Find number of solutions of $(x-1)^2+\lceil x \rceil=4$

Question

Find number of solutions of $$(x-1)^2+\lceil x \rceil=4$$

I could find that there are no integral solutions since , for $x$ an integer we have

$$(x-1)^2+x=4$$

$$x^2-x-3=0$$ which has Non integer roots.

How about Non integer solutions without graphing?

Answer 1

Since $x\le \lceil x\rceil\lt x+1$, we have to have $$x\le 4-(x-1)^2\lt x+1$$ giving $$x\in\bigg[\frac{1-\sqrt{13}}{2},-1\bigg)\cup\bigg(2,\frac{1+\sqrt{13}}{2}\bigg]$$ which implies $\lceil x\rceil=-1,3$.

For $\lceil x\rceil =-1$, we have $(x-1)^2=4-(-1)\implies x=1\pm\sqrt 5$. But $x=1+\sqrt 5$ does not satisfy $\lceil x\rceil =-1$.

For $\lceil x\rceil=3$, we have $(x-1)^2=4-3\implies x=0,2$. Both do not satisfy $\lceil x\rceil =3$.

Therefore, the only solution is $x=1-\sqrt 5$.

Answer 2

It's easy to check that $-2 < x < 3$. Now consider each of the cases $i < x \le i+1$, $i=-2 \ldots 2$. In the interval $i < x \le i+1$, $(x-1)^2 + \lceil x \rceil = (x-1)^2 + i+1$, which is monotonic in $x$ with limits $(i-1)^2 +i+1$ as $x \to i$ and $i^2+i+1$ at the right. Thus your function values are in the intervals $[3,8)$, $[1,4)$, $[1,2)$, $(2,3]$ and $(4,7]$ respectively for $i=-2,-1,0,1,2$. The only case where $4$ is in the interval is $i=-2$, so there is exactly one solution.