Question

If `x = (3 + sqrt(2))/(3 - sqrt(2)), y = (3 - sqrt(2))/(3 + sqrt(2))`, find x² + y² + xy.

Answer 1

Given: `x = (3 + sqrt(2))/(3 - sqrt(2)), y = (3 - sqrt(2))/(3 + sqrt(2))`

We are to find x² + y² + xy.

Stepwise calculation:

1. Rationalise x and y:

`x = (3 + sqrt(2))/(3 - sqrt(2)) xx (3 + sqrt(2))/(3 + sqrt(2))`

= `(3 + sqrt(2))^2/((3)^2 - (sqrt(2))^2`

= `(9 + 6sqrt(2) + 2)/(9 - 2)`

= `(11 + 6sqrt(2))/7`

Similarly for y:

`y = (3 - sqrt(2))/(3 + sqrt(2)) xx (3 - sqrt(2))/(3 - sqrt(2))`

= `(3 - sqrt(2))^2/(9 - 2)`

= `(9 - 6sqrt(2) + 2)/7`

= `(11 - 6sqrt(2))/7`

2. Calculate x²:

`x^2 = ((11 + 6sqrt(2))/7)^2`

= `((11)^2 + 2 xx 11 xx 6sqrt(2) + (6sqrt(2))^2)/49`

= `(121 + 132sqrt(2) + 72)/49`

= `(193 + 132sqrt(2))/49`

3. Calculate y²:

`y^2 = ((11 - 6sqrt(2))/7)^2`

= `(121 - 132sqrt(2) + 72)/49`

= `(193 - 132sqrt(2))/49`

4. Calculate xy:

`xy = ((3 + sqrt(2))/(3 - sqrt(2))) xx ((3 - sqrt(2))/(3 + sqrt(2)))`

xy = 1

5. Sum up x² + y² + xy:

x² + y² + xy

= `(193 + 132sqrt(2))/49 + (193 - 132sqrt(2))/49 + 1`

= `(193 + 132sqrt(2) + 193 - 132sqrt(2))/49 + 1`

= `386/49 + 1`

= `386/49 + 49/49`

= `435/49`