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What are the roots of $x^2-6x+\lfloor x \rfloor+7=0$, where $\lfloor x \rfloor$ is the greatest integer function?

Is there some way to solve the equation without graphing?

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A possible strategy is to solve $x^2-6x+n+7=0$ for all integers $n$, and find out when a solution $x_0$ (this will be a function of $n$, something square-root-ish) satisfies $n\leqslant x_0<n+1$. The problem reduces to $2$ polynomial inequalities of degree $2$ in the variable $n$. There will be some casework, though. –  barto Jun 7 at 9:37
    
Would inspection be acceptable ? –  Claude Leibovici Jun 7 at 9:37
    
@ClaudeLeibovici yes. –  shaurya gupta Jun 7 at 9:44
    
@barto $x_0$ would be a function of $n$ and $n$ itself is a function of $x$, so $x_0$ is a function of $x$...Would that give us the answer? –  shaurya gupta Jun 7 at 9:47
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Can I recommend you $\lfloor x\rfloor$ (\lfloor x \rfloor) ? –  Yves Daoust Jun 8 at 10:05

3 Answers 3

up vote 5 down vote accepted

$x-1 < [x] \le x$, so $x^2-5x+6<0$ while $x^2-5x+7\ge 0$, which furthur reduces to $2<x<3$. We know $[x]=2$, so $x^2-6x+9=0$ and $x=3$. But this contradicts $x<3$ and hence no solution.

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@shauryagupta, no, there is no solution, as I mentioned at the end. –  Magician Jun 7 at 10:13

Given that $[x]=x+\epsilon$ for some $\epsilon$ in range $[0\ 1[$, you should study the parametric equation $$x^2-5x+\epsilon+7=0.$$

Its discriminant is $25-4\epsilon-28$, always negative for the allowed $\epsilon$.

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Yep, just two complex. –  Yves Daoust Jun 7 at 10:12

The equation is equivalent to

$x=3\pm\sqrt{2-[x]}$

Suppose $a\leq x<a+1$, $a\in Z$

Then $[x]=a$ and

$x=3-\sqrt{2-a}\geq\\\geq 3-(2-a)=a+1$

or

$x=3+\sqrt{2-a}\geq a+1$

This is a contradiction and, hence there are no real solutions.

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