AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 15

2 marks · Medium difficulty · Short Answer

Solve the quadratic equation 3x^2 + 5x - 9 = 0, giving the solutions as decimals.

Practise this question

Question

Question 15 asks to solve 3x squared plus 5x minus 9 equals 0. A prompt states: Give your solutions as decimals. The question is worth 2 marks, followed by blank working lines and two answer spaces separated by 'and'.

Mark scheme

Show the mark scheme Mark scheme for question 15. M1 is awarded for substituting into the quadratic formula: (-5 ± sqrt(5^2 - 4 * 3 * -9)) / (2 * 3), or completing the square: ±sqrt((5/6)^2 + 3) - 5/6. A1 is awarded for solutions [1.08, 1.09] or 1.1 and [-2.76, -2.75] or -2.8.

How to answer it

Solving Quadratic Equations Using the Quadratic Formula

📋 What This Question Tests
  • Recognising the standard quadratic form: ax² + bx + c = 0
  • Recalling and substituting into the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a
  • Handling signs carefully when c is negative (i.e. -4ac becomes an addition)
  • Evaluating both solutions accurately and writing them as decimals

Question 15 (2 Marks)

Solve 3x² + 5x - 9 = 0. Give your solutions as decimals.

📐 Step-by-Step Calculation

  1. Identify the values of a, b, and c:
    Comparing 3x² + 5x - 9 = 0 to ax² + bx + c = 0 :
    a = 3 ,   b = 5 ,   c = -9
  2. Substitute into the quadratic formula:
    x = -5 ± √(5² - 4 × 3 × (-9)) 2 × 3
  3. Simplify the discriminant (inside the square root) and the denominator:
    5² - 4 × 3 × (-9) = 25 - (-108) = 25 + 108 = 133
    Denominator: 2 × 3 = 6
    x = -5 ± √133 6
  4. Calculate both decimal roots:
    • Using +: x = (-5 + √133) ÷ 6 = 1.0887... ≈ 1.09 (or 1.1)
    • Using -: x = (-5 - √133) ÷ 6 = -2.7554... ≈ -2.76 (or -2.8)
Mark Allocation:
• M1: Correct substitution seen, e.g. (-5 ± √(5² - 4 × 3 × -9)) / (2 × 3) or (-5 ± √133) / 6
• A1: Both decimal solutions correct: [1.08, 1.09] or 1.1 and [-2.76, -2.75] or -2.8

✅ Correct Answer

Answer: 1.09 and -2.76

Acceptable ranges:
• First value: 1.08 to 1.09 (or 1.1 to 1 d.p.)
• Second value: -2.76 to -2.75 (or -2.8 to 1 d.p.)
Solutions can be given in either order.

💡 Key Knowledge

  • Clue in the question: "Give your solutions as decimals" almost always signals that the quadratic cannot be factorised into whole brackets; use the formula!
  • Formula Recall: It is not always given on the front of the paper, so memorise:
    x = (-b ± √(b² - 4ac)) / (2a)
  • Negative constant: Since c = -9 , multiplying -4 × 3 × (-9) creates +108 .

🧠 Exam Technique

  • Write the formula with full substitution first: Even if you mistype something into your calculator later, writing the full substitution secures the M1 mark.
  • Use the fraction button [a b/c]: Enter the entire expression using the fraction template on your scientific calculator to avoid order-of-operation errors.
  • Use the replay arrow: Once you get the first decimal using + , tap the back arrow on your calculator, change the + to - , and press equals to get the second value.

❌ Common Errors (Examiner Traps)

  • Fraction line too short: Writing -5/6 ± √(133) or only putting the root over 6 scores M0. The whole numerator must be divided by 6.
  • Sign slip on -4ac: Calculating 25 - 108 = -83 instead of 25 - (-108) = 133 . Your calculator will say "Math ERROR" if you try to square root a negative number!
  • Only one answer: Providing only the positive root loses the final A1 mark.
  • Sign transposition: Writing -1.09 and +2.76 on the answer line after calculating them correctly in the working loses the A1 mark.
  • Trial and Improvement: Finding one or both values by trial and error receives 0 marks.

Topics

Algebra · 3.2.3 Solving equations and inequalities

Question and mark scheme from the AQA GCSE Mathematics examination, Paper 2 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.