Quadratic Equations in Context Revision Notes | GCSE AQA Higher Further Maths

1. Why Context Problems Need Equations

A rectangle's length is 3 cm more than its width. Its area is 40 cm². What are the dimensions? You cannot answer that with arithmetic alone — you need an equation. This is the core skill here: turning a written situation into algebra, then solving it.

Context problems appear across the GCSE paper in many forms: area and perimeter, consecutive integers, number puzzles, and motion. The equation that emerges is often quadratic, meaning it has an $x^2$ term. Your job is to build it, solve it, and then check whether each solution makes sense in the original situation.

The two main solving methods at GCSE are factorisation and the quadratic formula. Factorisation is faster when it works; the formula is the reliable backup for any quadratic. Both methods are covered here, applied directly to context problems.

2. Setting Up the Equation

Before you can solve anything, you need to translate the words into algebra. The process has three steps: choose a variable, write expressions for all unknown quantities in terms of that variable, then form an equation using the given condition (area, product, sum, etc.).

• Choose one unknown as your variable, usually $x$.
• Express every other unknown in terms of $x$ using the relationships given in the problem.
• Use a geometric or numerical condition (e.g. area = length × width) to write the equation.
• Expand and rearrange so the equation equals zero before solving.

Here is where many marks are dropped: students jump straight to solving without checking that their equation matches the problem. Always re-read the question after forming your equation and confirm each term has the right meaning.

3. Solving by Factorisation

Factorisation works when you can spot two integers that multiply to give $c$ and add to give $b$ in $x^2 + bx + c = 0$. Once you write the quadratic as $(x + p)(x + q) = 0$, each bracket gives a solution: $x = -p$ or $x = -q$.

Worked example: Area of a Rectangle — Solving by Factorisation

A rectangle has length $(x + 3)$ cm and width $x$ cm. Its area is 40 cm². Find the value of $x$.

1. $x(x + 3) = 40$ — Area = length × width. [M1 for forming the correct equation]
2. $x^2 + 3x = 40$ $x^2 + 3x - 40 = 0$ — Subtract 40 from both sides so the quadratic is in standard form.
3. $8 \times (-5) = -40$ and $8 + (-5) = 3$ ✓ — List factor pairs of 40: (1,40), (2,20), (4,10), (5,8). The pair 8 and −5 works.
4. $(x + 8)(x - 5) = 0$ — [M1 for correct factorisation]
5. $x + 8 = 0 \Rightarrow x = -8$ $x - 5 = 0 \Rightarrow x = 5$ — [A1 for both solutions]
6. $x = -8$ is rejected (width cannot be negative) $x = 5$ cm — Always interpret solutions in context. A length must be positive. [A1 for correct contextual conclusion]

Answer: $x = 5$ cm, so the rectangle is 5 cm wide and 8 cm long.

Notice that the algebra gave two valid mathematical solutions, but only one made sense as a length. This contextual rejection step is worth a mark on its own — examiners expect you to state why you discard a solution, not just ignore it.

4. Solving Using the Quadratic Formula

When the numbers in a context problem do not produce a neat factorisable quadratic, the quadratic formula is the reliable method. It works for every quadratic of the form $ax^2 + bx + c = 0$, regardless of whether the roots are integers or decimals.

Worked example: Consecutive Integers — Solving with the Quadratic Formula

The product of two consecutive positive integers is 56. Find the integers. Give your answer by solving the equation using the quadratic formula.

1. Let the smaller integer be $n$. The next consecutive integer is $n + 1$. — Consecutive integers differ by 1.
2. $n(n + 1) = 56$ — Product means multiply. [M1 for correct equation]
3. $n^2 + n = 56$ $n^2 + n - 56 = 0$ — Subtract 56 from both sides.
4. $a = 1,\quad b = 1,\quad c = -56$ — Read off the coefficients carefully, including the sign of $c$.
5. $n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-56)}}{2(1)}$ $n = \frac{-1 \pm \sqrt{1 + 224}}{2}$ $n = \frac{-1 \pm \sqrt{225}}{2}$ — $1 \times 1 = 1$; $4 \times 1 \times (-56) = -224$; subtracting a negative adds: $1 - (-224) = 225$. [M1 for correct substitution]
6. $\sqrt{225} = 15$ — $15 \times 15 = 225$ ✓
7. $n = \frac{-1 + 15}{2} = \frac{14}{2} = 7$ $n = \frac{-1 - 15}{2} = \frac{-16}{2} = -8$ — [A1 for both values]
8. $n = -8$ rejected (problem states positive integers) $n = 7$ — The two consecutive positive integers are 7 and 8. Check: $7 \times 8 = 56$ ✓ [A1 for correct contextual answer]

Answer: The two consecutive positive integers are 7 and 8.

5. A Variety of Context Types

GCSE questions set quadratic equations in many different situations. The algebra is the same each time — the skill is recognising which condition in the problem gives you the equation. Here are the most common context types you will encounter.

Worked example: Perimeter and Area Combined

A rectangle has perimeter 28 cm and area 45 cm². Find its length and width.

1. Perimeter $= 2(l + w) = 28$ $l + w = 14$ $l = 14 - w$ — Halving the perimeter gives the sum of length and width. Let width $= w$.
2. $l \times w = 45$ $(14 - w) \times w = 45$ — Replace $l$ with $14 - w$. [M1 for forming the equation]
3. $14w - w^2 = 45$ $0 = w^2 - 14w + 45$ — Rearrange so the quadratic equals zero. Multiply through by $-1$ to make the $w^2$ term positive.
4. Find two numbers that multiply to 45 and add to $-14$: $-9$ and $-5$ $(w - 9)(w - 5) = 0$ — $(-9) \times (-5) = 45$ and $(-9) + (-5) = -14$ ✓ [M1 for factorisation]
5. $w = 9$ or $w = 5$ — [A1 for both values]
6. If $w = 9$, then $l = 14 - 9 = 5$ If $w = 5$, then $l = 14 - 5 = 9$ — Both solutions are valid — they describe the same rectangle with dimensions 5 cm and 9 cm. Check: $5 \times 9 = 45$ ✓ and $2(5+9) = 28$ ✓ [A1 for correct dimensions]

Answer: The rectangle is 9 cm by 5 cm.

6. Deliberate Practice

Now that you have seen the methods in action, here are problems to work through independently. For each one, form the equation first, then solve it, then check your solutions against the context.

7. Exam Technique and Common Mistakes

Context problems are where method marks and accuracy marks can both be won or lost. The examiner is looking for a clear equation, correct algebraic working, and a conclusion that refers back to the original problem. Here is what separates full-mark answers from partial ones.

Following this sequence consistently means you will always earn the method mark for forming the equation, even if an arithmetic slip later costs an accuracy mark. Examiners award marks at each stage, so structured working protects your score throughout.