题目内容
解下列方程:
(1)(x+1)(x-3)=12
(2)3x2+2(x-1)=0.
(1)(x+1)(x-3)=12
(2)3x2+2(x-1)=0.
(1)整理得,x2-2x-15=0,
因式分解得,(x-5)(x+3)=0,
x-5=0或x+3=0,
解得x1=5,x2=-3;
(2)整理得,3x2+2x-2=0,
∵a=3,b=2,c=-2,△=b2-4ac=4+24=28,
x=
=
,
解得x1=
,x2=
.
因式分解得,(x-5)(x+3)=0,
x-5=0或x+3=0,
解得x1=5,x2=-3;
(2)整理得,3x2+2x-2=0,
∵a=3,b=2,c=-2,△=b2-4ac=4+24=28,
x=
-2±
| ||
| 2×3 |
-1±
| ||
| 3 |
解得x1=
-1+
| ||
| 3 |
-1-
| ||
| 3 |
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