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
问题描述:
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
答
(x1-x2)^2
=(x1+x2)^2-4x1*x2
=(-4)^2-4*2
=8
答
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