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College Algebra · Chapter 9

The Missing Corner

Factoring only works when a quadratic is willing to be factored, and most are not. Completing the square works every single time — but the instruction “take half of b and square it” sounds like a magic spell until you see where it comes from. It comes from an actual square. Build one below: lay x² + bx out as an area, cut the strip in half, and watch a small empty corner appear. That corner is the constant. Then step through five equations and solve them.

Build the square

Change b and watch the corner change

slide the coefficient of xb = 6
Completing the squarepick a problem, then step through it

Why half of b, and why squared

Do not memorize the rule — watch it fall out of one multiplication. Every perfect square trinomial is some (x + p)². Expand it slowly, without skipping the middle:
expand the exponent (x + p)² = (x + p)(x + p)
   = x·x  +  x·p  +  p·x  +  p·p
   = x² + 2px + p²

The x term appears twice — once from each parenthesis — so it comes out doubled. Now line that up against the expression you actually have, x² + bx. The middle terms must agree, so 2p = b, which means p = b/2. And once p is settled, the constant is not a choice — it is forced to be p² = (b/2)². That is the entire rule. You halve b because the middle term doubled; you square it because the last term is p times p.

The picture says the same thing. The strip of area bx has to be split so that half can go on the right and half on the bottom — a square needs both sides padded equally. Each piece is b/2 thick, so the corner they leave behind is b/2 by b/2, and its area is (b/2)². Geometry and algebra hand you the same number.

bhalf of badd (b/2)²becomes
211(x + 1)²
639(x + 3)²
−8−416(x − 4)²
552254(x + 52)²
−3−3294(x − 32)²

Two sign facts worth pinning down. The number you add is (b/2)², and a square is never negative — so you are always adding, even when b is negative. But the number that goes inside the parentheses is b/2 itself, sign and all: b = −8 gives (x − 4)², not (x + 4)².

Where the ± comes from. At the end you meet something like (x + 3)² = 16 and “take the square root of both sides.” Said precisely, √(A²) is not A — it is |A|, because the radical always returns the non-negative root. So the honest line is |x + 3| = 4, and a quantity whose distance from zero is 4 is either 4 or −4. That is the ±. It is not decoration, and dropping it silently deletes one of your two answers.

The routine. 1. Make the coefficient of x² equal to 1 — divide every term by a.  2. Move the constant to the other side.  3. Take half of b, then square it.  4. Add that to both sides.  5. Rewrite the left as (x + b/2)².  6. Square root both sides, keeping ±.  7. Isolate x.  8. Check.
Four traps.  • Adding (b/2)² to only the left side — that is a different equation, and its answers are not yours.  • Forgetting to divide the constant term too when a ≠ 1.  • Losing the ±.  • Reading (x + 1)² = −4 as solvable over the real numbers — no real number squares to a negative, so that equation has no real solution, and the method told you so instantly.

One reassurance about checking. Every step here is reversible — adding the same amount to both sides, dividing by a nonzero a, rewriting a trinomial in factored form. Completing the square therefore never invents extraneous solutions the way squaring both sides of a radical equation does, and there are no excluded values to exclude, since a polynomial is defined for every real x. Check anyway: the check catches arithmetic slips, and it is the only step that can.