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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.

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    $\begingroup$ If we start with either $a=b$ or $a=-b$ and square, we get $a^2=b^2$. Thus solutions to the latter include the solutions of BOTH of the former equations. Indeed, $x=0$ is a perfectly good solution to $2x-1 =-\sqrt {6x+1}$ $\endgroup$ – lulu Jun 10 '16 at 21:43
  • $\begingroup$ One might also take the viewpoint that algebra doesn't care about our arbitrary choice that $\sqrt{u}$ means the nonnegative square root of $u$, and gives us all the square roots. When $x = 0$, we do have that the LHS is a square root of $1$. This is basically what lulu said, I just like anthropomorphism :) $\endgroup$ – pjs36 Jun 10 '16 at 22:00
  • $\begingroup$ Fundamentally, it's because $x \mapsto x^2$ is not injective on the domain $\mathbb{R}$. $\endgroup$ – MathematicsStudent1122 Jun 11 '16 at 2:30
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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.

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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.

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The square root is an example of an operator that returns two results. You can have the positive square root or the negative square root. Squaring both sides caused the negative square root to be added to the solution set. You must therefore limit your domain to the positive square root and reject 0 as an answer.

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