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 ptoblem by inspection as $$\frac{x^2+6}{6}=[x] \implies x&gt;0.$$
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.
$$\left\lfloor x \right\rfloor\le x$$
$$-6\left\lfloor x \right\rfloor\ge -6x$$
$$x^2-6\left\lfloor x \right\rfloor+6\ge x^2-6x+6$$
$$0\ge x^2-6x+6$$
$$x\in[3-\sqrt{3},3+\sqrt{3}] \ \ \ \&amp; \ \ \left\lfloor x \right\rfloor\in\{1,2,3,4\}$$
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 $x^2-6\left\lfloor x \right\rfloor+6$
$$\sum \text{roots}=\sqrt{6}+\sqrt{12}+\sqrt{18} $$
$x^2 = 6[x] - 6= 6([x]-1)$.  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 $\sqrt 6 + \sqrt {12} = \sqrt 3(2+\sqrt 2)$.