How to extend "Veda's theorem" to equation of degree n with one variable?

How to extend "Veda's theorem" to equation of degree n with one variable?

For the univariate quadratic equation AX ^ 2 + BX + C = 0, if its two roots are X1 and X2, then X1 + x2 = - B / A, X1 * x2 = C / A for the univariate cubic equation AX ^ 3 + BX ^ 2 + CX + D = 0, if its three roots are x1, X2 and X3, then X1 + x2 + X3 = - B / a 1 / X1 + 1 / x2 + 1 / X3 = - C / D, X1 * x2 * X3 = - D / A for the univariate n
It is known that the function f (x) is an odd function defined on R. when x ≥ 0, f (x) = x (1 + x), the image of function f (x) is drawn and the analytic expression of function f (x) is obtained
∵ when x ≥ 0, f (x) = x (1 + x) = (x + 12) 2-14, f (x) is an odd function defined on R, ∵ when x < 0, - x > 0, f (- x) = - x (1-x) = (X-12) 2-14 = - f (x), ∵ f (x) = - (X-12) 2 + 14 ∵ f (x) = (x + 12) & nbsp; 2-14x ≥ 0 - (X-12) & nbsp; 2 + 14x < 0
Given the set a = empty set, B = {x | (x + 1) (x ^ 2 + 3x-4) = 0, X ∈ r}, a is really contained in C, C is really contained in B, find the set C satisfying the condition
B={x|（x+1）（x^2+3x-4）=0,x∈R}
B={x|（x+1）（x-1)(x+4）=0,x∈R}
B={-4,-1,1}
Because a is really contained in C, set a = empty set
So C is not an empty set
So C = {- 4} or C = {- 1} or C = {1} or C = {- 4, - 1} or C = {- 4,1} or C = {- 1,1}
（x+1）（x^2+3x-4）=0
(x+1)(x+4)(x-1)=0
x1=-1,x2=-4,x3=1
That is, B = {- 1, - 4,1}
A is really contained in C, C is really contained in B, the set C satisfying the condition
{-1,}{-4},{1},{-1,-4},{-1,1},{-4,1}
Weida's theorem for an equation of degree n of one variable is expressed as follows: as shown in the figure, but an equation: (x-1) (X-2) (x-3) = 0,
x1=1,x2=2,x3=3
X1x2 + x2x3 = 8, equation expansion x ^ 3-6x ^ 2 + 11x-6 = 0, 11 / 1 = 11, the two are not equal? Why?
That's the sum of the products of any two terms. You wrote one less term, which should be x1x2 + x1x3 + x2x3 = 2 + 3 + 6 = 11
It is known that the piecewise function f (x) is an odd function on R. when x > 0, f (x) = x2-2x + 3, the analytic expression of F (x) is obtained
Let - x ＞ 0, then x ＜ 0, f (- x) = x2 + 2x + 3, f (- x) = - f (x), f (x) = - x2-2x-3, f (x) = x2 − 2x + 3, (x ＞ 0) 0, (x = 0) − x2 − 2x − 3, (x ＜ 0)
Given the complete set u = R, a = {x | 2 ≤ x ＜ 4}, B = {x | 3x-7 ≥ 8-2x}, find: (1) a ∩ B; (2) (∁∪ a) ∩ B; (3) ∁∪ (a ∪ b)
(1) The inequality solutions in B are: 5x ≥ 15, that is, X ≥ 15, and the inequality solutions in B are: 5x ≥ 15, that is, X ≥ 3, X ≥ 15, that is, X ≥ 3, X ≥ 15, and the inequality solutions in B are: 5x ≥ 15, that is, X ≥ 15, X ≥ 15, that is, X ≥ 3, X ≥ 3, X ≥ 15, that is, X ≥ 3, X ≥ 3, \3, {B = (3,4,4,4,4) \\\ \\\\\\\\\\\\8745\\8745\\\\\\\\\b) = (∞, 2)
What is the Veda's theorem for multiple equation of one variable
Weida's theorem for multiple equation of one variable: in the quadratic equation of one variable ax ＾ 2 + BX + C = 0 (a ≠ 0 and B ＾ 2-4ac ≥ 0), two X1 and X2 have the following relations: X1 + x2 = - B / A; X1 * x2 = C / A
It is known that the function f (x) is an odd function defined on R, and when x > 0, the analytic expression of F (x) = x ^ 3 + 2x ^ 2-1
When X0
∴f(-x)=(-x)^3+2*(-x)^2-1
∵ f (x) is an odd function defined on R
∴f(x)=-f(x)
∴f(x)=x^3-2x^2+1,x0
f(x)={0,x=0
{x^3-2x^2+1,x
Given a = {x ∈ R | 3x + 2 ＞ 0}, B = {x ∈ R | (x + 1) (x-3) ＞ 0}, then a ∩ B = ()
A. （-∞，-1）B. （-1，−23）C. ﹙−23，3﹚D. （3，+∞）
Because B = {x ∈ R | (x + 1) (x-3) ＞ 0} = {x | x ＜ - 1 or X ＞ 3}, and set a = {x ∈ R | 3x + 2 ＞ 0} = {x | x ＞ − 23}, a ∩ B = {x | x ＞ 23} ∩ {x | x ＜ - 1 or X ＞ 3} = {x | x ＞ 3}, so D
A problem about Weida theorem of equation is not difficult~
Let 0, - 4 be the solution of the equation x & # 178; + 2 (a + 1) x + A & # 178; - 1 = 0
How to get the following conclusion from Weida theorem:
-2(a+1)=-4
a²-1=0
How do you get this equation set? Please, brother, it's better to have a text description, because I may not understand how a certain formula is obtained in some processes,
Weida theorem: in the quadratic equation AX ^ 2 + BX + C = 0 (Δ≥ 0), two X1 and X2 have the following relations: X1 + x2 = - B / A, x1 · x2 = C / A
Since 0 and - 4 are solutions of the equation, we can assume that X1 = 0 and X2 = - 4
So we have - 4 + 0 = - 2 (a + 1), 0 * - 4 = A & # 178; - 1