<s>[INST] Find the solution of $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$ Is anyone able to help me with the following equation concerned the floor function  $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$
I don't know how to deal with the floor terms properly.
 [/INST] From the definition of the floor function we have
[correction here]
$$\begin{align}
x^2-1 &< &\lfloor x^2 \rfloor &\le x^2 \\ 
3x-1 &< &\lfloor 3x \rfloor &\le 3x 
\end{align}$$
and so
$$\lfloor x^2 \rfloor - \lfloor 3x \rfloor + 2 > x^2-3x-1+2=x^2-3x+1$$
So all solutions to the floor equation must lie between the two roots of $x^2-3x+1=0$. That is
$$\frac{3-\sqrt5}{2}\le x \le \frac{3+\sqrt5}{2}$$
Now trying various cases:
$$\begin{align}
\lfloor 3x \rfloor=2,&\lfloor x^2 \rfloor=0&\implies x\in\left[\frac{2}{3},1\right)\cap[0,1)=\left[\frac{2}{3},1\right) \\[1em]
\lfloor 3x \rfloor=3,&\lfloor x^2 \rfloor=1&\implies x\in\left[1,\frac{4}{3}\right)\cap[1,\sqrt2)=\left[1,\frac{4}{3}\right) \\[1em]
\lfloor 3x \rfloor=4,&\lfloor x^2 \rfloor=2&\implies x\in\left[\frac{4}{3},\frac{5}{3}\right)\cap[\sqrt2,\sqrt3)=\left[\sqrt2,\frac{5}{3}\right) \\[1em]
\lfloor 3x \rfloor=5,&\lfloor x^2 \rfloor=3&\implies x\in\left[\frac{5}{3},2\right)\cap[\sqrt3,2)=[\sqrt3,2) \\[1em]
\lfloor 3x \rfloor=6,&\lfloor x^2 \rfloor=4&\implies x\in\left[2,\frac{7}{3}\right)\cap[2,\sqrt5)=[2,\sqrt5) \\[1em]
\lfloor 3x \rfloor=7,&\lfloor x^2 \rfloor=5&\implies x\in\left[\frac{7}{3},\frac{8}{3}\right)\cap[\sqrt5,\sqrt6)=\left[\frac{7}{3},\sqrt6\right) \\[1em]
\end{align}$$
So the full solution is:
$$x\in\left[\frac{2}{3},\frac{4}{3}\right)\cup\left[\sqrt2,\frac{5}{3}\right)\cup\left[\sqrt3,\sqrt5\right)\cup\left[\frac{7}{3},\sqrt6\right)$$
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