So I'm looking at the Key Concept "the nth Root" on p. 361.
I understand that When n is even that you have two real nth roots of (+)b. And when n is odd you only have one real nth root of (+)b. So I get the odd and even part about n.

What I don't understand is why it only tells you there are no real nth roots of (-)b, when n is even.

So my question would be; what happens when n is odd and b is negative? Does the fact that when b is negative, regardless the charge of n, mean that there will not be any real nth roots?

Question

So I'm looking at the Key Concept "the nth Root" on p. 361. 
I understand that When n is even that you have two real nth roots of (+)b. And when n is odd you only have one real nth root of (+)b. So I get the odd and even part about n.

What I don't understand is why it only tells you there are no real nth roots of (-)b, when n is even. 

So my question would be; what happens when n is odd and b is negative? Does the fact that when b is negative, regardless the charge of n, mean that there will not be any real nth roots?

anonymous · Answer

good question :)

anonymous · Answer

What I don't understand is why it only tells you there are no real nth roots of (-)b, when n is even. 

lets first think about this

anonymous · Answer

3x3 = 9 

so,  $\large   \sqrt[2]{9} = 3 $

anonymous · Answer

can u find any number which gives a negative number, when multiplied by itself ?

anonymous · Answer

I've been doing research on this and I think that it doesn't matter.

anonymous · Answer

Any number in this? Or in general?

anonymous · Answer

in general,  just find some real number $n$  wid below property :-

$\large   n \times  n$ = negative number

anonymous · Answer

a negative number times itself (which is negative) will be negative.. So would it be that you can have a negative -b and have roots when n is odd?? That's what im finding..

anonymous · Answer

multiplying a number by itself, fails to produce a negative numbebr.
so for negative numbers, $n$th roots are not defined when $n$ is 2, 4, 6...

anonymous · Answer

So what you're saying is that you can have a -b ONLY when n is odd.. RIGHT?! xD

anonymous · Answer

im saying, when $n$ is even, and $b$ is negative,  then $n$ th root is not defined in reals 

below thing doesnt exist in real numbers :-

$\large   \sqrt[n]{neg}$

anonymous · Answer

I get that. 

When n is odd and b is negative. You CAN have one real root. That's what my question was in the first place lol

anonymous · Answer

read ur question again,
ur first question was on why there are no even nth roots for negative numbers :o

anonymous · Answer

attachment

anonymous · Answer

Right cuz the book doesn't say anything about it not being possible to have roots when you have an odd nth power for a negative number. Its all good I figured it out.