$\sqrt {(3 + \sqrt 5 )} $ is equal to
$\sqrt 5 + 1$
$\sqrt 3 + \sqrt 2 $
$(\sqrt 5 + 1)/\sqrt 2 $
${1 \over 2}(\sqrt 5 + 1)$
${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $
${{\sqrt {6 + 2\sqrt 3 + 2\sqrt 2 + 2\sqrt 6 } - 1} \over {\sqrt {5 + 2\sqrt 6 } }}$
The equation $\sqrt {(x + 1)} - \sqrt {(x - 1)} = \sqrt {(4x - 1)} $, $x \in R$ has
${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} - {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }} = $
${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $