The equation 3x2+4ax+b=0  has atleast one root in [0,1]if :

Consider the function f(x)=x3+2ax2+bx Obviously f(x)  is continuous in [0,1] and differentiable in ]0,1[.Also f(0)=0  if f(1)=1+2a+b=0 , then all conditions of Rolle’s theorem are satisfied∴  f'(c)=0  for atleast one c in ]0,1[Hence, f'(x)=3x2+4ax+b=0  at least once in ]0,1[i.e., the equation 3x2+4ax+b=0  has atleast one root in ]0,1[.So 2a+b+1=0 .