If a, b, c are real numbers such that a+b+c=0, then the quadratic equation 3ax2+2bx+c=0 has

Let f1(x) denotes the quadratic expression f1(x)=3ax2+2bx+c, whose antiderivative be denoted by f(x)=ax3+bx2+cxNow f(x) being a polynomial in R, f(x) is continuous and differentiable on R. To apply Rolle's theorem.We observe that f(0)=0 and f(1)=a+b+c=0, by hypothesis. So there must exist at least one value of x, say x=α∈(0,1) such thatf1(α)=0⇔3aα2+2bα+c=0That is, f1(x)=3ax2+2bx+c=0 has at least one root in [0, 1]