The first statement says that the sum of the roots of this equation is equal to the value of the coefficient at the variable x (in this case, it is b), but with the opposite sign. It looks like this: x1 + x2 = −b.
The second statement is no longer connected with the sum, but with the product of the same two roots. This product is equated to the free coefficient, i.e. c. Or, x1 * x2 = c. Both of these examples are solved in the system.
Vieta's theorem greatly simplifies the solution, but has one limitation. A quadratic equation, the roots of which can be found using this technique, must be reduced. In the above equation of the coefficient a, the one in front of x2 is equal to one. Any equation can be reduced to a similar form by dividing the expression by the first coefficient, but this operation is not always rational.
Proof of the theorem
To begin with, you should remember how traditionally it is customary to look for roots. quadratic equation... The first and second roots are found, namely: x1 = (-b-√D) / 2, x2 = (-b + √D) / 2. Generally divisible by 2a, but, as already mentioned, the theorem can be applied only when a = 1.It is known from Vieta's theorem that the sum of the roots is equal to the second coefficient with a minus sign. This means that x1 + x2 = (-b-√D) / 2 + (-b + √D) / 2 = −2b / 2 = −b.
The same is true for the product of unknown roots: x1 * x2 = (-b-√D) / 2 * (-b + √D) / 2 = (b2-D) / 4. In turn, D = b2-4c (again with a = 1). It turns out that the result is as follows: x1 * x2 = (b2- b2) / 4 + c = c.
From the above simple proof, only one conclusion can be drawn: Vieta's theorem is fully confirmed.
Second formulation and proof
Vieta's theorem has another interpretation. More precisely, it is not an interpretation, but a wording. The fact is that if the same conditions are met as in the first case: there are two different real roots, then the theorem can be written in a different formula.This equality looks like this: x2 + bx + c = (x - x1) (x - x2). If the function P (x) intersects at two points x1 and x2, then it can be written as P (x) = (x - x1) (x - x2) * R (x). In the case when P has the second degree, and this is exactly what the original expression looks like, then R is a prime number, namely 1. This statement is true for the reason that otherwise the equality will not hold. The x2 factor when expanding parentheses must not exceed one, and the expression must remain square.
Three numbers 12x, x 2-5 and 4 in this order form an increasing arithmetic progression https://youtu.be/U0VO_N9udpI Choose the correct statement MATHEMATICS ZFTSh MIPT Moscow Institute of Physics and Technology ( State University) Correspondence physical and technical school. http://pin.it/9w-GqGp Find all the x, y, and z such that the numbers 5x + 3, y2 and 3z + 5 will form an arithmetic progression in that order. Find x and indicate the difference of this progression. Solve the system of equations Unified State Exam Mathematics... Video lessons. Divisibility of integers. Linear function. Divisibility problems. Vieta's theorem, converse theorem, Vieta's formulas. clever #students #equations #vietas_theorem #theorem Next, consider a theorem converse to Vieta's theorem. After that, we will analyze the solutions of the most typical examples. This proves the first relation of Vieta's theorem for the sum of the roots of a quadratic equation. Let's move on to the second. How to prove the converse theorem to Vieta's theorem? DOK-VO: x2 + px + f = 0 x2- (M + H) * x + M * H = 0 x2-Mx-Hx + M * H = 0 x (x-H) -M (x-H) = 0 (x-M) (x-H) = 0 x-M = 0 x-H = 0 x = M x = H CHTD. This is how we proved in profile class with a mathematical bias. Answers: help me understand the converse theorem to Vieta's theorem thanks for specific examples The converse theorem to Vieta's theorem helps to fulfill the solution: If the coefficient a is a number from which it is easy to extract Square root whole rational number, then the sum of x1 and x2 will be equal to the number Prove the converse theorem of Vieta's theorem - see how to complain about the proof of Vieta's theorem. Formulate and prove Vieta's theorem, as well as the converse theorem, apply theorems to solve equations and problems. Prove the converse theorem to Vieta's theorem. USE in mathematics for 100 points: secrets that are not talked about school teachers, derivative problems. Many applicants think that it is not necessary to prepare for the first fourteen problems, considering that they are very easy, but this is not so! Most of the examiners make the simplest arithmetic mistakes, thereby overshadowing great solution tasks of part C. Such situations are very common, therefore, you should not neglect the preparation for the first tasks, but prepare as in a sports training: if you are applying for 90-100 points, train to solve the first block in 20-25 minutes, if you are 70 -80 points - about 30 minutes, no more. An excellent way of training is a decision in the company of a tutor, in courses where certain conditions will be set: for example, you decide before the first mistake, after you hand over the work; another option - for every mistake you donate money to the general cashier. No matter how strange it may seem, we do not recommend the official website, since all the tests there are so mixed up that it is impossible to use it. The design of the tasks of Part C is important. If the decision is made inaccurately, then the course of solving the task will be incomprehensible, and therefore, the examiner will definitely pick on this and lower your score. It would seem that we talked about very simple things, but by adhering to our advice, you will ensure yourself a successful passing the exam! The secret links, which are described in the Master class, can be found here - these are links to Video courses for preparing for the exam. This result is called Vieta's theorem. For the given square trinomial 2 x px q Vieta's theorem looks like this: if there are roots, then the opposite theorem to Vieta's theorem also holds: if the numbers satisfy the conditions, then these numbers are the roots of the equation. The proof of this theorem is one of control questions Tasks. Sometimes, for brevity, both Vieta's theorems (direct and inverse) are simply called Vieta's theorem.
I. Vieta's theorem for the reduced quadratic equation.
The sum of the roots of the reduced quadratic equation x 2 + px + q = 0 is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term:
x 1 + x 2 = -p; x 1 ∙ x 2 = q.
Find the roots of the reduced quadratic equation using Vieta's theorem.
Example 1) x 2 -x-30 = 0. This is the reduced quadratic equation ( x 2 + px + q = 0), the second coefficient p = -1, a free member q = -30. First, make sure that the given equation has roots, and that the roots (if any) will be expressed in integers. For this, it is sufficient that the discriminant is the perfect square of an integer.
Find the discriminant D= b 2 - 4ac = (- 1) 2 -4 ∙ 1 ∙ (-30) = 1 + 120 = 121 = 11 2 .
Now, according to Vieta's theorem, the sum of the roots should be equal to the second coefficient taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:
x 1 + x 2 = 1; x 1 ∙ x 2 = -30. We need to choose two numbers so that their product is equal -30 , and the sum is unit... These are numbers -5 and 6 . Answer: -5; 6.
Example 2) x 2 + 6x + 8 = 0. We have the reduced quadratic equation with the second coefficient p = 6 and a free member q = 8... Let's make sure there are integer roots. Find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 ... The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us choose roots according to Vieta's theorem: the sum of the roots is equal to –P = -6, and the product of the roots is q = 8... These are numbers -4 and -2 .
In fact: -4-2 = -6 = -p; -4 ∙ (-2) = 8 = q. Answer: -4; -2.
Example 3) x 2 + 2x-4 = 0... In this reduced quadratic equation, the second coefficient p = 2 and the free term q = -4... Find the discriminant D 1 since the second coefficient is an even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, therefore, we do output: the roots of this equation are not integers and cannot be found by Vieta's theorem. This means that we will solve this equation, as usual, using the formulas (in this case, using the formulas). We get:
Example 4). Make a quadratic equation for its roots if x 1 = -7, x 2 = 4.
Solution. The required equation will be written in the form: x 2 + px + q = 0, and, on the basis of Vieta's theorem –P = x 1 + x 2=-7+4=-3 → p = 3; q = x 1 ∙ x 2=-7∙4=-28 ... Then the equation will take the form: x 2 + 3x-28 = 0.
Example 5). Make a quadratic equation for its roots if:
II. Vieta's theorem for the complete quadratic equation ax 2 + bx + c = 0.
The sum of the roots is minus b divided by a, the product of the roots is with divided by a:
x 1 + x 2 = -b / a; x 1 ∙ x 2 = c / a.
Example 6). Find the sum of the roots of a quadratic equation 2x 2 -7x-11 = 0.
One of the methods for solving a quadratic equation is to use VIETA formulas, which was named after FRANCOIS VIET.
He was a renowned lawyer and served under the French king in the 16th century. V free time studied astronomy and mathematics. He established a connection between the roots and the coefficients of a quadratic equation.
Advantages of the formula:
1 ... By applying a formula, you can quickly find a 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 in the formula to find the roots.
2 ... Without a solution, you can determine the signs of the roots, pick up the meanings of the roots.
3 ... Having solved the system of two records, it is easy to find the roots themselves. In the given 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 given quadratic equation is equal to the value of the third coefficient.
4 ... Using these roots, write down a quadratic equation, that is, solve the inverse problem. For example, this method is used to solve problems in theoretical mechanics.
5 ... It is convenient to apply the formula when the leading coefficient is equal to one.
Disadvantages:
1
... The formula is not universal.
Vieta's theorem grade 8
Formula
If x 1 and x 2 are the roots of the reduced quadratic equation x 2 + px + q = 0, then:
Examples of
x 1 = -1; x 2 = 3 - roots of the equation x 2 - 2x - 3 = 0.
P = -2, q = -3.
X 1 + x 2 = -1 + 3 = 2 = -p,
X 1 x 2 = -1 3 = -3 = q.
The converse theorem
Formula
If the numbers x 1, x 2, p, q are related by the conditions:
Then x 1 and x 2 are the roots of the equation x 2 + px + q = 0.
Example
Let's compose a quadratic equation for its roots:
X 1 = 2 -? 3 and x 2 = 2 +? 3.
P = x 1 + x 2 = 4; p = -4; q = x 1 x 2 = (2 -? 3) (2 +? 3) = 4 - 3 = 1.
The required equation is: x 2 - 4x + 1 = 0.
Between the roots and the coefficients of the quadratic equation, in addition to the root formulas, there are other useful relations that are set Vieta's theorem... In this article, we will give the formulation and proof of Vieta's theorem for a quadratic equation. Next, consider a theorem converse to Vieta's theorem. After that, we will analyze the solutions of the most typical examples. Finally, we write down Vieta's formulas defining the connection between the real roots algebraic equation degree n and its coefficients.
Page navigation.
Vieta's theorem, formulation, proof
The formulas for the roots of the quadratic equation a x 2 + b x + c = 0 of the form, where D = b 2 −4 a c, imply the relations x 1 + x 2 = −b / a, x 1 x 2 = c / a. These results are approved Vieta's theorem:
Theorem.
If x 1 and x 2 are the roots of the quadratic equation a x 2 + b x + c = 0, then the sum of the roots is equal to the ratio of the coefficients b and a, taken with the opposite sign, and the product of the roots is equal to the ratio of the coefficients c and a, that is, ...
Proof.
We will prove Vieta's theorem according to the following scheme: compose the sum and the product of the roots of the quadratic equation using the well-known root formulas, then transform the obtained expressions and make sure that they are equal to −b / a and c / a, respectively.
Let's start with the sum of the roots, compose it. Now we bring the fractions to a common denominator, we have. In the numerator of the resulting fraction, after which:. Finally, after 2, we get. This proves the first relation of Vieta's theorem for the sum of the roots of a quadratic equation. Let's move on to the second.
We compose the product of the roots of the quadratic equation:. According to the rule for multiplying fractions, last piece can be written as. Now we multiply the parenthesis by the parenthesis in the numerator, but it's faster to collapse this product by the difference of squares formula, So . Then, remembering, we perform the next transition. And since the discriminant of the quadratic equation corresponds to the formula D = b 2 −4 · a · c, then in the last fraction instead of D one can substitute b 2 −4 · a · c, we obtain. After opening the brackets and reducing similar terms, we come to a fraction, and its reduction by 4 · a gives. This proves the second relation of Vieta's theorem for the product of roots.
If we omit the explanations, then the proof of Vieta's theorem takes on a laconic form:
,
.
It remains only to note that when the discriminant is equal to zero, the quadratic equation has one root. However, if we assume that the equation in this case has two identical roots, then the equalities from Vieta's theorem also hold. Indeed, for D = 0 the root of the quadratic equation is equal, then and, and since D = 0, that is, b 2 −4 · a · c = 0, whence b 2 = 4 · a · c, then.
In practice, Vieta's theorem is most often used in relation to a reduced quadratic equation (with the leading coefficient a equal to 1) of the form x 2 + p x + q = 0. Sometimes it is formulated for quadratic equations of just this form, which does not limit the generality, since any quadratic equation can be replaced by an equivalent equation by dividing its both parts by a nonzero number a. Let us give the corresponding formulation of Vieta's theorem:
Theorem.
The sum of the roots of the reduced quadratic equation x 2 + p x + q = 0 is equal to the coefficient at x taken with the opposite sign, and the product of the roots is the free term, that is, x 1 + x 2 = −p, x 1 x 2 = q.
The converse of Vieta's theorem
The second formulation of Vieta's theorem, given in the previous paragraph, indicates that if x 1 and x 2 are the roots of the reduced quadratic equation x 2 + p x + q = 0, then the relations x 1 + x 2 = −p, x 1 x 2 = q. On the other hand, from the written relations x 1 + x 2 = −p, x 1 x 2 = q it follows that x 1 and x 2 are the roots of the quadratic equation x 2 + p x + q = 0. In other words, the opposite of Vieta's theorem is true. Let us formulate it in the form of a theorem and prove it.
Theorem.
If the numbers x 1 and x 2 are such that x 1 + x 2 = −p and x 1 x 2 = q, then x 1 and x 2 are the roots of the reduced quadratic equation x 2 + p x + q = 0.
Proof.
After replacing the coefficients p and q in the equation x 2 + p x + q = 0, their expressions in terms of x 1 and x 2, it is transformed into equivalent equation.
Substituting the number x 1 in the resulting equation instead of x, we have the equality x 1 2 - (x 1 + x 2) x 1 + x 1 x 2 = 0, which for any x 1 and x 2 is a true numerical equality 0 = 0, since x 1 2 - (x 1 + x 2) x 1 + x 1 x 2 = x 1 2 −x 1 2 −x 2 x 1 + x 1 x 2 = 0... Therefore, x 1 is a root of the equation x 2 - (x 1 + x 2) x + x 1 x 2 = 0, which means that x 1 is a root of the equivalent equation x 2 + p x + q = 0.
If the equation x 2 - (x 1 + x 2) x + x 1 x 2 = 0 substitute for x the number x 2, then we get the equality x 2 2 - (x 1 + x 2) x 2 + x 1 x 2 = 0... This is a valid equality, since x 2 2 - (x 1 + x 2) x 2 + x 1 x 2 = x 2 2 −x 1 x 2 −x 2 2 + x 1 x 2 = 0... Therefore, x 2 is also a root of the equation x 2 - (x 1 + x 2) x + x 1 x 2 = 0, and hence the equations x 2 + p x + q = 0.
This completes the proof of the theorem. converse theorem Vieta.
Examples of using Vieta's theorem
It's time to talk about the practical application of Vieta's theorem and its converse theorem. In this paragraph, we will analyze the solutions of several of the most typical examples.
We begin by applying a theorem converse to Vieta's theorem. It is convenient to use it to check if the given two numbers are the roots of a given quadratic equation. In this case, their sum and difference are calculated, after which the validity of the ratios is checked. If both of these relations are satisfied, then by virtue of a theorem inverse to Vieta's theorem, it is concluded that these numbers are the roots of the equation. If at least one of the relations is not satisfied, then these numbers are not the roots of the quadratic equation. This approach can be used when solving quadratic equations to check the found roots.
Example.
Which of the pairs of numbers 1) x 1 = −5, x 2 = 3, or 2), or 3) is a pair of roots of the quadratic equation 4 x 2 −16 x + 9 = 0?
Solution.
The coefficients of the given quadratic equation 4 x 2 −16 x + 9 = 0 are a = 4, b = −16, c = 9. According to Vieta's theorem, the sum of the roots of a quadratic equation should be equal to −b / a, that is, 16/4 = 4, and the product of the roots should be equal to c / a, that is, 9/4.
Now let's calculate the sum and product of the numbers in each of the three given pairs, and compare them with the values just obtained.
In the first case, we have x 1 + x 2 = −5 + 3 = −2. The resulting value is different from 4, so further verification can not be carried out, and according to the theorem inverse to Vieta's theorem, one can immediately conclude that the first pair of numbers is not a pair of roots of a given quadratic equation.
Let's move on to the second case. Here, that is, the first condition is fulfilled. We check the second condition:, the resulting value is different from 9/4. Consequently, the second pair of numbers is not a pair of roots of a quadratic equation.
The last case remains. Here and . Both conditions are met, so these numbers x 1 and x 2 are the roots of the given quadratic equation.
Answer:
The inverse theorem to Vieta's theorem can be used in practice to find the roots of a quadratic equation. Usually, whole roots of the reduced quadratic equations with integer coefficients are selected, since in other cases it is quite difficult to do this. In this case, they use the fact that if the sum of two numbers is equal to the second coefficient of the quadratic equation, taken with a minus sign, and the product of these numbers is equal to the free term, then these numbers are the roots of this quadratic equation. Let's look at this with an example.
Take the quadratic equation x 2 −5 x + 6 = 0. For the numbers x 1 and x 2 to be the roots of this equation, the two equalities x 1 + x 2 = 5 and x 1 x 2 = 6 must hold. It remains to find such numbers. In this case, it is quite simple to do this: such numbers are 2 and 3, since 2 + 3 = 5 and 2 · 3 = 6. Thus, 2 and 3 are the roots of this quadratic equation.
The converse theorem to Vieta's theorem is especially convenient to use to find the second root of a reduced quadratic equation when one of the roots is already known or obvious. In this case, the second root is found from any of the relations.
For example, let's take the quadratic equation 512 x 2 −509 x − 3 = 0. It is easy to see here that one is the root of the equation, since the sum of the coefficients of this quadratic equation is zero. So x 1 = 1. The second root x 2 can be found, for example, from the relation x 1 x 2 = c / a. We have 1 x 2 = −3 / 512, whence x 2 = −3 / 512. This is how we determined both roots of the quadratic equation: 1 and −3/512.
It is clear that the selection of roots is advisable only in the simplest cases. In other cases, to find the roots, you can apply the formulas for the roots of the quadratic equation through the discriminant.
One more practical use theorem, converse to Vieta's theorem, consists in drawing up quadratic equations for given roots x 1 and x 2. To do this, it is enough to calculate the sum of the roots, which gives the coefficient at x with the opposite sign of the reduced quadratic equation, and the product of the roots, which gives the free term.
Example.
Write a quadratic equation with the numbers −11 and 23 as roots.
Solution.
We set x 1 = −11 and x 2 = 23. Evaluate the sum and product of these numbers: x 1 + x 2 = 12 and x 1 x 2 = −253. Therefore, the indicated numbers are the roots of the reduced quadratic equation with the second coefficient −12 and an intercept of −253. That is, x 2 −12 x − 253 = 0 is the desired equation.
Answer:
x 2 −12 x − 253 = 0.
Vieta's theorem is very often used to solve problems related to the signs of the roots of quadratic equations. How is Vieta's theorem related to the signs of the roots of the reduced quadratic equation x 2 + p x + q = 0? Here are two relevant statements:
- If the intercept q is a positive number and if the quadratic equation has real roots, then either they are both positive or both are negative.
- If the free term q is a negative number and if the quadratic equation has real roots, then their signs are different, in other words, one root is positive and the other negative.
These statements follow from the formula x 1 x 2 = q, as well as the rules of multiplication of positive, negative numbers and numbers with different signs. Let's consider examples of their application.
Example.
R it is positive. Using the discriminant formula, we find D = (r + 2) 2 −4 1 (r − 1) = r 2 + 4 r + 4−4 r + 4 = r 2 +8, the value of the expression r 2 +8 is positive for any real r, thus D> 0 for any real r. Therefore, the original quadratic equation has two roots for any real values of the parameter r.
Now let's find out when the roots have different signs. If the signs of the roots are different, then their product is negative, and according to Vieta's theorem, the product of the roots of the reduced quadratic equation is equal to the free term. Therefore, we are interested in those values of r for which the free term r − 1 is negative. Thus, to find the values of r we are interested in, we need decide linear inequality r − 1<0 , откуда находим r<1 .
Answer:
at r<1 .
Vieta formulas
Above we talked about Vieta's theorem for a quadratic equation and analyzed the relations it claims. But there are formulas connecting the real roots and coefficients of not only quadratic equations, but also cubic equations, quadruple equations, and in general, algebraic equations degree n. They are called Vieta formulas.
Let us write Vieta's formulas for an algebraic equation of degree n of the form, in this case we assume that it has n real roots x 1, x 2, ..., x n (among them there may be coinciding ones): 
Get Vieta's formulas allows linear factorization theorem, as well as the definition of equal polynomials through the equality of all their corresponding coefficients. So the polynomial and its factorization into linear factors of the form are equal. Expanding the parentheses in the last product and equating the corresponding coefficients, we obtain Vieta's formulas.
In particular, for n = 2, we have the Vieta formulas for the quadratic equation that are already familiar to us.
For the cubic equation, Vieta's formulas are 
It remains only to note that on the left side of Vieta's formulas are the so-called elementary symmetric polynomials.
Bibliography.
- Algebra: study. for 8 cl. general education. institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M.: Education, 2008 .-- 271 p. : ill. - ISBN 978-5-09-019243-9.
- A. G. Mordkovich Algebra. 8th grade. At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., Erased. - M .: Mnemozina, 2009 .-- 215 p .: ill. ISBN 978-5-346-01155-2.
- Algebra and the beginning of mathematical analysis. Grade 10: textbook. for general education. institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; ed. A. B. Zhizhchenko. - 3rd ed. - M .: Education, 2010.- 368 p. : ill. - ISBN 978-5-09-022771-1.
