Solve the inequation: $\;$ $x^2 - 5 \left|x\right| + 6 < 0$
Given inequation: $\;$ $x^2 - 5 \left|x\right| + 6 < 0$ $\;\;\; \cdots \; (1)$
When $\;$ $x > 0$, $\;$ $\left|x\right| = + x$
Then $(1)$ becomes
$x^2 - 5x + 6 < 0$
i.e. $\;$ $\left(x - 3\right) \left(x - 2\right) < 0$
i.e. $\;$ $x - 3 < 0 \; \cap \; x - 2 > 0$ $\;\;$ OR $\;\;$ $x - 3 > 0 \; \cap \; x - 2 < 0$
i.e. $\;$ $x < 3 \; \cap \; x > 2$ $\implies$ $x \in \left(2, 3\right)$
OR $\;\;$ $x > 3 \; \cap \; x < 2$ $\;\;$ which is not possible
$\therefore \;$ When $x > 0$, solution of $(1)$ is $\;\;\;$ $x \in \left(2,3\right)$ $\;\;\; \cdots \; (2)$
When $\;$ $x < 0$, $\;$ $\left|x\right| = - x$
Then $(1)$ becomes
$x^2 + 5x + 6 < 0$
i.e. $\;$ $\left(x + 2\right) \left(x + 3\right) < 0$
i.e. $\;$ $x + 2 < 0 \; \cap \; x + 3 > 0$ $\;\;$ OR $\;\;$ $x + 2 > 0 \; \cap \; x + 3 < 0$
i.e. $\;$ $x < -2 \; \cap \; x > -3$ $\implies$ $x \in \left(-3, -2\right)$
OR $\;\;$ $x > -2 \; \cap \; x < -3$ $\;\;$ which is not possible
$\therefore \;$ When $x < 0$, solution of $(1)$ is $\;\;\;$ $x \in \left(-3, -2\right)$ $\;\;\; \cdots \; (3)$
$\therefore \;$ From $(2)$ and $(3)$, solution of $(1)$ is $\;$ $x \in \left(-3, -2\right) \; \cup \; \left(2, 3\right)$