Here's an elementary question on solving the following quadratic equation (well, it's not a quadratic until the square root is eliminated):

$$\sqrt{x+5} + 1 = x$$

Upon solving the above equation either using the method of factoring or the quadratic formula (after squaring both sides) you get $x = 4$ and $x = -1$. If you plug in $x = 4$ in the original equation, it checks out. However $x = -1$ doesn't work. You'll end up getting $3 = -1$ which is not true (in other words the LHS does not equal the RHS).

Is this still considered a solution/root of this particular equation? Does it have a special name?

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    $\begingroup$ I am baffled by the idea that one solution is better than the other. x=4 gives you a false equation if you choose the negative square root, and x= -1 gives you a false equation if you choose the positive square root. And vice versa. There is absolutely no algebraic reason to prefer one solution over the other. The idea that the positive square root is the one that "counts" is just some nonsense that a committee came up with for a junior high school textbook. It has no mathematical significance. $\endgroup$ Sep 2, 2011 at 1:08
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    $\begingroup$ @Marty: Choosing the positive square root for $\sqrt{a}$ (when $a$ is a positive real number) has some advantages and is not totally arbitrary. For example, it makes $\sqrt{a} \sqrt{b}=\sqrt{ab}$ true when $a$, $b$ are positive real numbers. If you chose the negative square root instead, you'd have to introduce a negative sign into that equation. It's like using the radian measure for trig functions or $e$ for the base of the natural logarithm: It simplifies things down the road. $\endgroup$
    – Ted
    Sep 2, 2011 at 4:04
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    $\begingroup$ @Marty: It's not that the positive square root is the only one that "counts", but that we need some definite way to refer to the the positive square root. The diagonal of a unit square is $\sqrt{2}$ and not $-\sqrt{2}$. The cosine of 36° is $(1+\sqrt{5})/4$ and not $(1-\sqrt{5})/4$. $\endgroup$ Sep 2, 2011 at 8:58
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    $\begingroup$ @Marty: It's not "some nonsense" that "a committee came up with for a junior high school textbook" (which is something which, to be frank, makes you sound more than a bit silly). It's a choice that needs to be made if you want to have a function, and your personal distaste is simply no match for just how mindbogglingly useful it is to have a well-defined, single valued function for this operation. $\endgroup$ Sep 2, 2011 at 13:09
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    $\begingroup$ @Marty: Besides the example already given to you by Dave Radcliffe? Suddenly you cannot express the length of the diagonal of a square in terms of the length of its side, or of the hypothenuse in terms of the other two sides. You cannot express solutions to certain equations. You cannot express values of useful functions like sine and cosine unambiguously. Etc. Etc. Etc. But, hey, if you don't use it and don't care, that's fine. Just don't pretend that it is all a conspiracy to annoy you and confuse poor innocent students. $\endgroup$ Sep 2, 2011 at 18:09

8 Answers 8


Generally speaking, the problem arises because squaring is not a "reversible" operation. That is, while it is true that if $a=b$ then $a^2=b^2$, it is not true that if $a^2=b^2$ then $a=b$. (For instance, even though $(-1)^2=1^2$, it does not follow that $-1=1$)

This is in contrast to other kinds of equation manipulations that we use routinely when we solve equations. For example, if $a=b$, then $a+k=b+k$, and conversely: if $a+k=b+k$, then $a=b$. So we can add to both sides of an equation (for instance, you can go from $\sqrt{x+5}+1 = x$ to $\sqrt{x+5}=x-1$ by adding $-1$ to both sides) without changing the solution set of the equation. Likewise, we can multiply both sides of an equation by a nonzero number, because $a=b$ is true if and only if $ka=kb$ is true when $k\neq 0$. We can also take exponentials (since $a=b$ if and only if $e^a=e^b$) and so on.

But squaring doesn't work like that, because it cannot be "reversed". If you try to reverse the squaring, you run into a rather big problem; namely, that $\sqrt{x^2}=|x|$, and is not equal to $x$.

So when you go from $\sqrt{x+5} = x-1$ to $(\sqrt{x+5})^2 = (x-1)^2$, you are considering a new problem. Anything that was a solution to the old problem ($\sqrt{x+5}=x-1$) is still a solution to the new one, but there may be (and in fact are) things that are solutions to the new problem that do not solve the old problem.

Any such solutions (solutions to the new problem that are not solutions to the original problem) are sometimes called "extraneous solutions". Extraneous means "coming from the outside". In this case, it's a solution that comes from "outside" the original problem.

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    $\begingroup$ Specifically, when you go from $\sqrt{x+5} = x-1$ to $\left(\sqrt{x+5}\right)^2 = (x-1)^2$ the solutions to the new problem that are not solutions to the old problem are solutions to $\sqrt{x+5} = -(x-1)$. Indeed, $x=-1$ satisfies $\sqrt{x+5} = \sqrt{4} = 2$ and $-(x-1) = -(-2) = 2$. $\endgroup$ Sep 2, 2011 at 8:50
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    $\begingroup$ Once you square, you can no longer know if you started with x or -x; Because squaring is not injective, information is "lost". $\endgroup$ Aug 15, 2016 at 7:33

It is often called an extraneous solution, or extraneous root. And it is not a solution of the original equation. This can be kind of confusing, since a black cat is still a cat. Think of an extraneous solution as a fake solution. (Thanks to the comment by Rahul Narain for this last formulation.)

The term extraneous solution occurs mainly in secondary school mathematics.

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    $\begingroup$ your black cat analogy doesn't make any sense either! {no offence} $\endgroup$
    – alok
    Sep 1, 2011 at 19:29
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    $\begingroup$ He just means that usually if we put an adjective on a noun, that does not change the essential nature of the noun. So a "red car" is a car, a "blue shirt" is a shirt, a "close friend" is a friend. But here, an "extraneous solution" is not a solution (that would be like saying that a "yellow hat" is not actually a hat). $\endgroup$ Sep 1, 2011 at 19:35
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    $\begingroup$ None taken! The point was that when one calls it an extraneous solution, that makes it sound as though it is a special kind of solution. Just like a black cat is a special kind of cat. However, what is called an extraneous solution is in fact not a solution. It should really be called an (extraneous) non-solution. $\endgroup$ Sep 1, 2011 at 19:37
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    $\begingroup$ "A white horse is not a horse" $\endgroup$
    – Mitch
    Sep 1, 2011 at 19:42
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    $\begingroup$ But a forged Picasso is not a Picasso, imitation crab is not crab, and so on... :) $\endgroup$
    – user856
    Sep 1, 2011 at 19:45

The square root sign always means the positive square root, so the only solution is $x=4$. You introduced the 'phantom' solution when you squared both sides.

  • $\begingroup$ I've heard of something called extraneous solution.What does that mean anyways ? the word solution in it bothers me! and what is this 'phantom' solution? $\endgroup$
    – alok
    Sep 1, 2011 at 19:25
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    $\begingroup$ "phantom" and "extraneous" would be synonymous in this context, @alok. $\endgroup$ Sep 1, 2011 at 22:29
  • $\begingroup$ Clearly $2 \ne -2$. But when you square both sides you get $4=4$. Sometimes squaring both sides turn false statements into true statements. So, even though you got an answer, your answer may not work for the original statement. $\endgroup$ Jul 27, 2021 at 22:05

Squaring both sides preserves equality (thus, if $a = b$, then $a^2 = b^2$), but squaring might not preserve inequality (for example, $2 \neq -2$, but their squares are equal). The "preserves equality" property means that you won't lose any solutions by squaring, but the "might not preserve inequality" means that you might gain solutions (i.e. the squared equation might have solutions that the pre-squared equation doesn't have).

In your case, $x = -1$ happens to be a situation where an inequality becomes an equality after squaring. In fact, when $x = -1$, then $\sqrt{x+5} = 2$ and $x - 1 = -2.$ Thus, $x = -1$ is not a solution to the pre-squared equation (because $2$ is not equal to $-2$), but $x = -1$ will be a solution to the squared equation (because the square of $2$ is equal to the square of $-2$). It's for this reason that you're told in high school algebra (or you should have been told) that you must check all solutions if at some point you squared both sides of an equation. Google (together) the phrase "extraneous solution" and "radical".


Added: Explanation of why squaring both sides of an equation may generate new solutions, called extraneous solutions, which are not solutions of the original equation.

Since $$A^{2}-B^{2}=(A-B)(A+B),$$ the equation $$A^{2}-B^{2}=0\Leftrightarrow A^{2}=B^{2}$$ means that $A=B$ or $A=-B.$ The roots of the equation $A^2=B^2$ are the roots of the equation $A=B$ and the roots of the equation $A=-B$. Since in general the roots of equation $A=-B$ are different from the roots of the equation $A=B$, when we pass from $A=B$ to $A^2=B^2$ we may generate other solutions.

This generalizes to $$ A^{n}-B^{n}=(A-B)(A^{n-1}+A^{n-2}B+\ldots +B^{n-1}). $$

Thus $$ \begin{equation} A^{n}=B^{n}\Leftrightarrow A^{n}-B^{n}=0 \tag{1} \end{equation}$$

is equivalent to $$ \begin{equation} A-B=0\Leftrightarrow A=B \tag{2} \end{equation}$$

or $$ \begin{equation} A^{n-1}+A^{n-2}B+\ldots +B^{n-1}=0 \tag{3} \end{equation}$$

The roots of equation $(1)$ are the roots of equation $(2)$ and the roots of equation $(3)$. In general the roots of equation $(3)$ are different from the roots of equation $(2)$. So when we pass from $(2)$ to $(1)$ we may generate new solutions that are not solutions of the original equation.

Theorem. Raising both sides of an equation $A=B$ to the $n^{\text{th}}$ power yields a new equation $A^n=B^n$, which has all the solutions of the given equation and may admit other solutions.


In short, applying a non $1$-to-$1$ operation like $\rm\:x\to x^2\:$ to an equation may enlarge its solution set, i.e. it may introduce extraneous solutions. For example, by squaring $\rm\: x\: =\: 1\:$ $\:\Rightarrow\:$ $\rm\:x^2 =\: 1\:$ which, besides the correct solution $\rm\:x=1\:,\:$ also yields the extraneous solution $\rm\:x = -1\:.\:$ The extraneous solution arises because the operation of squaring is not $1$-to-$1$, i.e. $\rm\:x^2 = y^2\:\: \not\Rightarrow\: x=y\:.\:$ Thus, although the inference $\rm\:x = y\:\Rightarrow\: x^2 = y^2\:$ is true, the reverse implication is not generally true. So any solutions obtained by applying this inference need not necessarily satisfy the initial equation.

But adding the same quantity $\rm\:c\:$ to both sides of an equation, or cancelling a nonzero quantity $\rm\:c\:$, are $1$-to-$1$ operations so they always preserve solution sets. More precisely we have the following equivalences $\rm\ x+c\: =\: y + c\:\iff\:x = y\:,\:$ and if $\rm\:c\ne 0\:$ then $\rm\: c\ x\: =\: c\ y\:\iff\: x = y\:$ (beware that the latter is true only in fields or domains, e.g. modulo $\:6:\ $ $\:2\cdot 3\ =\ 2\cdot 0\:,\:$ but $\rm\:3 \ne 0\:)\:.$

So, in the process of solving an equation, if you ever apply any non $1$-to-$1$ operation, then, to filter out extraneous solutions, you must explicitly check that the "solutions" satisfy the initial equation. A convenient way to to keep track of these matters is to explictly notate the equivalences as you apply each transformation to the equation - just as I did above with the arrows. Then, after obtaining your candidate solutions, glance back at the chain of arrows. If any arrow is not bidirectional then you must verify that each candidate solution does satisfy the initial equation (or that it satisfies some equivalent equation, e.g. any equation before the first $\:\Rightarrow\:$ in the chain).


I don't know if it have a special name, but it is not a solution of the original eqution. It is only a solution of $$-\sqrt{x+5}+1=x$$ which was introduced when squaring.


I discovered two more equations for which you can obtain extraneous solutions.Here they are,$$ln(x-2)+ln(2x-3)=2ln(x)$$ for which you get two values of $x$,i.e$x=6$,$x=1$ which is an extraneous solution. And the other one is $$e^{2x}-e^{x}-2=0$$ you get two values upon solving for $e^{x}$,one being $x$=ln2 and the other one being $e^{x}=-1$ which does not satisfy the given equation.


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