Completing the Square to Reveal a Maximum or Minimum

In this lesson you will: Rewrite x^2+bx+c as (x+p)^2+q by completing the square; Read the minimum or maximum value of the expression off that form; Handle a leading coefficient of -1, +/-2 or +/-3 by factoring it out first.

Half of b, squared

x^2+6x is almost a perfect square. It needs a 9: x^2+6x+9=(x+3)^2. The 9 is half of 6, squared, and that is the whole rule.

You cannot simply add 9 to an expression — that would change it. So add it and subtract it in the same line. x^2+6x+5=(x^2+6x+9)-9+5=(x+3)^2-4.

Why the new form answers a question the old one cannot

A square of a real number is never negative. So in (x+3)^2-4, the first part is at least 0, and the whole expression is at least -4. It hits -4 exactly when (x+3)^2=0, that is, when x=-3.

That single sentence is what A.SSE.3b is about: the completed-square form displays the minimum value of the expression and the input that produces it. The original form x^2+6x+5 hides both.

A leading coefficient other than 1

If a is not 1, factor it out of the first two terms before completing the square. Ohio keeps this gentle: a will be -1, +/-2 or +/-3, and b divides evenly by a, so no fractions appear.

2x^2+12x+5=2(x^2+6x)+5=2((x+3)^2-9)+5=2(x+3)^2-18+5=2(x+3)^2-13. Note what happened to the 9: it was inside the parentheses, so it left multiplied by 2.

When a is negative the square is subtracted, so the expression has a maximum instead of a minimum. -x^2+10x-7=-(x-5)^2+18 can never exceed 18.

Common mistake — Forgetting that the subtracted square gets multiplied too

Completing the square on 2x^2+12x+5, a student writes 2(x+3)^2+5-9=2(x+3)^2-4.

The 9 is correct, the (x+3)^2 is correct, and the subtraction feels like the same bookkeeping as always.

But the 9 was added inside a parenthesis that is multiplied by 2, so what actually got added to the expression was 18, and 18 is what must come back out. The correct form is 2(x+3)^2-13.

The same slip in reverse happens with a negative leading coefficient, where subtracting inside the parentheses adds outside it.

How to catch it: Substitute x=0 into both forms. The original gives 5; the student's version gives 2(9)-4=14, and the correct one gives 2(9)-13=5. One substitution settles it.