How to solve $2x^2-2\lfloor x \rfloor-1=0$

Question

How do I solve $2x^2-2\lfloor x \rfloor-1=0$?

I have tried setting $x=\lfloor x \rfloor + \{x\}$, where $\{x\}$ is the fractional part of $x$. Then, I tried $$2x^2-2\lfloor x \rfloor-1=0$$ $$2(\lfloor x \rfloor + \{x\})^2-2\lfloor x \rfloor-1=0$$ $$2\lfloor x \rfloor^2 + 4\lfloor x \rfloor\{x\}+2\{x\}^2-2\lfloor x \rfloor-1=0$$ but now I am stuck. How should I proceed?

Answer 1

Hint

$$2x^2-1=2\lfloor x \rfloor \leq 2x$$

Therefore $$2x^2-2x-1 \leq 0$$

Solve, and the solution will tell you that there are only few potential values for $\lfloor x \rfloor$. Solve for each.

Answer 2

Suppose $x = n + \delta$ where $n$ is an integer and $0 \le \delta < 1$

Then

\begin{align} 2x^2-2\lfloor x \rfloor-1&=0 \\ x^2 &= \lfloor x \rfloor + \dfrac 12 \\ x &= \sqrt{\lfloor x \rfloor + \dfrac 12} \\ n+\delta &= \sqrt{n + \dfrac 12} \\ \delta &= \sqrt{n + \dfrac 12} - n \\ \end{align}

\begin{array}{|c|c|c|} \hline n & \delta & x \\ \hline 0 & \dfrac{\sqrt 2}{2} & \dfrac{\sqrt 2}{2} \\ 1 & \dfrac{\sqrt 6}{2}-1 & \dfrac{\sqrt 6}{2} \\ 2 & \color{red}{\dfrac{\sqrt{10}}{2}-2 < 0} \\ \hline \end{array}

We have to stop at $n=2$ because, for $n \ge 2$, $\sqrt{n + \dfrac 12} - n < 0$

Answer 3

$$2x^2-1=2[x]~~~~(1).$$ $[x]$ is GIF/ Floor function We have $2[x]+1=2x^2 \ge 0 \implies [x] \ge 0 ~~~(2)$

Let $x=n+q$, $n$-is integer $n\ge 0$ and $0 \le q <1$, then we get $$2n^2+2q^2+4nq-1=2n~~~~(3)$$ $$\implies 2n^2-2n-1=-2q^2-4nq \le 0 \implies 2n^2-2n-1 \le 0 \implies \frac{1-\sqrt{3}}{2} \le n \le \frac{1+\sqrt{3}}{2}.$$ $$\implies n=0,1$$

Now put $n=0$ in (3) to get $q=\frac{\sqrt{6}-2}{2} >0, \implies x =1+q =\sqrt{\frac{3}{2}}.$

Next put $n=0$ in (3) to get $2q^2-1=0 \implies q=\frac{1}{\sqrt{2}}>0 \implies x=0+q=\frac{1}{\sqrt{2}}.$

So only two solutions $x=\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}$