+981 Hey, Rollingblade, we meet again!
Here is my solution from the same problem I posted.
We can designate \(\sqrt{a+x}=y\), which means that:
\(\sqrt{a-y}=x\)
From this system, we can square both.
\(a+x=y^2\) and \(a-y=x^2\)
When we subtract the second equation from the first, we get:
\(x+y=y^2-x^2\)
This also is:
\(x^2-y^2+x+y=(x+y)(x-y+1)=0\)
From this, we have two cases:
Case 1: x + y = 0
In this case, y = -x and \(x^2-x-a=0\)
Case 2: y = x + 1 and \(x^2+x+1-a=0\)
Solving for x in both equations, we get.
\(x_{1,2}=\frac{1}{2}\pm\sqrt{a+\frac{1}{4}}\)
And
\(x_{3,4}=-\frac{1}{2}\pm\sqrt{a-\frac{3}{4}}\)
I hope this helped,
Gavin
