<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] $$\left\lfloor x \right\rfloor\le x$$
$$-6\left\lfloor x \right\rfloor\ge -6x$$
$$x^2-6\left\lfloor x \right\rfloor+6\ge x^2-6x+6$$
$$0\ge x^2-6x+6$$
$$x\in[3-\sqrt{3},3+\sqrt{3}] \ \ \ \& \ \ \left\lfloor x \right\rfloor\in\{1,2,3,4\}$$
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 $x^2-6\left\lfloor x \right\rfloor+6$
$$\sum \text{roots}=\sqrt{6}+\sqrt{12}+\sqrt{18} $$
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