Cho phương trình \(2{x^2} - 13x - 6 = 0\) có \(2\) nghiệm là

Áp dụng định lí Vi – et ta có: \[\left\{ \begin{array}{l}{x_1} + {x_2} = \frac{{13}}{2}\\{x_1}{x_2} =  - 3\end{array} \right.\]

Ta có:

\[A = \left( {{x_1} + {x_2}} \right)\left( {{x_1} + 2{x_2}} \right) - x_2^2\]\[ = \left( {{x_1} + {x_2}} \right)\left[ {\left( {{x_1} + {x_2}} \right) + {x_2}} \right] - x_2^2\]\[ = \frac{{13}}{2}\left( {\frac{{13}}{2} + {x_2}} \right) - x_2^2\]\[ = \frac{{169}}{4} + \frac{{13}}{2}{x_2} - x_2^2\]

\[ \Rightarrow  - 2A =  - 2x_2^2 + 13{x_2} + \frac{{169}}{2}\]\[ \Leftrightarrow  - 2A = \left( { - 2x_2^2 + 13{x_2} - 6} \right) - \frac{{157}}{2}\]\[ \Leftrightarrow  - 2A =  - \frac{{157}}{2}\]\[ \Leftrightarrow A = \frac{{157}}{4}\]