Completing the Square and Vertex Form of a Quadratic

Q: How do you find the vertex by completing the square?

Rewrite the quadratic as a(x - h)^2 + k. The vertex is (h, k). For example (x - 3)^2 - 4 has vertex (3, -4).

Q: Why do you halve b when completing the square?

Because (x + b/2)^2 = x^2 + bx + b^2/4. Halving b gives the bracket whose expansion produces the right x term.

Completing the square as a picture of an actual square, the step-by-step method, how vertex form reveals the turning point and the range, and how it solves any quadratic equation.

3 min read Supports 4 lessons

The graph of any quadratic is a parabola, and every parabola is the basic $y = x^2$ shifted and stretched. Vertex form makes those shifts visible:

Vertex form

y = a(x - h)^2 + k

The vertex (turning point) is $(h, k)$ , and the axis of symmetry is the line $x = h$ .

Completing the square is the algebra that turns $x^2 + bx + c$ into this form.

Why it is called completing the square

Picture $x^2 + 6x$ as shapes: an $x$ by $x$ square plus a $6$ by $x$ strip. Cut the strip in half and attach one half to each side of the square. You almost have a bigger square, $(x + 3)$ on each side, but a $3 \times 3$ corner is missing. Add that corner, and subtract it again to stay honest.

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

The $9$ is the missing corner: $\left(\frac{6}{2}\right)^2$ .

Completing the square

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

Step by step

$x^2 - 6x + 5$
Halve the coefficient of $x$ : half of $-6$ is $-3$ .

1/4

Watch out

$(x - 3)^2 - 4$ has its vertex at $(3, -4)$ , not $(-3, -4)$ . The bracket $(x - h)$ shifts right by $h$ .

What vertex form tells you

Since $(x - 3)^2$ is never negative, $(x - 3)^2 - 4$ is never less than $-4$ . So the minimum value is $-4$ , reached at $x = 3$ , and the range is $y \ge -4$ . If $a$ is negative the parabola opens downwards and $k$ is a maximum instead.

Solving equations

Worked example

Solve $x^2 - 6x + 5 = 0$ by completing the square.

1
Complete the square.
$(x - 3)^2 - 4 = 0$

When a is not 1

Factor $a$ out of the $x^2$ and $x$ terms first, complete the square inside the bracket, then multiply back out.

2x^2 + 8x + 1 = 2(x^2 + 4x) + 1 = 2\big[(x + 2)^2 - 4\big] + 1 = 2(x + 2)^2 - 7

Key idea

Completing the square on $ax^2 + bx + c = 0$ in general is exactly how the quadratic formula is derived. It is also how you find the centre and radius of a circle from its equation.

Common questions

How do you find the vertex by completing the square?: Rewrite the quadratic as $a(x - h)^2 + k$ . The vertex is $(h, k)$ . For example $(x - 3)^2 - 4$ has vertex $(3, -4)$ .
Why do you halve b when completing the square?: Because $(x + \frac{b}{2})^2 = x^2 + bx + \frac{b^2}{4}$ . Halving $b$ gives the bracket whose expansion produces the right $x$ term.

Practise completing the square

The Completing the Square & Vertex Form lesson has worked examples and unlimited practice that adapts to you.

Open the lesson

Common questions

How do you find the vertex by completing the square?

Rewrite the quadratic as $a(x - h)^2 + k$ . The vertex is $(h, k)$ . For example $(x - 3)^2 - 4$ has vertex $(3, -4)$ .

Why do you halve b when completing the square?

Because $(x + \frac{b}{2})^2 = x^2 + bx + \frac{b^2}{4}$ . Halving $b$ gives the bracket whose expansion produces the right $x$ term.