Quadratic models and applications | Year 9

01 / LEARN

The idea, step by step

Define the variable and build the rule

Start by saying what the variable measures and which values are possible. A rectangle with width $x$ m and length $(x + 2)$ m has area $A = x (x + 2) = x^{2} + 2 x$ .

If the area is $48$ m $2$ , write $x^{2} + 2 x = 48$ , then move every term to one side before factorising.

x^{2} + 2 x - 48 = 0 \Rightarrow (x + 8) (x - 6) = 0

Interpret all the algebraic solutions

The equation above has roots $x = - 8$ and $x = 6$ . A negative width is impossible, so the rectangle is $6$ m by $8$ m. Check: $6 \times 8 = 48$ .

Sometimes both roots make sense. A height model can reach the same height on the way up and on the way down. State what each answer means rather than automatically discarding one.

Algebraic answer + context check = useful solution

Check the type and limits of the model

With equally spaced inputs, a linear table has constant first differences. A quadratic table has constant second differences. A model should match the pattern being described.

A supplied rule is an approximation, not a promise about every real situation. State the allowed inputs and do not use the rule outside that range without a reason. A turning point can show a maximum or minimum within the allowed range.

Compare equal input steps
x	0	1	2	3
Linear output	2	5	8	11
Quadratic output	1	4	9	16

A common trap

Keeping a negative length just because it solves the equation.

Instead: Check the meaning and allowed values of each variable. A negative root can be a valid algebraic solution but an impossible dimension.

02 / WORKED EXAMPLES

See the thinking

Example 1

Find a rectangle’s dimensions

A rectangle is $x$ cm wide and $(x + 3)$ cm long. Its area is $40$ cm $2$ . Find its dimensions.

Write $x (x + 3) = 40$ , then $x^{2} + 3 x - 40 = 0$ .
Factorise: $(x + 8) (x - 5) = 0$ , so $x = - 8$ or $x = 5$ .
Reject the negative width. With $x = 5$ , the length is $8$ . Check: $5 \times 8 = 40$ .

5

cm by

8

cm.

Example 2

Two times at one height

A simple height model is $h = t (6 - t)$ metres for $0 \leq t \leq 6$ seconds. When does the model give $h = 8$ m?

Set $t (6 - t) = 8$ , giving $t^{2} - 6 t + 8 = 0$ .
Factorise: $(t - 2) (t - 4) = 0$ , so $t = 2$ or $t = 4$ .
Both times lie in the allowed range. The model reaches $8$ m once before and once after its highest point.

2

seconds and

4

seconds.

Example 3

Read a maximum from a supplied rule

A fenced rectangle has width $w$ m and length $(10 - w)$ m. Its area is supplied as $A = 25 - {(w - 5)}^{2}$ , where $0 < w < 10$ . Find its greatest possible area.

A square cannot be negative, so ${(w - 5)}^{2} \geq 0$ .
The largest value of $A$ is $25$ , reached when $w - 5 = 0$ , so $w = 5$ .
The length is also $5$ m, which lies in the allowed range.

The greatest area is

25

2

, for a

5

m by

5

m square.

Check your understanding

A rectangle is $x$ m wide and $(x + 1)$ m long. Its area is $20$ m $2$ . Find its dimensions.

Show answer

$4$ m by $5$ m.

$x (x + 1) = 20$ gives $(x + 5) (x - 4) = 0$ . Reject $x = - 5$ , choose $x = 4$ , and check $4 \times 5 = 20$ .

03 / WORKSHEET

Your turn

6 questions · 24 marks

Worksheet PDF Worked solutions

Read the questions online

A rectangle is $x$ m wide and $(x + 2)$ m long. Its area is $48$ m $2$ . Form an equation and find its dimensions.
4 marks
A square picture has side length $x$ cm. A frame is $2$ cm wide all the way around it. The picture and frame together form a square of area $81$ cm $2$ . Find $x$ .
4 marks
A discount plan charges $(12 - n)$ dollars per item when $n$ items are bought. Here $n$ is a whole number from $1$ to $11$ . The total price is $R = n (12 - n)$ .
1. (a)Find $R$ for $n = 4$ .
2. (b)Find both values of $n$ for which $R = 35$ .
4 marks
A rectangle has perimeter $24$ m and area $32$ m $2$ . Let its width be $w$ m.
1. (a)Show that its length is $(12 - w)$ m.
2. (b)Find both roots of $w (12 - w) = 32$ .
3. (c)State the rectangle’s dimensions.
4 marks
A height model is $h = t (8 - t)$ metres for $0 \leq t \leq 8$ seconds.
1. (a)Find both times when $h = 12$ m.
2. (b)Explain why two times are possible in this model.
4 marks
One model is linear and the other is quadratic. For inputs $x = 0, 1, 2, 3$ , model A gives $1, 4, 9, 16$ and model B gives $2, 5, 8, 11$ .
1. (a)Which is quadratic? Explain using differences.
2. (b)Predict each model’s output for $x = 4$ .
4 marks

Read the worked solutions

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

A rectangle is $x$ m wide and $(x + 2)$ m long. Its area is $48$ m $2$ . Form an equation and find its dimensions.
4 marks
Use $x (x + 2) = 48$ [1].
Factorise $x^{2} + 2 x - 48 = (x + 8) (x - 6)$ [1].
Give roots $- 8, 6$ and reject the negative width [1].
State dimensions $6$ m by $8$ m [1].
$x^{2} + 2 x - 48 = 0$ . The rectangle is $6$ m by $8$ m.
A square picture has side length $x$ cm. A frame is $2$ cm wide all the way around it. The picture and frame together form a square of area $81$ cm $2$ . Find $x$ .
4 marks
The outside side length is $x + 4$ [1].
Use ${(x + 4)}^{2} = 81$ [1].
Find $x + 4 = \pm 9$ , so $x = 5$ or $x = - 13$ [1].
Reject the negative side length and give $5$ cm [1].
$x = 5$ cm.
A discount plan charges $(12 - n)$ dollars per item when $n$ items are bought. Here $n$ is a whole number from $1$ to $11$ . The total price is $R = n (12 - n)$ .
1. (a)Find $R$ for $n = 4$ .
2. (b)Find both values of $n$ for which $R = 35$ .
4 marks
(a) Substitute $n = 4$ and find $4 (8) = 32$ dollars [1].
(b) Form $n^{2} - 12 n + 35 = 0$ [1].
Factorise to $(n - 5) (n - 7) = 0$ [1].
Give both allowed values $5, 7$ [1].
1. (a)32 dollars.
2. (b) $n = 5$ or $n = 7$ .
A rectangle has perimeter $24$ m and area $32$ m $2$ . Let its width be $w$ m.
1. (a)Show that its length is $(12 - w)$ m.
2. (b)Find both roots of $w (12 - w) = 32$ .
3. (c)State the rectangle’s dimensions.
4 marks
(a) From $2 l + 2 w = 24$ , get $l + w = 12$ and $l = 12 - w$ [1].
(b) Factorise $w^{2} - 12 w + 32 = (w - 4) (w - 8)$ [1]; give $4, 8$ [1].
(c) Both choices describe a $4$ m by $8$ m rectangle [1].
1. (a) $l = 12 - w$ .
2. (b) $w = 4$ or $w = 8$ .
3. (c) $4$ m by $8$ m.
A height model is $h = t (8 - t)$ metres for $0 \leq t \leq 8$ seconds.
1. (a)Find both times when $h = 12$ m.
2. (b)Explain why two times are possible in this model.
4 marks
(a) Rearrange to $t^{2} - 8 t + 12 = 0$ [1].
Factorise to $(t - 2) (t - 6) = 0$ [1]; give both allowed times $2, 6$ seconds [1].
(b) Explain the upward and downward parts of the motion [1].
1. (a) $2$ seconds and $6$ seconds.
2. (b)It reaches $12$ m once on the way up and once on the way down.
One model is linear and the other is quadratic. For inputs $x = 0, 1, 2, 3$ , model A gives $1, 4, 9, 16$ and model B gives $2, 5, 8, 11$ .
1. (a)Which is quadratic? Explain using differences.
2. (b)Predict each model’s output for $x = 4$ .
4 marks
(a) Find first differences $3, 5, 7$ for A [1]; constant second differences $2, 2$ identify it as quadratic [1].
(b) Predict A gives $25$ [1] and B gives $14$ [1].
1. (a)Model A is quadratic: first differences $3, 5, 7$ give second differences $2, 2$ . Model B is linear.
2. (b)A: $25$ ; B: $14$ .

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Curriculum reference & lesson scope

AC9M9A05: Use linear and quadratic functions to model applied problems

Original, hand-friendly examples of area, height and financial rules. Students form or use a rule, find integer-root solutions, interpret units and feasible inputs, and discuss limitations. A full modelling investigation may involve collecting data and digital tools; these closed exercises do not claim to assess that whole process.

Official curriculum source ↗

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