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A11, A18Completing the square

Edexcel GCSE Maths (1MA1), Higher tier · Algebra

Practise Completing the square. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy2 marks
Write \(n^{2} - 6n - 12\) in the form \((n + p)^2 + q\) where \(p\) and \(q\) are integers.[2]
Show the answer and mark scheme
Answer: \((n - 3)^2 - 21\)
  • M1 for \((n - 3)^2\) or \((n - 3)^2 - 9 - 12\)
  • A1 for \((n - 3)^2 - 21\) cao

Worked solution: Halve the coefficient of \(n\): \(n^{2} - 6n - 12 = (n - 3)^2 - 9 - 12 = (n - 3)^2 - 21\)

Question 2Medium3 marks
(a) Write \(y^{2} + y - 12\) in the form \((y + a)^2 + b\) where \(a\) and \(b\) are fractions.[2]
(b) Hence write down the value of \(y\) for which \(y^{2} + y - 12\) is a minimum.[1]
Show the answer and mark scheme
(a) Answer: \((y + \frac{1}{2})^2 - \frac{49}{4}\)
  • M1 for \((y + \frac{1}{2})^2\) oe
  • A1 for \((y + \frac{1}{2})^2 - \frac{49}{4}\) oe

Worked solution: \(y^{2} + y - 12 = (y + \frac{1}{2})^2 - \frac{1}{4} - 12 = (y + \frac{1}{2})^2 - \frac{49}{4}\)

(b) Answer: \(y = -\frac{1}{2}\)
  • B1 for \(-\frac{1}{2}\) oe (ft their \(a\))

Worked solution: The minimum occurs when \(y + \frac{1}{2} = 0\), i.e. \(y = -\frac{1}{2}\).

Question 3Hard4 marks
Kim is finding the turning point of the curve \(y = x^2 - 8x + 21\).
She writes \(x^2 - 8x + 21 = (x - 4)^2 + 5\), so the turning point is \((-4, 5)\).
(a) Explain Kim's mistake and write down the correct coordinates of the turning point.[2]
(b) Find the set of values of \(k\) for which the equation \(x^2 - 8x + 21 = k\) has two different solutions.[2]
Show the answer and mark scheme
(a) Answer: The minimum is when \(x - 4 = 0\), i.e. \(x = 4\), not −4. The turning point is \((4, 5)\).
  • C1 for explaining that the minimum occurs when \((x - 4)^2 = 0\), so \(x = 4\) (Kim has the wrong sign)
  • B1 for \((4, 5)\)

Worked solution: \((x - 4)^2 \ge 0\) and it equals 0 when \(x = 4\). So the least value of \(y\) is 5, when \(x = 4\): the turning point is \((4, 5)\).

(b) Answer: \(k \gt 5\)
  • M1 for using the minimum value 5 (the line \(y = k\) must be above the turning point)
  • A1 for \(k \gt 5\)

Worked solution: \((x - 4)^2 + 5 = k\) gives \((x - 4)^2 = k - 5\). There are two different solutions when \(k - 5 \gt 0\), i.e. \(k \gt 5\).

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