Equation : $16x^{4}-\alpha x^{3}+(2\alpha+17)x^{2}-\alpha x+16=0$

Question

Find the values of the parameter $\alpha$ so

that the following equation :

$P(x)=16x^{4}-\alpha x^{3}+(2\alpha+17)x^{2}-\alpha x+16=0$

has four different real solutions, and they form a

geometric sequence

Actually I don't know how I solve but my try as

following :

let : $x_{1}=b$ , $x_{2}=bq$ , $x_{3}=bq^{2}$ and $x_{4}=bq^{3}$

Now : $P(x)=16(x-b)(x-bq)(x-bq^{2})(x-bq^{3})$

And comparing but I can't find any things!

If any one have any idea please drop here to see

Thanks!

Answer 1

hint

think about the sum and the product of the roots.

Answer 2

Recall that the product of the roots for that polynomial is 16.
Since they are in a geometric series, show they are
$1, 4^{1/3}, 4^{2/3}, 4$.
Now the sum of those roots is the negative of the coefficient of x$^3.$
Be sure to check the results for other possible series are
$-1, 4^{1/3}, -4^{2/3}, 4$ and
$1, -4^{1/3}, 4^{2/3}, -4$.

Answer 3

As you suggested already, we start by equating $$P(x)=16x^4−αx^3+(2α+17)x^2−αx+16\stackrel{!}{=}16(x−b)(x−b q)(x−bq^2)(x−bq^3).$$ Writing it all out this is equivalent to $$0 = 16 - 16 b^4 q^6 + (16 b^3 q^3 (1 + q + q^2 + q^3) - \alpha)x +(17 - 16 b^2 q (1 + q + 2 q^2 + q^3 + q^4) + 2 \alpha)x^2 + (16 b (1 + q + q^2 + q^3) - \alpha)x^3.$$

Therefore we have 4 necessary and sufficient conditions

\begin{align} b^4 q^6 &= 1 \\ \alpha &= 16 b^3 q^3 (1 + q + q^2 + q^3) \\ \alpha &= \frac{1}{2}\left[16 b^2 q (1 + q + 2 q^2 + q^3 + q^4) - 17 \right] \\ % \alpha &= 16 b (1 + q + q^2 + q^3). \end{align}

The second and fourth conditions imply $b^2q^3=1$, so we have the three independent equations

\begin{align} b^2 q^3 &= 1 \\ \alpha &= \frac{1}{2}\left[16 q^{-2} (1 + q + q^2 + q^3 + q^4)-1\right] \\ % \alpha &= 16 q^{-3/2} (1 + q + q^2 + q^3) \end{align}

Frankly speaking I'm not sure how to continue analytically at this point. But maybe this answer was useful anyway. In any case, Mathematica gives 4 solutions for the value of $\alpha$:

$$ \alpha = -6,\qquad \alpha = 2 (-1 + 3 \sqrt{2}),\qquad \alpha = 2 (-1 - 3 \sqrt{2}),\qquad \alpha = 170.$$