When do roots of three quadratic polynomials multiply to 1? Say I have a trio of quadratic polynomials $p_1,p_2,p_3$. Under what conditions will I have $r_1r_2r_3 = 1$ where each $p_i(r_i) = 0$?
In other words, when does the following nonlinear system have a solution?
$$ p_1(r_1) = 0, \ p_2(r_2) = 0, \ p_3(r_3) = 0, \ r_1r_2r_3 = 1 $$

First of all, I know that I can write out the three polynomials, solve them, multiply their coefficients, and set that equal to $1$ for a condition, but this turns out to be quite messy and includes $8$ different cases, corresponding to the choice of root from each polynomial. I'm hoping to find something a bit more elegant, if it exists.
I can see that this is equivalent to asking when there exists $r_1,r_2$ such that $p_3(r_1^{-1}r_2^{-1}) = 0$, or similarly for other combinations. However, this doesn't really change much as far as I can tell. Again, I can solve, invert, multiply, and substitute to get a condition, this time cleaner and with only $4$ cases, but it's still far messier than I'm hoping for.
This problem came up while looking for the conditions under which a bivariate quadratic has a factorization into two bivariate linears. I believe there is an equivalence between these two sets of conditions, so if there is a known condition for that, it should also be sufficient here.
I've poked around with this for quite a while trying to figure it out, but I haven't gotten anywhere. A particular set of equations I was working with is
$$ 6x^2−8x−1=0 \\ y^2−y−6=0 \\ z^2+3z+1=0 \\ xyz = 1 $$
I know that this does not have any solutions, but I can only show it by direct computation of the roots. This one isn't too bad since the $y$ equation has integer roots, but that's obviously not the case in general.
 A: Your question asks if you have three quadratic polynomials
$$ p_1(x)\!:=\!a_1 x^2+b_1 x+c_1, \;\;
   p_2(x)\!:=\!a_2 x^2+b_2 x+c_2, \;\;
   p_3(x)\!:=\!a_3 x^3+b_3 x+c_3 $$
with three pairs of roots
$$ p_1(r_1^+) = p_1(r_1^-) = 0, \quad
   p_2(r_2^+) = p_2(r_2^-) = 0, \quad
   p_3(r_3^+) = p_3(r_3^-) = 0  $$
where for $\,n=1,2,3,\,$
$$ r_n^{\,\pm} := \frac{-b_n\pm\sqrt{b_n^2-4a_n c_n}}{2a_n}, $$
then what is the condition that $\, r_1 r_2 r_3 = 1\,$ for some
choice of the roots as given in terms of the coefficients of the
three polynomials? The answer is given by a homogeneous degree
$12$ polynomial expanded out with $34$ monomial terms
$$ P := (a_1a_2a_3)^4 \prod_{i,j,k=\pm} 
 (1 - r_1^{\,i}\,r_2^{\,j}\,r_3^{\,k}) = 
 (a_1a_2a_3)^4 + \dots + (c_1c_2c_3)^4 $$ where the $\,\dots\,$
represents the other $32$ degree-$12$ monomial terms. I used a
computer algebra system to get the expansion. As stated in the
question

this turns out to be quite messy

and I don't think it can be simplified except for special cases,
but I have been wrong before, so maybe there is hope.
