(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.)