Let f(x) = ax3 + bx2 + cx ∴ f'(x) = 3ax2 + abx + c Clearly f(x) is continuous in [0, 1], derivable in (0, 1) and f(0) =0 = f(1) ∵ f (1) = a + b + c = 0 (given) ∴ By Rolle's theorem, there exists at least one real x ∊ (0, 1) such that f(x) = 0 , i.e 3ax2 + 2bx + c = 0 Hence, there exits at least one real root of 3ax2 + 2bx + c = 0 in (0, 1)