Question

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f\left(x\right)=a{x}^2+bx+c$, then evaluate:
(i) ${\alpha }^4+{\beta }^4$
(ii) $\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}$

Answer 1

$f\left(x\right)=a{x}^2+bx+c$

Given, $\alpha \&\beta$ are zeroes of $f\left(x\right)$

$\Rightarrow \alpha +\beta =\frac{-b}{a}$

$\Rightarrow \alpha \beta =\frac{c}{a}$

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

$={({(\alpha +\beta )}^2-2\alpha \beta )}^2-22{\left(\alpha \beta \right)}^2$

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

$={(\frac{b^2}{a^2}-\frac{2c}{a})}^2-2\frac{c^2}{a^2}$

$=\frac{b^4}{a^4}+\frac{4c^2}{a^2}-\frac{4b^{2}c}{a^3}-\frac{2c^2}{a^2}$

$=\frac{b^4}{a^4}+\frac{2c^2}{a^2}-\frac{4b^{2}c}{a^3}$

and, $\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}={(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })}^2-2\left(\frac{\alpha }{\beta }\right)\left(\frac{\beta }{\alpha }\right)$

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

Now, $\alpha +\beta =\frac{-b}{a}\Rightarrow {\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta =\frac{b^2}{a^2}-\frac{2c}{a}\to \left(1\right)$

$\alpha \beta =\frac{c}{a}\to \left(2\right)$

Divide both $\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

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

$=\frac{b^2-2ac}{ac}$

$=(\frac{b^2}{ac}-2)$

$\Rightarrow \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}={(\frac{\alpha }{\beta }-\frac{\beta }{\alpha })}^2-2={(\frac{b^2}{ac}-2)}^2-2$

$=\frac{b^4}{a^2{c}^2}+4-4\frac{b^2}{ac}-2$

$=\frac{b^4}{a^2{c}^2}-4\frac{b^2}{ac}+2$

Hence, the answer is $\frac{b^4}{a^2{c}^2}-4\frac{b^2}{ac}+2.$