10.2. Determine all values of the real parameter $a$ for which the equation

$$
\left|a x^{2}-6\right|=|2 a x|+|3 a| \text { has at least one real solution. }
$$
Solution: 1. Let $a=0$, then the equation has no solutions.

2. Let $a \neq 0$, then the equation is equivalent to

$\left|x^{2}-\frac{6}{a}\right|=2|x|+3 \Leftrightarrow\left[\begin{array}{c}x^{2}-\frac{6}{a}=2|x|+3, \\ \frac{6}{a}-x^{2}=2|x|+3\end{array} \Leftrightarrow\left[\begin{array}{l}(|x|-1)^{2}=4+\frac{6}{a}, \\ (|x|+1)^{2}=\frac{6}{a}-2 .\end{array}\right.\right.$

We observe that $(|x|-1)^{2} \geq 0, \forall x \in R$, and $\operatorname{Im}(|x|-1)^{2}=[0 ;+\infty)$.

We also observe that $(|x|+1)^{2} \geq 1, \forall x \in R$, and $\operatorname{Im}(|x|+1)^{2}=[1 ;+\infty)$.

The given equation will have at least one real solution if the following conditions are satisfied:

$\left[\begin{array}{l}4+\frac{6}{a} \geq 0, \\ \frac{6}{a}-2 \geq 1\end{array} \Leftrightarrow\left[\begin{array}{l}a \in\left(-\infty ;-\frac{3}{2}\right.\end{array}\right] \cup(0 ;+\infty), \Leftrightarrow a \in\left(-\infty ;-\frac{3}{2}\right] \cup(0 ;+\infty)\right.$.

Answer: $a \in\left(-\infty ;-\frac{3}{2}\right] \cup(0 ;+\infty)$