Calculate the roots of a quadratic equation using Vieta. Online calculator. Solving a quadratic equation
Vieta's theorem (more precisely, the theorem converse of the theorem Vieta) 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 the solution by Vieta’s theorem of the reduced quadratic equation. quadratic equation is an equation in which a, that is, the coefficient of x², 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 — positive numbers(since we added numbers of the same sign and got a positive number).
I.b. If -p — 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). 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 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.
Before moving on to Vieta's theorem, we introduce a definition. Quadratic equation of the form x² + px + q= 0 is called reduced. In this equation, the leading coefficient is equal to one. For example, the equation x² - 3 x- 4 = 0 is reduced. Any quadratic equation of the form ax² + b x + c= 0 can be reduced by dividing both sides of the equation by A≠ 0. For example, equation 4 x² + 4 x— 3 = 0 by dividing by 4 is reduced to the form: x² + x- 3/4 = 0. Let us derive the formula for the roots of the reduced quadratic equation; for this we use the formula for the roots of a general quadratic equation: ax² + bx + c = 0
Reduced equation x² + px + q= 0 coincides with a general equation in which A = 1, b = p, c = q. Therefore, for the given quadratic equation the formula takes the form: 
the last expression is called the formula for the roots of the reduced quadratic equation; it is especially convenient to use this formula when r- an even number. For example, let's solve the equation x² — 14 x — 15 = 0
In response, we write the equation has two roots.
For the reduced quadratic equation with positive, the following theorem holds.
Vieta's theorem
If x 1 and x 2 - roots of the equation x² + px + q= 0, then the formulas are valid:
x 1 + x 2 = — r
x 1 * x 2 = q, that is, the sum of the roots of the reduced 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.
Based on the formula for the roots of the above quadratic equation, we have:
Adding these equalities, we get: x 1 + x 2 = —r.
Multiplying these equalities, using the difference of squares formula we obtain:
Note that Vieta’s theorem is also valid when the discriminant is equal to zero, if we assume that in this case the quadratic equation has two identical roots: x 1 = x 2 = — r/2.
Without solving equations x² — 13 x+ 30 = 0 find the sum and product of its roots x 1 and x 2. this equation D= 169 – 120 = 49 > 0, so Vieta’s theorem can be applied: x 1 + x 2 = 13, x 1 * x 2 = 30. Let's look at a few more examples. One of the roots of the equation x² — px- 12 = 0 is equal x 1 = 4. Find coefficient r and the second root x 2 of this equation. By Vieta's theorem x 1 * x 2 =— 12, x 1 + x 2 = — r. Because x 1 = 4, then 4 x 2 = - 12, whence x 2 = — 3, r = — (x 1 + x 2) = - (4 - 3) = - 1. In answer we write down the second root x 2 = - 3, coefficient p = — 1.
Without solving equations x² + 2 x- 4 = 0 let’s find the sum of the squares of its roots. Let x 1 and x 2 - roots of the equation. By Vieta's theorem x 1 + x 2 = — 2, x 1 * x 2 = — 4. Because x 1²+ x 2² = ( x 1 + x 2)² - 2 x 1 x 2 then x 1²+ x 2² =(- 2)² -2 (- 4) = 12.
Let's find the sum and product of the roots of equation 3 x² + 4 x- 5 = 0. This equation has two different roots, since the discriminant D= 16 + 4*3*5 > 0. To solve the equation, we use Vieta’s theorem. This theorem has been proven for the given quadratic equation. So let's divide this equation by 3.
Therefore, the sum of the roots is equal to -4/3, and their product is equal to -5/3.
In general, the roots of the equation ax² + b x + c= 0 are related by the following equalities: x 1 + x 2 = — b/a, x 1 * x 2 = c/a, To obtain these formulas, it is enough to divide both sides of this quadratic equation by A ≠ 0 and apply Vieta’s theorem to the resulting reduced quadratic equation. Let's consider an example: you need to create a reduced quadratic equation whose roots x 1 = 3, x 2 = 4. Because x 1 = 3, x 2 = 4 - roots of quadratic equation x² + px + q= 0, then by Vieta’s theorem r = — (x 1 + x 2) = — 7, q = x 1 x 2 = 12. We write the answer as x² — 7 x+ 12 = 0. When solving some problems, the following theorem is used.
Theorem converse to Vieta's theorem
If the numbers r, q, x 1 , x 2 are such that x 1 + x 2 = — p, x 1 * x 2 = q, That x 1 And x 2- roots of the equation x² + px + q= 0. Substitute into the left side x² + px + q instead of r expression - ( x 1 + x 2), and instead q- work x 1 * x 2 . We get: x² + px + q = x² — ( x 1 + x 2) x + x 1 x 2 = x² - x 1 x - x 2 x + x 1 x 2 = (x - x 1) (x - x 2). Thus, if the numbers r, q, x 1 and x 2 are connected by these relations, then for all X equality holds x² + px + q = (x - x 1) (x - x 2), from which it follows that x 1 and x 2 - roots of the equation x² + px + q= 0. Using the theorem inverse to Vieta’s theorem, you can sometimes find the roots of a quadratic equation by selection. Let's look at an example, x² — 5 x+ 6 = 0. Here r = — 5, q= 6. Let's choose two numbers x 1 and x 2 so that x 1 + x 2 = 5, x 1 * x 2 = 6. Noticing that 6 = 2 * 3, and 2 + 3 = 5, by the theorem inverse to Vieta’s theorem, we obtain that x 1 = 2, x 2 = 3 - roots of the equation x² — 5 x + 6 = 0.
First, let's formulate the theorem itself: Let us have a reduced quadratic equation of the form x^2+b*x + c = 0. Let's say this equation contains roots x1 and x2. Then, according to the theorem, the following statements are valid:
1) The sum of the roots x1 and x2 will be equal to the negative value of the coefficient b.
2) The product of these same roots will give us the coefficient c.
But what is the given equation?
A reduced quadratic equation is a quadratic equation whose coefficient of the highest degree is equal to one, i.e. this is an equation of the form x^2 + b*x + c = 0. (and the equation a*x^2 + b*x + c = 0 is unreduced). In other words, to bring the equation to the given form, we must divide this equation by the coefficient of the highest power (a). The task is to bring this equation to the following form:
3*x^2 12*x + 18 = 0;
−4*x^2 + 32*x + 16 = 0;
1.5*x^2 + 7.5*x + 3 = 0; 2*x^2 + 7*x − 11 = 0.
Dividing each equation by the coefficient of the highest degree, we get:
X^2 4*x + 6 = 0; X^2 8*x − 4 = 0; X^2 + 5*x + 2 = 0;
X^2 + 3.5*x − 5.5 = 0.
As you can see from the examples, even equations containing fractions can be reduced to the given form.
Using Vieta's theorem
X^2 5*x + 6 = 0 ⇒ x1 + x2 = − (−5) = 5; x1*x2 = 6;
we get the roots: x1 = 2; x2 = 3;
X^2 + 6*x + 8 = 0 ⇒ x1 + x2 = −6; x1*x2 = 8;
as a result we get the roots: x1 = -2 ; x2 = -4;
X^2 + 5*x + 4 = 0 ⇒ x1 + x2 = −5; x1*x2 = 4;
we get the roots: x1 = −1; x2 = −4.
The meaning of Vieta's theorem
Vieta's theorem allows us to solve any quadratic reduced equation in almost seconds. At first glance, this seems to be a rather difficult task, but after 5 10 equations, you can learn to see the roots right away.
From the examples given, and using the theorem, it is clear how you can significantly simplify the solution of quadratic equations, because using this theorem, you can solve a quadratic equation practically without complex calculations and calculating the discriminant, and as you know, the fewer calculations, the more difficult it is to make a mistake, which is important.
In all examples, we used this rule based on two important assumptions:
The given equation, i.e. the coefficient of the highest degree is equal to one (this condition is easy to avoid. You can use the unreduced form of the equation, then the following statements will be valid x1+x2=-b/a; x1*x2=c/a, but it’s usually more difficult to solve :))
When an equation has two different roots. We assume that the inequality is true and the discriminant is strictly greater than zero.
Therefore, we can create a general solution algorithm using Vieta’s theorem.
General solution algorithm using Vieta's theorem
We reduce a quadratic equation to reduced form if the equation is given to us in unreduced form. When the coefficients in the quadratic equation, which we previously presented as given, turn out to be fractional (not decimal), then in this case our equation should be solved through the discriminant.
There are also cases when returning to the initial equation allows us to work with “convenient” numbers.
When studying methods for solving second-order equations in a school algebra course, the properties of the resulting roots are considered. They are currently known as Vieta's theorem. Examples of its use are given in this article.
Quadratic equation
The second order equation is the equality shown in the photo below.
Here the symbols a, b, c are some numbers called the coefficients of the equation under consideration. To solve an equality, you need to find values of x that make it true.
Note that since the maximum power to which x can be raised is two, then the number of roots in the general case is also two.
There are several ways to solve this type of equalities. In this article we will consider one of them, which involves the use of the so-called Vieta theorem.
Formulation of Vieta's theorem
At the end of the 16th century, the famous mathematician Francois Viète (French) noticed, while 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 shown in the photo in the previous section of the article, then mathematically this theorem can be written in the form of two equalities:
- r 2 + r 1 = -b / a;
- r 1 x r 2 = c / a.
Where r 1, r 2 is the value of the roots of the equation in question.
The above two equalities can be used to solve a number of different mathematical problems. The use of Vieta's theorem in examples with solutions is given in the following sections of the article.
Formulation and proof of Vieta's theorem for quadratic equations. Vieta's converse theorem. Vieta's theorem for cubic equations and equations of arbitrary order.
Quadratic equations
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 of , 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's find the roots of equation (1). To do this, apply the formula for the roots of a quadratic equation:
;
;
.
Find the sum of the roots:
.
To find the product, apply the formula:
.
Then
.
The theorem has been proven.
Proof two
If the numbers are the roots of the quadratic equation (1), then
.
Opening the parentheses.
.
Thus, equation (1) will take the form:
.
Comparing with (1) we find:
;
.
The theorem has been proven.
Vieta's converse 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).
Let's substitute (2) and (3) into (1):
.
We group the terms on the left side of the equation:
;
;
(4)
.
Let's substitute in (4):
;
.
Let's substitute in (4):
;
.
The equation holds. That is, the number is the root of equation (1).
The theorem has been proven.
Vieta's theorem for a complete quadratic equation
Now consider the complete quadratic equation
(5)
,
where , and are some numbers. Moreover.
Let's divide equation (5) by:
.
That is, we got the given equation
,
Where ; .
Then Vieta's theorem for a 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 cubic equation
In a similar way, we can establish connections between the roots of a cubic equation. Consider the cubic equation
(6)
,
where , , , are some numbers. Moreover.
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 equation of nth degree
In the same way, you can find connections between the roots , , ... , , for an equation of nth degree
.
Vieta's theorem for an equation of nth degree has the following form:
;
;
;
.
To obtain these formulas, we write the equation as follows:
.
Then we equate the coefficients for , , , ... , and compare the free term.
Used literature:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
CM. Nikolsky, M.K. Potapov et al., Algebra: textbook for 8th grade in general education institutions, Moscow, Education, 2006.
