Given that $w = e^{(2\pi/5)i}$, show that $w + \overline{w}$ is a root > of $z^2 + z - 1 = 0$ Hoping to get some help with the following problem

Given that $w = e^{(2\pi/5)i}$, show that $w + \overline{w}$ is a root
  of $z^2 + z - 1 = 0$.

I've noticed that $w + \overline{w} = 2Re(w)$, but I'm not sure on how to proceed from there. $2\pi/5$ is a weird angle, so substituting a cosine doesn't seem to make much sense.

$w + \overline{w} = 2\cos (2\pi/5)$, and $(2\cos (2\pi/5))^2 + 2\cos (2\pi/5) - 1 = 4\left(\dfrac{1+\cos (4\pi/5)}{2}\right) + 2\left(2\cos^2(\pi/5) - 1\right) - 1 = 2\left(1-\cos (\pi/5)\right) + 4\cos^2(\pi/5) - 2 - 1 = 4\cos^2(\pi/5) - 2\cos (\pi/5) - 1 = S$. 
We have: $\cos 36^\circ = \sin 54^\circ$ gives: $1-2x^2 = 3x - 4x^3$ with $x = \sin 18^\circ = \sin (\pi/10)$. Thus: $4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$. Since $0 < x < 1$, $4x^2 + 2x - 1 = 0$. Thus : $\cos (\pi/5) = 1 - 2x^2 = 1 - \dfrac{1-2x}{2} = \dfrac{1+2x}{2}$, and $\cos^2(\pi/5) = \left(\dfrac{1+2x}{2}\right)^2 = \dfrac{1+4x+4x^2}{4} = \dfrac{1+4x+1-2x}{4} = \dfrac{1+x}{2}$. Thus: $S = 4\left(\dfrac{1+x}{2}\right) - 2\left(\dfrac{1+2x}{2}\right) - 1 = 0$.
