<s>[INST] 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.
 [/INST] 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:
$$Q = \sqrt[3]{\frac{1}{2} \left(155+3 \sqrt{849}\right)}$$
$$S=\frac{1}{2\sqrt{3}} \sqrt{4+Q+\frac{16}{Q}}$$
$$ x= S \pm \frac{1}{2\sqrt{3a}} \sqrt{4 + \frac{1}{S} - 4 S^2} \simeq 0.5249, 1.4902$$
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