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. This is 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 be 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 in modulus is placed 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.
Decision.
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.
Decision.
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 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 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.
Decision.
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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One of the methods for solving a quadratic equation is the application VIETA formulas, which was named after FRANCOIS VIETE.
He was a famous lawyer, and served in the 16th century with the French king. In his free time he studied astronomy and mathematics. He established a connection between the roots and coefficients of a quadratic equation.
Advantages of the formula:
1 . By applying the formula, you can quickly find the solution. Because you do not need to enter the second coefficient into the square, then subtract 4ac from it, find the discriminant, substitute its value into the formula for finding the roots.
2 . Without a solution, you can determine the signs of the roots, pick up the values of the roots.
3 . Having solved the system of two records, it is not difficult to find the roots themselves. In the above quadratic equation, the sum of the roots is equal to the value of the second coefficient with a minus sign. The product of the roots in the above quadratic equation is equal to the value of the third coefficient.
4 . According to the given roots, write a quadratic equation, that is, solve the inverse problem. For example, this method is used in solving problems in theoretical mechanics.
5 . It is convenient to apply the formula when the leading coefficient is equal to one.
Disadvantages:
1
. The formula is not universal.
Vieta's theorem Grade 8
Formula
If x 1 and x 2 are the roots of the given quadratic equation x 2 + px + q \u003d 0, then:
Examples
x 1 \u003d -1; x 2 \u003d 3 - the roots of the equation x 2 - 2x - 3 \u003d 0.
P = -2, q = -3.
X 1 + x 2 \u003d -1 + 3 \u003d 2 \u003d -p,
X 1 x 2 = -1 3 = -3 = q.
Inverse theorem
Formula
If the numbers x 1 , x 2 , p, q are connected by the conditions:
Then x 1 and x 2 are the roots of the equation x 2 + px + q = 0.
Example
Let's make a quadratic equation by its roots:
X 1 \u003d 2 -? 3 and x 2 \u003d 2 +? 3 .
P \u003d x 1 + x 2 \u003d 4; p = -4; q \u003d x 1 x 2 \u003d (2 -? 3) (2 +? 3) \u003d 4 - 3 \u003d 1.
The desired equation has the form: x 2 - 4x + 1 = 0.
I. Vieta's theorem for the reduced quadratic equation.
The sum of the 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 \u003d-p; x 1 ∙ x 2 \u003d 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), the second coefficient p=-1, and the free term q=-30. First, make sure that the given equation has roots, and that the roots (if any) will be expressed as integers. For this, it is sufficient that the discriminant be the full square of an integer.
Finding the discriminant D=b 2 - 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .
Now, according to the Vieta 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 \u003d -30. We need to choose such two numbers so that their product is equal to -30 , and the sum is unit. These are the 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. Make sure that there are integer roots. Let's find the discriminant D1 D1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , so the roots of this equation are integers. We choose the roots according to the Vieta theorem: the sum of the roots is equal to –p=-6, and the product of the roots is q=8. These are the numbers -4 and -2 .
Actually: -4-2=-6=-p; -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 term q=-4. Let's find the discriminant D1, since the second coefficient is an even number. D1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of a number, so we do conclusion: the roots of this equation are not integers and cannot be found using Vieta's theorem. So, we solve this equation, as usual, according to the formulas (in this case, according to the formulas). We get:
Example 4). Write a quadratic equation using its roots if x 1 \u003d -7, x 2 \u003d 4.
Decision. The desired equation will be written in the form: x 2 +px+q=0, moreover, based on the Vieta theorem –p=x1 +x2=-7+4=-3 →p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x2 +3x-28=0.
Example 5). Write a quadratic equation using its roots if:
II. Vieta's theorem for the complete quadratic equation ax2+bx+c=0.
The sum of the roots is minus b divided by a, the product of the roots is with divided by a:
x 1 + x 2 \u003d -b / a; x 1 ∙ x 2 \u003d c / a.
Today it deserves to be sung in verse
On the properties of roots, Vieta's theorem.
Which is better, say, the constancy of this:
You multiplied the roots - and the fraction is ready
In the numerator with, in the denominator a.
And the sum of the roots is also a fraction
Even with a minus this fraction
What's the trouble
In numerators in, in the denominator a.
(From school folklore)
In the epigraph, the remarkable theorem of François Vieta is given not quite exactly. Indeed, we can write down a quadratic equation that has no roots and write down their sum and product. For example, the equation x 2 + 2x + 12 = 0 has no real roots. But, approaching formally, we can write down their product (x 1 x 2 \u003d 12) and the sum (x 1 + x 2 \u003d -2). Our the verses will correspond to the theorem with the caveat: "if the equation has roots", i.e. D ≥ 0.
The first practical application of this theorem is the compilation of a quadratic equation that has given roots. Second: it allows you to verbally solve many quadratic equations. At the development of these skills, first of all, attention is drawn to school textbooks.
Here we will consider more complex problems solved using the Vieta theorem.
Example 1
One of the roots of the equation 5x 2 - 12x + c \u003d 0 is three times larger than the second. Find with.
Decision.
Let the second root be x2.
Then the first root x1 = 3x2.
According to the Vieta theorem, the sum of the roots is 12/5 = 2.4.
Let's make the equation 3x2 + x2 = 2.4.
Hence x 2 \u003d 0.6. Therefore x 1 \u003d 1.8.
Answer: c \u003d (x 1 x 2) a \u003d 0.6 1.8 5 \u003d 5.4.
Example 2
It is known that x 1 and x 2 are the roots of the equation x 2 - 8x + p = 0, and 3x 1 + 4x 2 = 29. Find p.
Decision.
According to the Vieta theorem x 1 + x 2 = 8, and by condition 3x 1 + 4x 2 = 29.
Having solved the system of these two equations, we find the value x 1 \u003d 3, x 2 \u003d 5.
And therefore p = 15.
Answer: p = 15.
Example 3
Without calculating the roots of the equation 3x 2 + 8 x - 1 \u003d 0, find x 1 4 + x 2 4
Decision.
Note that according to the Vieta theorem x 1 + x 2 = -8/3 and x 1 x 2 = -1/3 and transform the expression
a) x 1 4 + x 2 4 = (x 1 2 + x 2 2) 2 - 2x 1 2 x 2 2 = ((x 1 + x 2) 2 - 2x 1 x 2) 2 - 2 (x 1 x 2) 2 \u003d ((-8/3) 2 - 2 (-1/3)) 2 - 2 (-1/3) 2 \u003d 4898/9
Answer: 4898/9.
Example 4
At what values of the parameter a is the difference between the largest and smallest roots of the equation
2x 2 - (a + 1) x + (a - 1) \u003d 0 is equal to their product.
Decision.
This is a quadratic equation. It will have 2 different roots if D > 0. In other words, (a + 1) 2 - 8 (a - 1) > 0 or (a - 3) 2 > 0. Therefore, we have 2 roots for all a, for except a = 3.
For definiteness, we assume that x 1 > x 2 and get x 1 + x 2 \u003d (a + 1) / 2 and x 1 x 2 \u003d (a - 1) / 2. Based on the condition of the problem x 1 - x 2 \u003d (a - 1) / 2. All three conditions must be met simultaneously. Consider the first and last equations as a system. It is easily solved by the algebraic addition method.
We get x 1 \u003d a / 2, x 2 \u003d 1/2. Let's check for what a the second equality will be fulfilled: x 1 x 2 \u003d (a - 1) / 2. Let's substitute the received values and we will have: а/4 = (а – 1)/2. Then, a = 2. It is obvious that if a = 2, then all conditions are satisfied.
Answer: when a = 2.
Example 5
What is the smallest value of a for which the sum of the roots of the equation
x 2 - 2a (x - 1) - 1 \u003d 0 is equal to the sum of the squares of its roots.
Decision.
First of all, let's bring the equation to the canonical form: x 2 - 2ax + 2a - 1 \u003d 0. It will have roots if D / 4 ≥ 0. Therefore: a 2 - (2a - 1) ≥ 0. Or (a - 1 ) 2 ≥ 0. And this condition is valid for any a.
We apply the Vieta theorem: x 1 + x 2 \u003d 2a, x 1 x 2 \u003d 2a - 1. We calculate
x 1 2 + x 2 2 \u003d (x 1 + x 2) 2 - 2x 1 x 2. Or after substituting x 1 2 + x 2 2 \u003d (2a) 2 - 2 (2a - 1) \u003d 4a 2 - 4a + 2. It remains to make an equality that corresponds to the condition of the problem: x 1 + x 2 \u003d x 1 2 + x 2 2 . We get: 2a \u003d 4a 2 - 4a + 2. This quadratic equation has 2 roots: a 1 \u003d 1 and a 2 \u003d 1/2. The smallest of them is -1/2.
Answer: 1/2.
Example 6
Find the relationship between the coefficients of the equation ax 2 + inx + c \u003d 0 if the sum of the cubes of its roots is equal to the product of the squares of these roots.
Decision.
We will proceed from the fact that this equation has roots and, therefore, Vieta's theorem can be applied to it.
Then the condition of the problem will be written as follows: x 1 3 + x 2 3 = x 1 2 x 2 2. Or: (x 1 + x 2) (x 1 2 - x 1 x 2 + x 2 2) \u003d (x 1 x 2) 2.
You need to convert the second factor. x 1 2 - x 1 x 2 + x 2 2 \u003d ((x 1 + x 2) 2 - 2x 1 x 2) - x 1 x 2.
We get (x 1 + x 2) ((x 1 + x 2) 2 - 3x 1 x 2) \u003d (x 1 x 2) 2. It remains to replace the sums and products of the roots through the coefficients.
(-b/a)((b/a) 2 – 3 c/a) = (c/a) 2 . This expression can easily be converted to the form b (3ac - b 2) / a \u003d c 2. The ratio is found.
Comment. It should be taken into account that the resulting relation makes sense to consider only after the other is fulfilled: D ≥ 0.
Example 7
Find the value of the variable a for which the sum of the squares of the roots of the equation x 2 + 2ax + 3a 2 - 6a - 2 \u003d 0 is the largest value.
Decision.
If this equation has roots x 1 and x 2, then their sum x 1 + x 2 \u003d -2a, and the product x 1 x 2 \u003d 3a 2 - 6a - 2.
We calculate x 1 2 + x 2 2 \u003d (x 1 + x 2) 2 - 2x 1 x 2 \u003d (-2a) 2 - 2 (3a 2 - 6a - 2) \u003d -2a 2 + 12a + 4 \u003d -2 (a – 3) 2 + 22.
It is now obvious that this expression takes on the largest value at a = 3.
It remains to check whether the original quadratic equation really has roots at a \u003d 3. We check by substitution and we get: x 2 + 6x + 7 \u003d 0 and for it D \u003d 36 - 28\u003e 0.
Therefore, the answer is: for a = 3.
Example 8
The equation 2x 2 - 7x - 3 \u003d 0 has roots x 1 and x 2. Find the triple sum of the coefficients of the given quadratic equation, the roots of which are the numbers X 1 \u003d 1 / x 1 and X 2 \u003d 1 / x 2. (*)
Decision.
Obviously, x 1 + x 2 \u003d 7/2 and x 1 x 2 \u003d -3/2. We compose the second equation by its roots in the form x 2 + px + q \u003d 0. To do this, we use the assertion inverse to the Vieta theorem. We get: p \u003d - (X 1 + X 2) and q \u003d X 1 X 2.
After substituting into these formulas, based on (*), then: p \u003d - (x 1 + x 2) / (x 1 x 2) \u003d 7/3 and q \u003d 1 / (x 1 x 2) \u003d - 2/3.
The desired equation will take the form: x 2 + 7/3 x - 2/3 = 0. Now we can easily calculate the tripled sum of its coefficients:
3(1 + 7/3 - 2/3) = 8. Answer received.
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Vieta's theorem
Let and denote the roots of the reduced quadratic equation
(1)
.
Then the sum of the roots is equal to the coefficient at taken with the opposite sign. The product of the roots is equal to the free term:
;
.
A note about multiple roots
If the discriminant of equation (1) is zero, then this equation has one root. But, in order to avoid cumbersome formulations, it is generally accepted that in this case, equation (1) has two multiple, or equal, roots:
.
Proof one
Let us find the roots of equation (1). To do this, apply the formula for the roots of the quadratic equation:
;
;
.
Finding the sum of the roots:
.
To find the product, we apply the formula:
.
Then
.
The theorem has been proven.
Proof two
If the numbers and are the roots of the quadratic equation (1), then
.
We open the brackets.
.
Thus, equation (1) will take the form:
.
Comparing with (1) we find:
;
.
The theorem has been proven.
Inverse Vieta theorem
Let there be arbitrary numbers. Then and are the roots of the quadratic equation
,
where
(2)
;
(3)
.
Proof of Vieta's converse theorem
Consider the quadratic equation
(1)
.
We need to prove that if and , then and are the roots of equation (1).
Substitute (2) and (3) into (1):
.
We group the terms of the left side of the equation:
;
;
(4)
.
Substitute in (4) :
;
.
Substitute in (4) :
;
.
The equation is fulfilled. That is, the number is the root of equation (1).
The theorem has been proven.
Vieta's theorem for the complete quadratic equation
Now consider the complete quadratic equation
(5)
,
where , and are some numbers. And .
We divide equation (5) by:
.
That is, we have obtained the above equation
,
where ; .
Then the Vieta theorem for the complete quadratic equation has the following form.
Let and denote the roots of the complete quadratic equation
.
Then the sum and product of the roots are determined by the formulas:
;
.
Vieta's theorem for a cubic equation
Similarly, we can establish connections between the roots of a cubic equation. Consider the cubic equation
(6)
,
where , , , are some numbers. And .
Let's divide this equation by:
(7)
,
where , , .
Let , , be the roots of equation (7) (and equation (6)). Then
.
Comparing with equation (7) we find:
;
;
.
Vieta's theorem for an nth degree equation
In the same way, you can find connections between the roots , , ... , , for the equation of the nth degree
.
Vieta's theorem for an nth degree equation has the following form:
;
;
;
.
To get these formulas, we write the equation in the following form:
.
Then we equate the coefficients at , , , ... , and compare the free term.
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of Mathematics for Engineers and Students of Higher Educational Institutions, Lan, 2009.
CM. Nikolsky, M.K. Potapov et al., Algebra: a textbook for the 8th grade of educational institutions, Moscow, Education, 2006.
