If x = 2 + √3, find x − 1⁄x.

Correct — B. Find 1⁄x by rationalising: multiply top and bottom by the conjugate 2 − √3.

1⁄x = (2 − √3) ⁄ [(2 + √3)(2 − √3)]
Denominator = 2² − (√3)² = 4 − 3 = 1
So 1⁄x = 2 − √3

x − 1⁄x = (2 + √3) − (2 − √3)
= 2 − 2 + √3 + √3 = 2√3 → option (b)

Why the others are wrong

• (a)
  √3 — √3 counts the root once. Subtracting 2 − √3 turns its −√3 into +√3, so the roots add: √3 + √3 = 2√3.
• (c)
  3√2 — 3√2 ≈ 4.24 is larger than x = 2 + √3 ≈ 3.73. Since 1⁄x is positive, x − 1⁄x must be smaller than x, and no √2 arises anywhere.
• (d)
  5√3 — 5√3 ≈ 8.66 is far above x ≈ 3.73. Subtracting the positive number 1⁄x from x must leave less than x, so this cannot be the value.

When a² − b = 1, the reciprocal of a + √b is simply its conjugate a − √b, because (a + √b)(a − √b) = a² − b. Here 2² − 3 = 1.

That makes x and 1⁄x easy to combine:
x + 1⁄x = 4, the roots cancel
x − 1⁄x = 2√3, the whole numbers cancel

Higher powers follow: x² + 1⁄x² = (x + 1⁄x)² − 2 = 16 − 2 = 14.

Key facts

• (2 + √3)(2 − √3) = 4 − 3 = 1, so 2 + √3 and 2 − √3 are reciprocals.
• For x = 2 + √3: x + 1⁄x = 4 and x − 1⁄x = 2√3.
• x² − 1⁄x² = (x + 1⁄x)(x − 1⁄x) = 4 × 2√3 = 8√3.

Common traps

• Writing 1⁄(2 + √3) as 1⁄2 + 1⁄√3: a reciprocal does not split across a sum.
• Opening −(2 − √3) as −2 − √3, which makes the answer 0.

No directly related past PYQ was found.