Question

if x = $\frac{ \sqrt{ \sqrt{6} + 2 } + \sqrt{ \sqrt{6} - 2 } }{ \sqrt{ \sqrt{6} + \sqrt{2} } }$

Find x^2​

Answer 1

if x = $\frac{ \sqrt{ \sqrt{6} + 2 } + \sqrt{ \sqrt{6} - 2 } }{ \sqrt{ \sqrt{6} + \sqrt{2} } }$

Answer 2

Step-by-step explanation:

Given:

$\frac{ \sqrt{5 + 2 \sqrt{6} } + \sqrt{5 - 2 \sqrt{6} } } { \sqrt{5 + \sqrt[]{6} } - \sqrt{5 - 2 \sqrt{6} } } = \sqrt{ \frac{x}{2} }$

To find:

$x {}^{2}$

Solution:

By squaring both sides we get,

${x}^{2} = \frac{( \sqrt{ \sqrt{6} + 2} + \sqrt{ \sqrt{6} - 2) {}^{2} } }{ ( \sqrt{ \sqrt{6} + 2) {}^{2} } } \\ \\ \Longrightarrow {x}^{2} = \frac{( \sqrt{ \sqrt{6} + 2 } ) {}^{2} + 2 \sqrt{ \sqrt{6 + 2} \times \sqrt{ \sqrt{6 - } }2 } }{ \sqrt{6} + \sqrt{2} }$

$= \frac{ \sqrt{6} + 2 + \sqrt{6} - 2 + 2 \sqrt{( \sqrt{6} + 2)( \sqrt{6} - 2} }{ \sqrt{6} + \sqrt{2} }$

$= \frac{2 \sqrt{6} + 2 \sqrt{( \sqrt{6) {}^{2} - {2}^{2} } } }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{2 \sqrt{6} + 2 \sqrt{6 - 4} }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{2 \sqrt{6} + 2 \sqrt{2} }{ \sqrt{6} + \sqrt{2} } \\ \\ = \frac{2 \sqrt{2} ( \sqrt{3 + 1)} }{ \sqrt{2}( \sqrt{3} + 1) } \\ \\ = 2$

$\Longrightarrow$ x^2 = 2.