MathDB

1/a < x_0<2/a , x_0 is zero with smallest abs. value f(x) = x^3-ax+1

Source: Netherlands - Dutch NMO 1980 p1

January 28, 2023

algebrapolynomialinequalities

Problem Statement

f(x) = x^3-ax+1

a \in R

has three different zeros in

R

. Prove that for the zero

x_o

with the smallest absolute value holds:

\frac{1}{a}< x_0 < \frac{2}{a}