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So \( rac{a + b}{a - b} = rac{3 \pm \sqrt{5}}{2} \). Letâs take one root (both valid due to symmetry). Set:
rac{a + b}{a - b} = rac{3 + \sqrt{5}}{2}
2(a + b) = (3 + \sqrt{5})(a - b)
2a + 2b = (3 + \sqrt{5})a - (3 + \sqrt{5})b
2a - (3 + \sqrt{5})a + 2b + (3 + \sqrt{5})b = 0
\Rightarrow a(-1 - \sqrt{5}) + b(5 + \sqrt{5}) = 0
Solve for \( rac{a}{b} \):
a b ( -\overline{1 + \sqrt{5}} + (5 + \sqrt{5}) ) = 0 \Rightarrow - rac{a}{b} - \sqrt{5} \cdot rac{a}{b} + 5 + \sqrt{5} = 0
Let \( r = rac{a}{b} = rac{2x}{y} \), then:
r - \sqrt{5} r + 5 + \sqrt{5} = 0 \Rightarrow r(-1 - \sqrt{5}) = - (5 + \sqrt{5})