Is there an algebraic closure for the quaternions? This post is a sequel of: 
Is the set of quaternions $\mathbb{H}$ algebraically closed? 
This answer shows that:
1.  $\mathbb{H}$ is algebraically closed for the polynomials of the form $\sum a_r x^r$
2.  It is not for the polynomials freely generated by $\mathbb{H}$ and $x$, because $xi+ix-j$ has no root.
Question: Is there an algebraic closure (for the case 2)?
If so:   What does it look like?   What's its dimension over $\mathbb{H}$? What's its matrix representations?  
 A: It looks like the answer is morally "no." Now, there is a formal closure for which we can solve free polynomials:  As in the case of fields, you take an inductive limit. Let $\Bbb H=R$ be our normed division algebra. Then
$$\overline{R}=\varinjlim_{[L:R]<\infty} L$$
where the inductive system is taken relative to inclusions of algebra extensions $L/R$ of finite dimension over $R$, each of the form
$$L_p=R\{x\}/(p(x))$$
where $p(x)$ is irreducible over $R$ and $R\{x\}$ is the polynomials freely generated as in case ($2$). This certainly has the required property that all polynomials in $R$ have a root in $\overline{R}$, and any other such object has a copy of this inside of it for purely formal reasons.
I note that the directed system so-defined is indeed a directed system--in fact a lattice--so this should go through unless I'm missing something obvious.
rschweib has noted that the result is no longer a division algebra, so this is really not ideal, but the "algebraic closure" property holds, and necessarily it's a minimal ring where this property can hold, so it seems this is the best we can hope for. However we also cannot force algebraicness of the result since $R\{x\}/(xi+ix-j)$ doesn't make $x$ algebraic appropriately in the sense that you want to mimic the field case's excellent definition that algebraicness means $F(\alpha)/F$ is finite dimensional as an algebra over $F$, which doesn't hold in this setting.
A: I don't think that there can be an associative $\Bbb{R}$-algebra $L$, containing $\Bbb{H}$ as a subring, such that the equation
$$xi+ix=j\qquad(1)$$
has a solution $x\in L$.
Multiplying $(1)$ by $i$ from the left gives us
$ixi+i^2x=ij$, or $ixi-ij=-i^2x$. As $i^2=-1$ and $ij=k$, this reads
$$
x=ixi-k.\qquad(2)
$$
On the other hand multiplying $(1)$ by $i$ from the right gives us
$xi^2+ixi=ji$, and using $i^2=-1, ji=-k$ this yields
$$
x=ixi+k.\qquad(3)
$$
The equations $(2)$ and $(3)$ together imply $k=-k$. As $k$ is a unit of $L$ this implies that $2=0$ in $L$, so $L$ cannot be an extension of $\Bbb{H}$.
