e 0\), and \(x

February 22, 2026

$e 0\), and \(x$

Related Articles

Multiply numerator and denominator by \(\sqrt{5} + 1\):
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 \(y
e y\) (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.

Trending Articles