EDEXCEL GCSE MATHS · HIGHER
Exam Practice: Completing the Square
Completing the square · Equations and inequalities · Lesson 4 of 7 · 16 marks · 30 minutes
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Instructions
• Answer all the questions.
• Write your answers in the spaces provided.
• The marks for each question are shown in brackets - use this as a guide to how much to write.
• The answers are on separate pages at the back. Attempt every question before you look at them.
• Answer all questions. Show your working.
Question 1 NON-CALCULATOR (2 marks)
Write x² + 6x + 5 in the form (x + a)² + b.
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(Total for Question 1 = 2 marks)
Question 2 NON-CALCULATOR (2 marks)
Write x² − 8x + 3 in the form (x + a)² + b.
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(Total for Question 2 = 2 marks)
Question 3 NON-CALCULATOR (3 marks)
Solve x² + 6x − 2 = 0. Give your answers in the form p ± √q.
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(Total for Question 3 = 3 marks)
Question 4 NON-CALCULATOR (3 marks)
The graph of y = x² − 4x + 7 has a minimum point. Find the coordinates of the minimum point.
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(Total for Question 4 = 3 marks)
Question 5 NON-CALCULATOR (3 marks)
Write 2x² + 12x + 7 in the form a(x + p)² + q.
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(Total for Question 5 = 3 marks)
Question 6 SHOW THAT (3 marks)
Show that x² + 4x + 5 is always positive for all values of x.
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(Total for Question 6 = 3 marks)
TOTAL FOR PAPER = 16 MARKS
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Answers and mark scheme
Check your answer only once you have written one.
Question 1 (2 marks)
(x + 3)² − 9 + 5 = (x + 3)² − 4.
• (x + 3)² M1
• (x + 3)² − 4 A1
Question 2 (2 marks)
(x − 4)² − 16 + 3 = (x − 4)² − 13.
• (x − 4)² M1
• (x − 4)² − 13 A1
Question 3 (3 marks)
(x + 3)² − 11 = 0, so (x + 3)² = 11 and x = −3 ± √11.
• (x + 3)² − 9 − 2 M1
• (x + 3)² = 11 M1
• −3 ± √11 A1
Question 4 (3 marks)
x² − 4x + 7 = (x − 2)² + 3, so the minimum point is (2, 3).
• (x − 2)² M1
• (x − 2)² + 3 M1
• (2, 3) A1
Question 5 (3 marks)
2(x² + 6x) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 11.
• 2(x² + 6x) M1
• 2(x + 3)² − 18 + 7 M1
• 2(x + 3)² − 11 A1
Question 6 (3 marks)
x² + 4x + 5 = (x + 2)² + 1. A square is never negative, so (x + 2)² ≥ 0 and the expression is at least 1, which is positive.
• (x + 2)² + 1 M1
• (x + 2)² ≥ 0 M1
• Conclusion C1