Examples of Vieta's theorem. Vieta's theorem for quadratic and other equations

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 interrelated, which is confirmed by Vieta’s theorem.

Vieta's theorem: The sum of the roots of the reduced 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. A free online solver will allow you to solve the equation online any complexity in 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.

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

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

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

Example 1. Solve the equation.

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

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

Answer: 2; 3.

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

write down the statement of Vieta's theorem

(*)

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

indicate in your answer the found roots of the equation.

Let's give more examples. .

Example 2: Solve the equation

Solution.

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

Answer: .

Example 2: Solve the equation

Example 3: Solve the equation

Let and be the roots of the given equation. Then, by Vieta’s theorem, we note that the product is negative. This means that the roots are of different signs. The sum of the roots is also a negative number. This means that the root with the largest modulus is negative. We select pairs of factors that give the product -10 (1 and -10; 2 and -5). The second pair of numbers adds up to -3. This means that the numbers 2 and -5 are the roots of this equation. Note that Vieta’s theorem can, in principle, be formulated for a complete quadratic equation: if quadratic equation has roots and , then the equalities , , are satisfied for them.

However, the application of this theorem is quite problematic, since in a complete quadratic equation at least one of the roots (if any, of course) is a fractional number. And working with selecting fractions is long and difficult. But still there is a way out. Consider the complete quadratic equation . Multiply both sides of the equation by the first coefficient A and write the equation in the form . Let us introduce a new variable and obtain the reduced quadratic equation, the roots of which and (if available) can be found using Vieta’s theorem. Then the roots of the original equation will be . Please note that it is very simple to create the auxiliary reduced equation: the second coefficient is preserved, and the third coefficient is equal to the product ac

. With a certain skill, students immediately create an auxiliary equation, find its roots using Vieta’s theorem, and indicate the roots of the given complete equation. Let's give examples. .

Example 4: Solve the equation Let's create an auxiliary equation .

Example 5: Solve the equation .

The auxiliary equation has the form . According to Vieta's theorem, its roots are . Finding the roots of the original equation .

And one more case when the application of Vieta’s theorem allows you to verbally find the roots of a complete quadratic equation. It is not difficult to prove that the number 1 is the root of the equation , if and only if. The second root of the equation is found by Vieta’s theorem and is equal to . One more statement: so that the number –1 is the root of the equation necessary and sufficient for. 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 .

Example 7. Solve the equation.

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

Examples of application of Vieta's theorem

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

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

Task 2. Solve the complete quadratic equation by passing to the auxiliary reduced quadratic equation.

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

Task 3. Solve a quadratic equation using the property.

First, let's formulate the theorem itself: Let us 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, according to the theorem, the following statements are valid:

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 same roots will give us the coefficient c.

But what is the given equation?

A reduced quadratic equation is a quadratic equation whose coefficient of the highest degree 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 unreduced). In other words, to bring the equation to the given form, we must divide this equation by the coefficient of the highest power (a). The task is to bring this equation to the following 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.

Dividing 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 you can see from the examples, even equations containing fractions can be reduced to the given 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.

The meaning of Vieta's theorem

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

From the examples given, and using the theorem, it is clear how you can significantly simplify the solution of quadratic equations, because using this theorem, you can solve a quadratic equation practically without 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 used this rule based on two important assumptions:

The given equation, i.e. the coefficient of 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 will be valid x1+x2=-b/a; x1*x2=c/a, but it’s usually more difficult to solve :))

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

Therefore, we can create a general solution algorithm using Vieta’s theorem.

General solution algorithm using Vieta's theorem

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

There are also cases when returning to the initial equation allows us to work with “convenient” numbers.

François Viète (1540-1603) – mathematician, creator of the famous Viète formulas

Vieta's theorem needed for quick solution quadratic equations (in simple words).

In more detail, then Vieta's theorem is that the sum of the roots of a given quadratic equation is equal to the second coefficient, which is taken with the opposite sign, and the product is equal to the free term. Any reduced quadratic equation that has roots has this property.

Using Vieta's theorem, you can easily solve quadratic equations by selection, so let's say “thank you” to this mathematician with a sword in his hands for our happy 7th grade.

Proof of Vieta's theorem

To prove the theorem, you can use well-known root formulas, thanks to which we will compose the sum and product of the roots of a quadratic equation. Only after this we can make sure that they are equal and, accordingly, .

Let's say we have an equation: . This equation has the following roots: and . Let us prove that , .

According to the formulas for the roots of a quadratic equation:

1. Find the sum of the roots:

Let's look at this equation, how we got it exactly like this:

= .

Step 1. Reducing the fractions to a common denominator, it turns out:

= = .

Step 2. We have a fraction where we need to open the brackets:

We reduce the fraction by 2 and get:

We have proved the relation for the sum of the roots of a quadratic equation using Vieta's theorem.

2. Find the product of the roots:

= = = = = .

Let's prove this equation:

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

Now let's remember the definition square root and consider:

= .

Step 3. Let us recall the discriminant of the quadratic equation: . Therefore, instead of D (discriminant), we substitute in the last fraction, then it turns out:

= .

Step 4. We open the brackets and reduce similar terms to the fraction:

Step 5. We shorten “4a” and get .

So we have proven the relation for the product of roots using Vieta’s theorem.

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

Theorem converse to Vieta's theorem

Using the theorem inverse to Vieta’s theorem, we can check whether our equation is solved correctly. To understand the theorem itself, you need to consider it in more detail.

If the numbers are like this:

And, then they are the roots of the quadratic equation.

Proof of Vieta's converse theorem

Step 1.Let us substitute expressions for its coefficients into the equation:

Step 2.Let's transform the left side of the equation:

Step 3. Let's find the roots of the equation, and for this we use the property that the product is equal to zero:

Or . Where it comes from: or .

Examples with solutions using Vieta's theorem

Example 1

Exercise

Find the sum, product and sum of squares of the roots of a quadratic equation without finding the roots of the equation.

Solution

Step 1. Let's remember the discriminant formula. We substitute our numbers for the letters. That is, , – this replaces , and . This implies:

It turns out:

Title="Rendered by QuickLaTeX.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 . !}

Let us express the sum of the squares of the roots through their sum and product:

Answer

7; 12; 25.

Example 2

Exercise

Solve the equation. However, do not use quadratic equation formulas.

Solution

This equation has roots whose discriminant (D) is greater than zero. Accordingly, according to Vieta’s theorem, the sum of the roots of this equation is equal to 4, and the product is 5. First, we determine the divisors 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 their sum is 4. This means that, according to the theorem inverse to Vieta’s theorem, they are the roots of this equation.

Answer

AND Example 4

Exercise

Write an equation where each root is twice the corresponding root of the equation:

Solution

According to Vieta's theorem, the sum of the roots of this equation is equal to 12, and the product = 7. This means that two roots are positive.

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

And the work.

By the theorem inverse to Vieta’s theorem, the new equation has the form:

Answer

The result is an equation, each root of which is twice as large:

So, we looked at how to solve the equation using Vieta's theorem. It is very convenient to use this theorem if you solve problems that involve the signs of the roots of quadratic equations. That is, if the free term in the formula is a positive number, and if the quadratic equation has real roots, then both of them can be either negative or positive.

And if the free term is a negative number, and if the quadratic equation has real roots, then both signs will be different. That is, if one root is positive, then the other root will only be negative.

Useful sources:

Dorofeev G.V., Suvorova S.B., Bunimovich E.A. Algebra 8th grade: Moscow “Enlightenment”, 2016 – 318 p.
Rubin A.G., Chulkov P.V. – textbook Algebra 8th grade: Moscow “Balass”, 2015 – 237 p.
Nikolsky S. M., Potopav M. K., Reshetnikov N. N., Shevkin A. V. – Algebra 8th grade: Moscow “Enlightenment”, 2014 – 300

Vieta's theorem inverse formula Vieta and examples with solutions for dummies updated: November 22, 2019 by: Scientific Articles.Ru

One of the methods for solving a quadratic equation is to use VIET formulas, which was named after FRANCOIS VIETTE.

He was a famous lawyer who served the French king in the 16th century. In his spare 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 a solution. Because there is no need to enter the second coefficient into the square, then subtract 4ac from it, find the discriminant, and substitute its value into the formula to find the roots.

2 . Without a solution, you can determine the signs of the roots and select the values of the roots.

3 . Having solved a 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 . Using these roots, write down a quadratic equation, that is, solve the inverse problem. For example, this method is used when solving problems in theoretical mechanics.

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

Flaws:

1 . The formula is not universal.