AQA GCSE Mathematics Paper 1 (Foundation), June 2025: Question 26

What this question tests

Forming an algebraic equation from geometric properties (Area of a rectangle = length × width).
Expanding binomials (multiplying two linear brackets) accurately.
Rearranging quadratics into the standard form: ax² + bx + c = 0 .
Solving quadratic equations by factorising, completing the square, or using the quadratic formula.
Contextual reasoning: rejecting a mathematically valid solution that produces an impossible physical length (negative length).

1 Set up the equation for area:
Area = (length) × (width)
(x + 2)(x - 5) = 120
2 Expand the brackets:
x² - 5x + 2x - 10 = 120
x² - 3x - 10 = 120
3 Rearrange to equal 0:
Subtract 120 from both sides:
x² - 3x - 130 = 0
4 Factorise the quadratic:
Find two numbers that multiply to -130 and add to -3:
The factors are +10 and -13.
(x + 10)(x - 13) = 0
So, x = -10 or x = 13
5 Interpret in context:
If x = -10 , width = -10 - 5 = -15 cm (impossible).
Therefore, x = 13 .

M1: Correct expansion of brackets with at least 3 terms correct from x² + 2x - 5x - 10 (or implied by x² - 3x + k ).
M1 (dep): Rearranging to equal zero: x² - 3x - 130 (= 0) .
M1: Correct method to solve their 3-term quadratic (e.g. factorising into (x + 10)(x - 13) , substitution into the formula, or stating roots -10 and 13).
A1: Final value of x = 13 only.

Special Case: A student who accidentally calculates using perimeter instead of area can score a maximum of SC1 for x = 31.5 .

Standard Form First: You cannot factorise to solve until the equation equals 0! Never try to factorise x² - 3x - 10 = 120 directly.
Spotting Factors of 130: 130 ends in 0, so divisible by 10 (10 × 13 = 130). Difference between 13 and 10 is 3, which matches the middle term.
Quadratic Formula Alternative:
If factorising feels tricky, use:
x = (-b ± √(b² - 4ac)) / (2a)
where a = 1, b = -3, c = -130:
x = (3 ± √(9 - 4(1)(-130))) / 2 = (3 ± √529) / 2 = (3 ± 23) / 2

Reject the negative root explicitly: The question asks for the value of x, not the solutions to the equation. Leaving your answer as x = 13 or -10 loses the final A1 mark.
Check your answer: Substitute x = 13 back into the side lengths:
Length = 13 + 2 = 15 cm
Width = 13 - 5 = 8 cm
Area = 15 × 8 = 120 cm² (Correct!)
Show all working: Even if you make an arithmetic slip solving the quadratic, you can still gain M marks for correct expansion and setting to 0.

Perimeter instead of Area: Writing 2(x + 2) + 2(x - 5) = 120 . Always read whether the question specifies area or perimeter.
Sign errors when expanding: Writing +10 instead of -10 when calculating (+2) × (-5) .
Premature factorisation: Attempting to factorise x² - 3x - 10 as (x - 5)(x + 2) = 120 and setting individual brackets equal to 120 (e.g. x - 5 = 120 ). A product only gives solutions when equal to zero!
Giving both roots: Writing x = 13, -10 on the answer line. Lengths cannot be negative in geometry.

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