Let $x^2-\lfloor 2x\rfloor=kx-2\lfloor x\rfloor $ have be four answers then what is range of $k$
My Try :
$$x=n+r : \ \ \ \ 0\leq r<1$$
$$x^2-\lfloor 2x\rfloor=kx-2\lfloor x\rfloor \\ (n+r)^2-\lfloor 2(n+r)\rfloor=k(n+r)-2n\\n^2+2nr+r^2-(2n+\lfloor 2r\rfloor)=kn+kr -2n\\n^2+2rn+r^2-kn-kr=\lfloor 2r\rfloor\\n^2+(2r-k)nr+r^2-kr=\lfloor 2r\rfloor$$
Now what ?
For $x$ in an interval $[n,n+0.5)$, where $n$ is an integer,$$x^2-kx=0.$$ For two solutions we therefore require $k$ to be in an interval of the form $[n,n+0.5)$.
For $x$ in an interval $[n+0.5,n+1)$, where $n$ is an integer,$$x^2-kx-1=0.$$ The two solutions $u,v$, say, satisfy $$uv=-1,u+v=k.$$
For $k$ to be the sum, we require $u$ and $v$ to be in intervals of the form $[n+0.5,n+0.75)$. One of these two solutions, $u$ say, must be less than one in modulus and is therefore in $[-0.5,-0.25)$ or $[0.5,0.75)$. Then $v$ is in $[2.5,2.75)$,$[3.5,3.75)$ or $[-1.5,-1.33)$. (Giving two DP answers.)
Finally, $k=v-\frac{1}{v}$ gives the following solutions for for $k$.
$[2.1,2.39)$, $[3.21,3.48)$ or $[-0.83,-0.58)$.