easy ratio question, need help please.

If I understand this correctly, you are saying that the rectangle with lenght x and width 1 can be partitioned into two unit squares and a "scaled down" rectangle similar to the orginal one. That is, the "scaled down" rectangle has sides of length $\alpha x$ and $\alpha$ for some $0< \alpha< 1$ . The we must have $\alpha x= 1$ , the width of the original rectangle and [tex]\alpha= x- 2[tex]. Putting that value of $\alpha$ into the first equation, $\alpha x= (x- 2)x= x^2- 2x= 1$ . That also gives your ratio, $\frac{x}{1}= \frac{1}{x- 2}$ Completing the square by adding 1 to both sides, $x^2- 2x+ 1= (x- 1)^2= 2$ , $x- 1= \pm\sqrt{2}$ so that $x= 1\pm\sqrt{1}$ . Since x, as the lenght of a side of a rectangle, must be positive, $x= 1+ \sqrt{2}$ , your first solution, is the only solution.

"Taking one square away" we still have $\alpha x= 1$ but now $x= 2+ \alpha$ or $\alpha= x- 2$ which reduces to the same thing.

Your "taking one square away" seems to be starting with a completly different problem in which the original rectangle is made up of only one square together with the "scaled down" rectangle.