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Vieta theorem for equation 5 degree. The formula of the Vieta Theorem, and the examples of the solution. Proof of the reverse Vieta Theorem

Municipal budgetary educational institution

"Secondary school №64" Bryansk

City Scientific and Practical Conference

"First steps in science"

Scientifically research

"Vieta theorem for third and fourth degree equations"

Mathematics

Performed: Student 11B class

Sharov Ilya Alekseevich

Scientific adviser:

mathematic teacher,

candidate Fiz.-Mat. Science

Bykov Sergey Valentinovich

Bryansk 2012.

Introduction ........................................................................... 3.

Goals and objectives ....................................................................... 4

Brief historical reference ................................................ 4

Quadratic equation …………………………………………………. five

Cubic equation ............................................................ 6.

Equation of the fourth degree ................................................ 7

Practical part ................................................................ nine

References ............................................................ 12

Appendix ..................................................................... 13.

Introduction

The main theorem of algebra argues that the field is an algebraic closed, in other words, that the equations of N-essentiality with complex coefficients (generally) over the field has exactly n complex roots. The equations of the third degree are solved by the Cordano formula. The equations of the fourth degree method of Ferrari. In addition, in the theory of algebra it is proved that if - the root of the equation, then It is also the root of this equation. For a cubic equation, the following cases are possible:

all three roots are valid;

two complex roots, one valid.

It follows that any cubic equation has at least one valid root.

For the fourth degree equation:

All four roots are different.

Two root valid, two - complex.

All four roots are complex.

This work is devoted to a thorough study of the Vieta's theorem: its wording, evidence, as well as solving problems with the use of this theorem.

The work done aims to assist the student of the 11th grade, which will have to pass the exam, as well as for young mathematicians who are not indifferent to the simplest and effective methods solutions in various fields of mathematics.

The Appendix to this work provides a collection of tasks for an independent solution and consolidate the new material under study by me.

This question cannot be left without attention, as it is important for mathematics for both science as a whole and for students and the solution of such tasks.

Goals and work tasks:

Get an analogue of the Vieta theorem for the third degree equation.

Prove an analogue of the Vieta theorem for the third degree equation.

Get an analogue of the Vieta theorem for the fourth degree equation.

Prove the analogue of the Viet's theorem for the fourth degree equation.

Consider applying these issues to solving practical tasks.

• Ensure in the practicality of the use of this theorem.

Deepen mathematical knowledge in the area of \u200b\u200bsolving equations.

Develop interest in mathematics.

Brief historical certificate

Rightfully worthy in verses be singing

On the properties of the roots of the Vieta Theorem ...

Francois Viet (1540-1603) - French mathematician. By profession a lawyer. In 1591, introduced letter not only for unknown values, but also for the coefficients of equations; Due to this, it was the first time for the first time the expression of the properties of equations and their roots with shared formulas. It owns the establishment of a uniform receipt of solutions of equations of the 2nd, 3rd and 4th degrees. Among the discovery itself, it has especially highly appreciated the establishment of the dependence between the roots and the coefficients of equations. For an approximate solution of equations with numerical coefficients, Viet suggested a method similar to the later method of Newton. In Trigonometry, Francois Vieta gave a complete solution of the problem of determining all elements of a flat or spherical triangle of three data, found important decomposition of COS nx and sin. nx in degrees cos. h. and sin. x. He first reviewed endless works. Writings of the Vieta are written by a difficult language and therefore got less dissemination at one time than they deserved .

Quadratic equation

To begin with, remember the formula of the Vieta for the equation of the second degree, which we learned in the school training program.

T.
eorn Vietafor a square equation (grade 8)

E.
if the roots of the square equation

i.e. the sum of the roots of the reduced 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.

Also, remember theorem, reverse Vieta Theorem:

If numbers - p. and q. Such, that

that is the roots of the equation

The Vieta Theorem is wonderful in that, not knowing the roots of square three decletes, we can easily calculate their sum and work, that is, the simplest symmetrical expressions.

The theorem of the Vieta allows you to guess the whole roots of the square three decar.

Cubic equation

Now let's turn directly to the formulation and solution of the cubic equation using the Vieta theorem.

Formulation

TO
the kills equation is the third order equation, the species

where a ≠ 0.

If a a \u003d 1., the equation is called a given cubic equation:

So, you need to prove that for the equation

the following theorem is valid:

p
ust roofs of this equation, then

Evidence

Imagine a polynomial

perform conversion:

So, we get that

Two The polynomial is then and only if their coefficients are equal to the corresponding degrees.

It means that

Q.E.D.

Now consider theorem, the reverse theorem of the Vieta for the third degree equation.

F.
ormulsion

E.
if the number is such that

The fourth degree equation

Now let's turn to the formulation and solving the equation of the fourth degree using the Vieta theorem for the fourth degree equation.

Formulation

W.
equality of the fourth degree - equation of type

g.
de a ≠ 0.

E.
silence a \u003d 1.then the equation is called the given

AND
so, we prove that for the equation

from
the following theorem is righteous: let the roots of this equation, then

Evidence

Imagine a polynomial

perform conversion:

So, we get that

We know that two polynomials are then and only if their coefficients are equal to the corresponding degrees.

It means that

Q.E.D.

Consider the theorem reverse Vieta theorem for the fourth degree equation.

Formulation

If the numbers are such that

then these numbers are roots equations

Practical part

Now consider solving problems, using the Viet's theorem for the third and fourth degree equations.

Task number 1

Answer: 4, -4.

Task number 2.

Answer: 16, 24.

To solve these equations, the Cardano formulas and the Ferrari method can be used, respectively, but using the Vieta theorem, we know the amount and the product of the roots of these equations.

Task number 3.

Make a third degree equation if it is known that the amount of the roots is 6, the pair product of the roots is 3, and the product -4.

Make an equation, get

Task number 4.

Make the equation of the third degree, if it is known that the amount of the roots is equal 8 , on the pair of the roots equal 4 , tripled work is equal 12 , and the work 20 .

Solution: Using the Formula of Vieta, we get

Make an equation, get

With the help of the Vieta Theorem, we easily accounted for equations on their roots. This is the most rational way to solve these tasks.

Task number 5.

where A, B, C is the Geron formulas.

We will reveal the brackets and transform the expression, we get

Z.
amethim that the guided expression is cubic expression. We use the Viet's theorem for the corresponding cubic equation, then we have that

Z.

naya that we get:

From solving this problem, it can be seen that the Vieta theorem is applicable to tasks from different regions of mathematics.

Conclusion

In this paper, the method of solving the equation of the third and fourth degrees was investigated using the Vieta Theorem. Disabled formulas are easy to use. In the course of the study, it became obvious that in some cases this method is more effective than the Cordano formula and the Ferrari method for the third and fourth degree equations, respectively.

Vieta Theorem was applied in practice. A number of tasks were solved, which helped better consolidate the new material.

This study was very interesting and informative for me. Having deepened his knowledge in mathematics, I opened a lot of interesting things and with pleasure I was engaged in this study.

But my study in the field of solving equations is not completed. In the future, I plan to study the solution of the N-essential equation using the Vieta Theorem.

I want to express great gratitude to my scientific leader, a candidate of physical and mathematical sciences, and the possibility of such an unusual research and constant attention in work.

Bibliography

Vinogradov I.M. Mathematical encyclopedia. M., 1977.

V. B. Lidsky, L. V. Ovsyannikov, A. N. Tulaykov, M. I. Shabunin. Tasks for elementary mathematics, fizmatlite, 1980.

poncele's theorem for Triangle ... g2 - degree Or ... Arc third Lunok less ... the equationgiving fourth ... mathematician F. Viet. mathematician ...

Scientific research work in mathematics

Research

... Scientific – researchwork by mathematics Geometry... theorem Poncel for Triangle ... g2 - degree Or ... Arc third Lunok less ... the equationgiving fourth ... mathematician F. Viet. Calculated in 1579. I am with 9 signs. Dutch mathematician ...

Book

... for equationthird and fourthdegree mathematics researchwork. The best scientists of France ...

Short essay of the history of mathematics 5th edition corrected

Book

... for Many later textbooks on Algeora. In it, the presentation was brought to the theory equationthird and fourthdegree ... theoretical and applied mathematics. Attention was paid to both teaching and researchwork. The best scientists of France ...

In mathematics there are special techniques with which many square equations are solved very quickly and without any discriminants. Moreover, with proper training, many begin to solve square equations orally, literally "at first glance."

Unfortunately, in the modern course of school mathematics, such technologies are almost not studied. And you need to know! And today we will look at one of these techniques - Vieta theorem. To begin with, we introduce a new definition.

The square equation of the form x 2 + bx + c \u003d 0 is called the above. Note: The coefficient at x 2 is 1. No other restrictions on the coefficients are not superimposed.

1. x 2 + 7x + 12 \u003d 0 is a given square equation;
2. x 2 - 5x + 6 \u003d 0 - also given;
3. 2x 2 - 6x + 8 \u003d 0 - But this is not the nifiga, since the coefficient at x 2 is 2.

Of course, any square equation of the type AX 2 + BX + C \u003d 0 can be made given - it is enough to divide all the coefficients to the number a. We can always do that, since it follows from the definition of the square equation, that A ≠ 0.

True, not always these transformations will be useful for finding roots. Just below, we make sure that it is necessary only when in the final square of the equation all the coefficients will be integer. In the meantime, consider the simplest examples:

A task. Convert a square equation to the above:

1. 3x 2 - 12x + 18 \u003d 0;
2. -4x 2 + 32x + 16 \u003d 0;
3. 1.5x 2 + 7.5X + 3 \u003d 0;
4. 2x 2 + 7x - 11 \u003d 0.

We divide each equation on the coefficient with a variable x 2. We get:

1. 3x 2 - 12x + 18 \u003d 0 ⇒ x 2 - 4x + 6 \u003d 0 - divided by all by 3;
2. -4x 2 + 32x + 16 \u003d 0 ⇒ x 2 - 8x - 4 \u003d 0 - divided into -4;
3. 1.5x 2 + 7.5X + 3 \u003d 0 ⇒ x 2 + 5x + 2 \u003d 0 - divided by 1.5, all coefficients became integer;
4. 2x 2 + 7x - 11 \u003d 0 ⇒ x 2 + 3.5x - 5.5 \u003d 0 - divided by 2. In this case, fractional coefficients arose.

As you can see, the presented square equations may have entire coefficients even if the initial equation contained the fraction.

Now we formulate the main theorem for which, in fact, the concept of a given square equation was introduced:

Vieta theorem. Consider the given square equation of the form x 2 + BX + C \u003d 0. Suppose that this equation has valid roots x 1 and x 2. In this case, the following statements are true:

1. x 1 + x 2 \u003d -b. In other words, the sum of the roots of the present square equation is equal to the coefficient with a variable X, taken with the opposite sign;
2. x 1 · x 2 \u003d c. The product of the roots of the square equation is equal to the free coefficient.

Examples. For simplicity, we will consider only the above square equations that do not require additional transformations:

1. x 2 - 9x + 20 \u003d 0 ⇒ x 1 + x 2 \u003d - (-9) \u003d 9; x 1 · x 2 \u003d 20; roots: x 1 \u003d 4; x 2 \u003d 5;
2. x 2 + 2x - 15 \u003d 0 ⇒ x 1 + x 2 \u003d -2; x 1 · x 2 \u003d -15; roots: x 1 \u003d 3; x 2 \u003d -5;
3. x 2 + 5x + 4 \u003d 0 ⇒ x 1 + x 2 \u003d -5; x 1 · x 2 \u003d 4; roots: x 1 \u003d -1; x 2 \u003d -4.

Vieta theorem gives us additional information about the roots of the square equation. At first glance, this may seem difficult, but even with a minimum training, you will learn to "see" the roots and literally guess them in seconds.

A task. Solve the square equation:

1. x 2 - 9x + 14 \u003d 0;
2. x 2 - 12x + 27 \u003d 0;
3. 3x 2 + 33x + 30 \u003d 0;
4. -7x 2 + 77x - 210 \u003d 0.

Let's try to write the coefficients on the theorem of Vieta and "Guess" the roots:

1. x 2 - 9x + 14 \u003d 0 is a given square equation.
   By the theorem, we have: x 1 + x 2 \u003d - (- 9) \u003d 9; x 1 · x 2 \u003d 14. It is easy to notice that the roots are numbers 2 and 7;
2. x 2 - 12x + 27 \u003d 0 - also given.
   By the Vieta Theorem: x 1 + x 2 \u003d - (- 12) \u003d 12; x 1 · x 2 \u003d 27. Hence the roots: 3 and 9;
3. 3x 2 + 33x + 30 \u003d 0 - this equation is not given. But we will correct it now, dividing both sides of the equation to the coefficient a \u003d 3. We obtain: x 2 + 11x + 10 \u003d 0.
   We decide on the VEETORE Theorem: x 1 + x 2 \u003d -11; x 1 · x 2 \u003d 10 ⇒ roots: -10 and -1;
4. -7x 2 + 77x - 210 \u003d 0 - again the coefficient at x 2 is not equal to 1, i.e. The equation is not given. We divide everything for the number a \u003d -7. We obtain: x 2 - 11x + 30 \u003d 0.
   By the Vieta Theorem: x 1 + x 2 \u003d - (- 11) \u003d 11; x 1 · x 2 \u003d 30; Of these equations, it is easy to guess the roots: 5 and 6.

From the above reasoning, it is seen how the Vieta theorem simplifies the solution of square equations. No computational computing arithmetic roots and fractions. And even discriminant (see the lesson "Solution of Square Equations") we did not need.

Of course, in all reflections, we proceeded from two important assumptions, which, generally speaking, are not always performed in real tasks:

1. The square equation is given, i.e. The coefficient at x 2 is 1;
2. The equation has two different roots. From the point of view of algebra, in this case discriminant D\u003e 0 - in fact, we initially assume that this inequality is true.

However, in typical mathematical tasks, these conditions are performed. If, as a result of the calculations, it turned out a "bad" square equation (the coefficient at x 2 is different from 1), it is easy to fix - take a look at the examples at the very beginning of the lesson. About the roots at all silent: what is this task in which there is no answer? Of course, the roots will be.

Thus, the general scheme of solving square equations on the Vieta Theorem looks like this:

1. Reduce the square equation to the given one if it has not yet been done in the condition of the problem;
2. If the coefficients in the specified square equation turned out fractional, solve through the discriminant. You can even return to the initial equation to work with more "convenient" numbers;
3. In the case of integer coefficients, we solve the equation on the Vieta Theorem;
4. If within a few seconds it did not manage to guess the roots, scored on the Vieta theorem and solve through the discriminant.

A task. Decide equation: 5x 2 - 35x + 50 \u003d 0.

So, in front of us, the equation that is not given, because The coefficient a \u003d 5. We divide all by 5, we get: x 2 - 7x + 10 \u003d 0.

All the coefficients of the square equation are integer - we will try to decide on the Vieta theorem. We have: x 1 + x 2 \u003d - (- 7) \u003d 7; x 1 · x 2 \u003d 10. In this case, the roots are guessed easily - this is 2 and 5. It is not necessary to count through the discriminant.

A task. Decide the equation: -5x 2 + 8x - 2.4 \u003d 0.

We look: -5x 2 + 8x - 2.4 \u003d 0 - this equation is not given, we divide both sides to the coefficient a \u003d -5. We obtain: x 2 - 1.6x + 0.48 \u003d 0 - Equation with fractional coefficients.

It is better to return to the initial equation and count through the discriminant: -5x 2 + 8x - 2.4 \u003d 0 ⇒ d \u003d 8 2 - 4 · (-5) · (-2.4) \u003d 16 ⇒ ... ⇒ x 1 \u003d 1.2; x 2 \u003d 0.4.

A task. Solve the equation: 2x 2 + 10x - 600 \u003d 0.

To begin with, we divide everything to the coefficient A \u003d 2. It turns out equation x 2 + 5x - 300 \u003d 0.

This is the reduced equation, on the Vieta theorem, we have: x 1 + x 2 \u003d -5; x 1 · x 2 \u003d -300. Guess the roots of the square equation in this case is difficult - personally, I seriously "hung" when I solved this task.

We'll have to look for roots through discriminant: d \u003d 5 2 - 4 · 1 · (-300) \u003d 1225 \u003d 35 2. If you do not remember the root from the discriminant, I simply note that 1225: 25 \u003d 49. Therefore, 1225 \u003d 25 · 49 \u003d 5 2 · 7 2 \u003d 35 2.

Now that the root of the discriminant is known, the equation will not be solved. We obtain: x 1 \u003d 15; x 2 \u003d -20.

These formulas are convenient to use to verify the correctness of the roots of the polynomial, as well as to compile a polynomial on the specified roots.

History

These identities implicitly present in the works of Francois Vieta. However, Viet considered only positive real roots, so he had no opportunity to write these formulas in general. : 138-139

Formulation

If a C 1, C 2, ..., C N (\\ DisplayStyle C_ (1), C_ (2), \\ LDOTS, C_ (N)) - Numerous roots

x n + a 1 x n - 1 + a 2 x n - 2 +. . . + a n, (\\ displaystyle x ^ (n) + a_ (1) x ^ (n - 1) + a_ (2) x ^ (n-2) + ... + a_ (n),)

(Each root takes the corresponding multiplicity of the number of times), then the coefficients a 1, ..., a n (\\ displaystyle a_ (1), \\ ldots, a_ (n)) expressed in the form of symmetric polynomials from the roots, namely:

a 1 \u003d - (C 1 + C 2 + ... + Cn) a 2 \u003d C 1 C 2 + C 1 C 3 + ... + C 1 CN + C 2 C 3 + ... + CN - 1 CNA 3 \u003d - (C 1 C 2 C 3 + C 1 C 2 C 4 + ... + CN - 2 CN - 1 CN) ... AN - 1 \u003d (- 1) N - 1 (C 1 C 2 ... CN - 1 + C 1 C 2 ... CN - 2 CN + ... + C 2 C 3... CN) AN \u003d (- 1) NC 1 C 2 ... CN (\\ TextStyle (\\ Begin (Matrix) A_ (1) & \u003d & - (C_ (1) + C_ (2) + \\ ldots + c_ (n)) \\\\ A_ (2) & \u003d & C_ (1) C_ (2) + C_ (1) C_ (3) + \\ LDOTS + C_ (1) C_ (N ) + c_ (2) c_ (3) + \\ ldots + c_ (n - 1) c_ (n) \\\\ A_ (3) & \u003d Δ (C_ (1) C_ (2) C_ (3) + C_ ( 1) C_ (2) C_ (4) + \\ LDOTS + C_ (N-2) C_ (N - 1) C_ (N)) \\\\ && \\ ldots \\\\ A_ (N-1) & \u003d & (- 1 ) ^ (n-1) (C_ (1) C_ (2) \\ ldots c_ (n - 1) + c_ (1) c_ (2) \\ ldots c_ (n-2) c_ (n) + \\ ldots + c_ (2) C_ (3) ... C_ (n)) \\\\ A_ (N) & \u003d & (- 1) ^ (n) C_ (1) C_ (2) \\ LDOTS C_ (N) \\ END (MATRIX )))

In other words, (- 1) k a k (\\ displaystyle (-1) ^ (k) a_ (k)) equal to the sum of all possible works from K (\\ DisplayStyle K) roots.

If the senior polynomial coefficient A 0 ≠ 1 (\\ DisplayStyle A_ (0) \\ NEQ 1), for the use of the formula of the Vieta, it is necessary to pre-divide all coefficients on A 0 (\\ DisplayStyle A_ (0)) (This does not affect the value of the numerous roots). In this case, the formula of the Vieta give an expression for the relationship of all coefficients to the older. From the last formula of the Vieta it follows that if the roots are numerous integer, then they are divisors of his free member, which is also integer.

Evidence

The proof is carried out by the consideration of the equality obtained by the decomposition of the polynomial on the roots, given that a 0 \u003d 1 (\\ displayStyle A_ (0) \u003d 1)

x n + a 1 x n - 1 + a 2 x n - 2 +. . . + an \u003d (x - C 1) (x - C 2) ⋯ (x - Cn) (\\ displaystyle x ^ (n) + a_ (1) x ^ (n - 1) + a_ (2) x ^ (n -2) + ... + a_ (n) \u003d (x-c_ (1)) (X-C_ (2)) \\ CDOTS (X-C_ (N)))

Equating coefficients with the same degrees X (\\ DisplayStyle X) (Uniqueness theorem), we get the formula of the Vieta.

Examples

Quadratic equation

If a X 1 (\\ DisplayStyle X_ (1)) and x 2 (\\ displaystyle x_ (2)) - roots of the square equation a x 2 + b x + c \u003d 0 (\\ displaystyle \\ ax ^ (2) + bx + c \u003d 0) T.

(x 1 + x 2 \u003d - bax 1 x 2 \u003d Ca (\\ displayStyle (\\ Begin (Cases) ~ x_ (1) + x_ (2) \u003d ~ - (\\ dfrac (b) (a)) \\\\ ~ x_ (1) x_ (2) \u003d ~ (\\ DFRAC (C) (A)) \\ End (Cases)))

In a particular case, if a \u003d 1 (\\ displaystyle a \u003d 1) (Limited form x 2 + p x + q \u003d 0 (\\ displaystyle x ^ (2) + px + q \u003d 0)), T.

(x 1 + x 2 \u003d - px 1 x 2 \u003d q (\\ displayStyle (\\ Begin (Cases) ~ x_ (1) + x_ (2) \u003d - p \\\\ ~ x_ (1) x_ (2) \u003d q \\ x 1, x 2, x 3 (\\ displaystyle x_ (1), x_ (2), x_ (3))

Cubic equation

If a - Cubic equation roots p (x) \u003d a x 3 + b x 2 + c x + d \u003d 0 (\\ displaystyle p (x) \u003d ax ^ (3) + bx ^ (2) + cx + d \u003d 0) T.I. Vieta Theorem

For a given square equation.the sum of the roots of the given square equation

x 2 + px + q \u003d 0 equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to a free member: x 1 + x 2 \u003d -p; x 1 ∙ x 2 \u003d Q.

find the roots of the specified square equation using the Vieta theorem.

Example 1) x 2 -X-30 \u003d 0.

This is the reduced square equation x 2 + px + q \u003d 0) ( , second coefficientP \u003d -1. , and free dickQ \u003d -30. first, make sure that this equation has a root, and that the roots (if any) will be expressed by integers. It is enough for the discriminant to be a complete square of an integer. We find discriminant

D. \u003d B 2 - 4AC \u003d (- 1) 2 -4 ∙ 1 ∙ (-30) \u003d 1 + 120 \u003d 121 \u003dNow, by the Vieta Theorem, the amount of roots should be equal to the second coefficient taken with the opposite sign, i.e. ( 11 2 .

-p. ), and the work is equal to a free member, i.e. (). Then: q.x 1 + x 2 \u003d 1; x 1 ∙ x 2 \u003d -30.

we need to pick up such two numbers so that their work is equal , and the amount - -30 unity . This is numbersand -5 Answer: -5; 6. 6 . Example 2) x 2 + 6x + 8 \u003d 0.

We have a given square equation with the second coefficient P \u003d 6. and free member Q \u003d 8. . We make sure there are integer roots. We find discriminantD 1. . Discriminant D 1 is a complete square of the number . Discriminant D 1 is a complete square of the number=3 2 -1∙8=9-8=1=1 2 So the roots of this equation are integers. We will select the roots on the Vieta Theorem: the amount of the roots is equal 1 -R \u003d -6. , and the product of the roots is equalIn fact: -4-2 \u003d -6 \u003d -r; -4 ∙ (-2) \u003d 8 \u003d Q. . We make sure there are integer roots. We find discriminantand -4 and -2 .

Answer: -4; -2. Example 3) x 2 + 2x-4 \u003d 0

. In this square equation, the second coefficientP \u003d 2. q \u003d -4.Q \u003d -30. . We find discriminantSince the second coefficient is an even number. . Discriminant D 1 is a complete square of the numberDiscriminant is not a complete square of the number, so we do . Discriminant D 1 is a complete square of the number=1 2 -1∙(-4)=1+4=5. output The roots of this equation are not integers and find them on the theorem of Vieta can not.: it means that this equation is solved, as usual, according to the formulas (in this case by formulas). We get: Example 4).

Make a square equation for its roots ifx 1 \u003d -7, x 2 \u003d 4. decision.

The desired equation will be recorded in the form: , moreover, on the basis of the Vieta theorem equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to a free member:-P \u003d x 1 + x 2 → p \u003d 3;=-7+4=-3 q \u003d x 1 ∙ x 2 . Then the equation will take the form:=-7∙4=-28 x 2 + 3x-28 \u003d 0. example 5).

Make a square equation on its roots, if:

II. Vieta theorem for a complete square equation aX 2 + BX + C \u003d 0.

The amount of the roots is minus b.divided by but, the product of the roots is equal fromdivided by

In square equations, there are a number of ratios. The main relationship between roots and coefficients. Also in square equations, a number of relations are operated, which are set by the Vieta theorem.

In this topic, we present the theorem of the Vieta and its proof for a square equation, theorem, the reverse theorem of the Vieta, we will analyze a number of examples of solving problems. We will pay special attention in the material to consider the formulas of the Vieta, which set the relationship between the valid roots of the algebraic equation of the degree N. and its coefficients.

Yandex.rtb R-A-339285-1

The wording and proof of the Vieta Theorem

Square equation root formula a · x 2 + b · x + c \u003d 0 species x 1 \u003d - b + d 2 · a, x 2 \u003d - b - d 2 · a, where D \u003d b 2 - 4 · a · c, establishes relations x 1 + x 2 \u003d - b a, x 1 · x 2 \u003d c a. This confirms the Vieta theorem.

Theorem 1.

In a square equation a · x 2 + b · x + c \u003d 0where x 1 and x 2 - roots, the amount of the roots will be equal to the ratio of coefficients B. and A.which was taken with the opposite sign, and the product of the roots will be equal to the attitude of the coefficients C. and A., i.e. x 1 + x 2 \u003d - b a, x 1 · x 2 \u003d c a.

Proof 1.

We offer you the following proof scheme: Take the formula of the roots, make up the Suma and the product of the roots of the square equation and then transform the received expressions in order to make sure that they are equal - B A. and C A. respectively.

We will make the amount of the roots x 1 + x 2 \u003d - b + d 2 · a + - b - d 2 · a. Let's give the fractions K. common denominator - b + d 2 · a + - b - d 2 · a \u003d - b + d + - b - d 2 · a. We will reveal the brackets in the numerator of the resulting fraction and give similar terms: - b + d + - b - d 2 · a \u003d - b + d - b - d 2 · a \u003d - 2 · b 2 · a. Sperate fraction on: 2 - b a \u003d - b a.

So we proved the first ratio of the Vieta theorem, which relates to the amount of the roots of the square equation.

Now let's go to the second ratio.

To do this, we need to make a product of the roots of the square equation: x 1 · x 2 \u003d - b + d 2 · a · - b - d 2 · a.

Recall the rule of multiplication of fractions and write down the last work as follows: - b + d · - b - d 4 · a 2.

We carry out the fraction in the numerator multiplication of the bracket to the bracket or we will use the square difference formula in order to convert this product faster: - b + d · - b - d 4 · a 2 \u003d - b 2 - d 2 4 · a 2.

We use the definition square root In order to carry out the following transition: - B 2 - D 2 4 · a 2 \u003d B 2 - D 4 · A 2. Formula D \u003d b 2 - 4 · a · c responds to the discriminator of the square equation, therefore, in the fraction instead D. can be substituted B 2 - 4 · A · C:

b 2 - D 4 · A 2 \u003d B 2 - (B 2 - 4 · A · C) 4 · A 2

We will open brackets, we give similar terms and obtain: 4 · A · C 4 · A 2. If you cut it on 4 · A., it remains C a. So we proved the second ratio of the Vieta theorem for the product of the roots.

Record the proof of the Vieta Theorem may have a very concise appearance if you lower the explanation:

x 1 + x 2 \u003d - b + d 2 · a + - b - d 2 · a \u003d - b + d + - b - d 2 · a \u003d - 2 · b 2 · a \u003d - ba, x 1 · x 2 \u003d - b + d 2 · a · - b - d 2 · a \u003d - b + d · - b - d 4 · a 2 \u003d - b 2 - d 2 4 · a 2 \u003d b 2 - d 4 · a 2 \u003d \u003d d \u003d b 2 - 4 · a · c \u003d b 2 - b 2 - 4 · a · C 4 · a 2 \u003d 4 · a · C 4 · a 2 \u003d Ca.

When the square equation is discriminated, the equation will have only one root. To be able to apply to such an equation theorem of the Vieta, we can assume that the equation during discriminant is zero, has two identical root. Indeed, for D \u003d 0. The root of the square equation is: - B 2 · A, then x 1 + x 2 \u003d - b 2 · a + - b 2 · a \u003d - b + (- b) 2 · a \u003d - 2 · b 2 · a \u003d - Ba and x 1 · x 2 \u003d - b 2 · a · - b 2 · a \u003d - b · - b 4 · a 2 \u003d b 2 4 · a 2, and since d \u003d 0, that is, B 2 - 4 · a · c \u003d 0, from where b 2 \u003d 4 · a · c, then b 2 4 · a 2 \u003d 4 · a · C 4 · a 2 \u003d Ca.

Most often, in practice, the Vieta Theorem is used in relation to the specified square equation of type x 2 + p · x + q \u003d 0where the senior coefficient A is equal to 1. In this regard, and formulate the theorem of the Vieta precisely for equations of this type. This does not limit the generality due to the fact that any square equation can be replaced by the equivalent equation. To do this, it is necessary to divide both of its parts by the number A, different from zero.

We give another formulation of the Vieta theorem.

Theorem 2.

The amount of the roots in the given square equation x 2 + p · x + q \u003d 0 will be equal to the coefficient with X, which is taken with the opposite sign, the product of the roots will be equal to a free member, i.e. x 1 + x 2 \u003d - p, x 1 · x 2 \u003d q.

Theorem, reverse theorem of Vieta

If you carefully look at the second wording of the Vieta theorem, you can see that for the roots x 1 and x 2given square equation x 2 + p · x + q \u003d 0 The relation x 1 + x 2 \u003d - p, x 1 · x 2 \u003d q will be valid. From these relations x 1 + x 2 \u003d - p, x 1 · x 2 \u003d q it follows that x 1 and x 2 - these are the roots of the square equation x 2 + p · x + q \u003d 0. So we come to the statement, which is the reverse Vieta theorem.

We propose to now issue this approval as a theorem and conduct its proof.

Theorem 3.

If numbers x 1 and x 2 Such, that x 1 + x 2 \u003d - p and x 1 · x 2 \u003d qT. x 1 and x 2 are the roots of the given square equation x 2 + p · x + q \u003d 0.

Proof 2.

Replacing coefficients P. and Q. on their expression through x 1 and x 2 Allows you to convert equation x 2 + p · x + q \u003d 0 In equivalent to it .

If the obtained equation substitute the number x 1 instead X.then we will get equality x 1 2 - (x 1 + x 2) · x 1 + x 1 · x 2 \u003d 0. This is equality with any x 1 and x 2 turns into faithful numerical equality 0 = 0 , as x 1 2 - (x 1 + x 2) · x 1 + x 1 · x 2 \u003d x 1 2 - x 1 2 - x 2 · x 1 + x 1 · x 2 \u003d 0. It means that x 1 - root of the equation x 2 - (x 1 + x 2) · x + x 1 · x 2 \u003d 0, So what x 1 is also the root of equivalent equation to him x 2 + p · x + q \u003d 0.

Substitution to equation x 2 - (x 1 + x 2) · x + x 1 · x 2 \u003d 0 numbers x 2instead of x allows you to get equality x 2 2 - (x 1 + x 2) · x 2 + x 1 · x 2 \u003d 0. This equality can be considered true, since x 2 2 - (x 1 + x 2) · x 2 + x 1 · x 2 \u003d x 2 2 - x 1 · x 2 - x 2 2 + x 1 · x 2 \u003d 0. Turns out that x 2 It is the root of the equation x 2 - (x 1 + x 2) · x + x 1 · x 2 \u003d 0and therefore equations x 2 + p · x + q \u003d 0.

Theorem, reverse theorem of Vieta, is proved.

Examples of using the Vieta Theorem

Let's now proceed to the analysis of the most typical examples on the topic. Let's start with the parsing of tasks that require the use of the theorem, the reverse theorem of the Vieta. It can be used to verify the numbers obtained during the calculations, on whether they are the roots of a given square equation. To do this, it is necessary to calculate their sum and difference, and then check the validity of the ratios x 1 + x 2 \u003d - b a, x 1 · x 2 \u003d a c.

The performance of both relations suggests that the numbers obtained during the calculations are roots of the equation. If we see that at least one of the conditions is not fulfilled, the data of the number cannot be the roots of the square equation given in the condition of the problem.

Example 1.

Which of the pairs of numbers 1) x 1 \u003d - 5, x 2 \u003d 3, or 2) x 1 \u003d 1 - 3, x 2 \u003d 3 + 3, or 3) x 1 \u003d 2 + 7 2, x 2 \u003d 2 - 7 2 is a pair of roots of the square equation 4 · x 2 - 16 · x + 9 \u003d 0?

Decision

Find the coefficients of the square equation 4 · x 2 - 16 · x + 9 \u003d 0.This is a \u003d 4, b \u003d - 16, c \u003d 9. In accordance with the theorem, the sum of the roots of the square equation must be equal to - B A., i.e, 16 4 = 4 and the product of the roots should be equal C A., i.e, 9 4 .

We check the numbers obtained by calculating the amount and the product of the numbers of three set pairs and comparing them with the obtained values.

In the first case x 1 + x 2 \u003d - 5 + 3 \u003d - 2. This value is different from 4, therefore, the check can not be continued. According to the theorem, the reverse theorem of the Vieta, it is possible to immediately conclude that the first pair of numbers is not the roots of this square equation.

In the second case x 1 + x 2 \u003d 1 - 3 + 3 + 3 \u003d 4. We see that the first condition is performed. But the second condition is not: x 1 · x 2 \u003d 1 - 3 · 3 + 3 \u003d 3 + 3 - 3 · 3 - 3 \u003d - 2 · 3. The value we got is different from 9 4 . This means that the second pair of numbers is not the roots of the square equation.

Let us turn to the consideration of the third pair. Here x 1 + x 2 \u003d 2 + 7 2 + 2 - 7 2 \u003d 4 and x 1 · x 2 \u003d 2 + 7 2 · 2 - 7 2 \u003d 2 2 - 7 2 2 \u003d 4 - 7 4 \u003d 16 4 - 7 4 \u003d 9 4. Both conditions are performed, which means that x 1 and x 2 are the roots of a given square equation.

Answer: x 1 \u003d 2 + 7 2, x 2 \u003d 2 - 7 2

We can also use the theorem, the reverse theorem of the Vieta, to select the roots of the square equation. The easiest way is the selection of whole roots of the given square equations with integer coefficients. You can consider other options. But it can significantly impede computation.

For the selection of roots, we use the fact that if the sum of two numbers is equal to the second coefficient of the square equation taken with a minus sign, and the product of these numbers is equal to a free member, then these numbers are roots of this square equation.

Example 2.

As an example, use a square equation x 2 - 5 · x + 6 \u003d 0. Numbers x 1 and x 2 may be the roots of this equation if two equalities are performed x 1 + x 2 \u003d 5 and x 1 · x 2 \u003d 6. We will select such numbers. These are numbers 2 and 3, as 2 + 3 = 5 and 2 · 3 \u003d 6. It turns out that 2 and 3 are the roots of this square equation.

Theorem, reverse theorem of Vieta, can be used to find the second root when first is known or obvious. For this, we can use the ratios x 1 + x 2 \u003d - b a, x 1 · x 2 \u003d c a.

Example 3.

Consider a square equation 512 · x 2 - 509 · x - 3 \u003d 0. It is necessary to find the roots of this equation.

Decision

The first root of the equation is 1, since the sum of the coefficients of this square equation is zero. Turns out that x 1 \u003d 1.

Now we find the second root. For this you can use the ratio x 1 · x 2 \u003d c a. Turns out that 1 · x 2 \u003d - 3 512From! x 2 \u003d - 3 512.

Answer: The roots of the challenge equation specified in the condition 1 and - 3 512 .

Select the roots using the theorem, the reverse theorem of the Vieta, is possible only in simple cases. In other cases, it is better to search with the formula of the roots of the square equation through the discriminant.

Thanks to the theorem, the reverse theorem of the Vieta, we can also make square equations on the available roots x 1 and x 2. To do this, we need to calculate the amount of the roots, which gives the coefficient when X. With the opposite sign of the specified square equation, and the product of the roots, which gives a free member.

Example 4.

Write a square equation, the roots of which are numbers − 11 and 23 .

Decision

We will take that x 1 \u003d - 11 and x 2 \u003d 23. The amount and product of these numbers will be equal: x 1 + x 2 \u003d 12 and x 1 · x 2 \u003d - 253. This means that the second coefficient is 12, a free member − 253.

Compile equation: x 2 - 12 · x - 253 \u003d 0.

Answer: x 2 - 12 · x - 253 \u003d 0.

We can use the Viet's theorem to solve tasks that are associated with the signs of the roots of square equations. The relationship between the theorem of the Vieta is associated with the signs of the roots of the given square equation x 2 + p · x + q \u003d 0 in the following way:

• if the square equation has a valid root and if a free dick q. is a positive number, then these roots will have the same sign "+" or "-";
• if the square equation has a root and if a free dick Q. It is a negative number, then one root will be "+", and the second "-".

Both of these statements are a consequence of formula x 1 · x 2 \u003d q and the rules of multiplication of positive and negative numbers, as well as numbers with different signs.

Example 5.

Are the roots of the square equation x 2 - 64 · x - 21 \u003d 0positive?

Decision

On the Vieta Theorem, the roots of this equation cannot be both positive, since equality should be performed for them x 1 · x 2 \u003d - 21. It is impossible with positive x 1 and x 2.

Answer: Not

Example 6.

Under what values \u200b\u200bof the parameter R. quadratic equation x 2 + (R + 2) · X + R - 1 \u003d 0there will be two valid roots with different signs.

Decision

Let's start with the fact that we find the meanings of what R., in which there will be two roots in the equation. We find discriminant and see what R. It will take positive values. D \u003d (R + 2) 2 - 4 · 1 · (R - 1) \u003d R 2 + 4 · R + 4 - 4 · R + 4 \u003d R 2 + 8. The value of the expression R 2 + 8 positive with any valid R.therefore discriminant will be larger with any valid R.. This means that the original square equation will have two roots at any actual values \u200b\u200bof the parameter R..

Now let's see when the roots will have different signs. This is possible if their work is negative. According to the Vieta Theorem, the product of the roots of the given square equation is equal to a free member. It means that the right solution will be the meanings R., in which a free member R - 1 is negative. We decide linear inequality R - 1< 0 , получаем r < 1 .

Answer: at R.< 1 .

Vieta formulas

There are a number of formulas that are applicable to the implementation of actions with roots and coefficients not only square, but also cubic and other types of equations. They are called Vieta Formulas.

For an algebraic equation N. species a 0 · x n + a 1 · x n - 1 +. . . + a n - 1 · x + a n \u003d 0 It is believed that the equation has N.valid roots x 1, x 2, ..., x n , among which may be coinciding:
x 1 + x 2 + x 3 +. . . + x n \u003d - a 1 a 0, x 1 · x 2 + x 1 · x 3 +. . . + x n - 1 · x n \u003d a 2 a 0, x 1 · x 2 · x 3 + x 1 · x 2 · x 4 +. . . + x n - 2 · x n - 1 · x n \u003d - a 3 a 0 ,. . . x 1 · x 2 · x 3 ·. . . · X n \u003d (- 1) N · a n a 0

Definition 1.

Get the formulas of the Vieta helps us:

• the theorem on the decomposition of the polynomial on linear multipliers;
• determination of equal polynomials through the equality of all their respective coefficients.

So, a polynomial A 0 · X n + A 1 · X n - 1 +. . . + a n - 1 · x + a n and its decomposition on linear factor of the form A 0 · (x - x 1) · (x - x 2) ·. . . · (X - x n) are equal.

If we reveal brackets in the last work and equate the appropriate coefficients, we get the formula of the Vieta. Taking N \u003d 2, we can obtain the formula of the wine for the square equation: x 1 + x 2 \u003d - a 1 a 0, x 1 · x 2 \u003d a 2 a 0.

Definition 2.

Vieta formula for cubic equation:
x 1 + x 2 + x 3 \u003d - a 1 a 0, x 1 · x 2 + x 1 · x 3 + x 2 · x 3 \u003d a 2 a 0, x 1 · x 2 · x 3 \u003d - a 3 A 0.

The left part of the recording of the formula of the Vieta contains the so-called elementary symmetric polynomials.

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