Checking solutions and domains | Year 10

01 / LEARN

The idea, step by step

What values are allowed?

The domain is the set of allowed inputs. Over the real numbers, a denominator cannot be zero and a square-root radicand cannot be negative. Combine all conditions: one expression can have more than one restriction. Simplification does not erase conditions from the original expression.

A model has context as well

A formula may accept negative or fractional values algebraically while its situation does not. Numbers of people are whole numbers; lengths are positive; a model for a ten-minute event may only apply for $0 \leq t \leq 10$ . After solving, substitute into the original equation and decide whether the answer is meaningful in context.

A common trap

Checking only the simplified expression at a value excluded by the original denominator.

Instead: Always check the original expression or equation and its stated domain. Equivalent simplified formulas agree only on the original allowed inputs.

02 / WORKED EXAMPLES

See the thinking

Example 1

Combine root and fraction restrictions

Find the real domain of $f (x) = \frac{\sqrt{x - 1}}{x - 4}$ .

The square root requires $x - 1 \geq 0$ , so $x \geq 1$ .
The denominator requires $x \neq 4$ .
Both conditions must hold simultaneously.

x \geq 1

, excluding

x = 4

Example 2

A valid equation but invalid length

A model gives a rectangle side $x$ and leads to $x = - 3$ . Can this be the rectangle length?

A real number $- 3$ can be used in many algebraic expressions.
A physical length must be positive.
Reject the candidate for the model and reconsider whether there are other valid solutions or whether the assumptions were suitable.

No: a rectangle length cannot be negative.

Check your understanding

Find the real domain of $g (x) = \sqrt{x + 5}$ .

Show answer

$x \geq - 5$

A real square root needs $x + 5 \geq 0$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

State the real domain of $f (x) = 1 / (x + 6)$ .
2 marks
State the real domain of $g (x) = \sqrt{3 - x}$ .
2 marks
State the real domain of $h (x) = \sqrt{x} / (x - 9)$ .
2 marks
Check whether $x = 4$ solves $3 / (x - 1) = 1$ .
2 marks
A bus hire cost is $C = 180 / n$ dollars per passenger, with at most 30 passengers. State a sensible domain.
3 marks
A calculation for $1 / (x - 2) = 1$ gives $x = 3$ . Another student says $x = 2$ is also possible because multiplying by $x - 2$ removes the denominator. Decide both claims.
3 marks

Read the worked solutions

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

State the real domain of $f (x) = 1 / (x + 6)$ .
2 marks
The denominator is zero only at $x = - 6$ .
All real $x$ except $- 6$ .
State the real domain of $g (x) = \sqrt{3 - x}$ .
2 marks
Require $3 - x \geq 0$ .
$x \leq 3$
State the real domain of $h (x) = \sqrt{x} / (x - 9)$ .
2 marks
The root requires nonnegative $x$ and the denominator excludes $9$ .
$x \geq 0$ , $x \neq 9$ .
Check whether $x = 4$ solves $3 / (x - 1) = 1$ .
2 marks
The input is allowed and $3 / (4 - 1) = 1$ .
Yes.
A bus hire cost is $C = 180 / n$ dollars per passenger, with at most 30 passengers. State a sensible domain.
3 marks
The head count is a positive whole number and cannot exceed the capacity.
$n \in {1, 2, \dots, 30}$ .
A calculation for $1 / (x - 2) = 1$ gives $x = 3$ . Another student says $x = 2$ is also possible because multiplying by $x - 2$ removes the denominator. Decide both claims.
3 marks
$1 / (3 - 2) = 1$ .
At $x = 2$ the original left side divides by zero.
$x = 3$ is valid; $x = 2$ is excluded.

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian Mathematics Curriculum

Year 10 algebra, including the approved fuller school-course treatment of rational expressions and formula restrictions; quadratic methods are in their own chapter.

Official curriculum source ↗

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