The reduced equation. Solution of square equations: Formula roots, examples

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 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, the concept of a negative number and general methods for solving square equations are lacking in clinox texts.

1.2 As accounted for and solved diofant 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 - x 2 \u003d 96

x 2 - 4 \u003d 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,

in 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

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:

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

In equation (1) coefficients except but may 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:

x 2 - 64x \u003d -768

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

x 2 - 64x + 32 2 \u003d -768 + 1024,

(x - 32) 2 \u003d 256,

x - 32 \u003d ± 16,

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

1.4 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. Ah 2 + C \u003d b. x.

2) "Squares are equal to the number", i.e. ah 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. Ah 2 + C \u003d b. x.

5) "Squares and roots are equal to the number", i.e. ah 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, that when solving an incomplete square equation of the first type

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 » (It is understood as 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.

1.5 square equations in Europe XIII. - XVII BB

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:

x 2 +. bX. \u003d C,

for all sorts of combinations of coefficient signs b. , from 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, 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.

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. ».

To understand Vieta, you should remember that BUT like 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 \u003d aB ,

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

x 1 \u003d a, x 2 \u003d 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.

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.

Objectives:

• Introduce the concept of a given square equation;
• "Open" Dependence between roots and coefficients of a given square equation;
• develop interest in mathematics, showing the example of the life of Vieta, that mathematics can be hobbies.

During the classes

1. Checking homework

№ 309 (g) x 1 \u003d 7, x 2 \u003d

№ 311 (g) x 1 \u003d 2, x 2 \u003d -1

№ 312 (d) No roots

2. Repetition of the material studied

Everyone has a table. Find the match between the left and right tables of the table.

Wondering wording	Literal expression
1. Square threechests	A. Ah 2 \u003d 0
2. Discriminant	B. Ah 2 + C \u003d 0, with< 0
3. An incomplete square equation having one root is equal to 0.	IN. D\u003e 0.
4. An incomplete square equation, one root of which 0, and the other is not equal to 0.	G. D.< 0
5. Not a complete square equation, the roots of which are equal to the module, but are opposed to the sign.	D. Ah 2 + WX + C \u003d 0
6. Not a complete square equation that does not have valid roots.	E. D \u003d in 2 + 4as
7. General view of the square equation.	J. x 2 + Rh + Q \u003d 0
8. Condition in which the square equation has two roots	Z. ah 2 + vx + with
9. Condition at which the square equation does not have roots	AND. Ah 2 + C \u003d 0, C\u003e 0
10. The condition in which the square equation has two equal roots	TO. Ah 2 + Vh \u003d 0
11. The reduced square equation.	L. D \u003d 0.

Right answers bring to the table.

1-s; 2nd; 3-A; 4-K; 5 B; 6th; 7-d; 8-in; 9-g; 10-l; 11th.

3. Fastening the material studied

Decide equations:

a) -5x 2 + 8x -3 \u003d 0;

Decision:

D \u003d 64 - 4 (-5) (- 3) \u003d 4,

x 1 \u003d x 2 \u003d \u003d a + in + c \u003d -5 + 8-3 \u003d 0

b) 2 x 2 + 6x - 8 \u003d 0;

Decision:

D \u003d 36 - 4 2 (-8) \u003d 100,

x 1 \u003d \u003d x 2 \u003d a + b + c \u003d 2 + 6-8 \u003d 0

c) 2009 x 2 + x - 2010 \u003d 0

Decision:

a + B + c \u003d 2009 + 1 + (-2010) \u003d 0, then x 1 \u003d 1 x 2 \u003d

4. Extension of the school courage

ah 2 + vkh + c \u003d 0, if a + b + c \u003d 0, then x 1 \u003d 1 x 2 \u003d

Consider the solution of equations

a) 2x 2 + 5x +3 \u003d 0

Decision:

D \u003d 25 -24 \u003d 1 x 1 \u003d x 2 \u003d a - B + C \u003d 2-5 + 3 \u003d 0

b) -4x 2 -5x -1 \u003d 0

Decision:

D \u003d 25 - 16 \u003d 9 x 1 \u003d - 1 x 2 \u003d A -B + C \u003d -4 - (- 5) - 1 \u003d 0

c) 1150x 2 + 1135x -15 \u003d 0

Decision:

a - B + C \u003d 1150-1135 + (- 15) \u003d 0 x 1 \u003d - 1 x 2 \u003d

ah 2 + vx + c \u003d 0, if A-B + C \u003d 0, then x 1 \u003d - 1 x 2 \u003d

5. New topic

Check your first task. What new concepts were outstretched. 11 - Well, i.e.

The reduced square equation - x 2 + Px + Q \u003d 0.

The theme of our lesson.
Fill the following table.
The left column themselves in the notebooks and one student at the board.
Solution equation ah 2 + WX + C \u003d 0
Right column, more prepared student at the board
Solution equation x 2 + px + q \u003d 0, at a \u003d 1, in \u003d p, c \u003d q

The teacher (if necessary) helps, the rest in the notebooks.

6. Practical part

X 2 - 6 h. + 8 = 0,	D \u003d 9 - 8 \u003d 1,	x 1 \u003d 3 - 1 \u003d 2	x 2 \u003d 3 + 1 \u003d 4
X 2 + 6 h. + 8 = 0,	D \u003d 9 - 8 \u003d 0,	x 1 \u003d -3 - 1 \u003d -4	x 2 \u003d -3 + 1 \u003d -2
X 2 + 20 h. + 51 = 0,	D \u003d 100 - 51 \u003d 49	x 1 \u003d 10 - 7 \u003d 3	x 2 \u003d 10 + 7 \u003d 17
X 2 - 20 h. – 69 = 0,	D \u003d 100 - 69 \u003d 31

According to the results of our calculations, fill the table.

No. equation	r	x 1+ x 2	q.	x 1 x 2
1	-6	6	8	8

Compare the results obtained with the coefficients of square equations.
What conclusion can be done?

7. Historical certificate

For the first time, the dependence between the roots and coefficients of the square equation was established by the famous French scientist Francois Viet (1540-1603).

Francois Viet was in a profession to a lawyer and worked for many years as an adviser to the king. And although mathematics was his hobby, or as the hobby say, thanks to his stubborn work, he achieved big results in it. Wiets in 1591 introduced alphabetic designation for unknown and coefficients of equations. What made it possible to record the main formulas of the roots and other properties of the equation.

The disadvantage of the Vieta algebra was that he recognized only positive numbers. To avoid negative solutions, it replaced the equation or searched for artificial techniques of solutions, which took a long time, complicated the solution and often led to errors.

Many different discoveries made Vieta, but he himself most appropriate to establish the relationship between the roots and coefficients of the square equation, that is, the dependence called the "Vieta Theorem".

We will consider this theorem in the next lesson.

8. Summary of Knowledge

Questions:

1. What equation is called the given square equation?
2. What formula can you find the roots of the given square equation?
3. What depends on the number of roots of the given square equation?
4. What is called the discriminant of the given square equation?
5. How are the roots of the present square equation and its coefficients?
6. Who installed this connection?

9. Homework

p. 4.5, №321 (b, e) №322 (a, g, g, s)

Fill the table.

The equation	Roots	The sum of the roots	Production of roots
X 2 - 8x + 7 \u003d 0	1 and 7.	8	7

Literature

CM. Nikolskyet al., "Algebra 8" Tutorial of the MGU-School series - M.: Enlightenment, 2007.

"That is, the equations of the first degree. In this lesson we will analyze what is called a square equation And how to solve it.

What is called a square equation

Important!

The degree of equation is determined by the greatest extent in which an unknown one is.

If the maximum degree in which the unknown is "2", it means that you are a square equation.

Examples of square equations

• 5x 2 - 14x + 17 \u003d 0
• -X 2 + x +

  1

  3

  = 0
• x 2 + 0.25x \u003d 0
• x 2 - 8 \u003d 0

Important! The general view of the square equation looks like this:

A x 2 + b x + c \u003d 0

"A", "B" and "C" - specified numbers.

• "A" is the first or senior coefficient;
• "B" - the second coefficient;
• "C" is a free member.

To find "a", "b" and "c" you need to compare your equation with a common view of the square equation "AX 2 + BX + C \u003d 0".

Let's take care of determining the coefficients "A", "B" and "C" in square equations.

5x 2 - 14x + 17 \u003d 0 -7x 2 - 13x + 8 \u003d 0 -X 2 + x +

The equation	Factors
a \u003d 5. b \u003d -14. c \u003d 17.
a \u003d -7. b \u003d -13 c \u003d 8.

1

3

= 0

• a \u003d -1.
• b \u003d 1.
• c \u003d.

  1

  3

x 2 + 0.25x \u003d 0

• a \u003d 1.
• b \u003d 0.25
• c \u003d 0.

x 2 - 8 \u003d 0

• a \u003d 1.
• b \u003d 0.
• c \u003d -8.

How to solve square equations

In contrast to linear equations for solving square equations, a special formula for finding roots.

Remember!

To solve the square equation you need:

• create a square equation to the total type "AX 2 + BX + C \u003d 0". That is, only "0" should remain in the right part;
• use the root formula:

Let's analyze on the example, how to apply the formula for finding the roots of the square equation. Let the square equation.

X 2 - 3x - 4 \u003d 0

The "X 2 - 3X - 4 \u003d 0" equation is already given to the total appearance of "AX 2 + BX + C \u003d 0" and does not require additional simplifications. To solve it, we have enough to apply the formula of finding the roots of the square equation.

We define the coefficients "A", "B" and "C" for this equation.

x 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d

With it, any square equation is solved.

In the formula "x 1; 2 \u003d" often replace the guided expression
"B 2 - 4AC" on the letter "D" and is called discriminant. The concept of discriminant is considered in more detail in the lesson "What is discriminant".

Consider another example of a square equation.

x 2 + 9 + x \u003d 7x

In this form, determine the coefficients "A", "B" and "C" is quite difficult. Let's first give the equation to the general type "AX 2 + BX + C \u003d 0".

X 2 + 9 + x \u003d 7x
x 2 + 9 + x - 7x \u003d 0
x 2 + 9 - 6x \u003d 0
x 2 - 6x + 9 \u003d 0

Now you can use the root formula.

X 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d
x 1; 2 \u003d
x \u003d.

6

2

x \u003d 3.
Answer: x \u003d 3

There are cases when there are no roots in square equations. This situation occurs when a negative number is under the root.

In modern society, the ability to perform actions with the equations containing the variable raised to the square can be useful in many areas of activity and is widely used in practice in scientific and technical developments. Evidence of this can serve the design of marine and river vessels, aircraft and missiles. With the help of such calculations, the trajectories of the movement of various bodies, including space objects. Examples with a solution of square equations are used not only in economic forecasting, in the design and construction of buildings, but also in the most ordinary everyday circumstances. They may be needed in tourist campaigns, in sports, in shopping stores and in other very common situations.

We break the expression on the components of multipliers

The degree of equation is determined by the maximum value of the degree in the variable, which contains this expression. In the event that it is 2, then such an equation is just called square.

If the language of the formulas is expressing, then the indicated expressions, no matter how they look, can always be caused by the form when the left part of the expression consists of three terms. Among them: AX 2 (that is, the variable erected into a square with its coefficient), BX (unknown without a square with its coefficient) and C (free component, that is, the usual number). All this in the right side is equal to 0. In the case when there is no one of its components of the terms, with the exception of AX 2, it is called an incomplete square equation. Examples with solving such tasks, the value of the variables in which it is easy to find, should be considered first.

If the expression appears in the form looks in such a way that two, more precisely, AX 2 and BX, the expression on the expression on the expression on the right side, is easiest to find a variable for brackets. Now our equation will look like this: x (ax + b). Next, it becomes obvious that or x \u003d 0, or the task is reduced to finding a variable from the following expression: AX + B \u003d 0. The specified dictated one of the multiplication properties. The rule says that the product of two factors gives as a result of 0 only if one of them is zero.

Example

x \u003d 0 or 8x - 3 \u003d 0

As a result, we obtain two roots of the equation: 0 and 0.375.

The equations of this kind can describe the movement of bodies under the influence of gravity, which began movement from a certain point adopted at the beginning of the coordinates. Here, the mathematical record takes the following form: Y \u003d V 0 T + GT 2/2. Substituting the necessary values, equating the right side 0 and finding possible unknowns, you can find out the time passing from the moment of the body's rise until its fall, as well as many other values. But we will talk about it later.

Decomposition of the expression on multipliers

The rule described above makes it possible to solve the specified tasks and in more complex cases. Consider examples with solving square equations of this type.

X 2 - 33x + 200 \u003d 0

This square triple is complete. To begin with, we transform the expression and decompose it for multipliers. They are obtained two: (x-8) and (x-25) \u003d 0. As a result, we have two roots 8 and 25.

Examples with solving square equations in grade 9 allow this method to find a variable in expressions not only the second, but even the third and fourth orders.

For example: 2x 3 + 2x 2 - 18x - 18 \u003d 0. With the decomposition of the right part of the multipliers with a variable, they are obtained three, that is, (x + 1), (x-3) and (x + 3).

As a result, it becomes obvious that this equation has three roots: -3; -one; 3.

Extract square root

Another case of the incomplete equation of the second order is the expression, in the language of the letters presented in such a way that the right side is built from the components of AX 2 and C. Here, for the value of the variable, the free member is transferred to the right side, and then a square root is extracted from both parts of equality. It should be noted that in this case the roots of the equation usually two. An exception can only be equal to equality, generally not containing the term C, where the variable is zero, as well as the options for expressions, when the right side turns out to be negative. In the latter case, the solutions do not exist at all, since the above action cannot be produced with roots. Examples of solutions of square equations of this type must be considered.

In this case, the roots of the equation will be -4 and 4.

Calculation of a land plot

The need for such calculations appeared in deep antiquity, because the development of mathematics in many respects in those distant times was due to the need to determine the most accuracy of the area and the perimeter of land plots.

Examples with solving square equations drawn up on the basis of tasks of this kind should be considered to us.

So, let's say there is a rectangular plot of land, the length of which is 16 meters more than the width. It should be found a length, width and perimeter of the site, if it is known that its area is equal to 612 m 2.

Starting a matter, first make the necessary equation. Denote by x the width of the site, then its length will be (x + 16). From the written it follows that the area is determined by the expression x (x + 16), which, according to the condition of our problem, is 612. This means that x (x + 16) \u003d 612.

The solution of complete square equations, and this expression is precisely such, cannot be carried out by the same way. Why? Although the left side of it still contains two factors, the product is not at all equal to 0, so other methods are used here.

Discriminant

First of all, we will produce the necessary conversion, then the appearance of this expression will look like this: x 2 + 16x - 612 \u003d 0. This means we got an expression in the form corresponding to the previously specified standard, where a \u003d 1, b \u003d 16, c \u003d -612.

This can be an example of solving square equations through discriminant. Here, the required calculations are made according to the scheme: d \u003d b 2 - 4ac. This auxiliary value does not just make it possible to find the desired values \u200b\u200bin the second order equation, it determines the number of possible options. In case D\u003e 0, there are two; When d \u003d 0, there is one root. In case D<0, никаких шансов для решения у уравнения вообще не имеется.

About roots and their formula

In our case, the discriminant is: 256 - 4 (-612) \u003d 2704. This suggests that the answer from our task exists. If you know, K, the solution of square equations must be continued using the formula below. It allows you to calculate the roots.

This means that in the case presented: x 1 \u003d 18, x 2 \u003d -34. The second version in this dilemma cannot be a solution, because the dimensions of the land can not be measured in negative values, it means x (i.e. the width of the site) is 18 m. From here, we calculate the length: 18 + 16 \u003d 34, and perimeter 2 (34+ 18) \u003d 104 (m 2).

Examples and objectives

We continue to study square equations. Examples and a detailed solution of several of them will be given further.

1) 15x 2 + 20x + 5 \u003d 12x 2 + 27x + 1

We transfer everything to the left part of equality, we will make a transformation, that is, we obtain the form of the equation that is called standard, and equalize it with zero.

15x 2 + 20x + 5 - 12x 2 - 27X - 1 \u003d 0

After folding like, we define the discriminant: d \u003d 49 - 48 \u003d 1. So, our equation will have two roots. We calculate them according to the above formula, which means that the first one of them is 4/3, and the second one.

2) Now reveal the riddles of another kind.

Find out, is there any roots here x 2 - 4x + 5 \u003d 1? To obtain a comprehensive response, we give a polynomial to the appropriate familiarity and calculate the discriminant. In the specified example, the solution of the square equation is not necessary, because the essence of the task is not at all this. In this case, D \u003d 16 - 20 \u003d4, which means there are really no roots.

Vieta theorem

Square equations are conveniently solved through the above formulas and discriminant when the square root is extracted from the last value. But it happens not always. However, there are many ways to obtain variables in this case. Example: solutions of square equations on the Vieta Theorem. She is named after which lived in the XVI century in France and made a brilliant career due to his mathematical talent and courtyards. Portrait of it can be seen in the article.

The pattern that the famous French noted was as follows. He proved that the roots of the equation in the amount are numerically equal to -p \u003d b / a, and their product corresponds to q \u003d c / a.

Now consider specific tasks.

3x 2 + 21x - 54 \u003d 0

For simplicity, we transform the expression:

x 2 + 7x - 18 \u003d 0

We use the Vieta theorem, it will give us the following: the amount of the roots is -7, and their work -18. From here, we obtain that the roots of the equation are numbers -9 and 2. Having made a check, make sure that these values \u200b\u200bof variables are really suitable in the expression.

Graph and Parabola equation

Concepts The quadratic function and square equations are closely connected. Examples of this have already been shown earlier. Now consider some mathematical riddles a little more. Any equation of the described type can be imagined. A similar dependence drawn in the form of a graph is called a parabola. Her various types are shown in the figure below.

Any parabola has a vertex, that is, the point from which its branches come out. In case a\u003e 0, they leave high in infinity, and when a<0, они рисуются вниз. Простейшим примером подобной зависимости является функция y = x 2 . В данном случае в уравнении x 2 =0 неизвестное может принимать только одно значение, то есть х=0, а значит существует только один корень. Это неудивительно, ведь здесь D=0, потому что a=1, b=0, c=0. Выходит формула корней (точнее одного корня) квадратного уравнения запишется так: x = -b/2a.

Visual images of functions help solve any equations, including square. This method is called graphic. And the value of the variable x is the coordinate of the abscissa at points where the graph of the graph is crossing from 0x. The coordinates of the vertices can be found according to the only given formula X 0 \u003d -B / 2A. And, substituting the resulting value to the initial equation of the function, you can learn Y 0, that is, the second coordinate of the pearabol vertex belonging to the ordinate axis.

Crossing the branches of parabola with the abscissa axis

Examples with solutions of square equations are very much, but there are general patterns. Consider them. It is clear that the intersection of the graph with the axis 0x at a\u003e 0 is only possible if 0 receives negative values. And for A.<0 координата у 0 должна быть положительна. Для указанных вариантов D>0. Otherwise D<0. А когда D=0, вершина параболы расположена непосредственно на оси 0х.

According to the chart, the parabolas can be defined and roots. The opposite is also true. That is, if you get a visual image of a quadratic function is not easy, you can equate the right side of the expression to 0 and solve the obtained equation. And knowing the intersection points with the 0x axis, it is easier to build a schedule.

From the history

With the help of equations containing the variable raised to the square, in the old days not only made mathematical calculations and determined the area of \u200b\u200bgeometric figures. Similar calculations of the ancient were needed for grand discoveries in the field of physics and astronomy, as well as to compile astrological forecasts.

As the modern science figures suggest, among the first solutions of square equations, residents of Babylon took up. It happened in four centuries before the onset of our era. Of course, their calculations in the root differed from now adopted and turned out to be much primitive. For example, Mesopotamian mathematicians had no idea about the existence of negative numbers. The strangers also had other subtleties from those who know any student of our time.

Perhaps even earlier scientists of Babylon, the solution of square equations, a sage of India Budhoyama was engaged. It happened in about eight centuries before the era of Christ. True, the equation of the second order, the methods of solving which he led was the most simultaneous. In addition to him, such questions were interested in old and Chinese mathematicians. In Europe, the square equations began to solve only in the early XIII century, but later they were used in their work such great scientists as Newton, Descartes and many others.

Square equations are studied in grade 8, so there is nothing difficult here. The ability to solve them is absolutely necessary.

The square equation is the equation of the form AX 2 + BX + C \u003d 0, where the coefficients A, B and C are arbitrary numbers, and a ≠ 0.

Before studying specific decision methods, we note that all square equations can be divided into three classes:

1. Do not have roots;
2. Have exactly one root;
3. Have two different roots.

This is an important difference between square equations from linear, where the root always exists and is unique. How to determine how many roots have an equation? For this there is a wonderful thing - discriminant.

Discriminant

Let the square equation AX 2 + BX + C \u003d 0. Then the discriminant is just the number d \u003d b 2 - 4ac.

This formula must be known by heart. Where she takes - now it does not matter. Other It is important: the discriminant sign can be determined how many roots has a square equation. Namely:

1. If D< 0, корней нет;
2. If D \u003d 0, there is exactly one root;
3. If D\u003e 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all on their signs, as for some reason, many consider. Take a look at the examples - and you will understand everything:

A task. How many roots are square equations:

1. x 2 - 8x + 12 \u003d 0;
2. 5x 2 + 3x + 7 \u003d 0;
3. x 2 - 6x + 9 \u003d 0.

We repel the coefficients for the first equation and find the discriminant:
a \u003d 1, b \u003d -8, c \u003d 12;
D \u003d (-8) 2 - 4 · 1 · 12 \u003d 64 - 48 \u003d 16

So, the discriminant is positive, so the equation has two different roots. Similarly, disassemble the second equation:
a \u003d 5; b \u003d 3; C \u003d 7;
D \u003d 3 2 - 4 · 5 · 7 \u003d 9 - 140 \u003d -131.

The discriminant is negative, no roots. The last equation remains:
a \u003d 1; b \u003d -6; C \u003d 9;
D \u003d (-6) 2 - 4 · 1 · 9 \u003d 36 - 36 \u003d 0.

The discriminant is zero - the root will be one.

Please note that for each equation the coefficients were discharged. Yes, it's a long time, yes, it's a tedious - but you will not confuse the coefficients and do not allow stupid mistakes. Choose yourself: speed or quality.

By the way, if you "fill the hand", after a while no longer need to write all the coefficients. Such operations you will be performed in your head. Most people begin to do so somewhere after 50-70 solved equations - in general, not so much.

Roots square equation

We now turn, actually, to the decision. If discriminant D\u003e 0, roots can be found by formulas:

The basic formula of the roots of the square equation

When D \u003d 0, you can use any of these formulas - it will be the same number that will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 - 2x - 3 \u003d 0;
2. 15 - 2x - x 2 \u003d 0;
3. x 2 + 12x + 36 \u003d 0.

First equation:
x 2 - 2x - 3 \u003d 0 ⇒ a \u003d 1; b \u003d -2; C \u003d -3;
D \u003d (-2) 2 - 4 · 1 · (-3) \u003d 16.

D\u003e 0 ⇒ The equation has two roots. Find them:

Second equation:
15 - 2x - x 2 \u003d 0 ⇒ a \u003d -1; b \u003d -2; C \u003d 15;
D \u003d (-2) 2 - 4 · (-1) · 15 \u003d 64.

D\u003e 0 ⇒ The equation again has two roots. We find them

\\ [\\ begin (Align) & ((x) _ (1)) \u003d \\ FRAC (2+ \\ SQRT (64)) (2 \\ Cdot \\ left (-1 \\ right)) \u003d - 5; \\\\ & ((x) _ (2)) \u003d \\ FRAC (2- \\ SQRT (64)) (2 \\ Cdot \\ left (-1 \\ right)) \u003d 3. \\\\ \\ END (ALIGN) \\]

Finally, the third equation:
x 2 + 12x + 36 \u003d 0 ⇒ a \u003d 1; b \u003d 12; C \u003d 36;
D \u003d 12 2 - 4 · 1 · 36 \u003d 0.

D \u003d 0 ⇒ The equation has one root. You can use any formula. For example, the first:

As can be seen from examples, everything is very simple. If you know the formula and be able to consider, there will be no problems. Most often, errors occur during substitution in the formula of negative coefficients. Here, again, the reception described above will help: look at the formula literally, paint every step - and very soon get rid of errors.

Incomplete square equations

It happens that the square equation is somewhat different from what is given in the definition. For example:

1. x 2 + 9x \u003d 0;
2. x 2 - 16 \u003d 0.

It is easy to see that in these equations there is no one of the terms. Such square equations are even easier than standard: they do not even need to consider discriminant. So, we introduce a new concept:

The AX 2 + BX + C \u003d 0 equation is called an incomplete square equation if B \u003d 0 or C \u003d 0, i.e. The coefficient with a variable x or the free element is zero.

Of course, a completely difficult case is possible when both of these coefficients are zero: b \u003d c \u003d 0. In this case, the equation takes the form AX 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.

Consider the remaining cases. Let b \u003d 0 be 0, then we obtain an incomplete square equation of the form AX 2 + C \u003d 0. We convert it a little:

Since the arithmetic square root exists only from a non-negative number, the latter equality makes sense exclusively at (-C / a) ≥ 0. Conclusion:

1. If in an incomplete square equation of the form AX 2 + C \u003d 0, inequality (-C / a) is performed ≥ 0, there will be two roots. The formula is given above;
2. If (-c / a)< 0, корней нет.

As you can see, the discriminant did not need - in incomplete square equations there are no complex computing. In fact, even it is not necessary to remember the inequality (-c / a) ≥ 0. It is enough to express the value of X 2 and see what stands on the other side of the equality sign. If there is a positive number - the roots will be two. If negative - the roots will not be at all.

Now we will understand with the equations of the form AX 2 + BX \u003d 0, in which the free element is zero. Everything is simple here: the roots will always be two. It is enough to decompose a polynomial to multipliers:

Multiplier for bracket

The work is zero, when at least one of the multipliers is zero. From here there are roots. In conclusion, we will analyze several such equations:

A task. Square square equations:

1. x 2 - 7x \u003d 0;
2. 5x 2 + 30 \u003d 0;
3. 4x 2 - 9 \u003d 0.

x 2 - 7x \u003d 0 ⇒ x · (x - 7) \u003d 0 ⇒ x 1 \u003d 0; x 2 \u003d - (- 7) / 1 \u003d 7.

5x 2 + 30 \u003d 0 ⇒ 5x 2 \u003d -30 ⇒ x 2 \u003d -6. No roots, because Square cannot be equal to a negative number.

4x 2 - 9 \u003d 0 ⇒ 4x 2 \u003d 9 ⇒ x 2 \u003d 9/4 ⇒ x 1 \u003d 3/2 \u003d 1.5; x 2 \u003d -1.5.