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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.

  • Higher
  • 6 key terms
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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.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Non-calculator 2 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \((x + 3)^2\) — M1
    • \((x + 3)^2 - 4\) — A1
  2. Question 2 Non-calculator 2 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \((x - 4)^2\) — M1
    • \((x - 4)^2 - 13\) — A1
  3. Question 3 Non-calculator 3 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \((x + 3)^2 - 9 - 2\) — M1
    • \((x + 3)^2 = 11\) — M1
    • \(-3 \pm \sqrt{11}\) — A1
  4. Question 4 Non-calculator 3 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \((x - 2)^2\) — M1
    • \((x - 2)^2 + 3\) — M1
    • \((2, 3)\) — A1
  5. Question 5 Non-calculator 3 marks

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

    Show answerHide answer

    Model answer

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

    Mark scheme

    • \(2(x^2 + 6x)\) — M1
    • \(2(x + 3)^2 - 18 + 7\) — M1
    • \(2(x + 3)^2 - 11\) — A1
  6. Question 6 Show that 3 marks

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

    Show answerHide answer

    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.

    Mark scheme

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

Quick check

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

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

    B: \((x + 5)^2 - 25\)

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

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

    1. A\((-2, 3)\)
    2. B\((-2, -3)\)
    3. C\((2, 3)\)
    4. D\((3, 2)\)
    Show answerHide answer

    C: \((2, 3)\)

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

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

    1. A\(-7\)
    2. B\(7\)
    3. C\(-4\)
    4. D\(4\)
    Show answerHide answer

    A: \(-7\)

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

  4. 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 \(x = -4\)
    Show answerHide answer

    D: \(x = 2\) or \(x = -4\)

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

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

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

    B: \((x - 3)^2 - 7\)

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

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

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

    C: \((-3, -4)\)

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

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