- #1
solakis1
- 422
- 0
Solve the following equation:
$[x]^2=[2x]-1$ where [x] is the floor value of the x real No
[sp] hint : start by puting x=n+b where n is an integer and $0\leq b<1$[/sp]
$[x]^2=[2x]-1$ where [x] is the floor value of the x real No
[sp] hint : start by puting x=n+b where n is an integer and $0\leq b<1$[/sp]