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I have written some code to solve for the roots of a 4th order polynomial, and in the process, I noticed that the resolvent cubic always has at least one positive real root.

I can't find anywhere that states that this is proven to be true, is it?

Edit: When I posted this, I didn't know there were several possible resolvent cubics. The one I am using is from Nonweiler's paper, Roots of Low Order Polynomial Equations, TOMS Algorithm 326.

Cubic Equation

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The answer is No. The cubic resolvent need not have a positive real root. To recall, given,

$$x^4+ax^3+bx^2+cx+d=0$$

then the four solutions are,

$$x_{1,2} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}+\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag1$$

$$x_{3,4} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}-\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag2$$

where $u$ is any non-zero root of the cubic resolvent,

$u^3-\tfrac{1}{4}(3 a^2 - 8 b)u^2+\tfrac{1}{48}\big((3 a^2 - 8 b)^2 - 16 (b^2 - 3 a c + 12 d)\big)u-\tfrac{1}{64}(\color{brown}{a^3 - 4 a b + 8 c})^2=0\tag3$

This is the solution used by Mathematica but I've simplified it. To answer your question, we can "cheat" and set $\color{brown}{a^3 - 4 a b + 8 c}= 0$ so $u_1=0$, and either of the non-zero roots must be employed. A quick search then reveals that using $a,b,c,d = 1,\;2,\;\frac{7}{8},\;\frac{1}{2}$ and the solutions of the resolvent are,

$$u_1,\,u_2,\,u_3 = 0,\;\tfrac{-13-\sqrt{101}}{8},\;\tfrac{-13+\sqrt{101}}{8}$$

none of which is a positive real.

However, we can modify your question by asking, "For real $a,b,c,d$, if the cubic resolvent has all non-zero roots, then is one root always positive?" That one remains to be settled.

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  • $\begingroup$ The product of the three roots of the cubic resolvent is ${1\over64}(a^3-4ab+8c)^2$, which is always non-negative, hence positive if the roots are all non-zero. But the product of three negative numbers is negative, as is the product of a negative number with a pair of complex conjugates. $\endgroup$ – Barry Cipra Mar 23 '16 at 23:26
  • $\begingroup$ @BarryCipra: Ah, yes, how could I have missed that. So this particular resolvent, if all roots are non-zero, has at least one positive root. Thanks. (I just remembered there are other possible resolvents.) $\endgroup$ – Tito Piezas III Mar 24 '16 at 0:24
  • $\begingroup$ @BarryCipra: Don Herbison-Evans' paper, "Finding Real Roots of Quartics" give five resolvents, only one of which (the one above) has a square constant term. $\endgroup$ – Tito Piezas III Mar 24 '16 at 1:18
  • $\begingroup$ @Tito Piezas III I posted the cubic I am actually using. $\endgroup$ – user5108_Dan Mar 25 '16 at 10:57
  • $\begingroup$ @user5108_Dan: The fact that both cubic resolvents have a square constant term is a clue that they are related. In fact, it is just $x = \frac{u}{4}$. Thus, the same conclusions apply for these two essentially equivalent resolvents: 1) if all roots are non-zero, then at least one is positive; 2) but if one root is zero, then all may be $\leq 0$. $\endgroup$ – Tito Piezas III Mar 25 '16 at 20:45

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