I was working on radical equations and I came across a few problems where I got answers that worked when I checked, but were not listed as solutions. My teacher's only explanation was, "just because." Here is one problem where the only solution is $1$.


How I solved it




$x= \{-2, 1\}$

Then plugging both numbers back in, I get

$1 = \sqrt{2-1}$

$1 = \sqrt{1}$

$1 = 1$


$-2 = \sqrt{2--2}$

$-2 = \sqrt{4}$

The square root of $4$ can be both $-2$ because $-2 \times -2 = 4$ and $2$. $1$ is the only solution listed and my teacher says that it's right.

What is the explanation for this? Why isn't $-2$ a solution for the problem?

  • 9
    $\begingroup$ "Just because" is a terrible response to anything math-related. $\endgroup$
    – anomaly
    Apr 17, 2015 at 15:20

5 Answers 5


This is due to notation. When we write $\sqrt{n}$ we mean only the positive square root of $n$. If we wished to include both negative and positive solutions, we would write $\pm\sqrt{n}$.

I know this can be irritating, but it is the convention that is used, since square root would not be a function if it gave multiple values.


The issue arises from the second line.

The statement: $ x = \sqrt{2-x} $

Is not equivalent to the statement: $ x^2 = 2 - x $

Think of it like this, although the first statement implies the second, the second does not imply the first (since you'd need to introduce the negative square root).

$ x=-2 $ Isn't actually a solution to your original equation, since you're only taking the positive square root. You just have to remember that when squaring, you can always introduce incorrect 'solutions' simply because A = B is not logically equivalent to A^2 = B^2.


When we use a radical, $\sqrt[n]{~~}$, we mean to take the principal $n^{th}$ root, which in the case of positive real numbers will always be a positive real number.

In other words $\sqrt{4} = 2$ and only $2$. $\sqrt{4}\neq -2$ even though $(-2)^2 = 4$

If we were to allow for multiple roots, we can use $(~~)^{\frac{1}{n}}$ to refer any element of the set of all $n^{th}$ roots, noting that this is a multivalued function (and thus not a function at all, since to be a function we require it have a single output for any given input).

Here we have $4^{\frac{1}{2}} \in \{2,-2\}$

In your specific example, $x = \sqrt{2-x}$, we have that $x$ being equal to the principle square root of something presumably positive, must itself also be positive, and so $x=1$ is the only solution (as shown by your algebra above).


The answer, as already the others has highlighted, is that $\sqrt{4} \neq -2$, but to aid you in the actual expression evaluation, consider that $\sqrt{x^2} \neq x$, but $\sqrt{x^2} = \left| x \right|$.


Omar already gave a good answer, but I wanted to expand on this part:

You just have to remember that when squaring, you can always introduce incorrect 'solutions' simply because A = B is not logically equivalent to A^2 = B^2.

If you have an equation, and you multiply both sides by another equation, you don't always get the same equation.

For example, $(x-2)=0$. If you multiply both sides by $(x+5)$, you'll get $(x-2)(x+5)=0$. The second equation has solutions $x=2,-5$, and the first has $x=2$ only.

In Omar's example, since $A=B$, he is multiplying both sides by $A$. In general, you can only multiply both sides by a non-zero constant to always preserve logical consistency. In your example, you're multiplying both sides by $(2-x)^{1/2}$. Since $(2-x)^{1/2}$ is not a non-zero constant, it does not always preserve logical consistency.

For example, you can multiply both sides by $4$, and it doesn't change the solutions to the equation. However, it will sometimes preserve the equation in something like the following example: $(x-4)^4=0$, if you multiply both sides by $(x-4)^2$, the result, $(x-4)^6=0$, will have the same set of solutions.

  • 1
    $\begingroup$ To go from $A=B$ to $A^2=B^2$ is just a property of equality, but getting from $A^2=B^2$ to $A=\pm B$ is not simply a matter of multiplying or dividing by $A$. The actual proof looks more like $$A^2-B^2=(A-B)(A+B)=0\implies A-B=0\lor A+B=0.$$ $\endgroup$ Apr 17, 2015 at 9:46
  • $\begingroup$ I don't know exactly what you mean by the action, "to go from". W.r.t. the two statements, $A=B$ and $A^2 = B^2$, they do not necessarily have the same solution set. $\endgroup$
    – user38858
    Apr 17, 2015 at 19:21
  • $\begingroup$ "To go from" here means that if $A=B$ is true, then $A^2=B^2$ is true, i.e. a forward implication. Note that this is actually the opposite order from that with which we solve equations; usually you start with the complicated equality and perform simplifications on it until you have a tautology. For example if you are asked to solve $A^2=B^2$ you can simplify it to $A=B$ and have a solution, since $A=B\implies A^2=B^2$. (You may not have found all solutions, but you have not introduced extraneous ones this way.) If you want to find all solutions, then all steps must be biconditional. $\endgroup$ Apr 17, 2015 at 19:53

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