Find the solution of $x^2-4=[x]$

Question

I am able to find the solution by using the help of graph. I know $x^2-4$ will cut $[x]$ only at $-2$ and $2$ and then I am able to find the answer.

I want to know, can we approach this question in any other way like replacing $x$ by $[x]$ and $\{x\}$, where $\{ \ \ \}$ is fractional part of $x$.

Then we can make relation in $[x]$ and $\{x\}$ and don't need to plot the graph roughly.

Answer 1

HINT

$x^2 = [x] + 4 \in \mathbb{N}$

Then:

$x^2 = [x] + 4 \ge x + 3 \Rightarrow x \ge [\ldots, +\infty)$

Also:

$x^2 = [x] + 4 \le x + 4 \Rightarrow x \in [\ldots, \ldots]$

Then you just need to consider $x$ from intersection.

Answer 2

(In this post, I'm assuming we're really talking about the floor of $x$, that is, the greatest integer at most $x$ — which I'm writing as $\lfloor x\rfloor$.)

One way we could solve this algebraically is to notice the floor of $x$ is bounded:

$$x-1\lt\lfloor x\rfloor\le x$$

So if we have $x^2-4=\lfloor x\rfloor$, then $x-1\lt x^2-4\le x$.

Solving the left inequality, we get the solution set:

$$A=\left(-\infty,\frac{1-\sqrt{13}}{2}\right)\cup\left(\frac{1+\sqrt{13}}{2},\infty\right)$$

Solving the right inequality, we get:

$$B=\left[\frac{1-\sqrt{15}}{2},\frac{1+\sqrt{15}}{2}\right]$$

Since these inequalities must be true simultaneously, the solutions are contained in the set:

$$A\cap B=\left[\frac{1-\sqrt{15}}{2},\frac{1-\sqrt{13}}{2}\right)\cup\left(\frac{1+\sqrt{13}}{2},\frac{1+\sqrt{15}}{2}\right]$$

From here, we can notice that the floor of the left region is $-2$ and the floor of the right region is $2$ (since $3<\sqrt{13}<\sqrt{15}<4$). Finding the solution to $x^2-4=-2$ that is in the left region gives us $x=-\sqrt2$; finding the solution to $x^2-4=2$ in the right region is $x=\sqrt6$.