Solve the equation $2x^2-[x]-1=0$ where $[x]$ is biggest integer not greater than $x$. I' ve tried with $x^2 = {[x]+1\over 2}$ so $x$ is a square root of half integer. And know? What to do with that?
 A: For a workable range, we can say that:
$$x-1 < [x] \le x$$
$$x< 2x^2 \le x+1$$
Solving this, we get:
$$x\in\left[\frac{-1}{2},0\right) \cup \left(\frac{1}{2},1\right]$$
Our roots must lie in this range. So for each of these intervals, substitute the value of $[x]$ and solve.


*

*In the first range, ie, $x\in\left[-\frac{1}{2},0\right)$, we have $[x] = -1$. So our equation reduces to:
$$2x^2+1-1=0$$
Giving us $x=0$. This does not lie inside our interval, so this is rejected.

*Similarly try for $x\in\left(\frac{1}{2},1\right)$. Here we have $[x] = 0$, so that,
$$2x^2 = 1$$
Giving $x = \frac{1}{\sqrt{2}}$.

*Lastly try for $x=1$. Here we have $[x] = 1$ and this too satisfies the equation.
In all we have two roots: $\frac{1}{\sqrt{2}}$ and $1$ 
A: Hint. Note that $x-1<\lfloor x\rfloor \leq x$ implies that
$$(2x+1)(x-1)=2x^2-x-1\leq 2x^2-\lfloor x\rfloor-1< 2x^2-(x-1)-1=x(2x-1)$$
A: Rewrite
$$x=\pm\sqrt{\frac{[x]+1}2}$$ and try increasing values for the integer $[x]$.


*

*$[x]=-1\to x=0$: incompabible;

*$[x]=0\to x=\pm\dfrac1{\sqrt 2}$: $x=\color{green}{\dfrac1{\sqrt2}}$ is a solution;

*$[x]=1\to x=\pm1$: $x=\color{green}{1}$ is a solution;

*$[x]=2\to x=\pm\sqrt{\dfrac32}$: from now on, the RHS is too small, no solution anymore.
A: we know $0\leq x-\lfloor x \rfloor <1 $
$$\quad{2x^2-\lfloor x \rfloor-1=0 \to \lfloor x \rfloor=2x^2-1
\\so\\0\leq x-(2x^2-1) <1\to \\
\begin{cases}0\leq x-(2x^2-1) \to & -(x-1)(2x+1)\geq 0 & (*)\\ x-(2x^2-1) <1 \to & x(1-2x)<0 & (**)\end{cases}
\\\begin{cases} (*) \to & x\in[-\frac12,1]\\  (**)\to &x\in (-\infty,0)\cup(\frac12,\infty) \end{cases}\\(*) \cap(*) =[-\frac12,0) \cup (\frac12,1]}$$ now :with respect to $[-\frac12,0) \cup (\frac12,1]$ floor of $x$ can be $\lfloor x \rfloor =-1,0,1$ so
 $$\quad{\lfloor x \rfloor =2x^2-1=-1,0,1 \\\lfloor x \rfloor =2x^2-1=-1 \implies 2x^2=0 \implies x=0 \text{ not acceptable}\\
\lfloor x \rfloor =2x^2-1=0 \implies 2x^2=1 \implies x=- \frac{1}{\sqrt 2} \text{ not acceptable} ,x=+ \frac{1}{\sqrt 2}\text{  acceptable}\checkmark\\
\lfloor x \rfloor =2x^2-1=1 \implies 2x^2=2 \implies x=- 1 \text{ not acceptable} ,x=1\text{  acceptable}\checkmark\\} $$ summary :the equation have two solution $\bf{x=1,\frac{1}{\sqrt 2}}$
