Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=x^2-x-4$, then evaluate: [4 MARKS]

(i) ${\alpha }^3+{\beta }^3$

(ii) $\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}$

Answer 1

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

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

$f\left(x\right)=x^2-x-4$

$\therefore \alpha +\beta =\frac{-b}{a}=\frac{1}{1}=1and\alpha \beta =\frac{c}{a}=\frac{-4}{1}=-4$

(i) We have,

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

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

$\Rightarrow {\alpha }^3+{\beta }^3=(1{)}^3-3(-4\left)\right(1)$ = 13


(ii) We have,

$\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}=\frac{{\alpha }^3+{\beta }^3}{(\alpha \beta {)}^3}=\frac{\frac{3abc-b^3}{a^3}}{{\left(\frac{c}{a}\right)}^3}$

$\Rightarrow \frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}=\frac{13}{(-4{)}^3}=\frac{-13}{64}$