SIMPLE AND COMPOUND SURDS

Simple Surd :

A surd having only one term is called a simple surd.

For example, 

√2, √3

Compound Surd :

An expression which contains addition or subtraction of two or more surds is called compound surd.

For example,

√2 + √5, √7 - √11 + √3

Note :

Compound surds can further be classified as binomial and trinomial surds.

Binomial Surd :

A compound surd which contains exactly two surds is called a binomial surd.

√2 + √3

Trinomial Surd :

A compound surd which contains exactly three surds is called a trinomial surd.

√7 - √11 + √3

Rules for Simple Surds

Rule 1 :

Product of two simple surds of same order.

√a x √b = √(a x b)

Rule 2 :

Division of two simple surds of same order.

√a/ √b = √(a/b)

Rule 3 :

Surds can be written in exponential form.

√a = a^1/2

ⁿ√a = a^1/n

Note :

1. Surds can not be added.

√a + √b ≠ √(a + b)

2. Surds can not be subtracted.

√a - √b ≠ √(a - b)

Conjugate Surds

Two binomial surds which are differ only in signs (+/–) between them are called conjugate surds.

For example,

√7 + √3 and √7 - √3 are conjugate to each other

3 + √2 and 3 - √2 are conjugate to each other

In a fraction, if the denominator is a binomial surd, then we can use its conjugate to rationalize the denominator. 

Comparing Simple Surds

Example 1 :

Which is greater ?

√4 or √6

Solution :

The above two surds have the same  order (i.e., 2).

To compare the above surds, we have to compare the radicands 4 and 6.

Clearly 6 is greater than 4.

So, √6 is greater than √4.

√6 > √4

Example 2 :

Which is greater ?

√2 or ³√3

Solution :

The above two surds have different orders. The are 2 and 3.

Using the least common multiple of the orders 2 and 3, we can convert them into surds of same order.

Least common multiple of (2 and 3) is 6.

√2 = ²√2 = ^3x2√(2³) = ⁶√8

³√3 = ^2x3√(3²) = ⁶√9

Now, the given two surds are expressed in the same order.

Compare the radicands :

9 > 8

Then,

⁶√9 > ⁶√8

Therefore,

³√3 > √2

Rationalizing the Denominator

In each case, rationalize the denominator :

Example 3 :

1/(5 + √3)

Solution :

The denominator is 5 + √3 and its conjugate is 5 - √3.

To rationalize the denominator of the given fraction, multiply both numerator and denominator by 5 - √3.

1/(5 + √3) = [1(5 - √3)]/[(5 + √3)(5 - √3)]

Using algebraic identity a² - b² = (a + b)(a - b),

= (5 - √3)/[(5² - (√3)²]

= (5 - √3)/(25 - 3)

= (5 - √3)/22

Example 4 :

1/(8 - 2√5)

Solution :

The denominator is 8 - 2√5 and its conjugate is 8 + 2√5.

To rationalize the denominator of the given fraction, multiply both numerator and denominator by 8 + 2√5.

1/(8 - 2√5) = [1(8 + 2√5)]/[(8 - 2√5)(8 + 2√5)]

Using algebraic identity a² - b² = (a + b)(a - b),

= (8 + 2√5)/[(8² - (2√5)²]

= (8 + 2√5)/(64 - 20)

= (8 + 2√5)/44

= [2(4 + √5)]/44

= (4 + √5)/22

Example 5 :

1/(√3 + √5)

Solution :

The denominator is √3 + √5 and its conjugate is √3 - √5.

To rationalize the denominator of the given fraction, multiply both numerator and denominator by √3 - √5.

1/(√3 + √5) = [1(√3 - √5)]/[(√3 + √5)(√3 - √5)]

Using algebraic identity a² - b² = (a + b)(a - b),

= (√3 - √5)/[(√3)² - (√5)²]

= (√3 - √5)/(3 - 5)

= (√3 - √5)/(-2)

= -(√3 - √5)/2

= (√5 - √)/2

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