Solving equations using Vieta's theorem examples. On the application of Vieta's theorem in solving quadratic equations. Formulation and proof of Vieta's theorem

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 the formulas (in this case, using the 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

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 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 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 compose the sum and product of the roots of the quadratic equation using known root formulas, then 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 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 a 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 intercept 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 that connect 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.

2.5 Vieta formula for polynomials (equations) of higher degrees

The formulas derived by Viète for quadratic equations are also true for polynomials of higher degrees.

Let the polynomial

P(x) = a 0 x n + a 1 x n -1 + … +a n

Has n different roots x 1, x 2..., x n.

In this case, it has a factorization of the form:

a 0 x n + a 1 x n-1 +…+ a n = a 0 (x – x 1)(x – x 2)…(x – x n)

Let's divide both sides of this equality by a 0 ≠ 0 and open the brackets in the first part. We get the equality:

x n + ()x n -1 + … + () = x n – (x 1 + x 2 + … + x n) x n -1 + (x 1 x 2 + x 2 x 3 + … + x n -1 x n)x n - 2 + … +(-1) n x 1 x 2 … x n

But two polynomials are identically equal if and only if the coefficients of the same powers are equal. It follows that the equality

x 1 + x 2 + … + x n = -

x 1 x 2 + x 2 x 3 + … + x n -1 x n =

x 1 x 2 … x n = (-1) n

For example, for polynomials of third degree

a 0 x³ + a 1 x² + a 2 x + a 3

We have identities

x 1 + x 2 + x 3 = -

x 1 x 2 + x 1 x 3 + x 2 x 3 =

x 1 x 2 x 3 = -

As for quadratic equations, this formula is called Vieta's formulas. The left-hand sides of these formulas are symmetric polynomials from the roots x 1, x 2 ..., x n of this equation, and the right-hand sides are expressed through the coefficient of the polynomial.

2.6 Equations reducible to quadratic (biquadratic)

Equations of the fourth degree are reduced to quadratic equations:

ax 4 + bx 2 + c = 0,

called biquadratic, and a ≠ 0.

It is enough to put x 2 = y in this equation, therefore,

ay² + by + c = 0

let's find the roots of the resulting quadratic equation

y 1,2 =

To immediately find the roots x 1, x 2, x 3, x 4, replace y with x and get

x² =

x 1,2,3,4 = .

If a fourth degree equation has x 1, then it also has a root x 2 = -x 1,

If has x 3, then x 4 = - x 3. The sum of the roots of such an equation is zero.

2x 4 - 9x² + 4 = 0

Let's substitute the equation into the formula for the roots of biquadratic equations:

x 1,2,3,4 = ,

knowing that x 1 = -x 2, and x 3 = -x 4, then:

x 3.4 =

Answer: x 1.2 = ±2; x 1.2 =

2.7 Study of biquadratic equations

Let's take the biquadratic equation

ax 4 + bx 2 + c = 0,

where a, b, c are real numbers, and a > 0. By introducing the auxiliary unknown y = x², we examine the roots of this equation and enter the results into the table (see Appendix No. 1)

2.8 Cardano formula

If we use modern symbolism, the derivation of the Cardano formula can look like this:

x =

This formula determines the roots of a general third-degree equation:

ax 3 + 3bx 2 + 3cx + d = 0.

This formula is very cumbersome and complex (it contains several complex radicals). It will not always apply, because... very difficult to fill out.

F ¢(xо) = 0, >0 (<0), то точка xоявляется точкой локального минимума (максимума) функции f(x). Если же =0, то нужно либо пользоваться первым достаточным условием, либо привлекать высшие производные. На отрезке функция y = f(x) может достигать наименьшего или наибольшего значения либо в критических точках, либо на концах отрезка . Пример 3.22. Найти экстремумы функции f(x) ...

List or select the most interesting places from 2-3 texts. Thus, we have examined the general provisions for creating and conducting elective courses, which will be taken into account when developing an elective course in algebra for grade 9 “Quadratic equations and inequalities with a parameter.” Chapter II. Methodology for conducting the elective course “Quadratic equations and inequalities with a parameter” 1.1. Are common...

Solutions from numerical calculation methods. To determine the roots of an equation, knowledge of the theories of Abel, Galois, Lie, etc. groups and the use of special mathematical terminology: rings, fields, ideals, isomorphisms, etc. are not required. To solve an algebraic equation of the nth degree, you only need the ability to solve quadratic equations and extract roots from a complex number. Roots can be determined by...

With units of measurement of physical quantities in the MathCAD system? 11. Describe in detail the text, graphic and mathematical blocks. Lecture No. 2. Linear algebra problems and solving differential equations in the MathCAD environment In linear algebra problems, there is almost always a need to perform various operations with matrices. The operator panel with matrices is located on the Math panel. ...

Any complete quadratic equation ax 2 + bx + c = 0 can be brought to mind x 2 + (b/a)x + (c/a) = 0, if you first divide each term by the coefficient a before x 2. And if we introduce new notations (b/a) = p And (c/a) = q, then we will have the equation x 2 + px + q = 0, which in mathematics is called given quadratic equation.

Roots of the reduced quadratic equation and coefficients p And q connected to each other. It's confirmed Vieta's theorem, named after the French mathematician Francois Vieta, who lived at the end of the 16th century.

Theorem. Sum of roots of the reduced quadratic equation x 2 + px + q = 0 equal to the second coefficient p, taken with the opposite sign, and the product of the roots - to the free term q.

Let us write these relations in the following form:

Let x 1 And x 2 different roots of the given equation x 2 + px + q = 0. According to Vieta's theorem x 1 + x 2 = -p And x 1 x 2 = q.

To prove this, let's substitute each of the roots x 1 and x 2 into the equation. We get two true equalities:

x 1 2 + px 1 + q = 0

x 2 2 + px 2 + q = 0

Let us subtract the second from the first equality. We get:

x 1 2 – x 2 2 + p(x 1 – x 2) = 0

We expand the first two terms using the difference of squares formula:

(x 1 – x 2)(x 1 – x 2) + p(x 1 – x 2) = 0

By condition, the roots x 1 and x 2 are different. Therefore, we can reduce the equality to (x 1 – x 2) ≠ 0 and express p.

(x 1 + x 2) + p = 0;

(x 1 + x 2) = -p.

The first equality has been proven.

To prove the second equality, we substitute into the first equation

x 1 2 + px 1 + q = 0 instead of the coefficient p, an equal number is (x 1 + x 2):

x 1 2 – (x 1 + x 2) x 1 + q = 0

Transforming the left side of the equation, we get:

x 1 2 – x 2 2 – x 1 x 2 + q = 0;

x 1 x 2 = q, which is what needed to be proved.

Vieta's theorem is good because Even without knowing the roots of a quadratic equation, we can calculate their sum and product .

Vieta's theorem helps determine the integer roots of a given quadratic equation. But for many students this causes difficulties due to the fact that they do not know a clear algorithm of action, especially if the roots of the equation have different signs.

So, the above quadratic equation has the form x 2 + px + q = 0, where x 1 and x 2 are its roots. According to Vieta's theorem, x 1 + x 2 = -p and x 1 · x 2 = q.

The following conclusion can be drawn.

If the last term in the equation is preceded by a minus sign, then the roots x 1 and x 2 have different signs. In addition, the sign of the smaller root coincides with the sign of the second coefficient in the equation.

Based on the fact that when adding numbers with different signs, their modules are subtracted, and the resulting result is preceded by the sign of the larger number in absolute value, you should proceed as follows:

determine the factors of the number q such that their difference is equal to the number p;
put the sign of the second coefficient of the equation in front of the smaller of the resulting numbers; the second root will have the opposite sign.

Let's look at some examples.

Example 1.

Solve the equation x 2 – 2x – 15 = 0.

Solution.

Let's try to solve this equation using the rules proposed above. Then we can say for sure that this equation will have two different roots, because D = b 2 – 4ac = 4 – 4 · (-15) = 64 > 0.

Now, from all the factors of the number 15 (1 and 15, 3 and 5), we select those whose difference is 2. These will be the numbers 3 and 5. We put a minus sign in front of the smaller number, i.e. sign of the second coefficient of the equation. Thus, we obtain the roots of the equation x 1 = -3 and x 2 = 5.

Answer. x 1 = -3 and x 2 = 5.

Example 2.

Solve the equation x 2 + 5x – 6 = 0.

Solution.

Let's check whether this equation has roots. To do this, we find a discriminant:

D = b 2 – 4ac = 25 + 24 = 49 > 0. The equation has two different roots.

Possible factors of the number 6 are 2 and 3, 6 and 1. The difference is 5 for the pair 6 and 1. In this example, the coefficient of the second term has a plus sign, so the smaller number will have the same sign. But before the second number there will be a minus sign.

Answer: x 1 = -6 and x 2 = 1.

Vieta's theorem can also be written for a complete quadratic equation. So, if the quadratic equation ax 2 + bx + c = 0 has roots x 1 and x 2, then the equalities hold for them

x 1 + x 2 = -(b/a) And x 1 x 2 = (c/a). However, the application of this theorem in a complete quadratic equation is quite problematic, because if there are roots, at least one of them is a fractional number. And working with selecting fractions is quite difficult. But still there is a way out.

Consider the complete quadratic equation ax 2 + bx + c = 0. Multiply its left and right sides by the coefficient a. The equation will take the form (ax) 2 + b(ax) + ac = 0. Now let's introduce a new variable, for example t = ax.

In this case, the resulting equation will turn into a reduced quadratic equation of the form t 2 + bt + ac = 0, the roots of which t 1 and t 2 (if any) can be determined by Vieta’s theorem.

In this case, the roots of the original quadratic equation will be

x 1 = (t 1 / a) and x 2 = (t 2 / a).

Example 3.

Solve the equation 15x 2 – 11x + 2 = 0.

Solution.

Let's create an auxiliary equation. Let's multiply each term of the equation by 15:

15 2 x 2 – 11 15x + 15 2 = 0.

We make the replacement t = 15x. We have:

t 2 – 11t + 30 = 0.

According to Vieta's theorem, the roots of this equation will be t 1 = 5 and t 2 = 6.

We return to the replacement t = 15x:

5 = 15x or 6 = 15x. So x 1 = 5/15 and x 2 = 6/15. We reduce and get the final answer: x 1 = 1/3 and x 2 = 2/5.

Answer. x 1 = 1/3 and x 2 = 2/5.

To master solving quadratic equations using Vieta's theorem, students need to practice as much as possible. This is precisely the secret of success.

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In mathematics, there are special techniques with which many quadratic equations can be solved very quickly and without any discriminants. Moreover, with proper training, many begin to solve quadratic equations orally, literally “at first sight.”

Unfortunately, in the modern course of school mathematics, such technologies are almost not studied. But you need to know! And today we will look at one of these techniques - Vieta's theorem. First, let's introduce a new definition.

A quadratic equation of the form x 2 + bx + c = 0 is called reduced. Please note that the coefficient for x 2 is 1. There are no other restrictions on the coefficients.

x 2 + 7x + 12 = 0 is a reduced quadratic equation;
x 2 − 5x + 6 = 0 - also reduced;
2x 2 − 6x + 8 = 0 - but this is not given at all, since the coefficient of x 2 is equal to 2.

Of course, any quadratic equation of the form ax 2 + bx + c = 0 can be reduced - just divide all the coefficients by the number a. We can always do this, since the definition of a quadratic equation implies that a ≠ 0.

True, these transformations will not always be useful for finding roots. Below we will make sure that this should be done only when in the final equation given by the square all the coefficients are integer. For now, let's look at the simplest examples:

Task. Convert the quadratic equation to the reduced equation:

3x 2 − 12x + 18 = 0;

−4x 2 + 32x + 16 = 0;

1.5x 2 + 7.5x + 3 = 0;

2x 2 + 7x − 11 = 0.

Let's divide each equation by the coefficient of the variable x 2. We get:

3x 2 − 12x + 18 = 0 ⇒ x 2 − 4x + 6 = 0 - divided everything by 3;
−4x 2 + 32x + 16 = 0 ⇒ x 2 − 8x − 4 = 0 - divided by −4;
1.5x 2 + 7.5x + 3 = 0 ⇒ x 2 + 5x + 2 = 0 - divided by 1.5, all coefficients became integers;
2x 2 + 7x − 11 = 0 ⇒ x 2 + 3.5x − 5.5 = 0 - divided by 2. In this case, fractional coefficients appeared.

As you can see, the above quadratic equations can have integer coefficients even if the original equation contained fractions.

Now let us formulate the main theorem, for which, in fact, the concept of a reduced quadratic equation was introduced:

Vieta's theorem. Consider the reduced quadratic equation of the form x 2 + bx + c = 0. Assume that this equation has real roots x 1 and x 2. In this case, the following statements are true:

x 1 + x 2 = −b. In other words, the sum of the roots of the given quadratic equation is equal to the coefficient of the variable x, taken with the opposite sign;

x 1 x 2 = c. The product of the roots of a quadratic equation is equal to the free coefficient.

Examples. For simplicity, we will consider only the above quadratic equations that do not require additional transformations:

x 2 − 9x + 20 = 0 ⇒ x 1 + x 2 = − (−9) = 9; x 1 x 2 = 20; roots: x 1 = 4; x 2 = 5;
x 2 + 2x − 15 = 0 ⇒ x 1 + x 2 = −2; x 1 x 2 = −15; roots: x 1 = 3; x 2 = −5;
x 2 + 5x + 4 = 0 ⇒ x 1 + x 2 = −5; x 1 x 2 = 4; roots: x 1 = −1; x 2 = −4.

Vieta's theorem gives us additional information about the roots of a quadratic equation. At first glance, this may seem difficult, but even with minimal training you will learn to “see” the roots and literally guess them in a matter of seconds.

Task. Solve the quadratic equation:

x 2 − 9x + 14 = 0;

x 2 − 12x + 27 = 0;

3x 2 + 33x + 30 = 0;

−7x 2 + 77x − 210 = 0.

Let’s try to write out the coefficients using Vieta’s theorem and “guess” the roots:

x 2 − 9x + 14 = 0 is a reduced quadratic equation.
By Vieta’s theorem we have: x 1 + x 2 = −(−9) = 9; x 1 · x 2 = 14. It is easy to see that the roots are the numbers 2 and 7;
x 2 − 12x + 27 = 0 - also reduced.
By Vieta's theorem: x 1 + x 2 = −(−12) = 12; x 1 x 2 = 27. Hence the roots: 3 and 9;
3x 2 + 33x + 30 = 0 - this equation is not reduced. But we will correct this now by dividing both sides of the equation by the coefficient a = 3. We get: x 2 + 11x + 10 = 0.
We solve using Vieta’s theorem: x 1 + x 2 = −11; x 1 x 2 = 10 ⇒ roots: −10 and −1;
−7x 2 + 77x − 210 = 0 - again the coefficient for x 2 is not equal to 1, i.e. equation not given. We divide everything by the number a = −7. We get: x 2 − 11x + 30 = 0.
By Vieta's theorem: x 1 + x 2 = −(−11) = 11; x 1 x 2 = 30; From these equations it is easy to guess the roots: 5 and 6.

From the above reasoning it is clear how Vieta’s theorem simplifies the solution of quadratic equations. No complicated calculations, no arithmetic roots or fractions. And we didn’t even need a discriminant (see lesson “Solving quadratic equations”).

Of course, in all our reflections we proceeded from two important assumptions, which, generally speaking, are not always met in real problems:

The quadratic equation is reduced, i.e. the coefficient for x 2 is 1;
The equation has two different roots. From an algebraic point of view, in this case the discriminant is D > 0 - in fact, we initially assume that this inequality is true.

However, in typical mathematical problems these conditions are met. If the calculation results in a “bad” quadratic equation (the coefficient of x 2 is different from 1), this can be easily corrected - look at the examples at the very beginning of the lesson. I’m generally silent about roots: what kind of problem is this that has no answer? Of course there will be roots.

Thus, the general scheme for solving quadratic equations using Vieta’s theorem is as follows:

Reduce the quadratic equation to the given one, if this has not already been done in the problem statement;
If the coefficients in the above quadratic equation are fractional, we solve using the discriminant. You can even go back to the original equation to work with more "handy" numbers;
In the case of integer coefficients, we solve the equation using Vieta’s theorem;
If you can’t guess the roots within a few seconds, forget about Vieta’s theorem and solve using the discriminant.

Task. Solve the equation: 5x 2 − 35x + 50 = 0.

So, we have before us an equation that is not reduced, because coefficient a = 5. Divide everything by 5, we get: x 2 − 7x + 10 = 0.

All coefficients of a quadratic equation are integer - let's try to solve it using Vieta's theorem. We have: x 1 + x 2 = −(−7) = 7; x 1 · x 2 = 10. In this case, the roots are easy to guess - they are 2 and 5. There is no need to count using the discriminant.

Task. Solve the equation: −5x 2 + 8x − 2.4 = 0.

Let's look: −5x 2 + 8x − 2.4 = 0 - this equation is not reduced, let's divide both sides by the coefficient a = −5. We get: x 2 − 1.6x + 0.48 = 0 - an equation with fractional coefficients.

It is better to return to the original equation and count through the discriminant: −5x 2 + 8x − 2.4 = 0 ⇒ D = 8 2 − 4 · (−5) · (−2.4) = 16 ⇒ ... ⇒ x 1 = 1.2; x 2 = 0.4.

Task. Solve the equation: 2x 2 + 10x − 600 = 0.

First, let's divide everything by the coefficient a = 2. We get the equation x 2 + 5x − 300 = 0.

This is the reduced equation, according to Vieta’s theorem we have: x 1 + x 2 = −5; x 1 x 2 = −300. It is difficult to guess the roots of the quadratic equation in this case - personally, I was seriously stuck when solving this problem.

You will have to look for roots through the discriminant: D = 5 2 − 4 · 1 · (−300) = 1225 = 35 2 . If you don't remember the root of the discriminant, I'll just note that 1225: 25 = 49. Therefore, 1225 = 25 49 = 5 2 7 2 = 35 2.

Now that the root of the discriminant is known, solving the equation is not difficult. We get: x 1 = 15; x 2 = −20.