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Maths · Equations and inequalities

Completing the square

Write quadratic expressions in the form (x + a) squared plus b, solve equations by completing the square, and find the turning point of a quadratic graph.

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  • 6 key terms
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Teacher resources

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Student handouts

The same files the students see, to print or hand out.

Warm-up

Answer each one, then check.

  1. 1

    Expand \((x + 3)^2\).

    Show answerHide answer

    \(x^2 + 6x + 9\)

  2. 2

    Expand \((x - 4)^2\).

    Show answerHide answer

    \(x^2 - 8x + 16\)

  3. 3

    Work out \(\sqrt{11}\) to 2 decimal places.

    Show answerHide answer

    3.32

  4. 4

    Half of 6 is what?

    Show answerHide answer

    3

  5. 5

    What is the turning point of \(y = x^2\)?

    Show answerHide answer

    \((0, 0)\)

Learning Objectives

  1. 1Write \(x^2 + bx + c\) in the form \((x + p)^2 + q\).
  2. 2Complete the square when the coefficient of \(x^2\) is not 1.
  3. 3Solve quadratic equations by completing the square.
  4. 4Find the turning point of a quadratic graph.

THE METHOD

\(x^2 + bx + c = \left(x + \dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2 + c\)

Halve the coefficient of \(x\), square it, and subtract it.

Completing the Square

Write \(x^2 + 6x + 5\) in the form \((x + p)^2 + q\).

Show the solutionHide the solution
  1. 1 Halve the coefficient of \(x\) \(6 \div 2 = 3\), so \((x + 3)^2\)
  2. 2 \((x + 3)^2\) expands to \(x^2 + 6x + 9\)
  3. 3 Adjust the constant: we need 5, not 9 \(5 - 9 = -4\)
  4. 4 Write the result \((x + 3)^2 - 4\)

Answer\((x + 3)^2 - 4\)

A Negative Coefficient

Write \(x^2 - 8x + 3\) in the form \((x + p)^2 + q\).

Show the solutionHide the solution
  1. 1 Halve \(-8\) \(-4\), so \((x - 4)^2\)
  2. 2 \((x - 4)^2\) expands to \(x^2 - 8x + 16\)
  3. 3 Adjust the constant \(3 - 16 = -13\)

Answer\((x - 4)^2 - 13\)

Reading the Completed Square Form

  • \(y = (x + 3)^2 - 4\)

    Turning point \((-3, -4)\): the sign inside the bracket is reversed.

  • Line of symmetry

    \(x = -3\).

  • Minimum value

    \(-4\), when \(x = -3\).

  • Never below the minimum

    \((x + 3)^2 \ge 0\), so \(y \ge -4\).

Solving by Completing the Square

Solve \(x^2 + 6x - 2 = 0\), leaving your answer in surd form.

Show the solutionHide the solution
  1. 1 Complete the square \((x + 3)^2 - 9 - 2 = 0\), so \((x + 3)^2 - 11 = 0\)
  2. 2 Rearrange \((x + 3)^2 = 11\)
  3. 3 Take the square root of both sides \(x + 3 = \pm\sqrt{11}\)
  4. 4 Solve \(x = -3 \pm \sqrt{11}\)

Answer\(x = -3 + \sqrt{11}\) or \(x = -3 - \sqrt{11}\)

A Coefficient of x² Not 1

Write \(2x^2 + 12x + 7\) in the form \(a(x + p)^2 + q\).

Show the solutionHide the solution
  1. 1 Take out the 2 from the x terms \(2(x^2 + 6x) + 7\)
  2. 2 Complete the square inside \(2\left[(x + 3)^2 - 9\right] + 7\)
  3. 3 Multiply out the 2 \(2(x + 3)^2 - 18 + 7\)
  4. 4 Simplify \(2(x + 3)^2 - 11\)

Answer\(2(x + 3)^2 - 11\)

A Proof

Show that \(x^2 + 4x + 5\) is always positive.

Show the solutionHide the solution
  1. 1 Complete the square \(x^2 + 4x + 5 = (x + 2)^2 + 1\)
  2. 2 A square is never negative \((x + 2)^2 \ge 0\)
  3. 3 Add 1 \((x + 2)^2 + 1 \ge 1\)

Answer\((x + 2)^2 + 1 \ge 1 > 0\), so the expression is always positive.

Complete the Square Race

Write each in the form \((x + p)^2 + q\), then state the turning point of its graph. (a) \(x^2 + 4x + 1\) (b) \(x^2 - 10x + 30\) (c) \(x^2 + 2x - 8\) (d) \(x^2 - 3x\).

1. Halve the coefficient of x.

2. Subtract its square.

3. Read off the turning point.

A good answer shows: (a) \((x + 2)^2 - 3\), turning point \((-2, -3)\). (b) \((x - 5)^2 + 5\), turning point \((5, 5)\). (c) \((x + 1)^2 - 9\), turning point \((-1, -9)\). (d) \(\left(x - \dfrac{3}{2}\right)^2 - \dfrac{9}{4}\), turning point \(\left(\dfrac{3}{2}, -\dfrac{9}{4}\right)\).

Can I...?

  1. 1Complete the square when \(a = 1\).
  2. 2Complete the square when \(a \ne 1\).
  3. 3Find the turning point from the completed form.
  4. 4Solve by completing the square.
  5. 5Give answers as surds.
  6. 6Use the form to find the minimum value.
  7. 7Prove a quadratic is always positive.
  8. 8Check by expanding.

Summary & Exam Focus

  • \(x^2 + bx + c = (x + \tfrac{b}{2})^2 + c - (\tfrac{b}{2})^2\).
  • Turning point of \(y = (x + p)^2 + q\) is \((-p, q)\).
  • Solve by rearranging to \((x + p)^2 = k\).
  • \((x + p)^2 \ge 0\) helps with proofs.

Exam focus

Write \(x^2 - 8x + 3\) in the form \((x + a)^2 + b\). (2 marks) (2 marks)

Halve the coefficient of \(x\) (keeping its sign) and square it. Then adjust the constant. Always check by expanding.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Completing the square
Rewriting a quadratic as a squared bracket plus a constant.
Turning point
The lowest or highest point of a quadratic graph.
Minimum value
The smallest value a function can take.
Line of symmetry
The vertical line through the turning point.
Surd
A root left in exact form, such as \(\sqrt{11}\).
Coefficient
The number multiplying a variable.

Questions and answers

12 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Non-calculator 2 marks Easier

Write \(x^2 + 6x + 5\) in the form \((x + a)^2 + b\).

Mark scheme — 2 marks available

  • \((x + 3)^2\) — M1
  • \((x + 3)^2 - 4\) — A1

Model answer

\((x + 3)^2 - 9 + 5 = (x + 3)^2 - 4\).

2. Exam question Non-calculator 2 marks Easier

Write \(x^2 - 8x + 3\) in the form \((x + a)^2 + b\).

Mark scheme — 2 marks available

  • \((x - 4)^2\) — M1
  • \((x - 4)^2 - 13\) — A1

Model answer

\((x - 4)^2 - 16 + 3 = (x - 4)^2 - 13\).

3. Exam question Non-calculator 3 marks Easier

Solve \(x^2 + 6x - 2 = 0\). Give your answers in the form \(p \pm \sqrt{q}\).

Mark scheme — 3 marks available

  • \((x + 3)^2 - 9 - 2\) — M1
  • \((x + 3)^2 = 11\) — M1
  • \(-3 \pm \sqrt{11}\) — A1

Model answer

\((x + 3)^2 - 11 = 0\), so \((x + 3)^2 = 11\) and \(x = -3 \pm \sqrt{11}\).

4. Exam question Non-calculator 3 marks Easier

The graph of \(y = x^2 - 4x + 7\) has a minimum point. Find the coordinates of the minimum point.

Mark scheme — 3 marks available

  • \((x - 2)^2\) — M1
  • \((x - 2)^2 + 3\) — M1
  • \((2, 3)\) — A1

Model answer

\(x^2 - 4x + 7 = (x - 2)^2 + 3\), so the minimum point is \((2, 3)\).

5. Exam question Non-calculator 3 marks Easier

Write \(2x^2 + 12x + 7\) in the form \(a(x + p)^2 + q\).

Mark scheme — 3 marks available

  • \(2(x^2 + 6x)\) — M1
  • \(2(x + 3)^2 - 18 + 7\) — M1
  • \(2(x + 3)^2 - 11\) — A1

Model answer

\(2(x^2 + 6x) + 7 = 2[(x + 3)^2 - 9] + 7 = 2(x + 3)^2 - 11\).

6. Exam question Show that 3 marks Easier

Show that \(x^2 + 4x + 5\) is always positive for all values of \(x\).

Mark scheme — 3 marks available

  • \((x + 2)^2 + 1\) — M1
  • \((x + 2)^2 \ge 0\) — M1
  • Conclusion — C1

Model answer

\(x^2 + 4x + 5 = (x + 2)^2 + 1\). A square is never negative, so \((x + 2)^2 \ge 0\) and the expression is at least 1, which is positive.

7. Multiple choice 1 mark Core

Complete the square: \(x^2 + 10x\).

  1. A \((x + 10)^2 - 100\)
  2. B \((x + 5)^2 - 25\) Correct
  3. C \((x + 5)^2 + 25\)
  4. D \((x - 5)^2 - 25\)

Why: \((x + 5)^2 - 25\).

8. Multiple choice 1 mark Core

What is the turning point of \(y = (x - 2)^2 + 3\)?

  1. A \((-2, 3)\)
  2. B \((-2, -3)\)
  3. C \((2, 3)\) Correct
  4. D \((3, 2)\)

Why: \((2, 3)\): the sign inside the bracket reverses.

9. Multiple choice 1 mark Core

What is the minimum value of \((x + 4)^2 - 7\)?

  1. A \(-7\) Correct
  2. B \(7\)
  3. C \(-4\)
  4. D \(4\)

Why: The square is at least 0, so the minimum is \(-7\).

10. Multiple choice 1 mark Core

Solve \((x + 1)^2 = 9\).

  1. A \(x = 2\) only
  2. B \(x = 8\)
  3. C \(x = -10\) or \(8\)
  4. D \(x = 2\) or \(x = -4\) Correct

Why: \(x + 1 = \pm 3\), so \(x = 2\) or \(x = -4\).

11. Multiple choice 1 mark Core

Complete the square: \(x^2 - 6x + 2\).

  1. A \((x + 3)^2 - 7\)
  2. B \((x - 3)^2 - 7\) Correct
  3. C \((x - 3)^2 + 7\)
  4. D \((x - 6)^2 - 34\)

Why: \((x - 3)^2 - 9 + 2 = (x - 3)^2 - 7\).

12. Multiple choice 1 mark Stretch

The graph of \(y = x^2 + 6x + 5\) has its turning point at...

  1. A \((3, -4)\)
  2. B \((-3, 4)\)
  3. C \((-3, -4)\) Correct
  4. D \((6, 5)\)

Why: \(y = (x + 3)^2 - 4\), so the turning point is \((-3, -4)\).