Completing the Square

Transform quadratics to reveal vertex form and find turning points

SPEC 2.3 • ALGEBRA AND FUNCTIONS

Why Complete the Square?

Completing the square is a powerful technique that transforms a quadratic from general form to vertex form.

General Form

y = a x^{2} + b x + c

Vertex Form

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

Key Benefits

Reveals the vertex — turning point is at

(h, k)

Shows max/min value — the value

k

is the maximum or minimum

Shows transformations — how the graph relates to

y = x^{2}

Determines range — for quadratic functions

The Method: When a = 1

For quadratics of the form

x^{2} + b x + c

, use this key result:

Key Formula

x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}

Remember

"Half the coefficient of x, then square it"

The number inside the bracket is

\frac{b}{2}

The number you subtract is

(\frac{b}{2})^{2}

Step-by-Step Process

Identify the coefficient of

x

(this is

b

)

Halve it to get

\frac{b}{2}

Write

(x + \frac{b}{2})^{2}

Subtract

(\frac{b}{2})^{2}

and add the constant

c

Simplify the constant terms

Worked Example01

Write $x^{2} + 6 x + 5$ in the form $(x + p)^{2} + q$ and state the minimum value.

1

Coefficient of

x

is 6, so

\frac{b}{2} = \frac{6}{2} = 3

2

Write the squared bracket:

(x + 3)^{2}

3

Subtract

3^{2} = 9

:

(x + 3)^{2} - 9

4

Add the constant (+5):

(x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4

5

Since

a = 1 > 0

, the parabola is U-shaped with a minimum.

6

Since

(x + 3)^{2} \geq 0

, the minimum value is

- 4

when

x = - 3

.

Answer: $(x + 3)^{2} - 4$ , minimum value = $- 4$

Worked Example02

Write $- x^{2} + 10 x - 21$ in the form $- (x + p)^{2} + q$ and state the maximum value.

1

Factor out

- 1

:

- 1 (x^{2} - 10 x) - 21

2

Complete the square inside:

x^{2} - 10 x = (x - 5)^{2} - 25

3

Substitute back:

- 1 [(x - 5)^{2} - 25] - 21

4

Distribute the

- 1

:

- (x - 5)^{2} + 25 - 21 = - (x - 5)^{2} + 4

5

Since

a = - 1 < 0

, the parabola is ∩-shaped with a maximum.

6

Since

- (x - 5)^{2} \leq 0

, the maximum value is

4

when

x = 5

.

Answer: $- (x - 5)^{2} + 4$ , maximum value = $4$

The Method: When a ≠ 1

For quadratics of the form

a x^{2} + b x + c

where

a \neq = 1

:

Method

Factor out $a$ from the first two terms

Complete the square inside the bracket

Distribute $a$ back through and simplify

Common Error

When you complete the square inside the bracket, remember that the constant you subtract is multiplied by $a$ when you expand back out.

a [(x + p)^{2} - p^{2}] = a (x + p)^{2} - a p^{2}

Worked Example03

Write $2 x^{2} + 12 x + 7$ in the form $a (x + p)^{2} + q$ and state the minimum value.

1

Factor out 2 from first two terms:

2 (x^{2} + 6 x) + 7

2

Complete the square inside:

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

3

Substitute back:

2 [(x + 3)^{2} - 9] + 7

4

Distribute the 2:

2 (x + 3)^{2} - 18 + 7 = 2 (x + 3)^{2} - 11

5

Since

a = 2 > 0

, the parabola is U-shaped with a minimum.

6

Minimum value is

- 11

when

x = - 3

.

Answer: $2 (x + 3)^{2} - 11$ , minimum value = $- 11$

Worked Example04

Write $- 3 x^{2} + 12 x + 5$ in the form $a (x + p)^{2} + q$ . Find the vertex and state the maximum value.

1

Factor out

- 3

:

- 3 (x^{2} - 4 x) + 5

2

Complete the square:

x^{2} - 4 x = (x - 2)^{2} - 4

3

Substitute:

- 3 [(x - 2)^{2} - 4] + 5

4

Expand:

- 3 (x - 2)^{2} + 12 + 5 = - 3 (x - 2)^{2} + 17

5

From

- 3 (x - 2)^{2} + 17

: vertex is at

(2, 17)

6

Since

a = - 3 < 0

, the parabola is ∩-shaped. Maximum value is

17

.

Answer: $- 3 (x - 2)^{2} + 17$ , Vertex: $(2, 17)$ , maximum = $17$

Finding Maximum and Minimum Values

The completed square form immediately tells us the maximum or minimum value:

When a > 0 (U-shape)

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

Minimum value = $k$

Occurs when

x = h

Since

(x - h)^{2} \geq 0

, the smallest value of

y

is

k

When a < 0 (∩-shape)

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

Maximum value = $k$

Occurs when

x = h

Since

a (x - h)^{2} \leq 0

, the largest value of

y

is

k

Key Insight

The vertex is always at

(h, k)

where the completed square form is

a (x - h)^{2} + k

Worked Example05

Find the maximum value of $10 + 8 x - 2 x^{2}$ .

1

Rewrite in standard order:

- 2 x^{2} + 8 x + 10

2

Factor out

- 2

:

- 2 (x^{2} - 4 x) + 10

3

Complete the square:

x^{2} - 4 x = (x - 2)^{2} - 4

4

Substitute:

- 2 [(x - 2)^{2} - 4] + 10

5

Expand:

- 2 (x - 2)^{2} + 8 + 10 = - 2 (x - 2)^{2} + 18

6

Since

a = - 2 < 0

, this is ∩-shaped with maximum at

k = 18

Maximum value = 18 (when

x = 2

)

Practice

Question 1: Write

x^{2} - 4 x + 7

in the form

(x + p)^{2} + q

and state the minimum value of the expression.

1

x^{2} - 4 x = (x - 2)^{2} - 4

2

x^{2} - 4 x + 7 = (x - 2)^{2} - 4 + 7 = (x - 2)^{2} + 3

3

Since

a = 1 > 0

and

(x - 2)^{2} \geq 0

, minimum value is

3

when

x = 2

Answer: $(x - 2)^{2} + 3$ , minimum value = 3

Practice

Question 2: Write

- 2 x^{2} + 8 x - 5

in completed square form and find the coordinates of the vertex and the maximum value.

1

Factor out

- 2

:

- 2 (x^{2} - 4 x) - 5

2

Complete the square:

x^{2} - 4 x = (x - 2)^{2} - 4

3

Substitute:

- 2 [(x - 2)^{2} - 4] - 5 = - 2 (x - 2)^{2} + 8 - 5 = - 2 (x - 2)^{2} + 3

4

Since

a = - 2 < 0

, this is ∩-shaped. Vertex:

(2, 3)

, maximum =

3

Answer: $- 2 (x - 2)^{2} + 3$ , Vertex: $(2, 3)$ , maximum = 3

Summary

Key Formulas

When

a = 1

When

a \neq = 1

x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}

Factor out

a

first, then complete the square inside

Key Results from Vertex Form

If

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

:

Vertex:

(h, k)

Axis of symmetry:

x = h

If $a > 0$ : minimum value is

k

If $a < 0$ : maximum value is

k

Exam Skills

Always show clear working for each step

Check your answer by expanding back to original form

State whether it's a maximum or minimum based on the sign of

a