Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=x^2+6x+2$, then evaluate:

$\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }$
[3 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 1 Mark

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial

$f\left(x\right)=x^2+6x+2$

$\therefore \alpha +\beta =\frac{-b}{a}=\frac{-6}{1}=-6and\alpha \beta =\frac{c}{a}=\frac{2}{1}=2$

We have

$\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=\frac{{\alpha }^3+{\beta }^3}{\left(\alpha \beta \right)}$

$=\frac{(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )}{\alpha \beta }$

$=\frac{{(-\frac{b}{a})}^3-3\left(\frac{c}{a}\right)(-\frac{b}{a})}{\frac{c}{a}}$

$=\frac{(-6{)}^3-3(2\left)\right(-6)}{2}$
= $=\frac{-180}{2}$

$\Rightarrow \frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }=-90$