GCSE Maths

Numbers
37 Topics | 1 Quiz
Integers
Prime Numbers and Composite Numbers
Negative Numbers
Comparing Numbers Using Symbols
The Square Numbers
The Laws of Indices
Negative Indices
The Cube Numbers
Triangular Numbers
The Fibonacci Sequence
Factors
Multiples
The Highest Common Factor
Fractions
Adding and Subtracting Fractions
Multiplication of Fractions
Division of Fractions
Fractional Indices
Decimal Places
Fractions and Decimals
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Vectors
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Vectors
Probability
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Introduction to probability
Finding the sum of probabilities
Conditional probability
Set Notation
Venn diagrams
The Complement of a Set
Statistics
2 Topics
Population and samples
Representing data
Algebra
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Algebraic Notation
Algebraic Vocabulary
Collecting Like Terms
Expanding Brackets and Simplifying
Expanding Double Brackets
The Difference of Two Squares
Substituting into Formulae
Subject of Formula
Factorisation
Factorising Quadratic Expressions
More on Factorisation
Equations
Quadratic Equations
Mathematical Formulae
Algebraic Equivalence and Proof
Functions
Algebra
Sequences
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Generating sequences
Nth Term Sequences
Sequences
Ratio, Proportion and Rates of Change
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Inverse Proportion
Conversion
Scale Factors
Percentages
Direct Proportion
Rate, Ratio and Proportion Examples
Ratios
Ratios
Ratio, Proportion and Rates of Change
Geometry and Measures
9 Topics
Loci and constructions
Properties of angles
Properties of polygons
Transformations
Enlargement
Circle theorems
Plane figure properties
Plans and elevations
Areas and Volumes in Map Scale Problems
Mensuration and Calculation
10 Topics
Map scales
Bearings
Area
Volume
Circles: arcs and sectors
Congruence and similarity
Pythagoras’ theorem
Sine and cosine rule
Area of a triangle
Standard units of measure
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Fractional Indices

GCSE Maths Numbers Fractional Indices

a^{\dfrac{1}{2}}=\sqrt{a} (square root).

a^{\dfrac{1}{3}}=\sqrt[3] {a} (cube root)

a^{\dfrac{1}{n}}=\sqrt[n] {a} (nth root)

a^{\dfrac{m}{n}}=\left( a^{m}\right) ^{\dfrac{1}{n}}=\sqrt[n] {a^{m}}

Example

Simplify the following:

1. 8^{\dfrac{2}{3}}

2. 16^{\dfrac{1}{4}}

3. 125^{-\dfrac{2}{3}}

4. 64^{-\dfrac{4}{3}}

5. 36^{-\dfrac{1}{2}}

6. 81^{\dfrac{5}{4}}

7. \left( \dfrac{2}{3}\right) ^{-2}

8. -8^{\dfrac{2}{3}}

9. -27^{-\dfrac{4}{3}}

Solutions:

1. 

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2. 16^{\dfrac{1}{4}}=\left( 2^{4}\right) ^{\dfrac{1}{4}}=2^{1}=2

3. 125^{-\dfrac{2}{3}}=\left( 5^{3}\right) ^{-\dfrac{2}{3}}=5^{-2}=\dfrac{1}{5^{2}}=\dfrac{1}{25}

4. 64^{-\dfrac{4}{3}}=\left( 4^{3}\right) ^{-\dfrac{4}{3}}=4^{-4}=\dfrac{1}{4^{4}}=\dfrac{1}{256}

5. 36^{-\dfrac{1}{2}}=\left( 6^{2}\right) ^{-\dfrac{1}{2}}=6^{-1}=\dfrac{1}{6}

6. 81^{\dfrac{5}{4}}=\left( 3^{4}\right) ^{\dfrac{5}{4}}=3^{5}=243

7. \left( \dfrac{2}{3}\right) ^{-2}=\dfrac{1}{\left( \dfrac{2}{3}\right) ^{2}}=\dfrac{1}{\dfrac{4}{9}}=1\div \dfrac{4}{9}=1\times \dfrac{9}{4}=\dfrac{9}{4}

  • Note that, \left( \dfrac{a}{b}\right) ^{-m}=\left( \dfrac{b}{a}\right) ^{n}

8. -8^{\dfrac{2}{3}}=\left[ \left( -2\right) ^{3}\right] ^{\dfrac{2}{3}}=\left( -2\right) ^{2}=4

9. -27^{-\dfrac{4}{3}}=\left[ \left( -3\right) ^{3}\right] ^{-\dfrac{4}{3}}=\left( -3\right) ^{-4}=\dfrac{1}{\left( -3\right) ^{4}}=\dfrac{1}{81}

Example

Evaluate:

1. { 16 }^{ \frac { 1 }{ 4 } }=({ { 2 }^{ 4 }) }^{ \frac { 1 }{ 4 } }={ 2 }^{ 1 }=2

  • We first write it as 2^4, then apply the power.

2. { 16 }^{ -\frac { 3 }{ 4 } }=({ { 2 }^{ 4 }) }^{ \frac { 3 }{ 4 } }={ 2 }^{ -2 }=\frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 8 }

3. { 125 }^{ -\frac { 1 }{ 3 } }=({ { 5 }^{ 3 }) }^{ -\frac { 1 }{ 3 } }={ 5 }^{ -1 }=\frac { 1 }{ { 5 }^{ 1 } } =\frac { 1 }{ 5 }

4. ({ \frac { 9 }{ 16 } })^{ -\frac { 3 }{ 2 } }=(({ { \frac { 3 }{ 4 } ) }^{ 2 }) }^{ -\frac { 3 }{ 2 } }={ (\frac { 3 }{ 4 } })^{ -3 }=\frac { 1 }{ { (\frac { 3 }{ 4 } ) }^{ 3 } }

  • =\frac { 1 }{ \frac { 27 }{ 64 } }
  • =\frac { 64 }{ 27 }

5. ({ 2\frac { 1 }{ 4 } })^{ 3 }={ (\frac { 9 }{ 4 } ) }^{ 3 }=\frac { 729 }{ 64 }

6. ({ 2\frac { 1 }{ 4 } ) }^{ -3 }=({ \frac { 9 }{ 4 } ) }^{ -3 }=({ \frac { 4 }{ 9 } ) }^{ 3 }=\frac { 64 }{ 729 }

7. ({ 2\frac { 1 }{ 4 } ) }^{ \frac { 3 }{ 2 } }=({ \frac { 9 }{ 4 } ) }^{ \frac { 3 }{ 2 } }=(({ { \frac { 3 }{ 2 } ) }^{ 2 } })^{ \frac { 3 }{ 2 } }={ (\frac { 3 }{ 2 } ) }^{ 3 }=\frac { 27 }{ 8 }

8. ({ 2\frac { 1 }{ 4 } ) }^{ -\frac { 3 }{ 2 } }=({ \frac { 9 }{ 4 } ) }^{ -\frac { 3 }{ 2 } }=(({ { \frac { 3 }{ 2 } ) }^{ 2 }) }^{ -\frac { 3 }{ 2 } }

  • ={ (\frac { 3 }{ 2 } ) }^{ -3 }
  • =({ \frac { 2 }{ 3 } ) }^{ 3 }
  • =\frac { 8 }{ 27 }

9. { 25 }^{ -\frac { 1 }{ 2 } }=({ { 5 }^{ 2 }) }^{ -\frac { 1 }{ 2 } }={ 5 }^{ -1 }=\frac { 1 }{ 5 }

10. { 49 }^{ -\frac { 3 }{ 2 } }=({ { 7 }^{ 2 }) }^{ -\frac { 3 }{ 2 } }={ 7 }^{ -3 }=\frac { 1 }{ { 7 }^{ 3 } } =\frac { 1 }{ 243 }

  • Note { a }^{ \frac { 1 }{ a } }=\sqrt [ n ]{ a } and { a }^{ \frac { m }{ n } }=\sqrt [ n ]{ { a }^{ m } } so 16^{ \frac { 1 }{ 2 } }=\sqrt { 16 } and \sqrt [ 3 ]{ 81 } ={ 81 }^{ \frac { 1 }{ 4 } }=3

11. \sqrt [ 3 ]{ { 125 }^{ 2 } } =({ { 125 }^{ 2 }) }^{ \frac { 1 }{ 3 } }={ 125 }^{ \frac { 2 }{ 3 } }={ 5 }^{ 3\times \frac { 2 }{ 3 } }={ 5 }^{ 2 }=25

12. \sqrt [ 3 ]{ { 343 }^{ -1 } } =({ { 343 }^{ -1 }) }^{ \frac { 1 }{ 3 } }=({ { 7 }^{ 3 }) }^{ -\frac { 1 }{ 3 } }={ 7 }^{ -1 }=\frac { 1 }{ 7 }

13. \sqrt { 25 } +\sqrt { 36 } +\sqrt { 49 } =5+6+7=18

14. \sqrt { 0.0064 } =\sqrt { \frac { 64 }{ 10000 } } =\frac { \sqrt { 64 } }{ \sqrt { 10000 } } =\frac { 8 }{ 100 } =0.08

15. \frac { { 2 }^{ \frac { 1 }{ 3 } }\times { 2 }^{ \frac { 4 }{ 3 } } }{ { 2 }^{ \frac { 2 }{ 3 } } }

  • Using the laws of indices, we know that { a }^{ n }\times { a }^{ m }={ a }^{ n+m } and { a }^{ n }\div { a }^{ m }={ a }^{ n-m }. So:
  • \frac { { 2 }^{ \frac { 1 }{ 3 } }\times { 2 }^{ \frac { 4 }{ 3 } } }{ { 2 }^{ \frac { 2 }{ 3 } } } ={ 2 }^{ \frac { 1 }{ 3 } +\frac { 4 }{ 3 } -\frac { 2 }{ 3 } }={ 2 }^{ \frac { 3 }{ 3 } }={ 2 }^{ 1 }=2

16. { 8 }^{ \frac { 2 }{ 3 } }\times { 8 }^{ -\frac { 1 }{ 3 } }={ 8 }^{ \frac { 2 }{ 3 } +(-3) }={ 8 }^{ \frac { 2 }{ 3 } -\frac { 1 }{ 3 } }={ 8 }^{ \frac { 1 }{ 3 } }=({ { 2 }^{ 3 }) }^{ \frac { 1 }{ 3 } }={ 2 }^{ 1 }=2

Example

This example is from paper 1 of a Cambridge exam, from Nov/Oct 2002.

Evaluate:

1. { 7 }^{ \frac { 1 }{ 4 } }-{ 6 }^{ \frac { 2 }{ 3 } }

2. 3\sqrt { 2 } \times 5\sqrt { 2 }

Solution:

1. 7\frac { 1 }{ 4 } =\frac { 29 }{ 4 } and 6\frac { 2 }{ 3 } =\frac { 20 }{ 3 }

Therefore, 7\frac { 1 }{ 4 } -6\frac { 2 }{ 3 } =\frac { 29 }{ 4 } -\frac { 20 }{ 5 } =\frac { 29\times 3 }{ 4\times 3 } -\frac { 20\times 4 }{ 3\times 4 }

=\frac { 87-80 }{ 12 }

=\frac { 7 }{ 12 }

2. 3\sqrt { 2 } \times 5\sqrt { 2 } =3\times 5\times \sqrt { 2 } \times \sqrt { 2 }

15\times ({ \sqrt { 2 } ) }^{ 2 }

15\times 2

=30

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