The first statement says that the sum of the roots of this equation is equal to the value of the coefficient of the variable x (in this case it is b), but with the opposite sign. Visually, it looks like this: x1 + x2 = −b.
The second statement is no longer connected with the sum, but with the product of the same two roots. This product is equated to a free coefficient, i.e. c. Or, x1 * x2 = c. Both of these examples are solved in the system.
Vieta's theorem greatly simplifies the solution, but has one limitation. A quadratic equation whose roots can be found using this technique must be reduced. In the above equation for the coefficient a, the one before x2 is equal to one. Any equation can be reduced to a similar form by dividing the expression by the first coefficient, but this operation is not always rational.
Proof of the theorem
To begin with, we should remember how, according to tradition, it is customary to look for the roots of a quadratic equation. The first and second roots are found, namely: x1 = (-b-√D)/2, x2 = (-b+√D)/2. Generally divisible by 2a, but, as already mentioned, the theorem can only be applied when a=1.It is known from Vieta's theorem that the sum of the roots is equal to the second coefficient with a minus sign. This means that x1 + x2 = (-b-√D)/2 + (-b+√D)/2 = −2b/2 = −b.
The same is true for the product of unknown roots: x1 * x2 = (-b-√D)/2 * (-b+√D)/2 = (b2-D)/4. In turn, D = b2-4c (again, with a=1). It turns out that the result is: x1 * x2 = (b2- b2)/4+c = c.
From the above simple proof, only one conclusion can be drawn: Vieta's theorem is completely confirmed.
Second formulation and proof
Vieta's theorem has another interpretation. To be more precise, it is not an interpretation, but a wording. The fact is that if the same conditions are met as in the first case: there are two different real roots, then the theorem can be written in a different formula.This equality looks like this: x2 + bx + c = (x - x1)(x - x2). If the function P(x) intersects at two points x1 and x2, then it can be written as P(x) = (x - x1)(x - x2) * R(x). In the case when P has the second degree, and this is exactly what the original expression looks like, then R is a prime number, namely 1. This statement is true for the reason that otherwise the equality will not hold. The coefficient x2 when opening brackets should not be more than one, and the expression should remain square.
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 \(x^2-5x+6=0\).
Decision
: 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 \(- one\). 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) \(x^2-88x+780=0\).
Decision
:
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.
Before proceeding to Vieta's theorem, we introduce a definition. Quadratic equation of the form x² + px + q= 0 is called reduced. In this equation, the leading coefficient is equal to one. For example, the equation x² - 3 x- 4 = 0 is reduced. Any quadratic equation of the form ax² + b x + c= 0 can be made reduced, for this we divide both sides of the equation by a≠ 0. For example, Equation 4 x² + 4 x- 3 \u003d 0 divided by 4 is reduced to the form: x² + x- 3/4 = 0. We derive the formula for the roots of the reduced quadratic equation, for this we use the formula for the roots of a general quadratic equation: ax² + bx + c = 0
Reduced Equation x² + px + q= 0 coincides with a general equation in which a = 1, b = p, c = q. Therefore, for the given quadratic equation, the formula takes the form: 
the last expression is called the formula of the roots of the reduced quadratic equation, it is especially convenient to use this formula when R- even number. For example, let's solve the equation x² - 14 x — 15 = 0
In response, we write the equation has two roots.
For a reduced quadratic equation with positive, the following theorem holds.
Vieta's theorem
If a x 1 and x 2 - roots of the equation x² + px + q= 0, then the formulas are valid:
x 1 + x 2 = — R
x 1 * x 2 \u003d q, that is, 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.
Based on the formula of the roots of the above quadratic equation, we have:
Adding these equalities, we get: x 1 + x 2 = —R.
Multiplying these equalities, using the difference of squares formula, we get:
Note that the Vieta theorem is also valid when the discriminant is zero, if we assume that in this case the quadratic equation has two identical roots: x 1 = x 2 = — R/2.
Not solving equations x² - 13 x+ 30 = 0 find the sum and product of its roots x 1 and x 2. this equation D\u003d 169 - 120 \u003d 49\u003e 0, so you can apply the Vieta theorem: x 1 + x 2 = 13, x 1 * x 2 = 30. Consider a few more examples. One of the roots of the equation x² — px- 12 = 0 is x 1 = 4. Find coefficient R and second root x 2 of this equation. According to Vieta's theorem x 1 * x 2 =— 12, x 1 + x 2 = — R. As x 1 = 4 then 4 x 2 = - 12, whence x 2 = — 3, R = — (x 1 + x 2) \u003d - (4 - 3) \u003d - 1. In response, we write down the second root x 2 = - 3, coefficient p = - 1.
Not solving equations x² + 2 x- 4 = 0 find the sum of the squares of its roots. Let be x 1 and x 2 are the roots of the equation. According to Vieta's theorem x 1 + x 2 = — 2, x 1 * x 2 = - 4. As x 1²+ x 2² = ( x 1 + x 2)² - 2 x 1 x 2 , then x 1²+ x 2 ² \u003d (- 2) ² -2 (- 4) \u003d 12.
Find the sum and product of the roots of equation 3 x² + 4 x- 5 \u003d 0. This equation has two different roots, since the discriminant D= 16 + 4*3*5 > 0. To solve the equation, we use the Vieta theorem. This theorem has been proved for the reduced quadratic equation. So let's divide this equation by 3.
Therefore, the sum of the roots is -4/3, and their product is -5/3.
In general, the roots of the equation ax² + b x + c= 0 are related by the following equalities: x 1 + x 2 = — b/a, x 1 * x 2 = c/a, To obtain these formulas, it suffices to divide both sides of this quadratic equation by a ≠ 0 and apply Vieta's theorem to the resulting reduced quadratic equation. Consider an example, you need to compose a given quadratic equation, the roots of which x 1 = 3, x 2 = 4. As x 1 = 3, x 2 = 4 are the roots of the quadratic equation x² + px + q= 0, then by the Vieta theorem R = — (x 1 + x 2) = — 7, q = x 1 x 2 = 12. In response, we write x² - 7 x+ 12 = 0. The following theorem is used in solving some problems.
Theorem inverse to Vieta's theorem
If numbers R, q, x 1 , x 2 are such that x 1 + x 2 = — p, x 1 * x 2 \u003d q, then x 1 and x2 are the roots of the equation x² + px + q= 0. Substitute in the left side x² + px + q instead of R expression - ( x 1 + x 2), but instead q- work x 1 * x 2 . We get: x² + px + q = x² — ( x 1 + x 2) x + x 1 x 2 \u003d x² - x 1 x - x 2 x + x 1 x 2 \u003d (x - x 1) (x - x 2). Thus, if the numbers R, q, x 1 and x 2 are related by these relations, then for all X equality x² + px + q = (x - x 1) (x - x 2), from which it follows that x 1 and x 2 - roots of the equation x² + px + q= 0. Using the theorem converse to Vieta's theorem, it is sometimes possible to find the roots of a quadratic equation by selection. Consider an example, x² - 5 x+ 6 = 0. Here R = — 5, q= 6. Pick two numbers x 1 and x 2 so that x 1 + x 2 = 5, x 1 * x 2 = 6. Noting that 6 = 2 * 3, and 2 + 3 = 5, by the theorem converse to Vieta's theorem, we obtain that x 1 = 2, x 2 = 3 - roots of the equation x² - 5 x + 6 = 0.
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 a 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 a 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 compose the sum and the product of the roots of the quadratic equation using the known root formulas, after that we 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 the 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?
Decision.
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, therefore, 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.
Decision.
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 free term 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 solve a 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.: Mnemozina, 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.
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.
Disadvantages:
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.
