\[2x^{2} - (b - 2)x + b + 4 = 0\]
\[D = (b - 2)^{2} - 4 \cdot 2 \bullet (b + 4) =\]
\[= b^{2} - 4b + 4 - 8b - 32 =\]
\[= b^{2} - 12b - 28 =\]
\[= (b - 14)(b + 2) > 0\]
\[x_{1} + x_{2} = \frac{b - 2}{2};\ \ \ x_{1} \cdot x_{2} = \frac{b + 4}{2}\ \]
\[x_{1}^{2} + x_{2}^{2} + 8x_{1}x_{2} = 9\]
\[\left( x_{1}^{2} + x_{2}^{2} \right) + 6x_{1}x_{2} = 9\]
\[\left( \frac{b - 2}{2} \right)^{2} + 6 \cdot \frac{b + 4}{2} = 9\]
\[b^{2} + 4 - 4b + 12b + 48 - 36 =\]
\[= 0\]
\[b^{2} + 8b + 16 = 0\]
\[(b + 4)^{2} = 0\]
\[b + 4 = 0\]
\[b = - 4.\]
\[Ответ:при\ b = - 4.\]