Question

已知方程(x-1)(x^2+8x-3)=0的三个根为x1,x2,x3，求x1x2+x2x3+x3x1的值

要过程

Answer 1

(X-1)(X²+8X-3)=0===>X-1=0或X²+8X-3=0===>X=1或X=±√17-4
===》=1×(√17-4)+(√17-4)(-√17-4)+(-√17-4)×1
=√17-4-√17-4-17+16
=-25+16
=-9

Answer 2

可知 x1 = 1,则x1x2+x2x3+x3x1 = x2+x3+x2x3 = -8 -3= -11

Answer 3

2楼说的没错