Question

If `x = (sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2))` and `y = (sqrt(3) - sqrt(2))/(sqrt(3) + sqrt(2))`, then find the value of x² + y².

Answer 1

`x = (sqrt(3) + sqrt(2))/(sqrt(3) - sqrt(2))` and `y = (sqrt(3) - sqrt(2))/(sqrt(3) + sqrt(2))`

(a + b)² = a² + 2ab + b²

Also `x = 1/y` or `y = 1/x`

Let a = x

b = y

(x + y)² = x² + 2xy + y²

But we know `y = 1/x`

`(x + 1/x)^2 = x^2 + 1/x^2 + 2 xx x xx 1/x`

`(x + 1/x)^2 = x^2 + 1/x^2 + 2`

`x^2 + 1/x^2 = (x + 1/x)^2 - 2`

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

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

= `((3 + 2sqrt(6) + 2 - 3 - 2sqrt(6) + 2)/((sqrt(3) - sqrt(2)) xx (sqrt(3) + sqrt(2))))^2 - 2`

Here the denominators form the expansion as

(a + b) × (a – b) = (a² – b²)

Here `a = sqrt(3)`

`b = sqrt(2)`

`a^2 = (sqrt(3))^2`

= 3

`b^2 = (sqrt(2))^2`

= 2

= `(10/(3 - 2))^2 - 2`

= 10² – 2

= 100 – 2

= 98