Evaluate √(12 + √(12 + √(12 + ⋯)))

Correct — B. The expression repeats inside its own first root, so call the whole value x.

The copy inside is x itself: x = √(12 + x)
Square both sides: x² = 12 + x
Rearrange: x² − x − 12 = 0

Factorise: (x − 4)(x + 3) = 0, so x = 4 or x = −3
A square root cannot be negative, so x = 4 → option (b)

Why the others are wrong

• (a)
  5 — 5 would need 5² = 12 + 5, but 25 ≠ 17. The nest that equals 5 is √(20 + √(20 + ⋯)), since 5² − 5 = 20.
• (c)
  6 — 6 would need 6² = 12 + 6, but 36 ≠ 18. It is the value of √(30 + √(30 + ⋯)), since 6² − 6 = 30.
• (d)
  2 — 2 would need 2² = 12 + 2, but 4 ≠ 14. It is also too small: the outer root is at least √12 ≈ 3.46.

An infinite nested radical contains a copy of itself, which turns it into an equation: x = √(a + x), so x² − x − a = 0.

When a = n(n − 1), the positive root is n. Here 12 = 4 × 3, so the value is 4. Partial values climb towards it: √12 ≈ 3.464, √(12 + 3.464) ≈ 3.932, √(12 + 3.932) ≈ 3.991.

Key facts

• √(a + √(a + ⋯)) = x solves x² − x − a = 0, taking the positive root.
• If a = n(n − 1), then √(a + √(a + ⋯)) = n. For a = 12 = 4 × 3, n = 4.
• √(a − √(a − ⋯)) solves x² + x − a = 0. For a = 12 it equals 3.

Common traps

• Keeping x = −3 from the quadratic. A square root is never negative, so only x = 4 fits.
• Mixing it up with √(12 − √(12 − ⋯)), which equals 3, not 4.

No directly related past PYQ was found.