Question: This is going to be extremely elementary given the caliber of questions being posted.
I was going over something basic and think I'm overlooking an extremely fundamental rule.

$$2x-1 = \sqrt{6x+1}$$

By inspection, the solution is x=5/2. However, to solve algebraically, say I square both sides.

That leaves me with.

$$4x^2 - 4x + 1 = 6x + 1$$ 
$$\implies 4x^2 - 10x = 0 \implies x(2x-5)=0$$ 

It would appear $x=0$ is now present along with the correct answer.

If someone could clarify for me why squaring out the squareroot sign adds a new (unwelcome) answer I would greatly appreciate. Especially since $x=0$ is not out of the domain of the original equation $x> -\frac 16$, which definitely includes $x=0$.

Is it because my original equation has a power of $1$, which means I have one solution, however, when I square it, I'm essentially adding a new "x" and upping the power to $2$, giving me $2$ solutions. Thus, if I were to keep upping the power, there would be more repeated solutions at $x=0$? Which would mean, in some senses to solve those equations for $x$, dividing both sides by $x$ to reduce it back to the original power of $1$ would mean $x$ cannot indeed be $0$ as a solution.

Regards.

Answer: 
Since the square of $2x-1$ is the same as the square of $1-2x$, squaring your initial equation you introduce also the solution of the equation
$$
1-2x=\sqrt{6x+1}
$$

In another way:
The equation
$$
2x-1=\sqrt{6x+1}
$$
require that $2x-1\ge 0$ because the square root is always a positive number, so the solution must be such that $x \ge \frac{1}{2}$ and this inequality selects the correct solution of the squared equation.
You do not have bi-implication all the way through. More correctly, you obtain
$$2x-1 = \sqrt{6x+1}$$
$$\Rightarrow 4x^2 - 4x + 1 = 6x + 1$$
$$\Leftrightarrow x(2x-5)=0$$ 

This only yields, that the solutions to first equation is a subset of the solutions to the last equation. In this case, you will see it is actually a strict subset.