y=1/3x[x²-3mx+3(m²-4)]有三个零点,要满足什么条件?
人气:180 ℃ 时间:2020-06-12 15:55:07
解答
很显然有一个零点是x=0
则x²-3mx+3(m²-4)=0有两个非零的实数根
则,△=b²-4ac=9m²-12(m²-4)>0
===> -3m²+48>0
===> m²<16
===> -4<m<4
且,x1*x2=c/a=3(m²-4)≠0
则,m≠±2
综上,-4<m<4,且m≠±2.3(m²-4)=0会怎么样?因为x1*x2=3(m²-4)=0则至少x1,、x2中有1个为零那么就与前面所说的x=0重复,无法满足有3个零点的要求!
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