Is the $\pm$ sign used when finding the root of a negative number?

If $$\sqrt{64}$$ is equal to $$\pm{}8$$, is $$-64$$ equal to $$\pm{}8i$$, or just $$8i$$?

• How do you define $\sqrt{\cdot}$? Do note that there is a difference between $\sqrt{a}$ and a number being a square root of $a$. Generally, $\sqrt{\cdot}$ is used to denoted the non-negative root function root, in which case $\sqrt{-64}$ meaningless. – Brian May 11 at 22:45

$$\sqrt{a}$$, for real $$a$$, is almost always defined to be the positive solution of the equation $$x^2 - a = 0$$, so $$\sqrt{64}$$ is $$8$$ and not $$\pm 8$$. The reason the square root only takes on one value is because it is a function and so each element in the domain can be mapped to at most one element in the codomain.
As for your second question, $$i$$ is defined to as a number satisfying the equation $$i^2 + 1 = 0$$ and so we can say that $$\sqrt{-1} = i$$.
Following from this, we have $$\sqrt{-64} = \sqrt{-1} \sqrt{64} = i\sqrt{64} = 8i$$.
• Is $i$ "positive"? – Deepak May 11 at 22:47
• This answer is slightly misleading since $i$ is not defined to be equal to $\sqrt{-1}$ (which is meaningless), but rather to be a number such that $i^2 = -1$. – Brian May 11 at 22:50
• Actually, there are two numbers whose square is $-1$; neither of them is "positive" or "negative", but one of them is called $i$ (or $+i$) and the other is called $-i$. Every nonzero (real or complex) number has two square roots. Just as $x^2-64=(x-8)(x+8)$, also $x^2-(-64)=(x-8i)(x+8i)$. – bof May 11 at 23:27
Well, in theory, the square root of any number should return both its negative and positive root. Meaning $$\sqrt{x^2}=\pm x$$. But if you think about the geometric meaning of the square root, it’s finding the side length which makes a square of area $$x^2$$. So some people think that the square root should only be a positive answer since length cannot be negative. In the case where we do want both the negative and positive root, here’s what you can do. $$\sqrt{-16}=\sqrt{-1} \cdot \sqrt{16}$$ $$\sqrt{-1}=i$$, $$\sqrt{16}=\pm 4$$, which gives us $$i \cdot \pm 4=\pm 4i$$