Surds

Subject: Optional Maths

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Overview

If the root of a rational number is an irrational number then the resulting number is called surds. This note gives the information about surds, an operation of surds and different forms of rationalization and square and root.
Surds

Surds

If the root of a rational number is an irrational number then the resulting number is called surds.
eg:\(\sqrt{4}\)
∴ Here, 2 is a rational number .When square root of 4 is taken out it becomes irrational.
Hence, \(\sqrt{4}\) is a surd.

Operation of surds.

Here we can learn the basic operations addition, subtraction, multiplication, division, finding square root and finding cube roots of irrational numbers .

Addition

We can add irrational numbers if they have the same order and the same radicand .
eg :\(\sqrt{2}\) + 3\(\sqrt{2}\) = 4\(\sqrt{2}\) [∴ a + 3a = 4a ]
Here,
\(\sqrt{2}\) is added with 3\(\sqrt{2}\) as the radicand is \(\sqrt{2}\) as well as the order is 3.
Again,
\(\sqrt{3}\) + \(\sqrt{2}\) = \(\sqrt{3}\) + \(\sqrt{2}\) [∴ a + b = a + b ]
Then, we can not add with add \(\sqrt{3}\) with \(\sqrt{2}\) as the radicand is different even though the order is same .
Simillary,
\(\sqrt[3]{5}\) + \(\sqrt{5}\) = \(\sqrt[3]{5}\) + \(\sqrt{5}\)
Here, addition is not positive as the order of \(\sqrt[3]{5}\) is 3 and \(\sqrt{5}\) is 2 even though the radicand are same .

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Subtraction

Here, we subtract irrational numbers if they have the same order and the same radicand same as in addition .

eg :

  • 4\(\sqrt {6}\) - \(\sqrt{6}\) = 4\(\sqrt {6}\) - \(\sqrt{6}\) [∴ a - b = a - b ]
  • 4\(\sqrt{5}\) - \(\sqrt{5}\) = 3\(\sqrt{5}\)
  • \(\sqrt[3]{2}\) - \(\sqrt{2}\) = \(\sqrt [3]{2}\) - \(\sqrt{2}\)

Multiplication

Here, multiplication is is possible with irrational numbers if the order of irrational number is same \(\sqrt{a}\) \(\times\) \(\sqrt{b}\) = \(\sqrt{ab}\)
eg :

  • \(\sqrt{2}\) \(\times\) \(\sqrt{3}\)
    = \(\sqrt{6}\)
  • The order of the both surds is 2, multiplication is possible .As the order is same , the radicand aremultiplied togeather to give the result
    \(\sqrt[3]{2}\) x \(\sqrt[3]{7}\)
    = \(\sqrt[3]{14}\)
    \(\sqrt{2}\) x \(\sqrt[3]{4}\)
    This can not be multiplied as the order are different .
  • \(\sqrt{2}\) x \(\sqrt{2}\)
    = \(\sqrt{4}\)

Division

We can divide the rationals if they have some order again

If p and q are real numbers .
\(\frac{\sqrt{p}}{\sqrt{q}}\) = \(\sqrt{\frac{p}{q}}\)

eg;

\(\sqrt{8}\)÷ \(\sqrt{2}\) = \(\sqrt{\frac{8}{2}}\) = \(\sqrt{4}\) = 2

Square and cube root of irrational number

Here the square and cube root of all irritational numbers are not possible .even we can find the square root and cube root of some numbers that may later turn to be rational or irrational.

eg;

  • \(\sqrt{4}\)
    Solution:
    = \(\sqrt{4}\)
    = \(\sqrt{2× 2}\)
    = \(\sqrt{(2)^2}\)
    = 22× \(\frac{1}{2}\)
    = 2 (Rational)
  • \(\sqrt[3]{27}\)
    = \(\sqrt[3]{3× 3× 3}\)
    = \(\sqrt[3]{(3)^3}\)
    = 33× \(\frac{1}{3}\)
    = 3 (Rational)

Rationalizing Factor . ( RF )

Let's see the example of following ;

Two irrationals are said to be rationalizing factors of each other if the product of these two irrationals is a rational number.

For example;

  • 5\(\sqrt{3}\) and 2\(\sqrt{3}\)
    5\(\sqrt{3}\)× 2\(\sqrt{3}\)
    = 10\(\sqrt{3×3}\)
    = 10× \(\sqrt{3}^2\)
  • 5\(\sqrt{3}\) and \(\sqrt{3}\)
    \(\sqrt{3}\)× \(\sqrt{3}\)
    = \(\sqrt{3×3}\)
    = 5 \(\sqrt{3^2}\)
    = 5× 3
    = 15

Rationalising factor for the surd of the form a - \(\sqrt{b}\) .

As we know , for the surd of form a-\(\sqrt{b}\) we need to multiply it with any another surd a + \(\sqrt{b}\) so that it turns it intoa perfect square and the irrational turns into rational a - \(\sqrt{b}\) and a + \(\sqrt{b}\) or \(\sqrt{a}\) + b and \(\sqrt{a}\) - b are the rationalizing factors for each other. these terms are called conjugates.

eg:

(a - \(\sqrt{b}\)× (a + \(\sqrt{b}\)
= (a - \(\sqrt{b}\) (a + \(\sqrt{b}\)
= (a)2 - (\(\sqrt{b}\)2
= a2 - b

Rationalization ;

The process of multiplying a given surd by its rationalizing factor to get a rational number as a product is known as a rationalization of a given surd.

eg :

\(\frac{1}{\sqrt{2}}\)
Solution:
= \(\frac{1}{\sqrt{2}}\)× \(\frac{\sqrt{2}}{\sqrt{2}}\) (multiplying both numerator and denominator by \(\sqrt{2}\)
= \(\frac{\sqrt{2}}{\sqrt{2×2}}\)
= \(\frac{\sqrt{2}}{\sqrt{4}}\)
= \(\frac{\sqrt{2}}{\sqrt{2^2}}\)
= \(\frac{\sqrt{2}}{2}\)
Hence, the denominatior is rational number.

Things to remember
  • If the root of a rational number is an irrational number then the resulting number is called surds.
  • Two irrationals are said to be rationalizing factors of each other if the product of these two irrationals is a rational number.
  • We can add irrational numbers if they have the same order and the same radicand.

 

  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Videos for Surds
Rationalisation Of A Surd / Radical : A Basic Introduction With Examples - Maths Arithmetic
Surds - the basics
Surds and Operations
What is a Rationalising Factor(Conjugate)?
Questions and Answers

Solution

\(\sqrt{3}\)

= \(\sqrt{3×3}\)

= 3 (rational)

Solution

\(\sqrt{9}\)

= \(\sqrt{3×3}\)

= \(\sqrt{(3)}\)2

=3

Solution

= \(\sqrt{25}\)

= \(\sqrt{5×5}\)

=  \(\sqrt{(5)}\)2

= 5

Solution

\(\frac{6\sqrt{3}}{2\sqrt{3}}\)

= \(\frac{6}{2}\)×\(\frac{\sqrt{3}}{\sqrt{3}}\)

= 3

Solution

( 3 + \(\sqrt{3}\) )  ( 2 + \(\sqrt{2}\))

= ( 3 + \(\sqrt{3}\)) ( 2 + \(\sqrt{2}\))

= 3 ( 2 + \(\sqrt{2}\)) + \(\sqrt{3}\) ( 2 + \(\sqrt{2}\))

=  6 + 3 \(\sqrt{2}\) + 2 \(\sqrt{3}\) + \(\sqrt{ 3 × 2 }\)

= 6 + 3\(\sqrt{2}\) + 2\(\sqrt{3}\) + \(\sqrt{6}\)

Solution

= ( 8 - 3 + 4 ) \(\sqrt{3}\)

= ( 12  - 3 ) \(\sqrt{3}\)

=  9 \(\sqrt{3}\)

Solution

= 16 \(\sqrt{3×3}\)

 =16 × \(\sqrt{(3)}\)2

 = 16 \(\times\) 3

  = 48

 

 Solution

\(\frac{1}{\sqrt{4}+1}\)

Here, the  denominator = \(\sqrt{4}\)+1

 Rationalizing factor = \(\sqrt{4}\)- 1

Multiplying the numerator and denominator by \(\sqrt{4}\) -1 , we get

 \(\frac{1}{\sqrt{4}+1}\) \(\times\)\(\frac{\sqrt{4}-1}{\sqrt{4}-1}\)

= \(\frac{\sqrt{4} - 1}{\sqrt{(4)}-(1)^2}\)

= \(\frac{\sqrt{4}-1}{3}\)

Hence, the denominator is rationalized. 

Solution

\(\frac{1}{\sqrt{4}+1}\)

Here, the  denominator = \(\sqrt{7}\)+1

 Rationalizing factor = \(\sqrt{7}\)- 1

Multiplying the numerator and denominator by \(\sqrt{7}\) -1 , we get

 \(\frac{1}{\sqrt{7}+1}\) \(\times\) \(\frac{\sqrt{7}- 1}{\sqrt{7}-1}\)

= \(\frac{\sqrt{7} - 1}{\sqrt{(7)}-(1)^2}\)

= \(\frac{\sqrt{7}-1}{6}\)

Hence, the denominator is rationalized. 

Solution

\(\frac{2}{\sqrt{5}}\)

= Here,  the denominator = \(\sqrt{5}\)

= Rationalizing factor (R.F) = \(\sqrt{5}\)

Multiplying the numerator and denominator by \(\sqrt{5}\), we get

= \(\frac{2}{\sqrt{5}}\)\(\times\)\(\frac{\sqrt{5}}{\sqrt{5}}\)

= \(\frac{2\sqrt{5}}{5}\)

Hence , the denominator is rationalized. 

Solution

\(\sqrt{50}\)

= \(\sqrt{5×5×2}\)

= \(\sqrt{(5)^2×2}\)

= 5\(\sqrt{2}\)

If we multiply 5\(\sqrt{2}\) by \(\sqrt{2}\) we will get the rational number. So, \(\sqrt{2}\) is the simplest rationalizing factor of  \(\sqrt{50}\). 

Solution

\(\sqrt{2}\)  + 2\(\sqrt{2}\)

= 3 \(\sqrt{2}\)

Solution

\(\sqrt{5}\)× \(\sqrt{3}\)

= \(\sqrt{15}\)

Multiplication is possible with irrational numbers if the order of irrational number is same. 

Solution

\(\sqrt{10}\) ÷ \(\sqrt{2}\)

= \(\frac{\sqrt{ 10}}{\sqrt{2}}\)

= \(\sqrt{5}\)

If the root of a rational number is an irrational number then the  resulting number is a surd. 

Quiz

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