You can't have a negative number under a radical that has an even index. We have an even index since we are dealing with the square root. Because of that we have to get the imaginary i involved. [tex]i^2=-1[/tex]. Keeping that in mind, let's rewrite our problem: [tex] \sqrt{(-1)(8)} [/tex]. If -1 equals i-squared, we can sub that in. Also, since 8 = 4*2 and 4 is a perfect square, let's break that down at the same time: [tex] \sqrt{i^2(4)(2)} [/tex]. i-squared is a perfect square which can be pulled out as a single i, and 4 is a perfect square which can be pulled out as a 2. We will leave a 2 under the radical. Here's your simplification: [tex]2i \sqrt{2} [/tex], choice C from above.
