Question

if x = root 3 + root 2/ root 3 - root 2 nd y =3 root - root 2 /root 3 = root 2 find the value of x square y square +xy

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

On Rationalizing the denominator we get,

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

Using the identities:

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$


$x = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2} )}^{2} + 2( \sqrt{3})( \sqrt{2} ) }{ {( \sqrt{3} )}^{2} - {( \sqrt{2}) }^{2} } \\ \\ x = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ x = 5 + 2 \sqrt{6} \\ \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator we get,

$y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

Using the identities:

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$

$y = \frac{ {( \sqrt{3}) }^{2} + {( \sqrt{2}) }^{2} - 2( \sqrt{3})( \sqrt{2} )}{ {( \sqrt{3} )}^{2} - {( \sqrt{2} )}^{2} } \\ \\ y = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ y = 5 - 2 \sqrt{6}$

Now,

Bro I guess you have forgot a sign in between x sq and y sq

So, for that I can solve putting both + and -

Here we go:

${x}^{2} + {y}^{2} + xy \\ \\ {(5 + 2 \sqrt{6} )}^{2} + {(5 - 2 \sqrt{6} )}^{2} + (5 + 2 \sqrt{6} )( 5 - 2 \sqrt{6} )$


Using same identities:


$( {(5)}^{2} + {(2 \sqrt{6}) }^{2} + 2(5)(2 \sqrt{6} )) + ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} - 2(5)(2 \sqrt{6})) + (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ = (25 + 24 + 20 \sqrt{6} ) + (25 + 24 - 20 \sqrt{6} ) + ( {(5)}^{2} - {(2 \sqrt{6} )}^{2} ) \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} + (25 - 24) \\ \\ = 98 + 1 \\ \\ = 99$

And if the question is,


${x}^{2} - {y}^{2} + xy \\ \\ = (5 + 2 \sqrt{6} )^{2} - (5 - 2 \sqrt{6} )^{2} + (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ = ( {(5)}^{2} + {(2 \sqrt{6} )}^{2} + 2(5)(2 \sqrt{6} )) - ( {(5)}^{2} + {(2 \sqrt{6}) }^{2} - 2(5)(2 \sqrt{6} )) - ( {(5)}^{2} - {(2 \sqrt{6}) }^{2} ) \\ \\ = (25 + 24 + 20 \sqrt{6} ) - (25 + 24 - 20 \sqrt{6} ) - (25 - 24) \\ \\ = 25 + 24 + 20 \sqrt{6} - 25 - 24 + 20 \sqrt{6} + 1 \\ \\ = 40 \sqrt{6} + 1$

Hope you have got your answer !!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺

Answer 2

Answer:

Value of x² + y² + xy is 99.

Step-by-step explanation:

Given: $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\:\:and\:\:y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

First,

$xy=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$

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

Now,

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

$=\frac{3+2-2\sqrt{3}\sqrt{2}}{3+2+2\sqrt{3}\sqrt{2}}=\frac{5-2\sqrt{6}}{5+2\sqrt{6}}$

$=\frac{5-2\sqrt{6}}{5+2\sqrt{6}}\times\frac{5-2\sqrt{6}}{5-2\sqrt{6}}=\frac{(5-2\sqrt{6})^2}{(5+2\sqrt{6})(5-2\sqrt{6})}=\frac{25+24-20\sqrt{6}}{25-24}$  

= 49 - 20√6

Now,

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

$=\frac{3+2+2\sqrt{3}\sqrt{2}}{3+2-2\sqrt{3}\sqrt{2}}=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{5+2\sqrt{6}}{5-2\sqrt{6}}\times\frac{5+2\sqrt{6}}{5+2\sqrt{6}}$

$=\frac{(5+2\sqrt{6})^2}{(5+2\sqrt{6})(5-2\sqrt{6})}=\frac{25+24+20\sqrt{6}}{25-24}$  

= 49 + 20√6

Finally,

x² + y² + xy

= ( 49 - 20√6 ) + ( 49 + 20√6 ) + 1

= 49 + 49 + 1 = 99

Therefore, Value of x² + y² + xy is 99.