Algebra

Completing the square

Q: How do you complete the square?

Halve the coefficient of x, write (x + \text{half})^2, then subtract the half squared and add the original constant.

Q: How do you find the turning point from completing the square?

For (x + p)^2 + q, the turning point is (-p, q). Change the sign of the number in the bracket; keep the sign of the number outside.

10 past-paper questions, easiest first. Try each one, then reveal the answer.

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Completing the square: what you need to know

Completing the square rewrites a quadratic as $(x + p)^2 + q$ . In that form you can read the turning point straight off, which is why it appears in grade 7–8 questions about sketching curves.

Key facts

$x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$
The turning point of $y = (x + p)^2 + q$ is $(-p, q)$ : change the sign inside the bracket only.
For $a(x + p)^2 + q$ the turning point is still $(-p, q)$ . It is a minimum if $a$ is positive and a maximum if $a$ is negative.
If the $x^2$ term has a coefficient (even $-1$ ), take it out of the $x^2$ and $x$ terms first: $2x^2 + 12x = 2(x^2 + 6x)$ .
To solve $(x + p)^2 + q = 0$ , rearrange to $(x + p)^2 = -q$ and square root both sides, remembering $\pm$ : e.g. $(x + 3)^2 = 5$ gives $x = -3 \pm \sqrt{5}$ .

Completing the square

If needed, factor out the coefficient of $x^2$ from the $x^2$ and $x$ terms.
Halve the coefficient of $x$ and write $(x + \text{half})^2$ .
Subtract the square of that half (multiplied by any factor you took out).
Tidy up the constants.

Worked examples

Example 1 Higher · grade 7

Write $x^2 - 6x + 11$ in the form $(x - a)^2 + b$ , and give the turning point of $y = x^2 - 6x + 11$ .

Show the solution

Half of $-6$ is $-3$ : $(x - 3)^2 = x^2 - 6x + 9$
$x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2$
Turning point: $(3, 2)$

Example 2 Higher · grade 8

Write $2x^2 + 12x + 5$ in the form $a(x + b)^2 + c$ and find the turning point of its graph.

Show the solution

Factor 2 from the $x$ terms: $2(x^2 + 6x) + 5$
$x^2 + 6x = (x + 3)^2 - 9$
$2[(x + 3)^2 - 9] + 5 = 2(x + 3)^2 - 18 + 5$
$= 2(x + 3)^2 - 13$
Turning point: $(-3, -13)$

Example 3 Higher · grade 8

Find the turning point of $y = 3 + 8x - 2x^2$ and say whether it is a maximum or a minimum.

Show the solution

Rewrite in order: $y = -2x^2 + 8x + 3$
Factor $-2$ from the $x$ terms: $-2(x^2 - 4x) + 3$
$x^2 - 4x = (x - 2)^2 - 4$
$-2[(x - 2)^2 - 4] + 3 = -2(x - 2)^2 + 8 + 3$
$= -2(x - 2)^2 + 11$
Turning point $(2, 11)$ . The $x^2$ coefficient is negative, so it is a maximum.

Common mistakes

Getting the sign of the turning point wrong: $(x + 3)^2 - 13$ gives $x = -3$ , not $3$ .
Forgetting to multiply the subtracted square by the factor taken out: in $2[(x + 3)^2 - 9]$ the $-9$ becomes $-18$ , and in $-2[(x - 2)^2 - 4]$ the $-4$ becomes $+8$ .
Adding the square instead of subtracting it.

Exam tips

Expand your completed-square form to check it matches the original quadratic.
If a question says 'hence' find the turning point, read it from your completed square rather than starting again.

Questions people ask

How do you complete the square?

Halve the coefficient of $x$ , write $(x + \text{half})^2$ , then subtract the half squared and add the original constant.

How do you find the turning point from completing the square?

For $(x + p)^2 + q$ , the turning point is $(-p, q)$ . Change the sign of the number in the bracket; keep the sign of the number outside.

Exam questions: Completing the square (10, easiest first)

Completing the square3 marksGrade 7CalculatorHigher

June 2024 · Paper 2 Higher · Q20

Find the coordinates of the turning point of a quadratic curve.

Show answer

Answer	$(3, -11)$	3 marks

Worked solution Checked against the mark scheme

$y = 2x^2 - 12x + 7 = 2(x^2 - 6x) + 7 = 2[(x-3)^2 - 9] + 7 = 2(x-3)^2 - 11$

The minimum is when $x = 3$ , giving $y = -11$ .

Turning point: $(3, -11)$

Watch out: Forgetting to multiply the $-9$ by 2, giving $(3, -2)$ .

Completing the square7 marksGrade 7CalculatorHigher

November 2023 · Paper 3 Higher · Q17

Complete the square and state the turning point, solve a quadratic with the formula, and spot an error in factorised solutions.

Show answer

(a(i))	$(x - 4)^2 - 13$	2 marks
(a(ii))	$(4, -13)$	1 mark
(b)	$x = 0.449$ or $x = -1.59$	3 marks
(c)	He ignored the 3: from $3k - 2 = 0$ the solution is $k = \frac{2}{3}$ , not 2	1 mark

Worked solution Checked against the mark scheme

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

(ii) Turning point $(4, -13)$

(b) $x = \dfrac{-8 \pm \sqrt{8^2 - 4(7)(-5)}}{2(7)} = \dfrac{-8 \pm \sqrt{204}}{14}$

$x = 0.449$ or $x = -1.59$ (3 s.f.)

Watch out: In (a)(ii), giving the turning point as (−4, −13).

Completing the square3 marksGrade 7CalculatorHigher

November 2022 · Paper 3 Higher · Q21

Complete the square to confirm the coordinates of a quadratic's turning point.

Show answer

Answer	$y = 4(x - 7)^2 - 196$ , so the turning point is $(7, -196)$	3 marks

Worked solution Checked against the mark scheme

$y = 4(x^2 - 14x)$

$x^2 - 14x = (x - 7)^2 - 49$

$y = 4\left((x - 7)^2 - 49\right) = 4(x - 7)^2 - 196$

The minimum is when $x - 7 = 0$ , so the turning point is $(7, -196)$

Watch out: Forgetting to multiply the $-49$ by 4.

Completing the square1 markGrade 7CalculatorHigher

November 2020 · Paper 2 Higher · Q17

State the turning point of a quadratic graph given in completed-square form.

Show answer

Answer	$(-12, -7)$	1 mark

Worked solution Checked against the mark scheme

For $y = (x + a)^2 + b$ the turning point is $(-a, b)$ .

Here $a = 12$ and $b = -7$ , so the turning point is $(-12, -7)$ .

Watch out: Giving $(12, -7)$ by not changing the sign of the 12.

Completing the square3 marksGrade 7No calculatorHigher

June 2019 · Paper 1 Higher · Q19

Complete the square on a quadratic, then use it to state the turning point of its graph.

Show answer

((i))	$a = 3,\ b = 8$	2 marks
((ii))	$(3, -8)$	1 mark

Worked solution Checked against the mark scheme

(i) Halve $-6$ to get $(x - 3)^2 = x^2 - 6x + 9$ .

$x^2 - 6x + 1 = (x - 3)^2 - 9 + 1 = (x - 3)^2 - 8$ , so $a = 3$ and $b = 8$ .

(ii) The turning point is $(a, -b) = (3, -8)$ .

Watch out: Giving $b = -8$ or a turning point of $(-3, -8)$ .

Completing the square2 marksGrade 7CalculatorHigher

Specimen · Paper 3 Higher · Q11

Complete the square on a quadratic expression to write it in the form (x + m)² + n.

Show answer

Answer	$(x + 1)^2 - 9$	2 marks

Worked solution Checked against the mark scheme

Halve the $x$ coefficient: $x^2 + 2x = (x + 1)^2 - 1$

So $x^2 + 2x - 8 = (x + 1)^2 - 1 - 8 = (x + 1)^2 - 9$

$m = 1$ , $n = -9$

Watch out: Writing (x + 1)² − 8, forgetting to subtract the 1.

Completing the square5 marksGrade 8CalculatorHigher

November 2024 · Paper 3 Higher · Q23

Complete the square to find the turning point of a quadratic involving a constant p, then sketch it with its y-intercept.

Show answer

Answer	U-shaped curve with minimum at $(3p,\ -9p^2 - 7)$ in the fourth quadrant and y-intercept $(0, -7)$	5 marks

Worked solution Checked against the mark scheme

$y = x^2 - 6px - 7 = (x - 3p)^2 - 9p^2 - 7$

Turning point (minimum): $(3p,\ -9p^2 - 7)$

When $x = 0$ , $y = -7$ , so the y-intercept is $(0, -7)$ .

Since $p > 0$ , $3p > 0$ and $-9p^2 - 7 < 0$ : the minimum is in the fourth quadrant. Sketch a U-shape through $(0, -7)$ with its lowest point there.

Watch out: Writing the turning point as (−3p, ...) by getting the sign in the bracket wrong.

Completing the square4 marksGrade 8No calculatorHigher

November 2021 · Paper 1 Higher · Q22

Complete the square on a quadratic with a negative squared term to find its turning point.

Show answer

Answer	$(3, 36)$	4 marks

Worked solution Checked against the mark scheme

$y = -3x^2 + 18x + 9 = -3(x^2 - 6x) + 9$

$= -3[(x - 3)^2 - 9] + 9$

$= -3(x - 3)^2 + 27 + 9 = -3(x - 3)^2 + 36$

Turning point: $(3, 36)$

Watch out: Forgetting to multiply the $-9$ by $-3$ , giving $(3, 0)$ .

Completing the square5 marksGrade 8No calculatorHigher

November 2019 · Paper 1 Higher · Q21

Sketch a quadratic curve, labelling its turning point and the exact points where it meets the axes.

Show answer

Answer	U-shaped curve with turning point $(2, -13)$ , $y$ -intercept $(0, -5)$ and $x$ -intercepts $\left(2 \pm \sqrt{\frac{13}{2}}, 0\right)$	5 marks

Worked solution Checked against the mark scheme

$y$ -intercept: $x = 0$ gives $y = -5$

Complete the square: $2x^2 - 8x - 5 = 2(x - 2)^2 - 8 - 5 = 2(x-2)^2 - 13$

Turning point: $(2, -13)$

$x$ -intercepts: $2(x-2)^2 = 13 \Rightarrow x = 2 \pm \sqrt{\frac{13}{2}}$

Sketch a U shape through these points.

Watch out: Forgetting to take out the factor of 2 before completing the square, giving (x − 4)².

Completing the square4 marksGrade 8No calculatorHigher

November 2018 · Paper 1 Higher · Q17

Use two points on a quadratic curve to find its equation, then find its turning point.

Show answer

Answer	$(2, -9)$	4 marks

Worked solution Checked against the mark scheme

At $(0, -5)$ : $b = -5$

At $(5, 0)$ : $0 = 25 + 5a - 5$ , so $a = -4$

$y = x^2 - 4x - 5 = (x - 2)^2 - 9$

Turning point: $(2, -9)$

Watch out: Getting the sign of the x-coordinate wrong when completing the square.