设集合A={x|x的平方+4x=0},B={x的平方+2（a+1）x的平方+a的平方-1=0},A并B=B,求实数a的值

A={x|x^2+4x=0}
={x| x(x+4)=0 }
={ -4,0 }
B={x| x^2+2(a+1)x + a^2-1 =0}
A∪B = B => A is subset of B
when x=0
=> a^2-1 =0
a = 1 or -1
when x= -4
16-8(a+1)+a^2-1=0
a^2-8a+7=0
a= 1 or 7
ie a=-1,1,7 #