<s>[INST] 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?
 [/INST] 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 $\bf{x=1,\frac{1}{\sqrt 2}}$
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