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Let's say that I have an incomplete quartic equation with real coefficients, which is $$x^4 - 3x^3 + ... - 10 = 0$$

And also given 2 complex roots, $a + 2i$ and $1 + bi$ where $a$ and $b$ are real numbers.

The problem asks the sum of the real roots, but firstly I don't know how to determine if the equation even has a real root.

I can't do Rule of Signs because obviously the polynomial is incomplete. Although I can do a (heavy) assumption that $1 + bi$ is the conjugate of $a + 2i$ or not, that still would give me $0$ or $2$ real roots.

How can I know if it has a real root or not?

EDIT : I got the second term wrong, it should be $-3x^3$ and not $-3x^2$ ! I have edited the original equation. Sorry!

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Edit : The OP changed $-3x^2$ to $-3x^3$.

Let $\alpha,\beta,\gamma,\omega$ be the four roots.

Then, by Vieta's formulas, $$\alpha+\beta+\gamma+\omega=3$$ $$\alpha\beta\gamma\omega=-10$$

Case 1 : If $(a,b)=(1,-2)$, then we may suppose that $\alpha=1+2i,\beta=1-2i$, so $$\gamma+\omega=1,\quad \gamma\omega=-2$$ So, $\gamma,\omega$ are the solutions of $x^2-x-2=(x-2)(x+1)=0$, i.e. $$\gamma,\omega=2,-1$$ giving that sum of real roots is $\color{red}{1}$.

Case 2 : If $(a,b)\not=(1,-2)$, then there is no real roots since four roots are $a\pm 2i,1\pm bi$ where $b\not=0$. However, as Tobias Kildetoft comments, this does not happen because there is no real a,b such that $(a^2+2^2)(1^2+b^2)=-10$.

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  • $\begingroup$ It is not possible for there to be no real roots with the information given (the constant term is negative). $\endgroup$ – Tobias Kildetoft Jul 8 '16 at 11:58
  • $\begingroup$ @TobiasKildetoft: Thanks! You are right. $\endgroup$ – mathlove Jul 8 '16 at 12:01
  • $\begingroup$ Oops, sorry @mathlove ! I got a slight mistake on the original equation. However I got the gist of your answer and solved it, but can you do a slight edit for future readers please? $\endgroup$ – possibility0 Jul 8 '16 at 12:27
  • $\begingroup$ @possibility0: Done. I was editing:) $\endgroup$ – mathlove Jul 8 '16 at 12:29
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Hint your assumption is true as complex roots are always in pair so indeed $1+bi$ is a conjugate of $a+2i$ then use product of roots is $-10$,sum of roots is $3$ to get the sum of real roots as $0$ now use that they aren't conjugate and use that roots are $1\pm bi,a\pm 2i$ to get the same answer

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  • $\begingroup$ He needs to use one more thing to rule out four complex roots (namely that the highest degree term is positive and the constant term is negative). $\endgroup$ – Tobias Kildetoft Jul 8 '16 at 11:32
  • $\begingroup$ Ya so if they aren't conjugate would be the other condition $\endgroup$ – Archis Welankar Jul 8 '16 at 11:43

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