题干

已知集合M={x|x2+ax+b=0,x∈R}.


(Ⅰ)若集合M是单元素集,求实数a,b满足的关系式;

(Ⅱ)若1,3∈M,求实数a,b的值.

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解:(Ⅰ)若集合M是单元素集,则方程只有一个解,则△=0,即△=a2﹣4b=0,则实数a,b满足的关系式为a2﹣4b=0.

(Ⅱ)若1,3∈M,则1,3是方程x2+ax+b=0的两个根,根据根与系数之间的关系得1+3=﹣a且1×3=b,解得a=﹣4,b=3

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