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

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 $x=1-\sqrt 5$.
