I am trying to solve when we have $a,b,c$ as just strictly positive numbers.
$2\sqrt{a-x}$ = $\sqrt{b-x}$+$\sqrt{c-x}$.
What is the approach solving for x?
In the actual proble, I was lucky to just plug in number, because I knew it would be a natural number, but if $x$ is a real number, I want to know how to solve it algebraically.
Squaring repeatedly both sides:
$4(a-x) = b+c-2x + 2 \sqrt{(b-x)(c-x)}$
$(4a-b-c-2x)^2 = 4(b-x)(c-x)$
Now you can solve this quadratic. In the end verify the solutions with original equation!
By making the both sides to the power $2$ and then some simple manipulations you have:
$x^2+(b+c)x+(\frac{2a-b-c}{2a})^2-bc=0$
and then: $$x=\frac{-(b+c)\pm\sqrt{(b+c)^2-4[(\frac{2a-b-c}{2a})^2-bc]}}{2}$$
