need help solving a quartic equation

Question

The question I am asking is to solve the equation $x^4-4x-1=0$, I need an exact answer. What I have done was found out that it equals $(x^2+1)^2 - 2(x+1)^2 =0$. Anybody help me, please?

Answer 1

So you know

$$\begin{align} x^4-4x-1&=(x^2+1)^2-2(x+1)^2\\ &=(x^2+1)^2-[\sqrt2(x+1)]^2. \end{align}$$

Factorise this difference of two squares to get two quadratic factors, hence the four roots.

Answer 2

HINT

Use $a^2-b^2=(a-b)(a+b)$ to get $2$ quadratic equations$(=0)$.