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#1**+2 **

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

GYanggg May 2, 2018