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oxed{rac{5 - \sqrt7}9} - Esdistancia

📅 March 17, 2026 👤 scraface

Mar 17, 2026

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So \( u = rac{3 + \sqrt{5}}{2} \) or \( u = rac{3 - \sqrt{5}}{2} \).
Now recall \( u = rac{x+y}{x-y} \). Let \( r = rac{x}{y} \), assumin...
e 0\). Then:
rac{x+y}{x-y} = rac{r+1}{r-1}.
Set this equal to each value of \(u\):
Case 1:** \( rac{r+1}{r-1} = rac{3 + \sqrt{5}}{2} \)
2(r+1) = (3 + \sqrt{5})(r - 1)
2r + 2 = (3 + \sqrt{5})r - (3 + \sqrt{5})
2r + 2 = (3 + \sqrt{5})r - 3 - \sqrt{5}
2r - (3 + \sqrt{5})r + 2 + 3 + \sqrt{5} = 0

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