Between the roots and the coefficients of the quadratic equation, in addition to the root formulas, there are other useful relationships that are given by Vieta's theorem. In this article, we will give a formulation and proof of Vieta's theorem for quadratic equation. Next, we consider a theorem converse to Vieta's theorem. After that, we will analyze the solutions of the most characteristic examples. Finally, we write down the Vieta formulas that define the connection between the real roots algebraic equation degree n and its coefficients.
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Vieta's theorem, formulation, proof
From the formulas of the roots of the quadratic equation a x 2 +b x+c=0 of the form , where D=b 2 −4 a c , the relations x 1 +x 2 = −b/a, x 1 x 2 = c/a . These results are confirmed Vieta's theorem:
Theorem.
If x 1 and x 2 are the roots of the quadratic equation a x 2 +b x+c=0, then the sum of the roots is equal to the ratio of the coefficients b and a, taken with the opposite sign, and the product of the roots is equal to the ratio of the coefficients c and a, that is, .
Proof.
We will prove the Vieta theorem according to the following scheme: we will compose the sum and product of the roots of the quadratic equation using the known root formulas, then we will transform the resulting expressions, and make sure that they are equal to −b/a and c/a, respectively.
Let's start with the sum of the roots, compose it. Now we bring the fractions to a common denominator, we have. In the numerator of the resulting fraction , after which : . Finally, after 2 , we get . This proves the first relation of Vieta's theorem for the sum of the roots of a quadratic equation. Let's move on to the second.
We compose the product of the roots of the quadratic equation:. According to the rule of multiplication of fractions, the last product can be written as. Now we multiply the bracket by the bracket in the numerator, but it is faster to collapse this product by difference of squares formula, So . Then, remembering , we perform the next transition . And since the formula D=b 2 −4 a·c corresponds to the discriminant of the quadratic equation, then b 2 −4·a·c can be substituted into the last fraction instead of D, we get . After opening the brackets and reducing like terms, we arrive at the fraction , and its reduction by 4·a gives . This proves the second relation of Vieta's theorem for the product of roots.
If we omit the explanations, then the proof of the Vieta theorem will take a concise form:
,
.
It remains only to note that when the discriminant is equal to zero, the quadratic equation has one root. However, if we assume that the equation in this case has two identical roots, then the equalities from the Vieta theorem also hold. Indeed, for D=0 the root of the quadratic equation is , then and , and since D=0 , that is, b 2 −4·a·c=0 , whence b 2 =4·a·c , then .
In practice, Vieta's theorem is most often used in relation to the reduced quadratic equation (with the highest coefficient a equal to 1 ) of the form x 2 +p·x+q=0 . Sometimes it is formulated for quadratic equations of just this type, which does not limit the generality, since any quadratic equation can be replaced by an equivalent equation by dividing both its parts by a non-zero number a. Here is the corresponding formulation of Vieta's theorem:
Theorem.
The sum of the roots of the reduced quadratic equation x 2 + p x + q \u003d 0 is equal to the coefficient at x, taken with the opposite sign, and the product of the roots is a free term, that is, x 1 + x 2 \u003d −p, x 1 x 2 \u003d q .
Theorem inverse to Vieta's theorem
The second formulation of the Vieta theorem, given in the previous paragraph, indicates that if x 1 and x 2 are the roots of the reduced quadratic equation x 2 +p x+q=0, then the relations x 1 +x 2 = −p, x 1 x 2=q. On the other hand, from the written relations x 1 +x 2 =−p, x 1 x 2 =q, it follows that x 1 and x 2 are the roots of the quadratic equation x 2 +p x+q=0. In other words, the assertion converse to Vieta's theorem is true. We formulate it in the form of a theorem, and prove it.
Theorem.
If the numbers x 1 and x 2 are such that x 1 +x 2 =−p and x 1 x 2 =q, then x 1 and x 2 are the roots of the reduced quadratic equation x 2 +p x+q=0.
Proof.
After replacing the coefficients p and q in the equation x 2 +p x+q=0 of their expression through x 1 and x 2, it is converted into an equivalent equation.
We substitute the number x 1 instead of x into the resulting equation, we have the equality x 1 2 −(x 1 + x 2) x 1 + x 1 x 2 =0, which for any x 1 and x 2 is the correct numerical equality 0=0, since 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. Therefore, x 1 is the root of the equation x 2 −(x 1 + x 2) x + x 1 x 2 \u003d 0, which means that x 1 is the root of the equivalent equation x 2 +p x+q=0 .
If in the equation x 2 −(x 1 + x 2) x + x 1 x 2 \u003d 0 substitute the number x 2 instead of x, then we get the equality x 2 2 −(x 1 + x 2) x 2 + x 1 x 2 =0. This is the correct equation because 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. Therefore, x 2 is also the root of the equation x 2 −(x 1 + x 2) x + x 1 x 2 \u003d 0, and hence the equations x 2 +p x+q=0 .
This completes the proof of the theorem converse to Vieta's theorem.
Examples of using Vieta's theorem
It's time to talk about the practical application of Vieta's theorem and its inverse theorem. In this subsection, we will analyze the solutions of several of the most typical examples.
We start by applying a theorem converse to Vieta's theorem. It is convenient to use it to check whether the given two numbers are the roots of a given quadratic equation. In this case, their sum and difference are calculated, after which the validity of the relations is checked. If both of these relations are satisfied, then, by virtue of the theorem converse to Vieta's theorem, it is concluded that these numbers are the roots of the equation. If at least one of the relations is not satisfied, then these numbers are not the roots of the quadratic equation. This approach can be used when solving quadratic equations to check the found roots.
Example.
Which of the pairs of numbers 1) x 1 =−5, x 2 =3, or 2), or 3) is a pair of roots of the quadratic equation 4 x 2 −16 x+9=0?
Solution.
The coefficients of the given quadratic equation 4 x 2 −16 x+9=0 are a=4 , b=−16 , c=9 . According to Vieta's 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 must be equal to c/a, that is, 9/4.
Now let's calculate the sum and product of the numbers in each of the three given pairs, and compare them with the values just obtained.
In the first case, we have x 1 +x 2 =−5+3=−2 . The resulting value is different from 4, so further verification can not be carried out, but by the theorem, the inverse of Vieta's theorem, we can immediately conclude that the first pair of numbers is not a pair of roots of a given quadratic equation.
Let's move on to the second case. Here, that is, the first condition is satisfied. We check the second condition: , the resulting value is different from 9/4 . Therefore, the second pair of numbers is not a pair of roots of a quadratic equation.
The last case remains. Here and . Both conditions are met, so these numbers x 1 and x 2 are the roots of the given quadratic equation.
Answer:
The theorem, the reverse of Vieta's theorem, can be used in practice to select the roots of a quadratic equation. Usually, integer roots of the given quadratic equations with integer coefficients are selected, since in other cases this is quite difficult to do. At the same time, they 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. Let's deal with this with an example.
Let's take the quadratic equation x 2 −5 x+6=0 . For the numbers x 1 and x 2 to be the roots of this equation, two equalities x 1 +x 2 \u003d 5 and x 1 x 2 \u003d 6 must be satisfied. It remains to choose such numbers. In this case, this is quite simple to do: such numbers are 2 and 3, since 2+3=5 and 2 3=6 . Thus, 2 and 3 are the roots of this quadratic equation.
The theorem, the reverse of Vieta's theorem, is especially convenient to apply to finding the second root of the reduced quadratic equation, when one of the roots is already known or obvious. In this case, the second root is found from any of the relations.
For example, let's take the quadratic equation 512 x 2 −509 x−3=0 . Here it is easy to see that the unit is the root of the equation, since the sum of the coefficients of this quadratic equation is zero. So x 1 =1 . The second root x 2 can be found, for example, from the relation x 1 x 2 =c/a. We have 1 x 2 =−3/512 , whence x 2 =−3/512 . So we have defined both roots of the quadratic equation: 1 and −3/512.
It is clear that the selection of roots is expedient only in the simplest cases. In other cases, to find the roots, you can apply the formulas of the roots of the quadratic equation through the discriminant.
Another practical application of the theorem, the inverse of Vieta's theorem, is the compilation of quadratic equations for given roots x 1 and x 2. To do this, it is enough to calculate the sum of the roots, which gives the coefficient of x with the opposite sign of the given quadratic equation, and the product of the roots, which gives the free term.
Example.
Write a quadratic equation whose roots are the numbers −11 and 23.
Solution.
Denote x 1 =−11 and x 2 =23 . We calculate the sum and product of these numbers: x 1 + x 2 \u003d 12 and x 1 x 2 \u003d −253. Therefore, these numbers are the roots of the given quadratic equation with the second coefficient -12 and the free term -253. That is, x 2 −12·x−253=0 is the desired equation.
Answer:
x 2 −12 x−253=0 .
Vieta's theorem is very often used in solving tasks related to the signs of the roots of quadratic equations. How is Vieta's theorem related to the signs of the roots of the reduced quadratic equation x 2 +p x+q=0 ? Here are two relevant statements:
- If the intercept q is a positive number and if the quadratic equation has real roots, then either they are both positive or both are negative.
- If the free term q is a negative number and if the quadratic equation has real roots, then their signs are different, in other words, one root is positive and the other is negative.
These statements follow from the formula x 1 x 2 =q, as well as the rules for multiplying positive, negative numbers and numbers with different signs. Consider examples of their application.
Example.
R is positive. According to the discriminant formula, we find D=(r+2) 2 −4 1 (r−1)= r 2 +4 r+4−4 r+4=r 2 +8 , the value of the expression r 2 +8 is positive for any real r , thus D>0 for any real r . Therefore, the original quadratic equation has two roots for any real values of the parameter r.
Now let's find out when the roots have different signs. If the signs of the roots are different, then their product is negative, and by the Vieta theorem, the product of the roots of the given quadratic equation is equal to the free term. Therefore, we are interested in those values of r for which the free term r−1 is negative. Thus, in order to find the values of r that are of interest to us, we need to decide linear inequality r−1<0 , откуда находим r<1 .
Answer:
at r<1 .
Vieta formulas
Above, we talked about Vieta's theorem for a quadratic equation and analyzed the relations it asserts. But there are formulas that connect the real roots and coefficients not only of quadratic equations, but also of cubic equations, quadruple equations, and in general, algebraic equations degree n. They are called Vieta formulas.
We write the Vieta formulas for an algebraic equation of degree n of the form, while we assume that it has n real roots x 1, x 2, ..., x n (among them there may be the same): 
Get Vieta formulas allows polynomial factorization theorem, as well as the definition of equal polynomials through the equality of all their corresponding coefficients. So the polynomial and its expansion into linear factors of the form are equal. Opening the brackets in the last product and equating the corresponding coefficients, we obtain the Vieta formulas.
In particular, for n=2 we have already familiar Vieta formulas for the quadratic equation .
For a cubic equation, the Vieta formulas have the form 
It only remains to note that on the left side of the Vieta formulas there are the so-called elementary symmetric polynomials.
Bibliography.
- Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
- Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemosyne, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
- Algebra and the beginning of mathematical analysis. Grade 10: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; ed. A. B. Zhizhchenko. - 3rd ed. - M.: Enlightenment, 2010.- 368 p. : ill. - ISBN 978-5-09-022771-1.
Three numbers 12x, x 2-5 and 4 in this order form an increasing arithmetic progression https://youtu.be/U0VO_N9udpI Choose the correct statement MATHEMATICS ZFTSH MIPT Moscow Institute of Physics and Technology (State University) Distance learning school of physics and technology http://pin.it/9w-GqGp Find all x, y and z such that the numbers 5x + 3, y2 and 3z + 5 form an arithmetic progression in that order. Find x and write the difference of this progression. Solve the system of equations Mathematics USE. Video lessons. Divisibility of integers. Linear function. divisibility problems. Vieta's theorem, inverse theorem, Vieta's formulas. clever #students #equations #vietas_theorem #theorem Next, we consider a theorem converse to Vieta's theorem. After that, we will analyze the solutions of the most characteristic examples. This proves the first relation of Vieta's theorem for the sum of the roots of a quadratic equation. Let's move on to the second. How to prove the converse theorem to Vieta's theorem? DOK-VO: x2 + px + f \u003d 0 x2- (M + H) * x + M * H \u003d 0 x2-Mx-Hx + M * H \u003d 0 x (x-H) -M (x-H) =0 (x-M) (x-H) =0 x-M=0 x-H=0 x=M x=H FTD. So we proved in the profile class with a mathematical bias. Answers: help me understand the Vieta inverse theorem thanks to specific examples The Vieta inverse theorem helps to solve: If the coefficient a is a number from which it is easy to extract the square root of a rational integer, then the sum of x1 and x2 will be equal to the number Prove the theorem inverse to the theorem Vieta - see how to complain about the proof of Vieta's theorem. Formulate and prove the Vieta theorem, as well as the inverse theorem, apply theorems to solve equations and problems. Prove a theorem converse to Vieta's theorem. USE in mathematics for 100 points: secrets that school teachers do not talk about, tasks for derivatives. Many applicants think that it is not necessary to prepare for the first fourteen tasks, believing that they are very easy, but this is not so! Most of the examiners make the simplest arithmetic mistakes, thereby overshadowing the excellent solution of the tasks of part C. Such situations are very common, therefore, you should not neglect the preparation for the first tasks, but prepare like in sports training: if you claim 90-100 points - train to solve the first block in 20-25 minutes, if by 70-80 points - somewhere around 30 minutes, no more. An excellent way to train is to solve in the company of a tutor, in courses where certain conditions will be set: for example, solve before the first mistake, then hand over the work; another option - for each mistake you hand over money to the general cash desk. No matter how strange it may seem, we do not recommend the official website, since all the tests there are so confused that it is impossible to use it. Formulation of tasks of part C is important. If the decision is made inaccurately, then the course of solving the task will not be clear, and therefore, the inspector will definitely find fault with this and lower your score. It would seem that we talked about very simple things, but following our advice, you will ensure your successful passing of the exam! The secret links, which are described in the Master class, can be found here - these are links to Video courses for preparing for the exam. The result obtained is called Vieta's theorem. For a reduced square trinomial 2 x px q, Vieta's theorem looks like this: if there are roots, then the theorem converse to Vieta's theorem also holds: if the numbers satisfy the conditions, then these numbers are the roots of the equation. The proof of this theorem is one of the control questions of the Assignment. Sometimes, for brevity, both Vieta's theorems (direct and inverse) are simply called Vieta's theorem.
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.
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.
I. Vieta's theorem for the reduced quadratic equation.
The sum of the roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term:
x 1 + x 2 \u003d-p; x 1 ∙ x 2 \u003d q.
Find the roots of the given quadratic equation using Vieta's theorem.
Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), the second coefficient p=-1, and the free term q=-30. First, make sure that the given equation has roots, and that the roots (if any) will be expressed as integers. For this, it is sufficient that the discriminant be the full square of an integer.
Finding the discriminant D=b 2 - 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .
Now, according to the Vieta theorem, the sum of the roots must be equal to the second coefficient, taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:
x 1 + x 2 =1; x 1 ∙ x 2 \u003d -30. We need to choose such two numbers so that their product is equal to -30 , and the sum is unit. These are the numbers -5 And 6 . Answer: -5; 6.
Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Make sure that there are integer roots. Let's find the discriminant D1 D1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , so the roots of this equation are integers. We choose the roots according to the Vieta theorem: the sum of the roots is equal to –p=-6, and the product of the roots is q=8. These are the numbers -4 And -2 .
Actually: -4-2=-6=-p; -4∙(-2)=8=q. Answer: -4; -2.
Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free term q=-4. Let's find the discriminant D1, since the second coefficient is an even number. D1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of a number, so we do conclusion: the roots of this equation are not integers and cannot be found using Vieta's theorem. So, we solve this equation, as usual, according to the formulas (in this case, according to the formulas). We get:
Example 4). Write a quadratic equation using its roots if x 1 \u003d -7, x 2 \u003d 4.
Solution. The desired equation will be written in the form: x 2 +px+q=0, moreover, based on the Vieta theorem –p=x1 +x2=-7+4=-3 →p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x2 +3x-28=0.
Example 5). Write a quadratic equation using its roots if:
II. Vieta's theorem for the complete quadratic equation ax2+bx+c=0.
The sum of the roots is minus b divided by A, the product of the roots is With divided by A:
x 1 + x 2 \u003d -b / a; x 1 ∙ x 2 \u003d c / a.
Example 6). Find the sum of the roots of a quadratic equation 2x2 -7x-11=0.