<s>[INST] 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.
 [/INST] 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

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