Ask your own question, for FREE!
Algebra 22 Online
OpenStudy (anonymous):

Write in simplified radical form with at most one radical sqrt cd^5 / ^6sqrt cd^2 Assume that all variables represent positive real numbers.

zepdrix (zepdrix):

\[\Large\rm \frac{\sqrt{cd^5}}{\sqrt[6]{cd^2}}\]Is this what the problem looks like?

OpenStudy (anonymous):

Yes @zepdrix

zepdrix (zepdrix):

hmm

zepdrix (zepdrix):

So what's the best approach here... hmm Do you know how to write roots as rational expressions? Example: \(\Large\rm \sqrt{x^3}=x^{3/2}\) We might need to do that here.

zepdrix (zepdrix):

\[\Large\rm \sqrt{cd^5}=c^{1/2}d^{5/2}\]Does that.. process.. look familiar or make sense? :d

OpenStudy (anonymous):

Yes, but the thing is they want atleast one radical. I got that answer though :)

zepdrix (zepdrix):

`At most` one radical. What'd you get for an answer? c:

zepdrix (zepdrix):

Oh the top thing is your answer?

OpenStudy (anonymous):

the same thing you got

zepdrix (zepdrix):

It's quite a few steps away from having a single root. Lemme show you :\

zepdrix (zepdrix):

\[\Large\rm \frac{\sqrt{cd^5}}{\sqrt[6]{cd^2}}=\frac{(cd^5)^{1/2}}{(cd^2)^{1/6}}=\frac{c^{1/2}d^{5/2}}{c^{1/6}d^{2/6}}\]Then apply rules of exponents lets us divide the c's, and the d's, separately,\[\Large\rm c^{\left(\frac{1}{2}-\frac{1}{6}\right)}d^{\left(\frac{5}{2}-\frac{6}{2}\right)}=c^{1/3}d^{13/6}=c^{1/3}\left(d^{13/2}\right)^{1/3}\]And finally we can write them under a single root,\[\Large\rm \sqrt[3]{c d^{13/2}}\]Something like that... :(

OpenStudy (agent0smith):

These are so awful.

zepdrix (zepdrix):

I mean I kind of understand.. it tests your knowledge of different exponent rules and radicals, but yah this seems kind of extreme...

zepdrix (zepdrix):

For this level of work at least

zepdrix (zepdrix):

I didn't explain all of the steps in there. Lemme know if I should back up :U

zepdrix (zepdrix):

Too much? did your head esplode?

OpenStudy (anonymous):

yeah, I'm dying with these. No one in my class understands these. We had to skip the whole section. I'll just try my best hehe Thanks.

zepdrix (zepdrix):

aw :3

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!