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

[solakis1]

Solve the following equation

[$x^2+1$]=[2x] ,where [x] is the floor value of x

 

[DaalChawal]

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
$x^2 - 2x + 1 = 0$ $\implies$ $x=1$

 

[solakis1]

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
$x^2 - 2x + 1 = 0$ $\implies$ $x=1$

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

 

[solakis1]

hint:
[sp] put $[x^2+1]=n=[2x]$[/sp]