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 a damn thing, 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 the definition of a quadratic equation implies 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 all coefficients in the final squared equation are integers. For now, let's look at some simple examples:
A 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.
A 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, no 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.
A 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.
A 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.
A 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.
2.5 Vieta formula for polynomials (equations) of higher degrees
The formulas derived by Vieta 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 distinct 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 parts of this equality by a 0 ≠ 0 and expand the brackets in the first part. We get the equality:
xn + ()xn -1 + ... + () = xn - (x 1 + x 2 + ... + xn) xn -1 + (x 1 x 2 + x 2 x 3 + ... + xn -1 xn)xn - 2 + … +(-1) nx 1 x 2 … xn
But two polynomials are identically equal if and only if the coefficients at the same powers are equal. It follows from this 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 the 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 the Vieta formulas. The left parts of these formulas are symmetric polynomials from the roots x 1 , x 2 ..., x n of the given equation, and the right parts are expressed in terms of the coefficient of the polynomial.
2.6 Equations reducible to squares (biquadratic)
Equations of the fourth degree are reduced to quadratic equations:
ax 4 + bx 2 + c = 0,
called biquadratic, moreover, a ≠ 0.
It is enough to put x 2 \u003d y in this equation, therefore,
ay² + by + c = 0
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 the equation of the fourth degree has x 1, then it also has a root x 2 \u003d -x 1,
If has x 3, then x 4 \u003d - x 3. The sum of the roots of such an equation is zero.
2x 4 - 9x² + 4 = 0
We substitute the equation into the formula for the roots of biquadratic equations:
x 1,2,3,4 = 
,
knowing that x 1 \u003d -x 2, and x 3 \u003d -x 4, then:
x 3.4 = 
Answer: x 1.2 \u003d ± 2; x 1.2 =
2.7 Study of biquadratic equations
Let's take the biquadratic equation
ax 4 + bx 2 + c = 0,
where a, b, c are real numbers, and a > 0. By introducing an auxiliary unknown y = x², we examine the roots of this equation, and enter the results in a table (see Appendix No. 1)
2.8 Cardano formula
If we use modern symbolism, then the derivation of the Cardano formula can look like this:
x = 
This formula determines the roots of the general equation of the third degree:
ax 3 + 3bx 2 + 3cx + d = 0.
This formula is very cumbersome and complex (it contains several complex radicals). It does not always apply, because. very difficult to complete.
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In this lecture, we will get acquainted with the curious relationships between the roots of a quadratic equation and its coefficients. These relationships were first discovered by the French mathematician Francois Viet (1540-1603).
For example, for the equation Зx 2 - 8x - 6 \u003d 0, without finding its roots, you can, using the Vieta theorem, immediately say that the sum of the roots is , and the product of the roots is
i.e. - 2. And for the equation x 2 - 6x + 8 \u003d 0 we conclude: the sum of the roots is 6, the product of the roots is 8; by the way, it is not difficult to guess what the roots are equal to: 4 and 2.
Proof of Vieta's theorem. The roots x 1 and x 2 of the quadratic equation ax 2 + bx + c \u003d 0 are found by the formulas
Where D \u003d b 2 - 4ac is the discriminant of the equation. Laying down these roots
we get
Now we calculate the product of the roots x 1 and x 2 We have
The second relation is proved:
Comment.
Vieta's theorem is also valid in the case when the quadratic equation has one root (i.e., when D \u003d 0), it's just that in this case it is considered that the equation has two identical roots, to which the above relations are applied.
The proven relations for the reduced quadratic equation x 2 + px + q \u003d 0 take a particularly simple form. In this case, we get:
x 1 \u003d x 2 \u003d -p, x 1 x 2 \u003d q
those. the sum of the roots of the given quadratic equation is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term.
Using the Vieta theorem, one can also obtain other relationships between the roots and coefficients of a quadratic equation. Let, for example, x 1 and x 2 be the roots of the reduced quadratic equation x 2 + px + q = 0. Then
However, the main purpose of Vieta's theorem is not that it expresses certain relationships between the roots and coefficients of a quadratic equation. Much more important is the fact that with the help of Vieta's theorem, a formula for factoring a square trinomial is derived, without which we will not do in the future.
Proof. We have
Example 1. Factorize the square trinomial 3x 2 - 10x + 3.
Solution. Having solved the equation Zx 2 - 10x + 3 \u003d 0, we find the roots of the square trinomial Zx 2 - 10x + 3: x 1 \u003d 3, x2 \u003d.
Using Theorem 2, we get
It makes sense instead to write Zx - 1. Then we finally get Zx 2 - 10x + 3 = (x - 3) (3x - 1).
Note that the given square trinomial can be factored without using Theorem 2, using the grouping method:
Zx 2 - 10x + 3 = Zx 2 - 9x - x + 3 =
\u003d Zx (x - 3) - (x - 3) \u003d (x - 3) (Zx - 1).
But, as you can see, with this method success depends on whether we can find a successful grouping or not, while with the first method success is guaranteed.
Example 1. Reduce fraction
Solution. From the equation 2x 2 + 5x + 2 = 0 we find x 1 = - 2,
From the equation x2 - 4x - 12 = 0 we find x 1 = 6, x 2 = -2. That's why
x 2 - 4x - 12 \u003d (x - 6) (x - (- 2)) \u003d (x - 6) (x + 2).
Now let's reduce the given fraction:
Example 3. Factorize expressions:
a) x4 + 5x 2 +6; b) 2x+-3
Solution. a) We introduce a new variable y = x 2 . This will allow us to rewrite the given expression in the form of a square trinomial with respect to the variable y, namely, in the form y 2 + bу + 6.
Having solved the equation y 2 + bу + 6 \u003d 0, we find the roots of the square trinomial y 2 + 5y + 6: y 1 \u003d - 2, y 2 \u003d -3. Now we use Theorem 2; we get
y 2 + 5y + 6 = (y + 2) (y + 3).
It remains to remember that y \u003d x 2, i.e., return to the given expression. So,
x 4 + 5x 2 + 6 \u003d (x 2 + 2) (x 2 + 3).
b) Let's introduce a new variable y = . This will allow you to rewrite the given expression in the form of a square trinomial with respect to the variable y, namely, in the form 2y 2 + y - 3. Having solved the equation
2y 2 + y - 3 \u003d 0, we find the roots of the square trinomial 2y 2 + y - 3:
y 1 = 1, y 2 = . Further, using Theorem 2, we obtain:
It remains to remember that y \u003d, i.e., return to the given expression. So,
The section concludes with some considerations, again connected with the Vieta theorem, or rather, with the converse assertion:
if the numbers x 1, x 2 are such that x 1 + x 2 \u003d - p, x 1 x 2 \u003d q, then these numbers are the roots of the equation
Using this statement, you can solve many quadratic equations orally, without using cumbersome root formulas, and also compose quadratic equations with given roots. Let's give examples.
1) x 2 - 11x + 24 = 0. Here x 1 + x 2 = 11, x 1 x 2 = 24. It is easy to guess that x 1 = 8, x 2 = 3.
2) x 2 + 11x + 30 = 0. Here x 1 + x 2 = -11, x 1 x 2 = 30. It is easy to guess that x 1 = -5, x 2 = -6.
Please note: if the free term of the equation is a positive number, then both roots are either positive or negative; this is important to consider when selecting roots.
3) x 2 + x - 12 = 0. Here x 1 + x 2 = -1, x 1 x 2 = -12. It is easy to guess that x 1 \u003d 3, x2 \u003d -4.
Please note: if the free term of the equation is a negative number, then the roots are different in sign; this is important to consider when selecting roots.
4) 5x 2 + 17x - 22 = 0. It is easy to see that x = 1 satisfies the equation, i.e. x 1 \u003d 1 - the root of the equation. Since x 1 x 2 \u003d -, and x 1 \u003d 1, we get that x 2 \u003d -.
5) x 2 - 293x + 2830 = 0. Here x 1 + x 2 = 293, x 1 x 2 = 2830. If you pay attention to the fact that 2830 = 283. 10, and 293 \u003d 283 + 10, then it becomes clear that x 1 \u003d 283, x 2 \u003d 10 (now imagine what calculations would have to be performed to solve this quadratic equation using standard formulas).
6) Let's compose a quadratic equation so that the numbers x 1 \u003d 8, x 2 \u003d - 4 serve as its roots. Usually in such cases they make up the reduced quadratic equation x 2 + px + q \u003d 0.
We have x 1 + x 2 \u003d -p, therefore 8 - 4 \u003d -p, that is, p \u003d -4. Further, x 1 x 2 = q, i.e. 8"(-4) = q, whence we get q = -32. So, p \u003d -4, q \u003d -32, which means that the desired quadratic equation has the form x 2 -4x-32 \u003d 0.
When studying ways to solve second-order equations in a school algebra course, consider the properties of the roots obtained. They are now known as Vieta's theorems. Examples of its use are given in this article.
Quadratic equation
The second order equation is an equality, which is shown in the photo below.
Here the symbols a, b, c are some numbers that are called the coefficients of the equation under consideration. To solve an equality, you need to find x values that make it true.
Note that since the maximum value of the power to which x is raised is two, then the number of roots in the general case is also two.
There are several ways to solve this type of equality. In this article, we will consider one of them, which involves the use of the so-called Vieta theorem.
Statement of Vieta's theorem
At the end of the 16th century, the famous mathematician Francois Viet (Frenchman) noticed, analyzing the properties of the roots of various quadratic equations, that certain combinations of them satisfy specific relationships. In particular, these combinations are their product and sum.
Vieta's theorem establishes the following: the roots of a quadratic equation, when summed, give the ratio of the linear to quadratic coefficients taken with the opposite sign, and when they are multiplied, they lead to the ratio of the free term to the quadratic coefficient.
If the general form of the equation is written as it is shown in the photo in the previous section of the article, then mathematically this theorem can be written as two equalities:
- r 2 + r 1 \u003d -b / a;
- r 1 x r 2 \u003d c / a.
Where r 1 , r 2 is the value of the roots of the considered equation.
These two equalities can be used to solve a number of very different mathematical problems. The use of the Vieta theorem in examples with a solution is given in the following sections of the article.
Vieta's theorem (more precisely, the theorem inverse to Vieta's theorem) allows us 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 easy if you think a little.
Now we will only talk about the solution of the reduced quadratic equation using the Vieta theorem. The reduced quadratic equation is an equation in which a, that is, the coefficient in front of x², is equal to one. Not given quadratic equations can also be solved using the Vieta theorem, but there already at least one of the roots is not an integer. They are harder to guess.
The theorem converse to Vieta's theorem says: 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 the Vieta theorem, only 4 options are possible. If you remember the course 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 (because only when multiplying numbers with the same signs, a positive number is obtained).
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 (they added numbers of the same sign, 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 modulo). Therefore, x1 + x2 shows how much the roots x1 and x2 differ, that is, how much one root is more than the other (modulo).
II.a. If -p is a positive number, (i.e. 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.
Consider the solution of quadratic equations according to 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 12. These are 1 and 12, 2 and 6, 3 and 4. The sum is 7 for the pair 3 and 4. Hence, 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 -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.
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