Almost any quadratic equation \can be converted to the form \ However, this is possible if you initially divide each term by a coefficient \before \ In addition, you can introduce a new notation:

\[(\frac (b)(a))= p\] and \[(\frac (c)(a)) = q\]

Due to this, we will have an equation \ called in mathematics a reduced quadratic equation. The roots of this equation and the coefficients are interconnected, which is confirmed by Vieta’s theorem.

Vieta's theorem: 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 the free term \

For clarity, let’s solve the following equation:

Let's solve this quadratic equation using the written rules. Having analyzed the initial data, we can conclude that the equation will have two different roots, since:

Now, from all the factors of the number 15 (1 and 15, 3 and 5), we select those whose difference is equal to 2. The numbers 3 and 5 fall under this condition. We put a minus sign in front of the smaller number. Thus, we obtain the roots of the equation \

Answer: \[ x_1= -3 and x_2 = 5\]

Where can I solve an equation using Vieta's theorem online?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

There are a number of relationships in quadratic equations. The main ones are the relationships between roots and coefficients. Also in quadratic equations there are a number of relationships that are given by Vieta’s theorem.

In this topic, we will present Vieta’s theorem itself and its proof for a quadratic equation, the theorem inverse to Vieta’s theorem, and analyze a number of examples of solving problems. In the material we will pay special attention to the consideration of Vieta’s formulas, which define the relationship between real roots algebraic equation degrees n and its coefficients.

Formulation and proof of Vieta's theorem

Formula for the roots of a quadratic equation a x 2 + b x + c = 0 of the form x 1 = - b + D 2 · a, x 2 = - b - D 2 · a, where D = b 2 − 4 a c, establishes relationships x 1 + x 2 = - b a, x 1 x 2 = c a. This is confirmed by Vieta's theorem.

Theorem 1

In a quadratic equation a x 2 + b x + c = 0, Where x 1 And x 2– roots, the sum of the roots will be equal to the ratio of the coefficients b And a, which was taken with the opposite sign, and the product of the roots will be equal to the ratio of the coefficients c And a, i.e. x 1 + x 2 = - b a, x 1 x 2 = c a.

Evidence 1

We offer you the following scheme for carrying out the proof: take the formula of roots, compose the sum and product of the roots of the quadratic equation and then transform the resulting expressions in order to make sure that they are equal -b a And c a respectively.

Let's make the sum of the roots x 1 + x 2 = - b + D 2 · a + - b - D 2 · a. Let's reduce the fractions to common denominator- b + D 2 · a + - b - D 2 · a = - b + D + - b - D 2 · a . Let's open the parentheses in the numerator of the resulting fraction and present similar terms: - b + D + - b - D 2 · a = - b + D - b - D 2 · a = - 2 · b 2 · a . Let's reduce the fraction by: 2 - b a = - b a.

This is how we proved the first relation of Vieta’s theorem, which relates to the sum of the roots of a quadratic equation.

Now let's move on to the second relationship.

To do this, we need to compose the product of the roots of the quadratic equation: x 1 · x 2 = - b + D 2 · a · - b - D 2 · a.

Let's remember the rule for multiplying fractions and write the last product as follows: - b + D · - b - D 4 · a 2.

Let's multiply a bracket by a bracket in the numerator of the fraction, or use the difference of squares formula to transform this product faster: - b + D · - b - D 4 · a 2 = - b 2 - D 2 4 · a 2 .

Let's use the definition square root in order to make the following transition: - b 2 - D 2 4 · a 2 = b 2 - D 4 · a 2. Formula D = b 2 − 4 a c corresponds to the discriminant of a quadratic equation, therefore, into a fraction instead of D can be substituted b 2 − 4 a c:

b 2 - D 4 a 2 = b 2 - (b 2 - 4 a c) 4 a 2

Let's open the brackets, add similar terms and get: 4 · a · c 4 · a 2 . If we shorten it to 4 a, then what remains is c a . This is how we proved the second relation of Vieta’s theorem for the product of roots.

The proof of Vieta's theorem can be written in a very laconic form if we omit the explanations:

x 1 + x 2 = - b + D 2 a + - b - D 2 a = - b + D + - b - D 2 a = - 2 b 2 a = - b a , x 1 x 2 = - b + D 2 · a · - b - D 2 · a = - b + D · - b - D 4 · a 2 = - b 2 - D 2 4 · a 2 = b 2 - D 4 · a 2 = = D = b 2 - 4 · a · c = b 2 - b 2 - 4 · a · c 4 · a 2 = 4 · a · c 4 · a 2 = c a .

When the discriminant of a quadratic equation is equal to zero, the equation will have only one root. To be able to apply Vieta's theorem to such an equation, we can assume that the equation, with a discriminant equal to zero, has two identical roots. Indeed, when D=0 the root of the quadratic equation is: - b 2 · a, then x 1 + x 2 = - b 2 · a + - b 2 · a = - b + (- b) 2 · a = - 2 · b 2 · a = - b a and x 1 · x 2 = - b 2 · a · - b 2 · a = - b · - b 4 · a 2 = b 2 4 · a 2 , and since D = 0, that is, b 2 - 4 · a · c = 0, whence b 2 = 4 · a · c, then b 2 4 · a 2 = 4 · a · c 4 · a 2 = c a.

Most often in practice, Vieta's theorem is applied to the reduced quadratic equation of the form x 2 + p x + q = 0, where the leading coefficient a is equal to 1. In this regard, Vieta’s theorem is formulated specifically for equations of this type. This does not limit the generality due to the fact that any quadratic equation can be replaced by an equivalent equation. To do this, you need to divide both of its parts by a number a different from zero.

Let us give another formulation of Vieta's theorem.

Theorem 2

Sum of roots in the given quadratic equation x 2 + p x + q = 0 will be equal to the coefficient of x, which is taken with the opposite sign, the product of the roots will be equal to the free term, i.e. x 1 + x 2 = − p, x 1 x 2 = q.

Theorem converse to Vieta's theorem

If you look carefully at the second formulation of Vieta’s theorem, you can see that for the roots x 1 And x 2 reduced quadratic equation x 2 + p x + q = 0 the following relations will be valid: x 1 + x 2 = − p, x 1 · x 2 = q. From these 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. So we come to a statement that is the converse of Vieta’s theorem.

We now propose to formalize this statement as a theorem and carry out its proof.

Theorem 3

If the numbers x 1 And x 2 are such that x 1 + x 2 = − p And x 1 x 2 = q, That x 1 And x 2 are the roots of the reduced quadratic equation x 2 + p x + q = 0.

Evidence 2

Replacing odds p And q to their expression through x 1 And x 2 allows you to transform the equation x 2 + p x + q = 0 into an equivalent .

If we substitute the number into the resulting equation x 1 instead of x, then we get the equality x 1 2 − (x 1 + x 2) x 1 + x 1 x 2 = 0. This is equality for any x 1 And x 2 turns into a true numerical equality 0 = 0 , because 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. This means that x 1– root of the equation x 2 − (x 1 + x 2) x + x 1 x 2 = 0, So what x 1 is also the root of the equivalent equation x 2 + p x + q = 0.

Substitution into equation x 2 − (x 1 + x 2) x + x 1 x 2 = 0 numbers x 2 instead of x allows us to obtain equality x 2 2 − (x 1 + x 2) x 2 + x 1 x 2 = 0. This equality can be considered true, 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. It turns out that x 2 is the 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.

The converse of Vieta's theorem has been proven.

Examples of using Vieta's theorem

Let's now begin to analyze the most typical examples on the topic. Let's start by analyzing problems that require the application of the theorem inverse to Vieta's theorem. It can be used to check numbers produced by calculations to see if they are the roots of a given quadratic equation. To do this, you need to calculate their sum and difference, and then check the validity of the relations x 1 + x 2 = - b a, x 1 · x 2 = a c.

The fulfillment of both relations indicates that the numbers obtained during the calculations are the roots of the equation. If we see that at least one of the conditions is not met, then these numbers cannot be the roots of the quadratic equation given in the problem statement.

Example 1

Which of the pairs of numbers 1) x 1 = − 5, x 2 = 3, or 2) x 1 = 1 - 3, x 2 = 3 + 3, or 3) x 1 = 2 + 7 2, x 2 = 2 - 7 2 is a pair of roots of a quadratic equation 4 x 2 − 16 x + 9 = 0?

Solution

Let's find the coefficients of the quadratic equation 4 x 2 − 16 x + 9 = 0. This is a = 4, b = − 16, c = 9. According to Vieta's theorem, the sum of the roots of a quadratic equation must be equal to -b a, that is, 16 4 = 4 , and the product of the roots must be equal c a, that is, 9 4 .

Let's check the obtained numbers by calculating the sum and product of numbers from three given pairs and comparing them with the obtained values.

In the first case x 1 + x 2 = − 5 + 3 = − 2. This value is different from 4, therefore, the check does not need to be continued. According to the theorem converse to Vieta's theorem, we can immediately conclude that the first pair of numbers are not the roots of this quadratic equation.

In the second case, x 1 + x 2 = 1 - 3 + 3 + 3 = 4. We see that the first condition is met. But the second condition is not: x 1 · x 2 = 1 - 3 · 3 + 3 = 3 + 3 - 3 · 3 - 3 = - 2 · 3. The value we got is different from 9 4 . This means that the second pair of numbers are not the roots of the quadratic equation.

Let's move on to consider the third pair. Here x 1 + x 2 = 2 + 7 2 + 2 - 7 2 = 4 and x 1 x 2 = 2 + 7 2 2 - 7 2 = 2 2 - 7 2 2 = 4 - 7 4 = 16 4 - 7 4 = 9 4. Both conditions are met, which means that x 1 And x 2 are the roots of a given quadratic equation.

Answer: x 1 = 2 + 7 2 , x 2 = 2 - 7 2

We can also use the converse of Vieta's theorem to find the roots of a quadratic equation. The simplest way is to select integer roots of the given quadratic equations with integer coefficients. Other options can be considered. But this can significantly complicate calculations.

To select roots, we 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.

Example 2

As an example, we use the quadratic equation x 2 − 5 x + 6 = 0. Numbers x 1 And x 2 can be the roots of this equation if two equalities are satisfied x 1 + x 2 = 5 And x 1 x 2 = 6. Let's select these numbers. These are numbers 2 and 3, since 2 + 3 = 5 And 2 3 = 6. It turns out that 2 and 3 are the roots of this quadratic equation.

The converse of Vieta's theorem can be used to find the second root when the first is known or obvious. To do this, we can use the relations x 1 + x 2 = - b a, x 1 · x 2 = c a.

Example 3

Consider the quadratic equation 512 x 2 − 509 x − 3 = 0. It is necessary to find the roots of this equation.

Solution

The first root of the equation is 1, since the sum of the coefficients of this quadratic equation is zero. It turns out that x 1 = 1.

Now let's find the second root. For this you can use the relation x 1 x 2 = c a. It turns out that 1 x 2 = − 3,512, where x 2 = - 3,512.

Answer: roots of the quadratic equation specified in the problem statement 1 And - 3 512 .

It is possible to select roots using the theorem inverse to Vieta’s theorem only in simple cases. In other cases, it is better to search using the formula for the roots of a quadratic equation through a discriminant.

Thanks to the converse of Vieta's theorem, we can also construct quadratic equations using the existing roots x 1 And x 2. To do this, we need to calculate the sum of the roots, which gives the coefficient for x with the opposite sign of the given quadratic equation, and the product of the roots, which gives the free term.

Example 4

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

Solution

Let's assume that x 1 = − 11 And x 2 = 23. The sum and product of these numbers will be equal: x 1 + x 2 = 12 And x 1 x 2 = − 253. This means that the second coefficient is 12, the free term − 253.

Let's make an equation: x 2 − 12 x − 253 = 0.

Answer: x 2 − 12 x − 253 = 0 .

We can use Vieta's theorem to solve problems that involve the signs of the roots of quadratic equations. The connection between Vieta's theorem is related to the signs of the roots of the reduced quadratic equation x 2 + p x + q = 0 as follows:

• if the quadratic equation has real roots and if the intercept term q is positive number, then these roots will have the same sign “+” or “-”;
• if the quadratic equation has roots and if the intercept term q is a negative number, then one root will be “+”, and the second “-”.

Both of these statements are a consequence of the formula x 1 x 2 = q and rules for multiplying positive and negative numbers, as well as numbers with different signs.

Example 5

Are the roots of a quadratic equation x 2 − 64 x − 21 = 0 positive?

Solution

According to Vieta’s theorem, the roots of this equation cannot both be positive, since they must satisfy the equality x 1 x 2 = − 21. This is impossible with positive x 1 And x 2.

Answer: No

Example 6

At what parameter values r quadratic equation x 2 + (r + 2) x + r − 1 = 0 will have two real roots with different signs.

Solution

Let's start by finding the values ​​of which r, for which the equation will have two roots. Let's find the discriminant and see at what r it will take positive values. D = (r + 2) 2 − 4 1 (r − 1) = r 2 + 4 r + 4 − 4 r + 4 = r 2 + 8. Expression value r 2 + 8 positive for any real r, therefore, the discriminant will be greater than zero for any real r. This means that the original quadratic equation will have two roots for any real values ​​of the parameter r.

Now let's see when the roots will take root different signs. This is possible if their product is negative. According to Vieta's theorem, the product of the roots of the reduced quadratic equation is equal to the free term. Means, the right decision there will be those values r, for which the free term r − 1 is negative. Let's decide linear inequality r − 1< 0 , получаем r < 1 .

Answer: at r< 1 .

Vieta formulas

There are a number of formulas that are applicable to carry out operations with the roots and coefficients of not only quadratic, but also cubic and other types of equations. They are called Vieta's formulas.

For an algebraic equation of degree n of the form a 0 · x n + a 1 · x n - 1 + . . . + a n - 1 x + a n = 0 the equation is considered to have n real roots x 1 , x 2 , … , x n, among which may be the same:
x 1 + x 2 + x 3 + . . . + x n = - a 1 a 0 , x 1 · x 2 + x 1 · x 3 + . . . + x n - 1 · x n = a 2 a 0 , x 1 · x 2 · x 3 + x 1 · x 2 · x 4 + . . . + x n - 2 · x n - 1 · x n = - a 3 a 0 , . . . x 1 · x 2 · x 3 · . . . · x n = (- 1) n · a n a 0

Definition 1

Vieta's formulas help us obtain:

• theorem on the decomposition of a polynomial into linear factors;
• determination of equal polynomials through the equality of all their corresponding coefficients.

Thus, the polynomial a 0 · x n + a 1 · x n - 1 + . . . + a n - 1 · x + a n and its expansion into linear factors of the form a 0 · (x - x 1) · (x - x 2) · . . . · (x - x n) are equal.

If we expand the parentheses in last work and equate the corresponding coefficients, we obtain Vieta’s formulas. Taking n = 2, we can obtain Vieta's formula for the quadratic equation: x 1 + x 2 = - a 1 a 0, x 1 · x 2 = a 2 a 0.

Definition 2

Vieta formula for cubic equation:
x 1 + x 2 + x 3 = - a 1 a 0 , x 1 x 2 + x 1 x 3 + x 2 x 3 = a 2 a 0 , x 1 x 2 x 3 = - a 3 a 0

The left side of the Vieta formula contains the so-called elementary symmetric polynomials.

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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 modern course In school mathematics, such technologies are almost never 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. Suppose 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 or 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 ah, 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, general scheme solving quadratic equations using Vieta’s theorem looks like this:

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.

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 converse theorem they work only with , that is, one whose coefficient in front 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, it is possible to 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 course mathematics have entire solutions.

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

Quadratic equation

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

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

Note that since the maximum power to which x can be raised is two, then the number of roots in the general case is also two.

There are several ways to solve this type of equalities. 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 Viète (French) noticed, while analyzing the properties of the roots of various quadratic equations, that certain combinations of them satisfy specific relationships. In particular, these combinations are their product and sum.

Vieta's theorem establishes the following: the roots of a quadratic equation, when summed, give the ratio of the linear to quadratic coefficients 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 view equation is written as shown in the photo in the previous section of the 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.

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