Year 10 Mathematics Core — Applications of Quadratic Equations

Cambridge Ch 5 — Section 5H  •  Sun 14 June 2026
📚 Also for this topic: 📄 Printable worksheet ✅ Solutions (answer key)

Today's lesson

Now we use the factorising skills from 5G to solve real problems. The hard part is turning words into a quadratic equation — and then checking which solution actually makes sense.

Learning intentions

Part 1 — The four steps (~6 min)

Key idea — applying quadratic equations

  1. Define a variable: "Let $x$ be …".
  2. Write an equation from the information given.
  3. Solve the equation by factorising.
  4. Choose the solution(s) that make sense in the context, and check they seem reasonable.
The rejecting step matters. Lengths, widths, times and ages cannot be negative. A quadratic gives two solutions, but often only one fits the real situation — say so and reject the other.

Part 2 — Lesson starter: the 10 cm² triangle (~8 min)

There are many base–height pairs that give a triangle of area $10\ \text{cm}^2$. Let's find the special one whose base is $1$ cm more than its height.

STARTER — set up and solve
A triangle has area $10\ \text{cm}^2$ and its base is $1$ cm more than its height. Find its height and base.
Let the height be $x$ cm, so the base is $(x+1)$ cm.
Area $=\tfrac12\times\text{base}\times\text{height}$:   $\tfrac12\,x(x+1)=10$
$x(x+1)=20 \;\Rightarrow\; x^2+x-20=0 \;\Rightarrow\; (x+5)(x-4)=0$
$\therefore x=-5$ or $x=4$.   A height can't be negative, so reject $x=-5$.
$\therefore$ height $=4$ cm, base $=5$ cm.   Check: $\tfrac12\times5\times4=10$ ✓

Part 3 — Finding dimensions (Example 1, ~14 min)

EXAMPLE 1 — Dimensions of a rectangle
The area of a rectangle is fixed at $28\ \text{m}^2$ and its length is $3$ m more than its width. Find the dimensions of the rectangle.
Let the width be $x$ m, so the length is $(x+3)$ m.
Area $=$ length $\times$ width:   $x(x+3)=28$
$x^2+3x-28=0 \;\Rightarrow\; (x+7)(x-4)=0$
$\therefore x+7=0$ or $x-4=0 \;\Rightarrow\; x=-7$ or $x=4$.
A width must be positive, so reject $x=-7$; choose $x=4$.
$\therefore$ the rectangle has width $4$ m and length $7$ m.   Check: $4\times7=28$ ✓

Now you try: A rectangle has area $48\ \text{m}^2$ and its length is $2$ m more than its width. Find its dimensions.   Answer: width $=6$ m, length $=8$ m (reject $x=-8$).

📺 Walkthrough: the $28\ \text{m}^2$ rectangle — set up $x(x+3)=28$, solve, then reject the negative width to find the real dimensions.

Building understanding — set up and solve.

  1. A rectangle has area $24\ \text{m}^2$ and its length is $5$ m more than its width. Find the dimensions.
  2. A rectangle has area $60\ \text{m}^2$ and its length is $4$ m more than its width. Find the dimensions.
  3. A rectangle has area $63\ \text{m}^2$ and its length is $2$ m less than its width. Find the dimensions.
a) Let width $=x$: $x(x+5)=24 \Rightarrow (x+8)(x-3)=0$, reject $-8$; width $3$ m, length $8$ m.
b) $x(x+4)=60 \Rightarrow (x+10)(x-6)=0$, reject $-10$; width $6$ m, length $10$ m.
c) Let width $=x$, length $=x-2$: $x(x-2)=63 \Rightarrow (x-9)(x+7)=0$, reject $-7$; width $9$ m, length $7$ m.

Part 4 — Number & other problems (~10 min)

EXAMPLE 2 — A number problem
A positive number is $12$ less than its own square. Find the number.
Let the number be $x$. "12 less than its square":   $x = x^2-12$
$0=x^2-x-12 \;\Rightarrow\; 0=(x-4)(x+3)$
$\therefore x=4$ or $x=-3$. The number is positive, so reject $x=-3$.
$\therefore$ the number is $\boxed{4}$.   Check: $4^2-12=4$ ✓

Practice 4.1 — write an equation, solve, and reject any invalid solution.

  1. Two consecutive whole numbers have a product of $56$. Find the numbers.
  2. A right-angled triangle has legs of length $x$ and $x+7$, and hypotenuse $13$. Find $x$.
  3. Two consecutive even integers have a product of $168$. Find the smaller one (positive case).
a) $x(x+1)=56 \Rightarrow (x-7)(x+8)=0$; positive case $x=7$, so $7$ and $8$.
b) $x^2+(x+7)^2=169 \Rightarrow x^2+7x-60=0 \Rightarrow (x-5)(x+12)=0$; reject $-12$, so $x=5$.
c) $x(x+2)=168 \Rightarrow (x-12)(x+14)=0$; positive case $x=12$, so $12$ and $14$.

Part 5 — Quick quiz (5 min)

Pick the correct answer for each, then click Mark.

Q1. The very first step in an application problem is to:

Q2. A rectangle has width $x$ and length $x+3$ and area $28\ \text{m}^2$. The equation is:

Q3. Solving a length problem gives $x=-7$ or $x=4$. You should:

Q4. For the $28\ \text{m}^2$ rectangle (width $4$ m), the length is:

Q5. Two consecutive whole numbers multiply to $56$. The equation $x(x+1)=56$ leads to:

Working program — Cambridge Ex 5H (p456)

After the quiz, open Cambridge Chapter 5 (page 456) and complete the following:

Set workExtension
Questions 1, 3–5, 7–10, 12, 13, 14Questions 15, 16

Always define your variable, then check your answer is reasonable in the context.

Exit ticket — write in your book

Before you pack up, write one sentence each:
  1. List the four steps for solving an application problem.
  2. Why might you reject one of the two solutions to a quadratic?
  3. Give an example of a quantity that cannot be negative.

— Mr Wong