0
$\begingroup$

So, I don't know if it's normal but this quartic seems long to solve...

When I use the ferrari method, I get the following reduced quartic :

$y^4 - {2883\over 128}y^2-{27103\over 512}y-{1809027\over 65536}=0$

Now, on paper, I did reduce the reduced quartic to a cubic form... But the numbers keep getting worser. Is it normal ? I don't have the courage to calculate all of this, if it correct :(

Thank yoU!

$\endgroup$
  • 1
    $\begingroup$ We invented computers to stop this insanity! But if you insist, keep going for another 1/2 hour then compare to the computers answer (which it'll generate in less than 30 seconds) $\endgroup$ – Zach466920 Apr 23 '15 at 22:47
  • $\begingroup$ Yes, I know about Wolfram Alpha ! It was killing me, so I just stopped ! $\endgroup$ – user108343 Apr 23 '15 at 22:51
  • $\begingroup$ Wolfram or the by hand method? If you insist on doing it by hand, just use newton's method (best of both worlds) $\endgroup$ – Zach466920 Apr 23 '15 at 22:54
  • $\begingroup$ Well, I began by hand and stopped after it was getting too long. I don't know Newton's method. $\endgroup$ – user108343 Apr 23 '15 at 23:02
  • $\begingroup$ Then go learn about it. It can approximate solutions to any polynomial equation, and more. $\endgroup$ – Zach466920 Apr 23 '15 at 23:05
-1
$\begingroup$

You can find:

$$4x^4-31x^3+21x=-18$$

$$x(4x^3-31x^2+21)=-18$$

$$4x^4-31x^3+21x+18=0$$

$$x((4x-31)x^2+21)+18=0$$

And the answer is very difficult/hard to find!

$\endgroup$

Your Answer

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