If $$2a + 3b + 6c = 0,\left( {a,b,c \in R} \right)$$      then the quadratic equation $$a{x^2} + bx + c = 0$$    has

Solution :
$$\eqalign{ & {\text{Let}}\,f\left( x \right) = \frac{{a{x^3}}}{3} + \frac{{b{x^2}}}{2} + cx \Rightarrow f\left( 0 \right) = 0\,{\text{and}}\,f\left( 1 \right) \cr & = \frac{a}{3} + \frac{b}{2} + c = \frac{{2a + 3b + 6c}}{6} = 0 \cr} $$
Also $$f(x)$$  is continuous and differentiable in [0, 1] and [0, 1[. So by Rolle’s theorem, $$f'\left( x \right) = 0.$$
i.e $$a{x^2} + bx + c = 0$$    has at least one root in [0, 1].