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Solve the following equation
[$x^2+1$]=[2x] ,where [x] is the floor value of x
[$x^2+1$]=[2x] ,where [x] is the floor value of x
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
$x^2 - 2x + 1 = 0$ $\implies$ $x=1$
Where do you base that assumptionso the upper limits and lower limits must be equal