Question

why is 4 in the root of -16 undefined ,if 5 in the root of -32= -2?

Answer 1

do you mean

sqrt(-16) = sqrt(-4^2) = 4*sqrt(-1)

AND

sqrt(-32) = sqrt(-2^5) = 2*2*sqrt(-2)

Answer 2

I think her question is in regards to `odd roots` vs `even roots`, i'll wait though, i think 55 will answer it for her :)

Answer 3

Pretty sure it is zepdrix.

Answer 4

The 4th root of -16 has to be a number that when you multiply 4 of the same number together you get 16. If you choose a positive number, for example 2, for the 4th root of -16, obviously when you multiply 4 of those positive numbers together, 2 * 2 * 2 * 2, that answer is +16, not negative 16. If you choose a negative number to be the 4th root of -16, let's say -2, then -2 * -2 * -2 * -2 = 16, also not -16. So no real number can be the 4th root of -16.
With the 5th root of -32, let -2 be it. Then -2 * -2 * -2 * -2 * -2 = -32. Since you have an odd number of negative signs, that answer is indeed negative, so the 5th root of
-32 is a real number.

Answer 5

Oh, so even roots with a negative under are undefined and odd roots with a negative under can be simplified?

Answer 6

This makes so much more sense now! Thank you~~

Answer 7

To get real answers, you can take an odd or even index root of a negative number but only an even index root of a positive number.

Answer 8

I am so grateful for this.