TITLE: 2 different patterns when solving seemingly same equation (Completing the square)
QUESTION [0 upvotes]: I'm working on my math skills and I encountered this problem that puzzles me. Let me give you an example of two equations and its workings:
$4x^2 - 36x + 81$
$4\cdot (x^2 - 9x) +81$
** $4\cdot(x^2 - 9/2x + 81/4) + 81 - 4\cdot 81/4$ **
$4\cdot(x - 9/2)^2 + 81 -81$
$4\cdot(x - 9/2)^2 + 0$
and the other one which is very similar:
$2x^2 + 3x - 2$
$2\cdot(x^2 + 3/2x) - 2$
** $2\cdot (x^2 + 3/2x + 9/16) - 2 - 2\cdot 9/16$ **
$2\cdot(x + 3/4)^2 - 2 - 9/8$
$2\cdot(x + 3/4)^2 - 25/8$
I marked the two lines that are confusing me and where I keep making the same mistake. Why in the first example I square numerator and denominator but in the second example I also square numerator but then multiply the denominator by 2 and then square it ? What gives ? What I'm missing ?

REPLY [2 votes]: \begin{align*}
(x - 9/2)^2 = x^2 - (9/2)x - (9/2)x + 81/4 &= x^2 - 9x + 81/4  \\
    &\neq x^2 - (9/2)x + 81/4  \text{.}
\end{align*}
So going from line 3 to line 4 in your first display is unjustifiable.
Given $x^2 - ax$, replace with 
$$  \rlap{\overbrace{\phantom{x^2 - (a/2)x - (a/2)x +(a/2)^2}}^{\text{binomial}}}x^2 - \underbrace{(a/2) x - (a/2) x}_{-a x} + \underbrace{(a/2)^2 - (a/2)^2}_{0} = (x-(a/2))^2 - (a/2)^2  \text{.}  $$
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