Solve the inequation: $\;$ $x^2 - \left|5x - 3\right| - x < 2$
Given inequation: $\;$ $x^2 - \left|5x - 3\right| - x < 2$ $\;\;\; \cdots \; (1)$
Case 1:
When $\;$ $5x - 3 > 0$
i.e. $\;$ when $\;$ $x > \dfrac{3}{5}$, $\;\;$ $\left|5x - 3\right| = 5x - 3$
Then $(1)$ becomes
$x^2 - \left(5x - 3\right) -x < 2$
i.e. $\;$ $x^2 - 6x + 1 < 0$ $\;\;\; \cdots \; (2)$
Solving the quadratic equation $\;$ $x^2 - 6x + 1 = 0$ $\;$ gives
$x = 3 + 2 \sqrt{2}$ $\;\;\; \cdots \; (3a)$ $\;$ OR $\;$ $x = 3 - 2 \sqrt{2}$ $\;\;\; \cdots \; (3b)$
$\therefore \;$ In view of $(3a)$ and $(3b)$, $(2)$ becomes
$\left[x - \left(3 + 2 \sqrt{2}\right)\right] \left[x - \left(3 - 2 \sqrt{2}\right)\right] < 0$
i.e. $\;$ $x - 3 - 2 \sqrt{2} < 0$ $\;$ AND $\;$ $x - 3 + 2 \sqrt{2} > 0$
OR $\;$ $x - 3 - 2 \sqrt{2} > 0$ $\;$ AND $\;$ $x - 3 + 2 \sqrt{2} < 0$
i.e. $\;$ $x < 3 + 2 \sqrt{2}$ $\;$ AND $\;$ $x > 3 - 2 \sqrt{2}$ $\implies$ $x \in \left(3 - 2 \sqrt{2}, 3 + 2 \sqrt{2}\right)$
OR $\;$ $x > 3 + 2 \sqrt{2}$ $\;$ AND $\;$ $x < 3 - 2 \sqrt{2}$ $\;$ which is not possible
$\therefore \;$ When $\;$ $5x - 3 > 0$, i.e. when $x > \dfrac{3}{5}$, $\;$ solution of $(1)$ is $\;$ $x \in \left(3 - 2 \sqrt{2}, 3 + 2 \sqrt{2}\right)$ $\;\;\; \cdots \; (4)$
Case 2:
When $\;$ $5x - 3 < 0$
i.e. $\;$ when $\;$ $x < \dfrac{3}{5}$, $\;$ $\left|5x - 3\right| = -5x + 3$
Then $(1)$ becomes
$x^2 - \left(-5x + 3\right) - x < 2$
i.e. $\;$ $x^2 + 4x - 5 < 0$
i.e. $\;$ $\left(x + 5\right) \left(x - 1\right) < 0$
i.e. $\;$ $x + 5 < 0$ $\;$ AND $\;$ $x - 1 > 0$
OR $\;$ $x + 5 > 0$ $\;$ AND $\;$ $x - 1 < 0$
i.e. $\;$ $x < -5$ $\;$ AND $\;$ $x > 1$ $\;\;$ which is not possible
OR $\;$ $x > -5$ $\;$ AND $\;$ $x < 1$ $\implies$ $x \in \left(-5, 1\right)$
$\therefore \;$ When $\;$ $5x - 3 < 0$, i.e. when $x < \dfrac{3}{5}$, $\;$ solution of $(1)$ is $\;$ $x \in \left(-5, 1\right)$ $\;\;\; \cdots \; (5)$
$\therefore \;$ In view of $(4)$ and $(5)$, solution of $(1)$ is $\;$ $x \in \left(-5, 3 + 2 \sqrt{2}\right)$