The polynomial is defined as:
$$ P(X)=\left(1+\left(\alpha+X\right)^2\right)\left(1+\left(\beta-X\right)^2\right)-\Delta. $$
Can I find positive $\alpha$, $\beta$ and $\Delta$ to make $P(X)=0\iff X=A$ for some $A$? $P(X)$ will be $P(X)=\left(X-A\right)^4$
EDIT:
In other words, how to solve for $\alpha$, $\beta$ and $\Delta$ this system of non-linear equalities.
\begin{align} \begin{cases} 2A = \beta-\alpha\\ 6A^2=\beta^2+2-4\alpha\beta+\alpha^2\\ -4A^3=2\alpha\beta^2+2\alpha-2\beta-2\beta\alpha^2\\ A^4=1+\beta^2+\alpha^2+\alpha^2\beta^2-\Delta \end{cases} \end{align}
Substitute $x:={\beta-\alpha\over2}+y$. Then your polynomial $P$ will assume the form $$q(y)=A+By^2+y^4\tag{1}$$ with $$B={1\over2}\bigl(4-(\alpha+\beta)^2\bigr)\ .$$ It follows that choosing $\alpha$ and $\beta$ such that $|\alpha+\beta|\leq2$ will guarantee that $q$ has a unique global minimum at $y=0$. Now choose $\Delta$ such that in $(1)$ we have $A=0$ as well.