MHB Polynomial with five roots: determine the roots of the equation x^5+ax^4+bx^3+cx^2+dx+e=0 as functions of a,d and e

[lfdahl]

I am so sorry for having posted this challenge/puzzle with a serious typo:

The roots of the equation should be functions of $a, d$ and $e$. In my old version I wrote $a, b$ and $e$.

I will see to, that future challenges are properly debugged before posting.For $e \ne 0$, determine the roots of the equation $x^5+ax^4+bx^3+cx^2+dx+e = 0$
as functions of $a, d$ and $e$, given that the equation has two roots whose product is $1$ and
two other roots whose product is $−1$.

 

[lfdahl]

Here´s the suggested solution:

Spoiler

Let the roots be $p, \frac{1}{p}, q, -\frac{1}{q}$ and $r$.

(1). $p \cdot \frac{1}{p} \cdot q \cdot( -\frac{1}{q})\cdot r = -e$ so $r = e$.

(2). $(p + \frac{1}{p}) + (q - \frac{1}{q})+r = -a$ so $(p + \frac{1}{p}) + (q - \frac{1}{q}) = -(a+e)$.

(3). $-\frac{e}{p}-ep-\frac{e}{q}+eq-1 = d$ so $(p + \frac{1}{p}) - (q - \frac{1}{q}) = -\frac{d+1}{e}$.

From this, we get:

$$p+\frac{1}{p} = \frac{1}{2}(-(a+e)-\frac{d+1}{e}) = -\frac{1}{2e}(e^2+ae+d+1) = A,$$

$$q - \frac{1}{q} = \frac{1}{2}(\frac{d+1}{e}-(a+e)) = -\frac{1}{2e}(e^2+ae-d-1) = B.$$

Then $p^2-Ap+1 =0$ or $p = \frac{1}{2}(A\pm \sqrt{A^2-4})$ and $\frac{1}{p}=A-p = \frac{1}{2}(A\mp \sqrt{A^2-4})$.
We may use the plus sign for $p$ and the minus sign for $\frac{1}{p}$.
Similarly, we have $q = \frac{1}{2}(B + \sqrt{B^2+4})$ and $\frac{1}{q} = \frac{1}{2}(B - \sqrt{B^2+4})$.