Question

If $\alpha ,\beta$ are the zeroes of the polynomial $x^2-2x+2$ find the value of
i) ${\alpha }^2+{\beta }^2$
ii) $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$
iii) ${\alpha }^3+{\beta }^3$
iv) $\frac{{\alpha }^2}{\beta }+\frac{{\beta }^2}{\alpha }$
v) ${\alpha }^4+{\beta }^4$

Answer 1

Sum of the roots$=\alpha +\beta =-\left(\frac{-2}{1}\right)=2$

Product of the roots$=\alpha \beta =\frac{c}{a}=2$

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

$={\left(2\right)}^2-2\times 2=4-4=0$

$\therefore {\alpha }^2+{\beta }^2=0$

$\left(ii\right)\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{0}{2}=0$ from $\left(i\right)$

$\therefore \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=0$

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

$={\left(2\right)}^3-3\times 2\left(2\right)=8-12=-4$ from $\left(i\right)$

$\therefore {\alpha }^3+{\beta }^3=-4$

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

$=\frac{{\alpha }^3+{\beta }^3}{\alpha \beta }=\frac{-4}{2}=-2$ from $\left(iii\right)$

$\left(v\right){\alpha }^4+{\beta }^4={({\alpha }^2+{\beta }^2)}^2-2{\alpha }^2{\beta }^2$

$=0-2{\left(\alpha \beta \right)}^2$ from $\left(i\right)$

$=-2{\left(2\right)}^2=-8$

$\therefore {\alpha }^4+{\beta }^4=-8$