# Solution to Practice Problem #1

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:

$\begin{array}{c} \boxed{\,x = 2\,} \end{array}$

$\begin{array}{c} 2 = \sqrt {2 + \sqrt {2 + \sqrt {2 + …} } } \end{array}$