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Home » Bigger Drawers, Bigger Savings—See How This Bed Solves Your Storage Problem! - Silent Sales Machine

Bigger Drawers, Bigger Savings—See How This Bed Solves Your Storage Problem! - Silent Sales Machine

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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Case 2:** \( rac{r+1}{r-1} = rac{3 - \sqrt{5}}{2} \)
Similarly:
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 + 5 - \sqrt{5} = 0
r = rac{\sqrt{5} - 5}{-1 + \sqrt{5}} = rac{-(5 - \sqrt{5})}{\sqrt{5}...
Multiply numerator and denominator by \(\sqrt{5} + 1\):
r = -rac{(5 - \sqrt{5})(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}...
Thus, \( r = \sqrt{5} \) or \( r = -\sqrt{5} \). These correspond to p...

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