What is going wrong in this "proof" of $0=1$?

Question

\begin{align} -20 &= -20\\ 16-36 &= 25-45\\ 4^2-4\times 9&=5^2-5\times 9\\ 4^2-4\times 9+81/4&=5^2-5\times 9+81/4\\ 4^2-4\times 9+(9/2)^2&=5^2-5\times 9+(9/2)^2\\ \end{align}

Considering the formula $a^2+2ab+b^2=(a-b)^2$, one has \begin{align} (4-9/2)^2&=(5-9/2)^2\\ \sqrt{(4-9/2)^2}&=\sqrt{(5-9/2)^2}\\ 4-9/2&=5-9/2\\ 4&=5\\ 4-4&=5-4\\ 0&=1 \end{align}

Answer 1

Here let me simplify your process:

$$ \begin{align} (-2)^2 = 4 &\implies \sqrt{(-2)^2} = \sqrt{2^2} \\ &\implies -2 = 2 \\ &\implies -2 + 2 = 2 +2 \\ &\implies 0 = 4 \end{align} $$

QED. Do you see the mistake?

Answer 2

$\sqrt {a^2}\ne a $. You should never have been taught that it does.

Instead $\sqrt {a^2}=|a| $ .

So....

$(4-9/2)^2=(5-9/2)^2$

$\sqrt {(4-9/2)^2}=\sqrt {(5-9/2)^2}$

$|4-9/2|=|5-9/2|$

$4 - 9/2 = \pm 5 \mp 9/2$

$4 = 5$ or $4 = -5 +2* 9/2=-5+9$

$4-4=5-4$ or $4-4 =-5-4+9$

$0=1$ or $0 = 0$