Discriminant, like square equations begin to study in the course of algebra in grade 8. It is possible to solve the square equation through the discriminant and using the Vieta theorem. The methodology for studying square equations, as well as the formulas of the discriminant, is unsuccessful to schoolchildren, as well as much in this education. Therefore pass school years, training in grade 9-11 replaces "Higher Education" and are still looking again - "How to solve a square equation?", "How to find the roots of the equation?", "How to find a discriminant?" and...

Formula discriminant

Discriminant D. square equation A * X ^ 2 + BX + C \u003d 0 is D \u003d B ^ 2-4 * A * C.
Roots (solutions) of the square equation depend on the discriminant sign (D):
D\u003e 0 - the equation has 2 different valid roots;
D \u003d 0 - the equation has 1 root (2 coinciding root):
D.<0 – не имеет действительных корней (в школьной теории). В ВУЗах изучают комплексные числа и уже на множестве комплексных чисел уравнение с отрицательным дискриминантом имеет два комплексных корня.
The formula for calculating the discriminant is quite simple, so many sites offer an online discriminant calculator. We have not figured out this kind of scripts, so who knows how to implement it please write to the post office This email address is protected from spam bots. You must have JavaScript enabled to view. .

General formula for finding the roots of the square equation:

Roots equations find by formula
If the coefficient with a variable in the square is paired, it is advisable to calculate the discriminant, but the fourth part of it
In such cases, the roots of the equation are found by the formula

The second way of finding roots is the Vieta Theorem.

The theorem is formulated not only for square equations, but also for polynomials. You can read this in Wikipedia or other electronic resources. However, to simplify, consider it part of it, which concerns the above square equations, that is, the equations of the form (A \u003d 1)
The essence of the formulas of the wine is that the amount of the roots of the equation is equal to the coefficient with a variable taken with the opposite sign. The product of the roots of the equation is equal to a free member. The formulas of the Vieta Theorem has a record.
The output of the formula of the Vieta is quite simple. Cut the square equation through simple multipliers
As you can see, everything ingenious is simultaneously simple. Effectively use the wine formula when the root difference in the module or the difference of roots modules is 1, 2. For example, the following equations on the Vieta theorem have roots




Up to 4 equations, the analysis should look as follows. The product of the equation equation is 6, therefore, the roots may be values \u200b\u200b(1, 6) and (2, 3) or pairs with the opposite sign. The amount of the roots is 7 (the coefficient with a variable with the opposite sign). From here we conclude that the solutions of the square equation are x \u003d 2; x \u003d 3.
It is easier to select the roots of the equation among the free member dividers, adjusting their sign in order to fulfill the formulas of Vieta. At the beginning, it seems difficult to do, but with practice on a number of square equations, such a technique will be more effective than the calculation of the discriminant and finding the roots of the square equation to the classical method.
As you can see the school theory of study of the discriminant and the methods of finding solutions of the equation is devoid of practical meaning - "Why schoolchildren square equation?", What is the physical meaning of the discriminant? ".

Let's try to figure out what does the discriminant describe?

The course of algebra is studying the functions, research schemes of the function and the construction of the graphics of the functions. Of all the functions, a parabola occupies an important place, the equation of which can be written as
So the physical meaning of the square equation is zero parabola, that is, the intersection points of the function with the axis of the abscissa OX
Properties Parabolas that are described below will ask you to remember. The time will come to pass exams, tests, or entrance exams and you will be grateful for the reference material. The sign with a variable in the square corresponds to whether the polebola branches will be on the schedule to go up (a\u003e 0),

or parabola branches down (a<0) .

The top of Parabola lies in the middle between the roots

The physical meaning of the discriminant:

If the discriminant is greater than zero (D\u003e 0) Parabola has two intersection points with OX axis.
If the discriminant is zero (d \u003d 0), then the parabol is in the top concerns the abscissa axis.
And the last case when the discriminant is less than zero (D<0) – график параболы принадлежит плоскости над осью абсцисс (ветки параболы вверх), или график полностью под осью абсцисс (ветки параболы опущены вниз).

Incomplete square equations

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 should be brought to standard. 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?), But 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 two. 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, A -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.

In this article we will look at the decision of incomplete square equations.

But first we repeat which equations are called square. Equation of the form ah 2 + BX + C \u003d 0, where x is a variable, and the coefficients A, B and with some numbers, and ≠ 0, called square. As we see the coefficient at x 2 is not zero, and therefore the coefficients at x or free member can be zero, in this case we get an incomplete square equation.

Incomplete square equations are three species:

1) if B \u003d 0, C ≠ 0, then ah 2 + c \u003d 0;

2) if b ≠ 0, c \u003d 0, then ah 2 + bx \u003d 0;

3) If B \u003d 0, C \u003d 0, then ah 2 \u003d 0.

• Let's understand how to solve equations of the form ah 2 + c \u003d 0.

To solve the equation by postponing a free member with the right-hand part of the equation, we get

ah 2 \u003d -C. Since a ≠ 0, then we split both parts of the equation at a, then x 2 \u003d -c / a.

If -s / a\u003e 0, the equation has two roots

x \u003d ± √ (-c / a).

If -c / a< 0, то это уравнение решений не имеет. Более наглядно решение данных уравнений представлено на схеме.

Let's try to figure out the examples how to solve such equations.

Example 1.. Decide equation 2x 2 - 32 \u003d 0.

Answer: x 1 \u003d - 4, x 2 \u003d 4.

Example 2.. Decide equation 2x 2 + 8 \u003d 0.

Answer: The solutions equation has no.

• We will understand how to solve equations of the form ah 2 + bx \u003d 0.

To solve the equation Ah 2 + BX \u003d 0, we will decompose it on multipliers, that is, we will bring it to the brackets x, we get x (ah + b) \u003d 0. The product is zero, if at least one of the multipliers is zero. Then or x \u003d 0, or ah + b \u003d 0. Resolving the equation ah + b \u003d 0, we get ah \u003d - b, where x \u003d - b / a. The equation of the form Ah 2 + BX \u003d 0, always has two roots x 1 \u003d 0 and x 2 \u003d - b / a. See how it looks like a solution to the solution of the equations of this species.

Secure our knowledge on a specific example.

Example 3.. Solve equation 3x 2 - 12x \u003d 0.

x (3x - 12) \u003d 0

x \u003d 0 or 3x - 12 \u003d 0

Answer: x 1 \u003d 0, x 2 \u003d 4.

• Third-type equations Ah 2 \u003d 0 Solved very simple.

If ah 2 \u003d 0, then x 2 \u003d 0. The equation has two equal roots x 1 \u003d 0, x 2 \u003d 0.

For clarity, consider the scheme.

We will be convinced when sampling Example 4 that the equations of this species are solved very simply.

Example 4. Solve equation 7x 2 \u003d 0.

Answer: x 1, 2 \u003d 0.

It is not always possible to immediately understand what kind of incomplete square equation we have to solve. Consider the following example.

Example 5. Solve equation

Multiply both parts of the equation on a common denominator, that is, at 30

Socil

5 (5x 2 + 9) - 6 (4x 2 - 9) \u003d 90.

Recall brackets

25x 2 + 45 - 24x 2 + 54 \u003d 90.

Let's give similar

We transfer 99 of the left part of the equation to the right, changing the sign to the opposite

Answer: No roots.

We dismantled how incomplete square equations are solved. I hope now you will not have difficulties with similar tasks. Be careful when determining the type of incomplete square equation, then you will succeed.

If you have questions about this topic, sign up for my lessons, we solve problems together.

the site, with full or partial copying of the material reference to the original source is required.

Consider the task. The base of the rectangle is more than a height of 10 cm, and its area is 24 cm². Find the height of the rectangle. Let be h. Santimeters - the height of the rectangle, then its base is equal ( h.+10) see the area of \u200b\u200bthis rectangle is equal to h.(h.+ 10) cm². Under the condition of the task h.(h.+ 10) \u003d 24. Revealing brackets and carrying the number 24 with the opposite sign into the left part of the equation, we get: h.² + 10. h.-24 \u003d 0. When solving this problem, an equation was obtained, which is called square.

The square equation is called the species equation

aX. ²+ bX.+c \u003d.0

where a, B, C - set numbers, and but≠ 0, and h. - Unknown.

Factors a, B, C Square equation is usually called this: a. - the first or senior coefficient, b. - second coefficient c. - Free member. For example, in our problem, the senior coefficient is 1, the second coefficient 10, free-member -24. The solution of many problems of mathematics and physics is reduced to solving square equations.

Solution of square equations

Full square equations. First of all, a given equation must be brought to standard aX.²+ bX.+ c \u003d.0. Let us return to our problem in which the equation can be written as h.(h.+ 10) \u003d 24 We give it to the standard form, open brackets h.² + 10. h.- 24 \u003d 0, solving this equation using the formula of the roots of the square equation of the general form.

The expression under the sign of the root in this formula is called discriminant D \u003d b.² - 4. aC

If D\u003e 0, the square equation has two different roots that can be found using the roots of the square equation.

If D \u003d 0, then the square equation has one root.

If D<0, то квадратное уравнение не имеет действительных корней, т. е. не имеет решения.

Substitute meanings to our formula but= 1, b.= 10, c.= -24.

we get D\u003e 0, therefore we will have two roots.

Consider an example where d \u003d 0, and the condition should turn out one root.

25x.² - 30. x.+ 9 = 0

Consider the example where D<0, при этом условии решения не должно быть.

2x.² + 3. x.+ 4 = 0

The number that stands under the root sign (discriminant) is negative, the answer will write this: the equation does not have valid roots.

Decision of incomplete square equations

Quadratic equation aX.² + bX.+ c.\u003d 0 called incomplete, if at least one of the coefficients b. or c. equal to zero. An incomplete square equation, there is an equation of one of the following types:

AX.² = 0,

AX.² + c.= 0, c.≠ 0,

AX.² + bX.= 0, b.≠ 0.

Consider several examples, solving equation

Sharing both parts of the equation to 5, we obtain the equation h.² \u003d 0, in response will be one root h.= 0.

Consider the view equation

3h.² - 27 \u003d 0

Dividing both parts by 3, we obtain equation h.² - 9 \u003d 0, or you can record it h.² \u003d 9, in response there will be two roots h.\u003d 3 I. h.= -3.

Consider the view equation

2h.² + 7 \u003d 0

Sharing both parts by 2, we obtain the equation h.² \u003d -7/2. This equation of valid roots does not have, because h.² ≥ 0 for any actual number h..

Consider the view equation

3h.² + 5. h.= 0

Decomposing the left part of the factory equation, we get h.(3h.+ 5) \u003d 0, the answer will be two roots h.= 0, h.=-5/3.

The most important thing in solving square equations, clarify the square equation to the standard form, to learn by heart the root formula of the square equation of the general type and not get confused in signs.

Bibliographic Description: Gasanov A. R., Kuramshin A. A., Yelkov A. A., Shirenkov N. V., Ulanov D. D., Smeleva O. V. Ways to solve square equations // Young scientist. - 2016. - №6.1. - S. 17-20..03.2019).



Our project is devoted to ways to solve square equations. Project goal: learn to solve square equations in ways that are not included in the school curriculum. Task: Find all possible ways to solve square equations and learn how to use them yourself and introduce classmates with these ways.

What is "square equations"?

Quadratic equation - equation of type aX.2 + BX + C \u003d 0where a., b., c. - Some numbers ( a ≠ 0.), x. - Unknown.

The numbers a, b, C are called the coefficients of the square equation.

• a is called the first coefficient;
• b is called a second coefficient;
• c - free member.

And who is the first "invented" square equations?

Some algebraic techniques for solving linear and square equations were known as 4,000 years ago in ancient Babylon. The ancient Babylonian clay plates found somewhere between 1800 and 1600 BC, are the earliest evidence of the study of square equations. On the same signs, methods for solving some types of square equations are presented.

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.

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.

Babylonian mathematics from about the IV century BC. Used the method of supplementing the square to solve equations with positive roots. About 300 BC. Euclide has come up with a more general geometric solution method. The first mathematician who found solutions of the equation with negative roots in the form of an algebraic formula was the Indian scientist Brahmagupta (India, VII century of our era).

Brahmagupta outlined the general rule of solving the square equations given to a single canonical form:

aX2 + BX \u003d C, A\u003e 0

In this equation, the coefficients may be negative. The brahmagupta rule essentially coincides with our.

In 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 is shine with its own overshadows, so the scientist is overshadowed with folk collections, offering and solving algebraic tasks." The tasks are often enjoyed in a poetic shape.

In algebraic treatise Al-Khorezmi The classification of linear and square equations is given. The author includes 6 species of equations, expressing them as follows:

1) "Squares are equal to roots", i.e. AH2 \u003d BX.

2) "Squares are equal to the number", i.e. Ah2 \u003d s.

3) "The roots are equal to the number", that is, AH2 \u003d p.

4) "Squares and numbers are equal to roots", i.e. Ah2 + C \u003d BX.

5) "Squares and roots are equal to the number", that is, AH2 + BX \u003d p.

6) "Roots and numbers are equal to squares", that is, BX + C \u003d\u003d AH2.

For al-Khorezmi, avoiding the use of negative numbers, members of each of these equations are foundated, 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 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 species of al-Korezmi, like all mathematics until the XVII century, does not take into account the zero solution, probably because in specific practical practical It does not matter tasks. When solving full square equations, al-chores on private numeric examples sets out the rules of decision, and then their geometrical evidence.

The forms of the solution of square equations for the sample al-Khorezmi in Europe were first set out in the "Book of Abaka" written in 1202g. Italian mathematician Leonard Fibonacci. The author developed independently some new algebraic examples of solving problems and the first in Europe approached the introduction of negative numbers.

This book contributed to the spread of algebraic knowledge not only in Italy, but also in Germany, France and other European countries. Many tasks from this book went almost into all European textbooks of the XIV-XVII centuries. General rule Solutions of the square equations given to the single canonical form x2 + BX \u003d C with all sorts of combinations of signs and coefficients B, C, was formulated in Europe in 1544. M. Stiffel.

Output of the formula for solving a square equation in general There is a Vieta, but Viet recognized only positive roots. Italian mathematicians Tartalia, Cardano, Bombelly Among the first in the XVI century. Given, in addition to positive, and negative roots. Only in the XVII century. Thanks to labor Girard, Descartes, Newton And other scientists a method for solving square equations takes a modern appearance.

Consider several ways to solve square equations.

Standard ways to solve square equations from the school program:

1. Decomposition of the left part of the factory equation.
2. Method of allocation of a full square.
3. Solution of square equations by the formula.
4. Graphic solution of the square equation.
5. Solving equations using the Vieta Theorem.

Let us dwell on the solution of the above and non-listed square equations on the Vieta theorem.

Recall that to solve the above square equations, it is sufficient to find two numbers such, the product of which is equal to a free member, and the amount is the second coefficient with the opposite sign.

Example.x. 2 -5x + 6 \u003d 0

It is necessary to find numbers whose work is 6, and the amount 5. Such numbers will be 3 and 2.

Answer: X. 1 \u003d 2, x 2 =3.

But this method can be used for equations with the first coefficient not equal one.

Example.3X. 2 + 2x-5 \u003d 0

Take the first coefficient and multiply it on a free term: x 2 + 2x-15 \u003d 0

The roots of this equation will be numbers, the product of which is - 15, and the amount is equal to - 2. These numbers are 5 and 3. To find the roots of the original equation, the roots obtained to divide the first coefficient.

Answer: X. 1 \u003d -5 / 3, x 2 =1

6. Solution of the equations by the method of "transit".

Consider the square equation Ah 2 + BX + C \u003d 0, where a ≠ 0.

Multiplying both parts by A, we obtain the equation a 2 x 2 + ABH + AC \u003d 0.

Let oh \u003d y, where x \u003d y / A; Then come to the equation in 2 + BY + AC \u003d 0, equivalent to this. Its roots in 1 and in 2 will find with the help of the Vieta theorem.

We finally obtain x 1 \u003d in 1 / a and x 2 \u003d y 2 / a.

In this method, the coefficient A is multiplied by a free member, no matter how "is transferred" to it, therefore it is called the "transit" method. 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.2x 2 - 11x + 15 \u003d 0.

"We will transfer the coefficient 2 to a free member and making a replacement to get the equation in 2 - 11U + 30 \u003d 0.

According to the reverse Vieta theorem

in 1 \u003d 5, x 1 \u003d 5/2, x 1 \u003d 2.5; in 2 \u003d 6, x 2 \u003d 6/2, x 2 \u003d 3.

Answer: H. 1 \u003d 2.5; H. 2 = 3.

7. Properties of the coefficients of the square equation.

Let the square equation ah 2 + BX + C \u003d 0, and ≠ 0.

1. If A + B + C \u003d 0 (i.e., the sum of the coefficients of the equation is zero), then x 1 \u003d 1.

2. If a - b + c \u003d 0, or b \u003d a + s, then x 1 \u003d - 1.

Example.345x 2 - 137x - 208 \u003d 0.

Since a + b + c \u003d 0 (345 - 137 - 208 \u003d 0), then x 1 \u003d 1, x 2 \u003d -208/345.

Answer: H. 1 \u003d 1; H. 2 = -208/345 .

Example.132x 2 + 247x + 115 \u003d 0

Because A-B + C \u003d 0 (132 - 247 + 115 \u003d 0), then x 1 \u003d - 1, x 2 \u003d - 115/132

Answer: H. 1 \u003d - 1; H. 2 =- 115/132

There are other properties of the square equation coefficients. But the icy use is more complicated.

8. Solution of square equations with a nomogram.

Fig 1. Nomogram

This is an old and currently forgotten way to solve square equations, placed on S.83 Collection: Brandis V.M. Four-digit mathematical tables. - M., Enlightenment, 1990.

Table XXII. Nomogram for solving the equation z 2 + pz + q \u003d 0. This nomogram allows, without solving the square equation, by its coefficients to determine the roots of the equation.

The curvilinear scale of the nomogram is constructed by formulas (Fig. 1):

Believed OS \u003d P, ED \u003d Q, OE \u003d A (all in cm), from Fig.1 of the similarity of triangles San and CDF. We get a proportion

where after substitutions and simplifications follow the equation z 2 + pz + q \u003d 0,moreover, the letter z. means a label of any point of the curvilinear scale.

Fig. 2 Solution of square equations using a nomogram

Examples.

1) for equation z. 2 - 9z + 8 \u003d 0 The nomogram gives the roots z 1 \u003d 8.0 and z 2 \u003d 1.0

Answer: 8.0; 1.0.

2) solutions with a nomogram equation

2z. 2 - 9z + 2 \u003d 0.

We divide the coefficients of this equation by 2, we obtain the equation z 2 - 4.5z + 1 \u003d 0.

The nomogram gives the roots z 1 \u003d 4 and z 2 \u003d 0.5.

Answer: 4; 0.5.

9. Geometric method of solving square equations.

Example.h. 2 + 10x \u003d 39.

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

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 complemented then to the new Square of ABSD, completing four equal squares in the corners, the side of each of them is 2.5, and the area is 6.25

Fig. 3 graphic method for solving equation x 2 + 10x \u003d 39

The ABCD square S can be represented as an amount of space: the initial square X 2, four rectangles (4 ∙ 2.5x \u003d 10x) and four attached squares (6.25 ∙ 4 \u003d 25), i.e. S \u003d x 2 + 10x \u003d 25. Replacing x 2 + 10x number 39, we obtain that S \u003d 39+ 25 \u003d 64, from where it follows that the side of the AVD square, i.e. Cut AB \u003d 8. For the desired side x of the original square we get

10. Solution of equations using the mouture theorem.

Theorem mow. The residue from the division of the polynomial P (x) on the twist of x - α is p (α) (that is, the value p (x) at x \u003d α).

If the number α is the root of the polynomial P (x), then this polynomial is divided into x -α without a residue.

Example.x²-4x + 3 \u003d 0

P (x) \u003d x²-4x + 3, α: ± 1, ± 3, α \u003d 1, 1-4 + 3 \u003d 0. We divide p (x) to (x - 1): (x²-4x + 3) / (x - 1) \u003d x-3

x²-4x + 3 \u003d (x - 1) (x-3), (x - 1) (x-3) \u003d 0

x - 1 \u003d 0; x \u003d 1, or x-3 \u003d 0, x \u003d 3; Answer: H.1 \u003d 2, x2 =3.

Output: The ability to quickly and rationally solve square equations is simply necessary for solving more complex equations, for example, fractional rational equations, equations of higher degrees, bic-duty equations, and in the senior school of trigonometric, indicative and logarithmic equations. Having studied all the found ways to solve square equations, we can advise classmates, except for standard methods, solving the method of transformation (6) and solving equations for the coefficient property (7), as they are more accessible to understanding.

Literature:

1. Bradis V.M. Four-digit mathematical tables. - M., Enlightenment, 1990.
2. Algebra Grade 8: Tutorial for 8 CL. general education. Institutions Makarychev Yu. N., Mindyuk N. G., Neshkov K. I., Suvorov S. B. Ed. S. A. Telikovsky 15th ed., Doraby. - M.: Enlightenment, 2015
3. https://ru.wikipedia.org/wiki/%D0%9A%D0%B2%D0%B0%D0%B4%D1%80%D0%B0%D1%82%D0%BD%D0%BE%D0. % B5_% D1% 83% D1% 80% D0% B0% D0% B2% D0% BD% D0% B5% D0% BD% D0% B8% D0% B5
4. Glaser G.I. History of mathematics at school. Manual for teachers. / Ed. V.N. Young. - M.: Enlightenment, 1964.