Hi, I think the answer of this question x=$\displaystyle \sqrt{1/2}$. But I am not sure for this answer. What do you think about for this problem ?
... a very easy alternative to find the solutions of the equation...
$\displaystyle x^{2} - \lfloor {x} \rfloor = \frac{1}{2}$ (1)
... is to set $\displaystyle \lfloor {x} \rfloor = k$ with k integer and verify if the solution of the 'standard' second order equation...
$\displaystyle x^{2} - k =\frac{1}{2}$ (2)
... is also solution of (1)...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
All right!... it is self evident that $\displaystyle \lfloor {x} \rfloor$ is an integer that we can indicate with k. Because $\displaystyle x^{2} \ge 0$ and $\displaystyle \frac{1}{2} >0$ and is...
$\displaystyle x^{2} - \lfloor {x} \rfloor = x^{2} - k = \frac{1}{2}$ (1)
... we have to suppose that is $\displaystyle k \ge 0$. Now we proceed 'step by step' to verify with increasing value of k starting from 0...
$\displaystyle k=0 \rightarrow x^{2} = \frac{1}{2} \rightarrow x= \sqrt{\frac{1}{2}} $ (2)
and because $\displaystyle \lfloor {x} \rfloor =0 =k$ , $\displaystyle x= \sqrt{\frac{1}{2}} $ is solution of (1)...
$\displaystyle k=1 \rightarrow x^{2} = \frac{3}{2} \rightarrow x= \sqrt{\frac{3}{2}} $ (3)
and because $\displaystyle \lfloor {x} \rfloor = 1 =k$ , $\displaystyle x= \sqrt{\frac{3}{2}} $ is also solution of (1)...
For k>1 there are no more solution , so that the (1) has two solutions... like a 'standard' second order equation ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$