题目
已知一元二次方程mx2+nx+(n-m)=0(m≠0),则判别式△=______,该方程根的情况是______.
提问时间:2020-10-28
答案
∵a=m,b=n,c=n-m,
∴△=n2-4m(n-m)=n2-4mn+4m2=(n-2m)2≥0,
∴方程有两个实数根.
故答案为:(n-2m)2,有两个实数根.
∴△=n2-4m(n-m)=n2-4mn+4m2=(n-2m)2≥0,
∴方程有两个实数根.
故答案为:(n-2m)2,有两个实数根.
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