The problem
Given the following figure
determine $x$.
My solution(s):
There are many possibilities to use similarities and the Pythagorean resulting more or less complicated systems of equations.
The simplest solution I could come up with was the following.
Because of the similarity of $AEF$ and $FCD$, we may say that $u=1/x$.
Based on the Pythagorean we have that $$(1+x)^2+(1+1/x)^2=100.$$
This equation has two positive real solutions. One of them is: $$x=\frac12\left(-1+\sqrt{101}+\sqrt{98-2\sqrt{101}}\right)\approx 8.937.$$ The other one is $\frac1{8.937}.$
(http://www.wolframalpha.com/input/?i=solve+(1%2Bx)%5E2%2B(1%2B1%2Fx)%5E2%3D100)
The question
The source of the problem above is an old (Hungarian) high scholl level problem book. So, I suspect that there must be a simpler solution not requiring the roots of a fourth order equations. Please, either prove that there is no simpler solution or show one.
I write down my thoughts here. If not satisfying you requirements, please inform me and I would delete this.
When solving $$ (1+x)^2 + \left(1 + \frac 1x\right)^2 = 100, $$ we first expand it to $$ x^2 + \frac 1{x^2} + 2 + 2x +\frac 2x = 100, $$ then notice that $$ x^2 +\frac 1{x^2} + 2 = \left(x + \frac 1x\right)^2, $$ we let $y = x + 1/x$, then $$ y^2 + 2y = 100, $$ which is easily solvable by the roots formula of quadratic equations. Then all knowledge we used is taught before college, and the question could be used as a high school level geometry exercise.