Suppose you have two equtaions: $$2xy + y^2 = 0$$ $$x^2 + 2xy + 1 = 0$$ Subtracting the second from the first yields $y^2 - x^2 - 1 = 0$. Isolating y, we discover that $y = \pm\sqrt{x^2 + 1}$. However, by inspection we can wee that the $2xy$ term in both equations must be negative, which means that a single value of x cannot have both a positive and negative corresponding y-value. (i.e. if x is positive, then y must be negative). It seems that the equation $y = \pm\sqrt{x^2 + 1}$ contains an "extraneous root", but I'm struggling to wrap my mind around how that could be. After all, I solved for y w.r.t. x without squaring both sides. Could anyone help me understand what is going on?

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The problem is that your reasoning isn't reversible. Your two equations together imply $y^2-x^2-1=0$, but the converse is not true: that equation does not imply your original system.

Compare your question to the following argument. The system $$x+y=1$$ $$2x+y=1$$ yields, by subtraction, that $x=0$. But this second equation allows $y$ to be arbitrary! So we get a whole lot of "extraneous solutions": $(0,y)$, for any $y$. The problem, of course, is that the equation $x=0$ does not, in turn, imply the original system. I take it you do not find this situation puzzling.

You generally lose information when you replace a system of equations with a linear combination of them. The combination will include the solutions of your original equation, but it likely will include non-solutions, as well.

The form of the reasoning is: "any solution of the original equations will be a solution of this equation, too." Yes, but then you've only found a superset containing the solution set. To characterize the solution set exactly, you must worry about the converse.


When you solve a system of equations (whether it be by elimination or other means), you cannot discard the individual equations you started with. The reason is that each one of these equations carry more information (restrictions for example) than the one equation you end up with.

This goes for other example as well. The function $f(x) = \sqrt{x} - \sqrt{x}$ is not equal to $0$. The reason being any $x < 0 $ is not in the domain of $f$. By eliminating $\sqrt{x}$ , you have lost a key piece of information about this function (its domain).

  • $\begingroup$ Of course you can discard the original equations -- so long as the new ones are equivalent. $\endgroup$ – symplectomorphic Feb 19 '16 at 2:39
  • $\begingroup$ @symplectomorphic I suppose a better choice of words would have been "you should not simply discard". But yes, I agree with you. $\endgroup$ – GaussTheBauss Feb 19 '16 at 2:45
  • $\begingroup$ Everything written here is true, but I'm not convinced it applies here; the domain of $y=\pm\sqrt{x^2+1}$ is all $x$ and this is indeed equivalent to $y^2-x^2-1=0$. $\endgroup$ – Milo Brandt Feb 19 '16 at 2:47
  • $\begingroup$ @MiloBrandt True. But the domain of solution of $2xy+y^2=0$ is not all of $\mathbb{R}^2$. $\endgroup$ – GaussTheBauss Feb 19 '16 at 2:54

I think the issue here is at least partially that you're assuming the rest of the solution will go something like:

We know that $y=\pm\sqrt{x^2+1}$. We'll use this to find the possible values for $x$, then substitute in for $y$.

and this is missing something: When we go to solve for $x$, we're going to need to know whether $y$ is positive or negative. That is, we need to split into cases $y=\sqrt{x^2+1}$ or $y=-\sqrt{x^2+1}$ and then solve for $x$ by substituting in $y$. This is to say that when you write:

After all, I solved for y w.r.t. x without squaring both sides.

you're missing something. You didn't solve for $y$. You reduced one statement to saying that $y$ was among two possibilities. You have to proceed to the end of the argument with each possibility separate, since you can't substitute "either this or that" in for $y$.

To be sure, what elimination does is we start with a system like $$2xy+y^2=0$$ $$x^2+2xy+1=0$$ And then you've arranged into an equivalent statement: $$y=\pm\sqrt{x^2+1}$$ $$x^2+2xy+1=0$$ - or whatever you take the second equation to be. This system is equivalent to the first since all of your steps are reversible (noting that you've correctly used $y=\pm\sqrt{z}$ as the equivalent to $y^2=z$). So, you aren't introducing extraneous roots - it's just that your formula for $x$ depends on the $\pm$, so you don't get to choose $x$, then choose the sign of $y$. Another way to say this is that your error is in this statement:

A more simple example might be a system like: $$y^2=1$$ $$x+y=0$$ You get that $y=\pm 1$, but you can't really substitute that into the second equation itself since it depends on a variable $\pm$.

  • $\begingroup$ Thanks for your reply. You are correct in that my intention was to continue the solution using a two-case substitution. You say that I have reduced to the point where y is "among two possibilities." The issue I'm having is I thought that $y = /pm/sqrt(x^2 + 1)$ implies that y is both of the two possibilities. As in, if x = 1 were a "possible value for x", then $y = /pm/sqrt(x^2 + 1)$ would imply that $(1, /sqrt(2))$ and $(1, -sqrt(2))$ were both solutions. This is clearly not the case in this system of equations, which is why I thought that it was an extraneous root. $\endgroup$ – Cam Feb 19 '16 at 3:32
  • $\begingroup$ @Cam Right; a better interpretation is "$y$ is one of these things" rather than "$y$ is both." It's maybe somewhat different from the one variable case (where $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ does mean both solves $ax^2+bx+c=0$) - the trick is integrating such things into further steps correctly. You might have some gains to think about the system $y^2=1$ and $x^2+y=-1$. It has only one (real) solution and is altogether quite well behaved, but demonstrates what's going on. $\endgroup$ – Milo Brandt Feb 19 '16 at 3:45
  • $\begingroup$ @Cam: I've answered the question directly in my answer. The solution set of the equation $y^2-x^2-1=0$ contains the solution set of your original system. This is not surprising: in combining the equations, you've lost information. $\endgroup$ – symplectomorphic Feb 19 '16 at 3:55

You have taken a square root. Whenever you take a square root you can introduce an extraneous root. Even for the simple system $x^2=y^2$, $x=y$ if you took a square root of the first equation you'd get an extraneous root. You must always consider all the equations as they all contain information about the solution. If you had gone about solving it via elimination or substitution (rather than merging the two equations together randomly like you did) then you would have not encountered this.

If you solve your equation fully (and use complex numbers) you get two pairs of roots: $$\left(\frac{1}{\sqrt3},-\frac{2}{\sqrt3}\right), \left(-\frac{1}{\sqrt3},\frac{2}{\sqrt3}\right),(i,0),(-i,0)$$ The second pair of roots does fit your reduced equation (with a small rearrangement), namely $x=\pm\sqrt{y^2-1}$.

  • 1
    $\begingroup$ -1 for "merging the two equations together randomly like you did." $\endgroup$ – symplectomorphic Feb 19 '16 at 2:33
  • $\begingroup$ @symplectomorphic If you dislike the phrase please edit it to something more suitable. The OP has joined the two equations together with no clear intend so to me it looks random. $\endgroup$ – Ian Miller Feb 19 '16 at 2:36
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    $\begingroup$ my objection is not the English, and the operation of eliminating the cross terms is obviously not random: it facilitates solution-finding. $\endgroup$ – symplectomorphic Feb 19 '16 at 2:37

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