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.
 A: 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:
$$x\in\left[\frac{2}{3},\frac{4}{3}\right)\cup\left[\sqrt2,\frac{5}{3}\right)\cup\left[\sqrt3,\sqrt5\right)\cup\left[\frac{7}{3},\sqrt6\right)$$
A: The solution of,
$$(1) \quad x^2-3\cdot x+2=0$$
Is, $x=1$ or $x=2$.
To find solutions to,
$$(2) \quad [x^2]-[3\cdot x]+2=0$$
We must solve six systems of equations. However, we'll solve eight for instructive purposes. The first is,
$$[x^2]=0$$
$$[3 \cdot x]=2$$
$$(a) \quad {2 \over 3} \le x \lt 1$$
The second is,
$$[x^2]=1$$
$$[3 \cdot x]=3$$
Which is solved similarly.
$$(b) \quad 1 \le x \lt {4 \over 3}$$
The third is,
$$[x^2]=2$$
$$[3 \cdot x]=4$$
Slightly trickier, we get,
$$(c) \quad \sqrt{2} \le x \lt {5 \over 3}$$
The fourth is,
$$[x^2]=3$$
$$[3 \cdot x]=5$$
$$(d) \quad \sqrt{3} \le x \lt 2$$
The fifth is,
$$[x^2]=4$$
$$[3 \cdot x]=6$$
$$(e) \quad 2 \le x \lt \sqrt{5}$$
The sixth is,
$$[x^2]=5$$
$$[3 \cdot x]=7$$
$$(f) \quad {7 \over 3} \le x \lt \sqrt{6}$$
The seventh is,
$$[x^2]=6$$
$$[3 \cdot x]=8$$
$$(g) \quad {8 \over 3} \le x \lt \sqrt{7}$$
Something goes wrong with this one. We must discard this solution because it is inconsistent. We're basically done.
The eighth is,
$$[x^2]=7$$
$$[3 \cdot x]=9$$
Which has an empty solution. We're done. We've covered all the cases. Higher numbers result in no solution. Lower numbers result in imaginary solutions.
Putting these together, it's clear that,
$${2 \over 3} \le x \lt {4 \over 3}$$
or
$$\sqrt{2} \le x \lt {5 \over 3}$$
or
$$\sqrt{3} \le x \lt \sqrt{5}$$
or
$${7 \over 3} \le x \lt \sqrt{6}$$
Solves, $(2)$. 
Why does this work? The main insight is that the floor function keeps things at integer values for suitable $x$. If we find these bounds in each of the floor terms, we can use this to solve the equation. We also saw that there are only a finite number of equations to solve, as only a few have intersecting solution sets. The ones that do, make up the solution to $(2)$.
See wolfram for the full solution and graph.
A: $$\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$$
\begin{matrix}
   x-1 &\lt &\lfloor x\rfloor &\lt &x+1\\
   x^2-1 &\lt &\lfloor x^2\rfloor &\lt &x^2+1\\
   \\
   3x-1 &\lt &\lfloor 3x\rfloor &\lt &3x+1\\
   -3x-1 &\lt &-\lfloor 3x\rfloor &\lt &-3x+1\\
   \\
   x^2-3x-2 &\lt &\lfloor x^2 \rfloor-\lfloor 3x\rfloor &\lt &x^2-3x+2\\
   x^2-3x-2 &\lt &-2 &\lt &x^2-3x+2\\
   -2 &\lt &-x^2+3x-2 &\lt &2\\
   -2 &\lt &x^2-3x+2 &\lt &2\\
   0 &\lt &x &\lt &3
\end{matrix}
The places where $\lfloor x^2\rfloor$ changes value are
$\{1, \sqrt 2, \sqrt 3, 2, \sqrt 5, \sqrt 6, \sqrt 7, \sqrt 8, 3 \}$
The places where $\lfloor 3x\rfloor$ changes value are
$\{\frac 13, \frac 23, 1, \frac 43, \frac 53, 2, \frac 73, \frac 83, 3 \}$
The sorted list of the union of these two sets is
$$\left\{\frac 13, \frac 23, 1, \frac 43, \sqrt 2, \frac 53,\sqrt 3 , 2,
  \sqrt 5, \frac 73, \sqrt 6, \sqrt 7, \frac 83, \sqrt 8, 3 \right\}$$
\begin{matrix}
   \text{breakpoint} & \lfloor x^2\rfloor &\lfloor 3x\rfloor & \lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2\\
   \frac 13 & 0 & 1 & 1  &\\
   \frac 23 & 0 & 2 & 0  & \checkmark\\
   1        & 1 & 3 & 0  & \checkmark\\
   \frac 43 & 1 & 4 & -1 &\\
   \sqrt 2  & 2 & 4 & 0  & \checkmark\\
   \frac 53 & 2 & 5 & -1 &\\
   \sqrt 3  & 3 & 5 & 0  & \checkmark\\
   2        & 4 & 6 & 0  & \checkmark\\
   \sqrt 5  & 5 & 6 & 1 &\\
   \frac 73 & 5 & 7 & 0 &\checkmark\\
   \sqrt 6  & 6 & 7 & 1 &\\
   \sqrt 7  & 7 & 7 & 2 &\\
   \frac 83 & 7 & 8 & 1 &\\
   \sqrt 8  & 8 & 8 & 2 &\\
   3        & 9 & 9 & 2 &\\
\end{matrix}
Hence the solution set is
$$ x \in 
   \left[ 2/3, 4/3 \right) \cup
   \left[ \sqrt 2, 5/3 \right) \cup
   \left[ \sqrt 3, \sqrt 5 \right) \cup
   \left[ 7/3, \sqrt 6 \right)
$$
A: For any $x$ that provide a solution for your problem, $x$ must satisfy these
two inequalities
$$
x^2-3x+2 \geq 0\\
x^2-3x+2<1
$$
From the first equation we see that $(x-2)(x-1)\geq 0$ so we have narrowed the solution space to $\mathbb{R}\setminus ]1,2[$.
The second equation gives us $x^2-3x+1<0$ which gives us the solution space $]\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}[$,  approximated to $]0.382,2.618[$
Knowing that both inequalities must be satisfied we get an intersection of
these two solution spaces, namely 
$$
x\in ]\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}[ \quad \setminus \quad ]1,2[\quad = \quad]\frac{3-\sqrt{5}}{2},1] \cup[2, \frac{3+\sqrt{5}}{2}[
$$
Which can be approximated to 
$$
x \in ]0.382,1]\cup [2,2.618[
$$
