y=a（x+3)(x-2/a)+5与x轴恒有两个交点吗（a不等于0）

解：y=a(x+3)(x-2/a)+5 =(x+3)(ax-2)+5 =ax^2+3ax-2x-6+5 =ax^2+(3a-2)x-1令ax^2+(3a-2)x-1=0∵抛物线y=ax^2+(3a-2)x-1与x轴恒有两个交点，∴关于x一元二次方程ax^2+(3a-2)x-1=0有两个不相等的实数根∴Δ＞0而（3a-2)^2+4a=(3a)^2-12a+4+4a =9a^2-8a+4 =9[a^2-(8/9)a]+4 =9[a^2-(8/9)a+(4/9)^2-(4/9)^2]+4 =9(a-4/9)^2-16/9+4 =9(a-4/9)^2+20/9不论a为何值9(a-4/9)^2+20/9≥20/9即Δ＞0抛物线y=a(x+3)(x-2/a)+5与x轴恒有两个交点。