Question

请问一道方程ax^2-2>=2x-ax (a属于全体实数)

Answer 1

原方程可化为ax^2+(a-2)x-2=0

(1)当a=0时,原方程化为2x+2=0,解得x=-1

(2)当a不等于0时,原方程化为(x+1)(ax-2)=0

1.当a=-2时,原方程化为(x+1)^2=0,解得x=-1

2.当a不等于-2时,原方程有两解,分别是x1=-1,x2=2/a.