Thew following steps will be useful to simplify radicals.
Step 1 :
Decompose the number inside the radical sign into prime factors.
Step 2 :
If you have square root (√), you have to take one number out of the square root for every two same numbers multiplied inside the radical.
If you have cube root (^{3}√), we have to take one number out of cube root for every three same numbers multiplied inside the radical.
If you have fourth root (^{4}√), you have to take one number out of fourth root for every four same numbers multiplied inside the radical.
Step 3 :
Simplify.
Examples :
Question 1 :
Simplify :
√20
Solution :
Decompose 20 into prime factors using synthetic division.
So, we have
√20 = √(2 ⋅ 2 ⋅ 5)
√20 = 2√5
Question 2 :
Simplify :
√121
Solution :
Decompose 121 into prime factors.
√121 = √(11 ⋅ 11)
√121 = 11
Question 3 :
Simplify :
√52
Solution :
Decompose 52 into prime factors using synthetic division.
So, we have
√52 = √(2 ⋅ 2 ⋅ 13)
√52 = 2√13
Question 4 :
Simplify :
√45
Solution :
Decompose 45 into prime factors using synthetic division.
So, we have
√45 = √(5 ⋅ 3 ⋅ 3)
√45 = 3√5
Question 5 :
Simplify :
^{3}√72
Solution :
Decompose 72 into prime factors using synthetic division.
So, we have
^{3}√72 = ^{3}√(2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3)
^{3}√72 = 2 ⋅ ^{3}√(3 ⋅ 3)
^{3}√72 = 2 ⋅ ^{3}√(3 ⋅ 3)
^{3}√72 = 2^{3}√9
Question 6 :
Simplify :
^{3}√40
Solution :
Decompose 40 into prime factors using synthetic division.
So, we have
^{3}√40 = ^{3}√(2 ⋅ 2 ⋅ 2 ⋅ 5)
^{3}√40 = 2^{3}√5
Question 7 :
Simplify :
^{3}√27
Solution :
Decompose 27 into prime factors using synthetic division.
So, we have
^{3}√27 = ^{3}√(3 ⋅ 3 ⋅ 3)
^{3}√27 = 3
Question 8 :
Simplify :
^{4}√243
Solution :
Decompose 243 into prime factors using synthetic division.
So, we have
^{4}√243 = √(3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3)
^{4}√243 = 3^{4}√3
Question 9 :
Simplify :
^{5}√288
Solution :
Decompose 288 into prime factors using synthetic division.
So, we have
^{5}√288 = ^{5}√(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3)
^{5}√288 = 2 ^{5}√3
Question 10 :
Simplify :
^{6}√320
Solution :
Decompose 320 into prime factors using synthetic division.
So, we have
^{6}√320 = ^{6}√(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5)
^{6}√320 = 2^{6}√5
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