Question

2.
if x= 5-2
$if \: x = 5 - 2 \sqrt{6 \: find \: the \:value \: of \: \sqrt{x + 1 \div \sqrt{x} } }$
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Answer 1

Answer:

$\huge \pink{ \sqrt{10} }$

Step-by-step explanation:

Given

$x = 5 - 2 \sqrt{6} \\ = 2 + 3 - 2 \sqrt{2} \sqrt{3} \\ = \sqrt{2} {}^{2} + \sqrt{3} {}^{2} - 2 \sqrt{2} \sqrt{3} \\ = ( \sqrt{3} - \sqrt{2} ) { }^{2} \\ \\ and \\ \\ \frac{1}{x} = \frac{1}{( \sqrt{3} - \sqrt{2}) {}^{2} } \times \frac{( \sqrt{3} + \sqrt{2} ) {}^{2} }{( \sqrt{3} + \sqrt{2}) {}^{2} } \\ \\ \frac{1}{x} = \frac{( \sqrt{3} + \sqrt{2} {}^{2} )}{(3 - 2) {}^{2} } =( \sqrt{3} + \sqrt{2} ) {}^{2}$

Now put this value in expression

$\sqrt{x + \frac{1}{x} } \\ \\ = \sqrt{( \sqrt{3} - \sqrt{2} ) {}^{2} + ( \sqrt{3} + \sqrt{2} ) {}^{2} } \\ \\ = \sqrt{5 - 2 \sqrt{6} +5 + 2 \sqrt{6} } \\ \\ = \sqrt{10}$