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water heater replacement cost 50 gallon - Deep Underground Poetry

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 r = -rac{(5 - \sqrt{5})(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = -rac{5\sqrt{5} + 5 - 5 - \sqrt{5}}{5 - 1} = -rac{4\sqrt{5}}{4} = -\sqrt{5}.

📌 Thus, \( r = \sqrt{5} \) or \( r = -\sqrt{5} \). These correspond to pairs \((x, y)\) such that \(x = \pm \sqrt{5} y\), with \(y

📌 e 0\), and \(x

📌 e y\) (to avoid division by zero in the original expression).

📌 Check that \(x = \sqrt{5} y\):

📌 Then \( rac{x+y}{x-y} = rac{\sqrt{5}+1}{\sqrt{5}-1} = rac{3+\sqrt{5}}{2} \), valid.

📌 Similarly for \(x = -\sqrt{5} y\):

📌 \( rac{-\sqrt{5}+1}{-\sqrt{5}-1} = rac{\sqrt{5}-1}{\sqrt{5}+1} = rac{(\sqrt{5}-1)^2}{5 - 1} = rac{6 - 2\sqrt{5}}{4} \), but we earlier derived it correctly as \( rac{3 - \sqrt{5}}{2} \), so it checks.

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