<s>[INST] The solution set of the equation $|\lfloor x|x|-7\rfloor|<5$ is? My solution approach :-
$|\lfloor x|x|-7\rfloor|<5$
When $x\geq 0$, then;
$|\lfloor x^2-7\rfloor|<5$
$\Rightarrow -5 \lt \lfloor x^2-7\rfloor \lt5$
After this I do know that I'll have to take another case in which $x\lt0$ but I am not able to get how to solve the floor function.
 [/INST] So far, when $x\geq 0$ you got here
$$
-5 \lt \lfloor x^2-7\rfloor \lt5
$$
This means $\lfloor x^2-7\rfloor \in \{-4,-3,-2,-1,0,1,2,3,4\}$ or in other words $x^2 - 7 \in [-4,5)$.
Therefore we have
$$
x^2 \in [3,12),
$$
or
$$
x\in[\sqrt{3},2\sqrt{3}).
$$
Note that $x\in(-2\sqrt{3}, -\sqrt{3}]$ is excluded because we only take $x\geq 0$.
For the case $x\leq 0$, we have
$$
-5 < \lfloor -x^2 - 7 \rfloor < 5.
$$
But $\lfloor -x^2 - 7 \rfloor \leq -7 < -5$, therefore there are no solutions for the case $x\leq 0$.
To sum it all up, the solution set of the equation $|\lfloor x|x|-7\rfloor|<5$ is $x\in[\sqrt{3},2\sqrt{3})$.
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