MULTIPLYING RADICAL EXPRESSIONS

When you multiply two radicals with same index, you can take the radical once and multiply the numbers inside the radical.

Examples :

 √a  √b = √(a  b)

 3√a 3b = 3(a  b)

In general,

 n√a  n√b = n√(a  b)

Example 1 :

Evaluate :

(√6)(√6) 

Solution :

= √6 ⋅ √6 

√(6 ⋅ 6)

= 6

Example 2 :

Evaluate :

(3√12)(√6)

Solution :

= 3√12  √6

= 3√(12 ⋅ 6) 

= 3√72

= 3√(6 ⋅ ⋅ 2)

= 3(6√2)

= 18√2

Example 3 :

Evaluate :

√6(√3 + √12)

Solution :

= √6(√3 + √12)

√6√3 + √6√12

= √(6 ⋅ 3) + √(6 ⋅ 12)

= √18 + √72

= √(3 ⋅ 3 ⋅ 2) + √(6 ⋅ 6 ⋅ 2)

= 3√2 + 6√2

= 9√2

Example 4 :

Evaluate :

(√2 + √7) (√5 - √6)

Solution :

= (√2 + √7) (√5 - √6)

= √2√5 + √2(-√6) + √7√5 + √7(-√6)

= √2√5 - √2√6) + √7√5 - √7√6

= √(2 ⋅ 5) - √(2 ⋅ 6) + √(7 ⋅ 5) - √(7 ⋅ 6)

= √10 - √12 + √35 - √42

= √10 - √(2 ⋅ 2 ⋅ 3) + √35 - √42

= √10 - 2√3 + √35 - √42

Example 5 :

Evaluate :

(2 - 3√5)(5 - √5)

Solution :

= (2 - 3√5)(5 - √5)

= 2(5) + 2(-√5) - 3√5(5) - 3√5(-√5)

= 10 - 2√5) - 15√5 + 3√(5 ⋅ 5)

= 10 - 17√5 + 3(5)

= 10 - 17√5 + 15

= 25 - 17√5

Example 6 :

Evaluate :

(5 + 4√3)(3 + √3)

Solution :

= (5 + 4√3) (3 + √3)

= 5(3) + 5(√3) + 4√3(3) + 4√3(√3)

= 15 + 5√3 + 12√3 + 4√(3 ⋅ 3)

= 15 + 17√3 + 4(3)

= 15 + 17√3 + 12

= 27 + 17√3

Example 7 :

Evaluate :

(5√24 + √3)(2 - √3)

Solution :

= (5√24 + √3)(2 - √3)

= (5√(2 ⋅ 2 ⋅ 2 ⋅ 3) + √3)(2 - √3)

= (10√(⋅ 3) + √3)(2 - √3)

= (10√6 + √3)(2 - √3)

= 10√6(2) + 10√6(-√3) + √3(2) + √3(-√3)

= 20√6 - 10√(6 ⋅ 3) + 2√3 - √(3 ⋅ 3)

= 20√6 - 10√(2 ⋅ 3 ⋅ 3) + 2√3 - 3

= 20√6 - 10(3√(2) + 2√3 - 3

= 20√6 - 30√2 + 2√3 - 3

Example 8 :

Evaluate :

(-3√3x + 4)(√3x - 5)

Solution :

= (-3√3x + 4)(√3x - 5)

= -3√3x(√3x) - 3√3x(-5) + 4(√3x) + 4(-5)

= -3√(3x ⋅ 3x) + 15√3x + 4√3x - 20

= -3(3x) + 19√3x - 20

= -9x + 19√3x - 20

Example 9 :

Evaluate :

(-4√28x)(√7x3)

Solution :

= (-4√28x)(√7x3)

= -4√(28x ⋅ 7x3)

= -4√(28x ⋅ 7x3)

= -4√(7 ⋅ 2  2 ⋅ 7 ⋅ x⋅ x2)

= -4(7 ⋅ 2 ⋅ x2)

= -4(14x2)

= -56x2

Example 10 :

Evaluate :

(√15x2)(√10x3)

Solution :

= (√15x2)(√10x3)

= √(15x2 ⋅ 10x3)

= √(5 ⋅ 3 ⋅ 2 ⋅ 5  x2 x⋅ x)

= 5x2√(⋅ 2 ⋅ x)

= 5x2√6x

Example 11 :

Evaluate :

Solution :

= (3√24)(3√9)

3√(24 ⋅ 9)

3√(2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3)

= 2 ⋅ 3

= 6

Example 12 :

Evaluate :

Solution :

= (3√250x3)(3√4y6)

3√(250x3  4y6)

3√(1000x3y6)

3√(10 ⋅ 10 ⋅ 10 ⋅ ⋅ x ⋅ x ⋅ y⋅ y⋅ y2)

= 10xy2

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Trigonometry Even and Odd Iidentities

    May 05, 24 12:25 AM

    ASTCnew.png
    Trigonometry Even and Odd Iidentities

    Read More

  2. SOHCAHTOA Worksheet

    May 03, 24 08:50 PM

    sohcahtoa39
    SOHCAHTOA Worksheet

    Read More

  3. Trigonometry Pythagorean Identities

    May 02, 24 11:43 PM

    Trigonometry Pythagorean Identities

    Read More