Question

If $\alpha ,\beta$ and $\gamma$ are zeroes of the polynomial $f\left(x\right)={px}^3+{qx}^2+rx+s$ then find the value of the ${\alpha }^2{+\beta }^2+{\gamma }^2$

Answer 1

$\alpha ,\beta ,\gamma$, are zeroes of $f\left(x\right)=p{x}^3+q{x}^2+rx+s.$

where a= p , b = q , c = r, d = s.

now, $\alpha +\beta +\gamma =\frac{-b}{a}$

$\Rightarrow \alpha +\beta +\gamma =\frac{-q}{p}$ __________________ (1)

and $\alpha \beta +\beta \gamma +\alpha \gamma =\frac{c}{q}.$

$\Rightarrow \alpha \beta +\beta \gamma +\alpha \gamma =\frac{r}{p}$ _________ (2)

and $\alpha \beta \gamma =\frac{-d}{q}$

i.e $\alpha \beta \gamma =\frac{-s}{p}$ ________ (3)

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

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

$=(\frac{q}{p}{)}^2-2\frac{r}{p}$ [from (1)and (2)]

$=\frac{q^2}{p^3}-\frac{2r}{p}$

$=\frac{q^2-2pr}{p^2}$