Three numbers 12x, x 2-5 and 4 in that order form an increasing arithmetic progression https://youtu.be/U0VO_N9udpI Choose the correct statement MATHEMATICS ZFTSH MIPT Moscow Institute of Physics and Technology (State University) Correspondence School of Physics and Technology. http://pin.it/9w-GqGp Find all x, y and z such that the numbers 5x + 3, y2 and 3z + 5 form an arithmetic progression in that order. Find x and indicate the difference of this progression. Solve the system of equations Mathematics of the Unified State Examination. Video lessons. Divisibility of integers. Linear function. Divisibility problems. Vieta's theorem, converse theorem, Vieta's formulas. clever #students #equations #vietas_theorem #theorem Next we consider the theorem converse to Vieta’s theorem. After this, we will analyze the solutions to the most typical examples. This proves the first relation of Vieta's theorem for the sum of the roots of a quadratic equation. Let's move on to the second. How to prove the converse of Vieta's theorem? DOK-VO: x2+px+f=0 x2-(M+N) *x+M*N=0 x2-Mx-Nx+M*N=0 x (x-N) -M (x-N) =0 (x-M) (x-N) =0 x-M=0 x-N=0 x=M x=N CTD. This is how we proved it in a specialized class with a mathematical bias. Answers: help to understand the inverse theorem of Vieta's theorem thanks to specific examples The theorem inverse of Vieta's theorem helps to solve the solution: If coefficient a is a number from which it is easy to extract the square root of a rational integer, then the sum of x1 and x2 will be equal to the number Prove the inverse theorem Vieta - see how to complain about the proof of Vieta's theorem. Formulate and prove Vieta’s theorem, as well as the converse theorem, and apply the theorems to solve equations and problems. Prove the converse of Vieta's theorem. Unified State Examination in mathematics for 100 points: secrets that school teachers don’t tell you, problems on derivatives. Many applicants think that they do not need to prepare for the first fourteen problems, thinking that they are very easy, but this is not so! Most test takers make the simplest arithmetic errors, thereby overshadowing the excellent solution to the problems of part C. Such situations occur very often, therefore, you should not neglect preparing for the first problems, but prepare as you would during a sports training: if you are applying for 90-100 points - practice solving the first block in 20-25 minutes, if for 70-80 points - about 30 minutes, no more. An excellent way to train is to solve in the company of a tutor, in courses where certain conditions will be set: for example, you solve before the first mistake, then hand in the work; Another option is that for every mistake you make, you donate money to the general cash register. No matter how strange it may seem, we do not recommend the official website, since all the tests there are so mixed up that it is impossible to use it. The formatting of Part C tasks is important. If the solution is not prepared carefully, then the progress of solving the task will be unclear, and therefore, the examiner will definitely find fault with this and reduce your score. It would seem that we talked about very simple things, but by following our advice, you will ensure that you successfully pass the Unified State Exam! The secret links discussed in the Master Class can be found here - these are links to Video courses for preparing for the Unified State Exam. The result obtained is called Vieta's theorem. For the reduced square trinomial 2 x px q, Vieta’s theorem looks like this: if there are roots, then the inverse of Vieta’s theorem also holds: if the numbers satisfy the conditions, then these numbers are the roots of the equation. The proof of this theorem is one of the control questions of the Assignment. Sometimes, for brevity, both Vieta's theorems (direct and inverse) are simply called Vieta's theorem.
One of the methods for solving a quadratic equation is to use VIET formulas, which was named after FRANCOIS VIETTE.
He was a famous lawyer who served the French king in the 16th century. In his spare 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 a solution. Because there is no need to enter the second coefficient into the square, then subtract 4ac from it, find the discriminant, and substitute its value into the formula to find the roots.
2 . Without a solution, you can determine the signs of the roots and select the values of the roots.
3 . Having solved a 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 . Using these roots, write down a quadratic equation, that is, solve the inverse problem. For example, this method is used when solving problems in theoretical mechanics.
5 . It is convenient to use the formula when the leading coefficient is equal to one.
Flaws:
1
. The formula is not universal.
Vieta's theorem 8th grade
Formula
If x 1 and x 2 are the roots of the reduced quadratic equation x 2 + px + q = 0, then:
Examples
x 1 = -1; x 2 = 3 - roots of the equation x 2 - 2x - 3 = 0.
P = -2, q = -3.
X 1 + x 2 = -1 + 3 = 2 = -p,
X 1 x 2 = -1 3 = -3 = q.
Converse theorem
Formula
If the numbers x 1, x 2, p, q are related by the conditions:
Then x 1 and x 2 are the roots of the equation x 2 + px + q = 0.
Example
Let's create a quadratic equation using its roots:
X 1 = 2 - ? 3 and x 2 = 2 + ? 3.
P = x 1 + x 2 = 4; p = -4; q = x 1 x 2 = (2 - ? 3 )(2 + ? 3 ) = 4 - 3 = 1.
The required equation has the form: x 2 - 4x + 1 = 0.
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.
References:
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.
Vieta's theorem is often used to check roots that have already been found. 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 of \(p\) 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 us, using , 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 solution process. 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 we solved the equation correctly.
This check can be done orally. It will take 5 seconds and will save you from stupid mistakes.
Vieta's converse 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 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 because it saves a lot of time.
Example . Solve the equation \(x^2-5x+6=0\).
Solution
: Using Vieta’s inverse theorem, we find 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 system \(x_1 \cdot x_2=6\). What two can the number \(6\) be decomposed into? On \(2\) and \(3\), \(6\) and \(1\) or \(-2\) and \(-3\), and \(-6\) and \(- 1\). The first equation of the system will tell you which pair to choose: \(x_1+x_2=5\). \(2\) and \(3\) are similar, since \(2+3=5\).
Answer
: \(x_1=2\), \(x_2=3\).
Examples
. Using the converse 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) \(x^2+3x-4=0\) – what factors does \(-4\) decompose into? \(-2\) and \(2\), \(4\) and \(-1\), \(1\) and \(-4\). What pairs of numbers add up to \(-3\)? Answer: \(1\) and \(-4\).
c) \(x^2+9x+20=0\) – what factors does \(20\) decompose into? \(4\) 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) \(x^2-88x+780=0\) – what factors does \(780\) decompose into? \(390\) and \(2\). Will they add up to \(88\)? No. What other multipliers does \(780\) have? \(78\) and \(10\). Will 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 the last example). You can immediately check whether their sum gives \(-p\).
Important! Vieta's theorem and the converse theorem only work with , that is, one for which the coefficient of \(x^2\) is equal to one. If we were initially given 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, since the coefficient of \(x^2\) is equal to \(2\). Let's get rid of it by dividing the entire equation by \(2\).
\(2x^2-4x-6=0\) \(|:2\)
\(x^2-2x-3=0\)
Ready. Now you can use both theorems.
Answers to frequently asked questions
Question:
Using Vieta's theorem, you can solve any ?
Answer:
Unfortunately no. If the equation does not contain integers 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 mathematics have integer solutions.
Quadratic function.
A function given by the formula y = ax2 + bx + c, where x and y are variables and a, b, c are given numbers, and a is not equal to 0.
called quadratic function
Selecting a complete square.
Derivation of the formula for the roots of a quadratic equation, conditions for their existence and numbers.
– discriminant of a quadratic equation.
Direct and inverse Vieta theorems.
Decomposition of a quadratic trinomial into linear factors.
Theorem. Let
x 1 and x 2 - roots of a square trinomialx 2 + px + q. Then this trinomial is decomposed into linear factors as follows:x 2 + px + q = (x - x 1) (x - x 2).Proof. Let's substitute instead
p And qtheir expressions throughx 1 and x 2 and use the grouping method:x 2 + px + q = x 2 - (x 1 + x 2 ) x + x 1 x 2 = x 2 - x 1 x - x 2 x + x 1 x 2 = x (x - x 1 ) - x 2 (x - x 1 ) = = (x - x 1 ) (x - x 2 ). The theorem has been proven.
Quadratic equation. Graph of a quadratic trinomial
Equation of the form
called a quadratic equation. The number D = b 2 - 4ac is the discriminant of this equation.
If
then the numbers
are the roots (or solutions) of a quadratic equation. If D = 0, then the roots are the same:
If D< 0, то квадратное уравнение корней не имеет.
Valid formulas:
— Vieta formulas; A
ax 2 + bx + c = a(x - x 1)(x - x 2) -
factorization formula.
The graph of the quadratic function (quadratic trinomial) y = ax 2 + bx + c is a parabola. The location of the parabola depending on the signs of the coefficient a and the discriminant D is shown in Fig.
The numbers x 1 and x 2 on the abscissa axis are the roots of the quadratic equation ax 2 + bx + + c = 0; coordinates of the vertex of the parabola (point A) in all cases
the point of intersection of the parabola with the ordinate axis has coordinates (0; c).
Like a straight line and a circle, a parabola divides a plane into two parts. In one of these parts, the coordinates of all points satisfy the inequality y > ax 2 + bx + c, and in the other, the opposite. We determine the inequality sign in the selected part of the plane by finding it at any point in this part of the plane.
Let's consider the concept of a tangent to a parabola (or circle). We will call the straight line y - kx + 1 tangent to a parabola (or circle) if it has one common point with this curve.
At the point of contact M(x; y), for a parabola the equality kx +1 = ax 2 + bx + c holds (for a circle - the equality (x - x 0) 2 + (kx + 1 - y 0) 2 - R 2). Equating the discriminant of the resulting quadratic equation to zero (since the equation must have a unique solution), we arrive at the conditions for calculating the tangent coefficients.