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 ? $\endgroup$ – Dietrich Burde May 19 '15 at 20:11
  • $\begingroup$ @DietrichBurde Yes, I'm supposed to find a closed form for the roots. $\endgroup$ – Henry Durham May 19 '15 at 20:12
  • $\begingroup$ For general case about quartic equation one can see Ferrari's solution $\endgroup$ – Alexey Burdin May 19 '15 at 20:22

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

| cite | improve this answer | |
  • 2
    $\begingroup$ You absolutely rock!! Thank you so much 😀 $\endgroup$ – Henry Durham May 19 '15 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 '15 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 '15 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 '15 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.