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

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Discussion Overview

The discussion revolves around solving the equation involving the floor function, specifically the equation [$x^2+1$]=[2x], where [x] denotes the floor value of x. Participants explore different approaches to find the values of x that satisfy this equation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant claims that the only solution is $x=1$, providing a method based on inequalities involving the floor function.
  • Another participant challenges the assumption that the upper and lower limits must be equal, asking for the supporting axioms or definitions.
  • A different perspective suggests that the solution set is actually $x\in[1/2,\sqrt{2}) \cup [3/2,\sqrt{3})$, proposing that whole intervals are valid solutions.
  • A hint is provided to set $[x^2+1]=n=[2x]$ as a potential approach to the problem.

Areas of Agreement / Disagreement

Participants express differing views on the solution to the equation. While one participant asserts that $x=1$ is the sole solution, others propose a broader set of intervals. The discussion remains unresolved with multiple competing views.

Contextual Notes

Participants have not fully clarified the assumptions underlying their approaches, particularly regarding the treatment of the floor function and the implications of the inequalities used in their reasoning.

solakis1
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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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