Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=x^2-5x+6$, then evaluate:

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

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=a{x}^2+bx+c$

$\therefore \alpha +\beta =-\frac{b}{a}=5and\alpha \beta =\frac{c}{a}=6$
We have $\alpha +\beta =5$
Hence $(\alpha +\beta {)}^2=25$
${\alpha }^2+{\beta }^2+2\alpha \beta =25$
${\alpha }^2+{\beta }^2=25-2\alpha \beta$
${\alpha }^2+{\beta }^2=25-2\left(6\right)$ ( $\because \alpha \beta =6$) = 13

We have,

$\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^4+{\beta }^4}{{\alpha }^2{\beta }^2}$
$\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2}{{\alpha }^2{\beta }^2}$
=$\frac{{13}^2-(2\times (6^2\left)\right)}{6^2}$
=$\frac{97}{36}$.
$\therefore \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{97}{36}$.