# Powers and roots > Edexcel International GCSE Mathematics A · 4MA1 > 来源: https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s1-powers-and-roots/ This guide covers all powers and roots content for Edexcel IGCSE Maths A (4MA1) specification point 1.4, including index laws, prime factorisation, HCF/LCM, and Higher tier surd manipulation for both calculator and non-calculator papers. **先修:** [Basic integer multiplication and division](https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s1-arithmetic-operations/); [Understanding of prime numbers](https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s1-prime-numbers/) ## 学习目标 - Identify and calculate square numbers, cube numbers and their respective roots - Apply index laws for positive, negative and zero integer powers - Express integers as products of prime factors to calculate HCF and LCM - (Higher only) Evaluate expressions with fractional and negative indices - (Higher only) Simplify surds and rationalise surd denominators ## Squares, Cubes and Their Roots (Foundation) Square numbers are the product of an integer multiplied by itself, while cube numbers are the product of an integer multiplied by itself twice. The square root of a number is the non-negative value that when squared gives the original number, and the cube root is the value that when cubed gives the original number (can be negative for negative inputs). **Square and Cube Rules** — Square number: $n^2 = n \times n$ for integer $n$. Square root: $\sqrt{n^2} = |n|$. Cube number: $n^3 = n \times n \times n$ for integer $n$. Cube root: $\sqrt[3]{n^3} = n$. **例题:** Calculate (a) $12^2$, (b) $\sqrt{169}$, (c) $5^3$, (d) $\sqrt[3]{-27}$ 1. (a) Multiply 12 by itself: $12 \times 12 = 144$ 2. (b) Find the positive integer that squares to 169: $13^2 = 169$, so $\sqrt{169} = 13$ 3. (c) Multiply 5 by itself three times: $5 \times 5 \times 5 = 125$ 4. (d) Find the integer that cubes to -27: $(-3)^3 = -27$, so $\sqrt[3]{-27} = -3$ > **考试提示:** Memorize the first 15 square numbers and first 10 cube numbers to save calculation time in the exam. ## Index Laws for Integer Powers (All Tiers) Index notation uses a base and a power (index) to represent repeated multiplication efficiently. The index laws let you simplify expressions with powers without expanding them fully, saving calculation time. **Integer Index Laws** — For any non-zero base $x$:
  1. $x^m \times x^n = x^{m+n}$
  2. $x^m \div x^n = x^{m-n}$
  3. $(x^m)^n = x^{mn}$
  4. $x^0 = 1$
  5. $x^{-n} = \frac{1}{x^n}$
**例题:** Simplify and evaluate (a) $3^4 \times 3^2$, (b) $7^5 \div 7^3$, (c) $(2^3)^4$, (d) $9^0$, (e) $4^{-2}$ 1. (a) Apply multiplication index law: $3^{4+2} = 3^6 = 729$ 2. (b) Apply division index law: $7^{5-3} = 7^2 = 49$ 3. (c) Apply power of a power index law: $2^{3 \times 4} = 2^{12} = 4096$ 4. (d) Apply zero index rule: $9^0 = 1$ 5. (e) Apply negative index rule: $4^{-2} = \frac{1}{4^2} = \frac{1}{16}$ > **考试提示:** Always confirm the base of the powers is identical before applying multiplication or division index laws. ## Prime Factorisation, HCF and LCM (All Tiers) Any positive integer can be written as a product of prime numbers raised to powers, called its prime factor form. This form is the fastest way to calculate the highest common factor (HCF) and lowest common multiple (LCM) of two or more numbers. **HCF and LCM Rules** — HCF: Product of the lowest power of each prime factor common to all numbers. LCM: Product of the highest power of every prime factor present in any of the numbers. **例题:** Write 72 and 90 as products of prime factors, then calculate their HCF and LCM. 1. Prime factorise 72 using factor tree: $72 = 2^3 \times 3^2$ 2. Prime factorise 90 using factor tree: $90 = 2^1 \times 3^2 \times 5^1$ 3. Calculate HCF: Take lowest power of common primes (2, 3): $2^1 \times 3^2 = 2 \times 9 = 18$ 4. Calculate LCM: Take highest power of all primes (2, 3, 5): $2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$ > **考试提示:** Use factor trees to break down large numbers into prime factors quickly, and cross-check your HCF divides both input numbers fully. ## Fractional and Negative Indices (Higher Only) For Higher tier, you will need to evaluate expressions with fractional indices, which combine powers and roots. The denominator of the fractional index is the root, and the numerator is the power you raise the result to. **Fractional Index Rule** — For positive base $x$: $x^{\frac{a}{b}} = (\sqrt[b]{x})^a = \sqrt[b]{x^a}$. Combine with the negative index rule for expressions of the form $x^{-\frac{a}{b}} = \frac{1}{x^{\frac{a}{b}}}$. **例题:** Evaluate (a) $8^{\frac{2}{3}}$, (b) $625^{-\frac{1}{2}}$, (c) $(\frac{1}{25})^{\frac{3}{2}}$ 1. (a) Take cube root first then square: $(\sqrt[3]{8})^2 = 2^2 = 4$ 2. (b) Apply negative index rule first, then square root: $\frac{1}{\sqrt{625}} = \frac{1}{25}$ 3. (c) Take square root first then cube: $(\sqrt{\frac{1}{25}})^3 = (\frac{1}{5})^3 = \frac{1}{125}$ > **考试提示:** Always compute the root first for fractional indices to keep numbers small and avoid arithmetic errors. ## Surds and Rationalising Denominators (Higher Only) A surd is an irrational root of a positive integer, e.g. $\sqrt{2}$ or $\sqrt{7}$. You will need to simplify surds and rationalise denominators (remove surds from the bottom of fractions) to give exact answers, rather than approximate decimal values. **Surd Rules** —
  1. $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
  2. $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$
  3. To rationalise a single surd denominator: Multiply numerator and denominator by the surd.
  4. To rationalise $a + b\sqrt{c}$ denominator: Multiply numerator and denominator by the conjugate $a - b\sqrt{c}$.
**例题:** (a) Simplify $\sqrt{8} + 3\sqrt{32}$, (b) Rationalise $\frac{2}{\sqrt{7}}$, (c) Rationalise $\frac{3}{2 + \sqrt{5}}$ 1. (a) Simplify each surd first: $\sqrt{8} = 2\sqrt{2}$, $3\sqrt{32} = 3 \times 4\sqrt{2} = 12\sqrt{2}$. Add like terms: $2\sqrt{2} + 12\sqrt{2} = 14\sqrt{2}$ 2. (b) Multiply numerator and denominator by $\sqrt{7}$: $\frac{2\sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{2\sqrt{7}}{7}$ 3. (c) Multiply numerator and denominator by conjugate $2 - \sqrt{5}$: $\frac{3(2 - \sqrt{5})}{(2 + \sqrt{5})(2 - \sqrt{5})} = \frac{6 - 3\sqrt{5}}{4 - 5} = -6 + 3\sqrt{5}$ > **考试提示:** Never leave a surd in the denominator unless the question explicitly allows it, or you will lose marks. ## 常见错误 - **错误做法:** Applying multiplication/division index laws to expressions with different bases, e.g. $2^3 \times 3^2 = 6^5$ - 原因: Index operations only work when the base of all terms is identical - 正确做法: Calculate each term separately before multiplying: $8 \times 9 = 72$ - **错误做法:** Writing $\sqrt{-16} = -4$, assuming negative numbers have real square roots - 原因: Squares of real numbers are always non-negative, so square roots of negative numbers are not real at this level - 正确做法: Only calculate cube roots of negative numbers; leave square roots of negative numbers as undefined for this syllabus - **错误做法:** Calculating the power first for fractional indices leading to very large numbers, e.g. $8^{5/3} = 32768^{1/3}$ - 原因: Calculating the power first creates unnecessarily large values that are hard to work with - 正确做法: Compute the root first to get a smaller number: $(\sqrt[3]{8})^5 = 2^5 = 32$ - **错误做法:** Using only common prime factors to calculate LCM, e.g. LCM of 72 and 90 = $2^3 \times 3^2 = 72$ - 原因: LCM must be divisible by both input numbers, so needs all primes from both factorisations - 正确做法: Use the highest power of every prime present in either number: LCM of 72 and 90 = $2^3 \times 3^2 \times 5 = 360$ - **错误做法:** Rationalising $\frac{1}{2 + \sqrt{3}}$ by multiplying only by $\sqrt{3}$ - 原因: This leaves a cross term with a surd still in the denominator - 正确做法: Multiply numerator and denominator by the conjugate $2 - \sqrt{3}$ to eliminate the surd via difference of squares ## 速查表 | Concept | Rule | Tier | | --- | --- | --- | | Square/Cube Calculations | $n^2 = n\times n$, $\sqrt{n^2}=\|n\|$, $n^3 = n\times n\times n$, $\sqrt[3]{n^3}=n$ | Foundation | | Integer Index Laws | $x^m \times x^n = x^{m+n}$, $x^m \div x^n = x^{m-n}$, $(x^m)^n = x^{mn}$, $x^0=1$, $x^{-n}=1/x^n$ | Foundation | | HCF via Prime Factors | Product of lowest power of common primes | Foundation | | LCM via Prime Factors | Product of highest power of all primes present | Foundation | | Fractional Indices | $x^{a/b} = (\sqrt[b]{x})^a$ | Higher | | Surd Simplification | $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ | Higher | | Rationalise Denominator | Multiply by surd (single) or conjugate ($a+b\sqrt{c}$) | Higher | ## 下一步 Now you have mastered powers and roots for Edexcel IGCSE Maths A, you can move on to algebraic index manipulation (section 2.1 of the specification), which applies the same index laws you have learned to expressions with variables. You should also practice applying these rules to real-world arithmetic problems, and work through tier-specific past paper questions to build speed and accuracy. For Higher tier students, make sure you are confident with surd manipulation before moving on to quadratic formula problems, which often require simplified exact surd answers. - [Quadratic Equations (Higher)](https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s2-quadratic-equations/) --- 来自 [OwlsPrep](https://www.owlsprep.com) —— A-Level / IB / AP / IGCSE 免费学习指南，依据官方考纲编写。原页面：https://www.owlsprep.com/zh/study/edexcel-igcse-math-a-s1-powers-and-roots/