Roots of a polynomial of degree 4

[Mathman23]

(*)$$p(x) = x^4 + ax^3 + bx^ 2 + ax + 1 = 0$$

where $$a,b \in \mathbb{C}$$

I would like to prove that a complex number x makes (*) true iff

$$s = x + x^{-1}$$ is a root of the $$Q(s) = s^2 + as + (b-2)$$

I see that that $$Q(x + x^{-1}) = \frac{p(x)}{x^2}$$

Then to prove the above do I then show that p(x) and Q((x + ^{-1}) shares roots?

Sincerely Yours
MM23

 

[arildno]

Yes. Note that Q(s) and p(x) must have coincident roots.

 

[HallsofIvy]

(*)$$p(x) = x^4 + ax^3 + bx^ 2 + ax + 1 = 0$$

where $$a,b \in \mathbb{C}$$

I would like to prove that a complex number x makes (*) true iff

$$s = x + x^{-1}$$ is a root of the $$Q(s) = s^2 + as + (b-2)$$

I see that that $$Q(x + x^{-1}) = \frac{p(x)}{x^2}$$

Then to prove the above do I then show that p(x) and Q((x + ^{-1}) shares roots?

Sincerely Yours
MM23

The onlly way a fraction can be 0 is if the numerator is 0. Isn't it obvious from $$Q(x + x^{-1}) = \frac{p(x)}{x^2}$$
that Q(x+ x^-1)= 0 if and only if p(x)= 0?