Question

$if x = < 3+\sqrt{8} > \: then \: find \: the \: value \: of \: \frac{ \sqrt{x + 1} }{ \sqrt{x} }$
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Answer 1

Step-by-step explanation:

$x = 3 + \sqrt{8} \\ \frac{ { \sqrt{x + 1} } }{ \sqrt{x} } \\ \frac{ \sqrt{3 + \sqrt{8} + 1 } }{ \sqrt{3 + \sqrt{8} } } \\ = \frac{ \sqrt{4 + \sqrt{8} } }{ \sqrt{3 + \sqrt{8} } } = \frac{ \sqrt{ {2}^{2} + 2 \sqrt{2} } }{ \sqrt{3 + 2 \sqrt{2} } } \\ = \frac{ \sqrt{2(2 + \sqrt{2}) } }{ \sqrt{3 + 2 \sqrt{2} } }$