Find the positive root of $100^{2}=x^{2}+ \left( \frac{100x}{100+x} \right)^{2}$

Question

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!

Answer 1

WA tell us that the positive root is $$ 50 (-1 + \sqrt 2 + \sqrt{2 \sqrt 2 - 1}) \approx 88.320 $$ WA also tells us that the minimal polynomial of this number over $\mathbb Q$ is $$ x^4 + 200 x^3 + 10000 x^2 - 2000000 x - 100000000 $$ and so there is no simpler answer.

On the other hand, substituting $x=50u$ in the minimal polynomial gives $$ 6250000 (u^4 + 4 u^3 + 4 u^2 - 16 u - 16) $$ Now this can factored into two reasonably looking quadratics: $$ u^4 + 4 u^3 + 4 u^2 - 16 u - 16 = (u^2 + (2 + 2 \sqrt 2) u + 4 \sqrt 2 + 4) (u^2 + (2 - 2 \sqrt 2) u - 4 \sqrt 2 + 4) $$ But all this is in hindsight...

Answer 2

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!