若抛物线Y=2X^2-4X+1与X轴两交点分别是（X1,0),(X2,0),则X!^2+X2^2=
问题描述：
若抛物线Y=2X^2-4X+1与X轴两交点分别是（X1,0),(X2,0),则X!^2+X2^2=
答
解析式 y = 2x^2 - 4x +1中,a = 2 ,b = - 4 ,c = 1
根据韦达定理 可得 x1 +x2 = -b/a = 2 ; x1 * x2 = c / a = 1/2
x1^2 + x2^2
= ( x1 + x2)^2 - 2 * x1*x2
= 4 - 2* (1/2)
=4 - 1
=3