Pre-Calculus · Quadratics & Vertex Form
Pre-Calc Diagnostic: Rewrite by Completing the Square (Stewart Problem 7)
Two complete-the-square problems from Stewart's Pre-Calculus Diagnostic Test — rewrite x² + x + 1 into vertex form when the linear coefficient is odd (so half of it is a fraction), and rewrite 2x² − 12x + 11 when the leading coefficient isn't 1 (so it has to be factored out first). Same geometric idea both times, two different traps. Taught the way we teach it at SOMATH on the Upper West Side.
Direct answer. To complete the square on x2 + x + 1, take half of the linear coefficient — half of 1 is 12 — square it to get 14, then add and subtract it: x2 + x + 14 + 1 − 14 = (x + 12)2 + 34. For 2x2 − 12x + 11, factor the leading 2 out of the first two terms first: 2(x2 − 6x) + 11, complete the square inside the parentheses with (−3)2 = 9, distribute the 2 carefully, and land on 2(x − 3)2 − 7. Both are now in vertex form a(x − h)2 + k and ready to be graphed, optimized, or integrated.
The one move behind both parts: half-it, square-it
Completing the square is a single geometric idea wearing different costumes. You start with x2 + bx — an x-by-x square with a b-by-x rectangle stuck to its side. You slice that rectangle down the middle so a b2-by-x strip sits on each open edge of the square. There's one corner missing: a b2-by-b2 tile. You add it to complete a full (x + b2)-by-(x + b2) square. Algebraically you added (b2)2, so you also subtract it to keep the expression equal. That's the whole technique. Once you see the picture, you never memorize the recipe again.
(a) Complete the square: x2 + x + 1
The linear coefficient is b = 1. Half of 1 is 12, and (12)2 = 14. So you add 14 to make a perfect square and subtract 14 to keep the expression equal:
x2 + x + 1 = x2 + x + 14 + 1 − 14 = (x + 12)2 + 34
The first three terms collapse into the perfect square (x + 12)2, and the leftover constant is 1 − 14 = 34. Final answer: (x + 12)2 + 34.
The trap here is the fraction. When b is odd, half of it is a fraction, and that fraction has to live inside the parentheses. A student who tries to clear it by multiplying both sides by 4 is treating an expression like an equation — and changing its value. Vertex form with a fraction inside is fine. In fact, the vertex of y = x2 + x + 1 is (−12, 34) — read straight off the boxed answer, no extra work.
(b) Complete the square: 2x2 − 12x + 11
The leading coefficient isn't 1, so you can't complete the square directly. Factor 2 out of the first two terms only — leave the constant 11 alone:
2x2 − 12x + 11 = 2(x2 − 6x) + 11
Inside the parentheses, b = −6. Half of −6 is −3, and (−3)2 = 9. Add 9 inside and immediately subtract 9 inside so nothing changes in value:
2(x2 − 6x + 9 − 9) + 11 = 2[(x − 3)2 − 9] + 11
Now distribute the 2 carefully — this is where students slip up. The 2 multiplies both pieces inside the bracket: 2 · (x − 3)2 stays as 2(x − 3)2, and 2 · (−9) = −18. Combine with the outside constant:
2(x − 3)2 − 18 + 11 = 2(x − 3)2 − 7
The parabola y = 2x2 − 12x + 11 opens upward with vertex at (3, −7). The 2 out front tells you it's narrower than y = x2 by a factor of 2 vertically.
The two traps students fall into
Trap 1: forgetting to subtract what you added. When you complete the square you add (b2)2 to make the perfect square. If you don't also subtract it, you've changed the value of the expression. The whole point is to rewrite it equivalently — same number, different form. Students who skip the subtraction step end up with a wrong vertex and a wrong parabola.
Trap 2: forgetting to multiply through when you factored out a leading coefficient. In part (b), when you added 9 inside the parentheses, you didn't actually add 9 to the whole expression — you added 2 · 9 = 18, because the 2 outside multiplies everything inside. That's why you subtract 9 inside the parentheses too, not 9 outside. The cleanest way to avoid this is to keep the −9 inside the brackets as we did, then distribute the 2 at the very end.
Why this technique matters past pre-calc
Completing the square shows up everywhere downstream. It's the proof of the quadratic formula — apply the technique to ax2 + bx + c = 0 in general and the formula falls out. It's how you graph any parabola: vertex form a(x − h)2 + k hands you the vertex (h, k) directly. It's how you put circles, ellipses, and hyperbolas into standard form so you can read off centers and radii. And in AP Calculus it's the trick that turns ∫1x2 + bx + c dx into a clean arctangent integral — but only after you complete the square in the denominator first. If a pre-calc student can do Problem 7 cold, they have one less thing to learn in Calc AB.
How SOMATH teaches completing the square on the Upper West Side
We teach the picture first, then the algebra. Every SOMATH pre-calc student spends one 60-minute session drawing the geometric square — the x-by-x tile, the two b2-by-x strips, the missing (b2)2 corner — before we ever write (b2)2 as a recipe. After that, the algebra is just bookkeeping. From our classroom on 226 W 79th St we run small groups of 3–5 students, every lesson on whiteboards, every step shown out loud. We don't hand out worksheet stacks — we hand out one well-chosen problem at a time and watch how the student thinks about it. Every new student gets a free 30-minute evaluation and a written diagnostic within 48 hours, even if they don't enroll.
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FAQ
What does it mean to complete the square?
Rewrite a quadratic ax2 + bx + c into vertex form a(x − h)2 + k. Geometrically you're literally finishing a square: an x-by-x square with two strips on its sides, plus the missing corner (b2)2. Add it, subtract it to stay equivalent, and you're done.
Why HALF the linear coefficient and squared?
Because (x + b2)2 = x2 + bx + (b2)2. The middle term is what you have. The square term is what you want. The constant (b2)2 is the bridge between them.
Why factor out the leading coefficient first?
Completing the square only works directly on x2 + bx with a leading 1. Factor a out of the first two terms only, then apply the technique inside, then distribute a back at the end.
When is completing the square actually used?
Proving the quadratic formula, finding the vertex of a parabola, putting conic sections into standard form, and as the substitution step in many Calc II integrals like ∫1x2 + bx + c dx.
Is a fraction inside the squared term a mistake?
No — if b is odd, half of it is a fraction and that fraction belongs inside the parentheses. (x + 12)2 + 34 is the correct simplified vertex form.
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