Revision - Surds

Production by

I Porter

2009

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Definition :

• If p and q are two integers with no common factors and , than a is an irrational number.

• Irrational number of the form are called SURDS.

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a ≠pq

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2, 3, 5, ....

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a

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21

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5

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Prove that is irrational.

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2 [Reproduction of Proof not required for assessment]

Let p and q be two integers with no common factors, such that is a rational number.

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2 =p

q

Then squaring both sides,

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2( )2

=p

q

⎛

⎝ ⎜

⎞

⎠ ⎟

2

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2 =p2

q2

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2q2 = p2rearrange 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 Dividing by 2.

q2 = 2m2 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,

were p and q are integers with no common factors.

Therefore cannot be a rational number, it must be an IRRATIONAL number..

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2

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2 =p

q

4

Surd Operations

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1) a × a = a or a( )2

= a

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2) a × b = ab

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3) c a ± d a = (c ± d) a

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4) 2 3 ± 5 7 = 2 3 ± 5 7

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5) ab

=ab

or a ÷ b =ab

If a, b, c and d are numbers and a > 0 and b > 0, then

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1) 5 × 5 = 5 or 5( )2

= 5

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2) 3 × 5 = (3× 5) = 15

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3) 2 3 + 5 3 = (2 + 5) 3 = 7 3

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4) c a ± d b = c a ± d b

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5) 53

=53

or 5 ÷ 3 =53

Surds can behave like numbers and/or algebra.

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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

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

Or any square factor:

72 = 36 x 2 72 = 9 x 8 72 = 4 x 18 72 = 4 x 9 x 2€

18 = 9 × 3

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24 = 4 × 6

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27 = 9 × 3

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48 = 16 × 3

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72 = 36 × 2

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72 = 9 × 8

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72 = 4 × 18

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72 = 4 × 9 × 2

Perfect Squares

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Examples: Simplify the following surds.

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a) 8

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b) 18

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c) 50

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4

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2

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9

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2

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25

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2

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2

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8 = 2 2

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18 = 3 2

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50 = 5 2

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3

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5

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d) 44

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e) 60

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f ) 80

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4

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11

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4

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15

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16

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5

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2

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44 = 2 11

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60 = 2 15

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80 = 4 5

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2

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4

N2

149

162536496481

100121144169196225

7

Exercise: Simplify each of the following surds.

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a) 72

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b) 20

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c) 27

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d) 96

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e) 300

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f ) 180

N2

149

162536496481

100121144169196225

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answer = 6 2

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answer = 2 5

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answer = 3 3

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answer = 4 6

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answer =10 3

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answer = 6 5

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Simplifying Surd Expressions.

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c a ± d a = (c ± d) a

Examples: Simplify the following

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a) 8 + 18

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= 4 × 2 + 9 × 2

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=2 × 2 + 3× 2

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= 2 + 3( ) × 2

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=5 2

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b) 75 − 20

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= 25 × 3 − 4 × 5

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=5 × 3 − 2 × 5

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=5 3 − 2 5

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c) 75 + 20 − 48 + 45

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= 25 × 3 + 4 × 5 − 16 × 3 + 9 × 5

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=5 × 3 + 2 × 5 − 4 × 3 + 3× 5

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= 5 − 4( ) × 3 + 2 + 3( ) × 5

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= 3 + 5 5

N2

149

162536496481

100121144169196225

9

Exercise: Simplify the following.

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1) 3 + 12 + 48

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2) 5 + 2 − 45 + 72

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3) 7 + 28 − 63

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4) 6 + 24 + 54

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5) 150 − 96 − 24

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6) 27 + 192 − 48

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7) 150 − 200

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8) 98 − 80 − 18

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9) 147 + 8 − 12 + 50

N2

149

162536496481

100121144169196225

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answer = 7 3

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answer = 7 2 − 2 5

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answer = 0

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answer = 0

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answer = 5 3 − 6 6

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answer = 7 3

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answer = 5 3 −10 2

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answer = 4 2 − 4 5

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answer = 5 3 + 7 2

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Simplifying Surd Expressions.

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a × a = a

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a × b = ab

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ab

=ab

Examples: Simplify the following

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a) 5 × 6

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b) 6 × 8

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c) 3 2 × 5 6

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= 30

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= 48

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=3× 5 × 2 × 6

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=15 12

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=15 × 4 × 3

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=15 × 2 × 3

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=30 3

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= 16 × 3

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=4 3

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d) 86

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= 4 × 23 × 2

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= 2 × 23 × 2

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= 23

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= 23

×33

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=2 33

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e) 3 25 6

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= 3× 25 × 2 × 3

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= 3× 25 × 2 × 3

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= 35 3

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= 35 3

×33

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=3 35 × 3

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= 35

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Exercise: Simplify each of the following (fraction must have a rational denominator).

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a) 6 × 8

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b) 5 3 × 4 2

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c) 2 3 × 6 3

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d) 3 8 × 18

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e) 3

2×

63

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f ) 8

3 2

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g) 2060

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h) 4 2

5

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i) 60 ×2745

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answer = 4 3

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answer = 20 6

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answer = 36

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answer = 36

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answer =2

2

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answer =23

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answer =26

=3

3

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answer =4 10

5

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answer = 6

12

Using the Distributive Law - Expanding Brackets.

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a(b ± c) = ab ± ac

Examples: Expand and simplify the following

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a) 3(4 + 2)

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=12 + 3 2

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b) 3(5 − 2)

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=5 3 − 6

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c) 2 3(6 + 3)

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=12 3 + 2 9

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=12 3 + 2 × 3

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=12 3 + 6

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d) 2 3( 6 − 5 3)

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=2 18 −10 9

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=2 × 9 × 2 −10 9

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=6 2 − 30

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e) 5 + 3( ) 4 − 3( )

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=5 4 − 3( ) + 3 4 − 3( )

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=20 − 5 3 + 4 3 − 3

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=17 − 3

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f ) 5 − 3( ) 4 − 3( )

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=5 4 − 3( ) − 3 4 − 3( )

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=20 − 5 3 − 4 3 + 3

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=23 − 9 3

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g) 5 − 2 3( ) 3+ 5 3( )

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=5 3+ 5 3( ) − 2 3 3+ 5 3( )

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=15 + 25 3 − 6 3 −10 × 3

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=19 3 −15

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Exercise: Expand the following (and simplify)

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a) − 2 4 + 5( )

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b) 4 3 1− 5 3( )

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c) 5 − 2( ) 1+ 2( )

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d) 5 3 + 2( ) 3 − 7 2( )

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e) 5 − 2( )2

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f ) 2 3 + 5 2( )2

Special cases - (Product of conjugates)

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a − b( ) a + b( ) = a( )2

− b( )2

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j) 5 − 2( ) 5 + 2( )

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k) 7 + 3( ) 7 − 3( )

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l) 1+ 2 3( ) 1− 2 3( )

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m) 5 2 − 3( ) 5 2 + 3( )

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n) 7 + 2 3( ) 7 − 2 3( )

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o) 3 5 − 2 3( ) 3 5 + 2 3( )

Difference of two squares.

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answer = −8 − 2 5

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answer = 4 3 − 60

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answer = 3+ 4 2

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answer =1− 34 6

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answer = 27 −10 2

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answer = 62 + 20 6

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answer = 23

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answer = 4

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answer = −11

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answer = 47

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answer = −5

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answer = 33

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Rationalising the Denominator - Using Conjugate

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a − b( ) a + b( ) = a( )2

− b( )2

Examples: In each of the following, express with a rational denominator.

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a) 5

3 2

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conjugate is of denominator is − 3 2 but we can use 2

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= 53 2

×22

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=5 23× 2

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=5 26

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b) 4

3− 2

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conjugate is of denominator is 3 + 2.

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= 43− 2

×3+ 23+ 2

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=4 3+ 2( )

3( )2

− 2( )2

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=12 + 4 29 − 2

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=12 + 4 27

[more examples next slide]

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More examples

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c) 5 2

3− 2

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= 5 23− 2

×3+ 23+ 2

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=5 2 3+ 2( )

3( )2

− 2( )2

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=15 2 +109 − 2

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=15 2 +107

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d) 3 2 + 55 2 + 5

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=3 2 + 55 2 + 5

×5 2 − 55 2 − 5

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=3 2 + 5( ) 5 2 − 5( )

5 2( )2

− 5( )2

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=30 − 3 10 + 5 10 − 550 − 5

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=25 + 2 1045

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the conjugate is 3+ 2

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the conjugate is 5 2 − 5

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Exercise: Express the following fraction with a rational denominator.

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a) 3

4 + 2

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b) 2 3

5 + 3

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c) 4 + 35 − 3

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d) 3 5 − 2 32 5 − 4 3

Harder Type questions.Express the following as a single fraction with a rational denominator.

Example

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a) 3

4 + 2+

23− 2

A possible method to solve this problems is to RATIONALISE the denominatorsof each fraction first, then combine the results.

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3

4 + 2×

4 − 2

4 − 2

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=3 4 − 2( )

4( )2

− 2( )2

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=12 − 3 2

16 − 2

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=12 − 3 2

14

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2

3− 2×

3 + 2

3 + 2

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=2 3+ 2( )

3( )2

− 2( )2

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=6 + 2 2

9 − 2

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=6 + 2 2

7€

=12 − 3 214

+6 + 2 2

7

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=12 − 3 214

+12 + 4 2

14

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=24 + 214

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answer =12 − 3 2

14

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answer =5 3 − 3

11

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answer =23+ 9 3

22

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answer =−3− 4 15

14

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Exercise: Express the following as a single fraction with a rational denominator.

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a) 5

3+ 2−

14 − 2

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b) 3

5 − 3+

24 − 3

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c) 2 + 33+ 3

−1− 35 − 3

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d) 3

5 − 2×

11+ 2

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e) 4 35 − 3

÷3

5 − 2 3

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f ) 2 5

10 − 15−

3 721− 14

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answer =26 −11 2

14

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answer =215 +109 3

286

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answer =27 + 23 3

66

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answer =−9 +12 2

23

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answer = −2 − 2 15

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answer = 5 3 − 5 2