已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p·q≠0),求p,q的值
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![已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p·q≠0),求p,q的值](/uploads/image/z/7305721-25-1.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%7Bx%7Cx2%2Bpx%2Bq%3D0%7D%2CB%3D%7Bx%7Cqx2%2Bpx%2B1%3D0%7D%2C%E5%90%8C%E6%97%B6%E6%BB%A1%E8%B6%B3%E8%B6%B3A%E2%88%A9B%E2%89%A0%E7%A9%BA%E9%9B%86%2CA%E2%88%A9CRB%3D%7B-2%7D%2C%28p%E5%B7%B2%E7%9F%A5%E9%9B%86%E5%90%88A%3D%7Bx%7Cx2%2Bpx%2Bq%3D0%7D%2CB%3D%7Bx%7Cqx2%2Bpx%2B1%3D0%7D%2C%E5%90%8C%E6%97%B6%E6%BB%A1%E8%B6%B3%E8%B6%B3A%E2%88%A9B%E2%89%A0%E7%A9%BA%E9%9B%86%EF%BC%8CA%E2%88%A9CRB%3D%7B-2%7D%EF%BC%8C%EF%BC%88p%C2%B7q%E2%89%A00%EF%BC%89%EF%BC%8C%E6%B1%82p%2Cq%E7%9A%84%E5%80%BC)
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已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p·q≠0),求p,q的值
已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p
已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p·q≠0),求p,q的值
已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p已知集合A={x|x2+px+q=0},B={x|qx2+px+1=0},同时满足足A∩B≠空集,A∩CRB={-2},(p·q≠0),求p,q的值
x=-2代入2+px+q=0得q=2p-4(1)
设A={-2,a},则a-2=-p即a=2-p(2),A∩B=a
所以q*a^2+pa+1=0,代入(1)(2)得
2p^3-13p^2+26p-15=0
解得p=1,2.5(舍)或3
所以p=1,q=-2或p=3,q=2