MHB Quartic polynomial has at least one real root

[Siron]

Hello,

Given the quartic:
$$16x^4-40ax^3+(15a^2+24b)x^2-18abx+3b^2 = 0$$
where $a,b$ are certain real constants. My question is if there is a (simple) condition on $a$ and/or $b$ such that the quartic has at least one real root.

Since the quartic has real coefficients the only possibilities for the roots are: 2 reals and 2 complex roots, 4 complex roots and 4 real roots. I found an article by E.L.Rees (Graphical discussion of the roots of a quartic equation) that describes the criteria regarding the nature of the roots for a general reduced quartic $x^4+qx^2+rx+s = 0$. It it not difficult to write the given quartic in this reduced form so this gives me quite a good idea whenever there is at least one real root (2 reals and 2 complex or 4 reals) in function of the discriminant and some conditions on the coefficients.

The expression for the discriminant of the given quartic is:
$$\frac{27}{4}b^6+\frac{27297}{1024}a^4b^4-\frac{729}{32}a^2b^5+\frac{10125}{8192}a^8b^2-\frac{6075}{512}a^6b^3$$
which in my opinion is not that nice to work with.

To summarize, using the discriminant I can derive the necessary conditions on the coefficients $a$ and $b$ to have at least one real root. However the expressions can be quite hard/ugly to work with so I'm wondering if there is no better or more efficient method here.

Thanks in advance!

 

[Ackbach]

I'm just shooting from the hip here, but could you use Descartes' Rule of Signs? Your polynomial is
$$16x^4-40ax^3+(15a^2+24b)x^2-18abx+3b^2 = 0.$$
If you could force an odd number of sign changes, then you'd be guaranteed a positive real root. Realistically, you can only control the middle three terms. On the other hand, if you examined the same polynomial with $x$ replaced by $-x$, which is
$$16x^4+40ax^3+(15a^2+24b)x^2+18abx+3b^2 = 0,$$
could you force an odd number of sign changes there?

 

[Siron]

I'm just shooting from the hip here, but could you use Descartes' Rule of Signs? Your polynomial is
$$16x^4-40ax^3+(15a^2+24b)x^2-18abx+3b^2 = 0.$$
If you could force an odd number of sign changes, then you'd be guaranteed a positive real root. Realistically, you can only control the middle three terms. On the other hand, if you examined the same polynomial with $x$ replaced by $-x$, which is
$$16x^4+40ax^3+(15a^2+24b)x^2+18abx+3b^2 = 0,$$
could you force an odd number of sign changes there?

Thanks for the reply Ackbach! I also thought about Descartes' Rule of Signs, but if I'm not mistaken it is impossible to force an odd number of sign changes since the first and last term are always positive. To solve the problem I thought it could be an idea to reduce the quartic to:
$$z^4+qz^2+rz+t = 0$$
by making the substitution $x = z + \frac{5}{8}a$ and where $q = \frac{-45}{32}a^2+\frac{3}{2}, r = \frac{-a}{32}(25a^2-24b)$ and $t = \frac{3}{16}b^2-\frac{15}{128}a^2b-\frac{375}{4096}a^4$. The latter is just applying Ferrari's idea of solving a quartic equation. The good news is that I now can force an odd number of sign changes for the reduced polynomial. The bad news is that it does not make the expressions easier (at least I think so).