Question

√(√(7+4√3)-√(7-4√3))

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,

$= > \sqrt{ \sqrt{7 + 4 \sqrt{ 3} } - \sqrt{7 - 4 \sqrt{ 3} } }$

Add/Subtract,

${ \sqrt{3} }^{2} = 3$

Therefore,,,,

$= > \sqrt{(7 + 4 \sqrt{3} + { \sqrt{3} }^{2} - 3) \sqrt{7 - 4 \sqrt{3 } } } \\ \\ \\ = > \sqrt{ { \sqrt{3} }^{2} + 4 \sqrt{3} + 4 - \sqrt{7 - 4 \sqrt{3} } }$

$= > \sqrt{( { \sqrt{3} }^{2} + 2 \sqrt{3}) + (2 \sqrt{3} + 4) - \sqrt{7 - 4 \sqrt{3} } } \\ \\ \\ = > \sqrt{ \sqrt{3} ( \sqrt{3} + 2) + 2( \sqrt{3} + 2) - \sqrt{7 - 4 \sqrt{3} } }$

$= > \sqrt{( \sqrt{3} + 2)( \sqrt{3} + 2) - \sqrt{7 - 4 \sqrt{3} } } \\ \\ \\ = > \sqrt{ {( \sqrt{3} + 2) }^{2} - ( \sqrt{7 - 4 \sqrt{3} )} }$

$= > \sqrt{ {( \sqrt{3 } + 2 })^{2} - 7 - 4 \sqrt{3} + { \sqrt{3} }^{2} - 3 } \\ \\ \\ = > \sqrt{ {( \sqrt{3} + 2) }^{2} - { \sqrt{3} }^{2} - 4 \sqrt{3} + 4} \\ \\ \\ = > \sqrt{ ( { \sqrt{3} + 2})^{2} - ( { \sqrt{3} }^{2} - 2 \sqrt{3} ) + ( - 2 \sqrt{3} + 4) }$

$= > \sqrt{ \sqrt{3} + 2 - \sqrt{3} ( \sqrt{3 } - 2) - 2( \sqrt{3} - 2)} \\ \\ \\ = > \sqrt{ \sqrt{3} + 2 - ( \sqrt{3} - 2)( \sqrt{3} - 2) } \\ \\ \\ = > \sqrt{ \sqrt{3} + 2 - {( \sqrt{3} }^{2} - 2) } \\ \\ \\ = > \sqrt{ \sqrt{3} + 2 - {(2 - \sqrt{3}) }^{2} }$

$= > \sqrt{ \sqrt{3} + 2 - (2 - \sqrt{3}) } \\ \\ \\ = > \sqrt{ \sqrt{3} + 2 - 2 + \sqrt{3} } \\ \\ \\ = > \sqrt{2 - 2 + \sqrt{3} + \sqrt{3} } \\ \\ \\ = > \sqrt{2 - 2 + 2 \sqrt{3} } \\ \\ \\ = > \sqrt{2 \sqrt{3} }$

Apply Radical method rule,,, that is,,,

$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b} \\ \\ Assuming \: \: that \: \: \: \: \: \: \: \: \: \: = > a \geqslant 0 \: \: ,b$

Therefore,,,,,

$= > \sqrt{2} \sqrt{ \sqrt{3} } \\ \\ \\ = > \sqrt{2} ( { {3}^{ \frac{1}{2} }) }^{ \frac{1}{2} }$

$= > \sqrt{2} \times {3}^{ \frac{1}{2} \times \frac{1}{2} } \\ \\ \\ = > \sqrt{2} \times {3}^{ \frac{1}{4} }$

Since,,,,,

$= > \frac{1}{ {a}^{n} } = \sqrt[n]{a}$

THEREFORE,,,,,

$= > > > \sqrt{2} \sqrt[4]{3}$

Which is the required solution for this query.

HOPE THIS HELPS AND CLEARS YOUR DOUBTS!!!!