Solve the inequation: $\;$ $x^2 - 7x + 12 < \left|x - 4\right|$
Given inequation: $\;$ $x^2 - 7x + 12 < \left|x - 4\right|$ $\;\;\; \cdots \; (1)$
Case 1:
When $\;$ $x - 4 > 0$
i.e. $\;$ when $\;$ $x > 4$, $\;\;$ $\left|x - 4\right| = x - 4$
Then $(1)$ becomes
$x^2 - 7x + 12 < x - 4$
i.e. $\;$ $x^2 - 8x + 16 < 0$
i.e. $\;$ $\left(x - 4\right)^2 < 0$
i.e. $\;$ $x - 4 < 0$ $\implies$ $x < 4$
$\therefore \;$ When $\;$ $x - 4 > 0$, i.e. when $x > 4$ $\;$ solution of $(1)$ is $\;$ $x < 4$ $\;\;\; \cdots \; (2)$
Case 2:
When $\;$ $x - 4 < 0$
i.e. $\;$ when $\;$ $x < 4$, $\;$ $\left|x - 4\right| = -x + 4$
Then $(1)$ becomes
$x^2 - 7x + 12 < -x + 4$
i.e. $\;$ $x^2 - 6x + 8 < 0$
i.e. $\;$ $\left(x - 2\right) \left(x - 4\right) < 0$
i.e. $\;$ $x - 2 < 0$ $\;$ AND $\;$ $x - 4 > 0$
OR $\;$ $x - 2 > 0$ $\;$ AND $\;$ $x - 4 < 0$
i.e. $\;$ $x < 2$ $\;$ AND $\;$ $x > 4$ $\;\;$ which is not possible
OR $\;$ $x > 2$ $\;$ AND $\;$ $x < 4$ $\implies$ $x \in \left(2, 4\right)$
$\therefore \;$ When $\;$ $x - 4 < 0$, i.e. when $x < 4$, $\;$ solution of $(1)$ is $\;$ $x \in \left(2, 4\right)$ $\;\;\; \cdots \; (3)$
$\therefore \;$ In view of $(2)$ and $(3)$, solution of $(1)$ is $\;$ $x \in \left(2, 4\right)$