# Nested radical: disprove

**Algebra**Level 4

(Identify the *incorrect* step.)

Is

\[\sqrt{x + \sqrt{x + \sqrt{x + \ldots }}} = \frac{-1 + \sqrt{4x + 1}}{2}\]?

Here is the proof ( with x > 0)

Let

\(\displaystyle y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots }}}\)

Step *1*:

Multiply "i" on both the side( \( i = \sqrt{-1}\))

\(\displaystyle yi = i\sqrt{x + \sqrt{x + \sqrt{x + \ldots }}}\)

Step *2*:

Take "i" inside the root

\(\displaystyle yi = \sqrt{xi^{2} + i^{2}\sqrt{x + \sqrt{x + \ldots }}}\)

Step *3*:

Take \(i^{2}\) inside the root

\(\displaystyle yi = \sqrt{xi^{2} + \sqrt{xi^{4} + i^{4}\sqrt{x + \ldots }}}\)

Step *4*:

Squaring both the sides

\(\displaystyle -y^{2} = -x + \sqrt{ x + \sqrt{x + \ldots}}\)

Step *5*:

Solving the above quadratic

\(\displaystyle -y^{2} = -x + y\)

\(\displaystyle y^{2} + y - x = 0\)

\[\boxed{ y = \frac{-1 + \sqrt{4x + 1}}{2}}\]( neglecting -ve sign as x > 0)

(Again, identify the *incorrect* step.)

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