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:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$

You have proceeded wrongly. We have $$3^x +3^{-x} =1 $$ $$\Rightarrow 3^{2x} +1 =3^x $$ $$\Rightarrow 3^{2x} -3^x +1=0$$ giving us $$3^x = \frac {1\pm \sqrt {3}i}{2}$$
Notice that the LHS is always real but we have an imaginary part on the RHS. This is enough evidence to suggest that the above equation has no solutions. Hope it helps. 
