Quadratic applications and model checks | Year 10

01 / LEARN

The idea, step by step

Define the variable before forming a rule

State what the variable measures and its unit. Translate each relationship separately, then combine them using area, revenue or another relevant quantity.

The equation may have two real roots but only one physically sensible answer. A model’s domain can also be narrower than all positive numbers.

Interpret and test the result

Use factorisation, a square form, the formula or a graph according to the information needed. A root describes when the model reaches a target; a vertex describes its maximum or minimum.

Check the answer in the original relationship and state assumptions. A quadratic fitted to a short interval should not automatically be trusted for all inputs or for later physical events.

A common trap

Keeping every positive root without checking the problem’s other limits.

Instead: Check all stated conditions, including time intervals, physical dimensions, integer counts and where the model is intended to apply.

02 / WORKED EXAMPLES

See the thinking

Example 1

An area fixes the dimensions

A rectangle is 5 m longer than it is wide. Its area is 84 m². Find its dimensions.

Let the positive width be $x$ m; the length is $(x + 5)$ m.
$x (x + 5) = 84$ , so $(x + 12) (x - 7) = 0$ .
Reject $x = - 12$ ; use $x = 7$ . Check $7 \times 12 = 84$ .

Width

7

m and length

12

Example 2

Revenue with changing demand

A model predicts that increasing a 20-dollar ticket by x dollars changes attendance from 100 to $(100 - 2 x)$ . Find the revenue-maximising whole-dollar increase for $0 \leq x \leq 50$ .

Revenue is $R = (20 + x) (100 - 2 x) = - 2 x^{2} + 60 x + 2000$ .
Complete the square: $R = - 2 {(x - 15)}^{2} + 2450$ .
The maximum is reached at $x = 15$ , which lies in the allowed domain. The model predicts 70 tickets at 35 dollars each.

Increase

15

; predicted maximum revenue

2450

Check your understanding

A width x satisfies $(x - 2) (x + 7) = 0$ and must be positive. What is the width?

Show answer

$x = 2$

The algebraic root −7 is not a positive width.

03 / WORKSHEET

Your turn

6 questions · 19 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

A rectangle is 3 cm longer than it is wide and has area 54 cm². Find its dimensions.
3 marks
A photo is 6 cm by 10 cm. A uniform border makes the total area 96 cm². Find the border width.
4 marks
A ball’s height is $h = - 5 t^{2} + 20 t + 1$ metres after t seconds. Find its maximum height and when it occurs.
3 marks
A model gives ticket revenue $R = p (80 - 2 p)$ dollars when price p dollars satisfies $0 \leq p \leq 40$ . Find the price giving greatest predicted revenue.
3 marks
A design width x metres must satisfy $x^{2} - 5 x + 6 = 0$ and $x > 2$ . Which root is allowed?
2 marks
A proposed height model is $h = - t^{2} + 6 t + 2$ . At t=4, a measurement is 9.5 m. Find the model’s error there and explain whether one check proves the model valid at all times.
4 marks

Read the worked solutions

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

A rectangle is 3 cm longer than it is wide and has area 54 cm². Find its dimensions.
3 marks
$x (x + 3) = 54$ , so $(x + 9) (x - 6) = 0$ .
The positive width is $6$ .
$6$ cm by $9$ cm.
A photo is 6 cm by 10 cm. A uniform border makes the total area 96 cm². Find the border width.
4 marks
$(6 + 2 x) (10 + 2 x) = 96$ .
$4 x^{2} + 32 x - 36 = 0$ , or $(x + 9) (x - 1) = 0$ .
The positive border width is $1$ .
$1$ cm.
A ball’s height is $h = - 5 t^{2} + 20 t + 1$ metres after t seconds. Find its maximum height and when it occurs.
3 marks
$h = - 5 {(t - 2)}^{2} + 21$ .
Maximum $21$ m at $t = 2$ s.
A model gives ticket revenue $R = p (80 - 2 p)$ dollars when price p dollars satisfies $0 \leq p \leq 40$ . Find the price giving greatest predicted revenue.
3 marks
$R = - 2 {(p - 20)}^{2} + 800$ .
The allowed price $p = 20$ gives 40 tickets.
Price $20$ ; revenue $800$ .
A design width x metres must satisfy $x^{2} - 5 x + 6 = 0$ and $x > 2$ . Which root is allowed?
2 marks
$(x - 2) (x - 3) = 0$ gives $2, 3$ .
The strict condition excludes $2$ .
$x = 3$ m.
A proposed height model is $h = - t^{2} + 6 t + 2$ . At t=4, a measurement is 9.5 m. Find the model’s error there and explain whether one check proves the model valid at all times.
4 marks
$h (4) = - 16 + 24 + 2 = 10$ .
$10 - 9.5 = 0.5$ .
Test more values in the intended interval and consider the physical assumptions.
Prediction $10$ m; overestimate $0.5$ m. One check does not establish validity for all times.

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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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