François Vieta (1540-1603) - mathematician, creator of the famous Vieta formulas
Vieta's theorem needed to quickly solve quadratic equations (in simple terms).
In more detail, t Vieta's theorem is the sum of the roots of a given quadratic equation equals the second coefficient, which is taken with the opposite sign, and the product is equal to the free term. This property has any given quadratic equation that has roots.
Using the Vieta theorem, you can easily solve quadratic equations by selection, so let's say “thank you” to this mathematician with a sword in his hands for our happy 7th grade.
Proof of Vieta's theorem
To prove the theorem, you can use the well-known root formulas, thanks to which we will compose the sum and product of the roots of the quadratic equation. Only after that we can make sure that they are equal and, accordingly, .
Let's say we have an equation: . This equation has the following roots: and . Let us prove that , .
According to the formulas of the roots of the quadratic equation:
1. Find the sum of the roots:
Let's analyze this equation, as we got it exactly like this:
= .
Step 1. We reduce the fractions to a common denominator, it turns out:
= = .
Step 2. We got a fraction where you need to open the brackets:
We reduce the fraction by 2 and get:
We have proved the relation for the sum of the roots of a quadratic equation using Vieta's theorem.
2. Find the product of the roots:
= = = = = .
Let's prove this equation:
Step 1. There is a rule for multiplying fractions, according to which we multiply this equation:
Now remember the definition square root and consider:
= .
Step 3. We recall the discriminant of the quadratic equation: . Therefore, instead of D (discriminant), we substitute in the last fraction, then we get:
= .
Step 4. Open the brackets and add like terms to fractions:
Step 5. We reduce "4a" and get.
So we have proved the relation for the product of roots according to Vieta's theorem.
IMPORTANT!If the discriminant is zero, then the quadratic equation has only one root.
Theorem inverse to Vieta's theorem
According to the theorem, the inverse of Vieta's theorem, we can check whether our equation is solved correctly. To understand the theorem itself, we need to consider it in more detail.
If the numbers are:
And then they are the roots of the quadratic equation.
Proof of Vieta's converse theorem
Step 1.Let us substitute expressions for its coefficients into the equation:
Step 2Let's transform the left side of the equation:
Step 3. Let's find the roots of the equation, and for this we use the property that the product is equal to zero:
Or . Where does it come from: or.
Examples with solutions by Vieta's theorem
Example 1
Exercise
Find the sum, product and sum of squares of the roots of a quadratic equation without finding the roots of the equation.
Solution
Step 1. Recall the discriminant formula. We substitute our numbers under the letters. That is, , is a substitute for , and . This implies:
It turns out:
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We express the sum of the squares of the roots through their sum and product:
Answer
7; 12; 25.
Example 2
Exercise
Solve the equation. In this case, do not use the formulas of the quadratic equation.
Solution
This equation has roots that are greater than zero in terms of the discriminant (D). Accordingly, according to the Vieta theorem, the sum of the roots of this equation is 4, and the product is 5. First, we determine the divisors of the number, the sum of which is 4. These are the numbers "5" and "-1". Their product is equal to - 5, and the sum - 4. Hence, according to the theorem, the converse of Vieta's theorem, they are the roots of this equation.
Answer
AND Example 4
Exercise
Write an equation where each root is twice the corresponding root of the equation:
Solution
By Vieta's theorem, the sum of the roots of this equation is 12, and the product = 7. Hence, the two roots are positive.
The sum of the roots of the new equation will be equal to:
And the work.
By a theorem converse to Vieta's theorem, the new equation has the form:
Answer
The result was an equation, each root of which is twice as large:
So, we looked at how to solve an equation using Vieta's theorem. It is very convenient to use this theorem if tasks are solved that are associated with the signs of the roots of quadratic equations. That is, if the free term in the formula is a positive number, and if there are real roots in the quadratic equation, then both of them can be either negative or positive.
And if the free term is a negative number, and if there are real roots in the quadratic equation, then both signs will be different. That is, if one root is positive, then the other root will only be negative.
Useful sources:
- Dorofeev G. V., Suvorova S. B., Bunimovich E. A. Algebra Grade 8: Moscow “Enlightenment”, 2016 – 318 p.
- Rubin A. G., Chulkov P. V. - textbook Algebra Grade 8: Moscow "Balass", 2015 - 237 p.
- Nikolsky S. M., Potopav M. K., Reshetnikov N. N., Shevkin A. V. – Algebra Grade 8: Moscow “Enlightenment”, 2014 – 300
Vieta's theorem, inverse Vieta formula and examples with solution for dummies updated: November 22, 2019 by: Scientific Articles.Ru
There are a number of relationships in quadratic equations. The main ones are the relations between roots and coefficients. Also, a number of relationships work in quadratic equations, which are given by the Vieta theorem.
In this topic, we present the Vieta theorem itself and its proof for a quadratic equation, a theorem converse to Vieta's theorem, and analyze a number of examples of solving problems. We will pay special attention in the material to the consideration of the Vieta formulas, which define the connection between the real roots of the algebraic equation of degree n and its coefficients.
Statement and proof of Vieta's theorem
The formula for the roots of a quadratic equation a x 2 + b x + c = 0 of the form x 1 \u003d - b + D 2 a, x 2 \u003d - b - D 2 a, where D = b 2 − 4 a c, establishes the ratio x 1 + x 2 \u003d - 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 x2- 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 \u003d - b a, x 1 x 2 = c a.
Proof 1
We offer you the following scheme for conducting the proof: we take the formula of the 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.
Compose the sum of the roots x 1 + x 2 \u003d - b + D 2 a + - b - D 2 a. Let's bring the fractions to a common denominator - b + D 2 · a + - b - D 2 · a = - b + D + - b - D 2 · a. Let's open the brackets in the numerator of the resulting fraction and give similar terms: - b + D + - b - D 2 a = - b + D - b - D 2 a = - 2 b 2 a . Reduce the fraction by: 2 - b a \u003d - b a.
So we have proved the first relation of Vieta's theorem, which refers to the sum of the roots of a quadratic equation.
Now let's move on to the second relation.
To do this, we need to compose the product of the roots of the quadratic equation: x 1 x 2 \u003d - b + D 2 a - b - D 2 a.
Recall the rule for multiplying fractions and write the last product as follows: - b + D · - b - D 4 · a 2 .
We will carry out the multiplication of the bracket by the bracket in the numerator of the fraction, or we will use the formula of the difference of squares in order to transform this product faster: - b + D · - b - D 4 · a 2 = - b 2 - D 2 4 · a 2 .
Let's use definition of a square root to carry out 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 the quadratic equation, therefore, into a fraction instead of D can be substituted b 2 − 4 a c:
b 2 - D 4 a 2 \u003d b 2 - (b 2 - 4 a c) 4 a 2
Let's open the brackets, give like terms and get: 4 · a · c 4 · a 2 . If we shorten it to 4 a, then c a remains. So we have proved the second relation of the Vieta theorem for the product of roots.
The record of the proof of Vieta's theorem can have a very concise form, if we omit the explanations:
x 1 + x 2 \u003d - b + D 2 a + - b - D 2 a \u003d - b + D + - b - D 2 a \u003d - 2 b 2 a \u003d - 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 .
When the discriminant of a quadratic equation is 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, at D=0 the root of the quadratic equation is: - b 2 a, then x 1 + x 2 \u003d - b 2 a + - b 2 a \u003d - b + (- b) 2 a \u003d - 2 b 2 a \u003d - b a and x 1 x 2 \u003d - b 2 a - b 2 a \u003d - b - b 4 a 2 \u003d b 2 4 a 2, and since D \u003d 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, the Vieta theorem is applied in relation 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 precisely for equations of this type. This does not limit the generality due to the fact that any quadratic equation can be replaced by an equivalent equation. To do this, it is necessary to divide both its parts by the number a, which is different from zero.
Let us give one more formulation of Vieta's theorem.
Theorem 2
The sum of the roots in the given quadratic equation x 2 + p x + q = 0 will be equal to the coefficient at 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 \u003d - p, x 1 x 2 \u003d q.
Theorem inverse to Vieta's theorem
If you look closely at the second formulation of Vieta's theorem, you can see that for the roots x 1 And x2 reduced quadratic equation x 2 + p x + q = 0 relations x 1 + x 2 = − p , x 1 · x 2 = q will be valid. From these relations x 1 + x 2 \u003d - p, x 1 x 2 \u003d q, it follows that x 1 And x2 are the roots of the quadratic equation x 2 + p x + q = 0. Thus we arrive at a statement which is the inverse of Vieta's theorem.
We now propose to formalize this statement as a theorem and carry out its proof.
Theorem 3
If numbers x 1 And x2 are such that x 1 + x 2 = − p And x 1 x 2 = q, That x 1 And x2 are the roots of the reduced quadratic equation x 2 + p x + q = 0.
Proof 2
Change of coefficients p And q to their expression through x 1 And x2 allows you to transform the equation x 2 + p x + q = 0 in 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 equality for any x 1 And x2 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.
Equation Substitution x 2 − (x 1 + x 2) x + x 1 x 2 = 0 numbers x2 instead of x allows you to get 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 x2 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.
The theorem converse to Vieta's theorem is proved.
Examples of using Vieta's theorem
Let's now proceed to the analysis of the most typical examples on the topic. Let's start with the analysis of problems that require the application of the theorem, the converse of Vieta's theorem. It can be used to check the numbers obtained in the course of calculations, whether 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 ratios x 1 + x 2 = - b a, x 1 x 2 = a c.
The fulfillment of both relations indicates that the numbers obtained in the course of 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 condition of the problem.
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 the quadratic equation 4 x 2 − 16 x + 9 = 0?
Solution
Find the coefficients of the quadratic equation 4 x 2 − 16 x + 9 = 0 . This is a = 4 , b = − 16 , c = 9 . In accordance with the Vieta theorem, the sum of the roots of the quadratic equation must be equal to -b a, that is, 16 4 = 4 , and the product of the roots should be equal to 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 , so you don't need to continue checking. According to the theorem, the inverse of 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 \u003d 1 - 3 3 + 3 \u003d 3 + 3 - 3 3 - 3 \u003d - 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 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 satisfied, which means that x 1 And x2 are the roots of the given quadratic equation.
Answer: x 1 \u003d 2 + 7 2, x 2 \u003d 2 - 7 2
We can also use the inverse of Vieta's theorem to find the roots of a quadratic equation. The easiest way is to select integer roots of the given quadratic equations with integer coefficients. Other options may also be considered. But this can significantly complicate the calculations.
To select the roots, we 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.
Example 2
As an example, we use the quadratic equation x 2 − 5 x + 6 = 0. Numbers x 1 And x2 can be the roots of this equation if the two equalities are satisfied x1 + x2 = 5 And x 1 x 2 = 6. Let's pick those numbers. These are the numbers 2 and 3 because 2 + 3 = 5 And 2 3 = 6. It turns out that 2 and 3 are the roots of this quadratic equation.
The inverse of Vieta's theorem can be used to find the second root when the first is known or obvious. For this we can use the ratios 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. We need to find the roots of this equation.
Solution
The first root of the equation is 1 because 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. To do this, you can use the ratio x 1 x 2 = c a. It turns out that 1 x 2 = − 3 512, where x 2 \u003d - 3 512.
Answer: the roots of the quadratic equation specified in the condition of the problem 1 And - 3 512 .
It is possible to select roots using the theorem converse to Vieta's theorem only in simple cases. In other cases, it is better to search using the formula of the roots of the quadratic equation through the discriminant.
Thanks to the converse theorem of Vieta, we can also form quadratic equations given the roots x 1 And x2. To do this, we need to calculate the sum of the roots, which gives the coefficient at x with the opposite sign of the reduced 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 accept that x 1 = − 11 And x2 = 23. The sum and product of these numbers will be equal to: x1 + x2 = 12 And x 1 x 2 = − 253. This means that the second coefficient is 12, the free term − 253.
We make an equation: x 2 - 12 x - 253 = 0.
Answer: x 2 - 12 x - 253 = 0 .
We can use the Vieta theorem to solve problems that are related to 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 free term q is positive number, then these roots will have the same sign "+" or "-";
- if the quadratic equation has roots and if the free 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
By 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 not possible with positive x 1 And x2.
Answer: No
Example 6
At what values of the parameter r quadratic equation x 2 + (r + 2) x + r − 1 = 0 will have two real roots with different signs.
Solution
Let's begin with find the values what r, for which the equation has two roots. Let us find the discriminant and see for what r he will accept positive values. D = (r + 2) 2 − 4 1 (r − 1) = r 2 + 4 r + 4 − 4 r + 4 = r 2 + 8. Expression value r2 + 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 will have different signs. This is possible if their product is negative. According to the Vieta theorem, the product of the roots of the reduced quadratic equation is equal to the free term. So the correct solution is those values r, for which the free term r − 1 is negative. We will decide linear inequality r − 1< 0 , получаем r < 1 .
Answer: at r< 1 .
Vieta formulas
There are a number of formulas that are applicable for performing operations with roots and coefficients of not only square, but also cubic and other types of equations. They are called Vieta 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, which may include the following:
x 1 + x 2 + x 3 + . . . + x n \u003d - 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
Get the Vieta formulas help us:
- theorem on decomposition of a polynomial into linear factors;
- definition of equal polynomials through the equality of all their corresponding coefficients.
So, 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 open the brackets in the last product and equate the corresponding coefficients, then we get the Vieta formulas. Taking n \u003d 2, we can get the Vieta formula for the quadratic equation: x 1 + x 2 \u003d - a 1 a 0, x 1 x 2 \u003d a 2 a 0.
Definition 2
Vieta formula for 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 formulas contains the so-called elementary symmetric polynomials.
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Any complete quadratic equation ax2 + bx + c = 0 can be brought to mind x 2 + (b/a)x + (c/a) = 0, if we first divide each term by the coefficient a before x2. And if we introduce new notation (b/a) = p And (c/a) = q, then we will have the equation x 2 + px + q = 0, which in mathematics is called reduced quadratic equation.
The roots of the reduced quadratic equation and the coefficients p And q interconnected. It's confirmed Vieta's theorem, named after the French mathematician Francois Vieta, who lived at the end of the 16th century.
Theorem. The sum of the 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.
We write these ratios in the following form:
Let x 1 And x2 various roots of the reduced equation x 2 + px + q = 0. According to Vieta's theorem x1 + x2 = -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
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 according to 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 by (x 1 - x 2) ≠ 0 and express p.
(x 1 + x 2) + p = 0;
(x 1 + x 2) = -p.
The first equality is proved.
To prove the second equality, we substitute into the first equation
x 1 2 + px 1 + q \u003d 0 instead of the coefficient p, its equal number is (x 1 + x 2):
x 1 2 - (x 1 + x 2) x 1 + q \u003d 0
Transforming the left side of the equation, we get:
x 1 2 - x 2 2 - x 1 x 2 + q \u003d 0;
x 1 x 2 = q, which was to be proved.
Vieta's theorem is good because, even without knowing the roots of the quadratic equation, we can calculate their sum and product .
Vieta's theorem helps to determine the integer roots of the 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 given quadratic equation has the form x 2 + px + q \u003d 0, where x 1 and x 2 are its roots. According to the Vieta theorem x 1 + x 2 = -p and x 1 x 2 = q.
We can draw the following conclusion.
If in the equation the last term 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 is the same as 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 sign of the larger number is put in front of the result, you should proceed as follows:
- determine such factors of the number q so 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 obtained 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 \u003d b 2 - 4ac \u003d 4 - 4 (-15) \u003d 64\u003e 0.
Now, from all the factors of the number 15 (1 and 15, 3 and 5), we select those whose difference is equal to 2. These will be the numbers 3 and 5. We put a minus sign in front of the smaller number, i.e. the sign of the second coefficient of the equation. Thus, we get the roots of the equation x 1 \u003d -3 and x 2 \u003d 5.
Answer. x 1 = -3 and x 2 = 5.
Example 2.
Solve the equation x 2 + 5x - 6 = 0.
Solution.
Let's check if this equation has roots. To do this, we find the discriminant:
D \u003d b 2 - 4ac \u003d 25 + 24 \u003d 49\u003e 0. The equation has two different roots.
The possible factors of the number 6 are 2 and 3, 6 and 1. The difference is 5 for a pair of 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 ax2 + bx + c = 0 has roots x 1 and x 2 , then they satisfy the equalities
x 1 + x 2 = -(b/a) And x 1 x 2 = (c/a). However, the application of this theorem in the full quadratic equation is rather problematic, since if there are roots, at least one of them is a fractional number. And working with the selection of 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 turns 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 the Vieta 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.
We make 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 change t = 15x. We have:
t 2 - 11t + 30 = 0.
According to the Vieta 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. Thus 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 the solution of quadratic equations using the Vieta 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 tricks with which many quadratic equations are solved very quickly and without any discriminants. Moreover, with proper training, many begin to solve quadratic equations verbally, literally "at a glance."
Unfortunately, in the modern course of school mathematics, such technologies are almost not studied. And you need to know! And today we will consider 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 at x 2 is equal to 1. There are no other restrictions on the coefficients.
- x 2 + 7x + 12 = 0 is the reduced quadratic equation;
- x 2 − 5x + 6 = 0 is also reduced;
- 2x 2 − 6x + 8 = 0 - but this is not given at all, since the coefficient at x 2 is 2.
Of course, any quadratic equation of the form ax 2 + bx + c = 0 can be made reduced - it is enough to divide all the coefficients by the number a . We can always do this, since it follows from the definition of a quadratic equation that a ≠ 0.
True, these transformations will not always be useful for finding roots. A little lower, we will make sure that this should be done only when in the final squared equation all the coefficients are integer. For now, let's look at some simple examples:
Task. Convert quadratic equation to reduced:
- 3x2 − 12x + 18 = 0;
- −4x2 + 32x + 16 = 0;
- 1.5x2 + 7.5x + 3 = 0;
- 2x2 + 7x − 11 = 0.
Let's divide each equation by the coefficient of the variable x 2 . We get:
- 3x 2 - 12x + 18 \u003d 0 ⇒ x 2 - 4x + 6 \u003d 0 - divided everything by 3;
- −4x 2 + 32x + 16 = 0 ⇒ x 2 − 8x − 4 = 0 - divided by −4;
- 1.5x 2 + 7.5x + 3 \u003d 0 ⇒ x 2 + 5x + 2 \u003d 0 - divided by 1.5, all coefficients became integer;
- 2x 2 + 7x - 11 \u003d 0 ⇒ x 2 + 3.5x - 5.5 \u003d 0 - divided by 2. In this case, fractional coefficients arose.
As you can see, the given quadratic equations can have integer coefficients even if the original equation contained fractions.
Now we 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 \u003d 0. Suppose that this equation has real roots x 1 and x 2. In this case, the following statements are true:
- x1 + x2 = −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 given 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 \u003d 5;
- x 2 + 2x - 15 = 0 ⇒ x 1 + x 2 = -2; x 1 x 2 \u003d -15; roots: x 1 = 3; x 2 \u003d -5;
- x 2 + 5x + 4 = 0 ⇒ x 1 + x 2 = −5; x 1 x 2 = 4; roots: x 1 \u003d -1; x 2 \u003d -4.
Vieta's theorem gives us Additional information about the roots of a quadratic equation. At first glance, this may seem complicated, 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:
- x2 − 9x + 14 = 0;
- x 2 - 12x + 27 = 0;
- 3x2 + 33x + 30 = 0;
- −7x2 + 77x − 210 = 0.
Let's try to write down the coefficients according to the Vieta theorem and "guess" the roots:
- x 2 − 9x + 14 = 0 is a reduced quadratic equation.
By the Vieta 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 is also reduced.
By the Vieta 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 fix this now by dividing both sides of the equation by the coefficient a \u003d 3. We get: x 2 + 11x + 10 \u003d 0.
We solve according to the Vieta theorem: x 1 + x 2 = −11; x 1 x 2 = 10 ⇒ roots: −10 and −1; - −7x 2 + 77x − 210 \u003d 0 - again the coefficient at 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 the Vieta 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 can be seen how Vieta's theorem simplifies the solution of quadratic equations. No complicated calculations arithmetic roots and fractions. And even the discriminant (see the lesson " Solving quadratic equations") We did not need.
Of course, in all our reflections, we proceeded from two important assumptions, which, generally speaking, are not always fulfilled in real problems:
- The quadratic equation is reduced, i.e. the coefficient at x 2 is 1;
- The equation has two different roots. From the point of view of algebra, in this case the discriminant D > 0 - in fact, we initially assume that this inequality is true.
However, in typical mathematical problems these conditions are met. If the result of the calculations is a “bad” quadratic equation (the coefficient at x 2 is different from 1), this is easy to fix - take a look at the examples at the very beginning of the lesson. I am generally silent about the roots: what kind of task is this in which there is no answer? Of course there will be roots.
Thus, the general scheme for solving quadratic equations according to the Vieta theorem is as follows:
- Reduce the quadratic equation to the given one, if this has not already been done in the condition of the problem;
- If the coefficients in the above quadratic equation turned out to be fractional, we solve through the discriminant. You can even go back to the original equation to work with more "convenient" numbers;
- In the case of integer coefficients, we solve the equation using the Vieta theorem;
- If within a few seconds it was not possible to guess the roots, we score on the Vieta theorem and solve through the discriminant.
Task. Solve the equation: 5x 2 − 35x + 50 = 0.
So, we have an equation that is not reduced, because coefficient a \u003d 5. Divide everything by 5, we get: x 2 - 7x + 10 \u003d 0.
All coefficients of the 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 \u003d 10. In this case, the roots are easy to guess - these are 2 and 5. You do not need to count through the discriminant.
Task. Solve the equation: -5x 2 + 8x - 2.4 = 0.
We look: −5x 2 + 8x − 2.4 = 0 - this equation is not reduced, we divide both sides by the coefficient a = −5. We get: x 2 - 1.6x + 0.48 \u003d 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 \u003d 0.4.
Task. Solve the equation: 2x 2 + 10x − 600 = 0.
To begin with, we divide everything by the coefficient a \u003d 2. We get the equation x 2 + 5x - 300 \u003d 0.
This is the reduced equation, according to the Vieta theorem we have: x 1 + x 2 = −5; x 1 x 2 \u003d -300. It is difficult to guess the roots of the quadratic equation in this case - personally, I seriously "froze" when I solved this problem.
We 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 \u003d 15; x 2 \u003d -20.
Vieta's theorem is often used to test already found roots. If you have found the roots, you can use the formulas \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\) to calculate the values \(p\) and \(q\ ). And if they turn out to be the same as in the original equation, then the roots are found correctly.
For example, let's use , solve the equation \(x^2+x-56=0\) and get the roots: \(x_1=7\), \(x_2=-8\). Let's check if we made a mistake in the process of solving. In our case, \(p=1\), and \(q=-56\). By Vieta's theorem we have:
\(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\) \(\Leftrightarrow\) \(\begin(cases)7+(-8)=-1 \\7\cdot(-8)=-56\end(cases)\) \(\Leftrightarrow\) \(\begin(cases)-1=-1\\-56=-56\end(cases)\ )
Both statements converged, which means that we solved the equation correctly.
This test can be done orally. It will take 5 seconds and save you from stupid mistakes.
Inverse Vieta theorem
If \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\end(cases)\), then \(x_1\) and \(x_2\) are the roots of the quadratic equation \(x^ 2+px+q=0\).
Or in a simple way: if you have an equation of the form \(x^2+px+q=0\), then by solving the system \(\begin(cases)x_1+x_2=-p \\x_1 \cdot x_2=q\ end(cases)\) you will find its roots.
Thanks to this theorem, you can quickly find the roots of a quadratic equation, especially if these roots are . This skill is important as it saves a lot of time.
Example . Solve the equation \(x^2-5x+6=0\).
Solution
: Using the inverse Vieta theorem, we get that the roots satisfy the conditions: \(\begin(cases)x_1+x_2=5 \\x_1 \cdot x_2=6\end(cases)\).
Look at the second equation of the \(x_1 \cdot x_2=6\) system. Into what two can the number \(6\) be decomposed? On \(2\) and \(3\), \(6\) and \(1\) or \(-2\) and \(-3\), and \(-6\) and \(- 1\). And which pair to choose, the first equation of the system will tell: \(x_1+x_2=5\). \(2\) and \(3\) are similar, because \(2+3=5\).
Answer
: \(x_1=2\), \(x_2=3\).
Examples
. Using the inverse of Vieta's theorem, find the roots of the quadratic equation:
a) \(x^2-15x+14=0\); b) \(x^2+3x-4=0\); c) \(x^2+9x+20=0\); d) \(x^2-88x+780=0\).
Solution
:
a) \(x^2-15x+14=0\) - what factors does \(14\) decompose into? \(2\) and \(7\), \(-2\) and \(-7\), \(-1\) and \(-14\), \(1\) and \(14\ ). What pairs of numbers add up to \(15\)? Answer: \(1\) and \(14\).
b) \(x^2+3x-4=0\) - into what factors does \(-4\) decompose? \(-2\) and \(2\), \(4\) and \(-1\), \(1\) and \(-4\). What pairs of numbers add up to \(-3\)? Answer: \(1\) and \(-4\).
c) \(x^2+9x+20=0\) – into what factors does \(20\) decompose? \(4\) and \(5\), \(-4\) and \(-5\), \(2\) and \(10\), \(-2\) and \(-10\ ), \(-20\) and \(-1\), \(20\) and \(1\). What pairs of numbers add up to \(-9\)? Answer: \(-4\) and \(-5\).
d) \(x^2-88x+780=0\) - into what factors does \(780\) decompose? \(390\) and \(2\). Do they add up to \(88\)? No. What other multipliers does \(780\) have? \(78\) and \(10\). Do they add up to \(88\)? Yes. Answer: \(78\) and \(10\).
It is not necessary to expand the last term into all possible factors (as in last example). You can immediately check whether their sum gives \(-p\).
Important! Vieta's theorem and converse theorem work only with , that is, one whose coefficient in front of \(x^2\) equal to one. If we initially have a non-reduced equation, then we can make it reduced by simply dividing by the coefficient in front of \ (x ^ 2 \).
For example, let the equation \(2x^2-4x-6=0\) be given and we want to use one of Vieta's theorems. But we can't, because the coefficient before \(x^2\) is equal to \(2\). Let's get rid of it by dividing the whole equation by \(2\).
\(2x^2-4x-6=0\) \(|:2\)
\(x^2-2x-3=0\)
Ready. Now we can use both theorems.
Answers to frequently asked questions
Question:
By Vieta's theorem, you can solve any ?
Answer:
Unfortunately no. If there are not integers in the equation or the equation has no roots at all, then Vieta's theorem will not help. In this case, you need to use discriminant
. Fortunately, 80% of the equations in school course mathematics have entire solutions.
