1

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.

Blue
  • 83,939
akira akira
  • 11
  • An alternative approach would be to follow the model in this answer. You could express the four roots as a function of $~a,~$ and proceed from there. – user2661923 Jul 29 '24 at 02:46
  • For what it's worth, I attempted to follow the approach suggested in my previous comment, to answer the question. Although this approach is theoretically valid, I found the math to be very ugly, so I abandoned the approach. – user2661923 Jul 30 '24 at 19:36

2 Answers2

1

HINT

Another way to approach it: \begin{align*} 16x^{4} - ax^{3} + (2a + 17)x^{2} - ax + 16 = 0 & \Longleftrightarrow 16\left(x^{2} + \frac{1}{x^{2}}\right) - a\left(x + \frac{1}{x}\right) + 2a + 17 = 0\\\\ & \Longleftrightarrow 16(y^{2} - 2) - ay + 2a + 17 = 0\\\\ & \Longleftrightarrow 16y^{2} - ay + 2a - 15 = 0\\\\ \end{align*}

Can you take it from here?

Átila Correia
  • 18,937
0

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 $a=170$.

Alexey Burdin
  • 5,299