Question

Need help with rationalization of the denominator.

Answer 1

\[\sqrt[4]{16/25x ^{7}}\]

Answer 2

assume all variables represent positive real numbers.

Answer 3

\(\large\color{black}{ \displaystyle \sqrt{\frac{16}{25x^7}} }\) this is the question?

Answer 4

\[\sqrt[4]{\frac{ 16 }{ 25x ^{7} }}\]

Answer 5

thats better

Answer 6

\(\large\color{black}{ \displaystyle \sqrt[4]{\frac{16}{25x^7}} }\)

Answer 7

yes

Answer 8

well, from that it is true that:

\(\large\color{black}{ \displaystyle \sqrt[4]{\frac{16}{25x^7}}~~~\Rightarrow ~~~\frac{\sqrt[4]{16}}{\sqrt[4]{25x^7}} }\)

Answer 9

then

\(\large\color{black}{ \displaystyle \frac{\sqrt[4]{16}}{\sqrt[4]{25x^7}}~~~\Rightarrow ~~~\frac{\sqrt[4]{16}\color{red}{\times\sqrt[4]{25x}}}{\sqrt[4]{25x^7}\color{red}{\times\sqrt[4]{25x}} } }\)

Answer 10

see what I am doing ?

Answer 11

yes, im just writing it down lol

Answer 12

so now your denominator is a perfect root

Answer 13

and you know that

\(\large\color{black}{ \displaystyle \sqrt[4]{25}=\sqrt[2]{5} =\sqrt{5}}\)
\(\large\color{black}{ \displaystyle \sqrt[4]{16}=\sqrt[2]{2^4} =2}\)

Answer 14

go for it...

Answer 15

alrighty one momnet

Answer 16

\[\frac{ 16\sqrt[3]{5x} }{ 2x }\]

Answer 17

@SolomonZelman

Answer 18

i think its incorrect... not sure

Answer 19

\(\large\color{black}{ \displaystyle \frac{\sqrt[4]{16}\color{red}{\times\sqrt[4]{25x}}}{\sqrt[4]{625x^{8}}} }\)

\(\large\color{black}{ \displaystyle \frac{\sqrt[4]{16}\color{red}{\times\sqrt[4]{25x}}}{5x^2} }\)

\(\large\color{black}{ \displaystyle \frac{2\color{red}{\times\sqrt[4]{25x}}}{5x^2} }\)

Answer 20

to be even more simple way....

Answer 21

\(\large\color{black}{ \displaystyle \frac{2\color{red}{\times\sqrt[4]{25x}}}{5x^2} }\)

\(\large\color{black}{ \displaystyle \frac{2\times\sqrt[4]{25}\sqrt[4]{x} }{5x^2} }\)



\(\large\color{black}{ \displaystyle \frac{2\times\sqrt[2]{5}\sqrt[4]{x} }{5x^2} }\)

same as,

\(\large\color{black}{ \displaystyle \frac{2\times\sqrt[]{5}\sqrt[4]{x} }{5x^2} }\)

Answer 22

\[\frac{ 2\sqrt[4]{5x} }{ 5x ^{2} }\]

Answer 23

25 in the 4th root on the top. it is 4th root (25x)

Answer 24

ok

Answer 25

I'm not sure how to structure the answer, this is the only question I don't know how to do...

Answer 26

you still there

Answer 27

@SolomonZelman

Answer 28

@sleepyjess need help structuring the answer..

Answer 29

??????????????????????

Answer 30

you want the steps?

Answer 31

yes, man I am having some serious connection issues... Sorry about that, I just need it step by step for future reference if thats alright.

Answer 32

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \sqrt[4]{\frac{16}{25x^7}} }}\)

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \frac{\sqrt[4]{16}}{\sqrt[4]{25x^7}} }}\)

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \frac{\sqrt[4]{2^4}}{\sqrt[4]{25x^7}} }}\)

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \frac{2}{\sqrt[4]{25x^7}} }}\)

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \frac{2}{\sqrt[4]{25}\sqrt[4]{x^7}} }}\)

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \frac{2\times \sqrt[4]{25}}{\sqrt[4]{25}\times \sqrt[4]{25}\sqrt[4]{x^7}} }}\)

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \frac{2\times \sqrt[4]{25}}{5\sqrt[4]{x^7}} }}\)

\(\large\color{black}{ \displaystyle \large\color{black}{ \displaystyle \frac{2\times \sqrt[4]{25}\times \sqrt[4]{x}}{5\sqrt[4]{x^7}\times \sqrt[4]{x}} }}\)

and then last step

Answer 33

\[\frac{ 2\sqrt[4]{25x} }{ 5x ^{2} }\]

Answer 34

is that right or do I need to square root the 25?

Answer 35

@SolomonZelman

Answer 36

you there ?

Answer 37

yes

Answer 38

I had to depart (cooking purposes)

Answer 39

ah ok :)
Anyway is that last question correct?

Answer 40

i mean my answer

Answer 41

\frac{ 2\sqrt[4]{25x} }{ 5x ^{2} }

Answer 42

\[\frac{ 2\sqrt[4]{25x} }{ 5x ^{2} } \]

Answer 43

there it is

Answer 44

@SolomonZelman

Answer 45

nvm I got it :)