We have the following system:
\(a+b=x\) (1)
\(ab=x\) (2)
\((a-b)(a+b)=x\) (3)
Substituting (1) into (3), we get: \((a-b)x=x\)
Dividing both sides by x, we find \(a-b=1\)
Solving for \(a\) in terms of b, we find: \(a=b+1\)
Substituting this into (2), we get: \(b(b+1)=x\)
Simplifying, we find: \(b^2+b=x\)
Substituting this in (3), we get: \(a^2-b^2=b^2+b\)
We also know that \(a = b+1\)
Solving this, we find: \({a={ {3 + \sqrt5} \over 2}}\) and \({b={{1 + \sqrt5} \over 2}}\)
However, these are interchangeable, so b can be either of these.