I would just like to confirm my approach to this equation:
$$\lfloor x - 1/2 \rfloor^2 = 25$$
Now square root both sides
$$\lfloor x - 1/2\rfloor = \pm5.$$
By the definition of floored value we have
$$\lfloor x - 1/2 \rfloor \leqslant x - 1/2 < \lfloor x - 1/2 \rfloor + 1.$$
Now, when the the floor expression is equal to $5$ we find that \begin{align*} 5 &\leqslant x - 1/2 < 5+1 \\ 11/2 &\leqslant x < 13/2 \end{align*} Similarly, if the expression is equal to $-5$ we find that \begin{align*} -5 &\leqslant x - 1/2 < -5 + 1 \\ -9/2 &\leqslant x < 7/2 \end{align*}
Now, can I say that the solution groups are: $[-9/2,-7/2)\cup[11/2,13/2)$?
Or is there something I'm missing, thank you.