If the radicands are same with the same index, then we call them as like radicals.

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

we call it as unlike radicals.

Note :

If we have composite numbers inside the radical sign, decompose it into product of prime factors.

Add or subtract. Assume all variables are positive. Answers must be simplified.

Problem 1 :

5√6 + 3√6

Solution :

The terms which are inside the radical sign is the same, so both are like radicals.

= 8√6

Problem 2 :

4√20 - 2√5

Solution :

The terms which are inside the radical sign are not same, immediately cannot decide they are not like radicals, because we can decompose 20.

√20 = √(2 ⋅ 2 ⋅ 5)

= 2√5

4√20 = 4(2√5) ==> 8√5

4√20 - 2√5 = 8√5 - 2√5

= 6√5

Problem 3 :

3√(32x2) + 5x√8

Solution :

= 3√(32x2) + 5x√8

Decomposing 32x2 and 8 as much as possible, we get

= 3√(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2x2) + 5x√(2 ⋅ 2 ⋅ 2)

= 3(2 ⋅ 2 ⋅ x) √2 + (5x ⋅ 2) √2

= 3(2 ⋅ 2 ⋅ x) √2 + (5x ⋅ 2) √2

= 12x√2 + 10x√2

=  22x√2

Problem 4 :

7√4x2 + 2√25x - √16x

Solution :

= 7√4x2 + 2√25x - √16x

Decomposing the radicands as much as possible.

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

= 7⋅2⋅x + 2⋅5√x - (2⋅ 2)√x

= 14x + 10√x - 4√x

= 14x + 6√x

Problem 5 :

5∛x2y + ∛27x5y

Solution :

= 5∛x2y + ∛(3 ⋅ 3 ⋅ 3 x5y4)

= 5∛x2y + 3xy∛x2y

Factoring ∛x2y, we get

= (5 + 3xy) ∛x2y

Problem 6 :

3√9y3 - 3y√16y + √25y3

Solution :

= 3√9y3 - 3y√16y + √25y3

= (3 ⋅ 3y)√y - (3y ⋅ 4)√y + 5y√y

= 9y√y - 12y√y+ 5y√y

= 14y√y - 12y√y

= 2y√y

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