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 answer x = -1, the correct answer is:

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

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