radicals

[Redanthony]

Simplify radicals. I can't seem to get them simplified enough. How do I know when the radical is simplified enough.
ex. /16n I simplified to 4/n and think I can further but don't know how to write it.

 

[tutor_joel]

this is correct if I'm reading your notation right.

sqrt(16n)

sqrt(4^2n) or sqrt(2^2*2^2*n)

= 4sqrt(n)

cannot be simplified further

 

[Mrspi]

Simplify radicals. I can't seem to get them simplified enough. How do I know when the radical is simplified enough.
ex. /16n I simplified to 4/n and think I can further but don't know how to write it.

A radical is simplified "enough" when what is left under the radical sign does not contain any factors that are powers of the indicated root.

If, for example, you are dealing with SQUARE (or "second")roots, then you are DONE when the expression under the radical sign does not contain any factors that are perfect squares.

sqrt(16 n) = sqrt(4[sup:52pwbo78]2[/sup:52pwbo78] * n), or sqrt(4[sup:52pwbo78]2[/sup:52pwbo78])*sqrt(n) = 4 sqrt(n) Now...we don't know anything about the factors of n, so we don't know if n has any perfect square factors. We're done.

sqrt(16n[sup:52pwbo78]3[/sup:52pwbo78]) = sqrt(4[sup:52pwbo78]2[/sup:52pwbo78]*n[sup:52pwbo78]2[/sup:52pwbo78]*n) = sqrt(4[sup:52pwbo78]2[/sup:52pwbo78]) * sqrt (n[sup:52pwbo78]2[/sup:52pwbo78]) * sqrt(n) = 4*n*sqrt(n)

Again, we don't know if n has an factors which are perfect squares, so we're done. 4n * sqrt(n) would be the simplest form of this radical.

If you are dealing with a different "index" on the radical, for example, a CUBE ROOT, you'd be looking for factors that are perfect cubes. For example, if you have

cuberoot(40 x[sup:52pwbo78]5[/sup:52pwbo78]), look for factors under the radical sign that are perfect cubes. 40 = 2[sup:52pwbo78]3[/sup:52pwbo78]*5, and x[sup:52pwbo78]5[/sup:52pwbo78] = x[sup:52pwbo78]3[/sup:52pwbo78]*x[sup:52pwbo78]2[/sup:52pwbo78], so

cuberoot(40 x[sup:52pwbo78]5[/sup:52pwbo78]) = cuberoot(2[sup:52pwbo78]3[/sup:52pwbo78]*5 * x[sup:52pwbo78]3[/sup:52pwbo78]*x[sup:52pwbo78]2[/sup:52pwbo78])

2[sup:52pwbo78]3[/sup:52pwbo78] is a perfect cube, as is x[sup:52pwbo78]3[/sup:52pwbo78]

cuberoot(2[sup:52pwbo78]3[/sup:52pwbo78] * x[sup:52pwbo78]3[/sup:52pwbo78]) * cuberoot(5 * x[sup:52pwbo78]2[/sup:52pwbo78]))

2*x *cuberoot(5 x[sup:52pwbo78]2[/sup:52pwbo78])

What is left under the cuberoot sign does not have any factors which are perfect cubes, so you're done.

 

[Redanthony]

Thank you because I though with the answer being 4 sq.rt n that the four would have to be simplified further and I din't how to achieve that task.