Many valid roots equation. Equations in higher mathematics. Organic numerous roots. Gorner scheme. The roots of the algebraic equation

The formulas of the roots of the square equation. Cases of valid, multiple and complex roots are considered. Decomposition of square three-shred multipliers. Geometric interpretation. Examples of determining the roots and decomposition of multipliers.

Content

See also: Solution of square equations online

Basic formulas

Consider a square equation:
(1) .
Roots square equation (1) are determined by formulas:
; .
These formulas can be combined like this:
.
When the roots of the square equation are known, the second degree polynomial can be represented as a work of the factors (decompose on multipliers):
.

Next, believe that - actual numbers.
Consider discriminant square equation:
.
If the discriminant is positive, then the square equation (1) has two different valid root:
; .
Then the decomposition of the square three decreases on the factors has the form:
.
If the discriminant is zero, then the square equation (1) has two multiple (equal) valid root:
.
Factorization:
.
If the discriminant is negative, then the square equation (1) has two comprehensively conjugated root:
;
.
Here - the imaginary unit;
And - the actual and imaginary parts of the roots:
; .
Then

.

Graphic interpretation

If build schedule function
,
which is parabola, then the point of intersection of the graph with the axis will be roots of the equation
.
When, the schedule crosses the abscissa axis (axis) at two points ().
When, the graph concerns the abscissa axis at one point ().
When, the schedule does not cross the abscissa axis ().

Useful formulas associated with the square equation

(F.1) ;
(F.2) ;
(F.3) .

The output of the formula for the roots of the square equation

We carry out transformations and apply formulas (F.1) and (F.3):

,
Where
; .

So, we got a formula for a polynomial of the second degree in the form:
.
From here it can be seen that the equation

performed at
and.
That is, the roots of the square equation are roots
.

Examples of determining the roots of the square equation

Example 1.

(1.1) .

.
Comparing with our equation (1.1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Since the discriminant is positive, the equation has two valid root:
;
;
.

From here we get a decomposition of a square three-stakes on multipliers:

.

Schedule function y \u003d 2 x 2 + 7 x + 3 Crosses the abscissa axis at two points.

We construct a function schedule
.
The schedule of this function is parabola. She places the abscissa axis (axis) at two points:
and.
These points are roots of the initial equation (1.1).

;
;
.

Example 2.

Find the roots of the square equation:
(2.1) .

We write the square equation in general form:
.
Comparing with the initial equation (2.1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Since the discriminant is zero, the equation has two multiple (equal) root:
;
.

Then the decomposition of three decisions on multipliers has the form:
.

Function graph y \u003d x 2 - 4 x + 4 Requests the abscissa axis at one point.

We construct a function schedule
.
The schedule of this function is parabola. It concerns the abscissa axis (axis) at one point:
.
This point is the root of the initial equation (2.1). Since this root enters the expansion of multipliers twice:
,
That such root is called multiple. That is, it is believed that there are two equal root:
.

;
.

Example 3.

Find the roots of the square equation:
(3.1) .

We write the square equation in general form:
(1) .
We rewrite the initial equation (3.1):
.
Compare C (1), we find the values \u200b\u200bof the coefficients:
.
We find discriminant:
.
Discriminant is negative. Therefore, there are no valid roots.

You can find complex roots:
;
;
.

Then

.

The function graph does not cross the abscissa axis. There are no valid roots.

We construct a function schedule
.
The schedule of this function is parabola. It does not intersect the abscissa axis (axis). Therefore, there are no valid roots.

There are no valid roots. Roings are integrated:
;
;
.

graphic solution solution

The equation has the appearance and its root - there "Iksova" coordinate point of intersection graphics of linear function with linear function schedule (abscissa axis):

It would seem that an example is so elementary that there is nothing more elementary here, however, it is possible to "squeeze" another unexpected nuance: to imagine the same equation in the form and construct graphs of functions:

Wherein, please do not confuse two concepts.: equation is an equation, and function - This is a feature! Functions only help Find the roots of the equation. Koi can be two, three, four and even infinitely a lot. The closest example in this sense is everyone known quadratic equation, the algorithm of the solution of which was awarded a separate paragraph "Hot" school formulas. And it is not by chance! If you know how to solve a square equation and know theorem of Pythagora, then, you can say, "the floor of the highest mathematics is already in his pocket" \u003d) exaggerated, of course, but not so far from the truth!

And therefore they will not be lazy and rewound some square equation on standard algorithm:

it means that the equation has two different valid Root:

It is easy to make sure that both found values \u200b\u200bare really satisfying this equation:

What if you suddenly forgotten the solution algorithm, and there is no funds / hand at hand? This situation may occur, for example, on the test or exam. Use the graphic method! And there are two ways: you can stop building parabola , thus finding out where it crosses the axis (if crosses generally). But it is better to do cunning: imagine an equation in the form, draw the schedules of simpler functions - and "ICS" coordinates Their points of intersection, as on the palm!

If it turns out that the direct is about parabola, the equation has two coincided (multiple) root. If it turns out that the direct does not cross the parabola, it means that there are no valid roots.

For this, of course, you need to be able to build charts of elementary functionsBut on the other hand, these skills are even a schoolboy.

And again - the equation is the equation, and the functions are functions that just helped Solve equation!

And here, by the way, it will be appropriate to remember another thing: if all the coefficients of the equation are multiplying to a non-zero number, its roots will not change.

So, for example, the equation It has the same roots. As the simplest "evidence" I submit a constant for brackets:
And painlessly remove it (I share both parts on "minus two"):

BUT! If we consider the function , then it is impossible to get rid of the constant! It is permissible, except for making a multiplier for brackets: .

Many underestimate the graphical decision method, considering it with something "unsolonged", and some are at all forget about such an opportunity. And it is in the root mistaken, since the construction of graphs sometimes just saves the situation!

Another example: Suppose you do not remember the roots of the simplest trigonometric equation :. General formula There are in school textbooks, in all directories for elementary mathematics, but they are not available to you. However, it is critical to solve the equation (otherwise "Two"). There is an exit! - Build features:

After that, we calmly write down the "ICS" coordinates of their intersection points:

The roots are infinitely a lot and their twisted record is adopted in the algebra:
where ( – many integers) .

And, not "departing from the cash register", a few words about the graphical method of solving inequalities with one variable. The principle is the same. So, for example, the solution of inequality is any "X", because The sinusoid almost completely lies in a straight line. The solution of inequality is a plurality of gaps, on which sinusoid sinusoids lie strictly above the direct (abscissa axis):

or, if shorter:

But the many solutions of inequality - emptyBecause no point of sinusoids lies above straight.

Is something not clear? Urgently stood lessons about sets and charts of functions!

We warm up:

Exercise 1

Solve graphically the following trigonometric equations:

Answers at the end of the lesson

As you can see, it is not necessary to sharpen formulas and directories to study the exact sciences! And moreover, this is a fundamentally vicious approach.

As I have already encouraged you at the very beginning of the lesson, complex trigonometric equations in the standard course of higher mathematics have to be solved extremely rarely. All complexity, as a rule, ends with equations like, the solution of which are two groups of roots originating from the simplest equations and . With the decision of the latter, do not worry, look in the book or find on the Internet \u003d)

The graphical solution method can help and in less trivial cases. Consider, for example, the following "different" equation:

Prospects for his decision look ... at all they do not look at, but it is worth only to present an equation in the form, build functions graphics And everything will be incredibly easy. The drawing is in the middle of the article about infinitely small features (opens on the next tab).

The same graphic method one can find out that the equation has already two roots, and one of them is zero, and the other, apparently, irrenionalen And belongs to the segment. This root can be calculated approximately, for example, by tangential. By the way, in some tasks, it happens, you need not to find roots, but to find out are they in general. And here too, the drawing can also help - if the graphs do not intersect, then there are no roots.

Rational roots of polynomials with whole coefficients.
Gorner scheme

And now I suggest you wrap your gaze into the Middle Ages and feel the unique atmosphere of classical algebra. For a better understanding of the material, I recommend at least a little familiar with complex numbers.

They are the most. Polynomials.

The object of our interest will be the most common polynomials of integer coefficients. Natural number called degree of polynomial, the number is the coefficient with a high degree (or just a senior coefficient)and coefficient - free member.

This polynomial I will be folded to designate through.

Roots of polynomial Called roots equation

I love iron logic \u003d)

For examples, we go to the very beginning of the article:

With the finding of the roots of polynomials of the 1st and 2nd degrees there are no problems, but as this task increases it is becoming more and more difficult. Although on the other hand - more and more interesting! And just this will be devoted to the second part of the lesson.

First, literally the floor of the theory screen:

1) according to the investigation the main theorem of algebra, the degree has exactly comprehensive roots. Some roots (or even everything) can be in particular valid. At the same time, among the valid roots, the same (multiple) roots can meet (minimum two, maximum pieces).

If some complex number is the root of the polynomial, then conjugate he is the number - also necessarily the root of this polynomial (Combined complex roots have a view).

The simplest example is a square equation that first met B8 (like) class, and which we finally "finished" in the subject complex numbers. I remind you: the square equation has either two different valid roots, or multiple roots, or conjugate complex roots.

2) from theorems Bezu It follows that if the number is the root of the equation, then the corresponding polynomial can be decomposed on multipliers:
where - a polynomial degree.

And again, our old example: because - the root of the equation, then. After that it is not difficult to get a good familiar "School" decomposition.

The consequence of the moutness theorem has a big practical value: if we know the root of the 3rd degree equation, we can present it as And from the square equation it is easy to find out the rest of the roots. If we know the root of the 4th degree equation, that is, the ability to decompose the left part into the work, etc.

And the question here is two:

Question first. How to find this very root? First of all, let's decide with its nature: in many challenges of higher mathematics it is necessary to find rational, in particular wholethe roots of polynomials, and in this connection further will be interested in us mainly .... ... they are so good, such fluffy, that they are right and I want to find! \u003d)

The first thing that suggests is the selection method. Consider, for example, the equation. The snag here in the free member - that if he were zero, then everything would be in Okra - we endure "X" for brackets and the roots themselves "fall out" to the surface:

But our free member is equal to the "Troika", and therefore we begin to substitute various numbers to the equation, claiming the title "root". First of all, the substitution of single values. Substitute:

Received invalid Equality, therefore, the unit "did not fit." Well, okay, we substitute:

Received faithful equality! That is, the value is the root of this equation.

To find the roots of the polynomial 3rd degree exist analytical method (so-called Cardano Formulas)But now we are interested in a somewhat different task.

Since - there is a root of our polynomial, then the polynomial can be represented as and arises Second question: How to find a "junior fellow"?

The simplest algebraic considerations suggest that it is necessary to divide for this. How to divide the polynomial to the polynomial? The same school method, which ordinary numbers share - "Stage"! I disassembled this method in the first examples of the lesson. Difficult limits, and now we will look at another way that got a name gorner scheme.

First, write the "senior" polynomial with all , including zero coefficients:
, after which it will bring these coefficients (strictly in order) to the top line of the table:

On the left record the root:

Immediately make a reservation that the Gorner scheme works and in the event that "Red" number not It is the root of the polynomial. However, we will not rush the events.

Sentence on top of the senior coefficient:

The process of filling the lower cells is like embroidery, where "minus one" is a kind of "needle", which permeates the next steps. The "demolished" number is multiplied by (-1) and add a number from the upper cell:

The found value is multiplied by the "red needle" and add the following coefficient of equation to the work:

And finally, the value received again "process" the "needle" and the upper coefficient:

Zero in the last cell tells us that the polynomial was divided into without residue (how it should be)At the same time, the decomposition coefficients are "removed" directly from the bottom line of the table:

Thus, from the equation, we switched to an equivalent equation and with two remaining roots everything is clear (In this case, conjugate complex roots are obtained).

Equation, by the way, can be solved and graphically: build "Lightning" and see that the schedule crosses the abscissa axis () at point. Or the same "cunning" reception - rewrite the equation in the form, blacksmith elementary graphics and detect the "oscus" coordinate of their intersection points.

By the way, the graph of any function-polynomial of the 3rd degree crosses the axis at least once, which means that the corresponding equation has at least one valid root. This fact is valid for any function-polynomial of odd.

And then I want to stay on an important momentwhich concerns terminology: polynomial and multicoon function – this is not the same! But in practice, it is often said, for example, about the "graphics of a polynomial", which, of course, negligence.

However, back to the Gunner scheme. As I recently mentioned, this scheme works for other numbers, but if the number not It is the root of the equation, then a non-zero additive appears in our formula (residue):

"Runting" according to the scheme of the city "unsuccessful" value. It is convenient to use the same table - write down the new "needle" on the left, demolish the senior coefficient from above (Left Green Arrow)and rushed:

To check with a bracket and we give similar terms:
, OK.

It is easy to see that the residue (Sixer) is exactly the value of the polynomial at. And in fact - so:
, and more pleasant - like this:

From the above calculation it is not difficult to understand that the Gorner scheme allows not only to decompose the polynomials into multipliers, but also to carry out a "civilized" selection of the root. I suggest you independently consolidate the calculation algorithm a small task:

Task 2.

Using the mountain scheme, find the whole root of the equation and decompose the corresponding polynomial to multipliers

In other words, here you need to consistently check the numbers 1, -1, 2, -2, ... - until the zero residue is "drew" in the last column. This will mean that the "needle" of this string is the root of the polynomial

Calculations are convenient to arrange in a single table. A detailed decision and answer at the end of the lesson.

The method of selection of roots is good for relatively simple cases, but if the coefficients and / or the degree of polynomial are large, the process can delay. And maybe there are some values \u200b\u200bfrom the same list 1, -1, 2, -2 and no sense to consider? And, moreover, the roots may also be fractional, which will lead to an absolutely not scientifically.

Fortunately, there are two powerful theorems that allow you to significantly reduce the bust of the values \u200b\u200bof "candidates" in rational roots:

Theorem 1. Consider unstable fraction where. If the number is the root of the equation, then the free member is divided into, and the senior coefficient is on.

In particularIf the elder coefficient is then this rational root - a whole:

And we begin to exploit the theorem just with this delicious particular:

Let's return to the equation. Since his senior coefficient, the hypothetical rational roots can be exclusively whole, and the free member must be able to share for these roots without a residue. And the "Troika" can be divided only to 1, -1, 3 and -3. That is, we have only 4 "candidates for roots." And, according to Theorem 1.Other rational numbers cannot be roots of this equation in principle.

In the "applicants" equation a little more: a free member is divided into 1, -1, 2, - 2, 4 and -4.

Note that numbers 1, -1 are "regulars" list of possible roots (obvious consequence of the theorem) And most the best choice For priority check.

Go to more informative examples:

Task 3.

Decision: Since the senior coefficient, the hypothetical rational roots can only be integer, while they must be free member divisors. "Minus forty" is divided into the following pairs of numbers:
- Total 16 "candidates."

And here immediately appears a tempting thought: whether it is impossible to cut off all negative or all positive roots? In some cases, you can! Formulate two signs:

1) if everything The coefficients of the polynomial are non-negative or all are non-positive, it cannot have positive roots. Unfortunately, this is not our case (now if we were given an equation - then yes, when substituting any value of the polynomial is strictly positive, and therefore all positive numbers (and irrational too) Can not be rooted equations.

2) if the coefficients at odd degrees are non-negative, and with all parts degrees (including a free member) - Negative, then the polynomial cannot have negative roots. Or "Mirror": the coefficients at odd degrees are non-positive, and with all that are positive.

This is our case! Little looking a little, it can be noted that when substituting in the equation of any negative "X", the left part will be strictly negative, and therefore negative roots disappear

Thus, 8 numbers remained for research:

Consistently "charge" them according to the horner scheme. I hope you have already mastered oral computing:

Good luck waited for us when testing "Two". Thus, there is a root of the considered equation, and

It remains to explore the equation . It is easy to do through the discriminant, but I will conduct an indicative check on the same scheme. First, we draw attention that a free member is 20, and therefore Theorem 1. From the list of possible roots, numbers 8 and 40 fall out, and values \u200b\u200bremain for study (the unit dropped according to the Gunner scheme).

Record the coefficients of the triple in the upper line of the new table and we start checking from the same "twos". Why? And because the roots can be more paints, please: - This equation has 10 identical roots. But do not get distracted:

And here, of course, I rushed a little, knowingly knowing that the roots are rational. After all, if they were irrational or complex, then I would shine unsuccessful checking of all the remaining numbers. Therefore, in practice, follow the discriminant.

Answer: Rational roots: 2, 4, 5

In the disassembled task, we were accompanied by luck, because: a) the negative values \u200b\u200bwere immediately falling off, and b) we very quickly found the root (and theoretically, they could check the entire list).

But actually the situation is much worse. I invite you to watch a fascinating game called "Last Hero":

Task 4.

Find rational roots equation

Decision: by Theorem 1. Numerals of hypothetical rational roots must satisfy the condition (We read "twelve divided by El"), and denominators - condition. Based on this, we get two list:

"List of El":
and "EM List": (good, here are Natural numbers).

Now make a list of all possible roots. First, "List of El" divide on. It is clear that the same numbers will turn out. For convenience, we bring them to the table:

Many fraraty declined, resulting in a value that already exist in the "list of heroes". We add only "newbies":

Similarly, we divide the same "List of El" on:

And finally, on

Thus, a team of participants in our game is equipped:

Unfortunately, the polynomial of this task does not satisfy the "positive" or "negative" feature, and therefore we cannot discard the upper or lower line. We will have to work with all the numbers.

How is your mood? Come on, above the nose - there is another theorem, which can be figuratively called the "killer theorem" .... ... "candidates", of course \u003d)

But first you need to scroll down the horner schema at least for one whole numbers. Traditionally, take a unit. In the upper line, we write the coefficients of the polynomial and everything as usual:

Since the four is clearly not zero, then the value is not the root of the polynomial under consideration. But she will help us very much.

Theorem 2. If at some whole The value of the polynomial is different from zero:, then its rational roots (if they are) Satisfy the condition

In our case, and therefore all possible roots must satisfy the condition (I call it condition number 1). This four will be the "killer" of many "candidates". As a demonstration, I will consider several checks:

Check the "candidate". To do this, artificially imagine it in the form of a fraction, where it is clearly seen that. We calculate the check difference :. Four is divided into "minus two":, which means that the possible root passed the test.

Check the value. Here and check difference is: . Of course, and therefore the second "subject" also remains in the list.

The project discusses the method of approximate finding the roots of the algebraic equation - the Lobachevsky-Haffe method. The work defines the idea of \u200b\u200bthe method, its computational scheme, found conditions for the applicability of the method. The implementation of the Lobachevsky-Greff method is given.

1 Theoretical Part 6

1.1 Problem Statement 6

1.2 Algebraic Equations 7

1.2.1 Basic concepts about Algebraic Equation 7

1.2.2 Kni Algebraic Equation 7

1.2.3 1.2.3 Polynomial Current Roots 9

1.3 Lobachevsky-Grefe method for an approximate solution of algebraic equations 11

1.3.1 Method Idea 11

1.3.2 Quadrings of roots 13

2.1 Task 1 16

2.2 Task 2 18

2.4 Analysis of the results obtained 20

List of links 23.

INTRODUCTION

The computational equipment of our days is a powerful means to actually perform the counting work. Due to this, in many cases it became possible to abandon the approximate interpretation of applied issues and move to solving problems in the exact formulation. The reasonable use of modern computing technology is not conceivable without the skillful application of the methods of approximate and numerical analysis.

Numerical methods are aimed at solving the tasks that arise in practice. The solution of the problem with numerical methods is reduced to arithmetic and logical actions above the numbers, which requires the use of computing equipment, such as table processors of modern office programs for personal computers.

The purpose of the discipline "numerical methods" is to search the most effective solution to a specific task.

The solution of equations - algebraic- represents one of the essential tasks of applied analysis, the need for which occurs in numerous and most diverse sections of physics, mechanics, technology and natural science in the broad sense of the word.

This course project is dedicated to one of the methods of solving algebraic equations - the Lobachevsky-Gherff method.

The purpose of this work is to consider the idea of \u200b\u200bthe Lobachevsky-Wherff method to solve algebraic, bring the computational scheme of finding the actual roots using MS Office Excel. The project examined the main theoretical issues related to the findings of the roots of algebraic equations, the Lobachevsky-Gherff method in the practical part of this work, the solutions of algebraic equations by Lobachevsky-Greff are given.

1 Theoretical part

1.1 Problem Statement

Let the set x of the elements x and the set Y with the elements y. Suppose, besides, the operator is defined on the set x, which puts in accordance with each element x from X some element y from y. Take some element

and put yourself in finding such elements

for which

is an image.

Such a task is equivalent to solving equation

(1.1)

For him, the following problems can be delivered.

The conditions for the existence of the solution of the equation.
The condition of the uniqueness of the solution of the equation.
The algorithm of the decision, following whom, it would be possible to find, depending on the target and conditions, accurately or approximately all solutions of equation (1.1), or any one solution, in advance, or any one of the existing ones.

Next, we will consider the equations in which X and Y will be numerical values, x, y - sets of their values, and the operator

There will be some feature. In this case, equation (1.1) can be written as

(1.2)

In the theory of numerical methods, they seek to build a computational process, with which it is possible to find the solution of equation (1.2) with a defined accuracy. The converging processes are especially important to solve the equation with any, an arbitrarily small error.

Our task is to find, generally speaking, approximate, element . For this purpose, an algorithm is developed, which gives the sequence of approximate solutions.

, and so that the ratio takes place

1.2 Algebraic equations

1.2.1 Basic concepts about the algebraic equation

Consider algebraic n-th equation degree

where coefficients are
- Actual numbers, and
.

Theorem 1.1 (Basic Algebra Theorem). The algebraic equation N-th degree (1.3) has exactly n roots, valid and complex, provided that each root is considered as many times as its multiplicity.

At the same time, they say that the root of equation (1.3) has multiplicity s if
,
.

The complex roots of equation (1.3) have the property of the pairness.

Theorem 1.2. If the coefficients of the algebraic equation (1.3) are valid, then the complex roots of this equation are in pairs of complex-conjugate, i.e. if a
(
- real numbers) there is the root of equation (1.3), multiplicity s, then
It is also the root of this equation and has the same multiplicity s.

Corollary. An algebraic equation of an odd degree with valid coefficients has at least one valid root.

1.2.2. The algebraic equation

If a

- Roots of equation (1.3), then for the left side decomposition
. (1.6)
By multiplying the benomes in formula (1.6) and equating the coefficients at the same degrees x in the left and right parts of equality (1.6), we obtain the relationship between the roots and coefficients of the algebraic equation (1.3):

(1.7)
If you take into account the digits of the roots, the decomposition (1.6) takes the form
,
where

-Dious roots equation (1) and

- their multiplicity, and

Derivative
It is expressed as follows:

where Q (x) is a polynomial such that

at k \u003d 1,2, ..., m

Therefore, polynomials

is the greatest common polynomial divider

and its derivative

and can be found using the Euclidea algorithm. Activate private

,
and get polynomials

with valid coefficients

, And 1, a 2, ..., a m, whose roots

Different.

Thus, the solution of the algebraic equation with multiple roots is reduced to the solution of an algebraic equation of lower order with various roots.

1.2.3 1.24 polynomial roots

The overall view of the number of valid roots of equation (1.3) on the interval (A, B) gives a graph

where roots

The abscissions of the intersection points of the graph with the OX axis.

We note some properties of the polynomial P (x):

If p (a) p (b)
If p (a) p (b)\u003e 0, then on the interval (A, B) exists even number Or there is no root of polynomial P (x).

The question of the number of valid roots of the algebraic equation at a given gap is solved by the assault method.

Definition. Let an ordered finite system of valid numbers other than zero:

,…,

(1.9)
They say that for a couple nearby standing elements

Systems (1.9) There is a change in the sign if these elements have opposite signs, i.e.

,
and there is no change to change if their signs are the same, i.e.

.
Definition. The total number of changes in the signs of all pairs of neighboring elements

Systems (1.9) is called the number of changes in the signs in the system (1.9).

Definition. For this polynomial P (x), the assault system is called the polynomial system

,…,

where
- taken with the opposite sign of the residue during the division of the polynomial on, - taken with the opposite sign of the residue during the division of the polynomial on, etc.

Remark 1. If the polynom has no multiple roots, then the last element of the assault system is different from zero a valid number.

Note 2. The elements of the assault system can be calculated with an accuracy of a positive numerical factor.

Denote by N (C) the number of changes in the storming system at x \u003d c, provided that the zero items of this system are crossed out.

Theorem 1.5. (assault theorem). If Polynina P (X) does not have multiple horses and
,
, then the number of his valid roots
At the interval
exactly equal to the number of lost changes of signs in the polynomial assault system
When moving OT.
before
.

.
Corollary 1. If

, then

Positive and number

Negative polynomial roots are respectively equal

.
Corollary 2. In order for all the roots of the polynomial P (x) degree n, which does not have multiple roots, were valid, it is necessary and enough to be carried out
.
Thus, in equation (1.3), all the roots will be valid then and only if:

With the help of the storming system, it is possible to separate the roots of an algebraic equation, breaking the interval (A, B) containing all the valid roots of the equation to the final number of partial intervals

such that

1.3 Lobachevsky-Haffe method for an approximate solution of algebraic equations

1.3.1 Idea method

Consider the algebraic equation (1.3).

Let's pretend that

, (1.15)
those. The roots are different in the module, and the module of each previous root is much larger than the subsequent module. In other words, suppose that the attitude of any two neighboring roots, counting in the order of declining their numbers, is the value, small by module:

, (1.16)

where
and - Small value. Such roots are called separated.

(1.17)
where , ,…, - Small modulo magnitude compared to one. Neglecting in the system (1.17) values
, We will have approximate ratios

(1.18)
Where to find roots

(1.19)
The accuracy of the roots in the system of equalities (1.20) depends on how small module is module In relations (1.16)

To achieve the root separation, based on equation (1.3), the converted equation is

, (1.20)
roots of which

,…,

are m-e degrees of roots

,…,

Equations (1.3).

If all the roots of equation (1.3) are different and their modules satisfy the condition (1.17), then with a sufficiently large M root ,, ..., equations (1.20) will be separated, because

for

.
Obviously, it is enough to build an algorithm for finding an equation whose roots will be the squares of the roots of a given equation. Then it will be possible to get the equation, the roots of which will be equal to the roots of the initial equation to the extent

1.3.2 Quadrings of roots

Polynomial (1.3) We write in the following form

And multiply it on a polynomial view

Then we get

Making a replacement

And multiplying on

will have
. (1.21)
The roots of the polynomial (1.21) are associated with the roots of the polynomial (1.3) the following ratio

.
Consequently, the equation you are interested in
,
whose coefficients are calculated by formula (1.22)

, (1.22)
where it is assumed that

for

Applying the KVADING process of the roots to the polynomial (1.3) sequentially (1.3), we get a polynomial

, (1.23)
in which

, etc.

At sufficiently large k, it can be achieved so that the system is performed for the roots of equation (1.23)

(1.24)
We define the number K for which the system (1.24) is performed with a given accuracy.

Suppose that the desired K has already been achieved and equality (1.24) are performed with accuracy accepted. We will do another conversion and find a polynomial

,
which also performed the system (1.24) at

Since by virtue of formula (1.22)

, (1.25)
then substituting (1.25) to the system (1.24), we obtain that the absolute values \u200b\u200bof the coefficients

should be in the accuracy taken equal to the squares of the coefficients

. The implementation of these equalities will indicate that the required value of K has already been achieved at the K-M step.

Thus, the quadrings of the roots of equation (1.3) should be discontinued if in the accepted accuracy in the right-hand side of formula (1.24) only the squares of the coefficients are stored, and the double amount of works will be below the accuracy boundary.

Then the valid roots of the equation are obtained by separated and their modules are in the formula

(1.26)
The root sign can be defined by a rough catch, substituting the values and
In equation (1.3).

2 Practical part

2.1 Task 1.

. (2.1)
First, we establish the number of valid and complex roots in equation (2.1). To do this, we use the assault theorem.

The assault system for equation (2.1) will have the following form:

Where do you get
Table 2.1.

Polynomial	Points on a valid axis
Polynomial
	+	+
	–	+
	–	–
	–	+
	–	–
The number of changes of the signs	1	3

Thus, we obtain that the number of valid roots in equation (2.1) is equal
,
those. Equation (2.1) contains 2 valid and two complex roots.

To find the roots of the equation, we use the Lobachevsky-Wherff method for a pair of complex-conjugate roots.

We produce quadrition of the roots of the equation. Calculations of coefficients were made according to the following formula

, (2.2)
where

, (2.3)
but
It is considered to be 0 when
.

The results of calculations with the eight significant numbers are shown in Table 2.2.

Table 2.2.

i.	0	1	2	3	4

	0	-3.8000000E + 01.	3.5400000E + 02.	3.8760000E + 03.	0
	1	4.3000000E + 01.	7.1500000E + 02.	4.8370000E + 03.	1.0404000E + 04.

	0	-1.4300000E + 03.	-3.9517400E + 05.	-1.4877720E + 07.	0
	1	4.1900000E + 02.	1.1605100E + 05.	8.5188490E + 06.	1.0824322E + 08.

	0	-2.3210200E + 05.	-6.9223090E + 09.	-2.5123467E + 13.	0
	1	-5.6541000E + 04.	6.5455256E + 09.	4.7447321E + 13.	1.1716594E + 16.

	0	-1.3091051E + 10.	5.3888712E + 18.	-1.5338253E + 26.	0
	1	-9.8941665E + 09.	4.8232776E + 19.	2.0978658E + 27.	1.3727857E + 32.

	0	-9.6465552E + 19.	4.1513541E + 37.	-1.3242653E + 52.	0
	1	1.4289776E + 18.	2.3679142E + 39.	4.3877982E + 54.	1.8845406E + 64.

	0	-4.7358285E + 39.	-1.2540130E + 73.	-8.9248610+103	0
	1	-4.7337865E + 39.	5.6070053E + 78.	1.9252683+109	3.5514932+128

	0	-1.1214011E + 79.	1.8227619+149	-3.9826483+207	0
	1	1.1194724E + 79.	3.1438509+157	3.7066582+218	1.2613104+257

As can be seen from Table 2.2 on the 7th Step Root , (Considering in descending order of modules) can be considered separated. The root modules are found by formula (1.27) and a rough attack determine their sign:

Since the converted coefficient when Changes the sign, this equation has complex roots, which are determined from equation (1.31) using formulas (1.29) and (1.30):

2.2 Task 2.

The Lobachevsky-Grefe method solve equation:
. (2.4)
To begin with, using the assault theorem, we define the number of valid and complex roots in equation (2.2).

For this equation, the assault system has the form

Where do you get

Table 2.3.

Polynomial	Points on a valid axis
Polynomial
	+	+
	–	+
	+	+
	–	+
	–	–
The number of changes of the signs	3	1

Thus, we obtain that the number of valid roots in equation (2.2) is equal

,
those. Equation (2.2) contains 2 valid and two complex roots.

For the approximate finding of the roots of the equation, we use the Lobachevsky-Haffe method for a pair of complex-conjugate roots.

We produce quadrition of the roots of the equation. Calculations of coefficients will produce according to formulas (2.2) and (2.3).

The results of calculations with the eight significant numbers are shown in Table 2.4

Table 2.4.
-1.8886934E + 24 4.6649263E + 47 i.
The relative error of the roots calculated by formula (1.28) is equal
,

2.4 Analysis of the results obtained

Of the equations obtained in solving equations (2.1) and (2.4), equations can be judged on the following features of the Lobachevsky-Gherde method.

With the help of the method under consideration, you can find all the roots of the polynomial with a sufficiently high accuracy, with a not large number of iterations.

The magnitude of the error received roots in high degree Depends on the fertile separation in the original polynomial, so, for example, in equation (2.1), the minimum difference between the roots varied by module is equal to
and
In equation (2.4), which results in the errors of various orders (4.52958089E-11 and 4.222,97,89E-06, respectively) with the same number of iterations.

Thus, the Lobachevsky-Greff method gives good accuracy during separated roots, and significantly loses with multiple or close roots in the module.

OUTPUT

Lobachevsky-Greff method, which was considered in this project, It has simple schema calculations and allows using Excel to find with great accuracy to the module of all roots of the algebraic equation,

Lobachevsky-Grefe method is one of the most effective methods Calculations, which, with a small number of iterations, gives a result with a fairly good accuracy, so the sphere of using this method in practice is very wide. The method can be used in constructing mathematical models of chemical and physical processes, in optimization methods.

List of links

1. V.P. Demidovich, I.A. Maroon. Basics of computational mathematics. - M.: Science, 1966.-664c.

2. VL. Zaguskin. Handbook on numerical methods of solving algebraic and transcendental equations. - M.: State Publishing House of Physics and Mathematics Literature, 1960.-216c.

3. V.I. Krylov, V.V. Bobkov, P.I. Monastic. Computational Methods of Higher Mathematics.-Minsk: Ex-school, 1972, t. 1.-584c.

4. A.G. Cute. The course of the highest algebra.-M.: Science, 1971, -432c.

5. Yu.I. Ryzhkov. Programming on Fortran PowerStation for engineers. Practical guide. - SPb.: Crown Print, 1999.-160s.

i.	0	1	2	3	4

	0	-9.20000E + 00.	-3.3300000E + 01.	1.3800000E + 02.	0

Page 1
Quadratic equations

In modern algebra, the square equation is called the view equation

where coefficients are
any actual numbers, and

An incomplete square equation is the species equation

Example a)

Thus, the equation has two roots:

Example b.)

Decision

The equation has two roots:

Example from)

Decision

The equation has two roots:

Example d.)

Decision

The equation has no valid roots.

Example e)

Decision

This equation is also an incomplete square equation, it always has one root

When solving square equations, various ways of decomposition of multipliers can be used. So when solving the equation b. A method for making a common factor was applied. There is another way - the grouping method.

Decision.

Answer:

The same equation can be solved by a variety of ways. Consider some of them on the example of a square equation.

I Method. Consider a square threechlen

Spread it on the factors in the grouping method, pre-presenting the term
as
Have

It means that the specified equation can be rewritten as

This equation has two roots:

II. method . Consider a square triple and decompose it for multipliers using a method of high-square isolation; Previously imagine the term 3 in the form of a difference
. Have

Using the formula for the difference of squares, we get

So, the roots are three

III method - graphic.

Consider a graphical method of solving equations

Decide equation

We construct a function schedule

The coordinates of the vertices:

Parabola axis - straight

Take two points on the abscissa axis, symmetrical relative to the parabola axis, such as the point
Find the value of the function at these points
Points
And the top of parabola
Build a function graph.

So, the roots of the equation are the abscissions of the points of intersection of parabola with the abscissa axis, i.e.

Consider another version of the graphical solution of the equation

We write an equation in the form

We will construct in one system of the coordinates of the graphs of functions

So, the roots of the equation are the abscissions of the intersection points of the constructed charts

The initial equation can be solved by several more methods, converting the equation
to sight
or to sight

Then the functions are introduced, the graphs build the abscissions of the intersection of graphs of constructed functions.

See Task 3 (Appendix1).

IV. method - With the formula of the roots of the square equation.

To solve a square equation of type
You can use the following algorithm:

As
This square equation has two roots. These roots find the formula

If b. - even number, i.e.
then

View equation
It is the reduced square equation.

If numbers
Such, that

These numbers are the roots of the equation.
With this statement, or rather approval, reverse theorem. Vieta can solve the above square equations.

So, the roots of the equation

If in equation
sum
then one root of the equation is always 1, and the other root is calculated by the formula.

In equation
The amount is therefore

See Task 4 (Appendix1).
Rational equations
If a
- rational expression, then equation
called a rational equation.

Example

Check the roots found:
those.

are roots of the original equation.

Example

I solve the equation by introducing a variable. Let be
This will reduce the equation in the form

From the equation
Find

Check the roots found

Insofar as
We have to solve two more equations:

and

The roots of the first equation are the numbers 1 and -4, the roots of the second equation - the number

Answer: 1, -4,

The introduction method of the new variable is also used in solving biquette equations.

View equation
called a biquette equation.

Example

We introduce a variable

Receive

Answer: 2, -2.

See tasks 5, 6, and 7 (Appendix1).
Irrational equations
If the variable is contained under the square root sign, then such an equation is called irrational.

Let us turn to the pages from the history of mathematics. The concept of irrational numbers was known to Pythagoreans. The theorem of Pythagora led mathematicians to the opening of incommensurable segments. They received a completely paradoxical statement: the diagonal length of the square cannot be measured by any natural number. This statement undermined the main thesis of their teachings: "Everything is there is a number."

The opening of incommensurability has shown that owning only rational numbers cannot be found the length of any segment. Therefore, many segments are significantly wider than the set of rational numbers. The Greeks decided to build mathematics not on the way to expand the concept of a number, which would lead them to the consideration of irrational numbers, but using geometric values. Unlike Pythagoreans, scientists of the Ancient East without any explanation used approximate numbers. So they recorded 1.41 instead
, and 3 instead of the number

Let us return to modern mathematics and consider ways to solve irrational equations.

Example:

The method of erection in both parts of the equation is the main method of solving irrational equations.

The method of building a square is simple, but sometimes leads to trouble.

Example:

But value
Being a root of a rational equation
It is not a root of a given irrational equation. Verification will confirm this statement.

Check:

The resulting expression does not make sense. Under the root of the even degree can not be a negative number.

Output:
foreign root

Posted IR. rational equation It does not have roots.

Example:

Check:

If a
that

- incorrect

If a
that

- incorrect

Conclusion: the specified irrational equation does not have roots.

So, the irrational equation solve the method of erecting both parts into a square; Deciding the resulting rational equation, it is necessary to make an inspection, extinguishing possible extraneous roots.

Example:

Check:

If a
that

- Faithful equality.

If a
that

- Faithful equality.

So, both found values \u200b\u200bare the roots of the equation.

Answer: 4; five.

Example:

This equation is solved by the introduction of a new variable.

Let be

Let's return to the source variable.

- right,

- incorrect.

See Task 8 (Appendix1).
A bit of theory
Definition. Two equations
and
Called equivalent if they have the same roots (or, in particular, if both equations do not have roots).

Usually, when solving the equation, it is trying to replace this equation to be easier, but equivalent to it. Such a replacement is called equivalent transformation of the equation.

The equivalent transformations of the equation are the following transformations:

1. Transferring members of the equation from one part of the equation to another with opposite signs.

For example, replacement equation
equation
There is equivalent transformation of the equation. This means that the equations
and
equivalent.

2. Multiplication or division of both parts of the equation per and the same different number from zero.

For example, replacement equation
equation
(both parts of the equation multiplied by 10) is equivalent to the equation conversion.

The following transformations are not equalized transformations of the equation:

1. Exemption from denominator containing variables.
For example, replacement equation
equation
There is an unequalization of the conversion of the equation. The fact is that the equation
has two roots: 2 and -2, and a given equation value
It cannot satisfy (the denominator turns to zero). In such cases they say so:
Foreign root.
2. Erect of both parts of the equation in the square.

If, in the process of solving the equation, one of the specified non-uniform transformations was used, then all the roots found should be checked to the source equation, since there may be extraneous roots.

Definition.

The area of \u200b\u200bdefinition of the equation
called many
Where
and
- field definition areas f. and g..

Example

Folding the fractions in the left side, we obtain the equation

equivalent source. The same equation in turn is equivalent to the system

Square equation has roots
Where
- Foreign root.

Consider the solution of the equation

Consequently, the initial equation is equivalent to the totality

or
or
or

Equations with a variable under the sign of the module
1. absolute number of numbers a. (denotes | a.| ) It is called the distance from the point depicting this number A on the coordinate direct, before the start of reference.

From the definition it follows that

The main properties of the module

Example

It is clear that there are two possibilities here:
or
Where it is easy to get

Answer:
or

Note that when solving the equations of the form

the most rational way - the transition to the aggregate

Example

Here, the above acceptance frees us from the need to find the surveys of the Square Square Three Streamed intervals with "unpleasant" roots.

We have:

Answer:
or
or

See Task 9 (Appendix1).
Equations with parameters
A bit of theory.

Students are found with the introduction of some concepts. For example, a direct proportionality function:

linear function:

linear equation:

quadratic equation:

Definition. The equation - appearance and a solution that depends on the values \u200b\u200bof one or several parameters is called the parameters equation.

Solve equation with parameters means

1. Find all system of parameter values \u200b\u200bunder which this equation has solutions.

2. Find all solutions for each parameter value system found, i.e. for unknown and parameters, its regions of permissible values \u200b\u200bshould be indicated.

Example:

Answer: If
That is no solution; example:
These equations are combined tasks, in the process of which the standard solutions solutions are being implemented, and the skills of working with the area of \u200b\u200bpermissible values \u200b\u200band the selection of roots are being formed and secured. These equations are intended as individual tasks for strong students.

The use of equations.

Navier-Stokes equations - system differential equations In private derivatives, describing the movement of a viscous fluid. The Navier-Stokes equations are one of the most important in hydrodynamics and are used in mathematical modeling of many natural phenomena and technical tasks. Named by the name of French Physics Louis Navier and British Mathematics George Stokes.

The system consists of the equation of the equation of the continuity.

One of the applications of the equation system is a description of the flows in the land mantle.

Variations of the equation are used to describe the movement of the air masses of the atmosphere in particular when the weather forecast is generated. The analysis of the solutions of the equation is the essence of one of the open problems, for which the decision of which the Mathematical Institute of Clai appointed a prize of 1 million US dollars. It is necessary to prove or disprove the existence of a global smooth solution to the Cauchy problem for three-dimensional Navier-Stokes equations.
List of used literature

Mordkovich A.G. Algebra. 7 CL: in two parts. Part 1: Tutorial for generalize. Institutions. - 5th ed. - M.: Mnemozina, 2002. - 160 p.: Il.
Mordkovich A.G. Algebra. 8 CL: in two parts. Part 1: Tutorial for generalize. Institutions. - 6th ed. - M.: Mnemozina, 2004. - 223 p.: Il.
A.G. Merzlyak, VB Polonsky, M.S. Yakir Algebraic simulator: allowance for schoolchildren and applicants "/ Ed. Merzlyak A.G., Polonsky V.B., Yakir M.S. - M.: Ilex, 2001 - 320s.
Krivonogov V.V. Non-standard tasks in mathematics: 5-11 classes. - M.: Publishing House "First September", 2002. - 224С.: IL.

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