Question

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

Answer 1

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

Step-wise calculation:

1. Let’s rationalise the denominator of x: 

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

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

= `(3 + sqrt(2))^2/7`

2. Expand numerator:

`(3 + sqrt(2))^2`

= `3^2 + 2 xx 3 xx sqrt(2) + (sqrt(2))^2`

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

= `11 + 6sqrt(2)`

Thus, \[ x = \frac{11 + 6\sqrt{2}}{7} \]

3. Now, find x²: 

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

`x^2 = (11 + 6sqrt(2))^2/49` 

4. Expand numerator of x²: 

`(11 + 6sqrt(2))^2`

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

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

= `193 + 132sqrt(2)`

5. Therefore, `x^2 = (193 + 132sqrt(2))/49`.