Solve the equation $2x^2-[x]-1=0$ where $[x]$ is biggest integer not greater than $x$.

Question

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?

Answer 1

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 $\bf{x=1,\frac{1}{\sqrt 2}}$

Answer 2

Hint. Note that $x-1<\lfloor x\rfloor \leq x$ implies that $$(2x+1)(x-1)=2x^2-x-1\leq 2x^2-\lfloor x\rfloor-1< 2x^2-(x-1)-1=x(2x-1)$$

Answer 3

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.

1. 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.

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

3. 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 $1$

Answer 4

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.