Given a quartic equation:


If no help should be obtained from calculators and computers, can the roots be found? If not, what is the nature of the roots? (Real, complex, positive, negative, zero, etc.)

It seems like by rational root theorem, this quartic equation does not have a rational root, but that's all I can manage, Vieta's doesn't help much for me either. (An elementary solution is very much preferred due to this being a secondary school contest problem)

Extra note: The original problem asked how many negative roots does the polynomial have. Thank you to whoever who reminded me about the unoriginality of the problem.

  • 3
    $\begingroup$ wolframalpha.com/input/… $\endgroup$
    – Kenny Lau
    Sep 14, 2017 at 14:25
  • $\begingroup$ do you need a symbolic solution? $\endgroup$ Sep 14, 2017 at 14:29
  • $\begingroup$ Look here for the general formula. $\endgroup$ Sep 14, 2017 at 14:30
  • $\begingroup$ Forget to mention, the problem appeared in a mathematics contest for secondary schools (referred to high schools sometimes), an elementary solution is very much preferred. $\endgroup$ Sep 14, 2017 at 14:31
  • 2
    $\begingroup$ @MathEnthusiast If is was for secondary school, then most probably there is a typo. The correct equation has a rational root, see for example here. $\endgroup$ Sep 14, 2017 at 14:33

3 Answers 3


It's easy to see that there are no negative roots, when $x<0$ the only negative term in $x^4-5x^3-4x^2-2017x+4$ is $-4x^2$ so for it to be $0$ it must be $$-4x^2+4<0,-4x^2-2017x<0,-4x^2+x^4<0$$ But those $3$ inequalities aren't satisfied for any negative $x$.

SIDE NOTE: One should probably use $\leq$ instead of $<$ however the edge cases clearly don't satisfy the inequalities either.


This quartic equation has no nice solution, which could be derived in a mathematics contest for secondary schools. The coefficient $2017$ seems to indicate it was from this year. Then the constant term is perhaps meant to be $-2019$, because then $$ x^4-5x^3-4x^2-2017x-2019 $$ has a rational root. Furthermore it is easy then to see that there is exactly one negative root, namely $x=-1$. And it would be a nice contest question for in two years.


suppose that the quartic has at least one negative root, $-a, a>0$.
$(a^2-2)^2+5a^3+2017a>0$ , contradiction
so no negative roots


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