YEAR 10 · QUADRATICS AND PARABOLAS · 14 MIN

Zero-product equations

Use the zero-product rule to solve a quadratic equation without losing a possible root.

In this lesson

Set an equation equal to zero before applying the rule.
Find and check all real roots of a factorised equation.

Before you start

Factorising quadratics
Solving linear equations

01 / LEARN

Understand the idea.

A product equal to zero

The zero-product rule applies when a product equals zero: at least one factor must be zero. Setting each variable factor to zero gives simpler equations whose solutions cover all possibilities.

The right-hand side matters. A product equal to a nonzero number does not let you set either factor to zero. Expand or rearrange until one side is zero, then factorise.

A B = 0 ⟹ A = 0 or B = 0

Keep both possibilities

An unknown factor may itself be zero. Dividing by $x$ assumes $x \neq 0$ and can erase a valid solution, so keep $x$ as a factor and solve its zero case.

Substitute every candidate into the original equation. For a length, time or other model, also apply the stated restrictions; an algebraic solution can be valid in the equation but unsuitable for the situation.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two bracket factors

Solve $(2 x - 3) (x + 4) = 0$ .

Worked solution

1
Set each variable factor equal to zero.
$2 x - 3 = 0 or x + 4 = 0$
The original product is zero, so at least one bracket is zero.
2
Solve the two linear equations.
$\begin{aligned} 2 x = 3 & ⟹ x = \frac{3}{2} \\ x + 4 = 0 & ⟹ x = - 4 \end{aligned}$
Solving each factor equation retains both possible roots.
3
Check both roots in the product.
$\begin{aligned} x = \frac{3}{2} : & (3 - 3) (\frac{3}{2} + 4) = 0 \\ x = - 4 : & (- 8 - 3) (- 4 + 4) = 0 \end{aligned}$
Each candidate makes at least one original factor zero.

Answer

$x = \frac{3}{2}$ or $x = - 4$

EXAMPLE 02

Do not divide away zero

Solve $3 x^{2} = 12 x$ .

Worked solution

1
Make one side zero.
$3 x^{2} - 12 x = 0$
Subtracting the same expression from both sides preserves the solutions.
2
Factor rather than divide by the unknown.
$3 x (x - 4) = 0$
Keeping the factor $x$ retains the possible root $x = 0$ .
3
Apply the zero-product rule.
$\begin{array}{c} x = 0 or x - 4 = 0 \\ x = 0 or x = 4 \end{array}$
The constant factor $3$ cannot be zero.
4
Substitute into the original equation.
$\begin{aligned} x = 0 : & 3 {(0)}^{2} = 12 (0) = 0 \\ x = 4 : & 3 {(4)}^{2} = 48 = 12 (4) \end{aligned}$
Both values satisfy the equation before rearrangement.

Answer

$x = 0$ or $x = 4$

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $(x - 2) (x + 5) = 0$ .

Show answer & reasoning

$x = 2$ or $x = - 5$

Setting $x - 2 = 0$ gives $x = 2$ ; setting $x + 5 = 0$ gives $x = - 5$ . Each value makes the original product zero, so both are solutions.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Solve $(x - 7) (x + 1) = 0$ .
2 marks
Solve $x^{2} + 2 x - 15 = 0$ .
2 marks
Solve $2 x^{2} - 5 x - 3 = 0$ .
3 marks
Solve $x^{2} = 6 x$ .
2 marks
Solve $2 (x - 3) (x + 1) = 10$ .
4 marks
A positive length $x$ satisfies $x (x + 3) = 28$ . Find $x$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $(x - 7) (x + 1) = 0$ .
2 marks
$x - 7 = 0$ or $x + 1 = 0$ .
$x = 7$ or $x = - 1$
Solve $x^{2} + 2 x - 15 = 0$ .
2 marks
$(x + 5) (x - 3) = 0$ .
$x = 3$ or $x = - 5$
Solve $2 x^{2} - 5 x - 3 = 0$ .
3 marks
$(2 x + 1) (x - 3) = 0$ .
$x = 3$ or $x = - \frac{1}{2}$
Solve $x^{2} = 6 x$ .
2 marks
$x (x - 6) = 0$ ; do not divide by $x$ .
$x = 0$ or $x = 6$
Solve $2 (x - 3) (x + 1) = 10$ .
4 marks
Divide by $2$ , expand and rearrange: $x^{2} - 2 x - 8 = 0$ .
$(x - 4) (x + 2) = 0$ .
$x = 4$ or $x = - 2$
A positive length $x$ satisfies $x (x + 3) = 28$ . Find $x$ .
3 marks
$x^{2} + 3 x - 28 = (x + 7) (x - 4) = 0$ .
The roots are $- 7$ and $4$ . A positive length requires $x = 4$ .
$x = 4$

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Curriculum reference & lesson scope

Australian Curriculum v9.0 and South Australian Curriculum

Year 10 quadratics, including the fuller school-course algebra requested for Ascent: non-monic factorisation, completing the square and the quadratic formula. These lessons consolidate Year 9 monic quadratics and extend the available solving methods.

Official curriculum source ↗

In this lesson

Understand the idea.

A product equal to zero

Keep both possibilities

See the method in action.

Two bracket factors

Set each variable factor equal to zero.

Solve the two linear equations.

Check both roots in the product.

Do not divide away zero

Make one side zero.

Factor rather than divide by the unknown.

Apply the zero-product rule.

Substitute into the original equation.

Try it yourself.

Check your understanding

Keep practising