\(D = {\left( {3a + 2} \right)^2}-4{a^2} = 5{a^2} + 12a + 4 > 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a\, \in \,\left( {-\infty ;-2} \right) \cup \left( {-\frac{2}{5};\infty } \right).\)
\({x_1} = \dfrac{{3a + 2 + \sqrt {5{a^2} + 12a + 4} }}{2};\,\,\,\,\,\,\,{x_2} = \dfrac{{3a + 2-\sqrt {5{a^2} + 12a + 4} }}{2}.\)
Видно, что \({x_1} > {x_2},\) поэтому:
\(\dfrac{{{x_1}}}{{{x_2}}} = \dfrac{{3a + 2 + \sqrt {5{a^2} + 12a + 4} }}{{3a + 2-\sqrt {5{a^2} + 12a + 4} }} = 9\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,27a + 18-9\sqrt {5{a^2} + 12a + 4} = 3a + 2 + \sqrt {5{a^2} + 12a + 4} \,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,5\sqrt {5{a^2} + 12a + 4} = 12a + 8\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{12a + 8 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{125{a^2} + 300a + 100 = 144{a^2} + 192a + 64}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \ge -\dfrac{2}{3},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{19{a^2}-108a-36 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \ge -\dfrac{2}{3},\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{a = 6,\,\,\,\,\,\,}\\{a = -\dfrac{6}{{19}}}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{a = 6,\,\,\,\,\,\,\,\,}\\{a = -\dfrac{6}{{19}}.}\end{array}} \right.\)
ОТВЕТ: -6/19; 6.