Question

if x is equal to root 3 + root 2 divided by root 3 minus root 2 Y is equal to root 3 minus root 2 / root 3 + root 2 find the value of x squared plus y squared minus xy

Answer 1

ans kitta h???????????????

Answer 2

Hey friend,
Here is the answer you were looking for:

x= √3 + √2 / √2 - √2

y = √3 - √2 / √3 + √2
$on \: rationalizing \: we \: get \\ \\ x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ x = \frac{ {( \sqrt{3}) }^{2} + ( { \sqrt{2} )}^{2} + 2 \times \sqrt{3} \times \sqrt{2} }{( {3})^{2} - ( {2})^{2} } \\ \\ x = \frac{3 + 2 + 2 \sqrt{6} }{3 - 2} \\ \\ x = 5 + 2 \sqrt{6} \\ \\ similarly \\ on \: rationalizing \: \\ y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{3} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ y = \frac{( { \sqrt{3} })^{2} + ( { \sqrt{2} )}^{2} - 2 \times \sqrt{3} \times \sqrt{2} }{( { \sqrt{3}) }^{2} - ( { \sqrt{2} )}^{2} } \\ \\ y = \frac{3 + 2 - 2 \sqrt{6} }{3 - 2} \\ \\ y = 5 - 2 \sqrt{6} \\ \\ {x}^{2} + {y}^{2} - xy \\ \\ = ( {5 + 2 \sqrt{6} )}^{2} + ( {5 - 2 \sqrt{6} )}^{2} \\ - (5 + 2 \sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ = (( {5})^{2} + ( {2 \sqrt{6} )}^{2} + 2 \times 5 \times 2 \sqrt{6}) \\ + (( {5})^{2} + ( {2 \sqrt{6} })^{2} - 2 \times 5 \times 2 \sqrt{6} ) \\ - (( {5})^{2} - ( {2 \sqrt{6} )}^{2} ) \\ \\ = (25 + 24 + 20 \sqrt{6} ) + (25 + 24 - 20 \sqrt{6} ) \\ - (25 - 24) \\ \\ = (49 + 20 \sqrt{6} ) + (49 - 20 \sqrt{6} ) - 1 \\ \\ = 49 + 20 \sqrt{6} + 49 - 20 \sqrt{6} - 1 \\ \\ = 98 - 1 \\ \\ = 97$

Hope this helps!!!

@Mahak24

Thanks...
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