# What does 3 sqrt 2

## How does partial root extraction work?

### Non-pullable roots

A root cannot be extracted if the exponent of the power below the root is not a multiple of the root exponent and is smaller than the root exponent.

\$ \ sqrt [3] {16} = \ sqrt [3] {4 ^ 2} \$

You can see that the value of the exponent is neither greater than the root exponent nor a multiple of the root exponent.

A root cannot be extracted if the exponent of the power below the root is not a multiple of the root exponent and is smaller than the root exponent.

\$ \ sqrt [2] {a} \$, \$ \ sqrt [3] {a ^ 2} \$

Non-pullable roots

\$ \ sqrt {29} \ approx \$ 5.38

\$ \ sqrt {23} \ approx \$ 4.79

\$ \ sqrt {67} \ approx 8.18 \$

### Partially pullable roots

Partially pullable roots are a special case that is crucial for us. They are characterized by the fact that the exponent of the power below the root is not a multiple of the root exponent, but is greater than the root exponent.

A root is partially extractable if the exponent of the power below the root is greater than the root exponent, but is not a multiple of the root exponent.

\$ \ sqrt [2] {a ^ 5} \$, \$ \ sqrt [3] {a ^ 8} \$

### Procedure for partial root extraction

When you partially pull the roots, you break them apart partially pullable root into a pullable and a non-pullable part. This means that you break the radicand under the root into a product of two numbers. You have to be able to pull the root of one of these numbers.

\$ \ sqrt {44} = \ sqrt {4 \ cdot 11} \$

Factors under the root

\$ \ sqrt {a \ cdot b} = \ sqrt {a} \ cdot \ sqrt {b} \$

\$ \ sqrt {44} = \ sqrt {4 \ cdot 11} = \ sqrt {4} \ cdot \ sqrt {11} = 2 \ cdot \ sqrt {11} \$

As you can see, we've transformed the partially extractable root into a product of an integer and a non-extractable root.

Partial root extraction

\$ \ sqrt {45} = \ sqrt {9 \ cdot 5} = \ sqrt {9} \ cdot \ sqrt {5} = 3 \ cdot \ sqrt {5} \$

\$ \ sqrt {8} = \ sqrt {4 \ cdot 2} = \ sqrt {4} \ cdot \ sqrt {2} = 2 \ cdot \ sqrt {2} \$

\$ \ sqrt [5] {128} = \ sqrt [5] {2 ^ 5 \ cdot 2 ^ 2} = \ sqrt [5] {2 ^ 5} \ cdot \ sqrt [5] {2 ^ 2} = 2 \ cdot \ sqrt [5] {4} \$

Test your newly learned knowledge now with our exercises! Good luck with it!