Question

已知A={x|x^2+(p+2)x+p-1=0,x∈R},A∩R﹢=Φ,求实数p的取值范围

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Answer 1

A={x|x^2+(p+2)x+p-1=0,x∈R},A∩R﹢=Φ：
△=(p+2)^2-4(p-1)=p^2+8>0,
∴[-(p+2)+√△]/2<=0,
∴√△<=p+2,
化为p+2>=0,△<=(p+2)^2,
∴p>=-2,-4(p-1)<=0,p>=1,
∴p>=1,为所求.