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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.

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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, the last product 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 of x taken with the opposite sign, and the product of the roots is the intercept, 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 an 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 to Vieta's theorem.

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 for multiplying 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 solve 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.

Before proceeding to Vieta's theorem, we introduce a definition. Quadratic equation of the form x² + px + q= 0 is called reduced. In this equation, the leading coefficient is one. For example, the equation x² - 3 x- 4 = 0 is reduced. Any quadratic equation of the form ax² + b x + c= 0 can be made reduced, for this we divide both sides of the equation by a≠ 0. For example, equation 4 x² + 4 x- 3 = 0 by dividing by 4 is reduced to the form: x² + x- 3/4 = 0. We derive the formula for the roots of the reduced quadratic equation, for this we use the formula for the roots of a quadratic equation of general form: ax² + bx + c = 0

Equation reduced x² + px + q= 0 coincides with the general equation in which a = 1, b = p, c = q. Therefore, for the given quadratic equation, the formula takes the form:

the last expression is called the formula for the roots of the reduced quadratic equation, it is especially convenient to use this formula when R- even number. For example, let's solve the equation x² - 14 x — 15 = 0

In response, we write down the equation has two roots.

For the reduced quadratic equation with positive, the following theorem is true.

Vieta's theorem

If x 1 and x 2 - roots of the equation x² + px + q= 0, then the following formulas are valid:

x 1 + x 2 = — R

x 1 * x 2 = q, that is, 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.

Based on the formula for the roots of the reduced quadratic equation, we have:

Adding these equalities, we get: x 1 + x 2 = —R.

Multiplying these equalities, using the formula for the difference of squares, we obtain:

Note that Vieta's theorem is also valid when the discriminant is zero, if we assume that in this case the quadratic equation has two identical roots: x 1 = x 2 = — R/2.

Without solving the equations x² - 13 x+ 30 = 0 find the sum and product of its roots x 1 and x 2. this equation D= 169 - 120 = 49> 0, so Vieta's theorem can be applied: x 1 + x 2 = 13, x 1 * x 2 = 30. Consider a few more examples. One of the roots of the equation x² — px- 12 = 0 equals x 1 = 4. Find coefficient R and the second root x 2 of this equation. By Vieta's theorem x 1 * x 2 =— 12, x 1 + x 2 = — R. Because x 1 = 4, then 4 x 2 = - 12, whence x 2 = — 3, R = — (x 1 + x 2) = - (4 - 3) = - 1. In response, write down the second root x 2 = - 3, coefficient p = - 1.

Without solving the equations x² + 2 x- 4 = 0 find the sum of the squares of its roots. Let x 1 and x 2 - the roots of the equation. By Vieta's theorem x 1 + x 2 = — 2, x 1 * x 2 = - 4. Because x 1 ² + x 2 ² = ( x 1 + x 2) ² - 2 x 1 x 2, then x 1 ² + x 2 ² = (- 2) ² -2 (- 4) = 12.

Find the sum and product of the roots of equation 3 x² + 4 x- 5 = 0. This equation has two different roots, since the discriminant D= 16 + 4 * 3 * 5> 0. To solve the equation, we use Vieta's theorem. This theorem is proved for the reduced quadratic equation. Therefore, we divide this equation by 3.

Therefore, the sum of the roots is -4/3, and their product is -5/3.

In the general case, the roots of the equation ax² + b x + c= 0 are related by the following equalities: x 1 + x 2 = — b / a, x 1 * x 2 = c / a, To obtain these formulas, it is enough to divide both sides of this quadratic equation by a ≠ 0 and apply Vieta's theorem to the resulting reduced quadratic equation. Consider an example, it is required to compose the reduced quadratic equation, the roots of which x 1 = 3, x 2 = 4. Because x 1 = 3, x 2 = 4 - the roots of the quadratic equation x² + px + q= 0, then by Vieta's theorem R = — (x 1 + x 2) = — 7, q = x 1 x 2 = 12. In response, write x² - 7 x+ 12 = 0. In solving some problems, the following theorem is applied.

The converse of Vieta's theorem

If the numbers R, q, x 1 , x 2 are such that x 1 + x 2 = — p, x 1 * x 2 = q, then x 1 and x 2- the roots of the equation x² + px + q= 0. Substitute in the left side x² + px + q instead of R expression - ( x 1 + x 2), and instead of q- work x 1 * x 2. We get: x² + px + q = x² — ( x 1 + x 2) x + x 1 x 2 = x² - x 1 x - x 2 x + x 1 x 2 = (x - x 1) (x - x 2). Thus, if the numbers R, q, x 1 and x 2 are related by these relations, then for all X equality holds x² + px + q = (x - x 1) (x - x 2), from which it follows that x 1 and x 2 - roots of the equation x² + px + q= 0. Using a theorem inverse to Vieta's theorem, it is sometimes possible to find the roots of a quadratic equation by selection. Consider an example, x² - 5 x+ 6 = 0. Here R = — 5, q= 6. Let's pick two numbers x 1 and x 2 so that x 1 + x 2 = 5, x 1 * x 2 = 6. Noticing that 6 = 2 * 3, and 2 + 3 = 5, by a theorem converse to Vieta's theorem, we obtain that x 1 = 2, x 2 = 3 - roots of the equation x² - 5 x + 6 = 0.

Vieta's theorem

Let and denote the roots of the reduced quadratic equation
(1) .
Then the sum of the roots is equal to the coefficient at, taken with the opposite sign. The product of the roots is equal to the free term:
;
.

A note on multiple roots

If the discriminant of equation (1) is equal to zero, then this equation has one root. But in order to avoid cumbersome formulations, it is generally accepted that in this case, equation (1) has two multiple, or equal, roots:
.

Proof one

Let us find the roots of equation (1). To do this, apply the formula for the roots of the quadratic equation:
;
;
.

We find the sum of the roots:
.

To find a work, apply the formula:
.
Then

.

The theorem is proved.

Proof of the second

If the numbers and are the roots of the quadratic equation (1), then
.
We expand the brackets.

.
Thus, equation (1) will take the form:
.
Comparing with (1) we find:
;
.

The theorem is proved.

Vieta's converse theorem

Let there be arbitrary numbers. Then and are the roots of the quadratic equation
,
where
(2) ;
(3) .

Proof of Vieta's converse theorem

Consider the quadratic equation
(1) .
We need to prove that if and, then u are the roots of equation (1).

Substitute (2) and (3) in (1):
.
We group the terms on the left side of the equation:
;
;
(4) .

Substitute in (4):
;
.

Substitute in (4):
;
.
The equation is fulfilled. That is, the number is the root of equation (1).

The theorem is proved.

Vieta's theorem for a complete quadratic equation

Now consider the complete quadratic equation
(5) ,
where, and there are some numbers. Moreover.

Let us divide equation (5) by:
.
That is, we got the reduced equation
,
where ; ...

Then Vieta's theorem for the complete quadratic equation has the following form.

Let and denote the roots of the complete quadratic equation
.
Then the sum and product of the roots are determined by the formulas:
;
.

Vieta's theorem for the cubic equation

In a similar way, we can establish connections between the roots of a cubic equation. Consider the cubic equation
(6) ,
where,,, are some numbers. Moreover.
Let's divide this equation into:
(7) ,
where , , .
Let,, be the roots of equation (7) (and equation (6)). Then

.

Comparing with equation (7) we find:
;
;
.

Vieta's theorem for an n-th degree equation

In the same way, you can find connections between the roots,, ...,, for an equation of the nth degree
.

Vieta's theorem for an equation of the nth degree has the following form:
;
;
;

.

To get these formulas, we write the equation in the following form:
.
Then we equate the coefficients at,,, ..., and compare the free term.

References:
I.N. Bronstein, K.A. Semendyaev, Handbook of Mathematics for Engineers and Students of Technical Institutions, "Lan", 2009.
CM. Nikolsky, M.K. Potapov et al., Algebra: a textbook for grade 8 educational institutions, Moscow, Education, 2006.

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. In his free time he 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.

Flaws:

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.