Question

please help me here​

Answer 1

$HEYA \: \\ \\ \sqrt{2 + \sqrt{3} } \: \: \: + \: \: \: \sqrt{2 - \sqrt{3} } = z \: \: let \: \\ \\ squaring \: on \: both \: sides \: we \: have \\ \\ 2 + \sqrt{3} + 2 - \sqrt{3} + 2 \sqrt{(2 + \sqrt{3} \times (2 - \sqrt{3}) } = z {}^{2} \\ \\ 4 + 2 \sqrt{(2 + \sqrt{3}) \times (2 - \sqrt{3} ) } = z {}^{2} \\ \\ 4 + 2 \sqrt{(2 {}^{2} - ( \sqrt{3}) {}^{2} } = z {}^{2} \\ becoz \: \: (a + b) \times (a - b) = a {}^{2} - b {}^{2} \\ \\ 4 + 2 \sqrt{(1)} = z {}^{2} \\ \\ z {}^{2} = 6 \\ \\ z = \sqrt{6} \\ \\ so \: \: \sqrt{2 + \sqrt{3} } + \sqrt{2 - \sqrt{3} } = \sqrt{6}$

Answer 2

Step-by-step explanation:

$\begin{lgathered}\sqrt{2 + \sqrt{3} } \: \: \: + \: \: \: \sqrt{2 - \sqrt{3} } = z \: \: let \: \\ \\ Squaring \: On \: Both \: Sides \: We \: Have \\ \\ 2 + \sqrt{3} + 2 - \sqrt{3} + 2 \sqrt{(2 + \sqrt{3} \times (2 - \sqrt{3}) } = z {}^{2} \\ \\ 4 + 2 \sqrt{(2 + \sqrt{3}) \times (2 - \sqrt{3} ) } = z {}^{2} \\ \\ 4 + 2 \sqrt{(2 {}^{2} - ( \sqrt{3}) {}^{2} } = z {}^{2} \\ \\ Because \\ \: \: (a + b) \times (a - b) = a {}^{2} - b {}^{2} \\ \\ 4 + 2 \sqrt{(1)} = z {}^{2} \\ \\ z {}^{2} = 6 \\ \\ z = \sqrt{6} \\ \\ so \: \: \sqrt{2 + \sqrt{3} } + \sqrt{2 - \sqrt{3} } = \sqrt{6}\end{lgathered}$