Completing the Square

Completing the square turns a quadratic like $x^2 + 6x + 2$ into vertex form, $(x + 3)^2 - 7$. Vertex form shows the vertex and the maximum or minimum value straight away, which is exactly what you need for optimization problems.

Key ideas

Perfect square trinomials

Expanding a squared binomial gives a pattern:

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

The constant $9$ is the square of half the $x$-coefficient: $\left(\tfrac{6}{2}\right)^2 = 9$. In general:

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

Completing the square when a = 1

To write $x^2 + bx + c$ in vertex form:

1. Take half of $b$ and square it.
2. Add and subtract that number right after the $x$ term (adding and subtracting the same number changes nothing).
3. Group the first three terms as a perfect square, and combine the constants.

$$\begin{aligned} x^2 + 6x + 2 &= x^2 + 6x + 9 - 9 + 2 \\ &= (x + 3)^2 - 7 \end{aligned}$$

When a ≠ 1

First factor $a$ out of the $x^2$ and $x$ terms only. Complete the square inside the bracket, then multiply the subtracted number by $a$ as you take it out.

Reading vertex form

$f(x) = a(x - h)^2 + k$ has vertex $(h, k)$.

• If $a \gt 0$, the parabola opens up, so $k$ is the minimum value.
• If $a \lt 0$, it opens down, so $k$ is the maximum value.

Watch the sign: $(x + 3)^2$ means $h = -3$.

Worked examples

Example 1: A simple case

Write $f(x) = x^2 + 6x + 2$ in vertex form, and give the vertex.

Solution. Half of $6$ is $3$, and $3^2 = 9$:

$$\begin{aligned} f(x) &= x^2 + 6x + 9 - 9 + 2 \\ &= (x + 3)^2 - 7 \end{aligned}$$

The vertex is $(-3, -7)$, and since $a = 1 \gt 0$, the minimum value is $-7$.

Example 2: An odd x-coefficient

Write $f(x) = x^2 - 5x + 1$ in vertex form.

Solution. Half of $-5$ is $-\tfrac{5}{2}$, and $\left(-\tfrac{5}{2}\right)^2 = \tfrac{25}{4}$:

$$\begin{aligned} f(x) &= x^2 - 5x + \tfrac{25}{4} - \tfrac{25}{4} + 1 \\ &= \left(x - \tfrac{5}{2}\right)^2 - \tfrac{21}{4} \end{aligned}$$

The vertex is $\left(\tfrac{5}{2}, -\tfrac{21}{4}\right)$.

Example 3: When a is greater than 1

Write $f(x) = 2x^2 - 12x + 7$ in vertex form, and give the minimum value.

Solution. Factor $2$ out of the first two terms only:

$$\begin{aligned} f(x) &= 2(x^2 - 6x) + 7 \\ &= 2(x^2 - 6x + 9 - 9) + 7 \\ &= 2\big((x - 3)^2 - 9\big) + 7 \\ &= 2(x - 3)^2 - 18 + 7 \\ &= 2(x - 3)^2 - 11 \end{aligned}$$

The vertex is $(3, -11)$. Since $a = 2 \gt 0$, the minimum value is $-11$, when $x = 3$.

Check by expanding: $2(x^2 - 6x + 9) - 11 = 2x^2 - 12x + 7$. ✓

Example 4: A negative value of a

Write $f(x) = -3x^2 - 12x + 1$ in vertex form, and give the maximum value.

Solution. Factor $-3$ out of the first two terms:

$$\begin{aligned} f(x) &= -3(x^2 + 4x) + 1 \\ &= -3(x^2 + 4x + 4 - 4) + 1 \\ &= -3\big((x + 2)^2 - 4\big) + 1 \\ &= -3(x + 2)^2 + 12 + 1 \\ &= -3(x + 2)^2 + 13 \end{aligned}$$

The maximum value is $13$, when $x = -2$.

Common mistakes

Adding the square without subtracting it. $x^2 + 6x + 2$ becomes $x^2 + 6x + 9 - 9 + 2$. If you only add $9$, you’ve changed the function.

Forgetting to multiply by $a$. In Example 3, the $-9$ inside the bracket becomes $-18$ when it comes out, because it’s multiplied by $2$.

Using $b$ instead of half of $b$. For $x^2 + 6x$, you add $3^2 = 9$, not $6^2 = 36$.

Factoring $a$ out of the constant too. Only the $x^2$ and $x$ terms go in the bracket. The constant waits outside.

Getting the vertex sign wrong. $(x + 3)^2 - 7$ has vertex $(-3, -7)$, not $(3, -7)$.

Practice

1. (Warm-up) What number should be added to $x^2 + 10x$ to make a perfect square? Write the result as a squared binomial.

Solution

$\left(\tfrac{10}{2}\right)^2 = 25$, and $x^2 + 10x + 25 = (x + 5)^2$.

2. (Warm-up) Write $x^2 - 8x + 3$ in vertex form.

Solution

$$x^2 - 8x + 16 - 16 + 3 = (x - 4)^2 - 13$$

3. (Warm-up) Write $x^2 + 2x - 5$ in vertex form, and give the vertex.

Solution

$$x^2 + 2x + 1 - 1 - 5 = (x + 1)^2 - 6$$

The vertex is $(-1, -6)$.

4. (Core) Write $x^2 + 3x - 1$ in vertex form.

Solution

Half of $3$ is $\tfrac{3}{2}$, and $\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}$:

$$x^2 + 3x + \tfrac{9}{4} - \tfrac{9}{4} - 1 = \left(x + \tfrac{3}{2}\right)^2 - \tfrac{13}{4}$$

5. (Core) Write $3x^2 + 18x + 20$ in vertex form, and give the minimum value.

Solution

$$\begin{aligned} 3(x^2 + 6x) + 20 &= 3(x^2 + 6x + 9 - 9) + 20 \\ &= 3(x + 3)^2 - 27 + 20 \\ &= 3(x + 3)^2 - 7 \end{aligned}$$

The minimum value is $-7$, when $x = -3$.

6. (Core) Write $-2x^2 + 8x - 3$ in vertex form, and give the maximum value.

Solution

$$\begin{aligned} -2(x^2 - 4x) - 3 &= -2(x^2 - 4x + 4 - 4) - 3 \\ &= -2(x - 2)^2 + 8 - 3 \\ &= -2(x - 2)^2 + 5 \end{aligned}$$

The maximum value is $5$, when $x = 2$.

7. (Core) Write $0.5x^2 - 3x + 4$ in vertex form.

Solution

$$\begin{aligned} 0.5(x^2 - 6x) + 4 &= 0.5(x^2 - 6x + 9 - 9) + 4 \\ &= 0.5(x - 3)^2 - 4.5 + 4 \\ &= 0.5(x - 3)^2 - 0.5 \end{aligned}$$

8. (Challenge) Solve $x^2 + 6x - 3 = 0$ by completing the square. Give exact answers.

Solution

$$\begin{aligned} x^2 + 6x + 9 - 9 - 3 &= 0 \\ (x + 3)^2 &= 12 \\ x + 3 &= \pm\sqrt{12} = \pm 2\sqrt{3} \\ x &= -3 \pm 2\sqrt{3} \end{aligned}$$

9. (Challenge) Complete the square on $f(x) = ax^2 + bx + c$ to show that the vertex has $x$-coordinate $-\dfrac{b}{2a}$.

Solution

$$\begin{aligned} f(x) &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c \\ &= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \end{aligned}$$

This is vertex form with $h = -\dfrac{b}{2a}$, so the vertex is at $x = -\dfrac{b}{2a}$.