设p:实数x满足x^2-4ax+3a^2

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设p:实数x满足x^2-4ax+3a^2
设p:实数x满足x^2-4ax+3a^2<0,a<0,q:实数x满足x^2-x-6<=0或x^2+2x-8>0且p是q的必要不充分条件,求a的取值范
求a的范围

设p:实数x满足x^2-4ax+3a^2
p：(x－a)(x－3a)>0,因a<0,则这个不等式的解集想x<3a或x>a；
q：x²－x－6≤0,即：(x－2)(x＋3)≤0,得：－3≤x≤2,及：x²＋2x－8>0,即：(x＋4)(x－2)>0,得：x<－4或x>2.从而有q：x<－4或x≥－3
因p是q的必要不充分条件,即q===>>>> p,则：－4≤3a且a>－3,解得：－3

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