What is (x sqrt{n sqrt{n sqrt{n ldots}}})?

Consider the expression:

(x sqrt{n sqrt{n sqrt{n ldots}}})

This infinite nested radical can be expressed as:

(x sqrt{n x})

To eliminate the square roots, we can square both sides of the equation:

(x^2 n x)

Rearranging this, we get a quadratic equation:

(x^2 - x - n 0)

Solving this quadratic equation using the quadratic formula:

(x frac{-b pm sqrt{b^2 - 4ac}}{2a})

where (a 1), (b -1), and (c -n).

Substitute these values into the quadratic formula:

(x frac{1 pm sqrt{1 4n}}{2})

Since (x) must be non-negative, we take the positive root:

(x frac{1 sqrt{1 4n}}{2})

Thus, the value of (sqrt{n sqrt{n sqrt{n ldots}}}) is:

(boxed{frac{1 sqrt{1 4n}}{2}})

Replacing (n) with a specific number for clarity

Let's replace (n) with 2 for simplicity:

(x sqrt{2 sqrt{2 sqrt{2 ldots}}})

Let (x_2 x). Squaring both sides, we get:

(x_2^2 2 x_2)

The quadratic equation is:

(x_2^2 - 2 x_2 0)

Factoring out (x_2), we get:

(x_2 (x_2 - 2) 0)

The factors for the quadratic equation are (x_2 2) and (x_2 0). Since (x) must be non-negative, we take:

(x_2 2)

Using the shortcut, the factor for such square roots can be written as:

(x frac{1 sqrt{1 4n}}{2})

Generalizing the Solution

To find a solution, we observe that: If there is a solution (x), then it must satisfy (x geq 0) And (x^2 n x)

This translates to:

(x frac{1 pm sqrt{1 4n}}{2})

The existence of a solution depends on the value of (n). For (n > 0), a valid solution exists:

(x frac{1 sqrt{1 4n}}{2})

When (n 0), (x 0) is a valid solution, but it is not typically considered the primary solution in this context.