Nested radical: disprove

(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

$\displaystyle y = \sqrt{x + \sqrt{x + \sqrt{x + \ldots }}}$

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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)

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(Again, identify the incorrect step.)