Question

(square root of 75b over c) plus (square root of 4b over 3c) plus (square root of 25b over c),

Answer 1

i started this in previous post. but unclear what problem is. if you want look at previous one

Answer 2

or not. i have to say it is somewhat annoying to start a problem and then see it posted again. but then again i am easily annoyed

Answer 3

Surely not...:-)

Answer 4

Why not take those common c's out from under the square root signs to start with?

Answer 5

http://openstudy.com/groups/mathematics?version=feed:get-live-help&referrer=mathematics&domain=tutorial.math.lamar.edu#/groups/mathematics/updates/4e1ef0b80b8b4841c1aa751c

Answer 6

i wasn't sure if the "c" are underneath radical or not. if so then rationalize denominator and add. if not then just add.

Answer 7

I can't get that link to work....

Answer 8

OK, got it now...

Answer 9

Ah, I see.

Answer 10

u there, Paul?

Answer 11

it is the previous post of this exact same problem. ok i must be getting grumpy so i will shut up and go do something else. but really, someone posts a problem, you spend some latex time asking for clarification, and then
\[\color{red}{\text{BAM HERE IT IS AGAIN!}}\]

Answer 12

Guess he's doing something else...

Answer 13

satellite?

Answer 14

U there now?

Answer 15

i am here now.

Answer 16

like telephone tag

Answer 17

u know how java applets stop working when you leave the page for another tab, would be good if the people icon would disappear...

Answer 18

ring ring. lets just do this
\[\sqrt{\frac{75b}{c}}=\sqrt{\frac{75b}{c}}\times \frac{\sqrt{c}}{\sqrt{c}}=\frac{\sqrt{75bc}}{c}\]

Answer 19

likewise
\[\sqrt{\frac{4b}{3c}}=\frac{2\sqrt{3bc}}{3c}\]

Answer 20

Guess he could manage the rest with that for an example.

Answer 21

Oh, you're doing it all, sorry.

Answer 22

and also
\[\sqrt{\frac{25b}{c}}=\frac{5\sqrt{bc}}{c}\]

Answer 23

tnx for the patience

Answer 24

final work is to write
\[\sqrt{75bc}=5\sqrt{3bc}\] get a least common denominator of c and 3c which is 3c, and add them up

Answer 25

so get
\[\frac{15\sqrt{3bc}+2\sqrt{3bc}+15\sqrt{bc}}{3c}\]

Answer 26

unless i made a mistake. first two terms in the numerator are the same so you can add them. third is not

Answer 27

okay thank you!