Is it allowed to solve this inequality $x|x-1|>-3$ by dividing each member with $x$?

Question

Is it allowed to solve this inequality $x|x-1|>-3$ by dividing each member with $x$? What if $x$ is negative?

My textbook provides the following solution:

Divide both sides by $x: $ $\frac { x | x - 1 | } { x } > \frac { - 3 } { x } ; \quad x \neq 0$

Simplify: $| x - 1 | > - \frac { 3 } { x } ; \quad x \neq 0$

Edit: provided textbook's solution

Answer 1

For $x\geq0$ this inequality is always true.

Assume that $x<0$, so $x=-y$ for some positive $y$ and we get $$y|\;\underbrace{y+1}_{>0}\;|<3\implies y(y+1)<3 ...$$

Answer 2

${}{}{}{}{}{}{}{}{}{}{}{}{}{}$

Answer 3

Because, the inequality holds for $x \ge 0$, let's consider the case $x<0$.
In this case, $|x-1|=-x+1$ and we have $-x^2+x+3>0$. The roots of the quadratic are $x_1=\frac{1-\sqrt{13}}{2},x_2=\frac{1+\sqrt{13}}{2}$ so the inequality also holds when $\frac{1-\sqrt{13}}{2} < x <0$