Calculate fractional powers and solve simple power equations
Evaluate roots and negative fractional powers · Solve simple radical and negative-power equations · Retain sign and domain conditions
Section E2.12 · Extended · Rational and radical equations
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Integer exponent laws
- Principal square roots
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Evaluate roots and negative fractional powers
- Solve simple radical and negative-power equations
- Retain sign and domain conditions
Root, power and reciprocal
For positive a, aᵐ⁄ⁿ = (ⁿ√a)ᵐ. Find the root then raise it to the numerator power; a negative exponent takes the reciprocal. Positive-base assumptions avoid ambiguities in real fractional powers.
Check a power equation's domain
For x³⁄² = 27 in the real principal-root convention, x ≥ 0 and (√x)³ = 27, so √x = 3 and x = 9. For x⁻³ = 1/8, x ≠ 0 and x³ = 8, giving x = 2.
Even integer powers can give two real roots: 2y⁴ = 32 gives y⁴ = 16, so y = 2 or −2. Both satisfy the original. An odd power gives a unique real root; a principal radical itself has a specified sign. Test candidates rather than automatically taking a positive root.
Sources for this lesson
Pause and explain
What are the real solutions of y⁴ = 16?
Worked example
Evaluate 27²⁄³ and 16⁻³⁄⁴.
Show the worked solution
- The cube root of 27 is 3; its square is 9.
- The fourth root of 16 is 2; cube it to 8 and take the reciprocal.
Answer 9 and 1/8.
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.
Evaluate 64²⁄³ and 81⁻¹⁄².
Give me a hint
Take the relevant root first.
Compare my reasoning
- ∛64 = 4, so its square is 16.
- √81 = 9; the negative exponent gives 1/9.
16 and 1/9.
Look for these in your work
- Interpret numerator and denominator
- Use a reciprocal for a negative exponent
Solve x³⁄² = 27 with real principal roots, and x⁻³ = 1/8.
Give me a hint
Record x ≥ 0 for the first and x ≠ 0 for the second.
Compare my reasoning
- (√x)³ = 27 gives √x = 3, so x = 9.
- x⁻³ = 1/x³ gives x³ = 8, so x = 2; both checks hold.
x = 9 in the first equation; x = 2 in the second.
Look for these in your work
- Keep each domain
- Check the original equations
Solve 2y⁴ = 32 over the real numbers. A model also requires y > 0; what remains then?
Give me a hint
Separate the equation from the contextual restriction.
Compare my reasoning
- y⁴ = 16 gives ±2, both valid algebraic solutions.
- The extra condition y > 0 keeps only 2.
Real solutions y = ±2; with y > 0, only y = 2.
Look for these in your work
- Include both even-power roots
- Apply the stated context afterwards
Common mistakes
- Do not divide the base by the denominator of the exponent.
- Even-power equations may have negative solutions.
What can you explain now?
What are the real solutions of y⁴ = 16?
Compare with the explanation
y = 2 and y = −2.
Both fourth powers equal 16; this is an equation, not just the principal fourth root.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
What are the real solutions of y⁴ = 16?
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.