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.