Vieta theorem formula, and examples of solutions. Vieta's theorem: examples of its use when working with quadratic equations

Formulation and proof of Vieta's theorem for quadratic equations. Vieta's converse theorem. Vieta's theorem for cubic equations and equations of arbitrary order.

Quadratic equations

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 two

If the numbers and are the roots of the quadratic equation (1), then
.
We open 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 equation of degree n

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

Vieta's theorem for equations 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.

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 reduced quadratic equation ( x 2 + px + q = 0), the second coefficient p = -1 and the free term 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 , therefore, 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 desired 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.

When studying methods of solving second-order equations in a school algebra course, the properties of the obtained roots are considered. They are now known as Vieta's theorem. Examples of its use are given in this article.

Quadratic equation

The second order equation is the equality, which is shown in the photo below.

Here the symbols a, b, c are some numbers that are called the coefficients of the equation under consideration. To solve an equality, you need to find the values ​​of x that make it true.

Note that since the maximum value of the degree to which x is raised is two, then the number of roots in general case is also equal to two.

There are several ways to solve this type of equality. In this article, we will consider one of them, which involves the use of the so-called Vieta theorem.

Formulation of Vieta's theorem

At the end of the 16th century, the famous mathematician François Viet (French) noticed, analyzing the properties of the roots of various quadratic equations, that certain combinations of them satisfy specific ratios. In particular, these combinations are their product and sum.

Vieta's theorem establishes the following: the roots of the quadratic equation, when they sum, give the ratio of the coefficients of the linear to the quadratic taken with the opposite sign, and when they are multiplied, they lead to the ratio of the free term to the quadratic coefficient.

If general form the equation is written as shown in the photo in previous section article, then mathematically this theorem can be written in the form of two equalities:

• r 2 + r 1 = -b / a;
• r 1 x r 2 = c / a.

Where r 1, r 2 is the value of the roots of the equation in question.

These two equalities can be used to solve a number of very different mathematical problems... The use of Vieta's theorem in examples with solutions is given in the following sections of the article.

The discriminant, like quadratic equations, begins to be studied in the course of algebra in the 8th grade. You can solve the quadratic equation through the discriminant and using Vieta's theorem. The method of studying quadratic equations, like the discriminant formulas, is rather unsuccessfully inculcated in schoolchildren, like much in real education. Therefore, school years pass, education in grades 9-11 replaces " higher education"and everyone is looking again - "How to solve a quadratic equation?", "How to find the roots of an equation?", "How to find the discriminant?" and...

Discriminant formula

The discriminant D of the quadratic equation a * x ^ 2 + bx + c = 0 is equal to D = b ^ 2–4 * a * c.
The roots (solutions) of the quadratic equation depend on the sign of the discriminant (D):
D> 0 - the equation has 2 different real roots;
D = 0 - the equation has 1 root (2 coinciding roots):
D<0 – не имеет действительных корней (в школьной теории). В ВУЗах изучают комплексные числа и уже на множестве комплексных чисел уравнение с отрицательным дискриминантом имеет два комплексных корня.
The formula for calculating the discriminant is quite simple, so many sites offer an online discriminant calculator. We have not figured out this kind of scripts yet, so who knows how to implement this, please write to the mail This email address is being protected from spambots. You need JavaScript enabled to view it. .

General formula for finding the roots of a quadratic equation:

We find the roots of the equation by the formula
If the coefficient of the variable squared is paired, then it is advisable to calculate not the discriminant, but its fourth part
In such cases, the roots of the equation are found by the formula

The second way to find roots is Vieta's Theorem.

A theorem is formulated not only for quadratic equations, but also for polynomials. You can read this on Wikipedia or other electronic resources. However, for simplicity, we will consider that part of it that concerns the reduced quadratic equations, that is, equations of the form (a = 1)
The essence of Vieta's formulas is that the sum of the roots of the equation is equal to the coefficient of the variable, taken with the opposite sign. The product of the roots of the equation is equal to the free term. Vieta's theorem is written in formulas.
The derivation of Vieta's formula is quite simple. Let's write the quadratic equation in terms of prime factors
As you can see, all ingenious is simple at the same time. It is effective to use the Vieta formula when the difference in the roots or the difference in the absolute values ​​of the roots is 1, 2. For example, the following equations by the Vieta theorem have roots




Up to 4 equations, the analysis should look like this. The product of the roots of the equation is 6, therefore the roots can be the values ​​(1, 6) and (2, 3) or pairs with the opposite sign. The sum of the roots is 7 (the coefficient of the variable with the opposite sign). Hence we conclude that the solutions of the quadratic equation are equal to x = 2; x = 3.
It is easier to select the roots of the equation among the divisors of the free term, correcting their sign in order to fulfill the Vieta formulas. At first it seems difficult to do, but with practice on a number of quadratic equations, this technique will be more effective than calculating the discriminant and finding the roots of the quadratic equation in the classical way.
As you can see, the school theory of studying the discriminant and ways of finding solutions to the equation is devoid of practical meaning - "Why do schoolchildren need a quadratic equation?", "What is the physical meaning of the discriminant?"

Let's try to figure it out what does the discriminant describe?

The algebra course teaches functions, function study charts, and function graphing. Of all the functions, an important place is occupied by a parabola, the equation of which can be written in the form
So the physical meaning of the quadratic equation is the zeros of the parabola, that is, the points of intersection of the graph of the function with the abscissa axis Ox
I ask you to remember the properties of parabolas that are described below. The time will come to pass exams, tests, or entrance exams and you will be grateful for the reference material. The sign at the variable in the square corresponds to whether the branches of the parabola on the graph go up (a> 0),

or a parabola with branches down (a<0) .

The vertex of the parabola lies in the middle between the roots

The physical meaning of the discriminant:

If the discriminant is greater than zero (D> 0), the parabola has two points of intersection with the Ox axis.
If the discriminant is zero (D = 0) then the parabola at the vertex touches the abscissa axis.
And the last case, when the discriminant is less than zero (D<0) – график параболы принадлежит плоскости над осью абсцисс (ветки параболы вверх), или график полностью под осью абсцисс (ветки параболы опущены вниз).

Incomplete quadratic equations

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 an equation of general form, in which a = 1, b = p, c = q. Therefore, for the reduced 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 be 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. The following theorem is used to solve some problems.

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 part 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 NS 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.