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Math · Algebra

Completing the Square

Rewrite ax² + bx + c as a(x − h)² + k — the vertex form — with every step shown.

1+ 6x + 5

Coefficient of x² (a ≠ 0).
Coefficient of x.
Constant term.
Example quadratics — tap to load
Vertex form
(x + 3)² − 4a(x − h)² + k

Vertex (h, k) = (−3, −4) — the axis of symmetry is x = −3.

Parabola y = ax² + bx + c — vertex at (-3, -4)
Parabola y = 1x² + 6x + 5, opening upward with a minimum at its vertex (-3, -4)y = 21y = -3.9896x = -8x = 2

Completing the square rewrites ax² + bx + c as a(x − h)² + k, where h = −b ÷ 2a and k = c − b² ÷ 4a. For x² + 6x + 5, h = −3 and k = −4, giving (x + 3)² − 4. This vertex form reveals the vertex directly: the turning point is (−3, −4).

What completing the square does

Completing the square takes a quadratic written in standard form, ax² + bx + c, and rewrites it as a perfect square plus a constant: a(x − h)² + k. The two forms describe the same parabola, but vertex form hands you the vertex (h, k) with no extra work — and that makes graphing, solving, and finding maxima or minima far easier.

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

the vertex sits at (h, k); the axis of symmetry is the vertical line x = h

Worked example

Rewrite x² + 6x + 5 in vertex form, so a = 1, b = 6, c = 5:

  1. 1
    Identify a, b, and c. For x² + 6x + 5 the coefficients are a = 1, b = 6, c = 5. Keep a factored out front if a ≠ 1.
  2. 2
    Find h = −b ÷ 2a. h = −6 ÷ (2×1) = −3. This is the x-coordinate of the vertex and the axis of symmetry.
  3. 3
    Find k = c − b² ÷ 4a. k = 5 − 6² ÷ (4×1) = 5 − 36 ÷ 4 = 5 − 9 = −4. This is the y-coordinate of the vertex.
  4. 4
    Assemble a(x − h)² + k. With h = −3 and k = −4: (x − (−3))² − 4 = (x + 3)² − 4. Check: (x + 3)² − 4 = x² + 6x + 9 − 4 = x² + 6x + 5. ✓

Reading the vertex form a(x − h)² + k

Every piece of the vertex form has a direct geometric meaning.

SymbolHow to get itWhat it tells you
h−b ÷ 2ax-coordinate of the vertex; the axis of symmetry is x = h
kc − b² ÷ 4ay-coordinate of the vertex; the minimum (a > 0) or maximum (a < 0) value
(h, k)combine the two abovethe vertex — the turning point of the parabola
athe leading coefficientopens upward if a > 0, downward if a < 0; sets the width
√(−k ÷ a)solve a(x − h)² + k = 0the distance from h to each root — the quadratic formula in disguise

Why it matters

Completing the square is the method that derives the quadratic formula: apply it to the general equation ax² + bx + c = 0 and solving a(x − h)² + k = 0 for x gives x = (−b ± √(b² − 4ac)) ÷ 2a. So every quadratic you solve with the formula is really a completed square underneath.

It also exposes the geometry at a glance. The vertex (h, k) is the parabola’s turning point, the line x = h is its axis of symmetry, and k is the extreme value — a minimum when a > 0 and a maximum when a < 0. The method only works when a ≠ 0; with a = 0 the expression is linear, not a parabola, and has no vertex.

What is vertex form?
Vertex form writes a quadratic as a(x − h)² + k. It is equivalent to standard form ax² + bx + c, but the vertex (h, k) is read off directly, so graphing and finding the maximum or minimum take no extra algebra.
How does completing the square help solve quadratics?
Once a quadratic is a(x − h)² + k, set it to zero and isolate the square: (x − h)² = −k ÷ a, then x = h ± √(−k ÷ a). Applying this to the general equation is exactly how the quadratic formula is derived.
How do I find h and k?
Use h = −b ÷ 2a and k = c − b² ÷ 4a. For x² + 6x + 5, h = −6 ÷ 2 = −3 and k = 5 − 36 ÷ 4 = −4, so the vertex is (−3, −4).
What happens when a is not 1?
Factor a out of the x² and x terms first. For 2x² − 8x + 3, h = 8 ÷ 4 = 2 and k = 3 − 64 ÷ 8 = −5, giving 2(x − 2)² − 5.
What is the axis of symmetry?
It is the vertical line x = h that splits the parabola into mirror halves. Because h = −b ÷ 2a, the axis of symmetry always passes through the vertex.
Why must a be non-zero?
With a = 0 there is no x² term, so ax² + bx + c is linear (bx + c) — a straight line with no vertex and no square to complete. Completing the square applies only to genuine quadratics.