设集合A={x|x2+2(a+1)x+a2-1=0},B={x|x2+4x=0}若A∩B=A,求实数a
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设集合A={x|x2+2(a+1)x+a2-1=0},B={x|x2+4x=0}若A∩B=A,求实数a
设集合A={x|x2+2(a+1)x+a2-1=0},B={x|x2+4x=0}若A∩B=A,求实数a
设集合A={x|x2+2(a+1)x+a2-1=0},B={x|x2+4x=0}若A∩B=A,求实数a
解得B={-4,0}
当-4∈A时,16-8a-8+a^2-1=0,解得a=1或7
当a=1时,A={x|x2+4x=0},解得A={-4,0}
当a=7时,A={x|x=x^2+16x+48},解得A={-12,-4}不和题舍去
当0∈A时,a^2-1=0,解得a=-1或1
当a=-1时,A={x|x2=0},解得A={0}
当a=1时,已证成立
综上所述,a=-1或1