MHB Solve the Floor Value of x in $x^2+1=2x$

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The equation $[x^2+1] = [2x]$, where [x] denotes the floor function, is discussed with an initial solution of $x=1$. The approach involves establishing inequalities based on the properties of the floor function, leading to the conclusion that $x^2 - 2x + 1 = 0$. However, further analysis reveals that the complete solution is actually the intervals $x \in [1/2, \sqrt{2}) \cup [3/2, \sqrt{3})$. The discussion highlights the need for a deeper understanding of the floor function's behavior in solving such equations.
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Solve the following equation

[$x^2+1$]=[2x] ,where [x] is the floor value of x
 
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I'm getting only $x=1$ as the answer.
Here is my approach please do tell me If I'm wrong.
we know that $x-1 \le [x] \lt x$ so
$x^2 \le [x^2+1] \lt x^2 + 1 $ and
$2x-1 \le [2x] \lt 2x$
so the upper limits and lower limits must be equal
$x^2 - 2x + 1 = 0$ $\implies$ $x=1$
 
DaalChawal said:
I'm getting only $x=1$ as the answer.
Here is my approach please do tell me If I'm wrong.
we know that $x-1 \le [x] \lt x$ so
$x^2 \le [x^2+1] \lt x^2 + 1 $ and
$2x-1 \le [2x] \lt 2x$
so the upper limits and lower limits must be equal
$x^2 - 2x + 1 = 0$ $\implies$ $x=1$
DaalChawal said:
so the upper limits and lower limits must be equal
Where do you base that assumption
I mean which axiom,definition ,theorem supports that assumption

The answer to the problem is : $x\in$[1/2,$\sqrt 2$)U[3/2,$\sqrt 3$)
Here we have that whole intervals is the answer
 
hint:
[sp] put $[x^2+1]=n=[2x]$[/sp]
 

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