Basic clarification: Solving equation by squaring, but end up getting non-existent answers 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.
 A: 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. 
A: Suppose we assume that $x = 1$.
Squaring both sides, we now have $x^2 = 1$, which is undeniably true.
From this we can conclude that $x = 1$ or $x = -1$. This is true too, since a disjunction of a true statement and a false statement is true. But the disjunction has lost information which was present in the original statement: you no longer know which of the two statements is true.
The point is that a polynomial equation is like an exclusive disjunction: it tells you that $x$ takes on exactly one of $n$ (the degree of the polynomial) possible values, but it doesn't tell you which.
A: 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.
A: You had it right but you chickened out on one of solutions. When you got $x(2x-5)=0$, dividing both sides by $x$ leaves $\quad 2x-5=0\implies 2x=5 \implies x=2.5\quad $ or you can solve it iswith the quadratic equation.
$$2x-1 = \sqrt{6x+1}$$
$$4x^2-4x+1=6x+1$$
$$4x^2-10x=0$$
$$x=\frac{10\pm\sqrt{10^2-4*4*0}}{2*4}=\frac{10\pm10}{8}=\frac{20}{8}=\frac{5}{2}=2.5\text{ or 0}$$
