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So \( rac{a + b}{a - b} = rac{3 \pm \sqrt{5}}{2} \). Let’s 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})b2a - (3 + \sqrt{5})a + 2b + (3 + \sqrt{5})b = 0\Rightarrow a(-1 - \sqrt{5}) + b(5 + \sqrt{5}) = 0Solve 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} = 0Let \( r = rac{a}{b} = rac{2x}{y} \), then:r - \sqrt{5} r + 5 + \sqrt{5} = 0 \Rightarrow r(-1 - \sqrt{5}) = - (5 + \sqrt{5})