Infinite nested radical evaluation: If x^2 = 6 + √(6 + √6 + √6 + … to infinity), find one value of x.

Introduction / Context:
Infinite nested radicals often stabilize to a constant that satisfies a self-referential equation. Recognizing this allows us to solve without approximations. Here the expression under the square root repeats identically, enabling a one-variable equation for the inner radical value, and then for x.

Given Data / Assumptions:

• x^2 = 6 + √(6 + √6 + √6 + …)
• Square roots are principal (nonnegative) values.

Concept / Approach:
Let S = √(6 + √6 + …), the infinite nested radical. Because the nesting is infinite and identical, S satisfies S = √(6 + S). Square both sides to eliminate the radical and solve for S. Then compute x from x^2 = 6 + S.

Step-by-Step Solution:

Verification / Alternative check:
Check: If x = 3, then x^2 = 9 = 6 + 3 = 6 + S, consistent. The question asks for one value; 3 is the standard principal value.

Why Other Options Are Wrong:

• 6, 5, 4, 2: None equal a valid value of x consistent with x^2 = 9 for this setup.

Common Pitfalls:
Setting S = 6 + S (forgetting the square root) or using S = 6 + √S. The correct self-reference is S = √(6 + S).

Final Answer:
3