BINOMIAL SURD

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

For example,

√2 + √5, √7 - √11 + √3, √7 - √11 + √3

Binomial Surd :

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

√2 + √3

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. 

Rationalizing the Denominator

In each case, rationalize the denominator :

Example 1 :

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 2 :

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 3 :

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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