Let us consider the quadratic equation


My question is: Find sufficient and necessary conditions in which $x=0$ is the only real zero of the above equation.

I have no idea to start

  • $\begingroup$ You mean real zero? $\endgroup$ – Nicky Hekster Apr 7 '17 at 10:54
  • $\begingroup$ @NickyHekster: Yes, corrected. $\endgroup$ – China Apr 7 '17 at 10:55
  • $\begingroup$ OK, if $x=0$ is a zero, then $v=0$. Hence you are left with a third degree equation if $a \neq 0$. But then it has always a real root. So $a=0$ and it boils down to a quadratic equation. Can you take it from there? $\endgroup$ – Nicky Hekster Apr 7 '17 at 10:57
  • $\begingroup$ @NickyHekster: But $a$ is not zero in my case. $\endgroup$ – China Apr 7 '17 at 10:58
  • 1
    $\begingroup$ No problem, next time show somewhat more of your efforts or thoughts. Otherwise your entry will be downvoted. Hope you learned something!!! $\endgroup$ – Nicky Hekster Apr 7 '17 at 11:08

Let $\alpha,\beta,\gamma$ be the other roots of the quartic $f(x) =ax^4+bx^3+cx^2+dx+v$

We have that $x=0$ is the only real root of the quartic. Therefore since complex roots come in pairs, we require that $x=0$ is a repeated root, either twice or four times.

Case $(1):$ $x=0$ repeated twice

Then $d=v=0\implies f(x)=x^2(ax^2+bx+c)=0\quad$We require that $g(x)=ax^2+bx+c$ has complex roots $\implies b^2-4ac<0$

Case $(2): x=0$ repeated four times

Then $b=c=d=v=0\implies f(x)=ax^4=0\implies a\in\mathbb{C}$


Denote the quartic function by $q(x)$.

For starters, if $x=0$ is a real zero you can factor it out and write $$q(x) = xp(x)$$ where $p(x)$ is a cubic. Now, any cubic has at least one real zero, so under the stated hypotheses this real zero must equal $0$. Thus $$q(x) = x^2 r(x)$$ where $r$ is quadratic.

Finally, you must have either $r(x) = x^2$, or $r(x)$ has no real zeros. Thus either $q(x) = x^4$ or $q(x) = x^2 r(x)$, where $r(x)$ has no zeros.


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