The solution is via Vieta online. Oral solution of quadratic equations and Vieta's theorem

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 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 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.

Vieta's theorem 8th grade

Formula
If x 1 and x 2 are the roots of the reduced quadratic equation x 2 + px + q = 0, then:

Examples
x 1 = -1; x 2 = 3 - roots of the equation x 2 - 2x - 3 = 0.

P = -2, q = -3.

X 1 + x 2 = -1 + 3 = 2 = -p,

X 1 x 2 = -1 3 = -3 = q.

Converse theorem

Formula
If the numbers x 1, x 2, p, q are related by the conditions:

Then x 1 and x 2 are the roots of the equation x 2 + px + q = 0.

Example
Let's create a quadratic equation using its roots:

X 1 = 2 - ? 3 and x 2 = 2 + ? 3.

P = x 1 + x 2 = 4; p = -4; q = x 1 x 2 = (2 - ? 3 )(2 + ? 3 ) = 4 - 3 = 1.

The required equation has the form: x 2 - 4x + 1 = 0.

Between the roots and coefficients of a quadratic equation, in addition to the root formulas, there are other useful relationships that are given Vieta's theorem. In this article we will give a formulation and proof of Vieta's theorem for a quadratic equation. Next we consider the theorem converse to Vieta’s theorem. After this, we will analyze the solutions to the most typical examples. Finally, we write down the Vieta formulas that define the relationship between the real roots algebraic equation degree n and its coefficients.

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Vieta's theorem, formulation, proof

From the formulas for the roots of the quadratic equation a·x 2 +b·x+c=0 of the form, where D=b 2 −4·a·c, the following relations follow: 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 carry out the proof of Vieta’s theorem according to the following scheme: we will compose the sum and product of the roots of the quadratic equation using 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 and make it up. Now we bring the fractions to a common denominator, we have . In the numerator of the resulting fraction, after which:. Finally, after on 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 multiplying fractions, the last product can be written as . Now we multiply a bracket by a bracket in the numerator, but it’s faster to collapse this product by square difference formula, So . Then, remembering, we perform the next transition. And since the discriminant of the quadratic equation corresponds to the formula D=b 2 −4·a·c, then instead of D in the last fraction we can substitute b 2 −4·a·c, we get. After opening the parentheses and bringing similar 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, the proof of Vieta’s theorem will take a laconic form:
,
.

It remains only to note that if 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 Vieta’s theorem also hold. Indeed, when D=0 the root of the quadratic equation is equal to , 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 leading 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 sides by a non-zero number a. Let us give the corresponding formulation of Vieta’s theorem:

Theorem.

The sum of the roots of the reduced quadratic equation x 2 +p x+q=0 is equal to the coefficient of x taken with the opposite sign, and the product of the roots is equal to the free term, that is, x 1 +x 2 =−p, x 1 x 2 = q.

Theorem converse to Vieta's theorem

The second formulation of Vieta’s 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 converse of Vieta’s theorem is true. Let's 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 with their expressions through x 1 and x 2, it is transformed into an equivalent equation.

Let us substitute the number x 1 instead of x into the resulting equation, and 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 represents 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 =0, which means 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 =0 substitute the number x 2 instead of x, we get the equality x 2 2 −(x 1 +x 2) x 2 +x 1 x 2 =0. This is a true equality, since 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 a root of the equation x 2 −(x 1 +x 2) x+x 1 x 2 =0, and therefore 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 converse theorem. In this section we will analyze solutions to several of the most typical examples.

Let's start by applying the theorem converse to Vieta's theorem. It is convenient to use to check whether given two numbers are 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 roots found.

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 a quadratic equation should be equal to −b/a, that is, 16/4=4, and the product of the roots should 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 ​​we just obtained.

In the first case we have x 1 +x 2 =−5+3=−2. The resulting value is different from 4, so no further verification can be carried out, but using the theorem inverse to Vieta’s theorem, one can immediately conclude that the first pair of numbers is not a pair of roots of the given quadratic equation.

Let's move on to the second case. Here, that is, the first condition is met. We check the second condition: the resulting value is different from 9/4. Consequently, the second pair of numbers is not a pair of roots of the quadratic equation.

There is one last case left. Here and . Both conditions are met, so these numbers x 1 and x 2 are the roots of the given quadratic equation.

Answer:

The converse of Vieta's theorem can be used in practice to find 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. In this case, 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 understand 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 must be satisfied: x 1 + x 2 =5 and x 1 ·x 2 =6. All that remains is to select 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 inverse to Vieta's theorem is especially convenient to use to find the second root of a given quadratic equation when one of the roots is already known or obvious. In this case, the second root can be 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 unity is the root of the equation, since the sum of the coefficients of this quadratic equation is equal to 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, from which x 2 =−3/512. This is how we determined both roots of the quadratic equation: 1 and −3/512.

It is clear that the selection of roots is advisable only in the simplest cases. In other cases, to find roots, you can use formulas for the roots of a quadratic equation through a discriminant.

Another practical application of the converse of Vieta's theorem is to construct quadratic equations given the 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 −11 and 23.

Solution.

Let's denote x 1 =−11 and x 2 =23. We calculate the sum and product of these numbers: x 1 +x 2 =12 and x 1 ·x 2 =−253. Therefore, the indicated numbers are the roots of the reduced quadratic equation with a second coefficient of −12 and a free term of −253. That is, x 2 −12·x−253=0 is the required equation.

Answer:

x 2 −12·x−253=0 .

Vieta's theorem is very often used when solving problems 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 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. Let's look at examples of their application.

Example.

R it is positive. Using 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. Consequently, 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 according to Vieta’s theorem, the product of the roots of the reduced 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, to find the values ​​of r we are interested in, we need solve 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 relationships it asserts. But there are formulas connecting the real roots and coefficients of not only quadratic equations, but also cubic equations, equations of the fourth degree, and in general, algebraic equations degree n. They are called Vieta's formulas.

Let us write the Vieta formula for an algebraic equation of degree n of the form, and we will assume that it has n real roots x 1, x 2, ..., x n (among them there may be coinciding ones):

Vieta's formulas can be obtained theorem on the decomposition of a polynomial into linear factors, 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 Vieta’s formulas.

In particular, for n=2 we have the already familiar Vieta formulas for a quadratic equation.

For a cubic equation, Vieta's formulas have the form

It remains only to note that on the left side of Vieta’s formulas there are the so-called elementary symmetric polynomials.

Bibliography.

• Algebra: textbook for 8th grade. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Mordkovich A. G. Algebra. 8th grade. In 2 hours. Part 1. Textbook for students of general education 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. 10th grade: textbook. for general education institutions: basic and profile. levels / [Yu. M. Kolyagin, M. V. Tkacheva, N. E. Fedorova, M. I. Shabunin]; edited by A. B. Zhizhchenko. - 3rd ed. - M.: Education, 2010.- 368 p. : ill. - ISBN 978-5-09-022771-1.

The discriminant, like quadratic equations, begins to be studied in an algebra course in the 8th grade. You can solve a quadratic equation through a discriminant and using Vieta's theorem. The method of studying quadratic equations, as well as discriminant formulas, is rather unsuccessfully taught to schoolchildren, like many things in real education. Therefore, school years pass, education in grades 9-11 is replaced by “higher education” and everyone is looking again - “How to solve a quadratic equation?”, “How to find the roots of the equation?”, “How to find the discriminant?” And...

Discriminant formula

The discriminant D of the quadratic equation a*x^2+bx+c=0 is equal to D=b^2–4*a*c.
The roots (solutions) of a quadratic equation depend on the sign of the discriminant (D):
D>0 – the equation has 2 different real roots;
D=0 - the equation has 1 root (2 matching roots):
D<0 – не имеет действительных корней (в школьной теории). В ВУЗах изучают комплексные числа и уже на множестве комплексных чисел уравнение с отрицательным дискриминантом имеет два комплексных корня.
The formula for calculating the discriminant is quite simple, so many websites offer an online discriminant calculator. We haven’t figured out this kind of scripts yet, so if anyone knows how to implement this, please write to us by email This email address is being protected from spambots. You must have JavaScript enabled to view it. .

General formula for finding the roots of a quadratic equation:

We find the roots of the equation using the formula
If the coefficient of a squared variable is paired, then it is advisable to calculate not the discriminant, but its fourth part
In such cases, the roots of the equation are found using the formula

The second way to find roots is Vieta's Theorem.

The theorem is formulated not only for quadratic equations, but also for polynomials. You can read this on Wikipedia or other electronic resources. However, to simplify, let’s consider the part that concerns the above quadratic equations, that is, equations of the form (a=1)
The essence of Vieta's formulas is that the sum of the roots of the equation is equal to the coefficient of the variable, taken with the opposite sign. The product of the roots of the equation is equal to the free term. Vieta's theorem can be written in formulas.
The derivation of Vieta's formula is quite simple. Let's write the quadratic equation through simple factors
As you can see, everything ingenious is simple at the same time. It is effective to use Vieta’s formula when the difference in modulus of the roots or the difference in the moduli of the roots is 1, 2. For example, the following equations, according to Vieta’s theorem, have roots




Up to equation 4, the analysis should look like this. The product of the roots of the equation is 6, therefore the roots can be the values ​​(1, 6) and (2, 3) or pairs with the opposite sign. The sum of the roots is 7 (the coefficient of the variable with the opposite sign). From here we conclude that the solutions to the quadratic equation are x=2; x=3.
It is easier to select the roots of the equation among the divisors of the free term, adjusting their sign in order to fulfill the Vieta formulas. At first, this seems difficult to do, but with practice on a number of quadratic equations, this technique will be more effective than calculating the discriminant and finding the roots of the quadratic equation in the classical way.
As you can see, the school theory of studying the discriminant and methods of finding solutions to the equation is devoid of practical meaning - “Why do schoolchildren need a quadratic equation?”, “What is the physical meaning of the discriminant?”

Let's try to figure it out What does the discriminant describe?

In the algebra course they study functions, schemes for studying functions and constructing a graph of functions. Of all the functions, the parabola occupies an important place, the equation of which can be written in the form
So the physical meaning of the quadratic equation is the zeros of the parabola, that is, the points of intersection of the graph of the function with the abscissa axis Ox
I ask you to remember the properties of parabolas that are described below. The time will come to take exams, tests, or entrance exams and you will be grateful for the reference material. The sign of the squared variable corresponds to whether the branches of the parabola on the graph will go up (a>0),

or a parabola with branches down (a<0) .

The vertex of the parabola lies midway between the roots

Physical meaning of the discriminant:

If the discriminant is greater than zero (D>0) the parabola has two points of intersection with the Ox axis.
If the discriminant is zero (D=0) then the parabola at the vertex touches the x-axis.
And the last case, when the discriminant is less than zero (D<0) – график параболы принадлежит плоскости над осью абсцисс (ветки параболы вверх), или график полностью под осью абсцисс (ветки параболы опущены вниз).

Incomplete quadratic equations

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 mathematics have integer solutions.

I. Vieta's theorem for the reduced quadratic equation.

Sum of 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 = -p; x 1 ∙x 2 =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), second coefficient p=-1, and the free member q=-30. First, let's make sure that this equation has roots, and that the roots (if any) will be expressed in integers. To do this, it is enough that the discriminant be a perfect square of an integer.

Finding the discriminant D=b 2 — 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .

Now, according to Vieta’s 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 =-30. We need to choose two numbers such that their product is equal to -30 , and the amount is unit. These are 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. Let's make sure that there are integer roots. Let's find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us select the roots using Vieta’s theorem: the sum of the roots is equal to –р=-6, and the product of the roots is equal to q=8. These are numbers -4 And -2 .

In fact: -4-2=-6=-р; -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 member q=-4. Let's find the discriminant D 1, since the second coefficient is an even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, so we do conclusion: The roots of this equation are not integers and cannot be found using Vieta’s theorem. This means that we solve this equation, as usual, using formulas (in this case, using formulas). We get:

Example 4). Write a quadratic equation using its roots if x 1 =-7, x 2 =4.

Solution. The required equation will be written in the form: x 2 +px+q=0, and, based on Vieta’s theorem –p=x 1 +x 2=-7+4=-3 → p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x 2 +3x-28=0.

Example 5). Write a quadratic equation using its roots if:

II. Vieta's theorem for a complete quadratic equation ax 2 +bx+c=0.

The sum of the roots is minus b, divided by A, the product of the roots is equal to With, divided by A:

x 1 + x 2 = -b/a; x 1 ∙x 2 =c/a.