Solve the inequation: $\;$ $\left|x^2 + x\right| - 5 < 0$
Given inequation: $\;$ $\left|x^2 + x\right| - 5 < 0$ $\;\;\; \cdots \; (1)$
Case 1:
When $\;$ $x^2 + x > 0$
i.e. $\;$ when $\;$ $x \left(x + 1\right) > 0$
i.e. $\;$ when $\;$ $x > 0 \; \cap \; x + 1 > 0$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x + 1 < 0$
i.e. $\;$ when $\;$ $x > 0 \; \cap \; x > -1$ $\;\;$ OR $\;\;$ $x < 0 \; \cap \; x < -1$
$\therefore \;$ when $\;$ $x > 0$ $\;$ OR $\;$ $x < -1$, $\;\;$ $\left|x^2 + x\right| = x^2 + x$
Then $(1)$ becomes
$x^2 + x - 5 < 0$ $\;\;\; \cdots \; (2)$
Solving the quadratic equation $\;\;$ $x^2 + x - 5 = 0$ $\;\;\; \cdots \; (3)$ $\;$ gives
$x = \dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times \left(-5\right)}}{2 \times 1}$
i.e. $\;$ $x = \dfrac{-1+ \sqrt{21}}{2}$ $\;\;\; \cdots \; (4a)$ $\;$ OR $\;$ $x = \dfrac{-1 - \sqrt{21}}{2}$ $\;\;\; \cdots \; (4b)$
In view of $(4a)$ and $(4b)$, $(2)$ can be written as
i.e. $\;$ $\left(x - \dfrac{1 - \sqrt{21}}{2}\right) \left(x - \dfrac{1 + \sqrt{21}}{2}\right) < 0$
i.e. $\;$ $x - \dfrac{1 - \sqrt{21}}{2} < 0 \; \cap \; x - \dfrac{1 + \sqrt{21}}{2} > 0$
OR $\;\;$ $x - \dfrac{1 - \sqrt{21}}{2} > 0 \; \cap \; x - \dfrac{1 + \sqrt{21}}{2} < 0$
i.e. $\;$ $x < \dfrac{1 - \sqrt{21}}{2} \; \cap \; x > \dfrac{1 + \sqrt{21}}{2}$ $\;\;$ which is not possible
OR $\;\;$ $x > \dfrac{1 - \sqrt{21}}{2} \; \cap \; x < \dfrac{1 + \sqrt{21}}{2}$ $\implies$ $x \in \left(\dfrac{1 - \sqrt{21}}{2}, \dfrac{1 + \sqrt{21}}{2}\right)$
$\therefore \;$ When $\;$ $x^2 + x > 0$, $\;$ solution of $(1)$ is $\;\;\;$ $x \in \left(\dfrac{1 - \sqrt{21}}{2}, \dfrac{1 + \sqrt{21}}{2}\right)$ $\;\;\; \cdots \; (5)$
Case 2:
When $\;$ $x^2 + x < 0$
i.e. $\;$ when $\;$ $x \left(x + 1\right) < 0$
i.e. $\;$ when $\;$ $x < 0 \; \cap \; x + 1 > 0$ $\;\;$ OR $\;\;$ $x > 0 \; \cap \; x + 1 < 0$
i.e. $\;$ when $\;$ $x < 0 \; \cap \; x > -1$ $\implies$ $x \in \left(-1, 0\right)$
OR $\;\;$ $x > 0 \; \cap \; x < -1$ $\;\;$ which is not possible
$\therefore \;$ when $\;$ $x \in \left(-1, 0\right)$, $\;\;$ $\left|x^2 + x\right| = -x^2 - x$
then $(1)$ becomes
$-x^2 - x - 5 < 0$
i.e. $\;$ $x^2 + x + 5 > 0$ $\;\;\; \cdots \; (6)$
Consider the quadratic equation $\;\;$ $x^2 + x + 5 = 0$ $\;\;\; \cdots \; (7)$
Discriminant of $(7)$ is $\;\;$ $\Delta = 1^2 - 4 \times 1 \times 5 = - 19 < 0$
$\implies$ Equation $(7)$ and hence inequality $(6)$ have no real solution.
$\therefore \;$ Solution of $(1)$ is
$x \in \left(\dfrac{1 - \sqrt{21}}{2}, \dfrac{1 + \sqrt{21}}{2}\right)$ $\;$ when $\;$ $x > 0$ $\;$ OR $\;$ $x < -1$