If $[x]^2+4\{x\}=2x$, then how many values of $x$ are possible?

Question

Let $\{x\}$ and $[x]$ denote the fractional and integral parts respectively of a real number x. If $[x]^2+4\{x\}=2x$, then how many values of $x$ are possible?

Let $x= f+I$ where $f$ is the fractional part and $ I$ is the integer part

$I^2+4f=2(f+I)$

$I^2+4f=2f+2I$

Now I have a problem in solving the below inequality

$0\leq(2I-I^2)/2\geq1$

Answer 1

Easy you can get that $$f = I-\frac{I^2}{2}$$ since we know that $0\leq f<1$ so $$0\leq I-\frac{I^2}{2}<1$$ If we focus first on $$0\leq I-\frac{I^2}{2}$$ we get that $0\leq I\leq 2$, so I can be $0,1,2$ if we try $I-\frac{I^2}{2}<1$ for those value we get that I can be 0,1 but not 2.
Now $$I-\frac{I^2}{2}<1$$ we see that it holds for every integer except 2.
So finally we have that $I=0,1$
If $I=0$ then $f=0$
If $I=1$ then $f=\frac{1}{2}$

Answer 2

The exact inequality is

$$0\le\frac{2i-i^2}2<1.$$

This can be rewritten

$$-1\le-(i-1)^2<1$$

and the only solutions are $i=0,1$, hence $x=0, 1.5$.