Difference between revisions of "MIE 2015/Day 1/Problem 2"

(Created page with "==Problem 2== The polynomial <math>x^3+ax^2+bx+c</math> has real roots <math>\alpha</math>, <math>-\alpha</math> and <math>\frac{1}{\alpha}</math>. Thus the value of the sum o...")
 
(No difference)

Latest revision as of 13:25, 11 January 2018

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),


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


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


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

Invalid username
Login to AoPS