Completing the square | Year 10

01 / LEARN

Understand the idea.

Build a perfect square

Completing the square builds a bracket whose expansion has the required linear term. For $x^{2} + b x$ , halve $b$ and square that half; add and subtract this amount to preserve the expression.

When the leading coefficient is not one, factor it from the quadratic and linear terms first. Any adjustment inside those brackets is multiplied by that outside coefficient when you simplify.

x^{2} + b x + c = {(x + \frac{b}{2})}^{2} + c - \frac{b^{2}}{4}

Use what the square reveals

In $a {(x - h)}^{2} + k$ , the square is nonnegative for real $x$ . If $a > 0$ , the minimum is $k$ at $x = h$ ; if $a < 0$ , the same point gives a maximum.

To solve an equation, isolate the square. A positive right-hand side gives two square-root choices, zero gives one repeated root, and a negative right-hand side gives no real solution.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An outside multiplier

Write $2 x^{2} - 12 x + 7$ in completed-square form.

Worked solution

1
Factor the first two terms.
$2 x^{2} - 12 x + 7 = 2 (x^{2} - 6 x) + 7$
The expression inside the brackets now has leading coefficient $1$ .
2
Build the square and compensate.
$\begin{aligned} x^{2} - 6 x & = x^{2} - 6 x + 9 - 9 \\ = {(x - 3)}^{2} - 9 \end{aligned}$
Half of $- 6$ is $- 3$ , whose square is $9$ .
3
Keep the outside multiplier.
$\begin{aligned} 2 [{(x - 3)}^{2} - 9] + 7 & = 2 {(x - 3)}^{2} - 18 + 7 \\ = 2 {(x - 3)}^{2} - 11 \end{aligned}$
The subtraction of $9$ inside the brackets becomes a subtraction of $18$ outside.
4
Check by expanding.
$\begin{aligned} 2 {(x - 3)}^{2} - 11 & = 2 x^{2} - 12 x + 18 - 11 \\ = 2 x^{2} - 12 x + 7 \end{aligned}$
The completed-square form is identical to the original expression.

Answer

$2 {(x - 3)}^{2} - 11$

EXAMPLE 02

Solve exactly

Solve $x^{2} - 4 x - 1 = 0$ by completing the square.

Worked solution

1
Move the constant to the other side.
$x^{2} - 4 x = 1$
Adding one to both sides keeps the equation balanced.
2
Add the missing square term to both sides.
$\begin{aligned} x^{2} - 4 x + 4 & = 1 + 4 \\ {(x - 2)}^{2} & = 5 \end{aligned}$
Half of $- 4$ is $- 2$ , and ${(- 2)}^{2} = 4$ .
3
Use both square roots.
$x - 2 = \sqrt{5} or x - 2 = - \sqrt{5}$
Both numbers square to $5$ .
4
Isolate the variable.
$x = 2 + \sqrt{5} or x = 2 - \sqrt{5}$
Adding two to each branch gives both exact solutions.

Answer

$x = 2 \pm \sqrt{5}$

03 / YOUR TURN

Try it yourself.

Check your understanding

Complete the square: $x^{2} + 8 x + 3$ .

Show answer & reasoning

${(x + 4)}^{2} - 13$

Half of $8$ is $4$ , so $x^{2} + 8 x + 3 = {(x + 4)}^{2} - 16 + 3 = {(x + 4)}^{2} - 13$ . The added and subtracted $16$ leave the value unchanged.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Complete the square: $x^{2} + 6 x + 5$ .
2 marks
Complete the square: $x^{2} - 10 x + 8$ .
2 marks
Complete the square: $3 x^{2} + 12 x + 1$ .
3 marks
Solve $x^{2} + 2 x - 4 = 0$ by completing the square.
3 marks
Write $x^{2} + 3 x + 1$ in completed-square form.
3 marks
Find the minimum value of $2 x^{2} + 8 x + 1$ and the $x$ -value where it occurs.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Complete the square: $x^{2} + 6 x + 5$ .
2 marks
$x^{2} + 6 x + 9 - 9 + 5 = {(x + 3)}^{2} - 4$ .
${(x + 3)}^{2} - 4$
Complete the square: $x^{2} - 10 x + 8$ .
2 marks
$x^{2} - 10 x + 25 - 25 + 8 = {(x - 5)}^{2} - 17$ .
${(x - 5)}^{2} - 17$
Complete the square: $3 x^{2} + 12 x + 1$ .
3 marks
$3 (x^{2} + 4 x) + 1 = 3 [{(x + 2)}^{2} - 4] + 1$ .
$3 {(x + 2)}^{2} - 11$
Solve $x^{2} + 2 x - 4 = 0$ by completing the square.
3 marks
${(x + 1)}^{2} = 5$ , so $x + 1 = \pm \sqrt{5}$ .
$x = - 1 \pm \sqrt{5}$
Write $x^{2} + 3 x + 1$ in completed-square form.
3 marks
Add and subtract $\frac{9}{4}$ : ${(x + \frac{3}{2})}^{2} + 1 - \frac{9}{4}$ .
${(x + \frac{3}{2})}^{2} - \frac{5}{4}$
Find the minimum value of $2 x^{2} + 8 x + 1$ and the $x$ -value where it occurs.
3 marks
$2 x^{2} + 8 x + 1 = 2 {(x + 2)}^{2} - 7$ .
The squared term is zero at $x = - 2$ and never negative.
Minimum $- 7$ at $x = - 2$ .

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Curriculum reference & lesson scope

Australian Curriculum v9.0 and South Australian Curriculum

Year 10 quadratics, including the fuller school-course algebra requested for Ascent: non-monic factorisation, completing the square and the quadratic formula. These lessons consolidate Year 9 monic quadratics and extend the available solving methods.

Official curriculum source ↗

In this lesson

Understand the idea.

Build a perfect square

Use what the square reveals

See the method in action.

An outside multiplier

Factor the first two terms.

Build the square and compensate.

Keep the outside multiplier.

Check by expanding.

Solve exactly

Move the constant to the other side.

Add the missing square term to both sides.

Use both square roots.

Isolate the variable.

Try it yourself.

Check your understanding

Keep practising