Solving $x^4-2 x^2-x+1 = 0$ for $x$ How could one solve the following, giving the answer as a closed form, not as an estimation:
$$\text{Solve }x^4-2 x^2-x+1 = 0\text{ for } x$$
Where $x$ is $\text{real.}$
I found this one particularly hard. Any help will be very much appreciated.

As others have mentioned, there are no rational solutions but you can find the exact solution through the formula for a quartic function (though why one would need it is another question). The fact that the coefficient of $x^3$ is zero makes things a bit easier, but not by much.
For: $$a x^4 + c x^2 + d x + e=0$$ 
The solutions are:
$$x_{1,2}=-S\pm \frac{1}{2}\sqrt{-4S^2+\frac{d-2cS}{Sa}}$$
$$x_{3,4}=+S\pm \frac{1}{2}\sqrt{-4S^2-\frac{d+2cS}{Sa}}$$
Where:
$$S=\frac{1}{2\sqrt{3a}}\sqrt{-2c+Q+\frac{\Delta_0}{Q}}$$
$$Q=\frac{1}{\sqrt[3]{2}}\sqrt[3]{\Delta_1+\sqrt{\Delta_1^2-4\Delta_0^3}}$$
$$\Delta_0=c^2+12ae,\ \ \Delta_1=2c^3+27ad^2-72ace$$
In your case, the two real solutions are $x_{3,4}$, and you can verify that:
$$Q = \sqrt[3]{\frac{1}{2} \left(155+3 \sqrt{849}\right)}$$
$$S=\frac{1}{2\sqrt{3}} \sqrt{4+Q+\frac{16}{Q}}$$
$$ x= S \pm \frac{1}{2\sqrt{3a}} \sqrt{4 + \frac{1}{S} - 4 S^2} \simeq 0.5249, 1.4902$$
