Question

Can someone help me with this?
√45n^5

Answer 1

\(\large\sqrt{45n^5}\) like this?

Answer 2

Yeah like that.

Answer 3

\(\color{black}{ \displaystyle \sqrt{45n^5}=\sqrt{9\cdot 5 \cdot n^4\cdot n} =\sqrt{9\cdot n^4}~\cdot ~\sqrt{5\cdot n}=\sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }\)

Answer 4

that is how i brake them down.

Answer 5

what is √9, can you tell me?

Answer 6

3?

Answer 7

yes

Answer 8

How did we lose an n from n^5?

Answer 9

we didn't, all I did is used the fact that

\(n^5=n^{4+1}=n^4\cdot n^1=n^4\cdot n\)

Answer 10

and of course, I rearranged them putting the terms that can be simplified in the beginning and then the terms that can't be simplified

Answer 11

\(\sqrt{n^4}\) is same as \(\sqrt[2]{n^4}\)
and that is =\(n^{4/2}={\rm to~what?}\)

Answer 12

Oh alright..

Answer 13

ok, did you get up to finding the \(\sqrt{n^4}\) ?

Answer 14

Nope don't understand what's going on this time.

Answer 15

\(\sqrt{n^4}=\sqrt[2]{n^4}\)

this is because whenever the root is not specificed, it is 2 (i.e. SQUARE root)

Answer 16

now, applying the rule: \(\sqrt[c]{a^b}=a^{b/c}\)

we get that \(\sqrt[2]{n^4}=n^{4/2}=n^{2/1}=n^2\)

Answer 17

Okay, didn't know that.

Answer 18

We had:

\(\color{black}{ \displaystyle \sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }\)

and now we had:

\(\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }\)

Answer 19

oh, now you know.... :0

Answer 20

Thanks.

Answer 21

\(\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }\)

is just same as
\(\color{black}{ \displaystyle3 n^2~\sqrt{5 n} }\)


\(\bf ... \)yw