An equation is extraneous if (at least to my understanding) it has no valid solutions. The example my math teacher used was $\sqrt{x}=-1$, citing this proof

$$\sqrt{x}=-1 \\ x=1$$

They then stated that $\sqrt{1} \ne -1$, and therefor the equation is extraneous. While this wasn't initially a problem, I seemed to accept it for some reason, I now realize that $\sqrt{1} = \pm 1$, and therefor -1 is a square root of 1, so why is the initial equation extraneous?

Am I missing something major here or was my math teacher wrong?

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    $\begingroup$ The function $\sqrt{x}$, when used in the real numbers, is defined as the non-negative solution $y$ to $x=y^2$. Since $-1$ is not non-negative, $\sqrt{x}=-1$ won't ever happen in the real numbers. $\endgroup$ – vrugtehagel Apr 25 '17 at 20:24
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    $\begingroup$ it is $(\pm 1)^2 =1$ and not $\sqrt 1=\pm 1$ $\endgroup$ – Surb Apr 25 '17 at 20:25
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    $\begingroup$ We say solutions are extraneous. $\endgroup$ – Sean Roberson Apr 25 '17 at 20:27
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    $\begingroup$ It is a matter of convention. The usual convention when dealing with the real numbers is that $\sqrt{}$ denotes the nonnegative square root. On the other hand, in the context of complex numbers $\sqrt{}$ is often taken to be a multivalued function. $\endgroup$ – Robert Israel Apr 25 '17 at 20:31
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    $\begingroup$ Note that the square root of a number's square is equal to the absolute value of that number. That is why, when you have $x^2=1$, the answer is $x=\pm 1$. But when you square a number's square root, the result is always that number, and not its absolute value. $\endgroup$ – Franklin Pezzuti Dyer Apr 25 '17 at 20:32

We talk about an extraneous or spurious solution when we solve an algebraic equation by raising both sides to some power. The map $x\mapsto x^n$ is not injective, hence once the original problem has been reduced to finding the roots of some polynomial, it is not granted that every root of such polynomial is indeed a solution of the original equation. For instance $$ \sqrt{x-1} = 7-x \tag{1}$$ implies $$ x-1 = (x-7)^2 \tag{2}$$ and $$ p(x) = (x-7)^2-(x-1) = x^2-15x+50 = 0 \tag{3} $$ but while $x=5$ is an actual solution of $(1)$, the other root of $p(x)$, i.e. $x=10$, is a spurious solution, because it fulfills $(2)$ but not $(1)$.

It is enough to recall that the very definition of the square root function over the real numbers:

Def. $\sqrt{x}$ is the only non-negative real number whose square equals $x$.

In particular the maximal domain of the square root function is the set of non-negative real numbers, and over such set the square root function is non-negative. So $\sqrt{1}=1$, not $\pm 1$.

Over the set of complex numbers, for any $z\neq 0$ there are two opposite numbers whose squares equal $z$: in such context we write $\sqrt{z}=\pm w$ by meaning that both $w$ and $-w$ are roots of the polynomial $q(t)=t^2-z$, i.e. we regard $\sqrt{\cdot}$ as a multi-valued function: not a function, strictly speaking.


I now realize that $\sqrt{1} = \pm 1$

You are mistaken. The square root symbol $\sqrt{\phantom0}$ denotes the positive square root function. Therefore $\sqrt1 = 1 \neq -1.$

A point of confusion is that we often look at equations such as $$ x^2 = 4 $$ and observe that they have two solutions, $x = \pm 2.$ But this does not say that $\sqrt4$ is "equal" to $\pm 2$; the actual solution is $$ x = \pm \sqrt4, $$ where $x$ is an unknown. The $\pm$ sign tells us that $x$ might be $2$ or might be $-2.$ We need the $\pm$ sign in front of $\sqrt4$ to tell us that, because $\sqrt4$ by itself is always $2,$ never $-2.$


Basically, the symbol $\sqrt{x}$ means that, if $x$ is a non-negative real number, then $\sqrt{x}$ is the single non-negative real number which becomes $x$ when you square it. This is because $\sqrt{x}$ is more useful when it's a function, so each $x$ input needs to have at most one output.

However, when you use the square root to solve an equation, for example $x^2 = 1$, you need to recognize that undoing a square means you could have started with a positive or a negative number, because squaring either will make it positive. So, we say $x = \pm 1$ in this case, but $\sqrt{1}$ is still just $1$.

  • $\begingroup$ "the single real number that become x when you square it"? $\endgroup$ – qbert Apr 25 '17 at 20:42
  • $\begingroup$ @EthanBolker: Thanks for catching that, I said it in my head but it didn't come out of my fingers! $\endgroup$ – AlexanderJ93 Apr 25 '17 at 20:43
  • $\begingroup$ It should still be "$\sqrt{x}$ is the non-negative real number which..." $\endgroup$ – Christopher.L Apr 25 '17 at 20:44
  • $\begingroup$ @qbert: I simply mean that it is unique (there is only one) and that $\sqrt{x}^2= \sqrt{x}\sqrt{x} = x$. $\endgroup$ – AlexanderJ93 Apr 25 '17 at 20:44
  • $\begingroup$ @MartinArgerami : Sorry, yes, I deleted that comment now, since AlexanderJ93 had just updated his answer, and I had not catched that yet. $\endgroup$ – Christopher.L Apr 25 '17 at 20:48

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