Solve the inequation: $\;$ $\left|x^2 - 4x\right| < 5$
Given inequation: $\;$ $\left|x^2 - 4x\right| < 5$ $\;\;\; \cdots \; (1)$
Case 1:
When $\;$ $x^2 - 4x > 0$
i.e. $\;$ when $\;$ $x \left(x - 4\right) > 0$
i.e. $\;$ when $\;$ $x > 0 \; \cap \; x - 4 > 0$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x - 4 < 0$
i.e. $\;$ when $\;$ $x > 0 \; \cap \; x > 4$ $\implies$ $x > 4$
OR $\;\;$ $x < 0 \; \cap \; x < 4$ $\implies$ $x < 0$
$\therefore \;$ When $\;$ $x < 0$ $\;$ OR $\;$ $x > 4$, $\;\;$ $\left|x^2 - 4x\right| = x^2 - 4x$
Then $(1)$ becomes
$x^2 - 4x < 5$
i.e. $\;$ $x^2 - 4x - 5 < 0$
i.e. $\;$ $\left(x - 5\right) \left(x + 1\right) < 0$
i.e. $\;$ $x - 5 < 0 \; \cap \; x + 1 > 0$ $\;\;$ OR $\;\;$ $x - 5 > 0 \; \cap \; x + 1 < 0$
i.e. $\;$ $x < 5 \; \cap \; x > -1$ $\implies$ $x \in \left(-1, 5\right)$
OR $\;\;$ $x > 5 \; \cap \; x < -1$ $\;\;$ which is not possible
$\therefore \;$ When $\;$ $x < 0$ $\;$ OR $\;$ $x > 4$, $\;$ solution of $(1)$ is $\;\;\;$ $x \in \left(-1, 5\right)$ $\;\;\; \cdots \; (2)$
Case 2:
When $\;$ $x^2 - 4x < 0$
i.e. $\;$ when $\;$ $x \left(x - 4\right) < 0$
i.e. $\;$ when $\;$ $x < 0 \; \cap \; x - 4 > 0$ $\;\;$ OR $\;\;$ $x > 0 \; \cap \; x - 4 < 0$
i.e. $\;$ when $\;$ $x < 0 \; \cap \; x > 4$ $\;\;$ which is not possible
OR $\;\;$ $x > 0 \; \cap \; x < 4$ $\implies$ $x \in \left(0, 4\right)$
$\therefore \;$ When $\;$ $x \in \left(0, 4\right)$, $\;\;$ $\left|x^2 - 4x\right| = -x^2 + 4x$
Then $(1)$ becomes
$-x^2 + 4x < 5$
i.e. $\;$ $x^2 - 4x + 5 < 0$
The equation $\;$ $x^2 - 4x + 5 = 0$ $\;$ does not have any real roots since its discriminant $\Delta$ is
$\Delta = \left(-4\right)^2 - 4 \times 1 \times 5 = 16 - 20 = -4 < 0$
$\therefore \;$ From $(2)$, solution of $(1)$ is $\;$ $x \in \left(-1, 5\right)$ $\;$ when $\;$ $x < 0$ $\;$ OR $\;$ $x > 4$