Answer:
-2
Step-by-step explanation:
[tex]\sqrt[n]{a}=b\iff b^n=a\\\\\sqrt[n]{a^n}=a\\\\\sqrt[5]{-a}=-\sqrt[5]{a}\\\\========================\\\\\sqrt[5]{-32}=\sqrt[5]{-2^5}=-\sqrt[5]{2^5}=-2[/tex]
[tex]\text{If is}\ 5\sqrt{-32},\ \text{then:}\\\\5\sqrt{-32}=5\sqrt{(16)(2)(-1)}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=5\cdot\sqrt{16}\cdot\sqrt2\cdot\sqrt{-1}\qquad\text{use}\ i=\sqrt{-1}\\\\=5\cdot4\cdot\sqrt2\cdot i=20\sqrt2\ i[/tex]
