Completing the Square

How high does a thrown ball climb before it drops back? What ticket price earns a concert the most money? Both answers sit at the very top (or bottom) of a quadratic curve, and completing the square is the move that pins that turning point down exactly — no guessing, no graph needed.

A jigsaw with a missing corner

Take the expression x^2 + 6x and — strange as it sounds — draw it. The x^2 is a square, x units on each side. The 6x is a strip, x tall and 6 wide. Snip the strip in half and lay one 3-wide piece along the right of the square and the other along the bottom. Look at what you've made: it is almost a bigger square of side x + 3 … except for a small square hole in the corner, 3 \times 3 = 9 in area.

Drop a 9 into that hole and the jigsaw is finished — the picture becomes a perfect square:

x^2 + 6x + 9 = (x + 3)^2

That is the whole trick, and the name is completely literal: we complete the square. It is probably the most geometric move in all of school algebra — a thousand-year-old scissors-and-paper idea that still powers the quadratic formula today. Of course, algebra doesn't let us just add a 9 that wasn't there — we'll have to pay it back. That bookkeeping is the rest of this page.

The picture, in general

The same construction works for any x^2 + bx, not just b = 6. Step through the figure below and watch each move. Notice why the halving is forced on us: the strip must be shared equally between two sides of the square — right and bottom — or the shape won't be square at all. Two equal shares of b means each share is \tfrac{b}{2}, and that is exactly why "half the x-coefficient" appears in every completed square you will ever write.

The final frame is the key identity. The corner we filled in has area \left(\tfrac{b}{2}\right)^2, and since we invented that area out of nothing, honesty demands we subtract it straight back off:

x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2

Every quadratic can be rewritten with a single squared bracket:

x^2 + bx + c = \left(x + \tfrac{b}{2}\right)^2 + c - \left(\tfrac{b}{2}\right)^2
The number inside the bracket is half the x-coefficient.
The correction term -\left(\tfrac{b}{2}\right)^2 cancels the extra area the square sneaks in.
If a \neq 1, factor a out of the x-terms first, then complete inside the bracket.

Worked example: x^2 + 6x + 2

Here is the whole routine, one honest step at a time.

Set the constant aside. Focus on x^2 + 6x; the +2 will wait patiently at the end.
Halve the x-coefficient. Half of 6 is 3, so the bracket is (x + 3)^2.
Subtract the square you smuggled in. Expanding gives (x + 3)^2 = x^2 + 6x + 9 — a 9 we never had. So x^2 + 6x = (x + 3)^2 - 9.
Bring the constant back. x^2 + 6x + 2 = (x + 3)^2 - 9 + 2 = (x + 3)^2 - 7

Always check by expanding: (x + 3)^2 - 7 = x^2 + 6x + 9 - 7 = x^2 + 6x + 2. ✓ Same expression, new outfit — one form wears its square on the outside.

A negative b changes nothing except the sign inside the bracket: for x^2 - 10x + 3, half of -10 is -5, so x^2 - 10x + 3 = (x - 5)^2 - 25 + 3 = (x - 5)^2 - 22.

The payoff: read the turning point straight off

Why bother? Because the completed form tells you where the parabola turns. Take our answer, (x + 3)^2 - 7, and think about what a square can do. A squared number is never negative — the smallest it can be is 0, and that happens at exactly one moment: when x + 3 = 0, i.e. x = -3. At that instant the whole expression bottoms out at 0 - 7 = -7. Move x anywhere else and the square turns positive, lifting the value back up.

So the parabola y = x^2 + 6x + 2 has its minimum point at (-3, -7), with a line of symmetry at x = -3 — no calculus, no table of values, just reading the two numbers off (x + 3)^2 - 7. In general (x + p)^2 + q turns at (-p, q): the bracket slides the curve sideways (mind the minus!), the tail number slides it up and down. Play with both below and watch the vertex obey.

The sliders start at p = 3,\; q = -7 — our worked example. Notice the bowl's lowest point sitting at (-3, -7), exactly where the completed square said it would.

Solving equations from the completed form

Completed-square form also solves the quadratic — even one that refuses to factorise. Solve x^2 + 6x + 2 = 0:

(x + 3)^2 - 7 = 0

(x + 3)^2 = 7

Now take the square root of both sides — remembering that 7 has two square roots, one positive and one negative:

x + 3 = \pm\sqrt{7} \qquad\Longrightarrow\qquad x = -3 \pm \sqrt{7}

Those are exact answers, written as surds — roughly -0.35 and -5.65, but the surd form is perfectly precise where the decimals are rounded. Exam questions that say "give your answer in exact form" are asking for exactly this. Forgetting the \pm throws away one of the two solutions — a parabola crossing the axis crosses it twice.

When a \neq 1: factor first

A leading coefficient spoils the picture — 2x^2 isn't a square of side x. The fix: factor the coefficient out of the x-terms first (leave the constant alone), then complete inside the bracket. Take 2x^2 + 8x + 5:

2x^2 + 8x + 5 = 2\left(x^2 + 4x\right) + 5

Inside the bracket, half of 4 is 2, so x^2 + 4x = (x + 2)^2 - 4. Substitute that in — and here is the step that catches people: the -4 is still inside the 2(\;\cdot\;), so it gets doubled on the way out:

2\left[(x + 2)^2 - 4\right] + 5 = 2(x + 2)^2 - 8 + 5 = 2(x + 2)^2 - 3

Check: 2(x+2)^2 - 3 = 2(x^2 + 4x + 4) - 3 = 2x^2 + 8x + 8 - 3 = 2x^2 + 8x + 5. ✓ And the payoff still works: the minimum of 2(x + 2)^2 - 3 is at (-2, -3) — the squared bracket is smallest (zero) at x = -2, leaving -3.

The square sneaks in extra. x^2 + 6x is not (x + 3)^2 — expanding the bracket smuggles in a +9 that was never there. Halve b, then subtract the square of that half, every single time: x^2 + 6x = (x + 3)^2 - 9.
The vertex sign flips. (x + 3)^2 - 7 has its minimum at x = -3, not +3 — the bracket is zero when x + 3 = 0. Plus inside the bracket means the graph slides to the left. If in doubt, ask "what makes the bracket zero?"
With a \neq 1, factor a out of the x-terms only — not the constant — and remember that the subtracted square is still inside the a(\;\cdot\;), so it gets multiplied by a when the brackets come off. Writing 2(x+2)^2 - 4 + 5 instead of 2(x+2)^2 - 8 + 5 is the single most common error on this topic.

Around 820 CE, in the House of Wisdom in Baghdad, the mathematician al-Khwarizmi solved the equation x^2 + 10x = 39 — and he did it with pictures, almost exactly the figure on this page. He drew the x^2 square, split the 10x into strips laid along its sides, and filled in the missing 5 \times 5 corner: adding 25 to both sides gives (x + 5)^2 = 64, so x + 5 = 8 and x = 3. For him this wasn't a metaphor — the square genuinely got completed, area by area. (Negative solutions had to wait a few more centuries; lengths can't be negative.)

His book's title, al-Kitāb al-mukhtaṣar fī ḥisāb al-jabr wa'l-muqābala — "the compendium on calculation by restoring and balancing" — gave us the very word algebra. And the trick never stopped paying: complete the square on the fully general ax^2 + bx + c = 0, once, with letters instead of numbers, and out pops the quadratic formula — it isn't a separate fact to memorise, it's this page's move done once and for all.

Where this fits

The move builds straight on factorising harder quadratics: when a quadratic refuses to factorise into whole-number brackets, completing the square always works — it never needs the constant to cooperate, because the leftover number simply absorbs whatever is left. That reliability is why it earns its keep three ways at once: it is the engine behind the quadratic formula, the fastest route to a parabola's turning point and line of symmetry, and the standard way to write exact (surd) solutions. Expanding undoes it, which is also your built-in answer check.

See it explained

Prefer to watch the whole routine performed live? Khan Academy walks through solving a quadratic by completing the square, step by step — a good second pass once you've worked the examples above yourself: