Question

try to evaluate \\(\sqrt5{-32}\\) without using a calculator. enter dne if the number is not real. the value is \\(\boxed{}\\)

Explanation:

Step1: Recall the definition of nth root

For an odd root \( \sqrt[n]{a} \), if \( n \) is odd, then \( \sqrt[n]{-a}=-\sqrt[n]{a} \) (since odd powers of negative numbers are negative and vice versa). Here, \( n = 5 \) (which is odd) and we need to find \( \sqrt[5]{-32} \).

Step2: Find the 5th root of 32

We know that \( 2^5=2\times2\times2\times2\times2 = 32 \). So, \( \sqrt[5]{32}=2 \).

Step3: Apply the property for odd roots

Using the property \( \sqrt[n]{-a}=-\sqrt[n]{a} \) with \( n = 5 \) and \( a = 32 \), we get \( \sqrt[5]{-32}=-\sqrt[5]{32}=- 2 \).

Answer:

\( -2 \)