Число 2/3 является корнем уравнения 6х^2+bх-3=0. Найдите значение b и второй корень уравнения.

\[x_{1} = \frac{2}{3}:\]

\[6x^{2} + bx - 3 = 0\]

\[\left\{ \begin{matrix}
x_{1} + x_{2} = - \frac{b}{6} \\
x_{1} \cdot x_{2} = - \frac{3}{6}\text{\ \ \ } \\
\end{matrix} \right.\ \]

\[\frac{2}{3} \cdot x_{2} = - \frac{1}{2}\]

\[\ x_{2} = - \frac{1}{2} \cdot \frac{3}{2}\text{\ \ }\]

\[x_{2} = - \frac{3}{4}.\]

\[\frac{2}{3} + \left( - \frac{3}{4} \right) = - \frac{b}{6}\]

\[\frac{8}{12} - \frac{9}{12} = - \frac{b}{6}\text{\ \ }\]

\[- \frac{1}{12} = - \frac{b}{6}\]

\[\frac{b}{6} = \frac{1}{12}\text{\ \ }\]

\[b = \frac{6}{12}\]

\[b = \frac{1}{2}.\]

\[Ответ:\ \ b = \frac{1}{2};\ \ x_{2} = - \frac{3}{4}.\]