题干

已知关于x的一元二次方程x2+(m+3)x+m+1=0.
⑴求证:无论m取何值,原方程总有两个不相等的实数根;
⑵若x1,x2是原方程的两根,且x1-x2=22,求m的值,并求出此时方程的两根.
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答案(点此获取答案解析)

解:(1)证明:∵△=(m+3)2-4(m+1)=(m+1)2+4
∵无论m取何值,(m+1)2+4恒大于0
∴原方程总有两个不相等的实数根
(2)∵x1,x2是原方程的两根
∴x1+x2=-(m+3),x1•x2=m+1…5分
∵|x1-x2

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