2021/025) Solve the following $\lfloor x\lfloor x\rfloor \rfloor = 10$ in positive x

 Let $\lfloor x \rfloor = n$

So we have $ n \le x < n+1$

So 

$ nx \le \lfloor x \lfloor x \rfloor \rfloor <  x(n+1) $

Ss $n^2 \ le nx $ and $(n+1)^2 > x(n+1)$ we have

$ n^2 \le \lfloor x \lfloor x \rfloor \rfloor <  (n+1)^2 $

Or

$ n^2 \le 10 < (n+1)^2 $

And $ 10 \le nx < 11$

Or n= 3 and putting n = 3 we have $\frac{10}{3} \le x < \frac{11}{3}$