Reverse Vieta Theorem for a Square Equation. Vieta theorem. Examples of solutions. Value Vieta Theorem

Virtually any square equation \\ can be converted to mind \\ However, this is possible if it is initially divided into each term to the ratio \\ before \\ besides, you can enter a new designation:

\\ [(\\ FRAC (B) (A)) \u003d p \\] and \\ [(\\ FRAC (C) (A)) \u003d Q \\]

Due to which we will have the equation \\ referred to in mathematics with a given square equation. The roots of this equation and coefficients \\ are interrelated, which is confirmed by the Vieta theorem.

Vieta Theorem: The amount of the roots of the given square equation \\ is equal to the second coefficient \\ taken with the opposite sign, and the product of the roots is a free member \\

For clarity, solve the equation of the following form:

We will solve this square equation using the prescribed rules. After analyzing the source data, it can be concluded that the equation will have two different roots, because:

Now, from all the multipliers of the number 15 (1 and 15, 3 and 5), we choose those whose difference is 2. Under this condition, numbers 3 and 5. Before a smaller number, put the "minus" sign. Thus, we obtain the roots of the equation \\

Answer: \\ [x_1 \u003d -3 and x_2 \u003d 5 \\]

Where can I solve the equation on the Vieta theorem online?

You can solve the equation on our website https: // Site. A free online solver will solve the online equation of any complexity in seconds. All you need to do is just enter your data in the solver. You can also watch the video instruction and learn how to solve the equation on our website. And if you have any questions, you can ask them in our Vkontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

When studying how to solve the second-order equations in the school year of algebra, consider the properties of the roots received. They are currently known under the name of the Vieta theorem. Examples of using it are given in this article.

Quadratic equation

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

Here, the symbols A, B, C are some numbers wearing the name of the coefficients of the equation under consideration. To solve equality, it is necessary to find such x values \u200b\u200bthat make it true.

Note that since the maximum value of the degree into which the X is erected is to two, then the number of roots in the general case is also two.

To solve this type of equations there are several ways. In this article, consider one of them, which involves the use of the so-called Vieta theorem.

The wording of the Vieta Theorem

At the end of the XVI, the well-known mathematician Francois Vieta (French) noticed by analyzing the properties of the roots of various square equations that their specific combinations satisfy specific ratios. In particular, these combinations are their work and amount.

The Vieta Theorem sets the following: The roots of the square equation under their sum give the ratio of the linear coefficients to the quadratic taken with the opposite sign, and when they are produced, they lead to the ratio of a free member to the quadratic coefficient.

If the general view of the equation is recorded as presented in the photo in the previous section of the article, then mathematically this theorem can be recorded in the form of 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 equation under consideration.

These equalities can be used to solve a number of various mathematical tasks. The use of the Vieta Theorem in the examples with the solution is given in the following sections of the article.

Francois Viet (1540-1603) - Mathematics, creator of famous formulas

Vieta theorem Need to quickly solve square equations (simple words).

If more details, then yeorema Vieta is the sum of the roots of this square equation equates to the second coefficient, which is taken with the opposite sign, and the work is equal to a free member. This property has any given square equation that has roots.

With the help of the Vieta Theorem, you can easily solve square equations by selecting, so let's say "thanks" to this math with a sword in the hands of our happy grade 7.

Proof of Vietorem Vieta

To prove to theorem, you can use the known root formulas, thanks to which we will make up the amount and product of the roots of the square equation. Only after that we can make sure that they are equal and, accordingly,.

Suppose we have an equation :. This equation has such roots: and. We prove that.

According to the roots of the square equation:

1. Find the amount of the roots:

We will analyze this equation, as we succeeded exactly:

= .

Step 1. We give a fraction for a common denominator, it turns out:

= = .

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

Reduce the fraction on 2 and get:

We have proved the ratio for the roots of the square equation on the Vieta theorem.

2. Find the product of the roots:

= = = = = .

Let us prove this equation:

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

Now I remember the definition of a square root and consider:

= .

Step 3.. Remember the discriminant of the square equation :. Therefore, in the last fraction instead of D (discriminant) we substitute, then it turns out:

= .

Step 4.. We reveal brackets and give such components to the fraction:

Step 5.. Reduce "4a" and get.

So we have proven the ratio for the product of the roots on the Vieta theorem.

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

Theorem, reverse theorem of Vieta

By theorem, the reverse theorem, the Vieta can be checked if our equation is solved correctly. To understand the theorem itself, you need to consider it in more detail.

If numbers and such:

And then they are the roots of the square equation.

Proof of the reverse Vieta Theorem

Step 1. We substitute the expression equation for its coefficients:

Step 2. We transform the left part of the equation:

Step 3.. We will find the roots of the equation, and for this we use the property of the equality of the work zero:

Or . From where it turns out: or.

Examples with solutions on Vieta Theorem

Example 1.

The task

Find the amount, the product and the sum of the squares of the roots of the square equation, without finding the roots of the equation.

Decision

Step 1. Recall the formula of the discriminant. We substitute our numbers under the letters. That is, it replaces, but. This implies:

It turns out:

Title \u003d "(! Lang: Rendered by QuickTextEx.com" height="13" width="170" style="vertical-align: -1px;">. Если дискриминант больше нуля, тогда у уравнения есть корни. По теореме Виета их сумма , а произведение . !}

Express the sum of the squares of the roots through their sum and the work:

Answer

7; 12; 25.

Example 2.

The task

Decide equation. In this case, do not apply the formulas of the square equation.

Decision

This equation has roots that are scratch more than zero by discriminator (D). Accordingly, according to the theorem, the sum of the roots of this equation is 4, and the product - 5. First, we determine the dividers of the number, the sum of which is equal to 4. These are the numbers "5" and "-1". Their product is equal to 5, and the amount - 4. So, by theorem, the reverse theorem of the Vieta, they are roots of this equation.

Answer

AND Example 4.

The task

Make the equation, each root of which is twice as a corresponding root of the equation:

Decision

By the theorem, the sum of the roots of this equation is 12, and the product \u003d 7. So, the two roots are positive.

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

And the work.

By theorem, the reverse theorem of the Vieta, the new equation is:

Answer

It turned out the equation, each root of which is twice as many:

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

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

Useful sources:

Dorofeyev G.V., Suvorova S. B., Baynovich E. A. Algebra Grade 8: Moscow "Education", 2016 - 318 p.
Rubin A. G., Chulkov P.V. - Tutorial Algebra Grade 8: Moscow "Balam", 2015 - 237 p.
Nikolsky S. M., Pokolov M. K., Reshetnikov N. N., Shevkin A. V. - Algebra Grade 8: Moscow "Education", 2014 - 300

Vieta theorem, reverse Vieta formula and examples with solutions for teapots Updated: November 22, 2019 by the author: Scientific articles.ru.

The Vieta Theorem is often used to check the roots already found. If you have found roots, you can use formulas \\ (\\ begin (Cases) x_1 + x_2 \u003d -p \\\\ x_1 \\ cdot x_2 \u003d q \\ END (Cases) \\) calculate the values \u200b\u200b\\ (p \\) and \\ (q \\ And if they are obtained by the same as in the source equation - it means the roots are found correctly.

For example, let us, using, solved the equation \\ (x ^ 2 + x-56 \u003d 0 \\) and the roots were obtained: \\ (x_1 \u003d 7 \\), \\ (x_2 \u003d -8 \\). Check if we were not mistaken in the process of solving. In our case \\ (p \u003d 1 \\), and \\ (q \u003d -56 \\). By Viet's theorem, we have:

\\ (\\ Begin (Cases) x_1 + x_2 \u003d -p \\\\ x_1 \\ cdot x_2 \u003d q \\ end (cases) \\ (\\ leftrightarrow \\) \\ (\\ begin (cases) 7 + (- 8) \u003d - 1 \\\\ 7 \\ CDOT (-8) \u003d - 56 \\ END (CASES) \\ (\\ leftrightarrow \\) \\ (\\ begin (Cases) -1 \u003d -1 \\\\ - 56 \u003d -56 \\ END (Cases) \\ Both statements agreed, it means that we solved the equation correctly.

Such a check can be performed orally. It will take 5 seconds and keeps you from stupid mistakes.

Reverse Vieta Theorem

If \\ (\\ begin (cases) x_1 + x_2 \u003d -p \\\\ x_1 \\ cdot x_2 \u003d q \\ end (Cases) \\), then \\ (x_1 \\) and \\ (x_2 \\) - the roots of the square equation \\ (x ^ 2 + px + Q \u003d 0 \\).

Or in a simple way: if you have a view equation \\ (x ^ 2 + px + q \u003d 0 \\), then solving the system \\ (\\ begin (Cases) x_1 + x_2 \u003d -p \\\\ x_1 \\ cdot x_2 \u003d q \\

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Due to this theorem, you can quickly pick up the roots of the square 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 \u003d 0 \\).

Decision : Using the reverse theorem of the West, we obtain that the roots satisfy the conditions: \\ (\\ begin (Cases) x_1 + x_2 \u003d 5 \\\\ x_1 \\ cdot x_2 \u003d 6 \\ end (Cases) \\).
Look at the second equation of the system \\ (x_1 \\ cdot x_2 \u003d 6 \\). What two can you decompose the number \\ (6 \\)? On \\ (2 \\) and \\ (3 \\), \\ (6 \\) and \\ (1 \\) or \\ (- 2 \\) and \\ (- 3 \\), and \\ (- 6 \\) and \\ (- one\\). And what pair to choose, the first system equation will prompt: \\ (x_1 + x_2 \u003d 5 \\). Side \\ (2 \\) and \\ (3 \\), since \\ (2 + 3 \u003d 5 \\).
Answer : \\ (x_1 \u003d 2 \\), \\ (x_2 \u003d 3 \\).

Examples . Using theorem, reverse theorem of Vieta, find the roots of the square equation:
a) \\ (x ^ 2-15x + 14 \u003d 0 \\); b) \\ (x ^ 2 + 3x-4 \u003d 0 \\); c) \\ (x ^ 2 + 9x + 20 \u003d 0 \\); d) \\ (x ^ 2-88x + 780 \u003d 0 \\).

Decision :
a) \\ (x ^ 2-15x + 14 \u003d 0 \\) - on what faults is unfolded \\ (14 \\)? \\ (2 \\) and \\ (7 \\), \\ (- 2 \\) and \\ (- 7 \\), \\ (- 1 \\) and \\ (- 14 \\), \\ (1 \\) and \\ (14 \\ What pairs of numbers in the amount will give \\ (15 \\)? Answer: \\ (1 \\) and \\ (14 \\).

b) \\ (x ^ 2 + 3x-4 \u003d 0 \\) - on what factors unfolding \\ (- 4 \\)? \\ (- 2 \\) and \\ (2 \\), \\ (4 \\) and \\ (- 1 \\), \\ (1 \\) and \\ (- 4 \\). What pairs of numbers in the amount will give \\ (- 3 \\)? Answer: \\ (1 \\) and \\ (- 4 \\).

c) \\ (x ^ 2 + 9x + 20 \u003d 0 \\) - on what faults is unfolded \\ (20 \\)? \\ (4 \\) and \\ (5 \\), \\ (- 4 \\) and \\ (- 5 \\), \\ (2 \\) and \\ (10 \u200b\u200b\\), \\ (- 2 \\) and \\ (- 10 \\ What pairs of numbers in the amount will give \\ (- 9 \\)? Answer: \\ (- 4 \\) and \\ (- 5 \\).

d) \\ (x ^ 2-88x + 780 \u003d 0 \\) - to which factors unfolded \\ (780 \\)? \\ (390 \\) and \\ (2 \\). Did they give \\ (88 \\)? Not. Which factors have \\ (780 \\)? \\ (78 \\) and \\ (10 \u200b\u200b\\). Did they give \\ (88 \\)? Yes. Answer: \\ (78 \\) and \\ (10 \u200b\u200b\\).

Optionally, the last term lay out to all possible multipliers (as in the last example). You can immediately check whether their sum \\ (- p \\).

Important! The Vieta Theorem and the reverse theorem work only C, that is, in which the coefficient before \\ (x ^ 2 \\) is equal to one. If we have initially not given the equation, we can make it given by simply separating to the coefficient facing \\ (x ^ 2 \\).

for example, let the equation \\ (2x ^ 2-4x-6 \u003d 0 \\) and we want to take advantage of one of the Vieta theorems. But we cannot, since the coefficient before \\ (x ^ 2 \\) is equal to \\ (2 \\). Get rid of it, dividing the entire equation on \\ (2 \\).

\\ (2x ^ 2-4x-6 \u003d 0 \\) \\ (|: 2 \\)
\\ (x ^ 2-2x-3 \u003d 0 \\)

Ready. Now you can use both theorems.

Answers to frequently asked questions

Question: On the Vieta Theorem, you can solve any?
Answer: Unfortunately no. If the equation is not integer or the equation does not have roots at all, the Vieta Theorem does not help. In this case, you need to use discriminant . Fortunately, 80% of the equations in the school course of mathematics have solutions.

In the eighth grade, students get acquainted with square equations and their solutions. At the same time, as experience shows, most students in solving full square equations apply only one method - the formula of the roots of the square equation. For students who speak well-owned oral account skills, this method is clearly irregular. Spelling square equations are often and in high schools, and there spend time on the calculation of the discriminant is simply sorry. In my opinion, when studying square equations, more time and attention should be paid to the use of the Vieta theorem (according to the program A.G. Mordkovich Algebra-8, to study the theme "Theorem of the Vieta. Decomposition of the square three decar on linear multipliers" is scheduled only two hours).

In most textbooks, algebra This theorem is formulated for the given square equation and states that if the equation has a root and, then the equality is performed for them.Then the assertion inverse to the Vieta Theorem is formulated, and a number of examples are offered to work out this topic.

Take specific examples And let's follow the logic of solutions using the Vieta theorem.

Example 1. Solve equation.

Suppose this equation has a root, namely, and. Then, on the Vieta Theorem, equality should be performed at the same time.

We note that the product of the roots is a positive number. So, the roots of the equation of one sign. And since the amount 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 the roots. Suppose the roots of the equation are whole positive numbers. Then get the right first equality can be possible in two ways (with an accuracy of the procedure of multipliers): or. Check for the proposed pairs of numbers feasibility of the second approval of the Vieta Theorem: . Thus, numbers 2 and 3 are satisfying both equalities, which means that the roots of a given equation are roots.

Answer: 2; 3.

We highlight the main stages of reasoning in solving the given square equation using the Vieta Theorem:

write a statement of Vietaorem

(*)

determine the signs of the roots of the equation (if the product and the amount of the roots are positive, then both roots are positive numbers. If the product of the roots is a positive number, and the amount of the roots is negative, then both roots are negative numbers. If the product is a negative number, then the roots have roots Different signs. At the same time, if the amount of the roots is positive, then the root is more than the module is a positive number, and if the amount of the roots is less than zero, the root is larger in the module - a negative number);
choose a pair of integers, the product of which gives the right first equality in the record (*);
from the found pairs of numbers to choose the pair, which when substituting in the second equality in the record (*) will give faithful equality;
specify in response found roots equation.

Let's give more examples.

Example 2. Decide equation .

Decision.

Let both be the roots of a given equation. Then, on the Vieta Theorem, we note that the work is positive, and the amount is a negative number. So, both roots are negative numbers. We select multiplier pairs giving a product 10 (-1 and -10; -2 and -5). The second pair of numbers in the amount gives -7. So, the numbers -2 and -5 are the roots of this equation.

Answer: -2; -5.

Example 3. Decide the equation .

Decision.

Let both be the roots of a given equation. Then, on the Vieta Theorem, we note that the work is negative. So roots - different signs. The amount of the roots is also a negative number. It means that the root is larger in the module - negative. We select pairs of multipliers giving a product -10 (1 and -10; 2 and -5). The second pair of numbers in the amount gives -3. So, numbers 2 and -5 are the roots of this equation.

Answer: 2; -5.

Note that the theorem of the Vieta in principle can be formulated for a complete square equation: if the square equation It has roots and, then equality are performed for them. However, the use of this theorem is rather problematic, since in a complete square equation at least one of the roots (with their presence, of course) is a fractional number. And working with the selection of fractions for a long time and difficult. But still there is a way out.

Consider a complete square equation . Multiply both parts of the equation for the first coefficient butand write the equation in the form . We introduce a new variable and we obtain the reduced square equation, the roots of which and (if available) can be found on the Vieta theorem. Then the roots of the source equation will be. We note that the auxiliary reduced equation is very simple: the second coefficient is preserved, and the third coefficient is equal to the work aC. With a certain skill, students immediately constitute the auxiliary equation, they find its roots along the Vieta theorem and indicate the roots of a given complete equation. We give examples.

Example 4. Decide the equation .

Make an auxiliary equation And on the theorem of Vieta we will find its roots. So, the roots of the source equation .

Answer: .

Example 5. Decide the equation .

The auxiliary equation is viewed. By the theorem of Vieta its roots. We find the roots of the source equation .

Answer: .

And one more case, when the use of the Vieta Theorem allows you to orally find the roots of a complete square equation. It is easy to prove that number 1 is the root of the equation then and only when. The second root of the equation is located on the Vieta theorem and is equal. Another statement: so that the number -1 was the root of the equation it is necessary and enough to. Then the second root of the equation on the Vieta Theorem is equal. Similar assertions can be formulated for the given square equation.

Example 6. Decide the equation.

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