Решение:
$a^{2} - ax - 2 x^{2} - 6 a + 3 x + 9 \left|x\right| = 0$
$\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ a^{2} - ax - 2 x^{2} - 6 a + 3 x - 9 x = 0\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x > 0 \\ a^{2} - ax - 2 x^{2} - 6 a + 3 x + 9 x = 0\end{matrix} \end{matrix}\right. \end{matrix}\right.$
$\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ a^{2} - ax - 2 x^{2} - 6 a - 6 x = 0\end{matrix} \end{matrix}\right ( 1 ) \\ \left\{\begin{matrix} \begin{matrix}x > 0 \\ a^{2} - ax - 2 x^{2} - 6 a + 12 x = 0\end{matrix} \end{matrix}\right ( 2 ) \end{matrix}\right.$
$\left(1\right) a^{2} - ax - 2 x^{2} - 6 a - 6 x = 0 ; D > 0$
$a^{2} - ax - 6 a - 6 x - 2 x^{2} = 0$
$a^{2} - \left(x + 6\right) a - 6 x - 2 x^{2} = 0$
$D = \left(x + 6\right)^{2} - 4 \cdot \left(- 6 x - 2 x^{2}\right) =$$x^{2} + 12 x + 36 + 24 x + 8 x^{2} =$$9 x^{2} + 36 x + 36 = \left(3 x + 6\right)^{2}$
$a_{1} = \frac{x + \cancel{6} - 3 x - \cancel{6}}{2} = \frac{x - 3 x}{2} = \frac{1}{2} x - \frac{3}{2} x$
$a_{2} = \frac{x + 6 + 3 x + 6}{2} = \frac{4 x + 12}{2} = 2 x + 6$
$\left(2\right) a^{2} - ax - 2 x^{2} - 6 a + 12 x = 0 ; D > 0$
$a^{2} - ax - 6 a - 2 x + 12 x = 0$
$a^{2} - \left(x + 6\right) a - 2 x^{2} + 12 x = 0$
$D = \left(x + 6\right)^{2} - 4 \cdot \left(- 2 x^{2} + 12 x\right) =$$x^{2} + 21 x + 36 + 8 x^{2} - 48 x =$$9 x^{2} - 36 x + 36 = \left(3 x - 6\right)^{2}$
$a_{1} = \frac{x + 6 - 3 x + 6}{2} = \frac{- 2 x + 12}{2} = 6 - x$
$a_{2} = \frac{x + 6 + 3 x - 6}{2} = \frac{4 x}{2} = 2 x$
$\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ \left[\begin{matrix} \begin{matrix}a = \frac{1}{2} x - \frac{3}{2} x \\ a = 2 x + 6\end{matrix} \end{matrix}\right.\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x > 0 \\ \left[\begin{matrix} \begin{matrix}a = 6 - x \\ a = 2 x\end{matrix} \end{matrix}\right.\end{matrix} \end{matrix}\right. \end{matrix}\right.$

$a \in \left(\left(1\right) ; \left(2\right)\right) \cup \left(\left(2\right) ; \left(3\right)\right) \cup \left(\left(3\right) ; \left(4\right)\right)$
$a = \frac{1}{2} x - \frac{3}{2} x$

x	-2	0
a	2	0

$a = 2 x + 6$

x	0	-1
a	6	4

$a = 6 - x$

x	1	2
a	5	4

$a = 2 x$

x	0	1
a	0	2

Ответ: $\left(0 ; 2\right) \cup \left(2 ; 4\right) \cup \left(4 ; 6\right)$

Источник: ФИПИ