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.


  • 3
    $\begingroup$ 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, 2016 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, 2016 at 22:00
  • $\begingroup$ Fundamentally, it's because $x \mapsto x^2$ is not injective on the domain $\mathbb{R}$. $\endgroup$ Jun 11, 2016 at 2:30

4 Answers 4


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.


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.


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.


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. $$2x-1 = \sqrt{6x+1}$$ $$4x^2-4x+1=6x+1$$ $$4x^2-10x=0$$ $$x=\frac{10\pm\sqrt{10^2-4*4*0}}{2*4}=\frac{10\pm10}{8}=\frac{20}{8}=\frac{5}{2}=2.5\text{ or 0}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.