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IGCSE · Cambridge (CIE) · 0606

Use substitution to solve transformed equations

Recognise a repeated expression, substitute a temporary variable, solve the quadratic and reverse the substitution within its domain.

Section 4 · Equations, inequalities and graphs

Cambridge IGCSE Additional Mathematics 0606 · 2028–2030 · Version 1 · 4

Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.

Before you begin

  • Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which substitution is natural in x⁴ − 5x² + 4 = 0?

What you will learn

  • Choose a substitution that turns an equation into a quadratic
  • Solve the resulting quadratic
  • Reject substituted roots that violate the substitution's range

Choose a repeated inner expression

Equations in x⁴ and x² can become quadratic by setting u = x². Equations with repeated ln x, eˣ or another expression can use the same strategy.

Solve the quadratic in u, then reverse the substitution. The temporary variable may have restrictions: x² ≥ 0, eˣ > 0 and ln x requires x > 0.

Check candidates in the original domain

For x⁴ − 5x² + 4 = 0, set u = x². Then u² − 5u + 4 = 0 gives u = 1 or 4; both are allowed because u ≥ 0.

For logarithmic substitutions, a quadratic root that makes ln x negative is not automatically invalid, but the resulting x must still be positive. Always test the original equation and domain.

Pause and explain

After solving u = x², which real u-values can correspond to x?

Put the idea to work

Worked example

Solve x⁴ − 5x² + 4 = 0 for real x.

Show the worked solution
  1. Set u = x², so u² − 5u + 4 = 0.
  2. Factor: (u − 1)(u − 4) = 0, giving u = 1 or 4.
  3. Return to x² = 1 or x² = 4 and take both real square roots.

Answer x = −2, −1, 1 or 2.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

Test this topic →

From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Solve x⁴ − 10x² + 9 = 0.

Give me a hint

Let u = x² and remember u cannot be negative.

Compare my reasoning
  1. Form u² − 10u + 9 = 0.
  2. Factor: (u − 1)(u − 9) = 0.
  3. Return to x² = 1 or x² = 9.

x = −3, −1, 1 or 3.

Look for these in your work

  • Substitution correct
  • Both u-values valid
  • Both signs of x included
2 · Independent

Solve (ln x)² − 3 ln x + 2 = 0 for x > 0.

Give me a hint

Use u = ln x, then exponentiate each solution.

Compare my reasoning
  1. Set u = ln x: u² − 3u + 2 = 0.
  2. Factor to get u = 1 or 2.
  3. Thus x = e or e², both positive.

x = e or e².

Look for these in your work

  • Quadratic solved
  • Logarithm reversed correctly
  • Domain checked
3 · Transfer

A population model produces 3^(2t) − 5·3^t + 4 = 0. Find its real time solutions t.

Give me a hint

Set u = 3^t and remember this quantity is positive.

Compare my reasoning
  1. Let u = 3^t > 0, giving u² − 5u + 4 = 0.
  2. Factor to get u = 1 or 4; both are positive.
  3. Therefore 3^t = 1 or 3^t = 4.

t = 0 or t = log₃4.

Look for these in your work

  • Repeated exponential identified
  • Positive roots retained
  • Logarithmic solution stated

Common mistakes

  • Stopping after solving for the temporary variable
  • Taking only positive square roots
  • Keeping substituted values that cannot occur in the original domain
Recall without your notes

What can you explain now?

If u = eˣ, what values can u take?

Compare with the explanation

u > 0

An exponential is strictly positive, so zero and negative substituted roots are impossible.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Describe substitution, the temporary variable's range and the final check in the original equation.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Mathematics — Additional

Course scope and assessment

All 14 topics in the Cambridge IGCSE Additional Mathematics 0606 syllabus have mapped teaching: Functions, Quadratic functions, Factors of polynomials, Equations/inequalities/graphs, Simultaneous equations, Logarithmic and exponential functions, Straight-line graphs, Coordinate geometry of the circle, Circular measure, Trigonometry, Permutations and combinations, Series, Vectors in two dimensions, and Calculus. Lessons include guided, independent and application practice with explained answers. No full timed Paper 1 or Paper 2, Cambridge marking or human specialist review is provided; self-checks do not certify a grade.

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