Question

find the root of \\(\sqrt5{-3125}\\).

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

\\(\bigcirc\\) a. \\(\sqrt5{-3125} = \square\\) (simplify your answer.)
\\(\bigcirc\\) b. the root is not a real number.

Explanation:

Step1: Recall the property of odd roots

For an odd root \( \sqrt[n]{a} \), if \( n \) is odd, then \( \sqrt[n]{-a}=-\sqrt[n]{a} \). Here, \( n = 5 \) (odd) and \( a = 3125 \).

Step2: Find the 5th root of 3125

We know that \( 5^5=5\times5\times5\times5\times5 = 3125 \). So, \( \sqrt[5]{3125}=5 \).

Step3: Apply the odd root property

Using the property \( \sqrt[5]{-3125}=-\sqrt[5]{3125} \), and since \( \sqrt[5]{3125} = 5 \), we get \( \sqrt[5]{-3125}=- 5 \).

Answer:

A. \( \sqrt[5]{-3125}=\boxed{-5} \)