Rationalizing trick: If x = (√3 + 1)/(√3 − 1) and y = (√3 − 1)/(√3 + 1), compute x^2 + y^2.

Introduction / Context:
This tests recognition that x and y are reciprocals when formed as complementary conjugate ratios. From xy = 1, compute x^2 + y^2 efficiently using the identity (x + 1/x)^2 = x^2 + 2 + 1/x^2.

Given Data / Assumptions:

• x = (√3 + 1)/(√3 − 1)
• y = (√3 − 1)/(√3 + 1)
• All radicals are principal.

Concept / Approach:
First, observe x·y = 1 (they are reciprocals). Compute S = x + 1/x directly using conjugates, then obtain x^2 + y^2 = x^2 + 1/x^2 = S^2 − 2.

Step-by-Step Solution:
Let a = √3, b = 1x = (a + b)/(a − b), 1/x = (a − b)/(a + b)S = x + 1/x = [(a + b)^2 + (a − b)^2]/(a^2 − b^2)= [a^2 + 2ab + b^2 + a^2 − 2ab + b^2]/(a^2 − b^2)= [2a^2 + 2b^2]/(a^2 − b^2) = 2(a^2 + b^2)/(a^2 − b^2)With a^2 = 3, b^2 = 1 ⇒ S = 2(3 + 1)/(3 − 1) = 8/2 = 4Thus x^2 + y^2 = S^2 − 2 = 16 − 2 = 14

Verification / Alternative check:
Numerically compute x and y, square, and sum to confirm 14.

Why Other Options Are Wrong:
They reflect arithmetic slips in evaluating S or in subtracting 2 to obtain x^2 + 1/x^2.

Common Pitfalls:
Forgetting that x·y = 1, which enables the shortcut; otherwise the computation becomes messy.

Final Answer:
14