To understand, how to add or subtract radicals, first we have to know about like radicals and unlike radicals.
Like radicals :
If the radicands are same with the same index, then we call them as like radicals.
Unlike radicals :
we call it as unlike radicals.
Note :
If we have composite numbers inside the radical sign, decompose it into product of prime factors.
We can add or subtract radicals, only if they are like radicals.
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√(32x^{2}) + 5x√8
Solution :
= 3√(32x^{2}) + 5x√8
Decomposing 32x^{2 }and 8 as much as possible, we get
= 3√(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2x^{2}) + 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√4x^{2} + 2√25x - √16x
Solution :
= 7√4x^{2} + 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∛x^{2}y + ∛27x^{5}y^{4 }
Solution :
= 5∛x^{2}y + ∛(3 ⋅ 3 ⋅ 3 x^{5}y^{4})
= 5∛x^{2}y + 3xy∛x^{2}y
Factoring ∛x^{2}y, we get
= (5 + 3xy) ∛x^{2}y
Problem 6 :
3√9y^{3} - 3y√16y + √25y^{3}
Solution :
= 3√9y^{3} - 3y√16y + √25y^{3}
= (3 ⋅ 3y)√y - (3y ⋅ 4)√y + 5y√y
= 9y√y - 12y√y+ 5y√y
= 14y√y - 12y√y
= 2y√y
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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