AQA AS Level Mathematics Paper 2, June 2025: Question 5

4 marks · Medium difficulty · Multi-step Problem

Use the substitution u = 4^x to solve the quadratic-form exponential equation 16(4^(2x)) - 33(4^x) + 2 = 0.

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Question

Question 5 asks: 'Use the substitution u = 4^x to solve the equation 16(4^(2x)) - 33(4^x) + 2 = 0.' The question is worth 4 marks.
Question text

5 Use the substitution

u = 4x

to solve the equation

16(42x) – 33(4x) + 2 = 0

[4 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 5 lists 4 marks: M1 for expressing in terms of u to form a quadratic with at least 16u^2 or -33u correct; A1 for obtaining roots u = 2 and u = 1/16; A1 for finding either x = 1/2 or x = -2; A1 for obtaining both solutions x = 1/2 and x = -2.

Q Marking instructions AO Marks Typical solution

5 Expresses 42x and 4x in terms 1.1a M1 u2 − u + =

16 33 2 0

of u 1

At least one of 16u2 or −33u u = 2 or u =

correct

x x 1

4 = 2 or 4 =

1 1.1b A1 1

Obtains 2 and x = or x = −2

16 2

ACF

Obtains x = or x = −2

2 1.1b A1

CSO

1 1.1b A1

Obtains x = and x = −2

CSO

Question 5 Total 4

How to answer it

Disguised Quadratic Equations: Exponential Substitution

AQA AS Level Mathematics • Algebra & Functions

What this question tests

  • Laws of Indices: Recognising that 4²ˣ = (4ˣ)² = u² to convert an exponential expression into polynomial form.
  • Quadratic Equations: Accurately setting up and solving a quadratic equation with leading coefficient > 1.
  • Inverting Exponential Expressions: Converting solutions for u back into solutions for x using negative and fractional powers of 4.

Question 5 Walkthrough (4 Marks)

Solve 16(4²ˣ) − 33(4ˣ) + 2 = 0 using u = 4ˣ

📐 Step-by-Step Calculation

Step 1: Apply the substitution
Given u = 4ˣ, using index laws: 4²ˣ = (4ˣ)² = u².
Substitute these terms into the equation:
16u² − 33u + 2 = 0
Step 2: Solve the quadratic for u
Factorise by finding factors of 16 × 2 = 32 that add to −33 (which are −32 and −1):
16u² − 32u − u + 2 = 0
16u(u − 2) − 1(u − 2) = 0
(16u − 1)(u − 2) = 0
Therefore: u = 2 or u = 1/16
Step 3: Substitute back to find x
• Case 1: 4ˣ = 2 → (2²)ˣ = 2¹ → 2²ˣ = 2¹ → 2x = 1 → x = 1/2
• Case 2: 4ˣ = 1/16 → 4ˣ = 1/(4²) → 4ˣ = 4⁻² → x = −2
Mark Breakdown:
• M1: Expresses 4²ˣ and 4ˣ in terms of u (at least one of 16u² or −33u correct).
• A1: Obtains u = 2 and u = 1/16 (or equivalent form).
• A1: Obtains one correct value of x (x = 1/2 or x = −2).
• A1: Obtains both correct values of x (x = 1/2 and x = −2) with no false roots.

✅ Final Answers

The solutions to the equation are:

x = 1/2  and  x = −2

Both solutions must be clearly stated in terms of x, not u.

💡 Key Knowledge

  • Index Power of a Power: (am)n = amn, meaning 4²ˣ = (4ˣ)² = u².
  • Fractional Indices: 41/2 = √4 = 2.
  • Negative Indices: 4⁻² = 1 / 4² = 1/16.
  • Alternatively, logarithms can be used: x = log₄(2) = 1/2 and x = log₄(1/16) = −2.

🧠 Exam Technique & Examiner Insight

  • Always Complete the Question: The most common place students drop the final 2 marks is stopping after finding u = 2 and u = 1/16. The original equation asks for x, not u!
  • Write Base 2 If Unsure: If solving 4ˣ = 2 is not immediately obvious, express 4 as 2² to get 2²ˣ = 2¹, making the linear equation 2x = 1 trivial to solve.
  • Check by Back-Substitution: Quick sanity check: 41/2 = 2, so 16(2²) − 33(2) + 2 = 64 − 66 + 2 = 0 ✓.

❌ Common Errors to Avoid

  • Prematurely Stopping: Leaving the answer as u = 2, u = 1/16 loses the last two A marks (50% of the question!).
  • Index Mistranslations: Writing 4²ˣ as 2u or u + 2 instead of u².
  • Sign / Power Slips: Solving 4ˣ = 2 as x = 2 (confusing 2² = 4 with 4² = 16), or 4ˣ = 1/16 as x = 2 instead of x = −2.

Topics

Pure Mathematics · B: Algebra and functions · F: Exponentials and logarithms

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