Completing the square
Algebra
- Higher
- Non-calculator
- 8 questions
- 29 marks
These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.
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Write $x^{2} + 6x + 2$ in the form $(x + a)^{2} + b$.
2 marks
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Write $x^{2} - 4x + 9$ in the form $(x + a)^{2} + b$.
2 marks
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Write $x^{2} + 8x - 3$ in completed-square form.
3 marks
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Solve $x^{2} + 6x + 2 = 0$ by completing the square.
3 marks
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The curve $y = x^{2} - 6x + 5$ has a minimum point. Find it by completing the square.
4 marks
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Write $2x^{2} + 12x + 7$ in the form $a(x + b)^{2} + c$.
4 marks
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Solve $2x^{2} - 8x + 3 = 0$ by completing the square. Give the exact answers.
5 marks
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For $y = x^{2} + 2x - 8$, write the completed-square form, write down the turning point, and solve $y = 0$.
6 marks
Mark scheme
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Answer. $(x + 3)^{2} - 7$
- $(x + 3)^{2} = x^{2} + 6x + 9$
- $(x + 3)^{2} - 7$
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Answer. $(x - 2)^{2} + 5$
- $(x - 2)^{2} = x^{2} - 4x + 4$
- $(x - 2)^{2} + 5$
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Answer. $(x + 4)^{2} - 19$
- $(x + 4)^{2} = x^{2} + 8x + 16$
- $(x + 4)^{2} - 19$
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Answer. $x = -3 \pm \sqrt{7}$
- $(x + 3)^{2} - 7 = 0$
- $x = -3 \pm \sqrt{7}$
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Answer. $(3, -4)$
- $y = (x - 3)^{2} - 9 + 5 = (x - 3)^{2} - 4$
- The minimum is at $(3, -4)$.
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Answer. $2(x + 3)^{2} - 11$
- $2(x^{2} + 6x) + 7 = 2\left((x + 3)^{2} - 9\right) + 7$
- $2(x + 3)^{2} - 11$
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Answer. $x = 2 \pm \dfrac{\sqrt{10}}{2}$
- $2\left((x - 2)^{2} - 4\right) + 3 = 0$
- $(x - 2)^{2} = \dfrac{5}{2}$
- $x = 2 \pm \sqrt{\dfrac{5}{2}} = 2 \pm \dfrac{\sqrt{10}}{2}$
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Answer. $(x + 1)^{2} - 9$, turning point $(-1, -9)$, $x = 2$ or $x = -4$
- $y = (x + 1)^{2} - 1 - 8 = (x + 1)^{2} - 9$.
- Turning point $(-1, -9)$.
- $(x + 1)^{2} = 9$, so $x + 1 = \pm 3$.
- $x = 2$ or $x = -4$.