Mathematics Standard • Year 11 • Module 1 • Lesson 7

Quadratic Formulas in Applied Contexts

Build fluency squaring values, substituting into supplied quadratic formulas and using brackets to avoid sign and decimal errors.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Evaluate each square (no calculator).

5² = ____________ 12² = ____________ (−4)² = ____________ (0.5)² = ____________

Q1.2 Circle the statement that is TRUE.

(a) v² means 2v (b) v² means v × v (c) v² means v + v (d) v² means v ÷ 2

Q1.3 A square has side s = 6 m. Use A = s² to find the area, including units. A = ____________ m²

Stuck? Revisit lesson § Squared Terms Change the Growth Pattern — v² means v × v, not double v.

2. Worked example — stopping-distance formula

Follow each line of working. Brackets are used around every substituted value.

Problem. Use D = 0.01v² + 0.3v to estimate the stopping distance when v = 50 km/h. D is in metres.

Step 1 — Substitute with brackets.

D = 0.01(50)² + 0.3(50)

Reason: brackets remind us to square the 50 first, then multiply by 0.01.

Step 2 — Square first (order of operations).

D = 0.01(2500) + 0.3(50)

Reason: (50)² = 2500. The square always happens before the multiplication.

Step 3 — Multiply each term, then add.

D = 25 + 15 = 40

Reason: 0.01 × 2500 = 25, and 0.3 × 50 = 15.

Conclusion. The estimated stopping distance is 40 m.

3. Faded example — fill in the missing steps

Use D = 0.01v² + 0.3v to estimate stopping distance at v = 60 km/h. Fill in each blank. 4 marks

Step 1 — Substitute with brackets:

D = 0.01(____)² + 0.3(____)

Step 2 — Square the 60:

D = 0.01(____________) + 0.3(60)

Step 3 — Multiply each term:

D = ____________ + ____________

Step 4 — Add and add units:

D = ____________ m

Conclusion sentence. The estimated stopping distance is ____________ m.

Stuck? Revisit lesson § Worked Example 1 — Use a stopping-distance formula.

4. Graduated practice — quadratic substitution

Show all working below each part. Always include units in the final answer.

Foundation — clean numbers, single substitution (4 questions)

Q	Problem	Answer
4.1 1	Use A = s² when s = 4 cm.
4.2 1	Use A = s² when s = 10 m.
4.3 1	Use D = 0.01v² + 0.3v when v = 20.
4.4 1	Use D = 0.01v² + 0.3v when v = 40.

Standard — typical HSC difficulty (6 questions)

Show at least one line of substitution with brackets and label the answer with units where relevant.

4.5 Use A = s² when s = 7.5 m. 2 marks

4.6 Use A = s² when s = 12 cm. 2 marks

4.7 Use D = 0.01v² + 0.3v when v = 50 km/h. Give D in metres. 2 marks

4.8 Use D = 0.01v² + 0.3v when v = 60 km/h. 2 marks

4.9 Use D = 0.01v² + 0.3v when v = 80 km/h. Round to the nearest metre. 2 marks

4.10 Use A = s² when s = 9.4 m. Round to 1 decimal place. 2 marks

Extension — compare and interpret (2 questions)

4.11 Use D = 0.01v² + 0.3v at v = 30 km/h and v = 60 km/h. (a) Calculate both D values. (b) State whether doubling the speed doubled the stopping distance, and explain why. 3 marks

4.12 Square gardens have side lengths s = 3 m, 6 m and 9 m. Use A = s² for each. Explain why tripling the side length more than triples the area. 3 marks

Stuck on 4.11/4.12? Squaring is multiplying the value by itself. If the input doubles, the squared term goes up by a factor of 4 — that is why the model grows faster than a linear one.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I squared the 4 first to get 16 (not 4 × 2 = 8).

For 4.3 I worked out 0.01 × 400 = 4, then added 0.3 × 20 = 6 to get 10 m.

For 4.4 I got 16 + 12 = 28 m (not 16 + 0.3 = 16.3).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Quick squares

5² = 25 12² = 144 (−4)² = 16 (0.5)² = 0.25.

Q1.2 — Meaning of v²

(b) v² means v × v.

Q1.3 — Square area

A = (6)² = 36 m².

Q3 — Faded stopping distance (v = 60)

Step 1: D = 0.01(60)² + 0.3(60).
Step 2: D = 0.01(3600) + 0.3(60).
Step 3: D = 36 + 18.
Step 4: D = 54 m.
Conclusion: The estimated stopping distance is 54 m.

Q4.1 — A = s² with s = 4 cm

A = (4)² = 16 cm².

Q4.2 — A = s² with s = 10 m

A = (10)² = 100 m².

Q4.3 — D at v = 20

D = 0.01(20)² + 0.3(20) = 0.01(400) + 6 = 4 + 6 = 10 m.

Q4.4 — D at v = 40

D = 0.01(40)² + 0.3(40) = 0.01(1600) + 12 = 16 + 12 = 28 m.

Q4.5 — A = s² with s = 7.5

A = (7.5)² = 7.5 × 7.5 = 56.25 m².

Q4.6 — A = s² with s = 12 cm

A = (12)² = 144 cm².

Q4.7 — D at v = 50

D = 0.01(50)² + 0.3(50) = 0.01(2500) + 15 = 25 + 15 = 40 m.

Q4.8 — D at v = 60

D = 0.01(60)² + 0.3(60) = 0.01(3600) + 18 = 36 + 18 = 54 m.

Q4.9 — D at v = 80, rounded to the metre

D = 0.01(80)² + 0.3(80) = 0.01(6400) + 24 = 64 + 24 = 88 m.

Q4.10 — A = s² with s = 9.4 m, 1 d.p.

A = (9.4)² = 88.36 ≈ 88.4 m².

Q4.11 — Compare v = 30 and v = 60

(a) D(30) = 0.01(900) + 9 = 9 + 9 = 18 m. D(60) = 0.01(3600) + 18 = 36 + 18 = 54 m.
(b) No — doubling speed (30 → 60) gave 3 × the stopping distance (18 → 54), not 2 ×. The v² term quadruples when v doubles, so the total grows faster than a constant-rate model.

Q4.12 — Square gardens

A(3) = 9 m². A(6) = 36 m². A(9) = 81 m². Tripling the side from 3 to 9 multiplied the area by 9 (not 3). When you square the side, both length and width grow by the factor, so the area grows by the factor squared.