Vedas (Vieta's, Fran? ois's, Senedra's)

Bigotiere) 1540 was born in poitiers, and 1603+02+03 died in Paris. He studied law in Pufa Street in his early years, then worked as a lawyer. 1567 became a member of parliament. In the war against Spain, he cracked the enemy's password for the government and won a high reputation. /kloc-One of the most influential mathematicians in France in the 6th century. He was the first to introduce algebraic symbols of systems and improved the theory of equations.

The content of Vieta's theorem.

In the unary quadratic equation ax 2+bx+c = 0 (a ≠ 0 and △ = b 2-4ac ≥ 0)

Let two roots be x 1 and x2.

Then x1+x2 =-b/a.

x 1*x2=c/a

Generalization of Vieta Theorem

Vieta's theorem can also be used for higher-order equations. Generally speaking, for an equation with n ∑ AIX I = 0.

Its roots are expressed as x 1, x2…, xn.

we have

∑xi=(- 1)^ 1*a(n- 1)/a(n)

∑xixj=(- 1)^2*a(n-2)/a(n)

…

πxi=(- 1)^n*a(0)/a(n)

Where ∑ is the sum and π is the quadrature.

If the unary quadratic equation

So, the root in the complex set is

Veda, a French mathematician, first discovered this relationship between the roots and coefficients of algebraic equations, so people called this relationship Vieta Theorem. History is very interesting. This theorem was obtained by David in16th century. The proof of this theorem depends on the basic theorem of algebra, but Gauss first demonstrated it in 1799.

From the basic theorem of algebra, we can deduce that any unary equation of degree n

There must be a root in a complex set. Therefore, the left end of the equation can be decomposed into the product of linear factors in the range of complex numbers:

Where is the root of the equation. Vieta's theorem is obtained by comparing the coefficients at both ends.

Vieta theorem is widely used in equation theory.

Proof of Vieta Theorem

Let x_ 1 and x_2 be two solutions of the unary quadratic equation ax 2+bx+c = 0.

According to the root formula, there are

x_ 1=[-b + -\sqrt

(b^2-4ac)]/2a，

therefore

x _ 1+x _ 2 =[-b+(-)\ sqrt(b^2-4ac)]/2a+[-b-

\sqrt (b^2-4ac)]/2a=-b/a

Application of Vieta Theorem Application of Vieta Theorem Application of Vieta Theorem::: 1 Knowing one root of the equation, find the other root and the unknown coefficient 2. Find the value of algebraic expression related to two roots of known equation 3. It is known that the two roots of the equation satisfy a certain relationship, and the value of the letter coefficient in Equation 4 is determined. Know the sum and product of two numbers and find these two numbers 5. Know the two roots of the equation x 1 , x2, find a new quadratic equation x2-(x1+x2) x+x1x2 = 0 6. The factor ax2+bx+c = a(x- x 1)(x- x2) in the real number range is decomposed by the root formula.

On the application of Vieta's theorem, Vieta's theorem and Vieta's theorem in solving problems. Vieta's theorem is an important theorem reflecting the relationship between the roots and coefficients of a quadratic equation with one variable. Looking at the examination questions of the senior high school entrance examination (competition) in recent years, we can find that the questions related to this theorem are common and the conditions are hidden. When students prove (solve) problems, they often can't see the conditions of Vieta's theorem implied in the questions, which leads to the occlusion of thinking, or the boring and tedious process of solving problems. Here are a few examples to talk about the application of Vieta theorem in solving problems for your reference.

1111,,,direct application of Vieta theorem If the known conditions or conclusions to be proved contain formulas in the form of A+B and AB, Vieta theorem can be directly applied. Example 1 In △ABC, where A, B and C are opposite sides of ∠A, ∠B and ∠C respectively. ②c d = B2-A2。 Analysis: Observing the conclusion to be proved, we can naturally associate it with Vieta's theorem, thus constructing a quadratic equation with one variable to prove it. Proof and proof: As shown in the figure, in △ABC and △ADC, there are A2 = B2+C2-2bcosa by cosine theorem; A2 = B2+D2-2BDCOSA (CD = BC = A)。 ∴ C2-2BCCOSA+B2-A2 = 0，D2-2BDCOSA+B2-A2 = 0。 Therefore, c and d are two roots of the equation X2-2BxCOSA+B2-A2 = 0. According to Vieta's theorem, C d = B2-A2. Example 2: A+A2- 1 = 0, B+B2- 1 = 0, a≠b, find the value analysis of AB+A+B: Obviously, it is known that the two formulas have the same form: X2+. It is easy to think of directly applying Vieta's theorem to solve it. Solution: A quadratic equation x2+x- 1 = 0 can be constructed with known things, and its two roots are A and B. According to Vieta's theorem, A+b =- 1, A+b =- 1.

Therefore, AB+A+B =-2. 2222,,, Invariant deformation first, Invariant deformation first, Invariant deformation first, then Vieta's theorem, Vieta's theorem, Vieta's theorem and Vieta's theorem are known conditions or conclusions. After invariant deformation or substitution, formulas in the form of A+B and A.B are constructed. Vieta's theorem can be considered. Example 3 If the real numbers x, y and z satisfy X = 6-Y and Z2 = XY-9. Proof: X = Y Proof: Convert the known binomial into X+Y = 6, XY = Z2+9. Vieta's theorem knows that x and y are equations U2-6U+(Z2+9). ∵z is a real number, ∴ Z2 = 0, that is, △ = 0. So the equation U2-6u+(Z2+9) = 0 has equal roots, so X = Y. From the known formula, it is easy to know that X and Y are two roots of T2+3t-8 = 0. From Vieta's theorem of three, three, three, three, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation, one-dimensional quadratic equation. Find the values of p and q and solve this equation. Solution: Let the two roots of x2+PX+Q = 0 be A and 2a, then the equation is given by.

A 2a = q, ② P2-4q = 1. ③ Substituting ① and ② into ③ gives (-3a) 2-4× x2=2 = 1, that is, 9a2-8a2 = 1, so A = 1. Or x 1 =- 1, x2 =-2. Example 6 Let the difference between two roots of equation x2+px+q = 0 be equal to the difference between two roots of equation x2+qx+p = 0. Proof: p = q or p+q =-4. Proof: Let the equation x2+px+. So there is α 2-2 α β+β 2 = α' 2-2 α' β'+β' 2. Thus, there is (α+β) 2-4 α β = (α'+β') 2-4 α' β'. ① If ② is substituted into ①, P2-4q = Q2-4p, which is P2. That is, p = q or p+q =-4.4444,,, on the topic of a quadratic equation with one variable having a common root, on the topic of a quadratic equation with one variable having a common root, we can consider using. And find this common root.

Solution: Let the common root of * * * be α, which is easy to know. The two roots of the original equation x2+MX-3 = 0 are α and-m -m-α；; X2-4x-(m- 1) = 0 is α and 4-α. According to Vieta's theorem, α (m+α) = 3, ① α (4-α) =-(m- 1). ② From ②, we get M = 1-4. ∴ α-3 = 0 means α = 3. Substitute α = 3 into ③ to get m =-2. So when m =-2, two known equations have a common root, and this common root is 3.