YEAR 9 · QUADRATICS AND PARABOLAS · 15 MIN

Zero-product principle and integer-root equations

Use a product equal to zero to find every integer solution of a quadratic equation.

In this lesson

Use the zero-product principle
Check all solutions in the original equation

Before you start

Factorising monic quadratics
Solving one-step equations

01 / LEARN

Understand the idea.

Why either factor can be zero

The zero-product principle says that a product of real numbers is zero only if at least one factor is zero. A factorised equation therefore splits into separate linear equations joined by "or".

For $(x - 2) (x + 5) = 0$ , either $x - 2 = 0$ or $x + 5 = 0$ . This gives $x = 2$ or $x = - 5$ . Each solution needs only one zero factor, since multiplying zero by any real number gives zero.

Put zero on one side first

First rearrange so one side is zero, then factorise the other side. The principle cannot be used directly on a product equal to a non-zero number: its factors need not equal that number individually.

Avoid dividing away an unknown factor that could be zero. In $x (x - 4) = 0$ , the factor $x$ gives a genuine solution $x = 0$ . Dividing by $x$ would exclude exactly that possibility and leave an incomplete answer.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

Solve $x^{2} - 3 x - 10 = 0$ .

Worked solution

1
Factorise the zero-side expression.
$x^{2} - 3 x - 10 = (x - 5) (x + 2)$
The bracket constants multiply to -10 and add to -3.
2
Set each factor equal to zero.
$x - 5 = 0 or x + 2 = 0$
A product equal to zero has at least one zero factor.
3
Solve both linear equations.
$x = 5 or x = - 2$
Each case provides a possible root.
4
Check in the original equation.
$\begin{aligned} 5^{2} - 3 (5) - 10 & = 0 \\ {(- 2)}^{2} - 3 (- 2) - 10 & = 0 \end{aligned}$
Both roots make the original quadratic equal zero.

Answer

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

EXAMPLE 02

Worked example

Solve $x^{2} = 6 x$ .

Worked solution

1
Subtract the right-hand side.
$x^{2} - 6 x = 0$
The zero-product principle requires zero on one side.
2
Factor out the common x.
$x (x - 6) = 0$
Both terms contain a factor of x.
3
Keep both zero-factor cases.
$x = 0 or x - 6 = 0$
Keeping the product avoids dividing away the zero solution.
4
State and check both roots.
$\begin{aligned} x & = 0 or x = 6 \\ 0^{2} & = 6 (0) \\ 6^{2} & = 6 (6) = 36 \end{aligned}$
Each root satisfies the original equation.

Answer

$x = 0$ or $x = 6$

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $(x + 2) (x - 7) = 0$ .

Show answer & reasoning

$x = - 2$ or $x = 7$

The product is zero, so $x + 2 = 0$ or $x - 7 = 0$ . Solving gives $x = - 2$ or $x = 7$ , with each value making one bracket zero.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Solve $(x - 4) (x + 6) = 0$ .
2 marks
Solve $x^{2} + 7 x + 12 = 0$ .
2 marks
Solve $x^{2} - 2 x - 15 = 0$ .
2 marks
Solve $x^{2} = 9 x$ .
2 marks
Solve $x^{2} + x = 20$ .
3 marks
A student divides $x^{2} = 5 x$ by $x$ and gives only $x = 5$ . Find the missing solution and explain.
3 marks

Read the worked solutions

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

Solve $(x - 4) (x + 6) = 0$ .
2 marks
$x - 4 = 0$ or $x + 6 = 0$ .
$x = 4, - 6$
Solve $x^{2} + 7 x + 12 = 0$ .
2 marks
$(x + 3) (x + 4) = 0$ .
$x = - 3, - 4$
Solve $x^{2} - 2 x - 15 = 0$ .
2 marks
$(x - 5) (x + 3) = 0$ .
$x = 5, - 3$
Solve $x^{2} = 9 x$ .
2 marks
$x^{2} - 9 x = x (x - 9) = 0$ .
$x = 0, 9$
Solve $x^{2} + x = 20$ .
3 marks
$x^{2} + x - 20 = (x + 5) (x - 4) = 0$ .
$x = 4, - 5$
A student divides $x^{2} = 5 x$ by $x$ and gives only $x = 5$ . Find the missing solution and explain.
3 marks
$x (x - 5) = 0$ gives $x = 0$ or $5$ .
Division by $x$ wrongly excludes $x = 0$ .
$x = 0$ is also a solution.

Checking saved progress…

Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

AC9M9A02, AC9M9A04–A06: monic quadratic expressions, integer-root equations, graphs and models.

Official curriculum source ↗

In this lesson

Understand the idea.

Why either factor can be zero

Put zero on one side first

See the method in action.

Worked example

Factorise the zero-side expression.

Set each factor equal to zero.

Solve both linear equations.

Check in the original equation.

Worked example

Subtract the right-hand side.

Factor out the common x.

Keep both zero-factor cases.

State and check both roots.

Try it yourself.

Check your understanding

Keep practising