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

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 $1$


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


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.

Tags:

Algebra Precalculus

Quadratics

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