question_answer

If the coefficient of z in the quadratic equation\[{{z}^{2}}+az+b=0\] is taken as 18 in place of 12, then its roots were found to be \[-\,16\] and\[-\,2\]. The roots of the original equation are:

A)  \[-\,12\,\,and\,\,-6\]                    

B)  \[-\,14\,\,and\,\,-4\]

C)  \[-\,16\,\,and\,\,-2\]                    

D)  \[-\,8\,\,and\,\,-4\]

E)  None of these