Solving equations using the Vieta theorem examples. Viet theorem, inverse viet formula and examples with solution for dummies

François Vieta (1540-1603) - mathematician, creator of the famous Vieta formulas

Vieta's theorem needed to quickly solve quadratic equations (in simple terms).

In more detail, t Vieta's theorem - this is the sum of the roots of this quadratic equation is equal to the second coefficient, which is taken with the opposite sign, and the product is equal to the free term. This property has any given quadratic equation that has roots.

Using the Vieta theorem, you can easily solve quadratic equations by selection, so let's say “thank you” to this mathematician with a sword in his hands for our happy 7th grade.

Proof of Vieta's theorem

To prove the theorem, you can use the well-known root formulas, thanks to which we will compose the sum and product of the roots of the quadratic equation. Only after that we can make sure that they are equal and, accordingly, .

Let's say we have an equation: . This equation has the following roots: and . Let us prove that , .

According to the formulas of the roots of the quadratic equation:

1. Find the sum of the roots:

Let's analyze this equation, as we got it exactly like this:

= .

Step 1. We reduce the fractions to a common denominator, it turns out:

= = .

Step 2. We got a fraction where you need to open the brackets:

We reduce the fraction by 2 and get:

We have proved the relation for the sum of the roots of a quadratic equation using Vieta's theorem.

2. Find the product of the roots:

= = = = = .

Let's prove this equation:

Step 1. There is a rule for multiplying fractions, according to which we multiply this equation:

Now we recall the definition of the square root and consider:

= .

Step 3. We recall the discriminant of the quadratic equation: . Therefore, instead of D (discriminant), we substitute in the last fraction, then we get:

= .

Step 4. Open the brackets and add like terms to fractions:

Step 5. We reduce "4a" and get.

So we have proved the relation for the product of roots according to Vieta's theorem.

IMPORTANT!If the discriminant is zero, then the quadratic equation has only one root.

Theorem inverse to Vieta's theorem

According to the theorem, the inverse of Vieta's theorem, we can check whether our equation is solved correctly. To understand the theorem itself, we need to consider it in more detail.

If the numbers are:

And then they are the roots of the quadratic equation.

Proof of Vieta's converse theorem

Step 1.Let us substitute expressions for its coefficients into the equation:

Step 2Let's transform the left side of the equation:

Step 3. Let's find the roots of the equation, and for this we use the property that the product is equal to zero:

Or . Where does it come from: or.

Examples with solutions by Vieta's theorem

Example 1

The task

Find the sum, product and sum of squares of the roots of a quadratic equation without finding the roots of the equation.

Solution

Step 1. Recall the discriminant formula. We substitute our numbers under the letters. That is, , is a substitute for , and . This implies:

It turns out:

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We express the sum of the squares of the roots through their sum and product:

Answer

7; 12; 25.

Example 2

The task

Solve the equation. In this case, do not use the formulas of the quadratic equation.

Solution

This equation has roots that are greater than zero in terms of the discriminant (D). Accordingly, according to the Vieta theorem, the sum of the roots of this equation is 4, and the product is 5. First, we determine the divisors of the number, the sum of which is 4. These are the numbers "5" and "-1". Their product is equal to - 5, and the sum - 4. Hence, according to the theorem, the converse of Vieta's theorem, they are the roots of this equation.

Answer

AND Example 4

The task

Write an equation where each root is twice the corresponding root of the equation:

Solution

By Vieta's theorem, the sum of the roots of this equation is 12, and the product = 7. Hence, the two roots are positive.

The sum of the roots of the new equation will be equal to:

And the work.

By a theorem converse to Vieta's theorem, the new equation has the form:

Answer

The result was an equation, each root of which is twice as large:

So, we looked at how to solve an equation using Vieta's theorem. It is very convenient to use this theorem if tasks are solved that are associated with the signs of the roots of quadratic equations. That is, if the free term in the formula is a positive number, and if there are real roots in the quadratic equation, then they can both be either negative or positive.

And if the free term is a negative number, and if there are real roots in the quadratic equation, then both signs will be different. That is, if one root is positive, then the other root will only be negative.

Useful sources:

1. Dorofeev G. V., Suvorova S. B., Bunimovich E. A. Algebra Grade 8: Moscow “Enlightenment”, 2016 – 318 p.
2. Rubin A. G., Chulkov P. V. - textbook Algebra Grade 8: Moscow "Balass", 2015 - 237 p.
3. Nikolsky S. M., Potopav M. K., Reshetnikov N. N., Shevkin A. V. – Algebra Grade 8: Moscow “Enlightenment”, 2014 – 300

Vieta's theorem, inverse Vieta formula and examples with solution for dummies updated: November 22, 2019 by: Scientific Articles.Ru

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

Content

See also: The roots of a quadratic equation

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 about multiple roots

If the discriminant of equation (1) is 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:
;
;
.

Finding the sum of the roots:
.

To find the product, we apply the formula:
.
Then

.

The theorem has been proven.

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 has been proven.

Inverse Vieta 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 and are the roots of equation (1).

Substitute (2) and (3) into (1):
.
We group the terms of 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 has been proven.

Vieta's theorem for the complete quadratic equation

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

We divide equation (5) by:
.
That is, we have obtained the above equation
,
where ; .

Then the Vieta 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 a cubic equation

Similarly, we can establish connections between the roots of a cubic equation. Consider the cubic equation
(6) ,
where , , , are some numbers. And .
Let's divide this equation by:
(7) ,
where , , .
Let , , be the roots of equation (7) (and equation (6)). Then

.

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

Vieta's theorem for an nth degree equation

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

Vieta's theorem for an nth degree equation 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 Higher Educational Institutions, Lan, 2009.
CM. Nikolsky, M.K. Potapov et al., Algebra: a textbook for the 8th grade of educational institutions, Moscow, Education, 2006.

See also:

In eighth grade, students are introduced to quadratic equations and how to solve them. At the same time, as experience shows, most students use only one method when solving complete quadratic equations - the formula for the roots of a quadratic equation. For students with good oral counting skills, this method is clearly irrational. Students often have to solve quadratic equations in high school, and there it is simply a pity to spend time calculating the discriminant. In my opinion, when studying quadratic equations, more time and attention should be paid to the application of the Vieta theorem (according to the program of A.G. Mordkovich Algebra-8, only two hours are planned to study the topic “Vieta Theorem. Decomposition of a square trinomial into linear factors”).

In most algebra textbooks, this theorem is formulated for a reduced quadratic equation and says that if the equation has roots and , then they satisfy the equalities , . Then a statement converse to Vieta's theorem is formulated, and a number of examples are offered to work on this topic.

Let's take specific examples and trace the logic of the solution on them using Vieta's theorem.

Example 1. Solve the equation.

Suppose this equation has roots, namely, and . Then, by Vieta's theorem, the equalities

Note that the product of the roots is a positive number. So, the roots of the equation have the same sign. And since the sum of the roots is also a positive number, we conclude that both roots of the equation are positive. Let's go back to the product of roots. Assume that the roots of the equation are positive integers. Then the correct first equality can be obtained in only two ways (up to the order of factors): or . Let's check for the proposed pairs of numbers the feasibility of the second assertion of the Vieta theorem: . Thus, the numbers 2 and 3 satisfy both equalities, and hence are the roots of the given equation.

Answer: 2; 3.

We single out the main stages of reasoning when solving the given quadratic equation using the Vieta theorem:

write down the assertion of Vieta's theorem

(*)

• determine the signs of the roots of the equation (If the product and the sum of the roots are positive, then both roots are positive numbers. If the product of the roots is a positive number, and the sum of the roots is negative, then both roots are negative numbers. If the product of the roots is a negative number, then the roots have different signs.Moreover, if the sum of the roots is positive, then the root with a greater modulus is a positive number, and if the sum of the roots is less than zero, then the root with a greater modulus is a negative number);
• select pairs of integers whose product gives the correct first equality in the notation (*);
• from the found pairs of numbers, choose the pair that, when substituted into the second equality in the notation (*), will give the correct equality;
• indicate in the answer the found roots of the equation.

Let's give some more examples.

Example 2: Solve the Equation .

Solution.

Let and be the roots of the given equation. Then by Vieta's theorem Note that the product is positive and the sum is negative. So both roots are negative numbers. We select pairs of factors that give the product of 10 (-1 and -10; -2 and -5). The second pair of numbers adds up to -7. So the numbers -2 and -5 are the roots of this equation.

Answer: -2; -5.

Example 3. Solve the equation .

Solution.

Let and be the roots of the given equation. Then by Vieta's theorem Note that the product is negative. So the roots are of different signs. The sum of the roots is also a negative number. Hence, the root with the greatest modulus is negative. We select pairs of factors that give the product -10 (1 and -10; 2 and -5). The second pair of numbers adds up to -3. So the numbers 2 and -5 are the roots of this equation.

Answer: 2; -5.

Note that the Vieta theorem can in principle be formulated for the complete quadratic equation: if the quadratic equation has roots and , then they satisfy the equalities , . However, the application of this theorem is rather problematic, since in the full quadratic equation at least one of the roots (if any, of course) is a fractional number. And working with the selection of fractions is long and difficult. But still there is a way out.

Consider the complete quadratic equation . Multiply both sides of the equation by the first coefficient but and write the equation in the form . We introduce a new variable and obtain a reduced quadratic equation , whose roots and (if any) can be found using the Vieta theorem. Then the roots of the original equation will be . Note that it is very easy to write the auxiliary reduced equation: the second coefficient is preserved, and the third coefficient is equal to the product ace. With a certain skill, students immediately compose an auxiliary equation, find its roots using the Vieta theorem and indicate the roots of the given complete equation. Let's give examples.

Example 4. Solve the equation .

Let's make an auxiliary equation and by Vieta's theorem we find its roots. So the roots of the original equation .

Answer: .

Example 5. Solve the equation .

The auxiliary equation has the form . By Vieta's theorem, its roots are . We find the roots of the original equation .

Answer: .

And one more case when the application of Vieta's theorem allows you to verbally find the roots of a complete quadratic equation. It is easy to prove that the number 1 is the root of the equation , if and only if. The second root of the equation is found by the Vieta theorem and is equal to . One more statement: so that the number -1 is the root of the equation necessary and sufficient to. Then the second root of the equation according to Vieta's theorem is equal to . Similar statements can be formulated for the reduced quadratic equation.

Example 6. Solve the equation.

Note that the sum of the coefficients of the equation is zero. So the roots of the equation .

Answer: .

Example 7. Solve the equation.

The coefficients of this equation satisfy the property (indeed, 1-(-999)+(-1000)=0). So the roots of the equation .

Answer: ..

Examples for the application of Vieta's theorem

Task 1. Solve the given quadratic equation using Vieta's theorem.

1.	6.	11.	16.
2.	7.	12.	17.
3.	8.	13.	18.
4.	9.	14.	19.
5.	10.	15.	20.

Task 2. Solve the complete quadratic equation using the transition to the auxiliary reduced quadratic equation.

1.	6.	11.	16.
2.	7.	12.	17.
3.	8.	13.	18.
4.	9.	14.	19.
5.	10.	15.	20.

Task 3. Solve a quadratic equation using the property.

Between the roots and the coefficients of the quadratic equation, in addition to the root formulas, there are other useful relationships that are given by Vieta's theorem. In this article, we will give a formulation and proof of Vieta's theorem for a quadratic equation. Next, we consider a theorem converse to Vieta's theorem. After that, we will analyze the solutions of the most characteristic examples. Finally, we write down the Vieta formulas that define the connection between the real roots algebraic equation degree n and its coefficients.

Page navigation.

Vieta's theorem, formulation, proof

From the formulas of the roots of the quadratic equation a x 2 +b x+c=0 of the form , where D=b 2 −4 a c , the relations x 1 +x 2 = −b/a, x 1 x 2 = c/a . These results are confirmed 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 the Vieta theorem according to the following scheme: we compose the sum and the product of the roots of the quadratic equation using the known root formulas, after that we transform the resulting 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 of multiplication of fractions, the last product can be written as. Now we multiply the bracket by the bracket in the numerator, but it is faster to collapse this product by difference of squares formula, So . Then, remembering , we perform the next transition . And since the formula D=b 2 −4 a·c corresponds to the discriminant of the quadratic equation, then b 2 −4·a·c can be substituted into the last fraction instead of D, we get . After opening the brackets and reducing like terms, we arrive at the 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 the Vieta theorem will take a concise 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 the Vieta theorem also hold. Indeed, for D=0 the root of the quadratic equation is , 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 the reduced quadratic equation (with the highest coefficient a equal to 1 ) of the form x 2 +p·x+q=0 . Sometimes it is formulated for quadratic equations of just this type, which does not limit the generality, since any quadratic equation can be replaced by an equivalent equation by dividing both its parts by a non-zero number a. Here is the corresponding formulation of Vieta's theorem:

Theorem.

The sum of the roots of the reduced quadratic equation x 2 + p x + q \u003d 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 \u003d −p, x 1 x 2 \u003d q .

Theorem inverse to Vieta's theorem

The second formulation of the Vieta 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 assertion converse to Vieta's theorem is true. We 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 of their expression through x 1 and x 2, it is converted into an equivalent equation.

We substitute the number x 1 instead of x into the resulting equation, 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 the correct 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 the root of the equation x 2 −(x 1 + x 2) x + x 1 x 2 \u003d 0, which means that x 1 is the root of the equivalent equation x 2 +p x+q=0 .

If in the equation x 2 −(x 1 + x 2) x + x 1 x 2 \u003d 0 substitute the number x 2 instead of x, then we get the equality x 2 2 −(x 1 + x 2) x 2 + x 1 x 2 =0. This is the correct equation because 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 the root of the equation x 2 −(x 1 + x 2) x + x 1 x 2 \u003d 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 inverse theorem. In this subsection, we will analyze the solutions of several of the most typical examples.

We start by applying a theorem converse to Vieta's theorem. It is convenient to use it to check whether 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 relations is checked. If both of these relations are satisfied, then, by virtue of the theorem converse 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 the quadratic equation must be equal to −b/a, that is, 16/4=4, and the product of the roots must 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, therefore, further verification can not be carried out, but by the theorem, the inverse of Vieta's theorem, we 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 satisfied. We check the second condition: , the resulting value is different from 9/4 . Therefore, 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 theorem, the reverse of Vieta's theorem, can be used in practice to select the roots of a quadratic equation. Usually, integer roots of the given quadratic equations with integer coefficients are selected, since in other cases this is quite difficult to do. At the same time, 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 deal with this with an example.

Let's 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, two equalities x 1 +x 2 \u003d 5 and x 1 x 2 \u003d 6 must be satisfied. It remains to choose such numbers. In this case, this is quite simple to do: 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 theorem, the reverse of Vieta's theorem, is especially convenient to apply to finding the second root of the 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 . Here it is easy to see that the unit 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 . So we have defined both roots of the quadratic equation: 1 and −3/512.

It is clear that the selection of roots is expedient only in the simplest cases. In other cases, to find the roots, you can apply the formulas of the roots of the quadratic equation through the discriminant.

Another practical application of the theorem, the inverse of Vieta's theorem, is the compilation of 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 of x with the opposite sign of the given quadratic equation, and the product of the roots, which gives the free term.

Example.

Write a quadratic equation whose roots are the numbers −11 and 23.

Solution.

Denote x 1 =−11 and x 2 =23 . We calculate the sum and product of these numbers: x 1 + x 2 \u003d 12 and x 1 x 2 \u003d −253. Therefore, these numbers are the roots of the given quadratic equation with the second coefficient -12 and the free term -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 in solving tasks 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 is 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. Consider examples of their application.

Example.

R is positive. According to 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 by the Vieta theorem, the product of the roots of the given 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, in order to find the values ​​of r that are of interest to us, we need to solve a 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 asserts. But there are formulas that connect the real roots and coefficients not only of quadratic equations, but also of cubic equations, quadruple equations, and in general, algebraic equations degree n. They are called Vieta formulas.

We write the Vieta formulas for an algebraic equation of degree n of the form, while we assume that it has n real roots x 1, x 2, ..., x n (among them there may be the same):

Get Vieta formulas allows polynomial factorization theorem, as well as the definition of equal polynomials through the equality of all their corresponding coefficients. So the polynomial and its expansion into linear factors of the form are equal. Opening the brackets in the last product and equating the corresponding coefficients, we obtain the Vieta formulas.

In particular, for n=2 we have already familiar Vieta formulas for the quadratic equation .

For a cubic equation, the Vieta formulas have the form

It only remains to note that on the left side of the Vieta formulas there are the so-called elementary symmetric polynomials.

Bibliography.

• Algebra: textbook for 8 cells. 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.
• Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A 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.: Enlightenment, 2010.- 368 p. : ill. - ISBN 978-5-09-022771-1.

Vieta's theorem is often used to test already found roots. If you have found the roots, you can use the formulas \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\) to calculate the values ​​\(p\) and \(q\ ). And if they turn out to be the same as in the original equation, then the roots are found correctly.

For example, let's use , solve the equation \(x^2+x-56=0\) and get the roots: \(x_1=7\), \(x_2=-8\). Let's check if we made a mistake in the process of solving. In our case, \(p=1\), and \(q=-56\). By Vieta's theorem we have:

\(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\) \(\Leftrightarrow\) \(\begin(cases)7+(-8)=-1 \\7\cdot(-8)=-56\end(cases)\) \(\Leftrightarrow\) \(\begin(cases)-1=-1\\-56=-56\end(cases)\ )

Both statements converged, which means that we solved the equation correctly.

This test can be done orally. It will take 5 seconds and save you from stupid mistakes.

Inverse Vieta theorem

If \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\), then \(x_1\) and \(x_2\) are the roots of the quadratic equation \(x^ 2+px+q=0\).

Or in a simple way: if you have an equation of the form \(x^2+px+q=0\), then by solving the system \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\ end(cases)\) you will find its roots.

Thanks to this theorem, you can quickly find the roots of a quadratic equation, especially if these roots are . This skill is important as it saves a lot of time.

Example . Solve the equation \(x^2-5x+6=0\).

Solution : Using the inverse Vieta theorem, we get that the roots satisfy the conditions: \(\begin(cases)x_1+x_2=5 \\x_1 \cdot x_2=6\end(cases)\).
Look at the second equation of the \(x_1 \cdot x_2=6\) system. Into what two can the number \(6\) be decomposed? On \(2\) and \(3\), \(6\) and \(1\) or \(-2\) and \(-3\), and \(-6\) and \(- one\). And which pair to choose, the first equation of the system will tell: \(x_1+x_2=5\). \(2\) and \(3\) are similar, because \(2+3=5\).
Answer : \(x_1=2\), \(x_2=3\).

Examples . Using the inverse of Vieta's theorem, find the roots of the quadratic equation:
a) \(x^2-15x+14=0\); b) \(x^2+3x-4=0\); c) \(x^2+9x+20=0\); d) \(x^2-88x+780=0\).

Solution :
a) \(x^2-15x+14=0\) - what factors does \(14\) decompose into? \(2\) and \(7\), \(-2\) and \(-7\), \(-1\) and \(-14\), \(1\) and \(14\ ). What pairs of numbers add up to \(15\)? Answer: \(1\) and \(14\).

b) \(x^2+3x-4=0\) - into what factors does \(-4\) decompose? \(-2\) and \(2\), \(4\) and \(-1\), \(1\) and \(-4\). What pairs of numbers add up to \(-3\)? Answer: \(1\) and \(-4\).

c) \(x^2+9x+20=0\) – into what factors does \(20\) decompose? \(4\) and \(5\), \(-4\) and \(-5\), \(2\) and \(10\), \(-2\) and \(-10\ ), \(-20\) and \(-1\), \(20\) and \(1\). What pairs of numbers add up to \(-9\)? Answer: \(-4\) and \(-5\).

d) \(x^2-88x+780=0\) - into what factors does \(780\) decompose? \(390\) and \(2\). Do they add up to \(88\)? No. What other multipliers does \(780\) have? \(78\) and \(10\). Do they add up to \(88\)? Yes. Answer: \(78\) and \(10\).

It is not necessary to decompose the last term into all possible factors (as in the last example). You can immediately check whether their sum gives \(-p\).

Important! Vieta's theorem and the converse theorem only work with , that is, one whose coefficient in front of \(x^2\) is equal to one. If we initially have a non-reduced equation, then we can make it reduced by simply dividing by the coefficient in front of \ (x ^ 2 \).

For example, let the equation \(2x^2-4x-6=0\) be given and we want to use one of Vieta's theorems. But we can't, because the coefficient before \(x^2\) is equal to \(2\). Let's get rid of it by dividing the whole equation by \(2\).

\(2x^2-4x-6=0\) \(|:2\)
\(x^2-2x-3=0\)

Ready. Now we can use both theorems.

Answers to frequently asked questions

Question: By Vieta's theorem, you can solve any ?
Answer: Unfortunately no. If there are not integers in the equation or the equation has no roots at all, then Vieta's theorem will not help. In this case, you need to use discriminant . Fortunately, 80% of the equations in the school math course have integer solutions.