8 < 0.

5. Solving equations by the method of "transit"

Consider a square equation aH2 + Vh+ c \u003d.0, where a ≠0. Multiplying both parts of it on but,we get the equation a2x2 +.aBX. + AC= 0.

Let be ah \u003d y,from h.\u003d; then come to the equation

u2.+ by + ac \u003d.0,

equivalent to this. His roots u1.and u2.we will find with the help of the Vieta theorem. Finally get x1\u003d https://pandia.ru/text/78/082/images/image012_77.gif "width \u003d" 24 "height \u003d" 43 "\u003e.

With this method coefficient butmultiplied by a free member, as if "moves" to him, so it is called the method of "transit".This method is used when you can easily find the roots of the equation using the Vieta theorem and, most importantly, when the discriminant is an accurate square.

1. Resolving equation 2x2 - 11x + 15 \u003d 0.

Decision."We will transfer" a coefficient 2 to a free member, as a result we obtain the equation

u2 - 11. w.+ 30 = 0.

According to the theorem of Vieta U1 \u003d 5, U2 \u003d 6, hence x1 \u003d https://pandia.ru/text/78/082/images/image014_69.gif "width \u003d" 16 height \u003d 41 "height \u003d" 41 "\u003e, t . e.

x1 \u003d 2.5 x2 \u003d 3.

Answer:2,5; 3.

6. Properties of square coefficientsequations

A. Let the square equation

ah2 + vx + with\u003d 0, where but ≠ 0.

1. If a + b + S.= 0 (i.e. the sum of the equation coefficients is zero), then x1 \u003d1, x2 \u003d.

2. If a - in + with= 0, orb. = but + c, then x1 \u003d -1, h.2 \u003d - https://pandia.ru/text/78/082/images/image016_58.gif "width \u003d" 44 height \u003d 41 "height \u003d" 41 "\u003e.

Answer:1; 184">

The following cases are possible:

Direct and Parabola can intersect at two points, the abscissions of the intersection points are roots of the square equation;

Direct and Parabola can concern (only one common point), i.e. the equation has one solution;

Direct and Parabola do not have common points, i.e. the square equation has no roots.

Examples.

1. Resolve graphically equation x2 - 3x - 4 \u003d 0 (Fig. 2).

Decision.We write an equation in the form x2 \u003d 3x + 4.

Let's build a parabola y \u003d x2.and straight y \u003d3 + 4. Straight w.\u003d 3x + 4 can be constructed by two points M (0; 4) and N (3; 13). Direct and Parabolas intersect at two points And to B.with abscissions x1\u003d - 1 and x2 \u003d 4.

Answer: x1\u003d - 1, x, \u003d 4.

8. Solution of square equations with a circulation and ruler

We offer the following method of finding the roots of the square equation

ah2+ vK+ from= 0

with the help of a circulation and a ruler (Fig.).

Suppose that the desired circle crosses the abscissa axis at points B.(x1;0) I. D.(x.2 ; 0) where x1and x2- Roots of the equation aH2 + Vh+from=0,
and passes through points A (0; 1) and from (0;) on the ordinate axis..gif "width \u003d" 197 "height \u003d" 123 "\u003e

So: 1) Build points https://pandia.ru/text/78/082/images/image023_40.gif "width \u003d" 171 "height \u003d" 45 "\u003e The circle crosses the axis OH at the point in (x1; 0), and d (x1 ; 0), where x1 and x2 - roots of the square equation AH2 + BX + C = 0.

2) Radius of the circle is equal to the ordinate center , the circle concerns the axis OH at point in (x1; 0), where xx- the root of the square equation.

3) The radius of the circle is less than the ordinate of the center of Left "\u003e

https://pandia.ru/text/78/082/images/image029_34.gif "width \u003d" 612 "height \u003d" 372 "\u003e 40" height \u003d "14"\u003e

https://pandia.ru/text/78/082/images/image031_28.gif "width \u003d" 612 "height \u003d" 432 src \u003d "\u003e

Where after the substitution and

Simplifications follows the equation z2 + pz + q \u003d 0, and the letter Z means a label of any point of the curvilinear scale.

10. Geometric method of solving square equations

In antiquity, when the geometry was more developed than algebra, the square equations were not solved algebraically, but geometrically. We give the famous example of algebra algebra by algebra.

And four attached squares, so. S \u003d x2 + 10x + 25. Replacing X2 + 10X by number 39, we obtain that S \u003d 39 + 25 \u003d 64, from where it follows that the side of the square Abcd., i.e. cut AU\u003d 8. For the desired side h.initial square get

Conclusion

We all know how to solve square equations, starting with school bench, until the end of the university. But in the school course of mathematics, the formulas of the roots of square equations are studied, with which any square equations can be solved. However, having studied this question about deeper, I was convinced that there are other ways to solve square equations that allow many equations very quickly and rationally.

Maybe math somewhere there in other dimensions, the eye is not visible, - all and we just get all the new facts from the holes with the worlds? ... God gaze; But it turns out that if the physicists, chemicals, economists or archaeologists need a new model of the world, this model can always be taken from the shelf, which three years ago put mathematics, or collect from parts lying on the same shelf. Perhaps these items will have to twist, caught up to each other, pollute, to sharpen a couple of new bushings of the theorems quickly; But the theory of the result will not only describe the actual situation, but also predicts the consequences! ...

Strange thing - this mind is the game that is always right ...

Literature

1. Alimov Sha., Ilyin Va. and others. Algebra, 6-8. Trial tutorial for high school grades 6-8. - M., Enlightenment, 1981.

2.Bradis mathematical tables for high school. Ed. 57th. - M., Enlightenment, 1990. P. 83.

3.Rotsky -Locking when teaching mathematics. Book for teacher. - M., Enlightenment, 1992.

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

5. Functions, equations and inequality. Manual for teacher. - M., Enlightenment, 1972.

6. Acrane B. C., Cute questions and tasks in mathematics. Ed. 4th, addition. - M., Higher School, 1973.

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

Review

to work student 11 class MOU "Sergievskaya average

comprehensive school"

Copsevskaya Rural Secondary School

10 ways to solve square equations

Leader: Patrikeva Galina Anatolyevna,

mathematic teacher

s.Kopievo, 2007.

1. The history of the development of square equations

1.1 Square equations in ancient Babylon

1.2 As accounted for and solved Diofant Square equations

1.3 square equations in India

1.4 Square equations in alcohise

1.5 Square equations in Europe XIII - XVII centuries

1.6 About Vieta Theorem

2. Methods for solving square equations

Conclusion

Literature

1. The history of the development of square equations

1 .1 Square EQsin ancient Babylon

By applying a modern algebraic record, we can say that in their clinox texts there are, except for incomplete, and such, for example, full square equations:

X. 2 + X. = ѕ; X. 2 - X. = 14,5

Despite the high level of development of algebra in Babylon, there are no concept of negative number and general methods for solving square equations in clinox texts.

1.2 As made and solved Diofant square equations.

When drawing up the Diofant equations to simplify the solution skillfully chooses unknown.

Here, for example, one of his tasks.

Task 11. "Find two numbers, knowing that their sum is 20, and the work is 96"

Hence the equation:

(10 + x) (10 - x) \u003d 96

100 - H. 2 = 96

h. 2 - 4 = 0 (1)

From here x \u003d 2.. One of the desired numbers is 12 , Other 8 . Decision x \u003d -2. It does not exist for diophanta, as Greek mathematics knew only positive numbers.

If we decide this task, choosing one of the desired numbers as an unknown, we will come to solve the equation

y (20 - y) \u003d 96,

w. 2 - 20U + 96 \u003d 0. (2)

It is clear that, choosing as an unknown game of the desired numbers, Diofant simplifies the decision; He can reduce the task of solving an incomplete square equation (1).

1.3 Square equations in India

Oh 2 + b.x \u003d s, a\u003e 0. (1)

In equation (1) coefficients except butmay be negative. The brahmagupta rule essentially coincides with our.

Here is one of the tasks of the famous Indian mathematics XII century. Bhaskara.

Task 13.

"Stating monkeys and twelve on Lianam ...

The power of the facing, having fun. Began to jump, hanging ...

They are in the square part of the eighth how many monkeys were,

In the glade was amused. Do you tell me, in this stack? "

The decision of Bhaskara testifies to the fact that he knew about the doubleness of the roots of square equations (Fig. 3).

The corresponding task 13 Equation:

(x./8) 2 + 12 = x.

Bhaskara writes under the guise of:

h. 2 - 64x \u003d -768

and to supplement the left part of this equation to the square adds to both parts 32 2 , getting then:

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

(x - 32) 2 = 256,

x - 32 \u003d ± 16,

h. 1 = 16, h. 2 = 48.

1.4 Square EQsal - Khorezmi

In the algebraic treatise al - Khorezmi gives the classification of linear and square equations. The author includes 6 species of equations, expressing them as follows:

1) "Squares are roots", i.e. oh 2 + C \u003d.b.x.

2) "Squares are equal to the number", i.e. oh 2 \u003d s.

3) "The roots are equal to the number", i.e. ah \u003d s.

4) "Squares and numbers are equal to roots", i.e. oh 2 + C \u003d.b.x.

5) "Squares and roots are equal to the number", i.e. Oh 2 + bX. \u003d s.

6) "Roots and numbers are equal to squares", i.e. bX. + C \u003d ah 2 .

al - Khorezmi, like all mathematics until the XVII century, takes into account the zero solution, probably because it does not matter in specific practical tasks. When solving complete square al-chores equations on private numeric examples, it sets out the rules of decision, and then geometrical evidence.

Task 14. "Square and number 21 are equal to 10 roots. Find the root » (Means the root of the equation x 2 + 21 \u003d 10x).

The Al-Khorezmi treatise is the first, which came to us the book in which the classification of square equations systematically set out and the formulas are given.

1.5 Square equations in EuropeXIII. - XVII BB

The general rule of solving the square equations given to the same canonical form:

h. 2 + bX. \u003d C,

for all sorts of combinations of coefficient signs b., fromit was formulated in Europe only in 1544 M. Stiffel.

1.6 About Vieta Theorem

(A +.b.) x - x 2 = aB,

h. 2 - (A +b.) x + ab. = 0,

h. 1 \u003d a, h. 2 = b..

Expressing the relationship between the roots and coefficients of the equations with common formulas recorded using symbols, the visiet has set uniformity in the methods of solving equations. At the same time, the symbolism of the Vieta is still far from the current species. He did not recognize the negative numbers and for this, when solving the equations, considered only cases when all the roots are positive.

2. Methods for solving square equations

Square equations are a foundation on which the majestic building of the algebra is resting. Square equations are widely used in solving trigonometric, indicative, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve square equations from school bench (grade 8), before the end of the university.

In the school course of mathematics are studied formulas of the roots of square equations, with which any square equations can be solved. In this case, there are other ways to solve square equations that allow many equations very quickly and rationally. There are ten ways to solve square equations. In detail in my work, I disassembled each of them.

1. Method : Decomposition of the left part of the factory equation.

Resolving equation

h. 2 + 10x - 24 \u003d 0.

Spatulate the left side of the factors:

h. 2 + 10x - 24 \u003d x 2 + 12x - 2x - 24 \u003d x (x + 12) - 2 (x + 12) \u003d (x + 12) (x - 2).

Consequently, the equation can be rewritten so:

(x + 12) (x - 2) \u003d 0

2. Method : Method of allocation of a full square.

Resolving equation h. 2 + 6x - 7 \u003d 0.

We highlight the full square in the left side.

To do this, write the expression x 2 + 6x in the following form:

h. 2 + 6x \u003d x 2 + 2 * x * 3.

In the resulting expression, the first term is the square of the number X, and the second - a double product X by 3. On this to get a full square, you need to add 3 2, since

x 2 +. 2 * x * 3 + 3 2 \u003d (x + 3) 2 .

We now transform the left part of the equation

h. 2 + 6x - 7 \u003d 0,

adding it and subtracting 3 2. We have:

h. 2 + 6x - 7 \u003dx 2 +. 2 * x * 3 + 3 2 - 3 2 - 7 \u003d (x + 3) 2 - 9 - 7 \u003d (x + 3) 2 - 16.

Thus, this equation can be written as:

(x + 3) 2 - 16 =0, (x + 3) 2 = 16.

Hence, x + 3 - 4 \u003d 0, x 1 \u003d 1, or x + 3 \u003d -4, x 2 = -7.

3. Method : Solution of square equations by the formula.

Multiply both parts of the equation

oh 2 + b.x + c \u003d 0, and? 0.

on 4a and consistently we have:

4A. 2 h. 2 + 4A.b.x + 4as \u003d 0,

((2ah) 2 + 2AH *b. + b. 2 ) - b. 2 + 4 aC = 0,

(2AX + B) 2 \u003d B. 2 - 4ac,

2AX + B \u003d ± V B 2 - 4ac,

2AX \u003d - B ± V B 2 - 4ac,

Examples.

but) Resolving equation: 4x 2 + 7x + 3 \u003d 0.

a \u003d 4,b. \u003d 7, C \u003d 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. for

b. 2 - 4 aC >0 , the equation oh 2 + b.x + c \u003d 0 It has two different roots.

b) Resolving equation: 4x 2 - 4x + 1 \u003d 0,

a \u003d 4,b. \u003d - 4, C \u003d 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 equation

oh 2 + b.x + c \u003d 0 has the only root

in) Resolving equation: 2x 2 + 3x + 4 \u003d 0,

a \u003d 2,b. \u003d 3, C \u003d 4,D. = b. 2 - 4 aC = 3 2 - 4 * 2 * 4 = 9 - 32 = - 13 , D. < 0.

This equation has no root.

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

the equation oh 2 + b.x + c \u003d 0 It does not have roots.

Formula (1) of the roots of the square equation oh 2 + b.x + c \u003d 0 Allows you to find roots anyone Square equation (if any), including the above and incomplete. Valid formula (1) is expressed as: the roots of the square equation are equal to the fraction, the numerator of which is equal to the second coefficient taken with the opposite sign, plus minus the root square from the square of this coefficient without the still-standing product of the first coefficient to the free member, and the denominator has a double-coefficient.

4. Method: Solving equations using the Vieta Theorem.

As you know, the reduced square equation has the form

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

His roots satisfy the Vieta theorem, which a \u003d 1. Has appearance

x. 1 x. 2 = q.,

x. 1 + x. 2 = - p.

From here you can draw the following conclusions (according to the coefficients p and q you can predict the signs of the roots).

For example,

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

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

For example,

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

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

5. Method: Solving equations by the method of "transit".

Consider a square equation

oh 2 + b.x + c \u003d 0,where but? 0.

Multiplying both parts by a, we get the equation

but 2 h. 2 + A.b.x + ac \u003d 0.

Let be ah \u003d uFrom! x \u003d y / a; then come to the equation

w. 2 + by + ac \u003d 0,

equivalent to this. His roots w. 1 and w. 2 We will find with the help of the Vieta theorem.

Finally get

h. 1 \u003d W. 1 /but and h. 1 \u003d W. 2 /but.

Example.

Resolving equation 2x 2 - 11x + 15 \u003d 0.

Decision. "We will transfer" a coefficient 2 to a free member, as a result we obtain the equation

w. 2 - 11th + 30 \u003d 0.

According to Viet theorem

w. 1 = 5 h. 1 = 5/2 x. 1 = 2,5

w. 2 = 6 x. 2 = 6/2 x. 2 = 3.

Answer: 2.5; 3.

6. Method: Properties of the coefficients of the square equation.

BUT. Let the square equation be given

oh 2 + b.x + c \u003d 0,where but? 0.

1) if, a +b. + C \u003d 0 (i.e., the sum of the coefficients is zero), then x 1 = 1,

h. 2 \u003d s / a.

Evidence. We divide both parts of the equation on a? 0, we obtain the reduced square equation

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

According to Viet theorem

x. 1 + x. 2 = - b./ a.,

x. 1 x. 2 = 1* c./ a..

By condition but -b. + c \u003d 0,from b. \u003d a + s.In this way,

x. 1 + X. 2 = - but + b / a \u003d -1 - C / A,

x. 1 x. 2 \u003d - 1 * (- C / A),

those. h. 1 = -1 and h. 2 = c./ a.that m required to prove.

Examples.

1) solve equation 345x 2 - 137x - 208 \u003d 0.

Decision.As a +.b. + C \u003d 0 (345 - 137 - 208 \u003d 0),that

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

Answer: 1; -208/345.

2) solutions equation 132x 2 - 247x + 115 \u003d 0.

Decision.As a +.b. + C \u003d 0 (132 - 247 + 115 \u003d 0), that

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

Answer: 1; 115/132.

B. If the second coefficient b. = 2 k. - even number, then the root formula

Example.

Resolving equation 3x2 - 14x + 16 \u003d 0.

Decision. We have: a \u003d 3,b. \u003d - 14, C \u003d 16,k. = -- 7 ;

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

Answer: 2; 8/3

IN. The reduced equation

h. 2 + RH +.q.= 0

coincides with the general view equation in which a \u003d 1., b. \u003d R. and c \u003d.q.. Therefore, for the reduced square equation of the formula roots

takes the View:

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

Example. Resolving equation h. 2 - 14x - 15 \u003d 0.

Decision.We have: h. 1,2 \u003d 7 ±

Answer: H. 1 \u003d 15; H. 2 = -1.

7. Method: Graphic solution of the square equation.

{!LANG-c4692f6a0af232a069a02e118a66fbc8!}

h. 2 + px. + q. = 0

transfer the second and third members on the right side, then we get

h. 2 = - px. - q..

We construct graphs of dependence y \u003d x 2 and y \u003d - px - q.

The first dependence schedule is parabola, passing through the origin of the coordinates. Graph of the second dependence -

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

{!LANG-96822d82155f107c1907f939552ed4c7!}

Direct and Parabola can touch (only one common point), i.e. The equation has one solution;

Direct and Parabola do not have common points, i.e. The square equation does not have roots.

Examples.

1) Grandically equation h. 2 - 3x - 4 \u003d 0 (Fig. 2).

Decision. We write an equation in the form H. 2 \u003d 3x + 4.

Let's build a parabola y \u003d x. 2 And straight y \u003d 3x + 4. Straight

y \u003d 3x + 4 can be built on two points M (0; 4) and

N. (3; 13) . Direct and Parabolas intersect at two points

BUT and IN With abscissions h. 1 = - 1 and h. 2 = 4 . Answer: H. 1 = - 1;

h. 2 = 4.

2) Resist graphically equation (Fig. 3) h. 2 - 2x + 1 \u003d 0.

Decision. We write an equation in the form h. 2 \u003d 2x - 1.

Let's build a parabola y \u003d x. 2 and straight y \u003d 2x - 1.

Straight y \u003d 2x - 1 Build on two points M (0; - 1)

and N.(1/2; 0) . Direct and Parabola intersect at the point BUT from

abscissa x \u003d 1.. Answer: x \u003d 1.

3) Grandically equation h. 2 - 2x + 5 \u003d 0(Fig. 4).

Answer. The equation h. 2 - 2x + 5 \u003d 0 No roots have.

8. Method: Solution of square equations with a circulation and line.

{!LANG-761429a419127126e52a809fbaeb0ab9!}

I propose the following method of finding the roots of the square equation oh 2 + b.x + c \u003d 0 With the help of a circulation and ruler (Fig. 5).

Suppose that the desired circle crosses the axis

abscissa at points In (x. 1 ; 0) and D. (H. 2 ; 0), Where h. 1 and h. 2 - Roots of the equation oh 2 + b.x + c \u003d 0and passes through points

A (0; 1)and C (0;c./ a.) On the axis of the ordinate. Then, by the theorem on the sequencing we have OB. * OD. = Oa. * OC.From! OC. = OB. * OD./ Oa.\u003d H. 1 h. 2 / 1 = c./ a..

The center of the circle is located at the point of intersection of perpendiculars Sf. and SKrestored in the middle of the chord AC and BD., so

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

2) We will conduct a circle with a radius SA;

3) abscissa points of intersection of this circle with axis Oh are roots of the original square equation.

It is possible three cases.

1) Radius of the circle more ordinate center (As > SK, or R. > a. + c./2 a.) , the circle crosses the axis oh at two points (Fig. 6, a) In (x. 1 ; 0) and D.(H. 2 ; 0) where h. 1 and h. 2 - roots of the square equation oh 2 + b.x + c \u003d 0.

{!LANG-e06e15bd9003a76abc07c7855d1bde63!}

Example.

Resolving equation h. 2 - 2x - 3 \u003d 0 {!LANG-808df9ffebae437d3a59344e9d8ed5c1!}

Decision.We define the coordinates of the center of the center of the circumference by formulas:

We carry out the SA radius circle, where A (0; 1).

Answer: h. 1 \u003d - 1; H. 2 = 3.

9. Method: Solution of square equations using nomograms.

{!LANG-24e6fd31ff76cea572f2787c22c390eb!}

Table XXII. Nomogram for solving the equation z. 2 + pz. + q. = 0 {!LANG-eaa7ce4ea16a2e7b32b5dfe61b9493ec!}

{!LANG-5d940a6e18b2dea6bcc7dfe3a6f0b5e7!}

Believed OS \u003d P,ED = q., O \u003d a{!LANG-1d68d1cd02dbf08aa3ea9dd9fc9e3a2b!} San and CDF.{!LANG-8498e1502043c9824ba54633d5ba3663!}

where after substitutions and simplifications follow the equation

z. 2 + pz. + q. = 0,

moreover, the letter z. means a label of any point of the curvilinear scale.

Examples.

1) For equation z. 2 - 9 z. + 8 = 0 {!LANG-a137a40b95dd8c512a7d6d84a8c4faff!}

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

2) {!LANG-05818d328738cfd4f0d79fb3c797af69!}

2 z. 2 - 9 z. + 2 = 0.

{!LANG-a7fbf5d47ab07e7e30f516ca9f466b66!}

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

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

3) For equation

z. 2 - 25 z. + 66 = 0

the P and Q coefficients go beyond the scale of the scale, perform the substitution z. = 5 t.{!LANG-a9e2d1bf8b13f21edad28d0487302f82!}

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

which we solve the nomogram and get t. 1 = 0,6 and t. 2 = 4,4, from z. 1 = 5 t. 1 = 3,0 and z. 2 = 5 t. 2 = 22,0.

10. Method: Geometric way of solving square equations.

Examples.

1) solve equation h. 2 + 10x \u003d 39.

In the original, this task is formulated as follows: "Square and ten roots are 39" (Fig.15).

Area S. {!LANG-7aa71393f7ceecb9144dc8698bdf19ba!} Abcd. can be represented as a sum of the square: the original square h. 2 {!LANG-ac42d332ad68f1e5cde51c9fc9bd5671!} {!LANG-9ad0039961ec403c0b781c47a99dfdad!} and four attached squares (6,25* 4 = 25) . S. = h. 2 + 10x + 25.Replacing

h. 2 + 10x Number 39 , I get that S. = 39 + 25 = 64 where it follows that the side of the square Abcd.. section AV \u003d 8.. For the desired side h. Initial square get

2) But, for example, as ancient Greeks solved the equation w. 2 + 6th - 16 \u003d 0.

Decisionpresented in Fig. 16, where

w. 2 {!LANG-d8d1a6b19bcd255c1e4296c6bde5c2a4!} w. 2 + 6U + 9 \u003d 16 + 9.

Decision. Expressions w. 2 + 6U + 9 and 16 + 9 {!LANG-65c28caaffcddd696a250eed42687919!} w. 2 + 6th - 16 + 9 - 9 \u003d 0 - the same equation. Where and get that y + 3 \u003d ± 5, or w. 1 \u003d 2, 2 = - 8 (Fig.16).

3) solve geometrically equation w. 2 - {!LANG-d0a16fb79fcdbd48f406da5cf3f6d80b!}

Converting the equation, get

w. 2 - {!LANG-0d9f882498e25a9f765d6d061ecab207!}

In fig. 17 Find "Images" of expressions w. 2 - {!LANG-0fcb7737e2326be666ffc5d802d3df70!} those. From the square of the square of the side, the square of the square of the side of the side is subtracted 3 . So if to express w. 2 - {!LANG-ceb32c324b343cb3b60fc8239a397d0a!} {!LANG-ccb5bd7e2124e7d99c0e6fe8173c4920!} 9 , then we get the square square with the side w. - 3 . Replacing expression w. 2 - {!LANG-ceb32c324b343cb3b60fc8239a397d0a!} {!LANG-2844ac619ada8400987955bffe55ec1e!}

we get: (y - 3) 2 = 16 + 9, those. {!LANG-e93e6ce5d1719acb1e6edabf9a44c5ac!}, or y - 3 \u003d ± 5, where w. 1 = 8 and W. 2 = - 2.

Conclusion

Square equations are widely used in solving trigonometric, indicative, logarithmic, irrational and transcendental equations and inequalities.

{!LANG-c78bc6be8d99bab463f5d1b824ab830e!}

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{!LANG-604f16bce9faf8e52a9fc509214a95a6!}

{!LANG-242b1032b9ded1d49176b925ca7d9ccc!}

Literature:

1. {!LANG-515fad93c4903bbb86d40453a990caf3!}

2. {!LANG-15d2b50c3b2d82b1755f9a7472fdc481!}

3. {!LANG-846357108871f9f50a5aa30dbe08ca55!}

4. {!LANG-acdbde56360ab2ba3cb00ab3e25c9157!}

5. {!LANG-faad4dfa0cf4566de35ae527fac77d84!}

6. {!LANG-2e837a532062db76e0a6a19a0d971352!}

7. {!LANG-06ecf7799b3b0188616884642312f609!}

Copsevskaya Rural Secondary School

10 ways to solve square equations

Leader: Patrikeva Galina Anatolyevna,

mathematic teacher

s.Kopievo, 2007.

1. The history of the development of square equations

1.1 Square equations in ancient Babylon

1.2 As accounted for and solved Diofant Square equations

1.3 square equations in India

1.4 Square equations in alcohise

1.5 Square equations in Europe XIII - XVII centuries

1.6 About Vieta Theorem

2. Methods for solving square equations

Conclusion

Literature

1. The history of the development of square equations

{!LANG-c4e6ff1643a1813c5cd42f223c093238!}

By applying a modern algebraic record, we can say that in their clinox texts there are, except for incomplete, and such, for example, full square equations:

X.2 + X.= ¾; X.2 - X.= 14,5

Despite the high level of development of algebra in Babylon, there are no concept of negative number and general methods for solving square equations in clinox texts.

{!LANG-e98d3c10167c301caff77641f7d8745a!}

When drawing up the Diofant equations to simplify the solution skillfully chooses unknown.

Here, for example, one of his tasks.

Task 11. "Find two numbers, knowing that their sum is 20, and the work is 96"

Hence the equation:

(10 + x) (10 - x) \u003d 96

100 - H. 2 = 96

h. 2 - 4 = 0 (1)

From here x \u003d 2.. One of the desired numbers is 12 , Other 8 . Decision x \u003d -2. It does not exist for diophanta, as Greek mathematics knew only positive numbers.

If we decide this task, choosing one of the desired numbers as an unknown, we will come to solve the equation

y (20 - y) \u003d 96,

w.2 {!LANG-88557ee5e9b33bb7756703ea3be0c654!}

It is clear that, choosing as an unknown game of the desired numbers, Diofant simplifies the decision; He can reduce the task of solving an incomplete square equation (1).

1.3 Square equations in India

oh2 + b.{!LANG-44b425d52673a0a6347582e907ab435f!}

In equation (1) coefficients except butmay be negative. The brahmagupta rule essentially coincides with our.

Here is one of the tasks of the famous Indian mathematics XII century. Bhaskara.

Task 13.

"Stating monkeys and twelve on Lianam ...

The power of the facing, having fun. Began to jump, hanging ...

They are in the square part of the eighth how many monkeys were,

In the glade was amused. Do you tell me, in this stack? "

The decision of Bhaskara testifies to the fact that he knew about the doubleness of the roots of square equations (Fig. 3).

The corresponding task 13 Equation:

(x./8) 2 + 12 = x.

Bhaskara writes under the guise of:

h.2 {!LANG-f702057f7a19578df40acaaa15fe69b9!}

and to supplement the left part of this equation to the square adds to both parts 32 2 , getting then:

h.2 {!LANG-71f4aa039d1afb6bd194a6c20c98d3f2!}2 = -768 + 1024,

(x - 32)2 = 256,

x - 32 \u003d ± 16,

h.1 {!LANG-71e8f15126508e52759593fdeece3870!}2 = 48.

{!LANG-01b1deb15bd9276a904a3f242c5b3f03!}

In the algebraic treatise al - Khorezmi gives the classification of linear and square equations. The author includes 6 species of equations, expressing them as follows:

1) "Squares are roots", i.e. Oh2 {!LANG-ead2bf89c06d8753656e67da2af95534!}b.x.

2) "Squares are equal to the number", i.e. Oh2 {!LANG-dfc04afa2a7059dcea9d44b0353f90f6!}

3) "The roots are equal to the number", i.e. ah \u003d s.

4) "Squares and numbers are equal to roots", i.e. Oh2 {!LANG-ead2bf89c06d8753656e67da2af95534!}b.x.

5) "Squares and roots are equal to the number", i.e. Oh2 + bX.{!LANG-dfc04afa2a7059dcea9d44b0353f90f6!}

6) "Roots and numbers are equal to squares", i.e.bX.{!LANG-24a677ec621ba5449a88c6b423a3b5dc!}2 .

{!LANG-f0bcaa34ea748e93e8c6148b1d2c9188!}

Task 14. "Square and number 21 are equal to 10 roots. Find the root » (Means the root of the equation x2 {!LANG-a1f6ea7c968f8760dd6f79f12baf2cfe!}

The Al-Khorezmi treatise is the first, which came to us the book in which the classification of square equations systematically set out and the formulas are given.

{!LANG-5b9b90893942d63c38b323fd4e547cc3!}XIII.- XVII{!LANG-14e4d4080fa57fef4f665459abbee794!}

{!LANG-e9ea86be542378a5ea0d4fbb010b9c17!}

The general rule of solving the square equations given to the same canonical form:

h.2 + bX.{!LANG-bcef50f8e468990832b2eeec848127e6!}

for all sorts of combinations of coefficient signs b., fromit was formulated in Europe only in 1544 M. Stiffel.

1.6 About Vieta Theorem

The theorem expressing the relationship between the coefficients of the square equation and its roots, which is the name of the Vieta, was formulated for the first time in 1591 as follows: "If B.+ D.multiplied by A.- A.2 well BD.T. A. equally IN And equal D.».

(A +.b.) x - x2 = aB,

h.2 {!LANG-6edf6890c07171b05ae5b40594106cbc!}b.) x + ab.= 0,

h.1 {!LANG-8653d49a81c8dea8912d23793cd07d48!}2 = b..

2. Methods for solving square equations

Square equations are a foundation on which the majestic building of the algebra is resting. Square equations are widely used in solving trigonometric, indicative, logarithmic, irrational and transcendental equations and inequalities. We all know how to solve square equations from school bench (grade 8), before the end of the university.

1. Method : Decomposition of the left part of the factory equation.

Resolving equation

h.2 {!LANG-f494a2e068934f32f343c18b792e6289!}.

Spatulate the left side of the factors:

h.2 {!LANG-81acf851991ff5db92a66988a0c17637!}2 {!LANG-c06ee5db639dfeb23124edd61cf6fee9!}

Consequently, the equation can be rewritten so:

(x + 12) (x - 2) \u003d 0

2. Method : Method of allocation of a full square.

Resolving equation h.2 {!LANG-21da244a62323f4b7e2c6fd56862827d!}.

We highlight the full square in the left side.

{!LANG-870be171af946f622fdf6dba6508cda9!}

h.2 {!LANG-4dd9289568a0f8f0e08fa482463a0ba4!}2 {!LANG-d81089758404457e8d029820ab5afb5f!}

{!LANG-ef098853b64b9abbf71a2e87b4cbcfee!}

{!LANG-dfd68e76bbf11a88befb76c0dd276d5b!} {!LANG-a2a64e626c2f2c6d131af52618fa7a55!}2 {!LANG-b037f7e0c54e3e0b83f67b9f33c36346!}2 .

We now transform the left part of the equation

h.2 {!LANG-21da244a62323f4b7e2c6fd56862827d!},

adding it and subtracting 32. We have:

h.2 {!LANG-2af27686f5b1d8f42a435ada39335717!}{!LANG-dfd68e76bbf11a88befb76c0dd276d5b!} {!LANG-a2a64e626c2f2c6d131af52618fa7a55!}2 - 3 2 {!LANG-914b9f1c34e1f4bfddf7bc0f0bf358b7!}2 {!LANG-44dfe0f26515b51e22e49b721faa59a9!}2 - 16.

Thus, this equation can be written as:

(x + 3)2 {!LANG-b06569f0a8078d080c06454e5956354b!}2 = 16.

Hence, x + 3 - 4 \u003d 0, x1 {!LANG-7c8e27a3d805251bba10069e88830250!}2 = -7.

3. Method :{!LANG-0ae9e68ef3cb309335bab798a3676f60!}

Multiply both parts of the equation

oh2 + b.x + c \u003d 0, and ≠ 0

on 4a and consistently we have:

4A.2 h.2 {!LANG-70851c0d9b3ac5d39bba418c1ceaaed8!}b.x + 4as \u003d 0,

((2ah)2 {!LANG-79a9a506da35bdf0b6caae1d023be52d!}b.+ b.2 ) - b.2 + 4 aC= 0,

(2AX + B)2 {!LANG-f6fb4981c93db9613f89ef5ecda407b3!}2 {!LANG-d0279fc8cc3d6611d574ccddc10b6fc5!}

2AX + B \u003d ± √ b2 {!LANG-d0279fc8cc3d6611d574ccddc10b6fc5!}

2AX \u003d - B ± √ b2 {!LANG-d0279fc8cc3d6611d574ccddc10b6fc5!}

Examples.

but){!LANG-d988794a5b7c426799169081a0f9d149!} 4x2 {!LANG-ae8b5af54009ac025faefcba95e5e74a!}

a \u003d 4,b.{!LANG-67396ced673fca6703d0e774f639738a!}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. for

b.2 - 4 aC>0 , the equation oh2 + b.x + c \u003d 0{!LANG-a27be0f053bb2689b329f562786b733c!}

b){!LANG-8d110e9fc36ae6225a7879d2b4d1e70c!} 4x2 {!LANG-ca96c6e63819ce1cc5aa876b4f4f3aec!}

a \u003d 4,b.{!LANG-628b4544147306efc1d8374a7f415b2c!}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 equation

oh2 + b.x + c \u003d 0{!LANG-45778589a2a7b55c5974abb028722232!}

in){!LANG-8d110e9fc36ae6225a7879d2b4d1e70c!} 2x2 {!LANG-0acef791eb3e2502e94ca7622c60c6c8!}

a \u003d 2,b.{!LANG-c0526833db39e4c4284f7d2ed0493c73!}D.= b.2 - 4 aC= 3 2 - 4 2 4 = 9 - 32 = - 13, D.< 0.

{!LANG-3eba5e1b4b85ba063f3deda3d9e15c7f!}
{!LANG-5ba6999a2763295fa8cbf9e11300ab30!}

This equation has no root.

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

the equation oh2 + b.x + c \u003d 0 It does not have roots.

Formula (1) of the roots of the square equation oh2 + b.x + c \u003d 0 Allows you to find roots anyone Square equation (if any), including the above and incomplete. Valid formula (1) is expressed as: the roots of the square equation are equal to the fraction, the numerator of which is equal to the second coefficient taken with the opposite sign, plus minus the root square from the square of this coefficient without the still-standing product of the first coefficient to the free member, and the denominator has a double-coefficient.

4. Method: {!LANG-14ddd4bcc52e3a103fa2c2425a12551e!}

As you know, the reduced square equation has the form

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

His roots satisfy the Vieta theorem, which a \u003d 1. Has appearance

/>x.1 x.2 = q.,

x.1 + x.2 = - p.

From here you can draw the following conclusions (according to the coefficients p and q you can predict the signs of the roots).

a) if a consolidated member q. The given equation (1) is positive ( q.> 0 ), the equation has two identical root signs and is the envy of the second coefficient p.. If a r< 0 then both root are negative if r< 0 , both root are positive.

For example,

x.2 – 3 x.+ 2 = 0; x.1 = 2 and x.2 = 1, as q.= 2 > 0 and p.= - 3 < 0;

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

b) if a free member q. The given equation (1) is negative ( q.< 0 ), the equation has two different on the root sign, and the root larger in the module will be positive if p.< 0 {!LANG-87ca8f4172234ee63c4dccc8c0581d4e!} p.> 0 .

For example,

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

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

5. Method: Solving equations by the method of "transit".

Consider a square equation

oh2 + b.x + c \u003d 0,where {!LANG-f12f4e78940b7478010c4b13b01fa899!}

Multiplying both parts by a, we get the equation

but2 h.2 {!LANG-1805b37066d451d87146e9513833657a!}b.x + ac \u003d 0.

Let be ah \u003d uFrom! x \u003d y / a; then come to the equation

w.2 + by{!LANG-9c2eca29d11fd879775bd887f8de9e84!}

equivalent to this. His roots w.1 and w.{!LANG-db9246bb2f994233747433ad11c72cc1!}

Finally get

h.1 {!LANG-a9e6de034b5660a9b883e934a6efd99f!}1 /butand h.1 {!LANG-a9e6de034b5660a9b883e934a6efd99f!}2 /but.

Example.

Resolving equation 2x2 {!LANG-70041768faf24e14a9250a5869846c5d!}

Decision. "We will transfer" a coefficient 2 to a free member, as a result we obtain the equation

w.2 {!LANG-ad17778d4ce6c713f899d887edd8131a!}

According to Viet theorem

/>/>/>/>/>w.1 \u003d 5 x1 = 5/2 x.1 = 2,5

w.2 = 6 x.2 = 6/2 x.2 = 3.

Answer: 2.5; 3.

6. Method: {!LANG-b21e2d96192e13ec4e9710361dcad64e!}

BUT. Let the square equation be given

oh2 + b.x + c \u003d 0,where {!LANG-f12f4e78940b7478010c4b13b01fa899!}

1) if, a +b.{!LANG-cbfcd028a8a711f0a9cb2d791e0c5f83!}1 = 1,

h.2 {!LANG-7c88ba024dce5108dcd5386746c2dcf1!}

Evidence. We divide both parts of the equation on a ≠ 0, we obtain the reduced square equation

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

{!LANG-0de73e82316cefedf4f9737e99c59f29!}

x.1 + x.2 = - b./ a.,

x.1 x.2 = 1 c./ a..

By condition but -b.{!LANG-918f1970b128e6740107bf66e1518896!}from b.{!LANG-60f6c12a2037f104bbf16bca5e76c31c!}In this way,

/>x.1 {!LANG-b744b3a4c620574ec79dedc93ba8c793!}2 = - but{!LANG-cbe3a7043fcf0917e261ddb7ce5a01f8!}

x.1 x.2 {!LANG-5a3731ca5aaf64914a42f7d118ef28d9!}

those. h.1 = -1 and h.2 = c./ a.that m required to prove.

Examples.

Resolving equation 345x2 {!LANG-f060895feeb09c4eb6b6719b7bc189b7!}

Decision.As a +.b.{!LANG-2d58fb476a07461e07c2b0b31b966cf1!}that

h.1 {!LANG-e30252c9e2e66016bd254e2b38db65ea!}2 = c./ a.= -208/345.

Answer: 1; -208/345.

2) solutions equation 132x2 {!LANG-5099696925bd66bd03d90cd04731d885!}

Decision.As a +.b.{!LANG-0ede508379e8e76f93d763bc3214cd80!}that

h.1 {!LANG-e30252c9e2e66016bd254e2b38db65ea!}2 = c./ a.= 115/132.

Answer: 1; 115/132.

B. {!LANG-9a2c595b267423594b7d2dcb3ca528d7!} b.= 2 k.- even number, then the root formula

{!LANG-3eba5e1b4b85ba063f3deda3d9e15c7f!}
{!LANG-690420ef93f089a77119c78c388cf7d2!}

Example.

Resolving equation 3x2 - 14x + 16 \u003d 0.

Decision. We have: a \u003d 3,b.{!LANG-5a42062d246a33f80db42ced1531d16a!}k.= - 7 ;

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

Answer: 2; 8/3

IN. {!LANG-cfaa70651883f8bcf4c48ee0bb541703!}

h.2 {!LANG-93768467211f918515a9b4670ae53bb7!}q.= 0

coincides with the general view equation in which a \u003d 1., b.{!LANG-6a526658fc9fa00942bbae2c354968f4!}and c \u003d.q.. Therefore, for the reduced square equation of the formula roots

takes the View:

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

Example.Resolving equation h.2 {!LANG-2058e521af8b239eff065616c62e5808!}

Decision.We have: h.1,2 {!LANG-109c59415a972bf96a296e2dafee519b!}

Answer: H.1 {!LANG-cacacacf426ed083c7ef0e929c00fa81!}2 = -1.

7. Method: {!LANG-7315203e0e349411abc9b32b546614f7!}

{!LANG-c4692f6a0af232a069a02e118a66fbc8!}

h.2 + px.+ q.= 0

transfer the second and third members on the right side, then we get

h.2 = - px.- q..

{!LANG-c9aafd6a4ccbc2dd7781b2a6b1027bd8!}

The first dependence schedule is parabola, passing through the origin of the coordinates. Graph of the second dependence -

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

{!LANG-96822d82155f107c1907f939552ed4c7!}

Direct and Parabola can touch (only one common point), i.e. The equation has one solution;

Direct and Parabola do not have common points, i.e. The square equation does not have roots.

Examples.

1) Grandically equation h.2 {!LANG-01b5e5b7dec92f50f79b12e827d55ed5!}{!LANG-45b1be7beff38a5e952df7c810aef88a!}

Decision.{!LANG-da666955d31165cea0f0ecb867569846!} h.2 {!LANG-e647bce0e0299da2e0f236837b8b63ec!}.

Let's build a parabola y \u003d x.2 and straight y \u003d 3x + 4. Straight

y \u003d 3x + 4{!LANG-cde85bc7e364bcdfce3fdd735e912cd5!} M (0; 4)and

N.(3; 13) . Direct and Parabolas intersect at two points

BUTand INwith abscissions h.1 = - 1 and h.2 = 4 . Answer : H.1 = - 1;

h.2 = 4.

2) {!LANG-a5eea6e42bfe92a6a6e508440b3eda4e!} h.2 {!LANG-ab00dc3432973307e0f7ebfa7af837ff!}.

Decision.We write an equation in the form h.2 {!LANG-7d8fe7f9f106956f1d4b6e9f33d9973f!}.

Let's build a parabola y \u003d x.2 and straight {!LANG-d8e0bd56edd145edd5d8847e13718314!}

Straight y \u003d 2x - 1{!LANG-6c69da338db435a6f0f445855d091be7!} M (0; - 1)

and N.(1/2; 0) . Direct and Parabola intersect at the point BUT{!LANG-91b2417d13d076fabf08a2684f817476!}

abscissa x \u003d 1.. Answer: {!LANG-4c136b51863011b6d4780458110a58a3!}

3) Grandically equation h.2 {!LANG-0a7e3bc3b6f0b6abfce251453cf3218e!}(Fig. 4).

Decision.We write an equation in the form h.2 {!LANG-1d5de88f7d0f19a1151860755b05e973!}. Let's build a parabola y \u003d x.2 and straight y \u003d 2x - 5. Straight y \u003d 2x - 5{!LANG-895884fc49fe2e0792e8cca16b6d24ea!}

Answer. The equation h.2 {!LANG-0a7e3bc3b6f0b6abfce251453cf3218e!} No roots have.

8. Method: {!LANG-bd4ec153c48ac9f5995f5e640cc5a348!}

I propose the following method of finding the roots of the square equation oh2 + b.x + c \u003d 0{!LANG-d8bd9cbd91bf09374d27449149f9f30d!}

Suppose that the desired circle crosses the axis

abscissa at points In (x.1 ; 0) and D.(H.2 ; 0), {!LANG-b4105a8365f96fbf4ef90b1000a2102e!} h.1 and h.2 {!LANG-223877438e98f30199d91cc816c53800!} oh2 + b.x + c \u003d 0and passes through points

A (0; 1)and {!LANG-06a8d97408b4d1e465b67f7d03095360!}c./ a.) {!LANG-0e14d3213357434eff01a994f5bca1df!} OB. OD.= Oa. OC.From! OC.= OB. OD./ Oa.\u003d H.1 h.2 / 1 = c./ a..

The center of the circle is located at the point of intersection of perpendiculars Sf.and SKrestored in the middle of the chord ACand BD., so

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

2) We will conduct a circle with a radius SA;

3) abscissa points of intersection of this circle with axis Oh are roots of the original square equation.

It is possible three cases.

1) Radius of the circle more ordinate center (As> SK, orR.> a.+ c./2 a.) {!LANG-e1724c4a220690cf4314fdfa653221a2!} In (x.1 ; 0) and D.(H.2 ; 0) where h.1 and h.2 - roots of the square equation oh2 + b.x + c \u003d 0.

2) Radius of the circle is equal to the ordinate center (As= SB., orR.= a.+ c./2 a.) {!LANG-f3cde32e7179dedb2b21d0a8284e9db1!} In (x.1 ; 0) {!LANG-a055e0788270272cc1352da13909b280!}

{!LANG-3eba5e1b4b85ba063f3deda3d9e15c7f!}
{!LANG-5ba6999a2763295fa8cbf9e11300ab30!}

{!LANG-b2efdf91e645433de8498866a6592c5d!}

Example.

Resolving equation h.2 {!LANG-26b349a1f3e0fc4ad33360db248b6d2b!}{!LANG-808df9ffebae437d3a59344e9d8ed5c1!}

Decision.We define the coordinates of the center of the center of the circumference by formulas:

We carry out the SA radius circle, where A (0; 1).

Answer:h.1 {!LANG-cdea095635389649929d942c090b6529!}2 = 3.

9. Method: {!LANG-b8764112131e2c68abc99c6db97d506e!}

{!LANG-24e6fd31ff76cea572f2787c22c390eb!}

Table XXII. Nomogram for solving the equation z.2 + pz.+ q.= 0 {!LANG-eaa7ce4ea16a2e7b32b5dfe61b9493ec!}

{!LANG-5d940a6e18b2dea6bcc7dfe3a6f0b5e7!}

Believed OS \u003d P,ED= q., O \u003d a{!LANG-1d68d1cd02dbf08aa3ea9dd9fc9e3a2b!} San and CDF.{!LANG-8498e1502043c9824ba54633d5ba3663!}

where after substitutions and simplifications follow the equation

z.2 + pz.+ q.= 0,

moreover, the letter z.{!LANG-0be637fd671f721c56904f6d5219e7ab!}

Examples.

1) {!LANG-24448a35b8b8c32ea5141d445385d767!} z.2 - 9 z.+ 8 = 0 {!LANG-431c7c1e5f9febeb6ce1ac97f000eb48!}

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

2) {!LANG-05818d328738cfd4f0d79fb3c797af69!}

2 z.2 - 9 z.+ 2 = 0.

{!LANG-a7fbf5d47ab07e7e30f516ca9f466b66!}

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

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

3) For equation

z.2 - 25 z.+ 66 = 0

the P and Q coefficients go beyond the scale of the scale, perform the substitution z.= 5 t.{!LANG-a9e2d1bf8b13f21edad28d0487302f82!}

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

which we solve the nomogram and get t.1 = 0,6 and t.2 = 4,4, from z.1 = 5 t.1 = 3,0 and z.2 = 5 t.2 = 22,0.

10. Method: {!LANG-1f160f8783265e54484811d95232ddd4!}

Examples.

1) solve equation h.2 {!LANG-9e38946151500871416173ee152b3436!}

In the original, this task is formulated as follows: "Square and ten roots are 39" (Fig.15).

Decision.{!LANG-d09cccce33a969291517852741dcba5f!}

Area S.{!LANG-7aa71393f7ceecb9144dc8698bdf19ba!} Abcd.{!LANG-cf01594bda6e343f6ee2c016a875bfbb!} h.2 {!LANG-ac42d332ad68f1e5cde51c9fc9bd5671!} (4 2.5x \u003d 10x){!LANG-35318bf34e128a238cd156d7f8130d35!} (6,25 4 = 25) . S.= h.2 {!LANG-0e33f7c6dbc39fceeeeea4c912f62634!}Replacing

h.2 {!LANG-6f8c3bf75524fcc1475446b073ed5a14!}{!LANG-3bc45b8941de22ba58f801122c427495!} 39 , I get that S.= 39 + 25 = 64 where it follows that the side of the square Abcd.. section AV \u003d 8.. For the desired side h.initial square get

2) But, for example, as ancient Greeks solved the equation w.2 {!LANG-839e298fe28a4532e0eb65ce1de65462!}.

Decisionpresented in Fig. 16, where

w.2 {!LANG-d757ba5e297863c407f150b3f6f94220!}2 {!LANG-1ec816e576eb4dae75301d857f2e3bdc!}

Decision. Expressions w.2 {!LANG-fe4ffc27e1fc3d2dda268338b3943b1d!} and 16 + 9 {!LANG-65c28caaffcddd696a250eed42687919!} w.2 {!LANG-b664e5e20971e29356cc74d3f62c8aea!} - the same equation. Where and get that y + 3 \u003d ± 5, or w.1 {!LANG-6a23872647a8ab4f56e15438c45fb42a!}2 = - 8 (Fig.16).

3) solve geometrically equation w.2 {!LANG-41b2955525125b9abb1dff90d7b0ad64!}

Converting the equation, get

w.2 {!LANG-bb5228538abe9b2dc3762d548bee8927!}

In fig. 17 Find "Images" of expressions w.2 {!LANG-f6bc238d6334d448589e0a707f6c4f80!}{!LANG-39e0db5106a2d6276cc5d0fcc60840ae!} 3 . So if to express w.2 {!LANG-6f6de5aeee1ba196d164cc5d3cd7adda!}{!LANG-ccb5bd7e2124e7d99c0e6fe8173c4920!} 9 , then we get the square square with the side y - 3.. Replacing expression w.2 {!LANG-6f6de5aeee1ba196d164cc5d3cd7adda!}{!LANG-2844ac619ada8400987955bffe55ec1e!}

we get: (y - 3)2 = 16 + 9, those. y - 3 \u003d ± √25, or y - 3 \u003d ± 5, where w.1 = 8 and w.2 = - 2.

Conclusion

Square equations are widely used in solving trigonometric, indicative, logarithmic, irrational and transcendental equations and inequalities.

{!LANG-088c0b7db8c6792509394f8992909c1e!}

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{!LANG-242b1032b9ded1d49176b925ca7d9ccc!}

Literature:

1. {!LANG-515fad93c4903bbb86d40453a990caf3!}

2. {!LANG-fd5869bc5aa09f7d1ee1a35ba2e8ff06!}

3. {!LANG-a32e2507e10f35023168d532be4ef3d3!}

4. {!LANG-1f532e01c0134832e892a5029469979a!}

5. {!LANG-7a727d3e2ce10ef62ea022939358ff56!}