TITLE: Find the positive root of $100^{2}=x^{2}+ \left( \frac{100x}{100+x} \right)^{2}$
QUESTION [5 upvotes]: I was struggling with this problem:
$$100^{2}=x^{2}+ \left( \frac{100x}{100+x} \right)^{2}$$
It came up when i was developing a solution to a geometry problem. I've already checked in Mathematica and the solution is right, according to the answer. The answer needs to be a real positive number because it's a measure of a segment.
I've tried factoring, manipulating algebraically, but i couldn't solve the resulting 4th degree polynomial. I appreciate if someone could help me. Thanks!

REPLY [4 votes]: Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$ shows that, if we have an equation:
$$a^{2}+b^{2}=k$$
Then, calling $c=\frac{ab}{a-b}$, we can manipulate algebraically the expression above:
$$(a-b)^{2}+2c(a-b)-k=0$$
For our problem, let $a=x$, $b=\frac{100x}{100+x}$ and $k=100^{2}$. It turns out that $c=100$ and $a-b=\frac{x^{2}}{100+x}$. So, if we put $u=\frac{x^{2}}{100+x}$, we will have:
$$u^{2}+200u-100^{2}=0$$
Which the only positive root is $100(\sqrt{2}-1)$. Solving the equation:
$$\frac{x^{2}}{100+x}=100(\sqrt{2}-1)$$
The only positive root is $50(\sqrt{2}-1+\sqrt{-1+2\sqrt{2}})$.
Thanks!