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

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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 following relations follow: 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 carry out the proof of Vieta's theorem according to the following scheme: we compose the sum and product of the roots of the quadratic equation using known root formulas, then 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 and make it up. Now we bring the fractions to a common denominator, we have . In the numerator of the resulting fraction, after which:. Finally, after on 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 multiplying fractions, the last product can be written as . Now we multiply a bracket by a bracket in the numerator, but it’s faster to collapse this product by square difference 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 instead of D in the last fraction we can substitute b 2 −4·a·c, we get. After opening the parentheses and bringing similar 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, the proof of Vieta’s theorem will take a laconic form:
,
.

It remains only to note that if 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, when D=0 the root of the quadratic equation is equal to , 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 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 type, which does not limit the generality, since any quadratic equation can be replaced by an equivalent equation by dividing both sides by a non-zero 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 equal to the free term, that is, x 1 +x 2 =−p, x 1 x 2 = q.

Theorem converse to 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 converse of Vieta’s theorem is true. Let's 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 with their expressions through x 1 and x 2, it is transformed into an equivalent equation.

Let us substitute the number x 1 instead of x into the resulting equation, and 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 represents 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 =0, which means 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 =0 substitute the number x 2 instead of x, we get the equality x 2 2 −(x 1 +x 2) x 2 +x 1 x 2 =0. This is a true 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 therefore 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 section we will analyze solutions to several of the most typical examples.

Let's start by applying the theorem converse to Vieta's theorem. It is convenient to use to check whether given two numbers are 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 roots found.

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 ​​we just obtained.

In the first case we have x 1 +x 2 =−5+3=−2. The resulting value is different from 4, so no further check can be carried out, but using the theorem inverse to Vieta’s theorem, one can immediately conclude that the first pair of numbers is not a pair of roots of the given quadratic equation.

Let's move on to the second case. Here, that is, the first condition is met. 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 the quadratic equation.

There is one last case left. Here and . Both conditions are met, so these numbers x 1 and x 2 are the roots of the given quadratic equation.

Answer:

The converse of Vieta's theorem can be used in practice to find 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. In this case, they use the fact that if the sum of two numbers is equal to the second coefficient of a 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 understand 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 must be satisfied: x 1 + x 2 =5 and x 1 · x 2 =6. All that remains is to select 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 inverse to Vieta's theorem is especially convenient to use to find the second root of a given quadratic equation when one of the roots is already known or obvious. In this case, the second root can be 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 unity is the root of the equation, since the sum of the coefficients of this quadratic equation is equal to 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, from which 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 roots, you can use formulas for the roots of a quadratic equation through a discriminant.

Another practical application of the converse of Vieta's theorem is to construct quadratic equations given the 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 −11 and 23.

Solution.

Let's denote x 1 =−11 and x 2 =23. We calculate 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 a second coefficient of −12 and a free term of −253. That is, x 2 −12·x−253=0 is the required equation.

Answer:

x 2 −12·x−253=0 .

Vieta's theorem is very often used when solving 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 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. Let's look at 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. Consequently, 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 relationships it asserts. But there are formulas that connect the real roots and coefficients of not only quadratic equations, but also cubic equations, equations of the fourth degree, and in general, algebraic equations degree n. They are called Vieta's formulas.

Let us write the Vieta formula for an algebraic equation of degree n of the form, and we will assume that it has n real roots x 1, x 2, ..., x n (among them there may be coinciding ones):

Vieta's formulas can be obtained theorem on the decomposition of a polynomial into linear factors, 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 Vieta’s formulas.

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

For a cubic equation, Vieta's formulas have the form

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

Bibliography.

• Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• Algebra and the beginning of mathematical analysis. 10th grade: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. - 3rd ed. - M.: Education, 2010.- 368 p. : ill. - ISBN 978-5-09-022771-1.

Vieta's theorem (more precisely, the theorem inverse to Vieta's theorem) allows you to reduce the time for solving quadratic equations. You just need to know how to use it. How to learn to solve quadratic equations using Vieta's theorem? It's not difficult if you think about it a little.

Now we will only talk about solving the reduced quadratic equation using Vieta’s theorem. A reduced quadratic equation is an equation in which a, that is, the coefficient of x², is equal to one. It is also possible to solve quadratic equations that are not given using Vieta’s theorem, but at least one of the roots is not an integer. They are more difficult to guess.

The inverse theorem to Vieta's theorem states: if the numbers x1 and x2 are such that

then x1 and x2 are the roots of the quadratic equation

When solving a quadratic equation using Vieta's theorem, only 4 options are possible. If you remember the line of reasoning, you can learn to find whole roots very quickly.

I. If q is a positive number,

this means that the roots x1 and x2 are numbers of the same sign (since only multiplying numbers with the same signs produces a positive number).

I.a. If -p is a positive number, (respectively, p<0), то оба корня x1 и x2 — положительные числа (поскольку складывали числа одного знака и получили положительное число).

I.b. If -p is a negative number, (respectively, p>0), then both roots are negative numbers (we added numbers of the same sign and got a negative number).

II. If q is a negative number,

this means that the roots x1 and x2 have different signs (when multiplying numbers, a negative number is obtained only when the signs of the factors are different). In this case, x1 + x2 is no longer a sum, but a difference (after all, when adding numbers with different signs, we subtract the smaller from the larger in absolute value). Therefore, x1+x2 shows how much the roots x1 and x2 differ, that is, how much one root is greater than the other (in absolute value).

II.a. If -p is a positive number, (that is, p<0), то больший (по модулю) корень — положительное число.

II.b. If -p is a negative number, (p>0), then the larger (modulo) root is a negative number.

Let's consider solving quadratic equations using Vieta's theorem using examples.

Solve the given quadratic equation using Vieta’s theorem:

Here q=12>0, so the roots x1 and x2 are numbers of the same sign. Their sum is -p=7>0, so both roots are positive numbers. We select integers whose product is equal to 12. These are 1 and 12, 2 and 6, 3 and 4. The sum is 7 for the pair 3 and 4. This means that 3 and 4 are the roots of the equation.

In this example, q=16>0, which means that the roots x1 and x2 are numbers of the same sign. Their sum is -p=-10<0, поэтому оба корня — отрицательные числа. Подбираем числа, произведение которых равно 16. Это 1 и 16, 2 и 8, 4 и 4. Сумма 2 и 8 равна 10, а раз нужны отрицательные числа, то искомые корни — это -2 и -8.

Here q=-15<0, что означает, что корни x1 и x2 — числа разных знаков. Поэтому 2 — это уже не их сумма, а разность, то есть числа отличаются на 2. Подбираем числа, произведение которых равно 15, отличающиеся на 2. Произведение равно 15 у 1 и 15, 3 и 5. Отличаются на 2 числа в паре 3 и 5. Поскольку -p=2>0, then the larger number is positive. So the roots are 5 and -3.

q=-36<0, значит, корни x1 и x2 имеют разные знаки. Тогда 5 — это то, насколько отличаются x1 и x2 (по модулю, то есть пока что без учета знака). Среди чисел, произведение которых равно 36: 1 и 36, 2 и 18, 3 и 12, 4 и 9 — выбираем пару, в которой числа отличаются на 5. Это 4 и 9. Осталось определить их знаки. Поскольку -p=-5<0, бОльшее число имеет знак минус. Поэтому корни данного уравнения равны -9 и 4.

In mathematics, there are special techniques with which many quadratic equations can be solved very quickly and without any discriminants. Moreover, with proper training, many begin to solve quadratic equations orally, literally “at first sight.”

Unfortunately, in the modern course of school mathematics, such technologies are almost not studied. But you need to know! And today we will look at one of these techniques - Vieta's theorem. First, let's introduce a new definition.

A quadratic equation of the form x 2 + bx + c = 0 is called reduced. Please note that the coefficient for x 2 is 1. There are no other restrictions on the coefficients.

1. x 2 + 7x + 12 = 0 is a reduced quadratic equation;
2. x 2 − 5x + 6 = 0 - also reduced;
3. 2x 2 − 6x + 8 = 0 - but this is not given at all, since the coefficient of x 2 is equal to 2.

Of course, any quadratic equation of the form ax 2 + bx + c = 0 can be reduced - just divide all the coefficients by the number a. We can always do this, since the definition of a quadratic equation implies that a ≠ 0.

True, these transformations will not always be useful for finding roots. Below we will make sure that this should be done only when in the final equation given by the square all the coefficients are integer. For now, let's look at the simplest examples:

Task. Convert the quadratic equation to the reduced equation:

1. 3x 2 − 12x + 18 = 0;
2. −4x 2 + 32x + 16 = 0;
3. 1.5x 2 + 7.5x + 3 = 0;
4. 2x 2 + 7x − 11 = 0.

Let's divide each equation by the coefficient of the variable x 2. We get:

1. 3x 2 − 12x + 18 = 0 ⇒ x 2 − 4x + 6 = 0 - divided everything by 3;
2. −4x 2 + 32x + 16 = 0 ⇒ x 2 − 8x − 4 = 0 - divided by −4;
3. 1.5x 2 + 7.5x + 3 = 0 ⇒ x 2 + 5x + 2 = 0 - divided by 1.5, all coefficients became integers;
4. 2x 2 + 7x − 11 = 0 ⇒ x 2 + 3.5x − 5.5 = 0 - divided by 2. In this case, fractional coefficients appeared.

As you can see, the above quadratic equations can have integer coefficients even if the original equation contained fractions.

Now let us formulate the main theorem, for which, in fact, the concept of a reduced quadratic equation was introduced:

Vieta's theorem. Consider the reduced quadratic equation of the form x 2 + bx + c = 0. Assume that this equation has real roots x 1 and x 2. In this case, the following statements are true:

1. x 1 + x 2 = −b. In other words, the sum of the roots of the given quadratic equation is equal to the coefficient of the variable x, taken with the opposite sign;
2. x 1 x 2 = c. The product of the roots of a quadratic equation is equal to the free coefficient.

Examples. For simplicity, we will consider only the above quadratic equations that do not require additional transformations:

1. x 2 − 9x + 20 = 0 ⇒ x 1 + x 2 = − (−9) = 9; x 1 x 2 = 20; roots: x 1 = 4; x 2 = 5;
2. x 2 + 2x − 15 = 0 ⇒ x 1 + x 2 = −2; x 1 x 2 = −15; roots: x 1 = 3; x 2 = −5;
3. x 2 + 5x + 4 = 0 ⇒ x 1 + x 2 = −5; x 1 x 2 = 4; roots: x 1 = −1; x 2 = −4.

Vieta's theorem gives us additional information about the roots of a quadratic equation. At first glance, this may seem difficult, but even with minimal training you will learn to “see” the roots and literally guess them in a matter of seconds.

Task. Solve the quadratic equation:

1. x 2 − 9x + 14 = 0;
2. x 2 − 12x + 27 = 0;
3. 3x 2 + 33x + 30 = 0;
4. −7x 2 + 77x − 210 = 0.

Let’s try to write out the coefficients using Vieta’s theorem and “guess” the roots:

1. x 2 − 9x + 14 = 0 is a reduced quadratic equation.
   By Vieta’s theorem we have: x 1 + x 2 = −(−9) = 9; x 1 · x 2 = 14. It is easy to see that the roots are the numbers 2 and 7;
2. x 2 − 12x + 27 = 0 - also reduced.
   By Vieta's theorem: x 1 + x 2 = −(−12) = 12; x 1 x 2 = 27. Hence the roots: 3 and 9;
3. 3x 2 + 33x + 30 = 0 - this equation is not reduced. But we will correct this now by dividing both sides of the equation by the coefficient a = 3. We get: x 2 + 11x + 10 = 0.
   We solve using Vieta’s theorem: x 1 + x 2 = −11; x 1 x 2 = 10 ⇒ roots: −10 and −1;
4. −7x 2 + 77x − 210 = 0 - again the coefficient for x 2 is not equal to 1, i.e. equation not given. We divide everything by the number a = −7. We get: x 2 − 11x + 30 = 0.
   By Vieta's theorem: x 1 + x 2 = −(−11) = 11; x 1 x 2 = 30; From these equations it is easy to guess the roots: 5 and 6.

From the above reasoning it is clear how Vieta’s theorem simplifies the solution of quadratic equations. No complicated calculations, no arithmetic roots and fractions. And we didn’t even need a discriminant (see lesson “Solving quadratic equations”).

Of course, in all our reflections we proceeded from two important assumptions, which, generally speaking, are not always met in real problems:

1. The quadratic equation is reduced, i.e. the coefficient for x 2 is 1;
2. The equation has two different roots. From an algebraic point of view, in this case the discriminant is D > 0 - in fact, we initially assume that this inequality is true.

However, in typical mathematical problems these conditions are met. If the calculation results in a “bad” quadratic equation (the coefficient of x 2 is different from 1), this can be easily corrected - look at the examples at the very beginning of the lesson. I’m generally silent about roots: what kind of problem is this that has no answer? Of course there will be roots.

Thus, the general scheme for solving quadratic equations using Vieta’s theorem is as follows:

1. Reduce the quadratic equation to the given one, if this has not already been done in the problem statement;
2. If the coefficients in the above quadratic equation are fractional, we solve using the discriminant. You can even go back to the original equation to work with more "handy" numbers;
3. In the case of integer coefficients, we solve the equation using Vieta’s theorem;
4. If you can’t guess the roots within a few seconds, forget about Vieta’s theorem and solve using the discriminant.

Task. Solve the equation: 5x 2 − 35x + 50 = 0.

So, we have before us an equation that is not reduced, because coefficient a = 5. Divide everything by 5, we get: x 2 − 7x + 10 = 0.

All coefficients of a quadratic equation are integer - let's try to solve it using Vieta's theorem. We have: x 1 + x 2 = −(−7) = 7; x 1 · x 2 = 10. In this case, the roots are easy to guess - they are 2 and 5. There is no need to count using the discriminant.

Task. Solve the equation: −5x 2 + 8x − 2.4 = 0.

Let's look: −5x 2 + 8x − 2.4 = 0 - this equation is not reduced, let's divide both sides by the coefficient a = −5. We get: x 2 − 1.6x + 0.48 = 0 - an equation with fractional coefficients.

It is better to return to the original equation and count through the discriminant: −5x 2 + 8x − 2.4 = 0 ⇒ D = 8 2 − 4 · (−5) · (−2.4) = 16 ⇒ ... ⇒ x 1 = 1.2; x 2 = 0.4.

Task. Solve the equation: 2x 2 + 10x − 600 = 0.

First, let's divide everything by the coefficient a = 2. We get the equation x 2 + 5x − 300 = 0.

This is the reduced equation, according to Vieta’s theorem we have: x 1 + x 2 = −5; x 1 x 2 = −300. It is difficult to guess the roots of the quadratic equation in this case - personally, I was seriously stuck when solving this problem.

You will have to look for roots through the discriminant: D = 5 2 − 4 · 1 · (−300) = 1225 = 35 2 . If you don't remember the root of the discriminant, I'll just note that 1225: 25 = 49. Therefore, 1225 = 25 49 = 5 2 7 2 = 35 2.

Now that the root of the discriminant is known, solving the equation is not difficult. We get: x 1 = 15; x 2 = −20.

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 who have good mental arithmetic skills, this method is clearly irrational. Students often have to solve quadratic equations even 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 Vieta’s theorem (according to the A.G. Mordkovich Algebra-8 program, only two hours are planned for studying the topic “Vieta’s Theorem. Decomposition of a quadratic trinomial into linear factors”).

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

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

Example 1. Solve the equation.

Let's say this equation has roots, namely, and . Then, according to Vieta’s theorem, the equalities must simultaneously hold:

Please note that the product of roots is a positive number. This means that the roots of the equation are of 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 return again to the product of roots. Let's assume that the roots of the equation are positive integers. Then the correct first equality can be obtained only in two ways (up to the order of the factors): or . Let us check for the proposed pairs of numbers the feasibility of the second statement of Vieta’s theorem: . Thus, the numbers 2 and 3 satisfy both equalities, and therefore are the roots of the given equation.

Answer: 2; 3.

Let us highlight the main stages of reasoning when solving the above quadratic equation using Vieta’s theorem:

write down the statement 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 larger root in modulus is a positive number, and if the sum of the roots is less than zero, then the larger root in 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, select the pair that, when substituted into the second equality in the notation (*), will give the correct equality;
• indicate in your answer the found roots of the equation.

Let's give more examples.

Example 2: Solve the equation .

Solution.

Let and be the roots of the given equation. Then, by Vieta’s theorem, we note that the product is positive, and the sum is a negative number. This means that both roots are negative numbers. We select pairs of factors that give a product of 10 (-1 and -10; -2 and -5). The second pair of numbers adds up to -7. This means that 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, we note that the product is negative. This means that the roots are of different signs. The sum of the roots is also a negative number. This means that the root with the largest 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. This means that the numbers 2 and -5 are the roots of this equation.

Answer: 2; -5.

Note that Vieta’s theorem can, in principle, be formulated for a complete quadratic equation: if quadratic equation has roots and , then the equalities , , are satisfied for them. However, the application of this theorem is quite problematic, since in a complete quadratic equation at least one of the roots (if any, of course) is a fractional number. And working with selecting 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 A and write the equation in the form . Let us introduce a new variable and obtain the reduced quadratic equation, the roots of which and (if available) can be found using Vieta’s theorem. Then the roots of the original equation will be . Please note that it is very simple to create the auxiliary reduced equation: the second coefficient is preserved, and the third coefficient is equal to the product ac. With a certain skill, students immediately create an auxiliary equation, find its roots using Vieta’s theorem, and indicate the roots of the given complete equation. Let's give examples.

Example 4: Solve the equation .

Let's create an auxiliary equation and using Vieta's theorem we will find its roots. This means that the roots of the original equation .

Answer: .

Example 5: Solve the equation .

The auxiliary equation has the form . According to Vieta's theorem, its roots are . Finding 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 not difficult 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 Vieta’s 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 of 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 by passing 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.

Vieta's theorem is often used to check roots that have already been found. 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 ​​of \(p\) 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 us, using , 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 solution process. 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 we solved the equation correctly.

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

Vieta's converse 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 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 because it saves a lot of time.

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

Solution : Using Vieta’s inverse theorem, we find 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 system \(x_1 \cdot x_2=6\). What two can the number \(6\) be decomposed into? On \(2\) and \(3\), \(6\) and \(1\) or \(-2\) and \(-3\), and \(-6\) and \(- 1\). The first equation of the system will tell you which pair to choose: \(x_1+x_2=5\). \(2\) and \(3\) are similar, since \(2+3=5\).
Answer : \(x_1=2\), \(x_2=3\).

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

c) \(x^2+9x+20=0\) – what factors does \(20\) decompose into? \(4\) 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) \(x^2-88x+780=0\) – what factors does \(780\) decompose into? \(390\) and \(2\). Will they add up to \(88\)? No. What other multipliers does \(780\) have? \(78\) and \(10\). Will they add up to \(88\)? Yes. Answer: \(78\) and \(10\).

It is not necessary to expand 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 for which the coefficient of \(x^2\) is equal to one. If we were initially given 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, since the coefficient of \(x^2\) is equal to \(2\). Let's get rid of it by dividing the entire equation by \(2\).

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

Ready. Now you can use both theorems.

Answers to frequently asked questions

Question: Using Vieta's theorem, you can solve any ?
Answer: Unfortunately no. If the equation does not contain integers 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 school mathematics have integer solutions.