If the roots of $${x^3} - 12{x^2} + 39x - 28 = 0$$      are in A.P. then their common difference is

Solution :
If roots are $$\alpha - k,\alpha ,\alpha + k$$    then
$$\eqalign{ & \alpha - k + \alpha + \alpha + k = 12 \cr & \Rightarrow \,\,\alpha = 4, \cr & \left( {\alpha - k} \right)\alpha \left( {\alpha + k} \right) = 28 \cr & \Rightarrow \,\,\left( {{4^2} - {k^2}} \right)4 = 28 \cr & \Rightarrow \,\,{k^2} = 9. \cr} $$