Completing the square

The method

1. Take out the coefficient of x² as a factor of the first two terms.
2. Halve the coefficient of x to get p inside the bracket.
3. Square that value and subtract it to keep the expression the same.
4. Tidy up the constants outside the bracket to get q.
5. Read the turning point straight off the completed form and check by expanding.

Mistakes that lose marks

• Changing the expression instead of subtracting the square of p to balance it.
• Forgetting to multiply the subtracted constant by a when the coefficient of x² is not 1.
• Reading the turning point as (p, q) instead of (−p, q).
• Losing the negative root when taking a square root.
• Leaving a fraction or surd unsimplified at the end.

7 exam-style questions

These are our own questions written in the style of the exam, not past-paper questions. The mark codes are M = method, A = accuracy and B = independent mark. Try each one, then open the solution.

1. Exam-style extended Write x² + 6x + 5 in the form (x + p)² + q. [2 marks]
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   1. (x + 3)² = x² + 6x + 9, which is 4 more than x² + 6x + 5 M1
   2. So x² + 6x + 5 = (x + 3)² − 4 A1

   Answer: (x + 3)² − 4

2. Exam-style extended Write x² − 8x + 3 in the form (x + p)² + q. [3 marks]
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   1. Half of −8 is −4, so start with (x − 4)² = x² − 8x + 16 M1
   2. Subtract 16 and then add the original 3: x² − 8x + 3 = (x − 4)² − 16 + 3 M1
   3. = (x − 4)² − 13 A1

   Answer: (x − 4)² − 13

3. Exam-style extended Write x² + 4x − 1 in the form (x + p)² + q, and hence solve x² + 4x − 1 = 0, giving exact answers. [4 marks]
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   1. (x + 2)² = x² + 4x + 4, so x² + 4x − 1 = (x + 2)² − 5 M1
   2. Completed square form: (x + 2)² − 5 A1
   3. Set (x + 2)² − 5 = 0, so (x + 2)² = 5 and x + 2 = ±√5 M1
   4. x = −2 ± √5 A1

   Answer: (x + 2)² − 5, so x = −2 ± √5

4. Exam-style extended The curve y = x² − 10x + 27 has a minimum point. Find the coordinates of the minimum point and state the minimum value of y. [4 marks]
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   1. Complete the square: y = (x − 5)² − 25 + 27 M1
   2. y = (x − 5)² + 2 A1
   3. The bracket is zero when x = 5, so the minimum point is (5, 2) A1
   4. The minimum value of y is 2 A1

   Answer: Minimum point (5, 2), minimum value y = 2

5. Exam-style extended Write 2x² + 8x + 5 in the form a(x + p)² + q. [3 marks]
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   1. Take out the 2 from the first two terms: 2(x² + 4x) + 5 M1
   2. Complete the square inside the bracket: 2((x + 2)² − 4) + 5 M1
   3. Multiply out: 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3 A1

   Answer: 2(x + 2)² − 3

6. Exam-style extended Solve 2x² + 8x + 5 = 0 by completing the square, giving your answers in surd form. [4 marks]
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   1. From the completed form: 2(x + 2)² − 3 = 0 M1
   2. So (x + 2)² = 3/2 and x + 2 = ±√(3/2) M1
   3. √(3/2) = √6 ÷ 2, so x = −2 + (√6 ÷ 2) A1
   4. The second root is x = −2 − (√6 ÷ 2) A1

   Answer: x = −2 ± (√6 ÷ 2)

7. Exam-style higher Solve 2x² − 10x + 7 = 0 by completing the square, giving your answers in the form (a ± √b) ÷ c. [4 marks]
   Show full solution

   1. Take out the 2: 2(x² − 5x) + 7 = 2((x − 5/2)² − 25/4) + 7 M1
   2. Multiply out: 2(x − 5/2)² − 25/2 + 7 = 2(x − 5/2)² − 11/2 M1
   3. Set equal to zero: (x − 5/2)² = 11/4, so x = 5/2 ± √11 ÷ 2 A1
   4. Write over one denominator: x = (5 ± √11) ÷ 2 A1

   Answer: x = (5 ± √11) ÷ 2

Where this is in the syllabus

Completing the square appears in Cambridge IGCSE Mathematics 0580 and Pearson Edexcel International GCSE Mathematics A 4MA1. For the official specifications and past papers, use the board's own website: Cambridge IGCSE Mathematics 0580 and Pearson Edexcel International GCSE Mathematics A 4MA1.

Questions students ask