Question

if x=√3+1/√3-1. y=√3-1/√3+1. find value of x2 +y2 + xy

Answer 1

Answer :

Given that,

x = (√3 + 1)/(√3 - 1)

and y = (√3 - 1)/(√3 + 1)

So, x + y

= (√3 + 1)/(√3 - 1) + (√3 - 1)/(√3 + 1)

= {(√3+1)(√3+1)+(√3-1)(√3-1)}/(√3-1)(√3+1)

= (3 + 2√3 + 1 + 3 - 2√3 + 1)/(3 - 1)

= 8/2

= 4

So, (x + y)²

= 4²

= 16

and xy

= (√3 + 1)/(√3 - 1) × (√3 - 1)/(√3 + 1)

= (3 - 1)/(3 - 1)

= 2/2

= 1

∴ x² + y² + xy

= (x + y)² - 2xy + xy

= (x + y)² - xy

= 16 - 1

= 15

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Answer 2

x=1/y
Therefore
xy=1
Rationalising x and y
x=2+root3, y=2-root3
x2=7+4root3,y2=7-4root3
Therefore
X2+y2+xy=14+1=15
This is required answer