$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong

Here's what your "proof" would look like correcting all the errors. As you can see, it's not nearly as impressive as a proof that 2+2=5.
$$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= -\sqrt{(4-\frac92)^2} +\frac92\\
&= -\sqrt{16 -2\times4\times\frac92 +(\frac92)^2} + \frac92\\
&= \left(-\sqrt{16 -36 + (\frac92)^2}\right) +\frac92\\
&= \left(-\sqrt {-20 +(\frac92)^2}\right) + \frac92\\
&= -\sqrt{25-45 +(\frac92)^2} +\frac92\\
&= -\sqrt {5^2 -2\times5\times\frac92 + (\frac92) ^2} + \frac92\\
&= -\sqrt {(5-\frac92)^2} +\frac92\\
&= -5 + \frac92 + \frac92 \\
&= -5+9\end{align}$$
For reference, the most serious mistake was in the 2nd line. In general, it's not true that $\sqrt{x^2} = x$, but rather $\sqrt{x^2} = |x|$. For $x=4-\frac92<0$, you need to keep track of the extra minus sign coming from the absolute value. Other than that, there were some obvious typos that I've corrected.
