Question

Please help medal+fanning!!! question down below
Find the exact value of the radical expression in simplest form.

Answer 1

\[c \sqrt{c}+12c \sqrt{c}-2c \sqrt{c}\]

Answer 2

treat it like
\[cx+12cx-2cx\] and combine like terms

Answer 3

How do you combine like terms?

Answer 4

@Directrix

Answer 5

I am attaching a blurb on combining like terms that may be of help.

Answer 6

Thank you

Answer 7

These terms you are adding are all like terms because they have the same "last name."

I'm going to write them so that will be clearer to see.

Answer 8

How do I add up this one does the c become 1. If so should i add it like this 12+2+1

Answer 9

(1) * c√ c + (12)* c√ c - 2* (c√ c) =

The common last name is (c√ c)

Answer 10

So you essentially have 1 + 12 - 2 = what and then add the last name of (c√ c) to that answer. Does that help?

Answer 11

Ok so the answer is 11(csqrtc)

Answer 12

\[11c \sqrt{c}\]

Answer 13

11c√ c

Answer 14

"last name" i like that
i will use it (steal it)

Answer 15

@satellite73 Consider it a gift. :)

Answer 16

Loll you two are funny. Thanks so much directrix could you help check one more to see if I got this

Answer 17

Okay.

Answer 18

Oh and was that the simplest form?

Answer 19

@Directrix

Answer 20

11c√ c is the simplest radical form. Are you "into" rational exponents? Usually, the instructions for a problem tell you what form in which to express the answer.

Answer 21

Find the exact value of the radical expression in simplest form. Yes, it is in simplest radical form and it is exact, too. On the button.

Answer 22

Loll ok thank you but what happens when the terms aren't so alike?

Answer 23

Then, you only combine the like terms.

Answer 24

\[\sqrt{p^6}+\sqrt{9p^6}+\sqrt{3p^6}-\sqrt{p^2}\]

Answer 25

What do I do with the unlike terms

Answer 26

First, simplify the square root of p^6, 9 p^6, and p^2. Those are perfect squares under the radical.

Answer 27

13p^6

Answer 28

minus p^2 is 13^4

Answer 29

@Directrix

Answer 30

What are you saying is the answer? I don't get what those posts mean in terms of the problem.

Answer 31

Oh i thought the answer was 13^4 after combing the first like terms giving me f13p^6

Answer 32

I will need to write the problem by hand and begin the solution that way.
Are you up for that? I don't care for LaTex myself.

Answer 33

Loll sure thing but whats latex?

Answer 34

LaTex is whatever program you used to write what you see on the attachment.

Answer 35

Oh the equation thing at the bottom below the comment I used

Answer 36

You can pick up from here. See the attachment.
Keep in mind: In square roots, it takes two of the same thing under the radical to get one of them out. That is not precise mathematics but the idea is what counts.

Answer 37

I'm a little confused on how you put the p3p3 instead of p6 @Directrix

Answer 38

13 times p cubed times the square root of p − 2 times p times the square root of p

13 times p cubed times the square root of p − 2p

7p3

7p3 − p

Answer 39

You may need to review the exponent properties. p^3 times p^3 = p^6.

Answer 40

I think it's a then

Answer 41

Here it is:

Answer 42

Thats the final answer? @Directrix

Answer 43

That's what I got. Do you have questions? Sorry for the delay but I had to complete a Geometry proof I had helped another asker begin. But, he became impatient and signed off before I could finish. I finished anyway.

Answer 44

@HelloGoodmorning

Answer 45

Lol No worries but do you see the four answers I put up one of those are the answers.
13 times p cubed times the square root of p − 2 times p times the square root of p

13 times p cubed times the square root of p − 2p

7p3

7p3 − p

Answer 46

Will you look at your text and check to be sure you LaTexed the third term correctly? I think that 3 is supposed to be a 9. I circled it on the atttachment.

Answer 47

Oops sorry the one where it say \[\sqrt{3p^6}\] is suppose to say

Answer 48

\[3\sqrt{p^6}\]

Answer 49

@Directrix

Answer 50

That's why the answer we got did not gee-haw with any of the options. Just a second.

Answer 51

@GodBlessYou Corrected problem and solution awaiting for your inspection and approval. :)

Answer 52

Lol I'm so sorry but there's no need to solve it I found out what the answer was Thanks so much for coming when I need you

Answer 53

Okay, we can close out this thread now.

Answer 54

Okay :D