Задание №19 — Уравнения с параметром

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

$\left[\begin{matrix} \left\{\begin{matrix} \begin{matrix}x \leq 0 \\ \left[\begin{matrix} \begin{matrix}a = 6 + x \\ = - 2 x\end{matrix} \end{matrix}\right.\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}x > 0 \\ \left[\begin{matrix} \begin{matrix}a = x \\ a = 6 - 2 x\end{matrix} \end{matrix}\right.\end{matrix} \end{matrix}\right. \end{matrix}\right.$
$a = 6 - 2 x$

Ответ: $( - \infty ; 0 \left]\right. \cap \left\{2\right\} \cup [ 6 ; + \infty )$

Источник: ФИПИ