Summary: I have two polynomial equations in four variables. How can I find solutions using constructible numbers? Can I use the fact that I know one (or a few closely related) such solutions to find more beyond these?
In my answer on constructing “double” circumscribed pentagons I characterized solutions to a geometric problem by two polynomial equations:
\begin{align*} 0 &= a^2b^2c - a^2b^2d + a^2bd^2 - a^2cd^2 + ac^2d^2 - bc^2d^2 \\&\quad - a^2b - 4abc - b^2c - ac^2 + bc^2 + a^2d + b^2d + 4bcd - ad^2 + cd^2 + a - d \\ 0 &= a^2bc + ab^2c - ab^2d - a^2cd + ac^2d - bc^2d + abd^2 - bcd^2 - ab - ac + bd + cd \end{align*}
Or, if you want to copy and paste them:
0 = a^2*b^2*c - a^2*b^2*d + a^2*b*d^2 - a^2*c*d^2 + a*c^2*d^2 - b*c^2*d^2 - a^2*b - 4*a*b*c - b^2*c - a*c^2 + b*c^2 + a^2*d + b^2*d + 4*b*c*d - a*d^2 + c*d^2 + a - d
0 = a^2*b*c + a*b^2*c - a*b^2*d - a^2*c*d + a*c^2*d - b*c^2*d + a*b*d^2 - b*c*d^2 - a*b - a*c + b*d + c*d
Four different permutations of the roots of $5x^4 - 10x^2 + 1$ solve these equations, e.g.
$$ a = -\sqrt{1+\frac2{\sqrt5}}\qquad b = -\sqrt{1-\frac2{\sqrt5}}\qquad c = \sqrt{1-\frac2{\sqrt5}}\qquad d = \sqrt{1+\frac2{\sqrt5}} $$
This solution corresponds to a geometric arrangement with a regular pentagon. But the question statement I was trying to answer was explicitly looking for a non-regular solution, so this specific solution is not acceptable. At the same time the original question was asking for a construction, so only a solution using constructible numbers would have a chance to satisfy that aspect.
In my original answer I drew a picture using
\begin{align*} a &= -1.2 & 5a+6 &= 0 \\ b &\approx -0.183 & 18032b^4 + 81385b^3 - 68156b^2 - 21040b - 1088 &= 0\\ c &= \phantom+0.5 & 2c-1 &= 0 \\ d &\approx \phantom+1.490 & 78608d^4 + 167620d^3 - 575824d^2 - 21430d + 368293 &= 0 \end{align*}
But the Galois group of those polynomials for $b$ and $d$ is the $S_4$ with order $4!=24$. Since that order is not a power of two, these algebraic numbers are proven not to be constructible. I could not provide a constructible answer, much less an actual construction.
Core question: How can I find a solution where all four variables are distinct constructible numbers, and at least one of them is not a root of $5x^4 - 10x^2 + 1$?
Ideally I would find some parametrization, where I can pick one rational number arbitrarily and derive all four variables from that. But even finding a single different solution should help understand this problem better.
Given how surprisingly hard some similar-looking problems around diophantine equations turn out to be, I wouldn't be surprised at all if this is a fairly hard problem. So I'm also prepared to accept answers providing evidence showing this to indeed be a hard problem. Maybe by showing the existence question to be a reasonably generic instance of a known NP-hard problem, or pointers at literature discussing similar problems.
I know usually I should show my own work for tackling the problem, but I don't have the first idea where to start. I attempted to isolate some additional relationships between the known regular pentagon arrangement, and combine that with the given equations using resultants. But the responses I got tended to either force the regular pentagon arrangement, or quickly lead to some numbers being equal to one another. I can't think of any systematic approach, just this hit and miss guesswork.
One thing that I did find out is that if I use an affine combination of the four permutations of the regular solution, they will all satisfy the second equation. But not the first, and I haven't found a way to leverage this observation.
Part 1.
Given your two equations, eliminating $d$ between them using resultants yields the quartic in $b$ in your other post,
$$ (-a^4c^2 - a^3c^3 + a^2c^4 - a^3c - 5a^2c^2 + 3ac^3 - 5ac + 2c^2 - 1)\,b^4 \\ + (a^4c^3 - 2a^3c^4 + a^4c + 2a^3c^2 - 4a^2c^3 - 3ac^4 + c^5 + 8a^2c - 12ac^2 + c^3 + 3a - 4c)\,b^3 \\ + (a^4c^4 - a^4c^2 + 7a^2c^4 - 3ac^5 - 4a^3c + 4a^2c^2 + 6ac^3 - 3c^4 - 3a^2 + 5ac - c^2)\,b^2 \\ + (a^4c^3 - 5a^3c^4 + 3a^2c^5 + a^4c - 10a^2c^3 + 6ac^4 + a^3 - a^2c - 6ac^2 + 3c^3 - c)\,b \\ + (a^4c^4 - a^3c^5 + 3a^3c^3 - 3a^2c^4 + 3a^2c^2 - 3ac^3 + ac - c^2) = 0 $$
To avoid cube roots, one way to do it is to find $(a,c)$ such that this quartic factors into two quadratics. I don't know your positivity requirements for $(a,c)$, but if sign is not an issue, then two such values are:
I. $(a,c) = (1/2,\,1)\,$ yielding $(b,d)$ as the appropriate roots of,
$$ 3 + 6 b + b^2 = 0\\ 1 - 6 d + 3 d^2 = 0$$
II. $(a,c) = (1/2,\,3)\,$ yielding $(b,d)$ as the appropriate roots of,
$$ -5 + 2 b + b^2 = 0\\ -1 + 2 d + 5 d^2 = 0 $$
since your $d$ from the other post is also just a quadratic. These particular $(a,b,c,d)$ should be constructible.
Part 2. (Added a day later)
We can prove that there are in fact infinitely many constructible $(a,b,c,d)$ such that $a<b<c<d.\,$ The OP's aforementioned system of two equations is,
\begin{align*} 0 &= a^2b^2c - a^2b^2d + a^2bd^2 - a^2cd^2 + ac^2d^2 - bc^2d^2 \\&\quad - a^2b - 4abc - b^2c - ac^2 + bc^2 + a^2d + b^2d + 4bcd - ad^2 + cd^2 + a - d \\ 0 &= a^2bc + ab^2c - ab^2d - a^2cd + ac^2d - bc^2d + abd^2 - bcd^2 - ab - ac + bd + cd \end{align*}
A solution to this was known as permutations of the roots of $5x^4 - 10x^2 + 1=0$, namely,
$$(a,b,c,d) \approx (-1.376381,\; -0.324919,\; 0.324919,\; 1.376381)$$
Note that $c= -b,\;d=-a$. Employing this same relationship satisfies the second equation and the first becomes the simpler,
$$a - a^3 - a^2 b + a^4 b + 3 a b^2 + a^3 b^2 + b^3 - a^2 b^3= 0$$
Using the substitution $a=x,\,b=nx,$ it transforms into the constructible quartic,
$$(n + n^2 - n^3)x^4 - (1 + n - 3n^2 - n^3)x^2 + 1 = 0$$
If $0<n<0.25$ and appropriate root $x$ chosen, then $a<b<c<d.$ For example, let $n=1/5=0.20$ and $x = -\sqrt{\frac{67+12\sqrt6}{29}} \approx -1.823163$ then,
$$(a,b,c,d) \approx (-1.823163,\; -0.364633,\; 0.364633,\; 1.823163)$$
where $n=b/a = 0.20,$ and so on for infinitely many $n$.
P.S. Note that the OP's original answer has $n=b/a = 1/\phi^3 \approx 0.23$ (with golden ratio $\phi$) so falls within the same range of $0<n<0.25$.