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Maths · Equations and inequalities

Solving quadratic equations 1

Solve quadratic equations by factorising, including the difference of two squares and equations that need rearranging first.

  • 6 key terms
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Teacher resources

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Student handouts

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Warm-up

Answer each one, then check.

  1. 1

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

    Show answerHide answer

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

  2. 2

    Factorise \(x^2 - 9\).

    Show answerHide answer

    \((x - 3)(x + 3)\)

  3. 3

    Solve \(x - 4 = 0\).

    Show answerHide answer

    \(x = 4\)

  4. 4

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

    Show answerHide answer

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

  5. 5

    What is the product of \(0\) and any number?

    Show answerHide answer

    0

Learning Objectives

  1. 1Solve \(x^2 + bx + c = 0\) by factorising.
  2. 2Solve equations of the form \(x^2 - a^2 = 0\).
  3. 3Rearrange to zero before factorising.
  4. 4Solve problems that lead to quadratic equations.

THE NULL FACTOR LAW

If two things multiply to give zero, at least one of them must be zero.

So \((x - 2)(x - 3) = 0\) means \(x = 2\) or \(x = 3\).

Solving by Factorising

Always get zero on one side first.

  1. 1 Rearrange

    Write the equation as \(ax^2 + bx + c = 0\)
  2. 2 Factorise

    Find two numbers that multiply to \(c\) and add to \(b\)
  3. 3 Set each bracket to zero

    Use the null factor law
  4. 4 Solve

    Two solutions, or one repeated solution

A Standard Quadratic

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

Show the solutionHide the solution
  1. 1 Two numbers with product 6 and sum 5 2 and 3
  2. 2 Factorise \((x + 2)(x + 3) = 0\)
  3. 3 Set each bracket to zero \(x + 2 = 0\) or \(x + 3 = 0\)

Answer\(x = -2\) or \(x = -3\)

With Negative Numbers

Solve \(x^2 - 2x - 15 = 0\).

Show the solutionHide the solution
  1. 1 Product \(-15\), sum \(-2\) \(-5\) and \(3\)
  2. 2 Factorise \((x - 5)(x + 3) = 0\)
  3. 3 Solve \(x = 5\) or \(x = -3\)

Answer\(x = 5\) or \(x = -3\)

Special Cases

  • Difference of two squares

    \(x^2 - 49 = 0\) factorises as \((x - 7)(x + 7)\), so \(x = 7\) or \(x = -7\).

  • No constant term

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

  • A perfect square

    \(x^2 - 6x + 9 = 0\) is \((x - 3)^2\), so there is one solution, \(x = 3\).

  • Do not divide by x

    Dividing \(x^2 = 7x\) by \(x\) loses the solution \(x = 0\).

Rearranging First

Solve \(x^2 = 3x + 10\).

Show the solutionHide the solution
  1. 1 Move everything to one side \(x^2 - 3x - 10 = 0\)
  2. 2 Factorise \((x - 5)(x + 2) = 0\)
  3. 3 Solve \(x = 5\) or \(x = -2\)

Answer\(x = 5\) or \(x = -2\)

A Rectangle Problem

A rectangle has length \((x + 3)\) cm and width \(x\) cm. Its area is 40 cm². Find \(x\), and the dimensions.

Show the solutionHide the solution
  1. 1 Area equation \(x(x + 3) = 40\)
  2. 2 Expand and rearrange \(x^2 + 3x - 40 = 0\)
  3. 3 Factorise \((x + 8)(x - 5) = 0\)
  4. 4 A length cannot be negative \(x = 5\), and \(x = -8\) is rejected

Answer\(x = 5\); the rectangle is 8 cm by 5 cm.

Solve and Check

Solve each equation by factorising, then substitute your answers back to check. (a) \(x^2 - 8x + 15 = 0\) (b) \(x^2 + 4x = 0\) (c) \(x^2 - 25 = 0\) (d) \(x^2 = 2x + 24\).

1. Rearrange to zero.

2. Factorise.

3. Check by substituting.

A good answer shows: (a) \((x - 3)(x - 5) = 0\): \(x = 3\) or 5. (b) \(x(x + 4) = 0\): \(x = 0\) or \(-4\). (c) \((x - 5)(x + 5) = 0\): \(x = \pm 5\). (d) \(x^2 - 2x - 24 = 0\), \((x - 6)(x + 4) = 0\): \(x = 6\) or \(-4\).

Can I...?

  1. 1Factorise \(x^2 + bx + c\).
  2. 2Use the null factor law.
  3. 3Solve a difference of two squares.
  4. 4Solve an equation with no constant term.
  5. 5Rearrange to zero first.
  6. 6Reject impossible solutions.
  7. 7Form a quadratic from a problem.
  8. 8Check by substituting.

Summary & Exam Focus

  • Make the equation equal zero before factorising.
  • Set each bracket to zero.
  • Difference of two squares: \(x^2 - a^2 = (x - a)(x + a)\).
  • Check solutions make sense in context.

Exam focus

A rectangle has length \((x + 3)\) cm and width \(x\) cm. The area of the rectangle is 40 cm². Work out the value of \(x\). (4 marks) (4 marks)

Form the equation, rearrange to zero, factorise, and reject any negative length.

Key terms

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

Quadratic equation
An equation whose highest power is \(x^2\).
Root
A solution of an equation; where a graph crosses the x-axis.
Factorise
Write as a product of brackets.
Null factor law
If \(ab = 0\), then \(a = 0\) or \(b = 0\).
Difference of two squares
\(a^2 - b^2 = (a - b)(a + b)\).
Perfect square
A quadratic that is a bracket squared.

Questions and answers

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

1. Exam question Non-calculator 3 marks Easier

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

Mark scheme — 3 marks available

  • Correct factorisation — M1
  • One solution — A1
  • Both solutions — A1

Model answer

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

2. Exam question Non-calculator 3 marks Easier

Solve \(x^2 - 2x - 15 = 0\).

Mark scheme — 3 marks available

  • Correct factorisation — M1
  • One solution — A1
  • Both solutions — A1

Model answer

\((x - 5)(x + 3) = 0\), so \(x = 5\) or \(x = -3\).

3. Exam question Non-calculator 2 marks Easier

Solve \(x^2 - 49 = 0\).

Mark scheme — 2 marks available

  • Factorising or \(x^2 = 49\) — M1
  • \(x = 7\) and \(x = -7\) — A1

Model answer

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

4. Exam question Non-calculator 2 marks Easier

Solve \(x^2 - 7x = 0\).

Mark scheme — 2 marks available

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

Model answer

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

5. Exam question Non-calculator 3 marks Easier

Solve \(x^2 = 3x + 10\).

Mark scheme — 3 marks available

  • Rearranging to zero — M1
  • Correct factorisation — M1
  • \(x = 5\) and \(x = -2\) — A1

Model answer

\(x^2 - 3x - 10 = 0\), so \((x - 5)(x + 2) = 0\) and \(x = 5\) or \(x = -2\).

6. Exam question Non-calculator 4 marks Easier

The diagram shows a rectangle. The length is \((x + 3)\) cm and the width is \(x\) cm. The area of the rectangle is 40 cm². Work out the value of \(x\).

A rectangle with length x plus 3 centimetres, width x centimetres and area 40 square centimetres.

Mark scheme — 4 marks available

  • Forming the equation — M1
  • Rearranging to zero — M1
  • Factorising — M1
  • \(x = 5\) — A1

Model answer

\(x(x + 3) = 40\), so \(x^2 + 3x - 40 = 0\) and \((x + 8)(x - 5) = 0\). \(x = 5\), since a length cannot be negative.

7. Multiple choice 1 mark Easier

Solve \((x - 2)(x - 3) = 0\).

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

Why: Set each bracket to zero: \(x = 2\) or \(x = 3\).

8. Multiple choice 1 mark Core

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

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

Why: Numbers with product 6 and sum \(-5\): \(-2\) and \(-3\).

9. Multiple choice 1 mark Core

Solve \(x^2 = 16\).

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

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

10. Multiple choice 1 mark Core

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

  1. A \(x = 0\) or \(x = -3\) Correct
  2. B \(x = 3\)
  3. C \(x = -3\) only
  4. D \(x = 0\) only

Why: \(x(x + 3) = 0\), so \(x = 0\) or \(x = -3\).

11. Multiple choice 1 mark Core

A quadratic graph crosses the x-axis at \(x = -1\) and \(x = 4\). Which is its equation?

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

Why: \(y = (x + 1)(x - 4) = x^2 - 3x - 4\).

12. Multiple choice 1 mark Stretch

Why should you not divide both sides of \(x^2 = 5x\) by \(x\)?

  1. A It makes the equation harder
  2. B You lose the solution \(x = 0\) Correct
  3. C It changes the sign
  4. D It gives a wrong answer for \(x = 5\)

Why: You lose the solution \(x = 0\). Rearrange and factorise instead: \(x(x - 5) = 0\).