Question

If x = root 5-2/root 5+2 and y = root 5-2/root 5 +2 ; find: [4]
(i) x2
(ii) y2 (iii) xy (iv) x2
+ y2
+ xy.

Answer 1

Step-by-step explanation:

Hey there !!

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Given, x = ( √5 – 2 ) / ( √5 + 2 ) and y = ( √5 + 2 ) / ( √5 – 2 )

x = ( √5 – 2 ) / ( √5 + 2 )

= ( √5 – 2 ) / ( √5 + 2 ) × ( √5 – 2 ) / ( √5 – 2 )

= ( √5 – 2 )² / [ (√5)² – (2)² ]

= ( 9– 4√5 ) / ( 5 – 4 )

= 9 – 4 √5

y = ( √5 + 2 ) / ( √5 – 2 )

= ( √5 + 2 ) / ( √5 – 2 ) × ( √5 + 2 ) / ( √5 + 2 )

= ( √5 + 2 )² / [ (√5)² – (2)² ]

= ( 9 + 4√5 ) / ( 5 – 4 )

= 9 + 4√5

Now, x² + xy + y²

= ( 9 – 4√5 )² + ( 9 – 4√5)( 9 + 4√5 ) + ( 9 + 4√5 )²

= ( 81 – 72√5 + 80 ) + [ 9( 9 + 4√5 ) –4√5( 9 + 4√5 ) ] + ( 81 + 72√5 + 80 )

= ( 161 – 72√5 ) + ( 81 + 13√5 – 13√5 – 80 ) + ( 161 + 72√5 )

= 161 – 72√5 + 1 + 161 + 72√5

= 161 + 1 + 161

= 323

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Hope it's satisfactory. (^-^)