Question

Write x^2 − 2x − 3 = 0 in the form (x − a)^2 = b, where a and b are integers.

Answer 1

X²-2x-3 = 0     let's start by grouping the "x"s

(x² - 2x + [?])² - 3 = 0

so, we have a missing fellow there, in order to make the group, a perfect square trinomial, namely to "complete the square", hmmm so the tell-tale fellow is the middle term.

from a perfect square trinomial we know that the middle term is a product of 2 and the "term on the left" and the "term on the right", like 

$\bf \begin{array}{cccccllllll} a^2& + &2 a b&+& b^2\\ \downarrow && &&\downarrow \\ a&& && b\\ &\to &( a + b)^2&\leftarrow \end{array}\qquad \qquad \quad \begin{array}{cccccllllll} a^2& - &2 a b&+& b^2\\ \downarrow && &&\downarrow \\ a&& && b\\ &\to &( a - b)^2&\leftarrow \end{array}$

$\bf \textit{therefore}\qquad 2(x)(\boxed{?})=\stackrel{middle~term}{2x}\implies \boxed{?}=\cfrac{2x}{2x}\implies \boxed{?}=1$

aha!!  so our missing fellow is 1.

now, let's keep in mind that all we're doing is borrowing from our very good friend Mr Zero, 0.  So if we add 1², we also have to subtract 1².

(x² - 2x + 1² - 1²) - 3 = 0

(x² - 2x + 1) -1 -3 =0

(x - 1)² - 4 = 0

(x - 1)² = 4