Question

if x is equal to root 3 + root 2 upon root 3 - root 2 and Y is equal to root 3 - root 2 upon root 3 + root 2 find the value of x square + Y square
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Answer 1

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ {x}^{2} + {y}^{2} = ( \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } )^{2} + ({ \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } })^{2}$

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

${(a + b)}^{4} = {a}^{4} + 4 {a}^{3} b + 6 {a}^{2} {b}^{2} + 4a {b}^{3} + {b}^{4} \\ {( \sqrt{3} + \sqrt{2} )}^{4} = { (\sqrt{3} )}^{4} + 4 {( \sqrt{3} )}^{3} ( \sqrt{2} )+ 6 {( \sqrt{3} )}^{2} {( \sqrt{2}) }^{2} + 4 (\sqrt{3}) {( \sqrt{2} )}^{3} + { \sqrt{2} }^{4} \\ = 9 + (4 \times 3 \sqrt{3} \times \sqrt{2} + (6 \times 3 \times 2) + (4 \times \sqrt{3} \times 2 \sqrt{2} ) + 4 \\ = 9 + 12 \sqrt{6} + 36 + 8 \sqrt{6} + 4 \\ = 49 + 20 \sqrt{6}$

${(a - b)}^{4} = {a}^{4} - 4 {a}^{3} b + 6 {a}^{2} {b}^{2} - 4a {b}^{3} + {b}^{4} \\ {( \sqrt{3} - \sqrt{2} )}^{4} = { (\sqrt{3} )}^{4} - 4 {( \sqrt{3} )}^{3} ( \sqrt{2} )+ 6 {( \sqrt{3} )}^{2} {( \sqrt{2}) }^{2} - 4 (\sqrt{3}) {( \sqrt{2} )}^{3} + {( \sqrt{2}) }^{4} \\ = 9 - (4 \times 3 \sqrt{3} \times \sqrt{2} ) + (6 \times 3 \times 2) - (4 \times \sqrt{3} \times 2 \sqrt{2} ) + 4 \\ = 9 - 12 \sqrt{6} + 36 - 8 \sqrt{6} + 4 \\ = 49 - 20 \sqrt{6}$

$= {( \sqrt{3} + \sqrt{2} ) }^{4} + {( \sqrt{3} - \sqrt{2} ) }^{4} \\ = (49 + 20 \sqrt{6} ) + (49 - 20 \sqrt{6} ) \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ = 49 + 49 \\ = 98$

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