题目内容
请选择适当的方法解下列一元二次方程.
(1)x(x+2)=(x+2)
(2)x2+2x-1=0
(3)x2-4x-12=0.
(1)x(x+2)=(x+2)
(2)x2+2x-1=0
(3)x2-4x-12=0.
(1)移项得:x(x+2)-(x+2)=0,
把左边分解因式得;(x+2)(x-1)=0,
则x+2=0,x-1=0,
解得;x1=-2,x2=1;
(2)a=1,b=2,c=-1,
x=
=
=-1±
;
故x1=
-1; x2=-
-1;
(3)把左边分解因式得;(x-6)(x+2)=0,
x-6=0,x+2=0,
解得:x1=6,x2=-2.
把左边分解因式得;(x+2)(x-1)=0,
则x+2=0,x-1=0,
解得;x1=-2,x2=1;
(2)a=1,b=2,c=-1,
x=
-b±
| ||
| 2a |
-2±2
| ||
| 2 |
| 2 |
故x1=
| 2 |
| 2 |
(3)把左边分解因式得;(x-6)(x+2)=0,
x-6=0,x+2=0,
解得:x1=6,x2=-2.
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