anonymous
  • anonymous
Simplify \[\sqrt[5]{-1/1024}\] This root is not a real number right
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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AccessDenied
  • AccessDenied
the root is not a real number only when the index is even your index is 5, so it is at least real
AccessDenied
  • AccessDenied
if it helps,you can just write this to break it apart. \[ \sqrt[5]{-1}\times\sqrt[5]{1/1024}~~~~-1~times~itself~5~times~is~negative!\\ -1\times\sqrt[5]{1024} \]
anonymous
  • anonymous
\[\frac{\sqrt[5]{-1}}{4}\]

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AccessDenied
  • AccessDenied
1/1024* in that last line
anonymous
  • anonymous
\[\frac{(-1*)\sqrt[5]{1}}{4}=\frac{-1*1}{4}= \frac{1}{4}*i\]
AccessDenied
  • AccessDenied
how did you get the i? there wasn't a square root of -1 in it
anonymous
  • anonymous
\[\sqrt{-1}=-1*\sqrt{1}=i*1 =i \]i is complex number (i if square nefgative)
AccessDenied
  • AccessDenied
The concept of the radical is to answer the question: "What to the exponent of (index) will get me (number)?" \[ \sqrt{-1} = \sqrt[2]{-1} \] What to the exponent TWO will get me -1? Well, we know that if we square any number, it becomes positive, so there isn't a real number that satisfies this. However, this is where i comes in to save the day. i is defined as the square root of -1. So, that will answer this question. Now, let's look at the other example: \[ \sqrt[5]{-1} \] So, what number to the exponent of FIVE will get me -1? x^5 = -1? Well, the fifth power can be negative if our 'x' is negative because it takes TWO -1's to cancel each-other out, so an odd number of -1's will make a negative. For x=-1... (-1)^5 := [(-1)(-1)][(-1)(-1)](-1) = 1*1 * -1 does = -1 So, the answer of the 5th root of -1 is -1.
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=%28%28-1%29%2F%281024%29%29%5E%281%2F5%29 http://www.mathway.com/answer.aspx?p=calg?p=SMB205:((-1)SMB10(1024))?p=22?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=Simplify
anonymous
  • anonymous
\[\sqrt{-1}=i\]
AccessDenied
  • AccessDenied
the square root of -1 is indeed i the fifth root of -1 is not i, since i^5 = i^2 i^2 i = -1*-1*i = i... it seems (to wolfram's mathworld) that there is more to the negatives under odd roots in the complex plane, but in my class (algebra II) we've always looked at it as "negatives under odd roots are real valued, negatives under even roots are complex valued"

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