Completing the square | Year 10

01 / LEARN

The idea, step by step

Build a perfect square

For x squared plus bx, halve b and square that half. Adding and subtracting the same amount leaves the expression unchanged.

If the coefficient of x squared is not one, factor it from the quadratic and linear terms first. Keep the outside multiplier when adjusting the constant.

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

Use what the square reveals

A real square cannot be negative. A positive multiple of a square therefore has its minimum when the square is zero. A negative multiple has its maximum there.

When solving a square equation, include both positive and negative square roots unless the square is zero. Keep an exact root until a decimal approximation is requested.

A common trap

Replacing $x^{2} + 6 x$ by ${(x + 3)}^{2}$ without subtracting 9.

Instead: The square expands to $x^{2} + 6 x + 9$ , so remove the extra $9$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

An outside multiplier

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

Factor $2$ from the first two terms: $2 (x^{2} - 6 x) + 7$ .
$x^{2} - 6 x = {(x - 3)}^{2} - 9$ .
So $2 [{(x - 3)}^{2} - 9] + 7 = 2 {(x - 3)}^{2} - 11$ .

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

Example 2

Solve exactly

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

$x^{2} - 4 x = 1$ . Add $4$ to both sides.
${(x - 2)}^{2} = 5$ .
$x - 2 = \pm \sqrt{5}$ .

x = 2 \pm \sqrt{5}

Check your understanding

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

Show answer

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

Half of $8$ is $4$ ; subtract its square $16$ and then add the original $3$ .

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF 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 ↗

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