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 $1$.
$x=\sqrt{2-x}$
How I solved it
$x^{2}=2-x$
$x^{2}+x-2=0$
$(x+2)(x-1)=0$
$x= \{-2, 1\}$
Then plugging both numbers back in, I get
$1 = \sqrt{2-1}$
$1 = \sqrt{1}$
$1 = 1$
and
$-2 = \sqrt{2–2}$
$-2 = \sqrt{4}$
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?
Best Answer
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.