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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.

Jyrki Lahtonen
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JohnWO
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  • Can $x$ be complex? – Kaster Aug 21 '13 at 18:18
  • @Kaster No, it should be real. – JohnWO Aug 21 '13 at 18:21
  • Just be sure that the solution is not trivial. – al-Hwarizmi Aug 21 '13 at 18:25
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    Well, based on what Mathematica shows, I don't think it can be done somehow shortcut. You need to apply general method of solving 4th order polynomial based on this Wiki article. – Kaster Aug 21 '13 at 18:28
  • It looks like none of the roots are rational. – rurouniwallace Aug 21 '13 at 18:32
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    A quick visualization of the equation (in the form $(x^2-1)^2=x$) indicates that there are two real roots, both positive and both simple, so there will also be two non-real roots. I don't see any clever way to get closed-form solutions, unless you count the solvability by radicals of the general quartic as clever. – Andreas Blass Aug 21 '13 at 18:33
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    @Ataraxia: since the lead coefficient is $1$, and none of the roots are integers, it is indeed the case that none of the roots are rational. – robjohn Aug 21 '13 at 18:35
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    One can even show that there is no expansion into product into to quadratics with integer coefficients – Norbert Aug 21 '13 at 18:42
  • Coincidentally $x \approx e^{e-\pi} \pi^{e-2}$ – JohnWO Aug 21 '13 at 18:53
  • Why does the "Solve" operation of Mathematica delivers an analytical solution? – al-Hwarizmi Aug 21 '13 at 19:11
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    @al-Hwarizmi Because up to degree 4 polynomials can be solved through a general formula involving only the coefficients, rational numbers and elementary operations (by which I mean addition, multiplication, integer powers and roots). This is not true anymore for polynomials of degree $5$ and higher (there is no general formula). – Daniel Robert-Nicoud Aug 21 '13 at 19:24
  • Exactly. Thanks; and that comes here: http://math.stackexchange.com/questions/785/is-there-a-general-formula-for-solving-4th-degree-equations – al-Hwarizmi Aug 21 '13 at 19:30

1 Answers1

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

Nathaniel Bubis
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