mx²+mx+4m+12=0
x=｛-m±√［m²-4m（4m+12）］｝/2m
= ［-m±√（-15m²-48m）］/2m
△=-15m²-48m
若要有解,必须m＜0
令［-m±√（-15m²-48m）］/2m＜0
∴ -m±√（-15m²-48m）＞0
±√（-15m²-48m）＞m
∵+√（-15m²-48m）＞0＞m
∴必须 -√（-15m²-48m）＞m
√（-15m²-48m）＜-m
-15m²-48m＜（-m）²
16m²+48m＞0
∵m≠0,
∴两边同除以m,得
16m+48＞0
m ＞-3
综上所述,当-3＜m＜0时,mx²+mx+4m+12=0的解至少有一个小于0