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.

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

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

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

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

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

Inverse Vieta theorem

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ready. Now we can use both theorems.

Answers to frequently asked questions

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

In this lecture, we will get acquainted with the curious relationships between the roots of a quadratic equation and its coefficients. These relationships were first discovered by the French mathematician Francois Viet (1540-1603).

For example, for the equation Зx 2 - 8x - 6 \u003d 0, without finding its roots, you can, using the Vieta theorem, immediately say that the sum of the roots is , and the product of the roots is
i.e. - 2. And for the equation x 2 - 6x + 8 \u003d 0 we conclude: the sum of the roots is 6, the product of the roots is 8; by the way, it is not difficult to guess what the roots are equal to: 4 and 2.
Proof of Vieta's theorem. The roots x 1 and x 2 of the quadratic equation ax 2 + bx + c \u003d 0 are found by the formulas

Where D \u003d b 2 - 4ac is the discriminant of the equation. Laying down these roots
we get

Now we calculate the product of the roots x 1 and x 2 We have

The second relation is proved:
Comment. Vieta's theorem is also valid in the case when the quadratic equation has one root (that is, when D \u003d 0), it's just that in this case it is considered that the equation has two identical roots, to which the above relations are applied.
The proven relations for the reduced quadratic equation x 2 + px + q \u003d 0 take a particularly simple form. In this case, we get:

x 1 \u003d x 2 \u003d -p, x 1 x 2 \u003d q
those. 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.
Using the Vieta theorem, one can also obtain other relationships between the roots and coefficients of a quadratic equation. Let, for example, x 1 and x 2 be the roots of the reduced quadratic equation x 2 + px + q = 0. Then

However, the main purpose of Vieta's theorem is not that it expresses certain relationships between the roots and coefficients of a quadratic equation. Much more important is the fact that with the help of Vieta's theorem, a formula for factoring a square trinomial is derived, without which we will not do in the future.

Proof. We have

Example 1. Factorize the square trinomial 3x 2 - 10x + 3.
Solution. Having solved the equation Zx 2 - 10x + 3 \u003d 0, we find the roots of the square trinomial Zx 2 - 10x + 3: x 1 \u003d 3, x2 \u003d.
Using Theorem 2, we get

It makes sense instead to write Zx - 1. Then we finally get Zx 2 - 10x + 3 = (x - 3) (3x - 1).
Note that the given square trinomial can be factored without using Theorem 2, using the grouping method:

Zx 2 - 10x + 3 = Zx 2 - 9x - x + 3 =
\u003d Zx (x - 3) - (x - 3) \u003d (x - 3) (Zx - 1).

But, as you can see, with this method success depends on whether we can find a successful grouping or not, while with the first method success is guaranteed.
Example 1. Reduce fraction

Solution. From the equation 2x 2 + 5x + 2 = 0 we find x 1 = - 2,

From the equation x2 - 4x - 12 = 0 we find x 1 = 6, x 2 = -2. That's why
x 2 - 4x - 12 \u003d (x - 6) (x - (- 2)) \u003d (x - 6) (x + 2).
Now let's reduce the given fraction:

Example 3. Factorize expressions:
a) x4 + 5x 2 +6; b) 2x+-3
Solution. a) We introduce a new variable y = x 2 . This will allow us to rewrite the given expression in the form of a square trinomial with respect to the variable y, namely, in the form y 2 + bу + 6.
Having solved the equation y 2 + bу + 6 \u003d 0, we find the roots of the square trinomial y 2 + 5y + 6: y 1 \u003d - 2, y 2 \u003d -3. Now we use Theorem 2; we get

y 2 + 5y + 6 = (y + 2) (y + 3).
It remains to remember that y \u003d x 2, i.e., return to the given expression. So,
x 4 + 5x 2 + 6 \u003d (x 2 + 2) (x 2 + 3).
b) Let's introduce a new variable y = . This will allow you to rewrite the given expression in the form of a square trinomial with respect to the variable y, namely, in the form 2y 2 + y - 3. Having solved the equation
2y 2 + y - 3 = 0, find the roots of the square trinomial 2y 2 + y - 3:
y 1 = 1, y 2 = . Further, using Theorem 2, we obtain:

It remains to remember that y \u003d, i.e., return to the given expression. So,

The section concludes with some considerations, again connected with the Vieta theorem, or rather, with the converse assertion:
if the numbers x 1, x 2 are such that x 1 + x 2 \u003d - p, x 1 x 2 \u003d q, then these numbers are the roots of the equation
Using this statement, you can solve many quadratic equations orally, without using cumbersome root formulas, and also compose quadratic equations with given roots. Let's give examples.

1) x 2 - 11x + 24 = 0. Here x 1 + x 2 = 11, x 1 x 2 = 24. It is easy to guess that x 1 = 8, x 2 = 3.

2) x 2 + 11x + 30 = 0. Here x 1 + x 2 = -11, x 1 x 2 = 30. It is easy to guess that x 1 = -5, x 2 = -6.
Please note: if the free term of the equation is a positive number, then both roots are either positive or negative; this is important to consider when selecting roots.

3) x 2 + x - 12 = 0. Here x 1 + x 2 = -1, x 1 x 2 = -12. It is easy to guess that x 1 \u003d 3, x2 \u003d -4.
Please note: if the free term of the equation is a negative number, then the roots are different in sign; this is important to consider when selecting roots.

4) 5x 2 + 17x - 22 = 0. It is easy to see that x = 1 satisfies the equation, i.e. x 1 \u003d 1 - the root of the equation. Since x 1 x 2 \u003d -, and x 1 \u003d 1, we get that x 2 \u003d -.

5) x 2 - 293x + 2830 = 0. Here x 1 + x 2 = 293, x 1 x 2 = 2830. If you pay attention to the fact that 2830 = 283. 10, and 293 \u003d 283 + 10, then it becomes clear that x 1 \u003d 283, x 2 \u003d 10 (now imagine what calculations would have to be performed to solve this quadratic equation using standard formulas).

6) Let's compose a quadratic equation so that the numbers x 1 \u003d 8, x 2 \u003d - 4 serve as its roots. Usually in such cases they make up the reduced quadratic equation x 2 + px + q \u003d 0.
We have x 1 + x 2 \u003d -p, therefore 8 - 4 \u003d -p, that is, p \u003d -4. Further, x 1 x 2 = q, i.e. 8"(-4) = q, whence we obtain q = -32. So, p \u003d -4, q \u003d -32, which means that the desired quadratic equation has the form x 2 -4x-32 \u003d 0.

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

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

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

Example 1. Solve the equation.

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

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

Answer: 2; 3.

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

write down the assertion of Vieta's theorem

(*)

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

Let's give some more examples.

Example 2: Solve the Equation .

Solution.

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

Answer: -2; -5.

Example 3. Solve the equation .

Solution.

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

Answer: 2; -5.

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

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

Example 4. Solve the equation .

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

Answer: .

Example 5. Solve the equation .

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

Answer: .

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

Example 6. Solve the equation.

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

Answer: .

Example 7. Solve the equation.

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

Answer: ..

Examples for the application of Vieta's theorem

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

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

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

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

Task 3. Solve a quadratic equation using the property.

When studying ways to solve second-order equations in a school algebra course, consider the properties of the roots obtained. They are now known as Vieta's theorems. Examples of its use are given in this article.

Quadratic equation

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

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

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

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

Statement of Vieta's theorem

At the end of the 16th century, the famous mathematician Francois Viet (Frenchman) noticed, 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 the general form of the equation is written as it is shown in the photo in the previous section of the article, then mathematically this theorem can be written as two equalities:

• r 2 + r 1 \u003d -b / a;
• r 1 x r 2 \u003d c / a.

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

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