What is $x$, if $3^x+3^{-x}=1$? I came across a really brain-racking problem. 

Determine $x$, such that $3^x+3^{-x}=1$.

This is how I tried solving it:
$$3^x+\frac{1}{3^x}=1$$
$$3^{2x}+1=3^x$$
$$3^{2x}-3^x=-1$$
Let $A=3^x$.
$$A^2-A+1=0$$
$$\frac{-1±\sqrt{1^2-4\cdot1\cdot1}}{2\cdot1}=0$$
$$\frac{-1±\sqrt{-3}}{2}=0$$
I end up with 
$$\frac{-1±i\sqrt{3}}{2}=0$$
which yields no real solution. And this is not the expected answer.
I'm a 7th grader, by the way. So, I've very limited knowledge on mathematics.
EDIT
I made one interesting observation.
$3^x+3^{-x}$ can be the middle term of a quadratic equation:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$

Note that we have
$$
3^x+3^{-x}=(3^{x/2}-3^{-x/2})^2+2
$$
Apply this to your equation, and you get
$$
(3^{x/2}-3^{-x/2})^2=-1
$$
Which means that $3^{x/2}-3^{-x/2}$ is imaginary.
