(1)x^2-ax+a = 0,討論△ = a^2 - 4a,
△>=0,a^2 - 4a>=0,即a∈(-∞,0]∪[4,+∞),有x = (a±√a^2 - 4a)/2
△0
x^2-ax+a = (x-a/2)^2 - a^2/4 + a > 0 ,即一元二次方程最低點(diǎn)大于0即可,
所以a - a^2/4 > 0,a∈(0,4)