Municipal budgetary educational institution
"Secondary school No. 64", Bryansk
City scientific and practical conference
"First Steps in Science"
Research work
"Vieta's theorem for equations of the third and fourth degree"
Mathematics
Completed by: 11b grade student
Shanov Ilya Alekseevich
Scientific adviser:
mathematic teacher,
Candidate of Physics and Mathematics Sciences
Bykov Sergey Valentinovich
Bryansk 2012
Introduction ………………………………………………………………… 3
Goals and objectives …………………………………………………………… 4
Brief historical background ………………………………………… 4
Quadratic equation …………………………………………………. 5
Cubic equation ……………………………………………………. 6
Equation of the fourth degree ………………………………………… 7
Practical part ……………………………………………………. 9
References ……………………………………………………… 12
Appendix …………………………………………………………… 13
Introduction
The Fundamental Theorem of Algebra states that a field is algebraically closed, in other words, that an nth degree equation with complex coefficients (generally) over a field has exactly n complex roots. Equations of the third degree are solved by Cordano's formula. Equations of the fourth degree by the Ferrari method. In addition to the fact that in the theory of algebra it is proved that if 
is the root of the equation, then 
is also the root of this equation. For a cubic equation, the following cases are possible:
all three roots are real;
two complex roots, one real.
This implies that any cubic equation has at least one real root.
For a fourth degree equation:
All four roots are different.
Two roots are real, two are complex.
All four roots are complex.
This work is devoted to a thorough study of the Vieta theorem: its formulation, proof, as well as solving problems using this theorem.
The work done is aimed at helping a 11th grade student who is about to pass the exam, as well as for young mathematicians who are not indifferent to simpler and more effective methods of solving in various areas of mathematics.
In the appendix to this work, a collection of problems is provided for independent solution and consolidation of the new material that I am studying.
This question cannot be ignored, since it is important for mathematics, both for science in general, and for students and those interested in solving such problems.
Goals and objectives of the work:
Obtain an analog of Vieta's theorem for an equation of the third degree.
Prove an analogue of Vieta's theorem for a third-degree equation.
Obtain an analog of Vieta's theorem for a fourth-degree equation.
Prove an analogue of Vieta's theorem for a fourth degree equation.
Consider the application of these questions to the solution of practical problems.
Verify the practicality of the application of this theorem.
Deepen mathematical knowledge in the field of solving equations.
Develop an interest in mathematics.
Brief historical background
Francois Viet (1540-1603) - French mathematician. By profession a lawyer. In 1591, he introduced letter designations not only for unknown quantities, but also for the coefficients of equations; thanks to this, it became possible for the first time to express the properties of equations and their roots by general formulas. He owns the establishment of a uniform method for solving equations of the 2nd, 3rd and 4th degrees. Among the discoveries, Viet himself especially appreciated the establishment of a relationship between the roots and coefficients of equations. For the approximate solution of equations with numerical coefficients, Viet proposed a method similar to Newton's later method. In trigonometry, Francois Viet gave a complete solution to the problem of determining all elements of a flat or spherical triangle from three data, found important expansions of cos nx and sin nx in powers of cos X and sin X. He considered infinite works for the first time. Vieta's writings are written in a difficult language and therefore received at one time less distribution than they deserved. .
Quadratic equation
To begin with, let's recall the Vieta formulas for the equation of the second degree, which we learned in the school curriculum.
T 
Vieta theoremfor quadratic equation (grade 8)
E 
if and are the roots of the quadratic equation then
i.e., the sum of the roots of the given quadratic equation is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term.
Also, remember the theorem converse to Vieta's theorem:
If numbers - p and q are such that
then and are the roots of the equation
Vieta's theorem is remarkable in that, without knowing the roots of a square trinomial, we can easily calculate their sum and product, that is, the simplest symmetric expressions.
Vieta's theorem allows you to guess the integer roots of a square trinomial.
cubic equation
Now let's proceed directly to the formulation and solution of the cubic equation using the Vieta theorem.
Wording
TO 
a ubic equation is a third-order equation, of the form
where a ≠ 0.
If a = 1, then the equation is called the reduced cubic equation:
So, we need to prove that for the equation
the following theorem holds true:
P 
let the roots of this equation, then
Proof
Imagine a polynomial
Let's do the transformations:
So we get that
Twopolynomials are equal if and only if their coefficients are equal at the corresponding powers.
It means that
Q.E.D.
Now consider the theorem, converse to the Vieta theorem for a third-degree equation.
F 
wording
E 
if the numbers are such that
Equation of the fourth degree
Now let's move on to setting and solving a fourth-degree equation using Vieta's theorem for a fourth-degree equation.
Wording
At 
equation of the fourth degree - an equation of the form
G 
de a ≠ 0.
E 
if a = 1, then the equation is called reduced
AND 
so, let us prove that for the equation
With 
the following theorem is true: let the roots of the given equation, then
Proof
Imagine a polynomial
Let's do the transformations:
So we get that
We know that two polynomials are equal if and only if their coefficients are equal at the corresponding powers.
It means that
Q.E.D.
Consider the theorem converse to the Vieta theorem for a fourth-degree equation.
Wording
If the numbers are such that
then these numbers are the roots of the equation
Practical part
Now consider solving problems using Vieta's theorems for equations of the third and fourth degree.
Task #1
Answer: 4, -4.
Task #2
Answer: 16, 24.
To solve these equations, you can use the Cardano formulas and the Ferrari method, respectively, but using the Vieta theorem, we know the sum and product of the roots of these equations.
Task #3
Write an equation of the third degree, if it is known that the sum of the roots is 6, the pairwise product of the roots is 3, and the product is -4.
Let's make an equation, we get
Task #4
Write an equation of the third degree, if it is known that the sum of the roots is equal to 8 , by the pair product of the roots is equal to 4 , tripled product is equal to 12 , and the product 20 .
Solution: using the Vieta formula, we get
Let's make an equation, we get
With the help of Vieta's theorem, we can easily compose equations by their roots. This is the most rational way to solve these problems.
Task #5
where a, b, c are Heron's formulas.
Let's open the brackets and transform the expression, we get
W 
Note that the radical expression is cubic expression. We use the Vieta theorem for the corresponding cubic equation, then we have that
W 
naya, what we get:
It can be seen from the solution of this problem that Vieta's theorem is applicable to problems from different areas of mathematics.
Conclusion
In this paper, a method for solving equations of the third and fourth degrees using the Vieta theorem was investigated. The formulas derived in the work 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 equations of the third and fourth powers, respectively.
Vieta's theorem has been applied in practice. A number of tasks were solved that helped to better consolidate the new material.
This study was very interesting and informative for me. Deepening my knowledge in mathematics, I discovered a lot of interesting things and was happy to do this research.
But my research in the field of solving equations is not over. In the future, I plan to study the solution of the n-th degree equation using Vieta's theorem.
I want to express my deep gratitude to my supervisor, candidate of physical and mathematical sciences, and the possibility of such an unusual study and constant attention to work.
Bibliography
Vinogradov I.M. Mathematical encyclopedia. M., 1977.
V. B. Lidsky, L. V. Ovsyannikov, A. N. Tulaikov, M. I. Shabunin. Problems in elementary mathematics, Fizmatlit, 1980.
... Scientific – researchWork on mathematics Geometry... theorem Poncelet for triangle... r2 - degree or... arc third moons smaller... the equation, giving fourth ... mathematician F. viet I calculated in 1579 with 9 signs. Dutch mathematician ...
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... for many later textbooks on algeora. In it, the presentation is brought to the theory equationthird and fourthdegree... theoretical and applied mathematics. Emphasis was placed on both teaching and researchwork. The best scientists of France...
One of the methods for solving a quadratic equation is the application VIETA formulas, which was named after FRANCOIS VIETE.
He was a famous lawyer, and served in the 16th century with the French king. In his free time he studied astronomy and mathematics. He established a connection between the roots and coefficients of a quadratic equation.
Advantages of the formula:
1 . By applying the formula, you can quickly find the solution. Because you do not need to enter the second coefficient into the square, then subtract 4ac from it, find the discriminant, substitute its value into the formula for finding the roots.
2 . Without a solution, you can determine the signs of the roots, pick up the values of the roots.
3 . Having solved the system of two records, it is not difficult to find the roots themselves. In the above quadratic equation, the sum of the roots is equal to the value of the second coefficient with a minus sign. The product of the roots in the above quadratic equation is equal to the value of the third coefficient.
4 . According to the given roots, write a quadratic equation, that is, solve the inverse problem. For example, this method is used in solving problems in theoretical mechanics.
5 . It is convenient to apply the formula when the leading coefficient is equal to one.
Flaws:
1
. The formula is not universal.
Formula
If x 1 and x 2 are the roots of the given quadratic equation x 2 + px + q \u003d 0, then:
Examples
x 1 \u003d -1; x 2 \u003d 3 - the roots of the equation x 2 - 2x - 3 \u003d 0.
P = -2, q = -3.
X 1 + x 2 \u003d -1 + 3 \u003d 2 \u003d -p,
X 1 x 2 = -1 3 = -3 = q.
Formula
If the numbers x 1 , x 2 , p, q are connected by the conditions:
Then x 1 and x 2 are the roots of the equation x 2 + px + q = 0.
Example
Let's make a quadratic equation by its roots:
X 1 \u003d 2 -? 3 and x 2 \u003d 2 +? 3 .
P \u003d x 1 + x 2 \u003d 4; p = -4; q \u003d x 1 x 2 \u003d (2 -? 3) (2 +? 3) \u003d 4 - 3 \u003d 1.
The desired equation has the form: x 2 - 4x + 1 = 0.
2.5 Vieta formula for polynomials (equations) of higher degrees
The formulas derived by Vieta for quadratic equations are also true for polynomials of higher degrees.
Let the polynomial
P(x) = a 0 x n + a 1 x n -1 + … +a n
Has n distinct roots x 1 , x 2 …, x n .
In this case, it has a factorization of the form:
a 0 x n + a 1 x n-1 +…+ a n = a 0 (x – x 1)(x – x 2)…(x – x n)
Let's divide both parts of this equality by a 0 ≠ 0 and expand the brackets in the first part. We get the equality:
xn + ()xn -1 + ... + () = xn - (x 1 + x 2 + ... + xn) xn -1 + (x 1 x 2 + x 2 x 3 + ... + xn -1 xn)xn - 2 + … +(-1) nx 1 x 2 … xn
But two polynomials are identically equal if and only if the coefficients at the same powers are equal. It follows from this that the equality
x 1 + x 2 + … + x n = -
x 1 x 2 + x 2 x 3 + … + x n -1 x n =
x 1 x 2 … x n = (-1) n
For example, for polynomials of the third degree
a 0 x³ + a 1 x² + a 2 x + a 3
We have identities
x 1 + x 2 + x 3 = -
x 1 x 2 + x 1 x 3 + x 2 x 3 =
x 1 x 2 x 3 = -
As for quadratic equations, this formula is called the Vieta formulas. The left parts of these formulas are symmetric polynomials from the roots x 1 , x 2 ..., x n of the given equation, and the right parts are expressed in terms of the coefficient of the polynomial.
2.6 Equations reducible to squares (biquadratic)
Equations of the fourth degree are reduced to quadratic equations:
ax 4 + bx 2 + c = 0,
called biquadratic, moreover, a ≠ 0.
It is enough to put x 2 \u003d y in this equation, therefore,
ay² + by + c = 0
find the roots of the resulting quadratic equation
y 1,2 = 
To immediately find the roots x 1, x 2, x 3, x 4, replace y with x and get
x2 =
x 1,2,3,4 = 
.
If the equation of the fourth degree has x 1, then it also has a root x 2 \u003d -x 1,
If has x 3, then x 4 \u003d - x 3. The sum of the roots of such an equation is zero.
2x 4 - 9x² + 4 = 0
We substitute the equation into the formula for the roots of biquadratic equations:
x 1,2,3,4 = 
,
knowing that x 1 \u003d -x 2, and x 3 \u003d -x 4, then:
x 3.4 = 
Answer: x 1.2 \u003d ± 2; x 1.2 =
2.7 Study of biquadratic equations
Let's take the biquadratic equation
ax 4 + bx 2 + c = 0,
where a, b, c are real numbers, and a > 0. By introducing an auxiliary unknown y = x², we examine the roots of this equation, and enter the results in a table (see Appendix No. 1)
2.8 Cardano formula
If we use modern symbolism, then the derivation of the Cardano formula can look like this:
x = 
This formula determines the roots of the general equation of the third degree:
ax 3 + 3bx 2 + 3cx + d = 0.
This formula is very cumbersome and complex (it contains several complex radicals). It does not always apply, because. very difficult to complete.
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