k为何值时,方程(k-1)x^2-(2k+3)x+(k+3)=0有实数根
问题描述:
k为何值时,方程(k-1)x^2-(2k+3)x+(k+3)=0有实数根
答
k-1≠0 k≠1
△=【-(2k+3)】^2-4(k-1)(k+3)≥0
解
k≥-21/4且 k≠1