Is there a name for this strange solution to a quadratic equation involving a square root?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.

Solution 4:

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".