Subject: Mathematics | Level: GCSE | Exam Board: OCR
Master the foundational rules of Indices and Surds to unlock higher-level algebra. This topic covers the manipulation of powers, roots, and exact values, essential for maximizing marks in both calculator and non-calculator papers.
Revision Notes & Key Concepts
Key Terms & Definitions
- Index (Power/Exponent)
- A number that indicates how many times a base number is multiplied by itself.
- Base
- The number that is being multiplied by itself when raised to a power.
- Surd
- An irrational number expressed as a root, which cannot be simplified to a rational number.
- Rationalise
- The process of converting an irrational denominator into a rational number.
- Conjugate
- A binomial formed by negating the second term of a binomial. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$.
- Reciprocal
- The multiplicative inverse of a number; 1 divided by the number.
Worked Examples
Worked Example
Question: Evaluate $27^{-\frac{2}{3}}$. (3 marks)
Solution: Step 1: Deal with the negative index by taking the reciprocal. $27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}}$ Step 2: Apply the fractional index rule (root first, then power). The denominator 3 means cube root, the numerator 2 means square. $\sqrt[3]{27} = 3$ Step 3: Apply the power. $3^2 = 9$ Final answer: $\frac{1}{9}$
Worked Example
Question: Simplify fully $\sqrt{75} + \sqrt{27}$. (3 marks)
Solution: Step 1: Simplify $\sqrt{75}$. The largest perfect square factor is 25. $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$ Step 2: Simplify $\sqrt{27}$. The largest perfect square factor is 9. $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ Step 3: Add the like surds. $5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}$ Final answer: $8\sqrt{3}$
Worked Example
Question: Rationalise the denominator and simplify $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$. (4 marks)
Solution: Step 1: Multiply numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + 1$. $\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$ Step 2: Expand the numerator. $(\sqrt{3} \times \sqrt{3}) + \sqrt{3} + \sqrt{3} + 1 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}$ Step 3: Expand the denominator using difference of two squares. $(\sqrt{3})^2 - 1^2 = 3 - 1 = 2$ Step 4: Combine and simplify the fraction. $\frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$ Final answer: $2 + \sqrt{3}$
Practice Questions
Question: Simplify $x^5 \times x^3 \div x^2$.
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Question: Evaluate $64^{\frac{1}{3}}$.
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Question: Simplify $\sqrt{48} + \sqrt{12}$.
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Question: Evaluate $(\frac{8}{27})^{-\frac{2}{3}}$.
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Question: Rationalise the denominator of $\frac{5}{3 - \sqrt{2}}$.
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