Completing the square | Year 11

01 / LEARN

Understand the idea.

Create a perfect square

Completing the square replaces a quadratic by a squared bracket plus a constant. For $x^{2} + b x$ , half the linear coefficient is $\frac{b}{2}$ ; its square is the amount needed to complete ${(x + \frac{b}{2})}^{2}$ .

Add and subtract the same amount to keep the expression unchanged. In $a {(x - h)}^{2} + k$ , the square is non-negative: if $a > 0$ , $k$ is the minimum; if $a < 0$ , $k$ is the maximum, provided $x = h$ is allowed.

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

Keep the outside coefficient

When the leading coefficient is not one, factor it from the quadratic and linear terms before completing the square. Any compensation inside that bracket is also multiplied by the outside coefficient. Keep this multiplication visible before combining constants.

To solve an equation using square form, isolate the squared bracket. A positive right-hand side gives two square-root branches; zero gives one distinct real value; a negative right-hand side gives no real solution.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Reveal a minimum

Complete the square in $x^{2} - 8 x + 11$ .

Worked solution

1
Find the missing constant for a perfect square.
${(\frac{- 8}{2})}^{2} = 16$
Half the linear coefficient is negative four, and its square is sixteen.
2
Add and subtract that constant.
$x^{2} - 8 x + 11 = (x^{2} - 8 x + 16) - 16 + 11$
The added and subtracted sixteen cancel, preserving the original expression.
3
Write the square and simplify.
$x^{2} - 8 x + 11 = {(x - 4)}^{2} - 5$
The first three terms form a perfect square and the remaining constants sum to negative five.

Answer

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

EXAMPLE 02

Leading coefficient two

Rewrite $2 x^{2} + 12 x + 5$ and find its minimum.

Worked solution

1
Factor the leading coefficient from the variable terms.
$2 x^{2} + 12 x + 5 = 2 (x^{2} + 6 x) + 5$
Completing the square inside a monic bracket keeps the coefficient two accounted for.
2
Complete the square inside the bracket.
$x^{2} + 6 x = {(x + 3)}^{2} - 9$
The square introduces nine, so nine must be subtracted inside the bracket.
3
Distribute the outside coefficient.
$\begin{aligned} 2 x^{2} + 12 x + 5 & = 2 [{(x + 3)}^{2} - 9] + 5 \\ = 2 {(x + 3)}^{2} - 13 \end{aligned}$
The compensating negative nine is multiplied by two before adding five.
4
Locate and evaluate the minimum.
$2 {(x + 3)}^{2} \geq 0, x = - 3 ⟹ y = - 13$
The squared term has minimum zero, reached at negative three, so the expression has minimum negative thirteen.

Answer

$2 {(x + 3)}^{2} - 13$ ; minimum -13 at x=-3.

03 / YOUR TURN

Try it yourself.

Check your understanding

Complete the square: $x^{2} + 10 x + 7$ .

Show answer & reasoning

${(x + 5)}^{2} - 18$

Half of ten is five, whose square is twenty-five. Thus $x^{2} + 10 x + 7 = {(x + 5)}^{2} - 25 + 7 = {(x + 5)}^{2} - 18$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Write $x^{2} + 4 x + 1$ in completed-square form.
2 marks
Write $x^{2} - 6 x + 2$ in completed-square form.
2 marks
Rewrite $3 x^{2} - 12 x + 8$ in square form.
3 marks
Solve ${(x - 2)}^{2} = 7$ exactly.
2 marks
Find the maximum of $- 2 {(x + 1)}^{2} + 9$ .
2 marks
Solve $x^{2} - 4 x - 1 = 0$ by completing the square.
3 marks

Read the worked solutions

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

Write $x^{2} + 4 x + 1$ in completed-square form.
2 marks
Add/subtract 4.
Group the perfect square.
${(x + 2)}^{2} - 3$
Write $x^{2} - 6 x + 2$ in completed-square form.
2 marks
Add/subtract 9.
Combine constants.
${(x - 3)}^{2} - 7$
Rewrite $3 x^{2} - 12 x + 8$ in square form.
3 marks
Use $3 (x^{2} - 4 x) + 8$ .
Replace the inside with ${(x - 2)}^{2} - 4$ .
$3 {(x - 2)}^{2} - 4$
Solve ${(x - 2)}^{2} = 7$ exactly.
2 marks
Take both roots.
Add 2.
$x = 2 \pm \sqrt{7}$
Find the maximum of $- 2 {(x + 1)}^{2} + 9$ .
2 marks
The square is at least zero.
Its negative multiple is at most zero.
9 at $x = - 1$ .
Solve $x^{2} - 4 x - 1 = 0$ by completing the square.
3 marks
Rewrite ${(x - 2)}^{2} - 5 = 0$ .
Isolate the square and take both roots.
$x = 2 \pm \sqrt{5}$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

In this lesson

Understand the idea.

Create a perfect square

Keep the outside coefficient

See the method in action.

Reveal a minimum

Find the missing constant for a perfect square.

Add and subtract that constant.

Write the square and simplify.

Leading coefficient two

Factor the leading coefficient from the variable terms.

Complete the square inside the bracket.

Distribute the outside coefficient.

Locate and evaluate the minimum.

Try it yourself.

Check your understanding

Keep practising