Question

If $\alpha, \beta$ are the roots of the quadratic equation $x^2 + px + q = 0$, then the values of $\alpha^{3}, \beta^{3}$ and $\alpha^{4}+\alpha^{2}\beta^{3}+\beta^{4}$ are respectively

• A. $3pq - p^3$ and $p^4 - 3p^2q + 3q^2$
• B. $-p(3q - p^2)$ and $(p^2 - q)(p^2 + 3q)$
• C. $pq - 4$ and $p^4 - q^4$
• D. $3pq - p^3$ and $(p^2 - q) (p^2 - 3q)$

Answer 1

Answer: (D)

$\because$ Sum of roots, $\alpha+\beta=-p$ and $\alpha \beta=q$
$\therefore \left(\alpha^{3}+\beta^{3}\right)=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$
$=(-p)^{3}-3 q(-p)$
$=-p^{3}+3 p q$
and $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}=\left(\alpha^{4}+\beta^{4}\right)+(\alpha \beta)^{2}$
$=\left(\alpha^{2}+\beta^{2}\right)^{2}-(\alpha \beta)^{2} $
$ = {\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]^{2}-(\alpha \beta)^{2}} $
$ =\left[(-p)^{2}-2 q\right]^{2}-3(q)^{2} $
$ =\left[p^{2}-2 q\right]^{2}-3 q^{2} $
$=p^{4}-4 p^{2} q+4 q^{2}-q^{2} $
$=p^{4}-4 p^{2} q+3 q^{2} $
$ =p^{4}-3 p^{2} q-p^{2} q+3 q^{2} $
$ =p^{2}\left(p^{2}-3 q\right)-q\left(p^{2}-3 q\right) $
$= \left(p^{2}-q\right)\left(p^{2}-3 q\right)$