First, let's formulate the theorem itself: Let's say we have a reduced quadratic equation of the form x^2+b*x + c = 0. Let's say this equation contains roots x1 and x2. Then, by the theorem, the following statements are admissible:

1) The sum of the roots x1 and x2 will be equal to the negative value of the coefficient b.

2) The product of these very roots will give us the coefficient c.

But what is the above equation?

A reduced quadratic equation is a quadratic equation, the coefficient of the highest degree, which is equal to one, i.e. this is an equation of the form x^2 + b*x + c = 0. (and the equation a*x^2 + b*x + c = 0 is not reduced). In other words, to reduce the equation to the reduced form, we must divide this equation by the coefficient at the highest degree (a). The task is to bring this equation to the reduced form:

3*x^2 12*x + 18 = 0;

−4*x^2 + 32*x + 16 = 0;

1.5*x^2 + 7.5*x + 3 = 0; 2*x^2 + 7*x − 11 = 0.

We divide each equation by the coefficient of the highest degree, we get:

X^2 4*x + 6 = 0; X^2 8*x − 4 = 0; X^2 + 5*x + 2 = 0;

X^2 + 3.5*x - 5.5 = 0.

As can be seen from the examples, even equations containing fractions can be reduced to the reduced form.

Using Vieta's Theorem

X^2 5*x + 6 = 0 ⇒ x1 + x2 = − (−5) = 5; x1*x2 = 6;

we get the roots: x1 = 2; x2 = 3;

X^2 + 6*x + 8 = 0 ⇒ x1 + x2 = −6; x1*x2 = 8;

as a result, we get the roots: x1 = -2; x2 = -4;

X^2 + 5*x + 4 = 0 ⇒ x1 + x2 = −5; x1*x2 = 4;

we get the roots: x1 = −1; x2 = −4.

Significance of Vieta's theorem

Vieta's theorem allows us to solve any given quadratic equation in almost seconds. At first glance, this seems like a rather difficult task, but after 5 10 equations, you can learn to see the roots right away.

From the above examples, and using the theorem, you can see how you can significantly simplify the solution of quadratic equations, because using this theorem, you can solve a quadratic equation with little or no complex calculations and calculating the discriminant, and as you know, the fewer calculations, the more difficult it is to make a mistake, which is important.

In all examples, we have used this rule based on two important assumptions:

The above equation, i.e. the coefficient at the highest degree is equal to one (this condition is easy to avoid. You can use the unreduced form of the equation, then the following statements x1+x2=-b/a; x1*x2=c/a will be valid, but usually it is more difficult to solve :))

When the equation will have two different roots. We assume that the inequality is true and the discriminant is strictly greater than zero.

Therefore, we can compose a general solution algorithm using Vieta's theorem.

General solution algorithm by Vieta's theorem

We bring the quadratic equation to the reduced form if the equation is given to us in the unreduced form. When the coefficients in the quadratic equation, which we previously presented as reduced, turned out to be fractional (not decimal), then in this case our equation should be solved through the discriminant.

There are also cases where going back to the original equation allows us to work with "convenient" numbers.

In mathematics, there are special tricks with which many quadratic equations are solved very quickly and without any discriminants. Moreover, with proper training, many begin to solve quadratic equations verbally, literally "at a glance."

Unfortunately, in the modern course of school mathematics, such technologies are almost not studied. And you need to know! And today we will consider 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 at x 2 is equal to 1. There are no other restrictions on the coefficients.

1. x 2 + 7x + 12 = 0 is the reduced quadratic equation;
2. x 2 − 5x + 6 = 0 is also reduced;
3. 2x 2 − 6x + 8 = 0 - but this is nothing reduced, since the coefficient at x 2 is 2.

Of course, any quadratic equation of the form ax 2 + bx + c = 0 can be made reduced - it is enough to 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. A little lower, we will make sure that this should be done only when in the final squared equation all the coefficients are integer. For now, let's look at some simple examples:

Task. Convert quadratic equation to reduced:

1. 3x2 − 12x + 18 = 0;
2. −4x2 + 32x + 16 = 0;
3. 1.5x2 + 7.5x + 3 = 0;
4. 2x2 + 7x − 11 = 0.

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

1. 3x 2 - 12x + 18 \u003d 0 ⇒ x 2 - 4x + 6 \u003d 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 \u003d 0 ⇒ x 2 + 5x + 2 \u003d 0 - divided by 1.5, all coefficients became integer;
4. 2x 2 + 7x - 11 \u003d 0 ⇒ x 2 + 3.5x - 5.5 \u003d 0 - divided by 2. In this case, fractional coefficients arose.

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

Now we 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 \u003d 0. Suppose that this equation has real roots x 1 and x 2. In this case, the following statements are true:

1. x1 + x2 = −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 given 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 \u003d 5;
2. x 2 + 2x − 15 = 0 ⇒ x 1 + x 2 = −2; x 1 x 2 \u003d -15; roots: x 1 = 3; x 2 \u003d -5;
3. x 2 + 5x + 4 = 0 ⇒ x 1 + x 2 = −5; x 1 x 2 = 4; roots: x 1 \u003d -1; x 2 \u003d -4.

Vieta's theorem gives us additional information about the roots of a quadratic equation. At first glance, this may seem complicated, 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. x2 − 9x + 14 = 0;
2. x 2 - 12x + 27 = 0;
3. 3x2 + 33x + 30 = 0;
4. −7x2 + 77x − 210 = 0.

Let's try to write down the coefficients according to the Vieta theorem and "guess" the roots:

1. x 2 − 9x + 14 = 0 is a reduced quadratic equation.
   By the Vieta 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 is also reduced.
   By the Vieta 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 fix this now by dividing both sides of the equation by the coefficient a \u003d 3. We get: x 2 + 11x + 10 \u003d 0.
   We solve according to the Vieta theorem: x 1 + x 2 = −11; x 1 x 2 = 10 ⇒ roots: −10 and −1;
4. −7x 2 + 77x − 210 \u003d 0 - again the coefficient at 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 the Vieta 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 can be seen how Vieta's theorem simplifies the solution of quadratic equations. No complicated calculations, no arithmetic roots and fractions. And even the discriminant (see the lesson " Solving quadratic equations") We did not need.

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

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

However, in typical mathematical problems these conditions are met. If the result of the calculations is a “bad” quadratic equation (the coefficient at x 2 is different from 1), this is easy to fix - take a look at the examples at the very beginning of the lesson. I am generally silent about the roots: what kind of task is this in which there is no answer? Of course there will be roots.

Thus, the general scheme for solving quadratic equations according to the Vieta theorem is as follows:

1. Reduce the quadratic equation to the given one, if this has not already been done in the condition of the problem;
2. If the coefficients in the above quadratic equation turned out to be fractional, we solve through the discriminant. You can even go back to the original equation to work with more "convenient" numbers;
3. In the case of integer coefficients, we solve the equation using the Vieta theorem;
4. If within a few seconds it was not possible to guess the roots, we score on the Vieta theorem and solve through the discriminant.

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

So, we have an equation that is not reduced, because coefficient a \u003d 5. Divide everything by 5, we get: x 2 - 7x + 10 \u003d 0.

All coefficients of the 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 \u003d 10. In this case, the roots are easy to guess - these are 2 and 5. You do not need to count through the discriminant.

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

We look: −5x 2 + 8x − 2.4 = 0 - this equation is not reduced, we divide both sides by the coefficient a = −5. We get: x 2 - 1.6x + 0.48 \u003d 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 \u003d 0.4.

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

To begin with, we divide everything by the coefficient a \u003d 2. We get the equation x 2 + 5x - 300 \u003d 0.

This is the reduced equation, according to the Vieta theorem we have: x 1 + x 2 = −5; x 1 x 2 \u003d -300. It is difficult to guess the roots of the quadratic equation in this case - personally, I seriously "froze" when I solved this problem.

We 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 \u003d 15; x 2 \u003d -20.

Before proceeding to Vieta's theorem, we introduce a definition. Quadratic equation of the form x² + px + q= 0 is called reduced. In this equation, the leading coefficient is equal to one. For example, the equation x² - 3 x- 4 = 0 is reduced. Any quadratic equation of the form ax² + b x + c= 0 can be made reduced, for this we divide both sides of the equation by A≠ 0. For example, Equation 4 x² + 4 x- 3 \u003d 0 divided by 4 is reduced to the form: x² + x- 3/4 = 0. We derive the formula for the roots of the reduced quadratic equation, for this we use the formula for the roots of a general quadratic equation: ax² + bx + c = 0

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

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

In response, we write the equation has two roots.

For a reduced quadratic equation with positive, the following theorem holds.

Vieta's theorem

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

x 1 + x 2 = — R

x 1 * x 2 \u003d q, that is, the sum of the roots of the given quadratic equation is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term.

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

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

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

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

Not solving equations x² - 13 x+ 30 = 0 find the sum and product of its roots x 1 and x 2. this equation D\u003d 169 - 120 \u003d 49\u003e 0, so you can apply the Vieta theorem: x 1 + x 2 = 13, x 1 * x 2 = 30. Consider a few more examples. One of the roots of the equation x² — px- 12 = 0 is x 1 = 4. Find coefficient R and second root x 2 of this equation. According to Vieta's theorem x 1 * x 2 =— 12, x 1 + x 2 = — R. Because x 1 = 4 then 4 x 2 = - 12, whence x 2 = — 3, R = — (x 1 + x 2) \u003d - (4 - 3) \u003d - 1. In response, we write down the second root x 2 = - 3, coefficient p = - 1.

Not solving equations x² + 2 x- 4 = 0 find the sum of the squares of its roots. Let x 1 and x 2 are the roots of the equation. According to Vieta's theorem x 1 + x 2 = — 2, x 1 * x 2 = - 4. Because x 1²+ x 2² = ( x 1 + x 2)² - 2 x 1 x 2 , then x 1²+ x 2 ² \u003d (- 2) ² -2 (- 4) \u003d 12.

Find the sum and product of the roots of equation 3 x² + 4 x- 5 \u003d 0. This equation has two different roots, since the discriminant D= 16 + 4*3*5 > 0. To solve the equation, we use the Vieta theorem. This theorem has been proved for the reduced quadratic equation. So let's divide this equation by 3.

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

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

Theorem inverse to Vieta's theorem

If numbers R, q, x 1 , x 2 are such that x 1 + x 2 = — p, x 1 * x 2 \u003d q, That x 1 And x2 are the roots of the equation x² + px + q= 0. Substitute in the left side x² + px + q instead of R expression - ( x 1 + x 2), but instead q- work x 1 * x 2 . We get: x² + px + q = x² — ( x 1 + x 2) x + x 1 x 2 \u003d x² - x 1 x - x 2 x + x 1 x 2 \u003d (x - x 1) (x - x 2). Thus, if the numbers R, q, x 1 and x 2 are related by these relations, then for all X equality x² + px + q = (x - x 1) (x - x 2), from which it follows that x 1 and x 2 - roots of the equation x² + px + q= 0. Using the theorem converse to Vieta's theorem, it is sometimes possible to find the roots of a quadratic equation by selection. Consider an example, x² - 5 x+ 6 = 0. Here R = — 5, q= 6. Pick two numbers x 1 and x 2 so that x 1 + x 2 = 5, x 1 * x 2 = 6. Noting that 6 = 2 * 3, and 2 + 3 = 5, by the theorem converse to Vieta's theorem, we obtain that x 1 = 2, x 2 = 3 - roots of the equation x² - 5 x + 6 = 0.

One of the methods for solving a quadratic equation is the application VIETA formulas, which was named after FRANCOIS VIETE.

He was a famous lawyer, and served in the 16th century with the French king. In his free time he studied astronomy and mathematics. He established a connection between the roots and coefficients of a quadratic equation.

Advantages of the formula:

1 . By applying the formula, you can quickly find the solution. Because you do not need to enter the second coefficient into the square, then subtract 4ac from it, find the discriminant, substitute its value into the formula for finding the roots.

2 . Without a solution, you can determine the signs of the roots, pick up the values ​​of the roots.

3 . Having solved the system of two records, it is not difficult to find the roots themselves. In the above quadratic equation, the sum of the roots is equal to the value of the second coefficient with a minus sign. The product of the roots in the above quadratic equation is equal to the value of the third coefficient.

4 . According to the given roots, write a quadratic equation, that is, solve the inverse problem. For example, this method is used in solving problems in theoretical mechanics.

5 . It is convenient to apply the formula when the leading coefficient is equal to one.

Flaws:

1 . The formula is not universal.

Vieta's theorem Grade 8

Formula
If x 1 and x 2 are the roots of the given quadratic equation x 2 + px + q \u003d 0, then:

Examples
x 1 \u003d -1; x 2 \u003d 3 - the roots of the equation x 2 - 2x - 3 \u003d 0.

P = -2, q = -3.

X 1 + x 2 \u003d -1 + 3 \u003d 2 \u003d -p,

X 1 x 2 = -1 3 = -3 = q.

Inverse theorem

Formula
If the numbers x 1 , x 2 , p, q are connected by the conditions:

Then x 1 and x 2 are the roots of the equation x 2 + px + q = 0.

Example
Let's make a quadratic equation by its roots:

X 1 \u003d 2 -? 3 and x 2 \u003d 2 +? 3 .

P \u003d x 1 + x 2 \u003d 4; p = -4; q \u003d x 1 x 2 \u003d (2 -? 3) (2 +? 3) \u003d 4 - 3 \u003d 1.

The desired equation has the form: x 2 - 4x + 1 = 0.

There are a number of relationships in quadratic equations. The main ones are the relations between roots and coefficients. Also, a number of relationships work in quadratic equations, which are given by the Vieta theorem.

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

Statement and proof of Vieta's theorem

The formula for the roots of a quadratic equation a x 2 + b x + c = 0 of the form x 1 \u003d - b + D 2 a, x 2 \u003d - b - D 2 a, where D = b 2 − 4 a c, establishes the ratio x 1 + x 2 \u003d - 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 x2- 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 \u003d - b a, x 1 x 2 = c a.

Proof 1

We offer you the following scheme for conducting the proof: we take the formula of the 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.

Compose the sum of the roots x 1 + x 2 \u003d - b + D 2 a + - b - D 2 a. Let's bring the fractions to a common denominator - b + D 2 · a + - b - D 2 · a = - b + D + - b - D 2 · a. Let's open the brackets in the numerator of the resulting fraction and give similar terms: - b + D + - b - D 2 a = - b + D - b - D 2 a = - 2 b 2 a . Reduce the fraction by: 2 - b a \u003d - b a.

So we have proved the first relation of Vieta's theorem, which refers to the sum of the roots of a quadratic equation.

Now let's move on to the second relation.

To do this, we need to compose the product of the roots of the quadratic equation: x 1 x 2 \u003d - b + D 2 a - b - D 2 a.

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

We will carry out the multiplication of the bracket by the bracket in the numerator of the fraction, or we will use the formula of the difference of squares in order to transform this product faster: - b + D · - b - D 4 · a 2 = - b 2 - D 2 4 · a 2 .

Let's use definition of a square root to carry out 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 the quadratic equation, therefore, into a fraction instead of D can be substituted b 2 − 4 a c:

b 2 - D 4 a 2 \u003d b 2 - (b 2 - 4 a c) 4 a 2

Let's open the brackets, give like terms and get: 4 · a · c 4 · a 2 . If we shorten it to 4 a, then c a remains. So we have proved the second relation of the Vieta theorem for the product of roots.

The record of the proof of Vieta's theorem can have a very concise form, if we omit the explanations:

x 1 + x 2 \u003d - b + D 2 a + - b - D 2 a \u003d - b + D + - b - D 2 a \u003d - 2 b 2 a \u003d - 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 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, at D=0 the root of the quadratic equation is: - b 2 a, then x 1 + x 2 \u003d - b 2 a + - b 2 a \u003d - b + (- b) 2 a \u003d - 2 b 2 a \u003d - b a and x 1 x 2 \u003d - b 2 a - b 2 a \u003d - b - b 4 a 2 \u003d b 2 4 a 2, and since D \u003d 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, the Vieta theorem is applied in relation 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 precisely 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, it is necessary to divide both its parts by the number a, which is different from zero.

Let us give one more formulation of Vieta's theorem.

Theorem 2

The sum of the roots in the given quadratic equation x 2 + p x + q = 0 will be equal to the coefficient at 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 \u003d - p, x 1 x 2 \u003d q.

Theorem inverse to Vieta's theorem

If you look closely at the second formulation of Vieta's theorem, you can see that for the roots x 1 And x2 reduced quadratic equation x 2 + p x + q = 0 relations x 1 + x 2 = − p , x 1 · x 2 = q will be valid. From these relations x 1 + x 2 \u003d - p, x 1 x 2 \u003d q, it follows that x 1 And x2 are the roots of the quadratic equation x 2 + p x + q = 0. Thus we arrive at a statement which is the inverse of Vieta's theorem.

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

Theorem 3

If numbers x 1 And x2 are such that x 1 + x 2 = − p And x 1 x 2 = q, That x 1 And x2 are the roots of the reduced quadratic equation x 2 + p x + q = 0.

Proof 2

Change of coefficients p And q to their expression through x 1 And x2 allows you to transform the equation x 2 + p x + q = 0 in 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 equality for any x 1 And x2 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. It 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.

Equation Substitution x 2 − (x 1 + x 2) x + x 1 x 2 = 0 numbers x2 instead of x allows you to get 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 x2 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 theorem converse to Vieta's theorem is proved.

Examples of using Vieta's theorem

Let's now proceed to the analysis of the most typical examples on the topic. Let's start with the analysis of problems that require the application of the theorem, the converse of Vieta's theorem. It can be used to check the numbers obtained in the course of calculations, whether 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 ratios x 1 + x 2 = - b a, x 1 x 2 = a c.

The fulfillment of both relations indicates that the numbers obtained in the course of 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 condition of the problem.

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 the quadratic equation 4 x 2 − 16 x + 9 = 0?

Solution

Find the coefficients of the quadratic equation 4 x 2 − 16 x + 9 = 0 . This is a = 4 , b = − 16 , c = 9 . In accordance with the Vieta 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 should be equal to 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 , so you don't need to continue checking. According to the theorem, the inverse of 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 \u003d 1 - 3 3 + 3 \u003d 3 + 3 - 3 3 - 3 \u003d - 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 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 satisfied, which means that x 1 And x2 are the roots of the given quadratic equation.

Answer: x 1 \u003d 2 + 7 2, x 2 \u003d 2 - 7 2

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

To select the roots, we 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.

Example 2

As an example, we use the quadratic equation x 2 − 5 x + 6 = 0. Numbers x 1 And x2 can be the roots of this equation if the two equalities are satisfied x1 + x2 = 5 And x 1 x 2 = 6. Let's pick those numbers. These are the numbers 2 and 3 because 2 + 3 = 5 And 2 3 = 6. It turns out that 2 and 3 are the roots of this quadratic equation.

The inverse of Vieta's theorem can be used to find the second root when the first is known or obvious. For this we can use the ratios 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. We need to find the roots of this equation.

Solution

The first root of the equation is 1 because 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. To do this, you can use the ratio x 1 x 2 = c a. It turns out that 1 x 2 = − 3 512, where x 2 \u003d - 3 512.

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

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

Thanks to the inverse theorem of Vieta, we can also form quadratic equations given the roots x 1 And x2. To do this, we need to calculate the sum of the roots, which gives the coefficient at x with the opposite sign of the reduced quadratic equation, and the product of the roots, which gives the free term.

Example 4

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

Solution

Let's accept that x 1 = − 11 And x2 = 23. The sum and product of these numbers will be equal to: x1 + x2 = 12 And x 1 x 2 = − 253. This means that the second coefficient is 12, the free term − 253.

We make an equation: x 2 - 12 x - 253 = 0.

Answer: x 2 - 12 x - 253 = 0 .

We can use the Vieta theorem to solve problems that are related to 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 in the following way:

• if the quadratic equation has real roots and if the free term q is a positive number, then these roots will have the same sign "+" or "-";
• if the quadratic equation has roots and if the free 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 multiplication rules for 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

By 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 not possible with positive x 1 And x2.

Answer: No

Example 6

At what values ​​of the parameter 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 what r, for which the equation has two roots. Let us find the discriminant and see for 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 r2 + 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 have different signs. This is possible if their product is negative. According to the Vieta theorem, the product of the roots of the reduced quadratic equation is equal to the free term. So the correct solution is those values r, for which the free term r − 1 is negative. We solve the linear inequality r − 1< 0 , получаем r < 1 .

Answer: at r< 1 .

Vieta formulas

There are a number of formulas that are applicable for performing operations with roots and coefficients of not only square, but also cubic and other types of equations. They are called Vieta 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, which may include the following:
x 1 + x 2 + x 3 + . . . + x n \u003d - 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

Get the Vieta formulas help us:

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

So, 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 open the brackets in the last product and equate the corresponding coefficients, then we get the Vieta formulas. Taking n \u003d 2, we can get the Vieta formula for the quadratic equation: x 1 + x 2 \u003d - a 1 a 0, x 1 x 2 \u003d a 2 a 0.

Definition 2

Vieta's formula for a 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 formulas contains the so-called elementary symmetric polynomials.

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