Question

if x=√3-√2÷√3+√2 and y= √3+√2÷√3-√2 find x^2+y^2+xy

Answer 1

X= (√3+√2)/(√3+√2)
=(√3-√2)^2/(√3^2 - √2^2)
=(3-2√6+2) / 1
=5 - 2√6
Similarly,
Y=5+2√6
Therefore, X+Y= 5-2√6+5+2√6
=10
And, XY= (5-2√6) (5+2√6)
=5^2 - (2√6)^2
=25-24
=1
Now,
X^2 + Y^2 + XY
= X^2 + Y^2 + 2XY - XY
=(X+Y)^2 - XY
=(10)^2 - 1
= 100-1
=99
answers is 99