View equation

Expression D. \u003d B. 2 - 4 AC. Call discriminant square equation. If a D. \u003d 0, the equation has one valid root; If D \u003e 0, the equation has two valid roots.
In the case when D. = 0 Sometimes they say that the square equation has two identical roots.
Using the designation D. \u003d B. 2 - 4 AC. , you can rewrite formula (2) as

If a B. \u003d 2 K. The formula (2) takes the form:

where K. \u003d B. / 2 .
The last formula is especially convenient in cases where B. / 2 - integer, i.e. coefficient B. - even number.
Example 1: Solve equation 2 X. 2 - 5 X. + 2 = 0 . Here a \u003d 2, b \u003d -5, c \u003d 2. Have D. \u003d B. 2 - 4 AC. = (-5) 2- 4*2*2 = 9 . As D. > 0 , the equation has two roots. Find them by formula (2)

so X. 1 \u003d (5 + 3) / 4 \u003d 2, x 2 =(5 - 3) / 4 = 1 / 2 ,
i.e X. 1 = 2 and X. 2 = 1 / 2 - The roots of the specified equation.
Example 2: Solve equation 2 X. 2 - 3 X. + 5 = 0 . Here a \u003d 2, b \u003d -3, c \u003d 5. We find discriminant D. \u003d B. 2 - 4 AC. = (-3) 2- 4*2*5 = -31 . As D. 0 The equation does not have valid roots.

Incomplete square equations. If in the square equation AX. 2 + BX. + C. =0 The second coefficient B. or free dick C. equal to zero, then the square equation is called incomplete. Incomplete equations are isolated because to find their roots, it is possible not to use the root formula of the square equation - it is easier to solve the equation by the method of decomposition of its left part of the factors.
Example 1: Solve equation 2 X. 2 - 5 X. = 0 .
Have X. (2 X. - 5) = 0 . So either X. = 0 either 2 X. - 5 = 0 , i.e X. = 2.5 . So the equation has two roots: 0 and 2.5
Example 2: Solve equation 3 X. 2 - 27 = 0 .
Have 3 X. 2 = 27 . Consequently, the roots of this equation - 3 and -3 .

Vieta theorem. If the reduced square equation X. 2 + px. + Q. =0 has valid roots, then their amount is equal - P. , and the work is equal Q. , i.e

x 1 + x 2 \u003d -p,
x 1 x 2 \u003d Q

(The sum of the roots of the given square equation is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to a free member).

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 decision square equations Find applications 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 right side, and then from both parts of equality is extracted square root. 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 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 later.

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 resulting 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.

Quadratic equations. Discriminant. Solution, examples.

Attention!
This topic has additional
Materials in a special section 555.
For those who are strongly "not very ..."
And for those who are "very ...")

Types of square equations

What is a square equation? What does it look like? In terms quadratic equation Keyword is "Square". It means that in the equation before Must be at the square in the square. Besides him, in the equation can be (and may not be!) Simply X (in the first degree) and just the number (free member). And there should be no ICS to a degree, more two.

Speaking by mathematical language, the square equation is the equation of the form:

Here a, b and with - some numbers. b and C. - all any, and but- anyone but zero. For example:

Here but =1; b. = 3; c. = -4

Here but =2; b. = -0,5; c. = 2,2

Here but =-3; b. = 6; c. = -18

Well, you understood ...

In these square equations, the left is present full set members. X Square with a coefficient but,x in the first degree with the coefficient b. and free dick with.

Such square equations are called full.

What if b. \u003d 0, What do we do? We have the X is the first degree disappear. From multiplication to zero it happens.) It turns out, for example:

5x 2 -25 \u003d 0,

2x 2 -6x \u003d 0,

- 2 + 4x \u003d 0

Etc. And if both coefficient, b. and c. equal to zero, it's still simpler:

2x 2 \u003d 0,

-0.3x 2 \u003d 0

Such equations where something is missing is called incomplete square equations. What is quite logical.) I ask you to notice that X is present in the square in all equations.

By the way, why but Can not be zero? And you substitute instead but Nolik.) We will disappear in the square! The equation will become linear. And it is already solved quite differently ...

That's all the main types of square equations. Full and incomplete.

Solution of square equations.

Solving full square equations.

Square equations are simply solved. According to formulas and clearly simple rules. At the first stage, a given equation must be brought to the standard form, i.e. To mind:

If the equation is given to you already in this form - the first stage is not needed.) The main thing is to correctly define all the coefficients, but, b. and c..

The formula for finding the roots of the square equation looks like this:

The expression under the sign of the root is called discriminant. But about it - below. As you can see, to find the ICA, we use only a, b and with. Those. The coefficients of the square equation. Just neatly substitute the values a, b and with In this formula and we consider. Substitute with your signs! For example, in equation:

but =1; b. = 3; c. \u003d -4. Here and write:

An example is practically solved:

This is the answer.

Everything is very simple. And what do you think it is impossible to make a mistake? Well, yes, how ...

The most common mistakes - confusion with signs of values a, b and with. Rather, not with their signs (where is there confused?), And with the substitution of negative values \u200b\u200bin the formula for calculating the roots. Here is a detailed entry of the formula with specific numbers. If there are problems with computing, do so!

Suppose you need to solve this one:

Here a. = -6; b. = -5; c. = -1

Suppose you know that you rarely have answers from the first time.

Well, do not be lazy. Write an excess line will take seconds 30. And the number of errors sharply cut. Here we write in detail, with all brackets and signs:

It seems incredibly difficult, so carefully paint. But it only seems. Try. Well, or choose. What is better, fast, or right? Also, I'll kick you. After a while, there will disappear so carefully to paint everything. Itself will be right. Especially if you apply practical techniques, which are described just below. This evil example with a bunch of minuses will be solved easily and without errors!

But, often, square equations look slightly different. For example, like this:

Find out?) Yes! it incomplete square equations.

Decision of incomplete square equations.

They can also be solved by the general formula. It is only necessary to correctly imagine what is equal to a, b and with.

Corrected? In the first example a \u003d 1; b \u003d4; but c.? There is no one at all! Well, yes, right. In mathematics, this means that c \u003d 0. ! That's all. We substitute in the zero formula instead c, And everything will turn out. Similarly, with the second example. Only zero here do not from, but b. !

But incomplete square equations can be solved much easier. Without any formulas. Consider the first incomplete equation. What can be done there in the left side? You can make the IS for brackets! Let's bring out.

And what from this? And the fact that the work is zero then, and only when some of the multipliers equals zero! Do not believe? Well, come up with two non-zero numbers, which will give zero with multiply!
Does not work? That's something ...
Consequently, you can confidently write: x 1 \u003d 0, x 2 \u003d 4.

Everything. This will be the roots of our equation. Both are suitable. When substituting any of them into the original equation, we obtain a faithful identity 0 \u003d 0. As you can see, the solution is much simpler than the general formula. I note, by the way, which X will be the first, and which second is absolutely indifferent. Convenient to record in a few, x 1 - what is less, and x 2 - What is more.

The second equation can also be solved simply. We carry 9 to the right side. We get:

It remains the root to extract out of 9, and that's it. It turns out:

Also two roots . x 1 \u003d -3, x 2 \u003d 3.

So all incomplete square equations are solved. Either by means of making a bracket, or by simply transferring the number to the right, followed by the extraction of the root.
It is extremely difficult to confuse these techniques. Simply because in the first case you will have to extract the root from XCA, which is somehow it is not clear, and in the second case, it is nothing for brackets ...

Discriminant. Discriminant formula.

Magic word discriminant ! A rare high school student did not hear the word! The phrase "decide through the discriminant" will instill confidence and encourages. Because it is not necessary to wait for the tricks from the discriminant! It is simple and trouble-free in circulation.) I remind you of the most general formula for solving any Square equations:

The expression under the sign of the root is called discriminant. Usually discriminant is indicated by the letter D.. Discriminant formula:

D \u003d b 2 - 4ac

And what is noteworthy expression? Why did it deserve a special name? In what meaning of discriminant? After all -b, or 2a. In this formula, they do not specifically call ... letters and letters.

The thing is what. When solving a square equation for this formula, it is possible total three cases.

1. Discriminant positive. This means that it is possible to extract the root. Good root is extracted, or bad - the question is different. It is important that it is extracted in principle. Then your square equation has two roots. Two different solutions.

2. The discriminant is zero. Then you get one solution. Since the zero subtracting in the numerator does not change anything. Strictly speaking, this is not one root, but two identical. But, in the simplified version, it is customary to talk about one solution.

3. The discriminant is negative. Of the negative number, the square root is not removed. Well, okay. This means that there are no solutions.

To be honest, with a simple solution of square equations, the concept of discriminant is not particularly required. We substitute the values \u200b\u200bof the coefficients in the formula, yes, we believe. It all happens everything, both two roots, and one, and not one. However, when solving more complex tasks, without knowing meaning and formula discriminant not enough. Especially - in equations with parameters. Such equations are the highest pilot on GIA and EGE!)

So, how to solve square equations Through the discriminant you remembered. Or learned that it is also not bad.) I know how to correctly determine a, b and with. Knowledge carefully substitute them in the root formula and carefully count the result. You understood that the key word is here - carefully?

And now take note of practical techniques that dramatically reduce the number of errors. The most that because of the inattention. ... for which then it happens hurt and hurt ...

Reception First . Do not be lazy before solving the square equation to bring it to the standard form. What does this mean?
Suppose, after all transformations, you received such an equation:

Do not rush to write the root formula! Almost probably, you confuse the coefficients a, b and s. Build an example correctly. First, X is in the square, then without a square, then a free dick. Like this:

And do not rush again! The minus in front of the ix in the square can be healthy to upset you. Forget it easy ... Get rid of a minus. How? Yes, as taught in the previous topic! It is necessary to multiply the entire equation on -1. We get:

But now you can safely record the formula for the roots, consider the discriminant and the example. Dore yourself. You must have roots 2 and -1.

Reception second. Check the roots! On the Vieta Theorem. Do not scare, I will explain everything! Check last thing the equation. Those. That we recorded the roots formula. If (as in this example) coefficient a \u003d 1., Check the roots easily. Enough to multiply them. There should be a free member, i.e. In our case -2. Note, not 2, and -2! Free dick with your sign . If it did not work, it means somewhere they have accumulated. Look for an error.

If it happened - it is necessary to fold the roots. Last and final check. Must happen the coefficient b. from opposite sign. In our case -1 + 2 \u003d +1. And coefficient b.which is in front of the ix, equal to -1. So, everything is right!
It is a pity that it is so simple for examples, where X is clean, with a coefficient a \u003d 1. But at least check in such equations! There will be less errors.

Taking third . If there are fractional coefficients in your equation, - get rid of fractions! Draw an equation for a common denominator, as described in the lesson "How to solve equations? Identical conversions". When working with fractions of the error, for some reason and climb ...

By the way, I promised an evil example with a bunch of minuses to simplify. You are welcome! Here it is.

In order not to be confused in the minuses, the equation on -1 is dominant. We get:

That's all! Decide - one pleasure!

So, summarize the topic.

Practical tips:

1. Before solving, we give a square equation to the standard form, build it right.

2. If a negative coefficient is worth a negative coefficient before X, eliminate its multiplication of the entire equation on -1.

3. If fractional coefficients are eliminating the fraction by multiplying the entire equation to the corresponding multiplier.

4. If X is in the square - clean, the coefficient is equal to one, the solution can be easily checked by the Vieta theorem. Do it!

Now it is possible to calculate.)

Solve equations:

8x 2 - 6x + 1 \u003d 0

x 2 + 3x + 8 \u003d 0

x 2 - 4x + 4 \u003d 0

(x + 1) 2 + x + 1 \u003d (x + 1) (x + 2)

Answers (in disorder):

x 1 \u003d 0
x 2 \u003d 5

x 1.2 \u003d2

x 1 \u003d 2
x 2 \u003d -0.5

x - any number

x 1 \u003d -3
x 2 \u003d 3

no solutions

x 1 \u003d 0.25
x 2 \u003d 0.5

Everything converges? Excellent! Square equations are not your headache. The first three turned out, and the rest - no? Then the problem is not in square equations. The problem is in identical transformations of equations. Stroll by reference, it is useful.

Not really gets? Or does not work at all? Then you need to help partition 555. There all these examples disassembled around the bones. Showing main Errors in solving. It is described, of course, the use of identical transformations in solving various equations. Helps a lot!

If you like this site ...

By the way, I have another couple of interesting sites for you.)

It can be accessed in solving examples and find out your level. Testing with instant check. Learn - with interest!)

You can get acquainted with features and derivatives.

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, it is said about such competitions as follows: "As the sun with glitter, the stars eclipses, so the scientist is eager than the fame of another in the people's 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 effect of mathematics, both countries of Islam and Ancient Greece, distinguished by both completeness and clarity of presentation. The author has developed independently some new algebraic examples 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.

Output of the formula for solving the square equation in general There is a 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 modern view. 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.

The challenges per square equation are studied in school Program and in universities. Beneath them understand the equations of the form a * x ^ 2 + b * x + c \u003d 0, where x - variable, a, b, c - constants; A.<>0. The task is to find the roots of the equation.

Geometric meaning of the square equation

The graph of the function, which is represented by the square equation is parabola. Solutions (roots) of the square equation are the points of intersection of the parabola with the abscissa axis (x). It follows from this that there are three possible cases:
1) Parabola has no intersection points with an abscissa axis. This means that it is in the upper plane with branches up or bottom with branches down. In such cases, the square equation has no valid roots (has two complex roots).

2) Parabola has one intersection point with the axis oh. Such a point is called the pearabol vertex, and the square equation in it acquires its minimum or maximum value. In this case, the square equation has one valid root (or two identical roots).

3) Last case In practice, there is more interesting - there are two points of intersection of parabola with the abscissa axis. This means that there are two valid equation root.

Based on the analysis of coefficients in the degrees of variables, it is possible to make interesting conclusions about the placement of parabola.

1) If the coefficient is the more zero, the parabola is directed upwards, if negative - the parabola branches are directed down.

2) If the coefficient b is greater than zero, then the top of the parabola lies in the left half-plane, if it takes a negative value - then in the right.

Output of the formula for solving a square equation

We transfer the constant from the square equation

per sign of equality, we get expression

Multiply both parts on 4a

To get the left of the full square add in both parts b ^ 2 and implement the transformation

From here to find

The formula of the discriminant and the roots of the square equation

The discriminant is called the value of the conditioned expression, it is positive, the equation has two valid roots calculated by the formula At a zero discriminant, the square equation has one solution (two coinciding root), which can be easily obtained from the above formula for d \u003d 0 with a negative discriminant of the equation of valid roots. However, to maintain the solutions of the square equation in the complex plane, and their value is calculated by the formula

Vieta theorem

Consider two roots of the square equation and construct on their basis the square equation. The record itself is easily followed by the Vieta Theorem itself: if we have a square equation of type the sum of its roots is equal to the P coefficient, taken with the opposite sign, and the product of the equation's roots is equal to the free term Q. The formulae record of the above will have seen in the classic equation of a constant A is different from zero, then all equation should be divided into it, and then apply the theorem of the Vieta.

Schedule of a square equation for multipliers

Let the task: decompose the square equation on multipliers. To fulfill it, we first solve the equation (we find the roots). Further, the roots found substituted in the decomposition formula of the square equation this task will be allowed.

Square equation

Task 1. Find the roots of the square equation

x ^ 2-26x + 120 \u003d 0.

Solution: We write the coefficients and substitute in the formula of the discriminant

The root of this value is 14, it is easy to find it with a calculator, or remember with frequent use, however, for convenience, at the end of the article, I will give you a list of numbers squares that can often meet with such tasks.
The foundation is substituted in the root formula

And get

Task 2. Solve equation

2x 2 + x-3 \u003d 0.

Solution: we have a complete square equation, we write out the coefficients and find the discriminant


According to famous formulas we find the roots of the square equation

Task 3. Solve equation

9x 2 -12x + 4 \u003d 0.

Solution: We have a complete square equation. Determine discriminant

We received a case when the roots coincide. Find the values \u200b\u200bof the roots by the formula

Task 4. Solve equation

x ^ 2 + x-6 \u003d 0.

Solution: In cases where there are small coefficients at x it is advisable to apply the theorem of the Vieta. According to her, we get two equations

From the second condition, we get that the work should be equal to -6. This means that one of the roots is negative. We have the following possible pair of solutions (-3; 2), (3; -2). Taking into account the first condition, the second pair of solutions reject.
Root equations are equal

Task 5. Find the lengths of the side of the rectangle, if its perimeter is 18 cm, and the area is 77 cm 2.

Solution: Half of the perimeter of the rectangle is equal to the sum of neighboring sides. Denote by x - most side, then 18-x is a smaller side. The area of \u200b\u200bthe rectangle is equal to the product of these lengths:
x (18-x) \u003d 77;
or
x 2 -18x + 77 \u003d 0.
We find the discriminant of the equation

Calculate the roots of the equation

If a x \u003d 11,that 18h \u003d 7, On the contrary, it is also true (if x \u003d 7, then 21-x \u003d 9).

Task 6. Square square 10x 2 -11x + 3 \u003d 0 Equations for multipliers.

Solution: Calculate the roots of the equation, for this we find discriminant

We substitute the value found in the root formula and calculate

Apply the decomposition formula of the square equation along the roots

The layout of the bracket will receive identity.

Square equation with parameter

Example 1. Under what values \u200b\u200bof the parameter but , Equation (A-3) x 2 + (3-A) x-1/4 \u003d 0 has one root?

Solution: a direct substitution of the value A \u003d 3 we see that it has no solution. Next, we use that at zero discriminant, the equation has one root of multiplicity 2. Drink discriminant

simplify it and equate to zero

Received a square equation on the parameter A, the solution of which is easy to obtain on the Vieta theorem. The amount of the roots is 7, and their work 12. Simple bust by installing that the numbers 3.4 will be rooted equations. Since the solution A \u003d 3, we already rejected at the beginning of the calculations, the only right will be - a \u003d 4.Thus, when A \u003d 4, the equation has one root.

Example 2. Under what values \u200b\u200bof the parameter but , the equation a (a + 3) x ^ 2 + (2a + 6) x-3a-9 \u003d 0has more than one root?

Solution: Consider first singular points, they will be values \u200b\u200ba \u003d 0 and a \u003d -3. When A \u003d 0, the equation will be simplified to the form 6x-9 \u003d 0; x \u003d 3/2 and there will be one root. When A \u003d -3, we obtain the identity 0 \u003d 0.
Calculate discriminant

and find values \u200b\u200band in which it is positive

From the first condition we will get a\u003e 3. For the second we find the discriminant and roots of the equation


We define the gaps where the function takes positive values. Figure point a \u003d 0 Get 3>0 . So, beyond the interval (-3; 1/3) the function is negative. Don't forget about the point a \u003d 0,this should be excluded because the initial equation in it has one root.
As a result, we obtain two intervals that satisfy the condition of the task

There will be many similar tasks in practice, try to deal with the tasks yourself and do not forget to consider the conditions that are mutually exclusive. Well read the formula for solving square equations, they are often needed when calculating in different tasks and sciences.