Let $r$ and $s$ denote the two real roots of $x^2 - x \sqrt{5} + 1 = 0.$  Then determine $r^8 + s^8.$
By Vieta's formulas, $r + s = \sqrt{5}$ and $rs = 1.$  Squaring the equation $r + s = \sqrt{5},$ we get
\[r^2 + 2rs + s^2 = 5,\]so $r^2 + s^2 = 5 - 2rs = 3.$  Squaring this equation, we get
\[r^4 + 2r^2 s^2 + s^4 = 9,\]so $r^4 + s^4 = 9 - 2r^2 s^2 = 9 - 2 = 7.$  Squaring once more, we get
\[r^8 + 2r^4 s^4 + s^8 = 49,\]so $r^8 + s^8 = 49 - 2r^4 s^4 = \boxed{47}.$