HOW TO MULTIPLY BINOMIAL RADICAL EXPRESSIONS

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Before using distributive property with radical expressions, we should be aware of 

  • How to multiply radical terms by a constant
  • How to multiply the two different radicands
  • How to multiply same radicands.
b× a = aba×b = aba×a = a2 = a

Like radicals :

If the radicands are same with the same index, then we call it as like radicals. We can combine only like radicals.

Unlike radicals :

  • Radicands are same, index are not same.
  • Index are same, radicands are not same.

we call it as unlike radicals.

Expand and simplify :

Problem 1 :

(1 + √2) (2 + √2)

Solution :

(1 + √2) (2 + √2)

= 1 2 + 1√2 + 2√2 + √2 √2

= 2 + 3√2 + 2

= 4 + 3√2

Problem 2 :

(2 + √3) (3 + √3)

Solution :

(2 + √3) (3 + √3)

= 2 3 + 2√3 + 3√3 + √3 √3

= 6 + 5√3 + 3

= 9 + 5√3

Problem 3 :

(√3 + 2) (√3 – 1)

Solution :

(√3 + 2) (√3 – 1)

 = √3 √3 - √3 1 + 2√3 – 2

= 3 - √3 + 2√3 – 2

= 1 + √3  

Problem 4 :

(4 - √2) (3 + √2)

Solution :

(4 - √2) (3 + √2)

= 4 3 + 4√2 - 3√2 - √2 √2

= 12 + √2 – 2

= 10 + √2

Problem 5 :

(1 + √3) (1 - √3)

Solution :

(1 + √3) (1 - √3)

= (1)2 – (√3)2

= 1 – 3

 = -2

Problem 6 :

(5 + √7) (2 - √7)

Solution :

(5 + √7) (2 - √7)

= 5 2 5√7 + 2√7 - √7 √7

= 10 - 3√7 – 7

= 3 - 3√7

Problem 7 :

(√5 + 2) (√5 – 3)

Solution :

(√5 + 2) (√5 – 3)

= √5 √5 - √5 3 + 2 √5 – 6

= 5 - 3√5 + 2√5 – 6

= -1 - √5    

Problem 8 :

(6 - √3) (2 + √3)

Solution :

(6 - √3) (2 + √3)

= 6 2 + 6√3 - 2√3 - √3 √3

= 12 + 4√3 – 3

= 9 + 4√3

Problem 9 :

(4 - √2) (3 - √2)

Solution :

(4 - √2) (3 - √2)

 = 4 3 – 4√2 - 3√2 + √2 √2

= 12 - 7√2 + 2

= 14 - 7√2

Problem 10 :

(4 - 3√3) (2 - √3)

Solution :

(4 - 3√3) (2 - √3)

= 4 2 - 4√3 - 3√3 2 + 3√3 √3

= 8 - 4√3 - 6√3 + 3 3

= 8 - 10√3 + 9

= 17 - 10√3

Problem 11 :

(-1 + 2√2) (2 - √2)

Solution :

(-1 + 2√2) (2 - √2)

= -1 2 + 1 √2 + 2√2 2 - 2√2 √2

= -2 + √2 + 4√2 – 2 2

= -2 + 5√2 – 4

= -6 + 5√2

Problem 12 :

(2√2 + 3) (2√2 + 5)

Solution :

(2√2 + 3) (2√2 + 5)

= 2√2 2√2 + 2√2 5 + 3 2√2 + 3 5

= 2 2 √2 √2 + 2 5√2 + 3 2√2 + 3 5

= 4 2 + 10√2 + 6√2 + 15 

= 8 + 10√2 + 6√2 + 15

= 23 + 16√2

Problem 13 :

The ratio of the length to the width of a golden rectangle is (1 + √5) : 2. The dimensions of the face of the Parthenon in Greece form a golden rectangle. What is the height h of the Parthenon?

multiply-binomials-radical-q1

Solution :

Think of the length and height of the Parthenon as the length and width of a golden rectangle. The length of the rectangular face is 31 meters. You know the ratio of the length to the height. Find the height h.

(1 + √5) : 2 = 31 : h

(1 + √5) / 2 = 31 / h

Doing cross multiplication, we get

h(1 + √5) = 31(2)

h = 62/(1 + √5)

Multiplying both numerator and denominator by the conjugate of the denominator, we get

= 62/(1 + √5) ⋅ [(1 -  √5) / (1 - √5)]

= 62(1 - √5)/(1 + √5)(1 - √5)

= 62(1 - √5)/(12 - √52)

= 62(1 - √5)/(-4)

= -15.5(1 - √5)

= -15.5(1 - 2.23)

= 15.5(1.23)

= 19.06

Problem 14 :

A rectangular walkway is √5 ft wide and 8√5 ft long. Find the perimeter of the walkway.

Solution :

Length = 8√5 ft and width =  √5 ft

Perimeter of walkway = 2(length + width)

= 2(8√5 + √5)

= 2(9 √5)

= 18 √5 ft

So, the perimeter is 18 √5 ft.

Problem 15 :

The process of eliminating a radical from the denominator of a radical expression is called _______________.

Solution :

The process of eliminating a radical from the denominator of a radical expression is called rationalizing the denominator.

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