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Mathematics 19 Online
OpenStudy (anonymous):

Simplify \[\sqrt[5]{-1/1024}\] This root is not a real number right

OpenStudy (accessdenied):

the root is not a real number only when the index is even your index is 5, so it is at least real

OpenStudy (accessdenied):

if it helps,you can just write this to break it apart. \[ \sqrt[5]{-1}\times\sqrt[5]{1/1024}~~~~-1~times~itself~5~times~is~negative!\\ -1\times\sqrt[5]{1024} \]

OpenStudy (anonymous):

\[\frac{\sqrt[5]{-1}}{4}\]

OpenStudy (accessdenied):

1/1024* in that last line

OpenStudy (anonymous):

\[\frac{(-1*)\sqrt[5]{1}}{4}=\frac{-1*1}{4}= \frac{1}{4}*i\]

OpenStudy (accessdenied):

how did you get the i? there wasn't a square root of -1 in it

OpenStudy (anonymous):

\[\sqrt{-1}=-1*\sqrt{1}=i*1 =i \]i is complex number (i if square nefgative)

OpenStudy (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.

OpenStudy (anonymous):

\[\sqrt{-1}=i\]

OpenStudy (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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