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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.
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Solution
`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
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