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Maths · Equations and graphs

Quadratic equations

Solve quadratic equations by factorising, by using the quadratic formula and by completing the square, and choose a suitable method.

  • 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

    Factorise \(x^2 + 5x + 6\).

    Show answerHide answer

    \((x + 2)(x + 3)\)

  2. 2

    Expand \((x + 3)(x - 1)\).

    Show answerHide answer

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

  3. 3

    What is \(\sqrt{49}\)?

    Show answerHide answer

    \(7\)

  4. 4

    If \(ab = 0\), what do you know?

    Show answerHide answer

    \(a = 0\) or \(b = 0\)

  5. 5

    What is the value of \(3^2\)?

    Show answerHide answer

    \(9\)

Learning Objectives

  1. 1Solve a quadratic by factorising.
  2. 2Solve a quadratic with the quadratic formula.
  3. 3Complete the square and use it to solve.
  4. 4Choose the most efficient method.

QUADRATIC EQUATIONS

A quadratic equation can be written \(ax^2 + bx + c = 0\) and usually has two solutions.

Rearrange so that one side is 0 before you use any method.

Choosing a Method

Use the quickest one that works.

  • Factorising

    Use it when: The quadratic factorises easily. Example: \(x^2 - 5x + 6 = 0\)

  • Formula

    Use it when: It does not factorise, or the answer is needed as a decimal. Example: \(x^2 + 4x - 7 = 0\)

  • Completing the square

    Use it when: You need exact surds or the turning point. Example: \(x^2 + 6x - 2 = 0\)

Factorising

Solve \(2x^2 + 5x - 3 = 0\).

Show the solutionHide the solution
  1. 1 Factorise \((2x - 1)(x + 3) = 0\)
  2. 2 Each bracket is 0 \(2x - 1 = 0\) or \(x + 3 = 0\)
  3. 3 Solve \(x = \tfrac{1}{2}\) or \(x = -3\)

Answer\(x = \tfrac{1}{2}\) or \(x = -3\)

The Quadratic Formula

Solve \(x^2 + 4x - 7 = 0\). Give your answers to 2 decimal places.

Show the solutionHide the solution
  1. 1 Formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
  2. 2 Substitute \(a = 1,\ b = 4,\ c = -7\) \(x = \dfrac{-4 \pm \sqrt{16 + 28}}{2}\)
  3. 3 Simplify \(x = \dfrac{-4 \pm \sqrt{44}}{2}\)
  4. 4 Evaluate \(x = 1.32\) or \(x = -5.32\)

Answer\(x = 1.32\) or \(x = -5.32\)

Completing the Square

Solve \(x^2 + 6x - 2 = 0\) by completing the square. Give your answers to 2 decimal places.

Show the solutionHide the solution
  1. 1 Complete the square \((x + 3)^2 - 9 - 2 = 0\)
  2. 2 Simplify \((x + 3)^2 = 11\)
  3. 3 Square root \(x + 3 = \pm\sqrt{11}\)
  4. 4 Solve \(x = -3 \pm \sqrt{11} = 0.32\) or \(-6.32\)

Answer\(x = 0.32\) or \(x = -6.32\)

Common Mistakes

Avoid these.

  • Not rearranging to 0

    Move everything to one side first.

  • Dividing by \(x\)

    Solving \(x^2 = 7x\) by dividing loses the solution \(x = 0\); factorise instead.

  • Signs in the formula

    Take care with \(-b\) and \(b^2 - 4ac\) when \(b\) or \(c\) is negative.

  • Only one solution

    Give both solutions.

Choose Your Method

Solve each and say which method you used. (a) \(x^2 - 7x + 10 = 0\) (b) \(x^2 + 2x - 5 = 0\) (2 d.p.) (c) \(x^2 = 9x\).

1. Look for factors first.

2. Otherwise use the formula.

A good answer shows: (a) \(x = 2\) or \(5\) (factorising). (b) \(x = -1 \pm \sqrt{6} = 1.45\) or \(-3.45\) (formula or completing the square). (c) \(x(x - 9) = 0\), so \(x = 0\) or \(9\).

Can I...?

  1. 1Rearrange to equal zero.
  2. 2Factorise and solve.
  3. 3Use the quadratic formula.
  4. 4Complete the square.
  5. 5Give answers to a given accuracy.
  6. 6Give both solutions.
  7. 7Avoid dividing by x.
  8. 8Choose a method.

Summary & Exam Focus

  • Factorise if you can.
  • Formula: \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
  • Complete the square: \((x + p)^2 + q\).
  • Every quadratic has up to two solutions.

Exam focus

Solve \(x^2 + 4x - 7 = 0\). Give your solutions correct to 2 decimal places. (3 marks) (3 marks)

Write the formula first, then substitute carefully, using brackets for negatives.

Key terms

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

Quadratic
An equation with a highest power of \(x^2\).
Factorise
Write as a product of brackets.
Root
A solution of an equation.
Discriminant
\(b^2 - 4ac\).
Completing the square
Writing \(x^2 + bx + c\) as \((x + p)^2 + q\).
Surd
A root that cannot be simplified to a whole number, e.g. \(\sqrt{11}\).

Questions and answers

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

1. Exam question Solve 2 marks Easier

Solve \(x^2 - 5x + 6 = 0\).

Mark scheme — 2 marks available

  • \((x - 2)(x - 3)\) — M1
  • 2 and 3 — A1

Model answer

\((x - 2)(x - 3) = 0\), so \(x = 2\) or \(x = 3\).

2. Exam question Solve 3 marks Easier

Solve \(2x^2 + 5x - 3 = 0\).

Mark scheme — 3 marks available

  • Factorises — M1
  • One correct solution — A1
  • Both correct — A1

Model answer

\((2x - 1)(x + 3) = 0\), so \(x = \tfrac{1}{2}\) or \(x = -3\).

3. Exam question Solve 3 marks Easier

Solve \(x^2 + 4x - 7 = 0\). Give your solutions correct to 2 decimal places.

Mark scheme — 3 marks available

  • Substitutes into the formula — M1
  • \(\dfrac{-4 \pm \sqrt{44}}{2}\) — A1
  • 1.32 and \(-5.32\) — A1

Model answer

\(x = \dfrac{-4 \pm \sqrt{44}}{2}\); \(x = 1.32\) or \(x = -5.32\).

4. Exam question Solve 4 marks Easier

(a) Write \(x^2 + 6x - 2\) in the form \((x + a)^2 + b\). (b) Hence solve \(x^2 + 6x - 2 = 0\). Give your answers correct to 2 decimal places.

Mark scheme — 4 marks available

  • \(a = 3\) — B1
  • \(b = -11\) — B1
  • \(x = -3 \pm \sqrt{11}\) — M1
  • 0.32 and \(-6.32\) — A1

Model answer

(a) \((x + 3)^2 - 11\). (b) \(x + 3 = \pm\sqrt{11}\), so \(x = 0.32\) or \(x = -6.32\).

5. Exam question Use the graph 2 marks Easier

The graph of \(y = x^2 - 4x + 1\) is shown. Use the graph to find estimates for the solutions of \(x^2 - 4x + 1 = 0\).

The graph of y equals x squared minus 4x plus 1 crossing the x-axis near 0.3 and 3.7.

Mark scheme — 2 marks available

  • One correct estimate — B1
  • Both correct — B1

Model answer

\(x \approx 0.3\) and \(x \approx 3.7\). Accept 0.2 to 0.4 and 3.6 to 3.8.

6. Exam question Solve 2 marks Easier

Solve \(x^2 = 7x\).

Mark scheme — 2 marks available

  • \(x(x - 7) = 0\) — M1
  • 0 and 7 — A1

Model answer

\(x^2 - 7x = 0\), \(x(x - 7) = 0\), so \(x = 0\) or \(x = 7\).

7. Multiple choice 1 mark Easier

How many solutions can a quadratic equation have?

  1. A Exactly one
  2. B Exactly three
  3. C Up to two Correct
  4. D Always four

Why: Up to two.

8. Multiple choice 1 mark Core

\((x - 4)(x + 1) = 0\) gives...

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

Why: \(x = 4\) or \(x = -1\).

9. Multiple choice 1 mark Core

The quadratic formula is...

  1. A \(\dfrac{b \pm \sqrt{b^2 + 4ac}}{2a}\)
  2. B \(\dfrac{-b \pm \sqrt{b^2 - 4ac}}{a}\)
  3. C \(-b \pm \sqrt{b^2 - 4ac}\)
  4. D \(\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Correct

Why: \(\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

10. Multiple choice 1 mark Core

\(x^2 + 6x\) completed the square is...

  1. A \((x + 6)^2\)
  2. B \((x + 3)^2 - 9\) Correct
  3. C \((x + 3)^2 + 9\)
  4. D \((x - 3)^2 - 9\)

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

11. Multiple choice 1 mark Core

To solve \(x^2 = 9x\) you should...

  1. A Divide both sides by \(x\)
  2. B Square root both sides
  3. C Rearrange to 0 and factorise Correct
  4. D Guess

Why: Rearrange to \(x^2 - 9x = 0\) and factorise; dividing by \(x\) would lose \(x = 0\).

12. Multiple choice 1 mark Stretch

The solutions of a quadratic are where its graph...

  1. A Crosses the x-axis Correct
  2. B Crosses the y-axis
  3. C Has its turning point
  4. D Is steepest

Why: Crosses the x-axis.