Suppose $a$ and $b$ are two positive real numbers such that the roots of cubic equation $x^3-ax+b=0$ are all real. Let $p$ be a root of this equation with minimal absolute value. What is the range of p. I tried to apply various sum and product of roots constraints but can't get it. Please help.
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1As written this equation is cubic, not quadratic. – Travis Willse May 19 '16 at 17:48
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1I have changed the title of the question, but change it back if this is not what you intended. – user84413 May 19 '16 at 17:52
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I am not sure what you want. Clearly it is easy to give a precise expression for the 3 roots, and to state which has the smallest absolute value. Do you simply want a handy rule of thumb? For example, the root with smallest absolute value has absolute value at most $\left|\frac{3b}{2a}\right|$. – almagest May 19 '16 at 18:38
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@almagest How did you get $|\frac{3b}{2a}|$ – Mathematics Nov 25 '17 at 19:11
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By Viète's theorem, the roots of such a polynomial have to be $\zeta_1,\zeta_2,\zeta_3=-(\zeta_1+\zeta_2)\in\mathbb{R}$ and
$$ \zeta_1\zeta_2(\zeta_1+\zeta_2)=b,\qquad \zeta_1\zeta_2-(\zeta_1+\zeta_2)^2=a\tag{1} $$ with the second condition representing an ellipse in the $(\zeta_1,\zeta_2)$ plane, with its minor and major axis on the $\zeta_1=\zeta_2$ and the $\zeta_1=-\zeta_2$ lines. It is not difficult to bound such ellipse into/around a rectangle and derive bounds for $|\zeta_1|,|\zeta_2|$ and $|\zeta_3|$.
Jack D'Aurizio
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