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Completing the square

Free Printable Completing the square PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 8+)

Download free completing the square worksheets (PDF), GCSE maths past paper questions and worked solutions for Completing the square (8D). This is a Grade 8+ algebra topic from the GCSE questions by topic collection for Foundation and Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from IGCSE MT, Maths Genie, Corbett Maths, Maths Made Easy, Naiker Maths, 1st Class Maths, Maths4EveryOne, Metatutor, Dr Austin, GCSE Maths Rocket, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our completing the square revision notes and worked examples with diagrams, try real GCSE past paper completing the square questions with step-by-step solutions, or download a PDF of all completing the square worksheet links.

Free Completing the square PDF Worksheets & Answers

Grade 8+ • Topic 8D
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GCSE Maths Grade 8+ Revision Notes & Method

Guide to Completing the square: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

To write x2 + bx + c in completed square form (x + p)2 + q, halve the coefficient of x to get p = b2, then subtract p2: (x + b2)2 − (b2)2 + c. The graph of y = (x + p)2 + q has its turning point (vertex) at (−p, q) and line of symmetry x = −p.

When the coefficient of x2 is not 1 (ax2 + bx + c), factorise a out of the first two terms before completing the square inside the square brackets, then multiply out by a at the end.

Example 1: Write x2 − 8x + 5 in the form (x − a)2 + b and state the turning point of y = x2 − 8x + 5.

5 x = 4 (4, −11)

Solution:

Halve −8 to get −4: (x − 4)2 − (−4)2 + 5
= (x − 4)2 − 16 + 5 = (x − 4)2 − 11
Minimum occurs when x − 4 = 0 ⇒ x = 4, y = −11
Answer: (x − 4)2 − 11; turning point (4, −11)

Example 2: Write 2x2 + 12x + 7 in the form a(x + b)2 + c.

Solution:

Factorise 2 from first two terms: 2[x2 + 6x] + 7
Complete square inside: 2[(x + 3)2 − 9] + 7
Multiply out by 2: 2(x + 3)2 − 18 + 7
Answer: 2(x + 3)2 − 11
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Completing the square Questions & Answers

Past Paper Example 1 GCSE Nov 2024 3H Q23 5 marks
Sketch the graph of
y = x2 - 6px - 7\quadwhere p > 0
showing the coordinates of the turning point, in terms of p, and the coordinates of the
intercept with the y-axis.
You must show all your working.
yxO
Worked Solution & Mark Scheme (5 marks)
y = (x – 3p)² – 9p² – 7 ⇒ turning point at (3p, –9p² – 7) in 4th quadrant since p > 0 M4
y-intercept at (0, –7) A1
yxO(0, –7)(3p, –9p² – 7)
Answer: Parabola sketched with y-intercept at (0, –7) and minimum turning point at (3p, –9p² – 7) in the 4th quadrant
Past Paper Example 2 IGCSE Jun 2025 2H Q25 6 marks
(a) Write 28 + 24x – 6x² in the form a – b(x – c)². (3) (b) Sketch the graph of y = 28 + 24x – 6x², showing the turning point and y-intercept. (3)
(a) Write 28 + 24x - 6x2 in the form a - b(x - c)2 where a, b and c are integers. [3 marks]
(b) On the axes below, sketch the graph of y = 28 + 24x - 6x2
Show clearly the coordinates of the turning point and the coordinates of the point of intersection of the graph with the y-axis.
[3 marks]
xyO
Worked Solution & Mark Scheme (6 marks)
Part (a) Solution:
28 – 6(x² – 4x) = 28 – 6[(x – 2)² – 4] = 52 – 6(x – 2)² M2 A1
Answer (a): 52 - 6(x - 2)2
Part (b) Solution:
Max at (2, 52), y-intercept at (0, 28) M2 A1
xyO(0, 28)(2, 52)
Answer (b): Inverted parabola with maximum at (2, 52) and y-intercept at (0, 28)
Past Paper Example 3 OCR Jun 2019 5H Q17 7 marks
(a) Write x2 + 8x + 3 in the form (x + a)2 - b.
(b) Sketch the graph of y = x2 + 8x + 3.
Show clearly the coordinates of any turning points and the y-intercept.
(a) Write x2 + 8x + 3 in the form (x + a)2 - b. [3 marks]
(b) Sketch the graph of y = x2 + 8x + 3.
Show clearly the coordinates of any turning points and the y-intercept.
[4 marks]
Oxy
Worked Solution & Mark Scheme (7 marks)
Part (a) Solution:
(x + 4)² − 16 + 3 = (x + 4)² − 13. M2 A1
Answer (a): (x + 4)2 - 13
Part (b) Solution:
Minimum at (−4, −13), y-intercept at (0, 3) M3 A1
Oxy(−4, −13)(0, 3)
Answer (b): U-shaped parabola with minimum at (−4, −13) and y-intercept at (0, 3)

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