Vieta theorem

Vieta's Theorem: In the unary quadratic equation AX 2+BX+C (A is not 0).

Let two roots be x and y.

Then x+y =-b/a.

xy=c/a

content analysis

1. A discriminant formula for the roots of a quadratic equation with one variable

Discriminant formula of roots of unary quadratic equation AX 2+BX+C = 0 (A ≠ 0) △ = B2-4ac

When △ > 0, the equation has two unequal real roots;

When △ = 0, the equation has two equal real roots.

When △ < 0, the equation has no real root.

2. The relationship between the roots and coefficients of a quadratic equation.

(1) if the two roots of the unary quadratic equation aX 2+bx+c = 0 (a ≠ 0) are x 1, x 2, then X 1+X2 =-B/A, x1* x2.

(2) If the two roots of the equation x2+px+q = 0 are x 1, x2, then x 1+x2=-P,

x 1x2=q

(3) The quadratic equation with x 1 and x2 as roots (the quadratic term coefficient is 1) is

x2-(x 1+x2)x+x 1x2=0。

3. Factorial decomposition of quadratic trinomial (formula method)

In the factorization of quadratic trinomial AX 2+BX+C, if the two roots of the equation AX 2+BX+C = 0 are X 1 and X 2, then AX 2+BX+C = A (X-X1) (X-X2).

Example: given two x 1, x2 where x 2-2x-3 = 0, find x 1 square +x2 square.

Solution 1: The roots of Equation 2 are-1 and 3, so x 1 square +x2 square = 10.

Solution 2: directly use Vieta's definition to understand the equation, x 1 square +x2 square = (x1+x2) 2-2x1* x2 = 4+6 =10.

If the equation is difficult to solve, the advantages of Vieta's theorem will come out.

Vieta Theorem-Theorem proves that x_ 1 and x_2 are two solutions of the unary quadratic equation AX 2+BX+C = 0, and x_ 1 \ge x_2. According to the root formula, there are

x _ 1 = \ frac {-b+ \ {b^2-4ac}},x_2=\frac{-b-\ {b^2-4ac}}

therefore

x _ 1+x _ 2 = \ frac {-b+ \ sqrt {b^2-4ac}+\ left(-b \ right)-\ sqrt {b^2-4ac}} =-\ frac

X _1x _ 2 = \ frac {\ left (-b+\ sqrt {b 2-4ac} \ right) \ left (-b-\ sqrt {b 2-4ac} \ right)} {\ left (2a \)

Edit this paragraph and go back to the table of contents. Vieta Theorem-Generalization of Theorem

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.

Edit this table of contents Vieta Theorem-Theorem The application of Vieta Theorem is an important theorem reflecting the relationship between the roots and coefficients of a quadratic equation in one variable. The questions related to this theorem are very common in the senior high school entrance examination (competition), and the conditions are relatively hidden. When students prove (solve) a problem, they often get blocked in thinking because they don't see the conditions of Vieta's theorem implied in the problem, or the solution is dull and the process is tedious. Let's take an example to talk about the application of Vieta's theorem in solving problems.

First, the direct application of Vieta's theorem.

If the known conditions or conclusions to be proved contain formulas in the form of A+B and A B, Vieta's theorem can be directly applied.

Example 1 In △ABC, A, B and C are opposite sides of ∠A, ∠B and ∠C respectively, D is a point on the side of AB, BC = DC, and let AD = D. 。

Verification:

( 1)c+d = 2 bcosa；

(2)c 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: As shown in the figure, in △ABC and △ADC, there are

a2 = B2+C2-2 bcco sa；

a2=b2+d2-2bdcosA(CD=BC=a)。

∴c2-2bccosA+b2-a2=0，

d2-2bdcosA+b2-a2=0。

So c and d are the two roots of the equation X 2-2 BxCOSA+B 2-A 2 = 0.

According to Vieta's theorem, there are

c+d=2bcosA，c d=b2-a2。

Example 2 A+A2- 1 = 0, B+B2- 1 = 0, a≠b, find the value of ab+a+B. 。

Analysis: Obviously, two formulas are known to have the same form: x2+x- 1 = 0. So a and b can be regarded as two roots of quadratic equation. After observing the structure of the formula to be solved, it is easy to think of directly applying Vieta theorem to solve it.

Solution: We can construct the unary quadratic equation X2+X- 1 = 0, and its two roots are A and B. 。

According to Vieta's theorem, a+b =- 1, a b =- 1.

So ab+a+b =-2.

Second, do not deform first, and then apply Vieta's theorem.

If the conditions or conclusions to be proved are known, the formulas in A+B and A B forms can be constructed by constant deformation or substitution, and Vieta's theorem can be considered.

Example 3 If the real numbers X, Y and Z satisfy X = 6-Y and Z2 = x=y.-9. Verification: x = y.

It is proved that the known formula is transformed into x+y = 6 and xy = z2+9.

Vieta's theorem knows that x and y are two roots of the equation U2-6u+(Z2+9) = 0.

∵x and Y are real numbers, ∴△ = 36-4Z2-36 ≥ 0.

Then z2≤0 and ∵z is a real number,

∴ Z2 = 0, that is, △ = 0.

Therefore, the equation U2-6u+(Z2+9) = 0 has equiroots, so x = y. 。

From the known formula, it is easy to know that X and Y are two roots of T2+3t-8 = 0, which are determined by Vieta's theorem.

Third, given the relationship (or coefficient relationship) between the two roots of a quadratic equation, we can consider using Vieta theorem to find the coefficient relationship (or the relationship between the two roots).

Example 5 It is known that the ratio of two roots of the equation x2+px+q = 0 is 1∶2, and the value of the discriminant of the equation is 1. Find the values of p and q and solve this equation.

Solution: Let x2+px+q = 0 be a and 2a, then Vieta's theorem, with.

a+2a=-P，①

a 2a=q，②

P2-4q= 1。 ③

Substituting ① and ② into ③ gives (-3a) 2-4× 2a2 = 1, which means 9a2-8a2 = 1, so a = 1.

The equation is x2-3x+2 = 0 or x2+3x+2 = 0.

The solution is x 1 = 1, x2 = 2, 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.

It is proved that two of the equation x2+px+q = 0 are α and β, and two of the equation x2+qx+p = 0 are α' and β'.

Judging from the meaning of the question, α-β = α'-β',

Therefore, α 2-2 α β+β 2 = α' 2-2 α' β'+β' 2.

Therefore, (α+β) 2-4 α β = (α'+β') 2-4 α' β'. ①

Substitute ② into ①, P2-4q = Q2-4p, that is, P2-Q2+4p-4q = 0, that is, (p+q) (p-q)+4 (p-q) = 0, that is, (p-q) (p+q+4) = 0.

So p-q = 0 or p+q+4 = 0,

That is, p = q or p+q =-4.

Fourth, regarding the problem that two quadratic equations with one variable have a common root, we can consider Vieta's theorem.

When 7m is the asking value, do the equations x2+MX-3 = 0 and x2-4x-(m- 1) = 0 have a common root? And find this common root.

Solution: Let the 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

M = 1-4α+α 2 in ② and ③.

Substituting ③ into ① gives α 3-3α 2+α 3 = 0,

That is, (α-3) (α 2+ 1) = 0.

∵ α 2+ 1 > 0, ∴ 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.