Surds SS2 Mathematics Lesson Note

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. 

Example 1

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 

Example 2

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

 (√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 

d) = 12√3 – 5√3 + 4√3 

          = 11√3 

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 

SURDS RATIONALISATION 

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

Example: Evaluate the following 

(a) 6/√3   (b)   3 ÷ √3 + √2 

Solution 

1. a) 6/√3

=   6 x √3             

 √3 x √3

=    6√3

         3

= 2√3

(b) 3√3 + √2                                                                          

  = 3 (√3 – √2)

(√3 + √2) (√3 – √2)                                                                                    

 = 3√3 – 3√2

(√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

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?

1. 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 is 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 – Q^2 = 24

          In turn, we have:

          Q^2 – 14Q + 24 = 0

          Q^2 – 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)

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) √35

5. Simplify √`128 – 4√8                                 (a) 0    (b) 1        (c) 2    (d) 3

6. Express  3√2 – √3

                    2√3 – √2 in the form √m/√n where m and n are whole numbers.                     

7. Express.  1

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