Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF
I have done this problem by inspection as $$\frac{x^2+6}{6}=[x] \implies x>0.$$
Let [x]=0, then $x$ is non real. Let $[x]=1$, then $x=0$ which contradicts. Let $[x]=2$, it gives $x=\sqrt{6}$, in agrrement. Similarly assuming $[x]=3,4$; we get correct roots as $\sqrt{12}$ and $\sqrt{18}$. But if er let $[x]=5$, it gives $x=\sqrt{24}$. which contradicts. So I get the sum of roots as $\sqrt{6}(1+\sqrt{2}+\sqrt{3}).$
The question is: Have I found all the real roots. In any case, what is(are) more appropriate method(s) of doing it.
 A: $$\left\lfloor x \right\rfloor\le x$$
$$-6\left\lfloor x \right\rfloor\ge -6x$$
$$x^2-6\left\lfloor x \right\rfloor+6\ge x^2-6x+6$$
$$0\ge x^2-6x+6$$
$$x\in[3-\sqrt{3},3+\sqrt{3}] \ \ \ \& \ \ \left\lfloor x \right\rfloor\in\{1,2,3,4\}$$
so let's consider the cases:
$1^{\circ}$ $\left\lfloor x \right\rfloor=1\Rightarrow x=0$ which is not a root of $x^2-6\left\lfloor x \right\rfloor+6$
$2^{\circ}$ $\left\lfloor x \right\rfloor=2\Rightarrow x=\sqrt{6} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$3^{\circ}$ $\left\lfloor x \right\rfloor=3\Rightarrow x=\sqrt{12} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$4^{\circ}$ $\left\lfloor x \right\rfloor=4\Rightarrow x=\sqrt{18} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$$\sum \text{roots}=\sqrt{6}+\sqrt{12}+\sqrt{18} $$
A: You have got all real solutions. Because in
$$\frac{x^2+6}{6}=[x]$$
LHS increases quadratically (parabolic) and RHS variies roughly as a line, so after $x=5$ the parabola leaves out the line.
Method I: You can use $$x-1 \le [x] \le x \implies x^2-6x+6 \le 0 ~~~~(1),~~~ x^2-6x+12 >0~~~(2)$$
(1) gives $3-\sqrt{3}< x \le 3+\sqrt{3}$ and $(2)$ is always true. So your choices of $[x]=2,3,4$ get justified and work well.
Method II: Let $$x=n+q,~~ n \in I^+, 0\le q< 1$$. Putting it in the equation you get
$$n^2-6n+6=-q^2-2nq \le 0 \implies n=2,3,4.$$
For $n=2$ get $$q=\pm \sqrt{6}-2 \implies q=\sqrt{6}-2>0 \implies x=2+\sqrt{6}-2= \sqrt{6}.$$ Similarly, you get other two roots.
Method III: Graphically, the LHS is a parabola and RHS is a starcase function. These two cut each other in the first quadrant at three points. See the Fig. below

A: replace $[x]=x-${$x$}
Thus as
$ 0\le ${$x$}$\le 1$
$0\le x^2-6x+6<6$
from here  we can find the interval in which x lies, and break the itervals to check individually.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{}$
\begin{align}
&x = \root{6\left\lfloor\,{x}\,\right\rfloor - 6} \implies \left\lfloor\,{x}\,\right\rfloor =
\left\lfloor\,
{\root{6\left\lfloor\,{x}\,\right\rfloor - 6}}
\,\right\rfloor
\\[5mm] \implies &\
\left\lfloor\,{x}\,\right\rfloor \leq
\root{6\left\lfloor\,{x}\,\right\rfloor - 6} <
\left\lfloor\,{x}\,\right\rfloor + 1
\\[5mm] \mbox{and}\quad &
\,\,\,\left\lfloor\,{x}\,\right\rfloor^{2} \leq
6\left\lfloor\,{x}\,\right\rfloor - 6 < \left\lfloor\,{x}\,\right\rfloor^{2} + 2\left\lfloor\,{x}\,\right\rfloor + 1
\end{align}
\begin{align}
&\mbox{which are equivalent to}\quad
\left\{\begin{array}{rcl}
\ds{\left\lfloor\,{x}\,\right\rfloor^{2} -
6\left\lfloor\,{x}\,\right\rfloor + 6} & \ds{\leq} & \ds{0}
\\[1mm]
\ds{\left\lfloor\,{x}\,\right\rfloor^{2} -
4\left\lfloor\,{x}\,\right\rfloor + 7} & \ds{>} & \ds{0}
\end{array}\right.
\\[5mm] &\ \implies
\underbrace{3 - \root{3}}_{\ds{\approx 1.2678}}\ \leq\ \left\lfloor\,{x}\,\right\rfloor\ \leq\
\underbrace{3 + \root{3}}_{\ds{\approx 4.7321}}
\\[5mm] &\ 
\bbx{\left\lfloor\,{x}\,\right\rfloor \in \braces{2,3,4}
\implies x \in \braces{\root{6},2\root{3},3\root{2}}}
\\ &
\end{align}

$\ds{\root{6} \approx 2.4495,\quad
2\root{3} \approx 3.4641,\quad
\root{6} \approx 4.2426}$.


A: $x^2 = 6[x] - 6= 6([x]-1)$.  Forget about the greatest aspect of GIF and concentrate on the integer aspect of GIF. $x^2$ is multiple of $6$ and $x = \pm\sqrt {6k}$ where $k=[x]-1$.
$x^2 \ge 0$ so $[x]-1\ge 0$ so $x\ge [x] \ge 1$ so $k \ge 1$.
And if $k \ge 6$ then $x =\sqrt{6k} \le \sqrt {k^2} = k$ so $[x]\le x \le k =[x]-1$ is impossible.
We can pick them off one by one.  If $k=1$ and $x=\sqrt 6$ then $[x]=2$ and $6[x]-6=6=x^2$.  So that's one solution.  If $k=2$ and $x = \sqrt{12}$ then $[x]=3$ and $6[x]-6=12$ so that's another.  If $k=4$ then $x=\sqrt{24}$ and $[x]=4=k = [x]-1$.  Impossible.  If $k=5$ then $x=\sqrt{30}$ and $[x]=5$ and $k\ne [x]-1$ so that's impossible.
$x = \sqrt 6$ or $\sqrt{12}$ and the sum of the roots is $\sqrt 6 + \sqrt {12} = \sqrt 3(2+\sqrt 2)$.
