Question

Rationalize the denominator.

index of 3√5/√4)

Answer 1

i can write this out if you want me to, i know its hard to understand

Answer 2

\[\sqrt[3]{\frac{5}{4}}\]?

Answer 3

yea if thats an index of 3, its kinda small

Answer 4

then multiply top and bottom by
\[\sqrt[3]{2}\] to get
\[\frac{\sqrt[3]{10}}{2}\]

Answer 5

i just have no clue what to do. how does the index of 3 affect it? am i supposed to multiply the numer and denom by radical 4?

Answer 6

no just
\[\sqrt[3]{2}\] because
\[\sqrt[3]{8}=2\]

Answer 7

where did you get the the radical 2?
it went from radical 4 to radical 2?

Answer 8

it is not 'radical 4' it is 'cubed root of 4'

Answer 9

ok so first step was simplify the 4 into a 2, then cube that number?

Answer 10

so it goes like this
\[\sqrt[3]{4}\times \sqrt[3]{2}=\sqrt[3]{8}=2\] because
\[2^3=8\]
that gets rid of the cubed root in the denominator

Answer 11

so how did you know to mult the 4 by 2? is it because its cubed?

Answer 12

no i did not "simplify the 4" it is the cubed root of 4. i wanted to multiply it by something to give a cube

Answer 13

so for example is the denom was 16, what would you do?

Answer 14

so for example if i had
\[\sqrt[3]{7}\] i would multiply by
\[\sqrt[3]{49}\] to get
\[\sqrt[3]{7^3}=7\]

Answer 15

\[\sqrt[3]{5/\sqrt{16}}\]

Answer 16

and if i had
\[\sqrt[3]{16}=\sqrt[3]{2^4}\] i would use
\[\sqrt[3]{2^2}\] to give
\[\sqrt[3]{2^6}=2^3\]