Question

设x1、x2是方程是2x^2+4x-3=0的两个根,利用根与系数的关系,求下列式的值:绝对值X1-X2绝对值

求详细步骤，急

Answer 1

x1+x2=-2
x1x2=-3/2
(x1-x2)²=(x1+x2)²-4x1x2=4+6=10
|x1-x2|=√10

Answer 2

∵x1、x2是方程是2x^2+4x-3=0的两个根
∴x1+x2=-2，x1x2=-3/2
|X1-X2|=√[（x1+x2)²-4x1x2]=√4+6=√10

Answer 3

(x1-x2)²=x1²+x2²-2x1x1=(x1+x2)²-4x1x2
x1+x2=-2
x1x2=-3/2
(x1+x2)²-4x1x2=4+6=10
|x1-x2|=√10

Answer 4

据韦达定理 X1+X2=-b/a是-2 X1*X2是c/a是-1.5 (X1+X2)^2=4 (X1-X2)^2=4+4*1.5=10 再开方即可