Solve for $x$

I have tried the Rational Root Theorem and found that there are no rational roots. Further, the polynomial $p(x)=(x^2-4)(x^2-2x)-2$ is irreducible since when I tried expanding it and writing it as a product of two quadratics, there were no integer solutions for the coefficients. I also depressed the quartic polynomial $p(x)$ hoping that the coefficient of $x$ would also vanish along with the coefficient of $x^3$, giving me a biquadratic. But that didn't happen. I also tried using substitutions, but none of them worked so far.

Any help will be appreciated.

  • $\begingroup$ It has four real roots, but are you supposed to find a formula ? $\endgroup$ May 19, 2015 at 20:11
  • $\begingroup$ @DietrichBurde Yes, I'm supposed to find a closed form for the roots. $\endgroup$
    – Henry
    May 19, 2015 at 20:12
  • $\begingroup$ For general case about quartic equation one can see Ferrari's solution $\endgroup$ May 19, 2015 at 20:22

1 Answer 1


$$(x^2-4)(x^2-2x)=2$$ $$\Rightarrow x^4-2x^3-4x^2+8x-2=0$$ $$\Rightarrow (x^2-x-1)^2-3(x-1)^2=0$$ $$\Rightarrow (x^2-x-1+\sqrt 3\ (x-1))(x^2-x-1-\sqrt 3\ (x-1))=0$$ $$\Rightarrow x^2+(\sqrt 3-1)x-1-\sqrt 3=0\ \ \text{or}\ \ x^2-(\sqrt 3+1)x-1+\sqrt 3=0$$ $$\Rightarrow x=\frac{-\sqrt 3+1\pm\sqrt{8+2\sqrt 3}}{2},\frac{\sqrt 3+1\pm\sqrt{8-2\sqrt 3}}{2}.$$

  • 2
    $\begingroup$ You absolutely rock!! Thank you so much 😀 $\endgroup$
    – Henry
    May 19, 2015 at 20:29
  • $\begingroup$ How did you go from the second to the third line in your solution? Did you assume an ansatz? $\endgroup$
    – John
    May 19, 2015 at 20:48
  • 4
    $\begingroup$ @John: I tried to find $a,b,c$ such that $x^4-2x^3-4x^2+8x-2=(x^2-x+a)^2-b(x+c)^2$. $\endgroup$
    – mathlove
    May 19, 2015 at 20:55
  • 3
    $\begingroup$ @John: it's a technique related to the general solution for a quartic equation. See here (equation 22 onwards). $\endgroup$
    – Zorawar
    May 19, 2015 at 22:52

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