Week 9 – SS1 First Term Further Mathematics Notes

 
 WEEK NINE
TOPIC: SURDS
CONTENT

• Rules of surds
• Basic Form of Surds
• Similar Surds
• Conjugate Surds
• Simplification of Surds
• Additional & Subtraction of Surds
• Multiplication and Division of Surds
• Rationalization of Surds
• Equality of Surds

Rules of Surds
Surds are irrational numbers. They are the root of rational numbers whose value cannot be expressed as exact fractions. Examples of surds are: √2, √7, √12, √18, etc.

1. √(a X b ) = √a X √ b
2. √(a / b ) = √a / √b
3. √(a + b ) ≠ √a + √b
4. √(a – b ) ≠ √a – √b

Basic Forms of Surds
√a is said to be in its basic form if A does not have a factor that is a perfect square. E.g. √6, √5, √3, √2 etc. √18 is not in its basic form because it can be broken into √ (9×2) = 3√2. Hence 3√2 is now in its basic form.

 Similar Surds
Surds are similar if their irrational part contains the same numerals e.g.

1. 3√n and 5√n
2. 6√2 and 7√2

 Conjugate Surds
Conjugate surds are two surds whose product result is a rational number.
(i)The conjugate of √3 – √5 is √3 + √5
The conjugate of -2√7 + √3 is 2√7 – √3
In general, the conjugate of √x + √y is √x – √y
The conjugate of √x – √y = √x + √y

 Simplification of Surds
Surds can be simplified either in the basic form or as a single surd.

 Examples
Simplify the following in its basic form (a) √45 (b) √98

 Solution
(a) √45 = √ (9 x 5) = √9 x √5 = 3√5
(b) √98 = √ (49 x 2) = √49 x √2 = 7√2

 Examples
Simplify the following as a single surd (a) 2√5 (b) 17√2

 Solution
(a) 2√5 = √4 x √5 = √ (4 x 5) = √20
(b) 17√2 = √289 x √2 = √ (289 x 2) = √578
Addition and Subtraction of Surds
Surds in their basic forms which are similar can be added or subtracted.

 Examples
Evaluate the following
(a)√32 + 3√8 (b) 7√3 – √75 (c) 3√48 – √75 + 2√12
Solution

1. (√32 + 3√8

= √ (16 x 2) + 3√ (4 x 2)
=4√2 + 6√2
= 10√2

  (b) 7√3 – √75
= 7√3 – √ (25 x 3)
=7√3 – 5√3 =2√2
(c) 3√48 – √75 + 2√12
= 3√ (16 x 3) – √ (25 x 3) + 2√ (4 x 3)
= 12√3 – 5√3 + 4√3
= 11√3

 Evaluation
1. Simplify the following (a) 5√ 12 – 3√ 18 + 4√72 + 2√75 (b) 3√2 – √32 + √50 + √98
2. Simplify the following as a single surd (i) 8√3 (ii) 13√2

 Multiplication and Division of Surds
Example: Evaluate the following (a) √45 x √28 (b) √24 /√50

 Solution
(a) √45 x √28
= √ (9 x 5) x √ (4 x 7)
= 3√5 x 2√7
= 3 x 2 x √ (5 x 7)
= 6√35
(b)√24 / √50
= √ (24 / 50)
= √ (12 / 25)
= √12 / √25
= √ (4 x 3) / 5
= 2√3 / 5

 Evaluation:
Simplify 1. √6 x (3 – √5) 2. (2√3 – √7)(2√3 + √7)
2. Multiply the following by their conjugate (a) √3 – 2√5 (b) 3√2 + 2√3

 Surds Rationalisation
Rationalisation of surds means multiplying the numerator and denominator by the denominator or by the conjugate of the denominator.

1. Example: Evaluate the following (a) 6/√3 (b)      3

√3 + √2

 
 Solution

1. 6/√3                     (b) 3

= 6 x √3 √3 + √2
√3 x √3 = 3 (√3 – √2)
= 6√3 (√3 + √2) (√3 – √2)
3 = 3√3 – 3√2
= 2√3 (√3)² – (√2)²
= 3√3 – 3√2
3 – 1
         = 3√3 – 3√2
             1
= 3(√3 -√2)
Equality of Surds
Given two surds i.e P + √m and q + √n if P +√m = q + √n then
√P – q = √n – m the L.H.S
Of the equation is a rational number while the L.H.S and R.H.S can only be equal if they are both equal to zero (0)
     P – q = 0
    :. P = q and n – m = 0 i.e.
√n = √m

 Examples:
Find the square root of the following?
a)    7 + 2√10    b) 14 – 4√6

 Solution
(a) Let the square root of 7 + 2 √10 be √m + √n
(√m + √n)² = 7 + 2√10
m +√2mn+ n = 7 + 2√10
m + n = 7 (1)
2√mn = 2√10
mn = 10
Squaring both surds we have
mn = 10 _______(ii)
m + n = 7 ______ (i)
m n = 10 _______ (ii)
From equation (1) m = 7 – n
Put m in (ii) we have
    (7 – n) n = 10
7n – n² = 10
In sum; n² – 7n + 10 = 0
    n² – 2n – 5n + 10 =0
    n (n – 2) – 5 (n – 2) = 0
    (n -5) (n – 2) = 0
n = 5 or 2
m = 7 – 2, where n = 2
    m = 5,
m = 7 – 5 , when n = 5
    m = 2
    m= 5 or 2
The square root of 7 + √10 are 5 & + 2.

 (b) Let the square root of 14 – 4√6 be √P – √Q
The (√P – √Q) ² =14 – 4√6    
P – 2√PQ + Q = 14 – 4 √6
P + Q = 14 …………………… (1)

  -2√PQ = – 4√6
-2 – 2
√PQ = 2√6 (squaring both sides)

  PQ = (2√6)²
PQ = 4 x 6 ……………………………….. (11)

  P + Q = 14 ………………………………… (1)

  PQ = 24 ……………………………………… (11)
From equation……………… (1) P = 14 – Q
Sub for p in equation ………………… (11)
(14 – Q) Q = 24
14Q – Q² = 24
In turn we have:
Q² – 14Q + 24 = 0
Q² – 12Q – 2Q + 24 = 0
Q (Q -12) – 2 (Q – 12) = 0
Q = 2 or 12
If P = 14 – Q ,when Q= 12
P = 14 – 12
P = 2
If P = 14 – Q when Q = 2
P = 14 – 2
= 12
(√12 – √2 )    = (2 √ 3 – √2)        and
(√2 – √12)     = (√2 –     2 √3)

 Evaluation:
1. Express 3√2 – √3 in the form √m where m and n are whole number.
2√3 – √2 √n

 2. Express 1

in the form p√5 + q√3, where p and q are rational numbers.
√5 +√3

 General Evaluation
1. Simplify 3x² x 4x³
6x⁷
2. Evaluate 23.97 x 0.7124
3.877 x 52.18
3. Solve 9^{(1 – x)} = (1/27) ^x+1
4. Log₈ (r² – 8r + 18) = 1/3
5. Simplify: 2√12 + 3√48 + √75

 Reading Assignment: Further Mathematics Project Book 1(New third edition).Chapter 3 pg.19-27

 Weekend Assignment
1. Expand (3√2 – 1) (3√2 + 1) (a) 16 (b) 20 (c) 17 (d) 24
2. Simplify √200 in its basic form (a) 10√2 (b) 5√4 (c) 2√10 (d) 2√50
3. Simplify 9/√3 (a) 3√2 (b) 3√3 (c) 1/3 (d) 2√2
4. Express 3√5 as a single surd (a) √40 (b) √55 (c) √45 (d) √35implify
5. Simplify √`128 – 4√8 (a) 0 (b) 1 (c) 2 (d) 3
Theory
1.Express 3√2 – √3 in the form √m where m and n are whole number.
2√3 – √2 √n
2.Express 1 in the form p√5 + q√3, where p and q are rational numbers.
√5 +√3