Solve Floor Equation 7: x^2=2x-1

[solakis1]

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]

 

[kaliprasad]

Spoiler: My Solution

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

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

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

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

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

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

 

[solakis1]

ok kaliprasad that's it good work