What is the value of √(6 + √(6 + √(6 + ⋯)))

Correct — A. The expression repeats forever, so call it x. What sits under the first root after the 6 is x again.

Set up: x = √(6 + x)
Square both sides: x² = 6 + x
Rearrange: x² − x − 6 = 0
Factor: (x − 3)(x + 2) = 0

A square root cannot be negative, so x ≠ −2
x = 3 → option (a)

Why the others are wrong

• (b)
  2 — 2 answers the minus version, √(6 − √(6 − ⋯)), where x² = 6 − x gives (x + 3)(x − 2) = 0. With plus signs, √(6 + 2) = √8 ≈ 2.83, not 2.
• (c)
  5 — √(6 + 5) = √11 ≈ 3.32, not 5. The value must satisfy x = √(6 + x), and 5 does not.
• (d)
  6 — 6 is the number under the root, not the value. √(6 + 6) = √12 ≈ 3.46, so x = 6 fails x = √(6 + x).

An infinite nested radical equals itself one layer down. Naming the whole value x turns the endless expression into a quadratic: x = √(6 + x), so x² − x − 6 = 0.

The quadratic has two roots, 3 and −2. The negative one is rejected because a square root is never negative.

Shortcut: if the number is k(k + 1), the plus version equals k + 1 and the minus version equals k. Here 6 = 2 × 3, giving 3 with plus signs and 2 with minus signs.

Key facts

• x = √(n + √(n + ⋯)) solves x² = n + x.
• If n = k(k + 1), √(n + √(n + ⋯)) = k + 1 and √(n − √(n − ⋯)) = k.
• A square root is never negative, so the negative root of the quadratic is rejected.

Common traps

• Swapping the plus and minus results. With plus signs the answer is the larger of the pair (3), with minus signs the smaller (2).
• Keeping x = −2 from the factorisation. A square root cannot be negative.

No directly related past PYQ was found.