Question

If $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f\left(x\right)=6x^2-x-7$, then evaluate: [3 MARKS]

(i) ${\alpha }^2+{\beta }^2$

(ii) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

Answer 1

Concept : 1 Mark
Application : 2 Marks

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

$f\left(x\right)=6x^2-x-7$

$\therefore \alpha +\beta =-\frac{b}{a}=\frac{1}{6}and\alpha \beta =\frac{c}{a}=\frac{-7}{6}$

(i) We have,

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

$\Rightarrow {\alpha }^2+{\beta }^2={\left(\frac{-b}{a}\right)}^2-\frac{2c}{a}$

$\Rightarrow {\alpha }^2+{\beta }^2={\left(\frac{1}{6}\right)}^2-2\times \frac{-7}{6}$

$\Rightarrow {\alpha }^2+{\beta }^2=\frac{1}{36}+\frac{7}{3}$

$\Rightarrow {\alpha }^2+{\beta }^2=\frac{85}{36}$


(ii) We have,

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{{\left(\frac{1}{6}\right)}^2-2\left(\frac{-7}{6}\right)}{\frac{-7}{6}}$

$\Rightarrow \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{\frac{85}{36}}{\frac{-7}{6}}$

$\therefore \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=-\frac{85}{42}$