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

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In summary, the equation $x^2+1=2x$, where $x$ is the floor value of $x$, has a solution of $x=1$. This is based on the fact that the upper and lower limits of $[x^2+1]$ and $[2x]$ must be equal, leading to the equation $x^2-2x+1=0$, which can be simplified to $x=1$. However, this is not the only solution, as $x$ can also be in the interval from 1/2 to the square root of 2, or from 3/2 to the square root of 3. This is based on the assumption that the upper and
  • #1
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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  • #2
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
$x^2 - 2x + 1 = 0$ $\implies$ $x=1$
 
  • #3
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
 
  • #4
hint:
[sp] put $[x^2+1]=n=[2x]$[/sp]
 

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

What is the floor value of x in the equation $x^2+1=2x?

The floor value of x in this equation refers to the largest integer that is less than or equal to the value of x. In this case, the floor value of x would be 1.

How do you solve for the floor value of x in this equation?

To solve for the floor value of x, you can rearrange the equation to get x on one side and a constant on the other side. In this case, it would be x^2 - 2x + 1 = 0. You can then use the quadratic formula or factor the equation to find the roots, which would be x = 1 and x = 1. Since the floor value is the largest integer less than or equal to x, the floor value would be 1.

Can the floor value of x be a negative number?

No, the floor value of x cannot be a negative number. The floor function always rounds down to the nearest integer, so the floor value of x will always be equal to or larger than 0.

What is the difference between the floor value and the ceiling value of x?

The floor value of x is the largest integer that is less than or equal to x, while the ceiling value of x is the smallest integer that is greater than or equal to x. In other words, the floor value rounds down to the nearest integer, while the ceiling value rounds up to the nearest integer.

Can the floor value of x be a decimal or fraction?

No, the floor value of x can only be a whole number or integer. The floor function always rounds down to the nearest integer, so it cannot produce a decimal or fraction as the result.

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