Answer: \(\displaystyle x=\frac32+\frac12\sqrt5\) or \(x=-1\).

Answer: \(\displaystyle x=\frac32+\frac12\sqrt5\) or \(x=-1\).

For

\((\sqrt5-1)x^2-2x-(\sqrt5+1)=0,\)

use the quadratic formula with

\(a=\sqrt5-1,\qquad b=-2,\qquad c=-(\sqrt5+1).\)

Then

\(x=\frac{2\pm\sqrt{(-2)^2-4(\sqrt5-1)(-(\sqrt5+1))}}{2(\sqrt5-1)}.\)

The discriminant is

\(4+4(\sqrt5-1)(\sqrt5+1)=4+4(5-1)=20.\)

So

\(x=\frac{2\pm2\sqrt5}{2(\sqrt5-1)} =\frac{1\pm\sqrt5}{\sqrt5-1}.\)

For the positive sign,

\(x=\frac{1+\sqrt5}{\sqrt5-1}.\)

Rationalising,

For the negative sign,

\(x=\frac{1-\sqrt5}{\sqrt5-1}=-1.\)