Question

Q.if x=root 3+root 2 divided by root 2- root 3 and if y=root 3 - root 2 divided by root 2 + root 3 then find the value of x square +y square - 10 xy.

Answer 1

$x = \sqrt{3} + \sqrt{2} \div \sqrt{2} - \sqrt{3}$
$= ( \sqrt{3} + \sqrt{2} )( \sqrt{2} + \sqrt{3} ) \div (\sqrt{2} - \sqrt{3} )( \sqrt{2} + \sqrt{3} )$
${( \sqrt{3} + \sqrt{2} ) }^{2} \div { \sqrt{2} }^{2} - { \sqrt{3} }^{2}$
$= (3 + 2 + 2 \sqrt{6} ) \div 2 - 3$
$= (5 + 2 \sqrt{6} ) \div - 1$
$= - (5 + 2 \sqrt{6} )$
${x}^{2} = {( - (5 + 2 \sqrt{6})) }^{2}$
${ (- 5 - 2 \sqrt{6}) }^{2}$
$= 25 + 24 + 20 \sqrt{6}$

now,
$y = \sqrt{3} - \sqrt{2} \div \sqrt{2} + \sqrt{3}$
$= ( \sqrt{3} - \sqrt{2} )( \sqrt{2} - \sqrt{3} ) \div ( \sqrt{2} + \sqrt{3} )( \sqrt{2} - \sqrt{3} )$
$= \sqrt{6} - 3 - 2 - \sqrt{6} \div { \sqrt{2} }^{2} - { \sqrt{3} }^{2}$
$= - 5 \div 2 - 3$
$= - 5 \div - 1$
$= 5$
now,
$10xy =10 ( \sqrt{3} + \sqrt{2} \div \sqrt{2} - \sqrt{3} ) \times ( \sqrt{3} - \sqrt{2} \div \sqrt{2} + \sqrt{3} )$
$= 10( { \sqrt{3} }^{2} - { \sqrt{2} }^{2} ) \div ( { \sqrt{2} }^{2} - { \sqrt{3} }^{2} )$
$= 10(3 - 2) \div (2 - 3)$
$10(1 \div - 1)$
$= 10 \times - 1$
$= - 10$
now,
x^2+y^2-10xy=
24+25+20root6+5-(-10)
=49+20root6+5+10
=49+15+20root6
=64+20root6

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