# C1.1 Algebra and functions 1

```AS-Level Maths:
Core 1
for Edexcel
C1.1 Algebra and
functions 1
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Using and manipulating surds
Using and manipulating surds
Contents
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Types of number
We can classify numbers into the following sets:
The set of natural numbers, :
Positive whole numbers {0, 1, 2, 3, 4 …}
The set of integers, :
Positive and negative whole numbers {0, ±1, ±2, ±3 …}
The set of rational numbers, :
Numbers that can be expressed in the form mn , where n and
m are integers. All fractions and all terminating and recurring
decimals are rational numbers; for example, ¾, –0.63, 0.2.
The set of real numbers, :
All numbers including irrational numbers; that is, numbers
that cannot be expressed in the form mn , where n and m are
integers. For example,  and 2 .
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Introducing surds
2, √3
3 or
5,
When the square root of a number, for example √2,
or √5,is
irrational, it is often preferable to write it with the root sign.
Numbers written in this form are called surds.
Can you explain why √1.69
1.69 is not a surd?
1.69 is not a surd because it is not irrational.
√1.69
169
1.69 =
100
169
=
100
13
=
10
= 1.3
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This uses the fact that
a
a
=
b
b
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Manipulating surds
When working with surds it is important to remember the
following two rules:
ab = a × b
and
a
a
=
b
b
Also:
a× a =a
You should also remember that, by definition, √aa means the
positive square root of a.
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Simplifying surds
We are often required to simplify surds by writing them in the
form a b .
We can do this using the fact that
For example:
ab = a × b .
Simplify 50 by writing it in the form a b .
Start by finding the largest square number that divides into 50.
This is 25. We can use this to write:
50 = 25× 2
= 25 × 2
=5 2
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Simplifying surds
Simplify the following surds by writing
b.
them in the form aa√b.
1) 45
45 = 9×5
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3) 3 40
2) 98
98 = 49× 2
3
40 = 3 8 × 5
= 9× 5
= 49 × 2
= 3 8×3 5
=3 5
=7 2
= 23 5
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Simplifying surds
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Adding and subtracting surds
Surds can be added or subtracted if the number under the
square root sign is the same. For example:
Simplify 45 + 80.
Start by writing 45 and 80 in their simplest forms.
45 = 9×5
80 = 16×5
= 9× 5
= 16 × 5
=3 5
=4 5
45 + 80 = 3 5 + 4 5 = 7 5
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Expanding brackets containing surds
Simplify the following:
1) (4  2)(1+ 3 2)
2) ( 7  2)( 7 + 2)
= 4 +12 2  2  6
= 7+ 2 7  2 7  2
= 11 2  2
=5
Problem 2) demonstrates the fact that (a – b)(a + b) = a2 – b2.
In general:
( a  b )( a + b )  a  b
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Rationalizing the denominator
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Rationalizing the denominator
When a fraction contains a surd as the denominator we
usually rewrite it so that the denominator is a rational number.
This is called rationalizing the denominator. For example:
5
Simplify the fraction
.
2
In this example we rationalize the denominator by multiplying
the numerator and the denominator by 2.
× 2
5
5 2
=
2
2
× 2
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Rationalizing the denominator
Simplify the following fractions by
rationalizing their denominators.
2
1)
3
2)
× 3
2
2 3
=
3
3
× 3
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2
5
3
3)
4 7
× 5
2
=
5
× 5
× 7
10
5
3
3 7
=
28
4 7
× 7
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Rationalizing the denominator
When the denominator involves sums of differences between
surds we can use the fact that
(a – b)(a + b) = a2 – b2
to rationalize the denominator. For example:
Simplify
1
5 2
=
1
.
5 2
1
5 2
×
5 +2
5 +2
5 +2
=
54
= 5 +2
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Rationalizing the denominator
More difficult examples may include surds in both the
numerator and the denominator. For example:
2 3 1
.
Simplify
3 +1
2 3  1 (2 3  1)( 3  1)
=
3 +1
( 3 +1)( 3  1)
73 3
=
3 1
73 3
=
2
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Working:
2

3 1

3 1
= 6  2 3 3 + 1
=7 3 3
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The index laws
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Index notation
Simplify:
a to the power of 5
a × a × a × a × a = a5
a5 has been written using index notation.
The number a is
called the base.
an
The number n is called
the index, power or
exponent.
In general:
n of these
an = a × a × a × … × a
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Index notation
Evaluate the following:
0.62 = 0.6 × 0.6 = 0.36
34 = 3 × 3 × 3 × 3 = 81
(–5)3 = –5 × –5 × –5 = –125
When we raise a
negative number to
an odd power the
27 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128
(–1)5 = –1 × –1 × –1 × –1 × –1 = –1
(–4)4 = –4 × –4 × –4 × –4 = 256
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When we raise a
negative number to
an even power the
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The multiplication rule
When we multiply two terms with the
same base the indices are added.
For example:
a4 × a2 = (a × a × a × a) × (a × a)
=a×a×a×a×a×a
= a6 = a (4 + 2)
In general:
am × an = a(m + n)
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The division rule
When we divide two terms with the
same base the indices are subtracted.
For example:
a5
÷
a2
a×a×a×a×a
=
a×a
= a3 = a (5 – 2)
2
4p6
÷
2p4
4×p×p×p×p×p×p
=
= 2p2 = 2p(6 – 4)
2×p×p×p×p
In general:
am ÷ an = a(m – n)
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The power rule
When a term is raised to a power and the result
raised to another power, the powers are multiplied.
For example:
(y3)2 = y3 × y3
(pq2)4 = pq2 × pq2 × pq2 × pq2
= (y × y × y) × (y × y × y)
= p4 × q (2 + 2 + 2 + 2)
= y6 = y3×2
= p4 × q8
= p4q8 = p1×4q2×4
In general:
(am)n = amn
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Using index laws
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Zero and negative indices
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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The zero index
Any number or term divided
by itself is equal to 1.
Look at the following division:
y4 ÷ y4 = 1
But using the rule that xm ÷ xn = x(m – n)
y4 ÷ y4 = y(4 – 4) = y0
That means that
y0 = 1
In general:
a0 = 1
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(for all a ≠ 0)
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Negative indices
Look at the following division:
b2
÷
b4
=
b×b
b×b×b×b
1
1
=
= 2
b×b
b
But using the rule that am ÷ an = a(m – n)
b2 ÷ b4 = b(2 – 4) = b–2
That means that
b–2 =
1
b2
In general:
a–n =
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1
an
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Negative indices
Write the following using fraction notation:
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1)
u–1 =
1
u
2)
2n–4 =
2
n4
3) x2y–3 =
x2
y3
4) 5a(3 –
b)–2 =
This is the
reciprocal of u.
5a
(3  b)2
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Negative indices
Write the following using negative indices:
2
1) = 2a–1
a
x3
2) 4 = x3y–4
y
p2
3)
= p2(q + 2)–1
q+2
3m
4) 2
= 3m(n2 – 5)–3
3
(n  5)
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Fractional indices
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Fractional indices
Indices can also be fractional. For example:
1
2
What is the meaning of a ?
Using the multiplication rule:
1
2
1
2
a × a =a
1+1
2 2
= a1
=a
a= a × a
But
1
2
So
a = a
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1
2
a is the square
root of a.
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Fractional indices
Similarly:
1
3
1
3
1
3
a ×a ×a = a
1+1+1
3 3 3
= a1
=a
a = 3 a × 3 a ×3 a
But
1
3
1
3
So
a =3a
a is the cube
root of a.
In general:
1
n
a =na
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Fractional indices
2
3
What is the meaning of a ?
2
3
We can write a as a
2 31
.
Using the rule that (am)n = amn, we can write
2
3
1
3
a  (a )  3 a 2
2
3
We can also write a as a
1 2
3
2
3
2
.
1
3
a  (a )2  ( 3 a )2
In general:
m
n
n
a = a
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m
or
m
n
a =
 a
n
m
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Fractional indices
Evaluate the following:
1) 16
5
4
2) (0.125)
5
4
4
5
16 = ( 16)
 32
3) 36
 32
(0.125) = 8
2
3
36
 32
 32
=
1
( 36 )3
= 25
= ( 3 8)2
= 32
= 22
1
= 3
6
=4
=
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1
216
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Summary of the index laws
Here is a summary of the index laws for all rational exponents:
a m × a n = a ( m+n )
m
n
a ÷a =a
m n
(a ) = a
( m n )
mn
1
a =a
a = 1 (for a  0)
0
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a
n
1
= n
a
1
2
a = a
1
n
n
a = a
m
n
n
a = a
m
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Solving equations involving indices
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Solving equations involving indices
We can use the index laws to solve certain types of equation
involving indices. For example:
Solve the equation 25x = 1255 – x.
25x = 1255 – x
(52)x = (53)5 – x
52x = 53(5 – x)
2x = 3(5 – x)
2x = 15 – 3x
5x = 15
x=3
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Examination-style questions
Contents
Using and manipulating surds
Rationalizing the denominator
The index laws
Zero and negative indices
Fractional indices
Solving equations involving indices
Examination-style questions
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Examination-style question 1
6+ 3
Show that
can be written in the form a + b 2 where
6 3
a and b are integers. Hence find the values of a and b.
Multiplying top and bottom by 6 + 3 gives



3 
6+ 3
6
6+
So
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 = 6+2 6 3 +3
63
3
6+ 3
9 + 2 18
18 can be written as 3 2.
=
3
9+6 2
=
3
= 3+2 2
a = 3 and b = 2
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Examination-style question 2
a) Express 32x in the form 2ax where a is an integer to be
determined.
b) Use your answer to part a) to solve the equation
x
32 = 2
x2
32 = 25
So
32x = (25)x
Using the rule that (am)n = amn
32x = 25x
b) Using the answer from part a) this equation can be written as
5x
x2
2 =2
5x = x2
5x – x2 = 0
x (5 – x) = 0
x = 0 or x = 5
a)
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