Question

已知(x²+x)(x²+x-3)=-2，求x的值

用初三上学期一元二次方程的知识解

Answer 1

如图

Answer 2

图

Answer 3

(x²+x)(x²+x-3)=-2
(x²+x)²-3(x²+x)+2=0
(x²+x-2)(x²+x-1)=0
(x-1)(x+2)(x²+x+¼- 5/4)=0
(x-1)(x+2)[(x+½)² -5/4]=0
(x-1)(x+2)(x+½ +√5/2)(x+½ -√5/2)=0
x=1或x=-2或x=-(1+√5)/2或x=(√5-1)/2

Answer 4

(x²+x)(x²+x-3)=-2
(x²+x)²-3(x²+x)=-2
(x²+x)²-3(x²+x)+2=0
(x²+x-2)(x²+x-1)=0
(x+2)(x-1)(x²+x+1/4-5/4)=0
(x+2)(x-1)[(x+1/2)²-5/4]=0
(x+2)(x-1)(x+1/2+√5/2)(x+1/2-√5/2）=0
x1=-2 x2=1 x3=-1/2-√5/2 x4=-1/2+√5/2

Answer 5

设a=x方