The given equation is symmetric, which means its roots come in reciprocal pairs. We want all four roots to be in geometric progression, which leaves us with a form of 1/r^3, 1/r, r, r^3.
So then the quartic with those roots is (x-1/r^3)*(x-1/r)*(x-r)*(x-r^3)
= x^4 - (r^3+r+1/r+1/r^3)x^3 + ((r^3+1/r^3)*(r+1/r)+2)x^2 - (r^3+r+1/r+1/r^3)x + 1 = 0.
Divide the original equation through by 16 and then equate coefficients to get a/16 = r^3+r+1/r+1/r^3
and (2a+17)/16 = (r^3+1/r^3)*(r+1/r)+2
I'm going to substitute q=r+1/r. One thing to note is that r is real only when q>=2.
Then we have a/16 = q^3-2q and (2a+17)/16 = q^4-3q^2+2
Combining these equations we can form a quartic in q: q^4-2q^3-3q^2+4q+15/16 = 0. This has four real roots q=5/2, q=-3/2, q=1/2+1/sqrt(2), and q=1/2-1/sqrt(2). However we need q>=2 for r to come out as a real value so that discards most of the roots leaving only q=5/2.
With q=5/2 then r = 2 or 1/2 and a = 16*((5/2)^3-2*(5/2)) = 170. Both choices of r describe the same geometric sequence differing only by ascending or descending.
The roots are 8, 2, 1/2, and 1/8 of the equation 16x^4-170x^3+357x^2-170x+16=0.
Solution
