Last updated: 2026-05-20

Completing the square

Completing the square transforms a quadratic expression $ax^2 + bx + c$ into vertex form $a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola. The transformation isolates the squared term so the coordinates of the vertex can be read directly from the rewritten expression.

The completing-the-square method

Starting from $ax^2 + bx + c$:

Step 1 — Find h (the x-coordinate of the vertex):

$$h = -\frac{b}{2a}$$

Step 2 — Find k (the y-coordinate of the vertex):

$$k = c - \frac{b^2}{4a}$$

Step 3 — Write in vertex form:

$$ax^2 + bx + c = a(x - h)^2 + k$$

Why this works

Factor $a$ from the first two terms and add the square completion term inside:

$$a\!\left(x^2 + \frac{b}{a}x\right) + c = a\!\left(x + \frac{b}{2a}\right)^{\!2} - \frac{b^2}{4a} + c$$

The term $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$ is added inside the bracket (gaining $a \cdot \frac{b^2}{4a^2} = \frac{b^2}{4a}$) and immediately subtracted outside, keeping the expression equivalent. Substituting $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$ gives $a(x - h)^2 + k$.

Worked example

Rewrite $x^2 - 6x + 11$.

Here $a = 1$, $b = -6$, $c = 11$.

$$h = -\frac{-6}{2 \cdot 1} = 3$$ $$k = 11 - \frac{(-6)^2}{4 \cdot 1} = 11 - 9 = 2$$ $$x^2 - 6x + 11 = (x - 3)^2 + 2$$

Check: expand $(x - 3)^2 + 2 = x^2 - 6x + 9 + 2 = x^2 - 6x + 11$ ✓

The vertex is $(3, 2)$, the axis of symmetry is $x = 3$, and the minimum value is $2$.

The vertex and direction of opening

Property	Value
Vertex	$(h,\, k)$
Axis of symmetry	$x = h$
Minimum (if $a > 0$)	$y = k$ at $x = h$
Maximum (if $a < 0$)	$y = k$ at $x = h$

When $a > 0$, the parabola opens upward (∪ shape) and the vertex is the lowest point. When $a < 0$, it opens downward (∩ shape) and the vertex is the highest point. The magnitude $|a|$ controls the width: larger $|a|$ makes the parabola narrower; smaller $|a|$ makes it wider.

Deriving the quadratic formula

Completing the square also yields the quadratic formula. Set $ax^2 + bx + c = 0$, convert to vertex form, and solve for $x$:

$$a(x - h)^2 + k = 0 \implies (x - h)^2 = -\frac{k}{a}$$ $$x - h = \pm\sqrt{-\frac{k}{a}}$$

Substituting $h = -\frac{b}{2a}$ and $k = c - \frac{b^2}{4a}$ and simplifying:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

This is the quadratic formula — completing the square is the proof that the formula works for any quadratic.

Applications

• Finding the vertex without graphing.
• Solving quadratic equations when factoring is difficult.
• Proving the quadratic formula from first principles.
• Analyzing parabolas in physics and engineering (projectile motion, optics, antennas).
• Calculus — recognizing the vertex form simplifies integration of rational functions.

Frequently Asked Questions (FAQ)

Why complete the square?

Completing the square rewrites ax² + bx + c into the form a(x − h)² + k, where the vertex (h, k) is visible by inspection. This reveals the minimum (or maximum) of the parabola directly, without differentiation. It also provides the cleanest derivation of the quadratic formula and makes graphing straightforward because you know the axis of symmetry (x = h) and the extreme value (y = k) immediately.

How is completing the square related to the quadratic formula?

The quadratic formula is derived by completing the square on the general form ax² + bx + c = 0. Dividing by a gives x² + (b/a)x + c/a = 0. Adding and subtracting (b/(2a))² to create a perfect square trinomial yields (x + b/(2a))² = (b² − 4ac)/(4a²). Taking the square root of both sides and solving for x produces x = (−b ± √(b² − 4ac))/(2a) — the quadratic formula.

What does the vertex (h, k) tell me?

The vertex is the turning point of the parabola. When a > 0, the parabola opens upward and the vertex is the minimum point: the smallest value of ax² + bx + c is k, reached at x = h. When a < 0, the parabola opens downward and (h, k) is the maximum. The axis of symmetry is the vertical line x = h — the parabola is a mirror image about this line.

How does the sign of a affect the parabola?

When a > 0, the parabola opens upward (∪ shape) and has a minimum at the vertex. When a < 0, it opens downward (∩ shape) and has a maximum at the vertex. The magnitude |a| controls how wide or narrow the parabola is: larger |a| makes it steeper; smaller |a| makes it wider. When a = 0, the expression reduces to a straight line and completing the square is undefined.