There are a number of relationships in quadratic equations. The main ones are the relationships between roots and coefficients. Also in quadratic equations there are a number of relationships that are given by Vieta’s theorem.
In this topic, we will present Vieta’s theorem itself and its proof for a quadratic equation, the theorem inverse to Vieta’s theorem, and analyze a number of examples of solving problems. Special attention in the material we will focus on Vieta’s formulas, which define the relationship between real roots algebraic equation degrees n and its coefficients.
Formulation and proof of Vieta's theorem
Formula for the roots of a quadratic equation a x 2 + b x + c = 0 of the form x 1 = - b + D 2 · a, x 2 = - b - D 2 · a, where D = b 2 − 4 a c, establishes relationships x 1 + x 2 = - b a, x 1 x 2 = c a. This is confirmed by Vieta's theorem.
Theorem 1
In a quadratic equation a x 2 + b x + c = 0, Where x 1 And x 2– roots, the sum of the roots will be equal to the ratio of the coefficients b And a, which was taken with the opposite sign, and the product of the roots will be equal to the ratio of the coefficients c And a, i.e. x 1 + x 2 = - b a, x 1 x 2 = c a.
Evidence 1
We offer you the following scheme for carrying out the proof: take the formula of roots, compose the sum and product of the roots of the quadratic equation and then transform the resulting expressions in order to make sure that they are equal - b a And c a respectively.
Let's make the sum of the roots x 1 + x 2 = - b + D 2 · a + - b - D 2 · a. Let's reduce the fractions to common denominator- b + D 2 · a + - b - D 2 · a = - b + D + - b - D 2 · a . Let's open the parentheses in the numerator of the resulting fraction and present similar terms: - b + D + - b - D 2 · a = - b + D - b - D 2 · a = - 2 · b 2 · a . Let's reduce the fraction by: 2 - b a = - b a.
This is how we proved the first relation of Vieta’s theorem, which relates to the sum of the roots of a quadratic equation.
Now let's move on to the second relationship.
To do this, we need to compose the product of the roots of the quadratic equation: x 1 · x 2 = - b + D 2 · a · - b - D 2 · a.
Let's remember the rule for multiplying fractions and write the last product as follows: - b + D · - b - D 4 · a 2.
Let's multiply a bracket by a bracket in the numerator of the fraction, or use the difference of squares formula to transform this product faster: - b + D · - b - D 4 · a 2 = - b 2 - D 2 4 · a 2 .
Let's use the definition square root in order to make the following transition: - b 2 - D 2 4 · a 2 = b 2 - D 4 · a 2. Formula D = b 2 − 4 a c corresponds to the discriminant of a quadratic equation, therefore, into a fraction instead of D can be substituted b 2 − 4 a c:
b 2 - D 4 a 2 = b 2 - (b 2 - 4 a c) 4 a 2
Let's open the brackets, add similar terms and get: 4 · a · c 4 · a 2 . If we shorten it to 4 a, then what remains is c a . This is how we proved the second relation of Vieta’s theorem for the product of roots.
The proof of Vieta's theorem can be written in a very laconic form if we omit the explanations:
x 1 + x 2 = - b + D 2 a + - b - D 2 a = - b + D + - b - D 2 a = - 2 b 2 a = - b a , x 1 x 2 = - b + D 2 · a · - b - D 2 · a = - b + D · - b - D 4 · a 2 = - b 2 - D 2 4 · a 2 = b 2 - D 4 · a 2 = = D = b 2 - 4 · a · c = b 2 - b 2 - 4 · a · c 4 · a 2 = 4 · a · c 4 · a 2 = c a .
With the discriminant of the quadratic equation equal to zero the equation will have only one root. To be able to apply Vieta's theorem to such an equation, we can assume that the equation, with a discriminant equal to zero, has two identical roots. Indeed, when D=0 the root of the quadratic equation is: - b 2 · a, then x 1 + x 2 = - b 2 · a + - b 2 · a = - b + (- b) 2 · a = - 2 · b 2 · a = - b a and x 1 · x 2 = - b 2 · a · - b 2 · a = - b · - b 4 · a 2 = b 2 4 · a 2 , and since D = 0, that is, b 2 - 4 · a · c = 0, whence b 2 = 4 · a · c, then b 2 4 · a 2 = 4 · a · c 4 · a 2 = c a.
Most often in practice, Vieta's theorem is applied to the reduced quadratic equation of the form x 2 + p x + q = 0, where the leading coefficient a is equal to 1. In this regard, Vieta’s theorem is formulated specifically for equations of this type. This is without loss of generality due to the fact that any quadratic equation can be replaced by an equivalent equation. To do this, you need to divide both of its parts by a number a different from zero.
Let us give another formulation of Vieta's theorem.
Theorem 2
Sum of roots in the given quadratic equation x 2 + p x + q = 0 will be equal to the coefficient of x, which is taken with the opposite sign, the product of the roots will be equal to the free term, i.e. x 1 + x 2 = − p, x 1 x 2 = q.
Theorem converse to Vieta's theorem
If you look carefully at the second formulation of Vieta’s theorem, you can see that for the roots x 1 And x 2 reduced quadratic equation x 2 + p x + q = 0 the following relations will be valid: x 1 + x 2 = − p, x 1 · x 2 = q. From these 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. So we come to a statement that is the converse of Vieta’s theorem.
We now propose to formalize this statement as a theorem and carry out its proof.
Theorem 3
If the numbers x 1 And x 2 are such that x 1 + x 2 = − p And x 1 x 2 = q, That x 1 And x 2 are the roots of the reduced quadratic equation x 2 + p x + q = 0.
Evidence 2
Replacing odds p And q to their expression through x 1 And x 2 allows you to transform the equation x 2 + p x + q = 0 into an equivalent .
If we substitute the number into the resulting equation x 1 instead of x, then we get the equality x 1 2 − (x 1 + x 2) x 1 + x 1 x 2 = 0. This is equality for any x 1 And x 2 turns into a true numerical equality 0 = 0 , because 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. It means that x 1- root of the equation x 2 − (x 1 + x 2) x + x 1 x 2 = 0, So what x 1 is also the root of the equivalent equation x 2 + p x + q = 0.
Substitution into equation x 2 − (x 1 + x 2) x + x 1 x 2 = 0 numbers x 2 instead of x allows us to obtain equality x 2 2 − (x 1 + x 2) x 2 + x 1 x 2 = 0. This equality can be considered true, 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. It turns out that x 2 is the root of the equation x 2 − (x 1 + x 2) x + x 1 x 2 = 0, and hence the equations x 2 + p x + q = 0.
Theorem, converse of the theorem Vieta, proven.
Examples of using Vieta's theorem
Let's now begin to analyze the most typical examples on the topic. Let's start by analyzing problems that require the application of the theorem inverse to Vieta's theorem. It can be used to check numbers produced by calculations to see if they are the roots of a given quadratic equation. To do this, you need to calculate their sum and difference, and then check the validity of the relations x 1 + x 2 = - b a, x 1 · x 2 = a c.
The fulfillment of both relations indicates that the numbers obtained during the calculations are the roots of the equation. If we see that at least one of the conditions is not met, then these numbers cannot be the roots of the quadratic equation given in the problem statement.
Example 1
Which of the pairs of numbers 1) x 1 = − 5, x 2 = 3, or 2) x 1 = 1 - 3, x 2 = 3 + 3, or 3) x 1 = 2 + 7 2, x 2 = 2 - 7 2 is a pair of roots of a quadratic equation 4 x 2 − 16 x + 9 = 0?
Solution
Let's find the coefficients of the quadratic equation 4 x 2 − 16 x + 9 = 0. This is a = 4, b = − 16, c = 9. According to Vieta's theorem, the sum of the roots of a quadratic equation must be equal to - b a, that is, 16 4 = 4 , and the product of the roots must be equal c a, that is, 9 4 .
Let's check the obtained numbers by calculating the sum and product of numbers from three given pairs and comparing them with the obtained values.
In the first case x 1 + x 2 = − 5 + 3 = − 2. This value is different from 4, therefore, the check does not need to be continued. According to the theorem converse to Vieta's theorem, we can immediately conclude that the first pair of numbers are not the roots of this quadratic equation.
In the second case, x 1 + x 2 = 1 - 3 + 3 + 3 = 4. We see that the first condition is met. But the second condition is not: x 1 · x 2 = 1 - 3 · 3 + 3 = 3 + 3 - 3 · 3 - 3 = - 2 · 3. The value we got is different from 9 4 . This means that the second pair of numbers are not the roots of the quadratic equation.
Let's move on to consider the third pair. Here x 1 + x 2 = 2 + 7 2 + 2 - 7 2 = 4 and x 1 x 2 = 2 + 7 2 2 - 7 2 = 2 2 - 7 2 2 = 4 - 7 4 = 16 4 - 7 4 = 9 4. Both conditions are met, which means that x 1 And x 2 are the roots of a given quadratic equation.
Answer: x 1 = 2 + 7 2 , x 2 = 2 - 7 2
We can also use the converse of Vieta's theorem to find the roots of a quadratic equation. The simplest way is to select integer roots of the given quadratic equations with integer coefficients. Other options can be considered. But this can significantly complicate calculations.
To select roots, we 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.
Example 2
As an example, we use the quadratic equation x 2 − 5 x + 6 = 0. Numbers x 1 And x 2 can be the roots of this equation if two equalities are satisfied x 1 + x 2 = 5 And x 1 x 2 = 6. Let's select these numbers. These are numbers 2 and 3, since 2 + 3 = 5 And 2 3 = 6. It turns out that 2 and 3 are the roots of this quadratic equation.
The converse of Vieta's theorem can be used to find the second root when the first is known or obvious. To do this, we can use the relations x 1 + x 2 = - b a, x 1 · x 2 = c a.
Example 3
Consider the quadratic equation 512 x 2 − 509 x − 3 = 0. It is necessary to find the roots of this equation.
Solution
The first root of the equation is 1, since the sum of the coefficients of this quadratic equation is zero. It turns out that x 1 = 1.
Now let's find the second root. For this you can use the relation x 1 x 2 = c a. It turns out that 1 x 2 = − 3,512, where x 2 = - 3,512.
Answer: roots of the quadratic equation specified in the problem statement 1 And - 3 512 .
It is possible to select roots using the theorem inverse to Vieta’s theorem only in simple cases. In other cases, it is better to search using the formula for the roots of a quadratic equation through a discriminant.
Thanks to the converse of Vieta's theorem, we can also construct quadratic equations using the existing roots x 1 And x 2. To do this, we need to calculate the sum of the roots, which gives the coefficient for x with the opposite sign of the given quadratic equation, and the product of the roots, which gives the free term.
Example 4
Write a quadratic equation whose roots are numbers − 11 And 23 .
Solution
Let's assume that x 1 = − 11 And x 2 = 23. The sum and product of these numbers will be equal: x 1 + x 2 = 12 And x 1 x 2 = − 253. This means that the second coefficient is 12, the free term − 253.
Let's make an equation: x 2 − 12 x − 253 = 0.
Answer: x 2 − 12 x − 253 = 0 .
We can use Vieta's theorem to solve problems that involve the signs of the roots of quadratic equations. The connection between Vieta's theorem is related to the signs of the roots of the reduced quadratic equation x 2 + p x + q = 0 in the following way:
- if the quadratic equation has real roots and if the intercept term q is a positive number, then these roots will have the same sign “+” or “-”;
- if the quadratic equation has roots and if the intercept term q is a negative number, then one root will be “+”, and the second “-”.
Both of these statements are a consequence of the formula x 1 x 2 = q and rules for multiplying positive and negative numbers, as well as numbers with different signs.
Example 5
Are the roots of a quadratic equation x 2 − 64 x − 21 = 0 positive?
Solution
According to Vieta’s theorem, the roots of this equation cannot both be positive, since they must satisfy the equality x 1 x 2 = − 21. This is impossible with positive x 1 And x 2.
Answer: No
Example 6
At what parameter values r quadratic equation x 2 + (r + 2) x + r − 1 = 0 will have two real roots with different signs.
Solution
Let's start by finding the values of which r, for which the equation will have two roots. Let's find the discriminant and see at what r it will take positive values. D = (r + 2) 2 − 4 1 (r − 1) = r 2 + 4 r + 4 − 4 r + 4 = r 2 + 8. Expression value r 2 + 8 positive for any real r, therefore, the discriminant will be greater than zero for any real r. This means that the original quadratic equation will have two roots for any real values of the parameter r.
Now let's see when the roots have different signs. This is possible if their product is negative. According to Vieta's theorem, the product of the roots of the reduced quadratic equation is equal to the free term. This means that the correct solution will be those values r, for which the free term r − 1 is negative. Let's decide linear inequality r − 1< 0 , получаем r < 1 .
Answer: at r< 1 .
Vieta formulas
There are a number of formulas that are applicable to carry out operations with the roots and coefficients of not only quadratic, but also cubic and other types of equations. They are called Vieta's formulas.
For an algebraic equation of degree n of the form a 0 · x n + a 1 · x n - 1 + . . . + a n - 1 x + a n = 0 the equation is considered to have n real roots x 1 , x 2 , … , x n, among which may be the same:
x 1 + x 2 + x 3 + . . . + x n = - a 1 a 0 , x 1 · x 2 + x 1 · x 3 + . . . + x n - 1 · x n = a 2 a 0 , x 1 · x 2 · x 3 + x 1 · x 2 · x 4 + . . . + x n - 2 · x n - 1 · x n = - a 3 a 0 , . . . x 1 · x 2 · x 3 · . . . · x n = (- 1) n · a n a 0
Definition 1
Vieta's formulas help us obtain:
- theorem on the decomposition of a polynomial into linear factors;
- determination of equal polynomials through the equality of all their corresponding coefficients.
Thus, the polynomial a 0 · x n + a 1 · x n - 1 + . . . + a n - 1 · x + a n and its expansion into linear factors of the form a 0 · (x - x 1) · (x - x 2) · . . . · (x - x n) are equal.
If we expand the parentheses in last work and equate the corresponding coefficients, we obtain Vieta’s formulas. Taking n = 2, we can obtain Vieta's formula for the quadratic equation: x 1 + x 2 = - a 1 a 0, x 1 · x 2 = a 2 a 0.
Definition 2
Vieta's formula for the cubic equation:
x 1 + x 2 + x 3 = - a 1 a 0 , x 1 x 2 + x 1 x 3 + x 2 x 3 = a 2 a 0 , x 1 x 2 x 3 = - a 3 a 0
The left side of the Vieta formula contains the so-called elementary symmetric polynomials.
If you notice an error in the text, please highlight it and press Ctrl+Enter
2.5 Vieta's formula for polynomials (equations) 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 equal degrees 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 – 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. ...
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 and 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, general scheme solving quadratic equations using Vieta’s theorem looks like this:
- 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.
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 proven.
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 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.
website, when copying material in full or in part, a link to the source is required.
Vieta's theorem (more precisely, the theorem inverse to Vieta's theorem) allows you to reduce the time for solving quadratic equations. You just need to know how to use it. How to learn to solve quadratic equations using Vieta's theorem? It's not difficult if you think about it a little.
Now we will only talk about solving the reduced quadratic equation using Vieta’s theorem. A reduced quadratic equation is an equation in which a, that is, the coefficient of x², is equal to one. It is also possible to solve quadratic equations that are not given using Vieta’s theorem, but at least one of the roots is not an integer. They are more difficult to guess.
The inverse theorem to Vieta's theorem states: if the numbers x1 and x2 are such that
then x1 and x2 are the roots of the quadratic equation
When solving a quadratic equation using Vieta's theorem, only 4 options are possible. If you remember the line of reasoning, you can learn to find whole roots very quickly.
I. If q is a positive number,
this means that the roots x1 and x2 are numbers of the same sign (since only multiplying numbers with the same signs produces a positive number).
I.a. If -p is a positive number, (respectively, p<0), то оба корня x1 и x2 — положительные числа (поскольку складывали числа одного знака и получили положительное число).
I.b. If -p is a negative number, (respectively, p>0), then both roots are negative numbers (we added numbers of the same sign and got a negative number).
II. If q is a negative number,
this means that the roots x1 and x2 have different signs (when multiplying numbers, a negative number is obtained only when the signs of the factors are different). In this case, x1 + x2 is no longer a sum, but a difference (after all, when adding numbers with different signs, we subtract the smaller from the larger in absolute value). Therefore, x1+x2 shows how much the roots x1 and x2 differ, that is, how much one root is greater than the other (in absolute value).
II.a. If -p is a positive number, (that is, p<0), то больший (по модулю) корень — положительное число.
II.b. If -p is a negative number, (p>0), then the larger (modulo) root is a negative number.
Let's consider solving quadratic equations using Vieta's theorem using examples.
Solve the given quadratic equation using Vieta's theorem:
Here q=12>0, so the roots x1 and x2 are numbers of the same sign. Their sum is -p=7>0, so both roots are positive numbers. We select integers whose product is equal to 12. These are 1 and 12, 2 and 6, 3 and 4. The sum is 7 for the pair 3 and 4. This means that 3 and 4 are the roots of the equation.
In this example, q=16>0, which means that the roots x1 and x2 are numbers of the same sign. Their sum is -p=-10<0, поэтому оба корня — отрицательные числа. Подбираем числа, произведение которых равно 16. Это 1 и 16, 2 и 8, 4 и 4. Сумма 2 и 8 равна 10, а раз нужны отрицательные числа, то искомые корни — это -2 и -8.
Here q=-15<0, что означает, что корни x1 и x2 — числа разных знаков. Поэтому 2 — это уже не их сумма, а разность, то есть числа отличаются на 2. Подбираем числа, произведение которых равно 15, отличающиеся на 2. Произведение равно 15 у 1 и 15, 3 и 5. Отличаются на 2 числа в паре 3 и 5. Поскольку -p=2>0, then the larger number is positive. So the roots are 5 and -3.
q=-36<0, значит, корни x1 и x2 имеют разные знаки. Тогда 5 — это то, насколько отличаются x1 и x2 (по модулю, то есть пока что без учета знака). Среди чисел, произведение которых равно 36: 1 и 36, 2 и 18, 3 и 12, 4 и 9 — выбираем пару, в которой числа отличаются на 5. Это 4 и 9. Осталось определить их знаки. Поскольку -p=-5<0, бОльшее число имеет знак минус. Поэтому корни данного уравнения равны -9 и 4.