anonymous
  • anonymous
Can someone help me with this? √45n^5
Mathematics
schrodinger
  • schrodinger
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SolomonZelman
  • SolomonZelman
\(\large\sqrt{45n^5}\) like this?
anonymous
  • anonymous
Yeah like that.
SolomonZelman
  • SolomonZelman
\(\color{black}{ \displaystyle \sqrt{45n^5}=\sqrt{9\cdot 5 \cdot n^4\cdot n} =\sqrt{9\cdot n^4}~\cdot ~\sqrt{5\cdot n}=\sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }\)

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SolomonZelman
  • SolomonZelman
that is how i brake them down.
SolomonZelman
  • SolomonZelman
what is √9, can you tell me?
anonymous
  • anonymous
3?
SolomonZelman
  • SolomonZelman
yes
anonymous
  • anonymous
How did we lose an n from n^5?
SolomonZelman
  • SolomonZelman
we didn't, all I did is used the fact that \(n^5=n^{4+1}=n^4\cdot n^1=n^4\cdot n\)
SolomonZelman
  • SolomonZelman
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
SolomonZelman
  • SolomonZelman
\(\sqrt{n^4}\) is same as \(\sqrt[2]{n^4}\) and that is =\(n^{4/2}={\rm to~what?}\)
anonymous
  • anonymous
Oh alright..
SolomonZelman
  • SolomonZelman
ok, did you get up to finding the \(\sqrt{n^4}\) ?
anonymous
  • anonymous
Nope don't understand what's going on this time.
SolomonZelman
  • SolomonZelman
\(\sqrt{n^4}=\sqrt[2]{n^4}\) this is because whenever the root is not specificed, it is 2 (i.e. SQUARE root)
SolomonZelman
  • SolomonZelman
now, applying the rule: \(\sqrt[c]{a^b}=a^{b/c}\) we get that \(\sqrt[2]{n^4}=n^{4/2}=n^{2/1}=n^2\)
anonymous
  • anonymous
Okay, didn't know that.
SolomonZelman
  • SolomonZelman
We had: \(\color{black}{ \displaystyle \sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }\) and now we had: \(\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }\)
SolomonZelman
  • SolomonZelman
oh, now you know.... :0
anonymous
  • anonymous
Thanks.
SolomonZelman
  • SolomonZelman
\(\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }\) is just same as \(\color{black}{ \displaystyle3 n^2~\sqrt{5 n} }\) \(\bf ... \)yw

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