Question

$\sqrt{5 \sqrt{5 \sqrt{5 \sqrt{5 } } } }$
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Answer 1

$\huge\tt\green{Solution}$

$\sqrt{5\sqrt{5 \sqrt{5 \sqrt{5} }}} \\ \\ Let\: x = \sqrt{5\sqrt{5\sqrt{5} } } \\ \\ \sf\implies\: x = (\sqrt{5 + x}) \\ \\ Squaring \: both \: sides \\ \\ \sf\implies\: {x}^{2} = {(\sqrt{5 + x})}^{2} \\ \\ \sf\implies\: {x}^{2} = 5 + x \\ \\ \sf\implies\: {x}^{2} - x - 5 =0 </p><p>\\ \\ using \: squaring\: method \\ \\ {x}^{2} - 2x × \frac{1}{2} = 5 \\ \\ Adding {(\frac{1}{2})}^{2} on \: both\: sides \\ \\ {x}^{2} - 2x × \frac{1}{2} + {(\frac{1}{2})}^{2} = 5 + {(\frac{1}{2})}^{2} \\ \\ {(x - \frac{1}{2})}^{2} = 5 + \frac{1}{4} \\ \\ {(x - \frac{1}{2})}^{2} = \frac{21}{4} \\ \\ x - \frac{1}{2} = \sqrt{\frac{21}{4}} \\ \\ x - \frac{1}{2} = \frac{\sqrt{21}}{2} \\ \\ Taking \: + \\ \\ x = \frac{\sqrt{21}}{2} + \frac{1}{2} \\ \\x = \frac{\sqrt{21} + 2}{2}\\ \\ x = \sqrt{21} \\ \\ taking \: - \\ \\ x = -\frac{\sqrt{21}}{2} + \frac{1}{2} \\ \\ x = \frac{-\sqrt{21} + 2}{2} \\ \\ x = -\sqrt{21}$