# How come there is only one solution to $\sqrt{110-n} = n$

Given: $\sqrt{110-n} = n$

It follows that:

$110 - n = n^2$

$n^2 + n - 110 = 0$

$(n + 11)(n - 10)=0$

If $n = 10$, $\sqrt{110 - 10} = \sqrt{100} = 10$, it checks out.

If $n = -11$, $\sqrt{110 - (-11)} = \sqrt{121} = 11$, it doesn't

However!

I thought that $\sqrt{x}$ gives you two possible numbers, $+\sqrt{x}$ and $-\sqrt{x}$, so in the example above, $\sqrt{121} = -11$ is a valid root.

Can someone explain what's going on here? Why aren't they both valid?

• Note that your sentence: "I thought that $\sqrt x$ gives you two possible numbers $+\sqrt x$ and $-\sqrt x$", is not correct. $\sqrt x$ has 2 solutions $(+y)$ and (or) $(-y)$ where $y^2 = x$. Sep 12, 2012 at 4:44

When you squared, you introduced a second solution. Consider a much simpler problem: $\sqrt 4=n$. This has the unique solution $n=2$. If you square both sides, however, you get $4=n^2$, which has two solutions, $n=2$ and $n=-2$. The second is not a solution of $\sqrt4=n$, because the square root symbol by convention means the non-negative square root: when you use it, you’re implicitly choosing one of the solutions to the squared equation. Similarly, there are two integers $n$ with the property that $110-n=n^2$, $10$ and $-11$, but when you write $n=\sqrt{110-n}$, you’re forcing $n$ to be non-negative and thereby eliminating one of the two solutions to $110-n=n^2$.
It just has to do with the fact that $(-n)^2 = (n)^2$ does not imply that $-n = n$.
The $\sqrt{n}$ can only be positive by definition.
You can pinpoint the error by working backwards, finding the first equation which doesn't hold for the value $\rm\:n = -11.\:$ You'll find that all equations except the first hold true for $\rm\:n= -11,\:$ so your inference from the first to the second equation must be erroneous. The problem is that you applied to the first equation a non $1$-to-$1$ operation (squaring), which generally enlarges its solution set, introducing extraneous solutions.
This method of isolating errors works quite generally. It is especially useful when debugging proofs on abstract objects, since the error may become simpler to locate after specializing to more concrete objects. Above, evaluating all equations at $\rm\,n=-11\,$ has the effect of specializing the arithmetic of polynomial and algebraic functions (in $\rm\,n)\,$ to the simpler arithmetic of integers, making the error much easier to locate. See here for further discussion.