Hfrhyu

I was struggling with this problem:

$$100^2=x^2+ left( frac100x100+x right)^2$$

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!

WA tell us that the positive root is
$$
50 (-1 + sqrt 2 + sqrt2 sqrt 2 - 1) approx 88.320
$$
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...

Thanks for all the colaboration. Due to @labbhattacharjee 's comment, i was able to solve the problem analytically. Solve the equation $x^2+frac9x^2(x+3)^2=27$ shows that, if we have an equation:

$$a^2+b^2=k$$

Then, calling $c=fracaba-b$, we can manipulate algebraically the expression above:

$$(a-b)^2+2c(a-b)-k=0$$

For our problem, let $a=x$, $b=frac100x100+x$ and $k=100^2$. It turns out that $c=100$ and $a-b=fracx^2100+x$. So, if we put $u=fracx^2100+x$, we will have:

$$u^2+200u-100^2=0$$

Which the only positive root is $100(sqrt2-1)$. Solving the equation:

$$fracx^2100+x=100(sqrt2-1)$$

The only positive root is $50(sqrt2-1+sqrt-1+2sqrt2)$.

Thanks!