Question

$if \: x = \binom{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } and \: y = \binom{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } find \: {x}^{2} + {y}^{2}$
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Answer 1

GIVEN :-

• x =[( √3+√2)/(√3-√2)]
• y = [( √3 -√2)/(√3+√2)]

TO FIND :-

• The value of x² + y².

SOLUTION :-

value of x,

$\tt{x = \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } }\\ \\ \tt{on \: \: rationalising \: \: the \: \: denominator } \\ \\ \tt{ \implies \dfrac{ (\sqrt{3} + \sqrt{2} )( \sqrt{3} + \sqrt{2}) }{( \sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} )}} \\ \\ \implies \dfrac{( \sqrt{3} + \sqrt{2} {)}^{2} }{( \sqrt{3} {)}^{2} - ( \sqrt{2} {)}^{2} } \\ \\ \implies \tt \dfrac{( \sqrt{3} {)}^{2} + ( \sqrt{2} {)}^{2} + 2( \sqrt{3})( \sqrt{2} ) }{3 - 2} \\ \\ \implies \tt \dfrac{3 + 2 + 2 \sqrt{6} }{1} \\ \\ \implies \tt \red {5 + 2\sqrt{6}}$

Value of y,

$\tt{y = \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } }\\ \\ \tt{on \: \: rationalising \: \: the \: \: denominator } \\ \\ \tt{ \implies \dfrac{ (\sqrt{3} - \sqrt{2} )( \sqrt{3} - \sqrt{2}) }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} )}} \\ \\ \implies \dfrac{( \sqrt{3} - \sqrt{2} {)}^{2} }{( \sqrt{3} {)}^{2} - ( \sqrt{2} {)}^{2} } \\ \\ \implies \tt \dfrac{( \sqrt{3} {)}^{2} + ( \sqrt{2} {)}^{2} - 2( \sqrt{3})( \sqrt{2} ) }{3 - 2} \\ \\ \implies \tt \dfrac{3 + 2 - 2 \sqrt{6} }{1} \\ \\ \implies \tt \red {5 - 2\sqrt{6}}$

• value of x²

$\tt {x}^{2} = (5 + 2 { \sqrt{6} })^{2} \\ \\ \tt = [( {5}^{2}) + 2 \times 5 \times 2 \sqrt{6} + ( 2 \sqrt{6 } ) {}^{2} ] \\ \\ \tt = (25 + 20 \sqrt{6 } + 24) \\ \\ \large{ \boxed{\tt = 49 + 20 \sqrt{6} }}$

• value of y²

$\tt {x}^{2} = (5 - 2 { \sqrt{6} })^{2} \\ \\ \tt = [( {5}^{2}) - 2 \times 5 \times 2 \sqrt{6} + ( 2 \sqrt{6 } ) {}^{2} ] \\ \\ \tt = (25 - 20 \sqrt{6 } + 24) \\ \\ \large{ \boxed{\tt = 49 - 20 \sqrt{6} }}$

• value of x² + y²

$\implies \bf \: 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} \\ \\ \implies \bf \: 49 + 49 + 20 \sqrt{6} - 20 \sqrt{6} \\ \\ \large{\boxed{\tt \implies \: 98}}$

Hence the value of x²+ y² is 98.