Question

Find the exact value of the radical expression in simplest form.

Answer 1

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

Answer 2

No idea /:

Answer 3

I thought I had a answer, but I looked at choice but none were even close. Here is choices;

Answer 4

7p^3 - p
7p^3
\[13p^3 \sqrt{p} - 2p \sqrt{p}\]
\[13p^3 \sqrt{p} - 2p\]

Answer 5

@johnweldon1993

Answer 6

Hint*

\[\large \sqrt{p^6} = \sqrt{p^2}\sqrt{p^2}\sqrt{p^2}\]

Answer 7

Turn all p^6 to those?

Answer 8

Sure we can do that...just try not to get lost...

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

So if we do that...we will have

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

Wow that was a hassle lol....okay now...what does √(p^2) equal??

Answer 9

\[\sqrt{p}\sqrt{p}\]? I think lol

Answer 10

Well...yeah...but when you do that out what do you get? √p times √p = p right?

Answer 11

Yes

Answer 12

Alright...so this can all be simplified...

Answer 13

\[12\sqrt{8p^2}\] ?

Answer 14

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

becomes

\[\large p \times p \times p + \sqrt{9}\space p \times p \times p + 3 \space p \times p \times p - p\]

which now becomes...

\[\large p^3 + 3p^3 + 3p^3 - p\]

and finally we add 3p^3 + 3p^3 and 1p^3 to get 7p^3

so all we have left is...

\[\large \ 7p^3 - p\]

make sense?

Answer 15

Oh, you put into one p. I see it now. Makes more sense than it did when I tried to do it in my head :p

Answer 16

lol yeah that was a hassle but glad you see it now! :D lol

Answer 17

Random question if you have: \[\sqrt{11} + \sqrt{11} = \sqrt{22}\] Does it work like that?

Answer 18

No no no remember like with the last one...when you are adding or subtracting a number being multiplied to the same square root ...you add the outside numbers and leave the square roots alone...

\[\large \sqrt{11} + \sqrt{11}\]

We can agree that if you put a 1 in front...it changes nothing...so

\[\large 1\sqrt{11} + 1\sqrt{11}\]

and to solve that...you add the 2 numbers in front...and leave the square root alone...

\[\large 1\sqrt{11} + 1\sqrt{11} = (1 + 1)\sqrt{11} = 2\sqrt{11}\]

Answer 19

Just making sure, let me put what you said in my notes :D

Answer 20

Want me to write some general rules for you to write down?

Answer 21

2 more questions on my quiz. woot. and sure!

Answer 22

Alright here we go lol...

Answer 23

Summarize as much as you can :P

Answer 24

\[\large \sqrt{ab} = \sqrt{a}\sqrt{b}\]
\[\large \sqrt{\frac{ a }{ b }} = \frac{ \sqrt{a} }{ \sqrt{b} }\]
\[\large \sqrt{a}\sqrt{a} = a\]
\[\large \sqrt{a^n} = (\sqrt{a})^n \space \text{or} \space a^{n/2}\]
\[\large x\sqrt{a} + y\sqrt{a} = (x + y)\sqrt{a} \]
\[\large x\sqrt{a} - y\sqrt{a} = (x - y)\sqrt{a} \]

Answer 25

That should give you a good headstart lol....there's other things like rationalizing the denominator and stuff like that...but you can learn that a bit later...

Answer 26

@johnweldon1993 sorry laptop froze again .-. and thanks!

Answer 27

Lol man you gotta get that fixed lol...no problem!

Answer 28

Getting a new laptop soon :P