Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

$x^2 = 6[x] - 6= 6([x]-1)$. 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 $\sqrt 6 + \sqrt {12} = \sqrt 3(2+\sqrt 2)$.

Answer 4

$$\left\lfloor x \right\rfloor\le x$$ $$-6\left\lfloor x \right\rfloor\ge -6x$$ $$x^2-6\left\lfloor x \right\rfloor+6\ge x^2-6x+6$$ $$0\ge x^2-6x+6$$ $$x\in[3-\sqrt{3},3+\sqrt{3}] \ \ \ \& \ \ \left\lfloor x \right\rfloor\in\{1,2,3,4\}$$ 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 $x^2-6\left\lfloor x \right\rfloor+6$

$$\sum \text{roots}=\sqrt{6}+\sqrt{12}+\sqrt{18} $$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{}$ \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}

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

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