In the second-degree equation x^2-14x+m = 0, the constant m is a posit

\({x^2} - 14x + m = 0\,\,,\,\,\,m \ge 1\,\,{\mathop{\rm int}}\)

\(? = B - A\,\,\,\,\left( {A < B\,\,{\rm{roots}}} \right)\,\,\left( * \right)\)

\(\left( 1 \right)\,\,{\rm{equation}}\,\,{\rm{known}}\,\,\,\,\, \Rightarrow \,\,\,{\rm{the}}\,\,2\,\,{\rm{roots}}\,\,A,B\,\,\,{\rm{known}}\,\,\,\,\mathop \Rightarrow \limits^{A\, < \,B} \,\,\,\,\,A\,\,{\rm{unique}}\,,B\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)


\(\left( 2 \right)\,\,\left\{ \matrix{\\
A + B = 14 \hfill \cr \\
A < B\,\,{\rm{primes}} \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{inspection}}} \,\,\,\,\,\left( {A,B} \right) = \left( {3,11} \right)\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)


The correct answer is (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.