Question

If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c$, find the values of
(1) $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$
(2) ${\alpha }^4{\beta }^7+{\alpha }^7{\beta }^4$
(3) ${(\frac{\alpha }{\beta }-\frac{\beta }{\alpha })}^2$

Answer 1

Given that $\alpha ,\beta$ are the roots of equation $a{x}^2+bx+c=0$

So we have $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

1. $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{{\left(\alpha \beta \right)}^2}=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{{\left(\alpha \beta \right)}^2}=\frac{b^2-2ca}{c^2}$

2. ${\alpha }^4{\beta }^7+{\alpha }^7{\beta }^4={\left(\alpha \beta \right)}^4({\alpha }^3+{\beta }^3)$

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

$=\frac{c^4}{a^4}\times (-\frac{b}{a})\times \left(\frac{b^2-3ac}{a^2}\right)$

$=\frac{b{c}^4(3ac-b^2)}{a^7}$

3. ${(\frac{\alpha }{\beta }-\frac{\beta }{\alpha })}^2={\left(\frac{(\alpha +\beta )(\alpha -\beta )}{\alpha \beta }\right)}^2=\frac{b^2}{c^2}\times \left(\frac{b^2-4ac}{a^2}\right)$