Question

if x=√3+√2/√3-√2 and y=√3-√2/√3+√2, then find the value of x2+y2-10xy

Answer 1

$\bold{ x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } } \\ \\ \\ \mathbf{ \underline{By \: rationalization}} \\ \\ \\ \bold{ \: x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \: \: \: \: \: \: \: \: \: \rightarrow x = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \: \: \: \: \: \: \: \: \: \rightarrow \: x = 5 + 2 \sqrt{6} } \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$

$\bold{y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } } \\ \\ \\ \bold{ \: \underline{ By \: rationalization}} \\ \\ \\ \bold{y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }\times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \: \: \: \: \: \: \: \: \: \rightarrow \: y = \frac{3 + 2 - 2 \sqrt{6}}{3 - 2} \: \: \: \: \: \: \: \: \rightarrow \: y = 5 - 2 \sqrt{6} \: \: \: \: \: \: \: ...(ii)}$


$\bold{ \: xy \: = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \: \: \: \: \: \: \: \: \rightarrow \: xy = 1 \: \: \: \: \: \: \: \: \: \: ...(iii) }$





$\bold{ \underline{ Putting the value(s) from ( i ) , ( ii ) ( iii ) ,}}$



x^2 + y^2 - 10xy


=> ( 5 + 2√6 )^2 + ( 5 - 2√6 )^2 - 10( 1 )


=> 25 + 24 + 20√6 + 25 + 24 - 20√6 - 10


=> 2( 25 + 24 ) - 10


=> 2( 25 + 24 - 5 )


=> 2( 44 )


=> 88

Answer 2

$\text {Your answer !!}$

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ rationalising \: the \: denominator.... \\ \\ = > x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > x = \frac{( { \sqrt{3} + \sqrt{2} )}^{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ = > x = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2} \\ \\ = > x = \frac{3 + 2 + 2 \sqrt{6} }{1} \\ \\ = > x = 5 + 2 \sqrt{6} \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ rationalising \: the \: denominator.... \\ \\ = > y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ = > y = \frac{ {( \sqrt{3} - \sqrt{2} ) }^{2} }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ = > y = \frac{( { \sqrt{3} ) }^{2} + {( \sqrt{2} ) }^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2} \\ \\ = > y = \frac{3 + 2 - 2 \sqrt{6} }{1} \\ \\ = > y = 5 - 2 \sqrt{6} \\ \\ = > xy = 5 + 2 \sqrt{6} \times 5 - 2 \sqrt{6} \\ \\ = > xy = 5(5 - 2 \sqrt{6} ) + 2 \sqrt{6} (5 - 2 \sqrt{6} ) \\ \\ = > xy = 25 - 10 \sqrt{6} + 10 \sqrt{6} - 24 \\ \\ = > xy = 25 - 24 \\ \\ = > xy = 1 \\ \\ = > {x}^{2} + {y}^{2} - 10xy \\ \\ = > {(5 + 2 \sqrt{6} )}^{2} + {(5 - 2 \sqrt{6} )}^{2} - 10 \times 1 \\ \\ = > {(5)}^{2} + {(2 \sqrt{6} )}^{2} + 2 \times 5 \times 2 \sqrt{6} + {(5)}^{2} + {(2 \sqrt{6} )}^{2} - 2 \times 5 \times 2 \sqrt{6} - 10 \\ \\ = > 25 + 24 + 20 \sqrt{6} + 25 + 24 - 20 \sqrt{6} - 10 \\ \\ = > 49 + 49 - 10 \\ \\ = > 98 - 10 \\ \\ = > 88$

$\textbf {Hence, 88 is your answer !!}$

$\text { Thanks !! }$