What rule governs $x^4=10,000$ having complex solutions? Given
$$x^2=100$$
$$x=\pm 10$$
This is acquired by taking the square root of both sides. So given:
$$x^4=10,000$$
$$x=\pm 10$$
This is done by taking the fourth root of both sides. But, I miss two complex solutions:
$$x=\pm 10i$$
My question is how do I avoid making mistakes like this? In this instance, I did not know that there were complex answers as well. How was I supposed to recognize this? I know that looking at the value of the discriminant is one option but since problems like the former are so easy the latter problem seemed intuitive. Am I supposed to calculate the discriminant for each function like a paranoid madman or is there some general rule involving complex roots that I am unaware of?
 A: The Fundamental Theorem of Algebra states that a polynomial of $n^\text{th}$ degree has exactly $n$ complex solutions. Hence, the polynomial equation of degree $4$
$$x^4 - 10000 = 0$$
has exactly $4$ complex solutions, namely $\pm 10$ and $\pm 10i,$ which can be found via factoring (difference of squares) or realizing that $x^4 = 10000$ implies that $x^2$ must be $\pm 100$, then getting $4$ solutions that way by factoring.

Note that in your comment about the equation
$$x^2 = 100$$
taking the square root of both sides to get
$$x = \pm 10$$
is incorrect. The square root function gives only the principal, or positive root. That would leave you with just $x = 10$. Similar to how we rewrote $x^4 = 10000$ as $x^4 - 10000 = 0$ and factored, we must do $x^2 - 100 = 0$ and factor to get $x = \pm 10$.
A: Put yourself in the complex plane and rewrite the original equation as $z^4=10000, z\in \mathbb C$. Expressing $z$ as $re^{i\theta}$ leads to $r=10, \theta=\frac{k\pi}{2}, k\in \mathbb Z$. 
Thus, you have $x=10, 10i, -10$, or $-10i$.
(I assume you're familiar with some complex algebra, then you could think of these points as on a circle centered at $z=0$)
A: One way is to factor the whole thing down:
$$\begin{align} x^4 &= 10,000\\ x^4-10^4&=0\\ \color{blue}{(x^2+10^2)} \color{red}{(x^2-10^2)}&=0\\ \color{blue}{(x+10i)(x-10i)}\color{red}{(x+10)(x-10)}&=0\end{align}$$
A: We know that we have to look for $4$ solutions by Fundamental Theorem of Algebra including multiplicity.
$$x^4=10,000$$
$$x^4-10000=0$$
$$(x^2-100)(x^2+100)=0$$
$$(x-10)(x+10)(x+10i)(x-10i)=0$$
Another way is to solve this is 
$$x^4=10000\exp(2n\pi i)$$
$$x=10\exp\left(\frac{2n\pi i}{4}  \right), n=0,1,2,3$$
A: Here is a way of taking roots from both side, if you do not want to use factoring method.
$x^4=10000$
$\sqrt{x^4}=\sqrt{10000}$
$|x^2|=100$, $x^2=100$   or $-100$ 
Since a radical with even index has the assumption of the principal/positive root, you take the absolute value of $x^2$. 
$x^2=100$, $\sqrt {x^2}=\sqrt{100}$, $|x|=10$, $x=10, -10$
$x^2=-100$, $\sqrt {x^2}=\sqrt{-100}$, $|x|=10i$, $x=10i, -10i$
