Download Surds - Warilla High School

Revision - Surds
1
1
1
4
3
5


2
1


0
1
2
 

Production by
I Porter
2009
3 2
5
Definition : a
•
•
p
If p and q are two integers with no common factors and a 
, than a is an
q
irrational number.
Irrational number of the form
2, 3, 5, .... are called SURDS.



1
1
1
4
3
5


2
1


0
1
2
 

3 2
5
2
Prove that 2 is irrational.
[Reproduction of Proof not required for assessment]
Let p and q be two integers with no common factors, such that

Then squaring both sides,
 2
2
p 2
  
q 
2

p
is a rational number.
q

p2
q2
2q 2  p 2
rearrange
2

Now, the left side is an even number, this implies that p2 is also an even
number, which implies that p is divisible by 2. Let p = 2m, where m is
an integer. Hence, p2 = 4m2.

2q2 = 4m2
q2 = 2m2
Dividing by 2.
Now, the right side is an even number, this implies that q2 is also an even
number, which implies that q is divisible by 2.
But, this contradict our assumption that p and q have NO common factors..
Hence, we cannot write,
p
2
were p and q are integers with no common factors.
q
Therefore

2 cannotbe a rational number, it must be an IRRATIONAL number..
3
Surd Operations
If a, b, c and d are numbers and a > 0 and b > 0, then
1)
a aa
2)
a  b  ab
or

a  a
2

5  5
1)
5 5 5
2)
3  5  (3 5)  15
or
2
3) c a  d a  (c  d) a

3) 2 3  5 3  (2  5) 3  7 3
4) c a  d b  c a  d b

4) 2 3  5 7  2 3  5 7
5)
a
a

b
b
or
a
a  b  
b

5)
5
5

3
3
or
5 3
5
3
Surds can behave like numbers and/or algebra.

4
Simplifying Surds
To simplify a surd expression, we need to (if possible) write the given surd
number as a product a perfect square, n2, and another factor. It is important to
use the highest perfect square factor but not essential. If the highest perfect
square factor is not used first off, then the process needs to be repeated
(sometimes it faster to use a smaller factor).
Generate the perfect square number:
Order
1
2
3
4
5
6
7
8
9
10
n2
12
22
32
42
52
62
72
82
92
102
Value
1
4
9
16 25 36 49 64 81
100
Perfect Squares
Factorise the surd number using the largest perfect square :
18 = 9 x 3
24 = 4 x 6
27 = 9 x 3
48 = 16 x 3
18  9  3
24  4  6
27  9  3
48  16  3
Or any square factor:
72 = 36 x 2

72  36  2
72 = 9 x 8

72  9  8
72 = 4 x 18

72  4  18
72 = 4 x 9 x 2
72  4  9  2
5

Examples: Simplify the following surds.
a)
8
4
b)
2

 8  2 2


2
11
25




60
4



2
50  5 2
15
f)

60  2 15
80
16

5
4
2

N2
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
5
18  3 2
e)
2
 44  2 11
50



44
4
c)
3


9
2
d)
18




80  4 5
6
Exercise: Simplify each of the following surds.
a)
72
answer  6 2
b)

20
answer  2 5

d)
96
answer  4 6
c)

answer  3 3
f)
300
answer  10 3 

N2
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225

e)

27
180
answer  6 5

7
Simplifying Surd Expressions.
c a  d a  (c  d) a
N2
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
Examples: Simplify the following
a)
8  18
 25  3  4  5
 2  2  3 2
 5 32 5
 2  3  2

5 2

c)
75  20
 4 2 9 2



b)

75  20  48  45

 5 32 5


 25  3  4  5  16  3  9  5
 5  3  2  5  4  3  3 5

 5  4  3  2  3  5

 35 5

8


Exercise: Simplify the following.
1)
3  12  48
2)
answer  7 3
4)

6  24  54

answer  0

7)
150  200

answer  5 3 10 2


5  2  45  72
3)
answer  7 2  2 5
5)

150  96  24
N2
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
7  28  63
answer  0

6)
answer  5 3  6 6
27  192  48
answer  7 3

8)
98  80  18
answer  4 2  4 5

9)
147  8  12  50
answer  5 3  7 2


9
Simplifying Surd Expressions.
a aa
a
a

b
b
a  b  ab
Examples: Simplify the following
a)
5 6
c)
 30






b)
3 2 
5 6
 3 5  2  6

6  8
 15 12

 15  4  3
 48 
 15  2  3 
 16  3
 30 3
 4 3 

8
6
d)




3 2
5 6
e)

4 2
3 2
2  2

3 2
2


3
2
3

 
3
3
2 3 

3



3 2
5 2  3
3 2
5 2  3
3

5 3


3
5 3

3
3
3 3
53
3

5


10

Exercise: Simplify each of the following (fraction must have a rational denominator).
a)
6 8
b) 5 3  4 2
answer  4 3

3 8  18

e)
answer  36


g)
20
60
answer  36
answer  20 6

d)
c) 2 3  6 3
3
6

2
3
answer 

h)

2
3
answer 

6
3


4 2
5
answer 


f)
2
2

8
3 2
answer 
i)

4 10
5

60 
2
3
27
45
answer  6

11
Using the Distributive Law - Expanding Brackets. a(b  c)  ab  ac
Examples: Expand and simplify the following
a)

d) 2 3( 6  5 3)
3(4  2)
f)
 12  3 2
 2 18 10 9

3(5  
2)
 5 3  6 
 2  9  2 10 9

 6 2  30

c) 2 3(6  3)

 12 3  2 9
b)
e)


5  34  3

 54  3 34  3

 12 3  2  3

 12 3  6


 20  5 3  4 3  3

 17  3






5  34  3
 54  3 34  3
 20  5 3  4 3  3
 23  9 3
g)
5  2 33 5 3
 53 5 3 2 33 5 3
 15  25 3  6 3 10  3
 19 3 15
12




Exercise: Expand the following (and simplify)
a)  24  5 
b) 4 31 5 3
answer  8  2 5
answer  4 3  60
d ) 5 3  2  3 7 2 
answer  1 34 6
c) 5  2 1 2 

j) 5  2 5  2 

e) 5  2 
2
answer  27 10 2
 of conjugates)
Special cases - (Product

f ) 2 3  5 2 
2

answer  62  20 6
a  ba  b a   b

2
k)  7  3 7  3

answer  4
answer  23
answer  3 4 2
2
Difference of two squares.
l) 1 2 31 2 3
answer  11
m) 5 2  35 2  3 n)  7  2 3 7 2 3 o) 3 5  2 33 5  2 3


answer  5
answer  47
answer  33



13


a  ba  b a   b
2
Rationalising the Denominator - Using Conjugate
2
Examples: In each of the following, express with a rational denominator.

a)
5
conjugate is of denominator
is  3 2 but we can use 2
3 2
5
2
  
3 2
2
5 2

3 2
5 2

6


b)
4
conjugate is
is 3 2.
3 2
4
3 2
 

3 2 3 2


of denominator
43 2 
3   2 
2
2
12  4 2
92
12  4 2

7





[more examples next slide]
14




More examples
c)
5 2
the conjugate is
3 2
5 2 3 2
  
3 2 3 2
5 2 3 2 

2
2
3   2 

15 2 10
92
15 2 10

7
3 2
d)

3
5
3

5
3





2 5
the conjugate is
2 5
2  5 5 2  5

2 5 5 2 5
2  5 5 2  5 
5 2 5
5 2   5 
2
2

30  3 10  5 10  5
50  5

25  2 10
45

15
Exercise: Express the following fraction with a rational denominator.
3
2 3
4 3
a)
b)
c)
d)
4 2
5 3
5 3
3 52 3
2 54 3
12  3 2
5 3 3
23 9 3
3 4 15
answer 
answer 
answer 
14
11
22
14



Harder Type questions.
Express the following as a single fraction with a rational denominator.



A
possible
method
to
solve
this problems is to RATIONALISE the denominators
Example
of each fraction first, then combine the results.
3
2
3
4 2
2
3 2
a)



4  2 3 2
4 2 4 2
3 2 3 2
34  2 
23  2 
12  3 2 6  2 2




4    2 
3   2 
14
7


12  3 2
62 2
12  3 2 12  4 2




16  2
92


14
14
12  3 2
62 2
24  2



14
7


14
answer 
2

2
2
2

16


Exercise: Express the following as a single fraction with a rational denominator.
a)
5
1

3 2 4  2
answer 
26 11 2
14

3
1
d)

5  2 1 2
answer 
b)
3
2
5 3 4 3

answer 

e)
c)
215 109 3
286
4 3
3 

5 3
52 3
9 12 2

23
answer  2 2 15


2  3 1 3

3 3 5  3
answer 
f)
27  23 3
66
2 5
3 7

10  15
21  14
answer  5 3  5 2
17