5
$\begingroup$

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.

$\endgroup$
3
  • 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
    Commented 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
    Commented Jun 10, 2016 at 22:00
  • $\begingroup$ Fundamentally, it's because $x \mapsto x^2$ is not injective on the domain $\mathbb{R}$. $\endgroup$ Commented Jun 11, 2016 at 2:30

4 Answers 4

8
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .