<s>[INST] Finding all real $a$ such that $16x^4-(a)x^3+(2a+17)x^2−(a)x+16=0$ has four distinct roots in geometric progression 
Determine all real values of the parameter, $a$, for which the equation  $$16x^4−(a)x^3+(2a+17)x^2−(a)x+16=0$$ has exactly four distinct real roots that form a geometric progression?

I noticed that the coefficients are symmetric: namely, the first coefficient is the same as the fifth one, the second is the same as the fourth, and the third is the same as the third.
I don’t know how to proceed using Vieta's formula.
 [/INST] Let's do like derivation of Vieta's formulas rather than the formulas themselves.
Let $b,\,bq,\,bq^2,\,bq^3$ be the roots of $$p(x)=(x-b)(x-bq)(x-bq^2)(x-bq^3)$$ $$\hbox{and }f(x)=x^4−\frac{a}{16}x^3+\frac{2a+17}{16}x^2−\frac{a}{16}x+1.$$
It can be verified that $$p(x)=b^4 q^6 - b^3 (q^3 + q^2 + q + 1) q^3 x + \\
b^2 (q^4 + q^3 + 2 q^2 + q + 1) q x^2 - b (q^3 + q^2 + q + 1) x^3 + x^4$$
But $p(x)$ and $f(x)$ are same degree polynomials with highest degree coefficient $1$ and the same roots $\Rightarrow$ they have the same coefficients.
$$\begin{cases}
b^4 q^6=1\\
16 b^3 (q^3 + q^2 + q + 1) q^3=a\\
16 b^2 (q^4 + q^3 + 2 q^2 + q + 1) q = 2a+17\\
16 b (q^3 + q^2 + q + 1) = a
\end{cases}$$
Hence $a=170$.
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