2021/097) Solve the following equation: $[x]^2=[2x]-1$ where [x] is the floor value of the x real No

Let x = n + r where n is the integer part and r is the fractional part

we have

$\lfloor x \rfloor ^2 = 2(x+r) - 1$

so $n^2 = 2n -1$ when $ r < \frac{1}{2}$ or $n^2 = 2n$  and $ \frac{1}{2} \le   r <  1$

$n^2 = 2n -1$ when $ r < \frac{1}{2}$

gives $n^2 - 2n + 1 = 0$ or $(n-1)^2 =0 $ or n= 1 giving $ 1 \le x < 1.5$

$n^2 = 2n$  and $\frac{1}{2} \le   r <  1$

gives n = 0 or 2  giving $ .5 \le x < 1$ or  giving $ 2.5 \le x < 3$

combining them we have  $ .5 \le x < 1. 5 $ or  $ 2.5 \le x < 3$