\[\boxed{\mathbf{717\ (717).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[x_{1} = \frac{1}{3}\]
\[x_{2} = ?\]
\[6x² - bx + 4 = 0\ \ \ \ |\ :6\]
\[x² - \frac{b}{6}x + \frac{2}{3} = 0\]
\[\left\{ \begin{matrix} x_{2} + \frac{1}{3} = \frac{b}{6} \\ \frac{1}{3}x_{2} = \frac{2}{3}\text{\ \ \ } \\ \end{matrix} \right.\ \]
\[x_{2} = \frac{2 \cdot 3}{3 \cdot 1} = 2.\]
\[2 + \frac{1}{3} = \frac{b}{6}\text{\ \ \ }\]
\[\frac{7}{3} = \frac{b}{6}\text{\ \ }\]
\[3b = 42\]
\[b = 14\]
\[Ответ:x_{2} = 2;\ \ b = 14.\]
\[\boxed{\mathbf{71}\mathbf{7}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[ax^{2} + bx + c = 0\]
\[D = b^{2} - 4ac\]
\[При\ условии,\ если\ a\ и\ \text{c\ }имеют\ \]
\[разные\ знаки:например,\ ( - a)\ \]
\[и\ с,\ то\ D = b^{2} + 4ac > 0.\]
\[Следовательно,\ в\ уравнении\ \]
\[2\ корня.\]