What does completing the square actually do?

It rewrites ax² + bx + c as a(x − h)² + k, so the variable appears once instead of twice. That single change is what makes everything else easy: with x in only one place the expression can be rearranged directly to solve for it, and the numbers h and k turn out to be the axis of symmetry and the turning value of the graph. The value of the expression is never altered — the piece added to complete the square is subtracted again in the same line.

Why bother when the quadratic formula already exists?

Three reasons. The formula gives roots and nothing else, while the completed form also gives the vertex, the axis and the range — which is what most real questions actually want. It keeps working when there are no real roots, where the formula stops being useful. And it is the derivation of the formula, so understanding it means you can rebuild the formula rather than hope you have remembered it correctly. Beyond quadratics it also reappears constantly: in integration, in conic sections and in the normal distribution.

Where does the quadratic formula come from?

From exactly these four steps, run on letters. Pull a out of the first two terms, add and subtract (b/2a)², close the bracket into a square, and you have a(x + b/2a)² + (c − b²/4a). Set that to zero, rearrange to (x + b/2a)² = (b² − 4ac)/4a², take the square root and subtract b/2a. The result is x = (−b ± √(b² − 4ac))/2a. Nothing else is involved, which is also why the discriminant is the expression under the root: it is the quantity that decides whether that square root exists.

What are h and k, and why are they worth more than the roots?

h is where the parabola turns and k is the value it turns at, so together they are the vertex and h alone is the axis of symmetry. If a is positive the curve opens upward and k is the smallest value the expression can take; if a is negative it is the largest. That answers the questions people usually have — the minimum cost, the maximum height, the range of possible outputs — none of which the roots address. The roots only tell you where the expression happens to equal zero, which is often not the interesting number at all.

How do I complete the square when a is not 1?

Factor a out of the first two terms only, leaving the constant outside the bracket: ax² + bx + c becomes a(x² + (b/a)x) + c. Complete the square inside the bracket as usual, using half of b/a. The one step people get wrong is forgetting that the piece subtracted has to be multiplied by a on the way out of the bracket, because it was added inside something that gets multiplied by a. Dividing the whole equation by a instead also works when you are solving rather than rewriting, but it changes the expression, so it is only valid when there is an equals sign.

Does it still work when there are no real roots?

Yes, and this is where it earns its keep. Factoring gives you nothing and the formula gives you a square root of a negative number, but completing the square still returns the vertex, the axis and the extreme value. It also explains the situation rather than just reporting it: x² + 2x + 5 becomes (x + 1)² + 4, and solving would require a square to equal −4. Since a real number squared is never negative, the curve simply never reaches the axis — which you can see as well as prove, because the vertex sits four units above it.

Why is the number you add always (b/2a)²?

Because of how squaring a bracket works. (x + m)² expands to x² + 2mx + m², so to match a middle term of (b/a)x you need 2m = b/a, which makes m = b/2a. The constant that belongs with it is then m², or (b/2a)². Put another way, the middle coefficient is always twice the number inside the bracket, so halving it recovers that number and squaring it recovers the missing constant. In the diagram this is literally the corner square: the two strips each have width b/2a, so the gap they leave is b/2a by b/2a.

How is the discriminant connected to the vertex?

Directly: k = −D/4a, where D = b² − 4ac. So when a is positive, a negative discriminant forces k above zero and the curve stays above the axis, while a positive discriminant pushes the vertex below it and the curve must cross twice. A discriminant of zero puts the vertex exactly on the axis, which is the repeated-root case where the curve touches and turns back. The two quantities are not separate facts about a quadratic; they are the same fact scaled by −4a.

Where does this get used outside a maths exam?

More often than its reputation suggests. Any optimisation with a quadratic cost or revenue model is answered by the vertex, so the profit-maximising price falls straight out of h. Projectile motion under gravity is a quadratic in time, so peak height is k and the time it occurs is h. Integrals with a quadratic under a root are completed first, because that is what turns them into a standard form. And the normal distribution is derived by completing the square in the exponent — the step that turns a messy exponential into a recognisable bell.

Completing the Square Calculator

Complete the square on any quadratic exactly, and see the four steps carried out on your own numbers rather than described. Completing the square is not a technique that sits alongside the quadratic formula. It is where the formula comes from. Run the same four steps on ax² + bx + c using letters i...

ax² + bx + c = a(x − h)² + k, h = −b/2a, k = c − b²/4a

Run these four steps on letters instead of numbers and the last line is the quadratic formula. They are not two methods.

THE QUADRATIC

x²+x+= 0

COMPLETED SQUARE

(x − 3)² − 4

Expanded again it is x² − 6x + 5 — the original, exactly. Checked on every keystroke.

VERTEX

(3, −4)

AXIS

x = 3

MINIMUM

−4

Quantity	Exact value	Where it comes from
h	3	−b / 2a
k	−4	c − b² / 4a
(b/2a)²	9	the corner that completes the square
Discriminant	16	b² − 4ac, after clearing denominators
Roots	1, 5	h ± √(−k/a)

These four steps are the quadratic formula. Carry them out on ax² + bx + c with letters rather than the numbers above and the final line reads x = (−b ± √(b² − 4ac)) / 2a. Nothing else is needed to derive it, and nothing else was ever used to: the formula is the general answer to this procedure, which is why every quadratic yields to it. Being taught them as two separate things to remember is a quirk of curriculum rather than of mathematics.

Why the vertex is worth more than the roots. The completed form gives the minimum value −4 at x = 3 directly, with no calculus and no guessing, and the range follows immediately: the expression takes every value at or above −4 and none below it. That is the answer to most real questions asked of a quadratic — the cheapest price, the highest point, the best setting — none of which the roots tell you.

THE SQUARE

The corner is the only thing the method does. Everything else is bookkeeping: you add 9 to finish the square, then subtract it again so the expression still means what it did before.

Live calculation · updates as you type

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HOW TO USE

1
Type the three coefficients a, b and c. Whole numbers, decimals and fractions written as 3/4 are all read exactly, and an empty box counts as zero. If a is zero there is no square to complete, because dividing by a is the first thing the method does.
2
Read the completed form, then the line beneath it. That line multiplies the square back out and compares all three coefficients with what you entered, so the result is checked rather than merely asserted.
3
Take h and k as the answer to most questions. h is the axis of symmetry and k is the minimum or maximum value, both exact fractions — that is the range, the turning point and the best-case value, none of which the roots tell you.
4
Use the four-line derivation on the right as a worked template. The same four steps on letters rather than your numbers produce the quadratic formula, which is the fastest way to stop treating them as separate things to remember.

REFERENCE FORMULAS

Quantity	Formula	What it is for
Vertex form	ax² + bx + c = a(x − h)² + k	The whole method in one line. Everything below follows from it.
The axis	h = −b / 2a	Half the middle coefficient, negated. The axis of symmetry and the x of the vertex.
The extreme value	k = c − b² / 4a	The minimum when a > 0, the maximum when a < 0. Equivalently k = −D/4a.
The piece you add	(b / 2a)²	Added and subtracted in the same line, so the value never changes.
Discriminant	D = b² − 4ac = −4ak	Its sign is the sign of −k/a, which is why the vertex tells you the root count.
Quadratic formula	x = (−b ± √(b² − 4ac)) / 2a	What these four steps produce when you run them on letters rather than numbers.
Roots from vertex form	x = h ± √(−k/a)	The same answer, read straight off the completed square. Real exactly when −k/a ≥ 0.
Sum and product of roots	r₁ + r₂ = −b/a, r₁r₂ = c/a	Vieta’s relations. The sum is 2h, because the roots straddle the axis evenly.

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Last updated: August 3, 2026 · Exact rational arithmetic · the completed form is expanded back and checked against the input on every keystroke