# MIE 2015/Day 1/Problem 2

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## Problem 2

The polynomial $x^3+ax^2+bx+c$ has real roots $\alpha$, $-\alpha$ and $\frac{1}{\alpha}$. Thus the value of the sum of $b+c^2+ac+\frac{b}{c^2}$ is:

(a) $-2$

(b) $-1$

(c) $0$

(d) $1$

(e) $2$

## Solution

By Girard's relations (also called Vieta's formulas or Newton's identities),

$\begin{cases}a=-\frac{1}{\alpha}\\b=-\alpha^2\\c=\alpha\end{cases}$

$b+c^2+ac+\frac{b}{c^2}$

$-\alpha^2+\alpha^2-\frac{1}{\alpha}\cdot\alpha+\frac{-\alpha^2}{\alpha^2}=\boxed{-2}~~~~\boxed{\text{A}}$