If ax ` 2 + BX + C = 0 (a ≠ 0) has two roots, which are 2,3, then find the root of Cx ` 2 + BX + a = 0

If ax ` 2 + BX + C = 0 (a ≠ 0) has two roots, which are 2,3, then find the root of Cx ` 2 + BX + a = 0

Ax ^ 2 + BX + C = o the roots of two real numbers are 2 and 3
By Weida theorem
2+3=-b/a
2*3=c/a
So B = - 5A, C = 6A
cx^2+bx+a=0
6ax^2-5ax+a=0
6x^2-5x+1=0
(3x-1)(2x-1)=0
x=1/3,x=1/2

What is the Weida theorem in mathematics? How to use it? In which problems?

In the 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
Generally, for an equation of degree n ∑ AIX ^ i = 0
Its roots are recorded as x1, X2 ,Xn
We have
∑Xi=(-1)^1*A(n-1)/A(n)
∑XiXj=(-1)^2*A(n-2)/A(n)
…
∏Xi=(-1)^n*A(0)/A(n)
Where ∑ is the sum and Π is the product
If the quadratic equation of one variable
The root in the complex set is, then
Weida, a French mathematician, first discovered this relationship between the roots and coefficients of algebraic equations. Therefore, people call this relationship Weida's theorem. History is interesting. Weida obtained this theorem in the 16th century. The proof of this theorem depends on the basic theorem of algebra. However, the basic theorem of algebra was first made by Gauss in 1799
From the basic theorem of algebra, we can deduce: any equation of degree n with one variable
Therefore, the left end of the equation can be decomposed into a product of factors in the complex range
Where is the root of the equation. By comparing the coefficients at both ends, we obtain the Veda theorem
Weida's theorem is widely used in equation theory
Proof of theorem
Let X_ 1,x_ 2 is two solutions of the quadratic equation AX ^ 2 + BX + C = 0, and let x_ 1 \ge x_ 2. According to the root formula, there are
x_ 1=\frac{-b + \sqrt {b^2-4ac}},x_ 2=\frac{-b - \sqrt {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 \right)^2} =\frac

Middle school mathematics algebra problems, Weida theorem class
We know that X1 and X2 are the two real roots of the equation x ^ 2-x-9 = 0,
Find the value of the algebraic formula X1 ^ 3 + 7X2 ^ 2 + 3x2-66

X1 + x2 = 1, X1 * x2 = 9 this problem can be reduced to power, because x ^ 2-x-9 = 0, so X1 ^ 2-x1-9 = 0, X1 ^ 2 = X1 + 9x2 ^ 2-x2-9 = 0, X2 ^ 2 = x2 + 9x1 ^ 3 = X1 * (x1 + 9) = x1 ^ 2 + 9x1 = X1 + 9 + 9x1 = 10x1 + 97x2 ^ 2 = 7 (x2 + 9) = 7X2 + 63x1 ^ 3 + 7X2 ^ 2 + 3x2-66 = 10x1 + 9 + 7X2 + 63 + 3x2-66 = 1