Since the pattern repeats indefinitely, we can use substitution. This is an example of a nested radical.

\[\begin{array}{c}

x = \sqrt {2 + \sqrt {2 + \sqrt {2 + …} } }
\end{array}\]

So, the part within the parentheses below is also equal to *x*.

\[\begin{array}{c}

x = \sqrt {2 + \left( {\sqrt {2 + \sqrt {2 + \sqrt {2 + …} } } } \right)}
\end{array}\]

Using substitution we get:

\[\begin{array}{c}

x = \sqrt {2 + x}
\end{array}\]

Next, square both sides:

\[\begin{array}{c}

{x^2} = 2 + x
\end{array}\]

Rearrange:

\[\begin{array}{c}

{x^2} – x – 2 = 0
\end{array}\]

Factor:

\[\begin{array}{c}

\left( {x – 2} \right)\left( {x + 1} \right) = 0
\end{array}\]

We get two possible answers:

\[\begin{array}{c}

x = 2\,\,\,\,\,\,x = \, – 1
\end{array}\]

Eliminating the erroneous result x = -1, the correct answer is:

\[\begin{array}{c}

\boxed{\,x = 2\,}
\end{array}\]

\[\begin{array}{c}

2 = \sqrt {2 + \sqrt {2 + \sqrt {2 + …} } }
\end{array}\]