How do i simplify? problem attached

Question

anonymous · Answer

$$(\sqrt{-2})^9$$

anonymous · Answer

i would like someone to show me how to work it out rather than giving me the answer

anonymous · Answer

Well if you are rooting the a negative number, you'll need the imaginary root of negative 1. Remember we can't take the square root of negative numbers, so the root of a negative number would need to be an imaginary number.

$$so \sqrt{-2} = \sqrt{2*-1} = \sqrt{2}\sqrt{-1}$$

we can call the square root of -1 as (i)
$$= \sqrt{2} (i)$$

now we take this to the 9th power. We can write it as:

$$[\sqrt{2}(i)]^2*[\sqrt{2}(i)]^2*[\sqrt{2}(i)]^2*[\sqrt{2}(i)]^2*\sqrt{2}(i)$$

now root 2 squared is just 2, and (i) squared is just -1 so:
$$2(-1)*2(-1)*2(-1)*2(-1)*\sqrt{2}(i)$$

that simplifies to:

$$-2*-2*-2*-2*\sqrt{2}(i) = 16*\sqrt{2}(i)$$

$$= 16\sqrt{2}(i)$$

anonymous · Answer

awesome!