Question

已知X1,X2是方程x^2-4x+2=0的两根,韦达定理求（X2-X2)^2的值

Answer 1

X1,X2是方程x^2-4x+2=0的两根,则

X1 + X2 = 4

X1X2 = 2

（X2-X2)^2 = (X1 + X2)^2 - 4X1X2 = 4^2 - 4*2 = 8