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Taking the square root at the end leaves you with two solutions (+/-). The one shown is the wrong one, the other one will lead to exactly what we started with.
You know what, it's impossible to solve this formula, because #9 really doesn't exist. That's the problem we have in this World, we have made it imperfect. Number 9 is a clown & a joker number.
This is a good one, I've read it like five times and still haven't noticed anything wrong.
But to answer an imperfect answer to your formula, 3.14141414....
You see, tell your collective audience that you will always get different answers....#9 is an illusion number that never is us, imperfect human that brought about this Clown number, the Misleading number into existence. E= MC2N2D4 Absolute Relativity....not a theory E=MC2
+Ondřej Hruška should have used brain before posting ... :-) if the outcome is impossible, there's a multiplication by zero somewhere ...
amir yb
In the 7th line it should be (+/) (3-x)  = (+/) (pi-x).  Then, using opposite possibilities for the (+/-) on both sides of the equation returns the first line as a value for a x. 
Titus L
→ 3-x = x-π 【not 3-x=3-π】
∵ 3<x<π
[ ∵ 3< (3+π)/2 <π ] ^_^
The error occurs when you remove the powers of 2.  The reason for this is that [abs(x)]^2=x^2, but abs(x) might not equal x.

Example (-2)^2 = 2^2, but -2 doesn't equal 2.
You see how nice we all get along.
the step removing power is wrong. 
1. 3.14 >x > 3.07  (because of 1st step)
2. so when we removing power,
 (3-x)^2 becomes x-3,  and (pi-x)^2 becomes pi -x
3. so, x-3 = pi -x 
4. 2x = pi +3 
5. x = pi+3 / 2
Edited to fix confusing grammar.

Just to elaborate, because +Kevin Battisti is correct.

sqrt(x^2) does not equal x.  As such,

sqrt(x^2) = abs(x).  

So that means the second last line should read

abs(3 - x) = abs(pi - x)

This results in two possibilities.

3 - x = pi - x

OR (this being the key word)

3 - x = - (pi - x)

It is the second possibility that applies here as solving it just gets

x = (pi + 3) / 2 which is just how x was defined in the first place.
+Mike Aben clearly you didn't understand the math that I presented or the point I was trying to make. In a mathematical proof of contradiction, all I need is one example of a contradiction (which I provided). 
+Kevin Battisti Sorry, when I was saying this is correct, the "this" was what you said.  I was providing the elaboration for others.

I understood what you said completely.
You have chosen the wrong solution when took the square roots.
The other one is
2x=pi + 3
the cause for error is simple. The wrong square root is taken. -(sqrt(LHS))=+(sqrt(RHS)), which returns the original expression. +Alexander N. Benner  is correct
it's not allowed to add x² on both sides of an equation.Only multiply or divide.
Good one, it took me a little while. The problem is the removing of the squares
So abs(x-3)=abs(3-x)=abs(pi-x).
I guess it doesn't matter. They both have the same value.
So there is no proof at all that pi=3, but x is halfway between 3 and pi.
i think the square root should have been replaced with the step of moving the (pi-x)^2 to the LHS and difference of 2 squares.
I think it is like this he drew out a conclusion that pie=3 then pie-3=0 so he multiplied wid 0 on both sides....
if you ask wolfram alpha if (3-x)^2-(pi-x)^2=(3-x)-(pi-x), its comes back with false, so the mistake must be in the power reduction.
+Andre Wemans actually i am pretty sure u can add and subtract to both sides of an equation, think about completing the square.
All you can infer from x^2 = y^2 is that EITHER x= y OR x=- y. The last step is fallacious.
I've probably just been sucked in by trolls. 
X = (pi + 3)/2 > 3
So where you have:
(3 - x)^2 = (pi - x)^2
The LHS: 3 - x is actually negative.
So by taking it's square root you need to take into account an imaginary part. 
Cutely a ÷0 where 0 = π-3.
(The other solution is the original.)
Where in the proof did we establish that 2πx = 9?  

It just sort of came out of nowhere on the left side of step 5.  

Then we used this baseless assumption to switch out the 9 on the right side of step 5 with 2πx.
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