Evaluate: (√11 + √7)² − (√11 − √7)²

Correct — A.
Expand the first square: (√11 + √7)² = 11 + 7 + 2√77 = 18 + 2√77
Expand the second: (√11 − √7)² = 11 + 7 − 2√77 = 18 − 2√77

Subtract: the 18s cancel, and 2√77 − (−2√77) = 4√77
Shortcut: (a + b)² − (a − b)² = 4ab = 4 × √11 × √7 = 4√77
So the value is 4√77 → option (a)

Why the others are wrong

• (b)
  2√77 — 2√77 is the cross term of one square, not the difference. Subtracting (18 − 2√77) flips its sign, so the cross terms add: 2√77 + 2√77 = 4√77.
• (c)
  6√77 — 6√77 has no source in the working. Each square carries a cross term of ±2√77, so the difference is exactly 2√77 − (−2√77) = 4√77.
• (d)
  8√77 — 8√77 doubles the true value, as if the identity gave 8ab. It gives 4ab: (a + b)² − (a − b)² = 4ab, and 4 × √11 × √7 = 4√77.

Two squaring identities sit behind this: (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².

Subtract the second from the first and a² and b² cancel, leaving (a + b)² − (a − b)² = 4ab. Adding them instead gives 2(a² + b²).

With surds, the product is taken under one root: √11 × √7 = √77, and 77 = 7 × 11 has no square factor to pull out.

Key facts

• (a + b)² − (a − b)² = 4ab
• (a + b)² + (a − b)² = 2(a² + b²)
• √m × √n = √(mn), so √11 × √7 = √77

Common traps

• Treating (√11 − √7)² as 11 − 7, which drops the cross term −2√77
• Subtracting 18 − 2√77 without flipping the sign of −2√77, which leaves 0

No directly related past PYQ was found.