If x = √6 + 2 and y = √6 − 2, then what is the value of (x⁄y + y⁄x)² − 3?

Correct — D. Do not divide the surds. Use x⁄y + y⁄x = (x² + y²)⁄(xy).

xy = (√6 + 2)(√6 − 2) = 6 − 4 = 2

x + y = 2√6, so
x² + y² = (x + y)² − 2xy
= 24 − 4 = 20

x⁄y + y⁄x = 20⁄2 = 10

10² − 3 = 100 − 3 = 97 → option (d)

Why the others are wrong

• (a)
  22 — 22 is 5² − 3, and 5 is (x² + y²)⁄(2xy) — the average of x⁄y and y⁄x, not their sum. The 2 does not belong in the denominator.
• (b)
  35 — 35 needs (x⁄y + y⁄x)² = 38, which is not a perfect square. With xy = 2 and x² + y² = 20 that bracket is the whole number 10.
• (c)
  42 — 42 needs (x⁄y + y⁄x)² = 45, so x⁄y + y⁄x would be 3√5. The values here make it rational, and equal to 10.

x and y are conjugate surds, chosen so that the two quantities a symmetric expression needs are both small whole numbers.

xy = (√6)² − 2² = 2 by the difference of squares, and x + y = 2√6, so x² + y² = (x + y)² − 2xy = 24 − 4 = 20.

Any expression that is unchanged when x and y swap places can be rebuilt from those two numbers, and x⁄y + y⁄x = (x² + y²)⁄(xy) is exactly such an expression.

Rationalising x⁄y and y⁄x separately reaches the same 10, but through two surd multiplications instead of none.

Key facts

• x⁄y + y⁄x = (x² + y²)⁄(xy) for any non-zero x and y.
• With x = √6 + 2 and y = √6 − 2, xy = (√6)² − 2² = 2.
• x² + y² = (x + y)² − 2xy = (2√6)² − 4 = 20.
• So x⁄y + y⁄x = 10 and the expression equals 10² − 3 = 97.

Common traps

• Rationalising each fraction separately, which is slower and invites sign slips.
• Using (x² + y²)⁄(2xy), the average, in place of the sum.
• Subtracting 3 before squaring, which gives 49.

No directly related past PYQ was found.