Classwork Series and Exercises {Mathematics- SS2}: Surds

Introduction

Where does the word SURD come from?

Around 820AD the famous Persian Mathematician al-Khwarizmi saw irrational numbers (namely numbers that cannot be written as fractions) as “inaudible numbers”.
The LATIN for “inaudible” is “SURDUS” and so from this comes our use of the word SURD.
We now use SURD for a number which we cannot find the square root of.

The positive integerth root of a positive rational number is defined a surd.

In arithmetic, surds is a mathematical concept, returns irrational numbers. It is often come in mathematics including problems. Although, dealing surds is slightly complex but understanding various types of surds and properties simplifies the dealing the surds.

z⁺ÖQ⁺

Note: The square root is represented by Ö

Algebraic Expression

It can be understood by an algebraic analysis.

Integers and Rational numbers are denoted by Z and Q respectively in mathematics. Both integers and rational numbers have positive and negative numbers but surds are only formed by positive integers and positive rational numbers. Therefore, assume positive integer numbers and positive rational numbers are denoted by Z+ and Q+ respectively.

According to our definition of surd, a surd can be expressed in an algebraic form given here.

z⁺ÖQ⁺

It is mathematically known as follows.

(Q⁺)^1/z+

A surd is also called as radical.

• The symbol  z⁺√ is called radical sign.
• The positive integers (z⁺) is also called the order of the surd.
• The positive rational number (Q⁺⁾ is also called radicand.

Examples:

√6, ⁵√5, ⁴√3, ¹¹√4, ⁵√2, …………….. ⁿ√a

Alternatively, these surds are originally known as follows respectively.

(6)^1/2, (5)^1/3, (3)^1/4, (11)^1/4, (2^1/5, …………….. (a)^1/n

Rules for Basic Operations with Surds

The rules for the four basic operation s with surds for any positive numbers a and b, are:

(i) Öa – Öb ≠ Öa – b        eg  Ö9 – Ö4 = 3 – 2 = 1 ≠ Ö9 – 4 = Ö5

(ii) Öa + Öb ≠ Öa + b      eg  Ö9 + Ö4 = 3 + 2 = 5 ≠ Ö9 + 4 = Ö13

(iii) Öa x Öb = Öa x b      eg  Ö25 x Ö16 = 5 x 4 = 20 = Ö25 x 16 = Ö400 = 20

(iv) Öa ¸ Öb = Öa/b        eg  Ö25 ¸ Ö16 = 5/4 = Ö25/16 =5/4

Check the above four rules by doing the following exercises. Compare the following pairs of expression – work them out looking up the tables where necessary.

1  a.  Ö16 x Ö9; Ö16 x 9

1. Ö25 x Ö9; Ö25 x 9
2. Ö3 x Ö2; Ö3 x 2

2  a. Ö16/Ö4; Ö16/4

1. Ö75/Ö3 ; Ö75/3

3  a. Ö9 + Ö4; Ö9 + 4

1. Ö16 + Ö4; Ö16 + 4

4  a. Ö7 – Ö5;  Ö7 – 5

1. Ö25 – Ö9; Ö25 – 9

Hints

For multiplication of surds, the following procedure is useful:

(i) Simplify expression by using simpler surds, if possible

(ii) Group rational numbers and surds separately,

(ii) If the denominator has any surds, rationalise the denominator by multiplying both numerator and denominator by a conjugate surd, and

(iv) Simply against the whole expression, if necessary.

Rationalisation of the Denominator

We often express a radical fraction, such as Ö2/Ö3, in a form that has a rational denominator.

Knowing that Ö5 x Ö5 = 5, we multiply the radical fraction Ö2/Ö5 by Ö5/Ö5. This leaves the value of the fraction unchanged, but changes the denominator.

Ö2/Ö5 = Ö2/Ö5 x Ö5/Ö5 {\Ö5/Ö5 = 1}

Example: Rationalise the denominator of 3/2Ö7

3/2Ö7 x Ö7/Ö7

3Ö7/2Ö7 = 3Ö7/2 x 7

                 = 3Ö7/14

Express 3/Ö7 – 5/Ö2 as a single fraction with a rational denominator

Solution:

3/Ö7 x Ö7/Ö7 – 5/Ö2 x Ö2/Ö2

3Ö7/7 – 5Ö2/2

6Ö7 – 35Ö2/14

This cannot be simplified further since they have unlike surds.

Evaluation of Expressions involving Surds

Example 1

Evaluate 5Ö3/2Ö50

Solution

5Ö3/2Ö50 = 5Ö3/2Ö25 x 2 = 5Ö3/2 x 5Ö2 = Ö3/2Ö2 Rationalising the denominator, we have

                   = Ö3/2Ö2 x Ö2/Ö2

                   = Ö6/2 x 2 = 1/4 Ö6

Expressions involving surds can be worked out like other mathematical expressions. The order of operations like the order involving “brackets”, off, “addition”, “subtraction”, “multiplication” and “division” treated earlier.

Example

Simplify the following expressions:

(i) (5Ö3 – 2Ö7)(Ö27 + 3Ö7)  (ii) 1 + Ö2/Ö5 + Ö3 + 1 – Ö2/Ö5 – Ö3

Solution

(i) (5Ö3 – 2Ö7)(Ö27 + 3Ö7) = 5Ö3Ö27 + (5Ö3)(3Ö7) – 2(Ö7)(Ö27) – 2 x 3(Ö7)²

                                              = 5Ö81 + 15Ö21 – (2Ö7)(3Ö3) – 42

                                              = 5 x 9 + 15Ö21 – 6Ö21 – 42

                                              = 45 – 42 + 9Ö21

                                              = 3 + 9Ö21

(ii) 1 + Ö2/Ö5 + Ö3 + 1 – Ö2/Ö5 – Ö3 = (1 + Ö2)(Ö5 – Ö3) + (1 – Ö2)(Ö5 + Ö3)/(Ö5 + Ö3)(Ö5 – Ö3)

= (Ö5 – Ö3 + Ö2Ö5 – Ö2Ö5 – Ö2Ö3) + (Ö5 + Ö3 – Ö2Ö5 – Ö2Ö3)/ 5 – 3

= Ö5 – Ö3 + Ö10 – Ö6 + Ö5 + Ö3 – Ö10 – Ö6/2

= 2Ö5 – 2Ö6/2 = 2(Ö5 – Ö6)/2 = Ö5 – Ö6

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

1. Simplify this expression Ö18 A. 6Ö3 B. 3Ö2 C. 3Ö6  D. 2Ö3
2. Simplify this expression Ö225 A. 14 B. 13 C. 15  D. 17
3. 2Ö2 – 3/3 + Ö8 A. 12Ö2 – 17 B. 8Ö2 – 17 C. 12Ö6 – 15  D. 5Ö14 – 17
4. Simplify this expression Ö12 – Ö27 A. Ö3 B. 3 C. -Ö3  D. -Ö5
5. Rationalise the denominator of Ö8 – Ö10/Ö6 A. 2Ö3 – Ö15/3 B. 3Ö3 – Ö13/3 C. Ö3 – Ö17/2  D. 3Ö3 – Ö15/3

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