If the graph of f(x)=2x3+ax2+bx(a,b∈N) cuts the x -axis at three distinct points, then minimum value of a+bis

f(x)=2x3+ax2+bx;a,b∈N=x2x2+ax+b Given f(x) cuts x -axis at 3 distinct points then 2x2+ax+b must have 2 distinct real roots ∴    a2−8b>0⇒    a2>8b Since a,b∈N minimum value of a+b is 4 when a=3 and b=1