解下列一元二次方程: (1)(x-2)2=2x-4; (2)2x2-4x-1=0.
问题描述:
解下列一元二次方程:
(1)(x-2)2=2x-4;
(2)2x2-4x-1=0.
答
(1)∵x2-4x+4=2x-4
∴x2-6x+8=0
(x-2)(x-4)=0
解得:x1=2,x2=4;
(2)∵b2-4ac=
=2
16+8
>0,
6
.∴x=
,4±2
6
4
故x1=
, x2=2+
6
2
.2−
6
2