Solutions of $\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$ 
:Solutions of:
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0$$

My try :
$$\lfloor 2x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac12 \rfloor  $$
$$\lfloor 3x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac13 \rfloor +\lfloor x+\dfrac23 \rfloor $$
$$\lfloor 4x \rfloor =\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor $$
Now :
$$\lfloor 2x \rfloor \times \lfloor 3x \rfloor + \lfloor 4x \rfloor=0 $$
$$(\lfloor x \rfloor)^2+(\lfloor x \rfloor)(\lfloor x+\dfrac13 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac23 \rfloor)+(\lfloor x \rfloor)(\lfloor x+\dfrac12 \rfloor)+(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac13 \rfloor)\\(\lfloor x+\dfrac12 \rfloor)(\lfloor x+\dfrac23 \rfloor) +\lfloor x \rfloor +\lfloor x+\dfrac14 \rfloor +\lfloor x+\dfrac24 \rfloor +\lfloor x+\dfrac34 \rfloor=0$$
Now what ?
 A: Just do it in 6 cases:
$x = [x]+\{x\}$

Case 1: $\{x\} < 1/4$

Then $$[2x]\times[3x] + [4x] = 2[x]3[x] + 4[x] = 0$$ so $[x] = 0$ of $[x] = -\frac 32$.  So $[x]$ is an integer $[x] = 0$ and $0 \le x < \frac 14$.

Case 2: $1/4 \le \{x\} < 1/3$

Then $$[2x]\times[3x] + [4x] = 2[x]3[x] + 4[x]+ 1 =  6[x]^2 + 4[x] + 1 = 0.$$
$[x]$ must be negative but then $6[x]^2 +1 > 4|[x]| $ so that is impossible.

Case 3: $1/3 \le \{x\} < 1/2$

then $$[2x]\times[3x] + [4x] =2[x](3[x] + 1) + 4[x] + 1 = 6[x]^2 + 6[x] + 1 = 0.$$  Then $[x]^2 + [x] = -\frac 16$ which is not an integer so impossible.

Case 4: $\frac 12 \le \{x\} < 2/3$

then $$[2x]\times[3x] + [4x] = (2[x] + 1)(3[x] + 1) + 4[x] + 2 = 6[x]^2 + 9[x] + 3 = 0$$ so $2[x]^3 + 3[x] + 1 = 0$ so $(2[x] + 1)([x] + 1) = 0$ so $[x] = -1$ or $[x ] = -\frac 12$.  So $[x] = -1$ and $- \frac 12 \le x < -1/3$.

Case 5: $\frac 23 \le \{x\} < 3/4$

then $$[2x]\times[3x] + [4x] = (2[x] + 1)(3[x] + 2) + 4[x] + 2= 6[x]^2 + 11[x] + 4 = 0$$
$$[x] = \frac {-11 \pm \sqrt{121 - 4\times4\times6}}{12} = \frac {-11 \pm 6}{12} \not \in \mathbb Z.$$

Case 6: $\frac 34 \le \{x\} < 1$

then $$[2x]\times[3x] + [4x] = (2[x] + 1)(3[x] + 2) + 4[x] + 3= 6[x]^2 + 11[x] + 5 = 0$$
$$[x] = \frac {-11 \pm \sqrt{121 - 4\times4\times5}}{10} = \frac {-11 \pm \sqrt{41}}{12} \not \in \mathbb Z.$$
So $- \frac 12 \le x < -\frac 13$ or $0 \le x < \frac 14$
A: HINT: 
If $0<b < 1$ then $\lfloor x + b \rfloor = \lfloor x \rfloor$ if  $x-\lfloor 
 x\rfloor + b < 1$ or $\lfloor x \rfloor + 1$ otherwise
you might have to look at the different cases that arise
