In eighth grade, students are introduced to quadratic equations and how to solve them. At the same time, as experience shows, most students, when solving complete quadratic equations, use only one method - 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 us assume that the roots of the equation are integers positive numbers. 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 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 some 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.

Answer: -2; -5.

Example 3: Solve the equation .

Solution.

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.

Answer: 2; -5.

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 an 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 and using Vieta's theorem we will find its roots. This means that the roots of the original equation .

Answer: .

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 .

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 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 given 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 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.

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 the converse theorem only work with , that is, one for which the coefficient 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, you can 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.

Vieta's theorem (more precisely, the theorem converse of the theorem Vieta) allows you to reduce the time for solving quadratic equations. You just need to know how to use it. How to learn to solve quadratic equations using Vieta's theorem? It's not difficult if you think about it a little.

Now we will only talk about solving the reduced quadratic equation using Vieta’s theorem. A reduced quadratic equation is an equation in which a, that is, the coefficient of x², is equal to one. It is also possible to solve quadratic equations that are not given using Vieta’s theorem, but at least one of the roots is not an integer. They are harder to guess.

The inverse theorem to Vieta's theorem states: if the numbers x1 and x2 are such that

then x1 and x2 are the roots of the quadratic equation

When solving a quadratic equation using Vieta's theorem, only 4 options are possible. If you remember the line of reasoning, you can learn to find whole roots very quickly.

I. If q is a positive number,

this means that the roots x1 and x2 are numbers of the same sign (since only multiplying numbers with the same signs produces a positive number).

I.a. If -p is a positive number, (respectively, p<0), то оба корня x1 и x2 — положительные числа (поскольку складывали числа одного знака и получили положительное число).

I.b. If -p is a negative number, (respectively, p>0), then both roots are negative numbers (we added numbers of the same sign and got a negative number).

II. If q is a negative number,

this means that the roots x1 and x2 have different signs (when multiplying numbers, a negative number is obtained only when the signs of the factors are different). In this case, x1+x2 is no longer a sum, but a difference (after all, when adding numbers with different signs we subtract the smaller from the larger modulo). Therefore, x1+x2 shows how much the roots x1 and x2 differ, that is, how much one root is greater than the other (in absolute value).

II.a. If -p is a positive number, (that is, p<0), то больший (по модулю) корень — положительное число.

II.b. If -p is a negative number, (p>0), then the larger (modulo) root is a negative number.

Let's consider solving quadratic equations using Vieta's theorem using examples.

Solve the given quadratic equation using Vieta's theorem:

Here q=12>0, so the roots x1 and x2 are numbers of the same sign. Their sum is -p=7>0, so both roots are positive numbers. We select integers whose product is equal to 12. These are 1 and 12, 2 and 6, 3 and 4. The sum is 7 for the pair 3 and 4. This means that 3 and 4 are the roots of the equation.

In this example, q=16>0, which means that the roots x1 and x2 are numbers of the same sign. Their sum is -p=-10<0, поэтому оба корня — отрицательные числа. Подбираем числа, произведение которых равно 16. Это 1 и 16, 2 и 8, 4 и 4. Сумма 2 и 8 равна 10, а раз нужны отрицательные числа, то искомые корни — это -2 и -8.

Here q=-15<0, что означает, что корни x1 и x2 — числа разных знаков. Поэтому 2 — это уже не их сумма, а разность, то есть числа отличаются на 2. Подбираем числа, произведение которых равно 15, отличающиеся на 2. Произведение равно 15 у 1 и 15, 3 и 5. Отличаются на 2 числа в паре 3 и 5. Поскольку -p=2>0, then the larger number is positive. So the roots are 5 and -3.

q=-36<0, значит, корни x1 и x2 имеют разные знаки. Тогда 5 — это то, насколько отличаются x1 и x2 (по модулю, то есть пока что без учета знака). Среди чисел, произведение которых равно 36: 1 и 36, 2 и 18, 3 и 12, 4 и 9 — выбираем пару, в которой числа отличаются на 5. Это 4 и 9. Осталось определить их знаки. Поскольку -p=-5<0, бОльшее число имеет знак минус. Поэтому корни данного уравнения равны -9 и 4.

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

Vieta's theorem needed to quickly solve 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 we recall the definition of square root and calculate:

= .

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. Open the parentheses and add 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 . It follows from this:

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 – 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:

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

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