Zero-product equations | Year 10

01 / LEARN

The idea, step by step

A product equal to zero

If two real numbers multiply to zero, at least one of them is zero. This lets a quadratic equation become two simpler linear equations.

An equation such as a product equal to ten is not ready for this rule. Rearrange it first, then factorise the resulting expression.

A B = 0 ⟹ A = 0 or B = 0

Keep both possibilities

When x is a factor, one solution may be x=0. Dividing both sides by x would remove that solution. Factor instead.

Finally substitute each candidate into the original equation. In an application, also check that the value fits the stated domain.

A common trap

Dividing by x before checking whether x can be zero.

Instead: Factor out x and apply the zero-product rule so that the zero root is retained.

02 / WORKED EXAMPLES

See the thinking

Example 1

Two bracket factors

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

Set $2 x - 3 = 0$ or $x + 4 = 0$ .
The first equation gives $x = \frac{3}{2}$ ; the second gives $x = - 4$ .
Each value makes one bracket zero.

x = \frac{3}{2}

x = - 4

Example 2

Do not divide away zero

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

Rearrange: $3 x^{2} - 12 x = 0$ .
Factorise: $3 x (x - 4) = 0$ .
Thus $x = 0$ or $x - 4 = 0$ .

x = 0

x = 4

Check your understanding

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

Show answer

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

Either bracket may equal zero; the signs reverse when each linear equation is solved.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF 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 ↗

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