Viewing as

Teaching this? The teacher view adds 2 files, the mark schemes and the model answers.

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
All boards

Download the full pack · 3 files

Warm-up

Answer each one, then check.

1

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

Show answerHide answer

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

Factorise \(x^2 - 9\).

Show answerHide answer

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

Solve \(x - 4 = 0\).

Show answerHide answer

\(x = 4\)
4

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

Show answerHide answer

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

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

Show answerHide answer

0

Learning Objectives

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

Roots on a Graph

The graph of y equals x squared minus 5x plus 6 crossing the x-axis at 2 and 3.

Solving by Factorising

Always get zero on one side first.

1 Rearrange
Write the equation as \(ax^2 + bx + c = 0\)
2 Factorise
Find two numbers that multiply to \(c\) and add to \(b\)
3 Set each bracket to zero
Use the null factor law
4 Solve
Two solutions, or one repeated solution

A Standard Quadratic

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

Show the solutionHide the solution

1 Two numbers with product 6 and sum 5 2 and 3
2 Factorise \((x + 2)(x + 3) = 0\)
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 Product \(-15\), sum \(-2\) \(-5\) and \(3\)
2 Factorise \((x - 5)(x + 3) = 0\)
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 Move everything to one side \(x^2 - 3x - 10 = 0\)
2 Factorise \((x - 5)(x + 2) = 0\)
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 Area equation \(x(x + 3) = 40\)
2 Expand and rearrange \(x^2 + 3x - 40 = 0\)
3 Factorise \((x + 8)(x - 5) = 0\)
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...?

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

Practice questions

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

Question 1 Non-calculator 3 marks

Solve \(x^2 + 5x + 6 = 0\).
Show answerHide answer
Model answer

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

Mark scheme
- Correct factorisation — M1
- One solution — A1
- Both solutions — A1
Question 2 Non-calculator 3 marks

Solve \(x^2 - 2x - 15 = 0\).
Show answerHide answer
Model answer

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

Mark scheme
- Correct factorisation — M1
- One solution — A1
- Both solutions — A1
Question 3 Non-calculator 2 marks

Solve \(x^2 - 49 = 0\).
Show answerHide answer
Model answer

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

Mark scheme
- Factorising or \(x^2 = 49\) — M1
- \(x = 7\) and \(x = -7\) — A1
Question 4 Non-calculator 2 marks

Solve \(x^2 - 7x = 0\).
Show answerHide answer
Model answer

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

Mark scheme
- \(x(x - 7)\) — M1
- \(x = 0\) and \(x = 7\) — A1
Question 5 Non-calculator 3 marks

Solve \(x^2 = 3x + 10\).
Show answerHide answer
Model answer

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

Mark scheme
- Rearranging to zero — M1
- Correct factorisation — M1
- \(x = 5\) and \(x = -2\) — A1
Question 6 Non-calculator 4 marks

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 \(x\).
Show answerHide answer
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.

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

Quick check

Solve \((x - 2)(x - 3) = 0\).
1. A\(x = -2\) or \(x = -3\)
2. B\(x = 5\)
3. C\(x = 2\) or \(x = 3\)
4. D\(x = 6\)
Show answerHide answer

C: \(x = 2\) or \(x = 3\)

Set each bracket to zero: \(x = 2\) or \(x = 3\).
Factorise \(x^2 - 5x + 6\).
1. A\((x + 2)(x + 3)\)
2. B\((x - 2)(x - 3)\)
3. C\((x - 1)(x - 6)\)
4. D\((x + 1)(x - 6)\)
Show answerHide answer

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

Numbers with product 6 and sum \(-5\): \(-2\) and \(-3\).
Solve \(x^2 = 16\).
1. A\(x = 4\)
2. B\(x = 8\)
3. C\(x = -4\)
4. D\(x = 4\) or \(x = -4\)
Show answerHide answer

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

\(x = 4\) or \(x = -4\).
Solve \(x^2 + 3x = 0\).
1. A\(x = 0\) or \(x = -3\)
2. B\(x = 3\)
3. C\(x = -3\) only
4. D\(x = 0\) only
Show answerHide answer

A: \(x = 0\) or \(x = -3\)

\(x(x + 3) = 0\), so \(x = 0\) or \(x = -3\).
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\(y = (x + 1)(x - 4)\)
4. D\(y = (x - 1)(x - 4)\)
Show answerHide answer

C: \(y = (x + 1)(x - 4)\)

\(y = (x + 1)(x - 4) = x^2 - 3x - 4\).
Why should you not divide both sides of \(x^2 = 5x\) by \(x\)?
1. AIt makes the equation harder
2. BYou lose the solution \(x = 0\)
3. CIt changes the sign
4. DIt gives a wrong answer for \(x = 5\)
Show answerHide answer

B: You lose the solution \(x = 0\)

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

Downloads

Free to keep, print and annotate.

Download all 3 as a zip

Something here looks wrong?

Tell us what and we will go and look. It goes to whoever writes these pages, nobody else, and we do not ask who you are — so there is nothing to sign and nothing comes back to you.