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 ${\alpha }^4+{\alpha }^2{\beta }^2+{\beta }^4$ are respectively

• A. $3pq-p^3$ and $p^4-3p^{2}q+3q^2$
• 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

The correct option is D $3pq-p^3$ and $(p^2-q)(p^2-3q)$

$\because$ Sum of roots, $\alpha +\beta =-p$ and $\alpha \beta =q$

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

$={(-p)}^3-3q(-p)$

$=-p^3+3pq$

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

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

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

$={[{(-p)}^2-2q]}^2-3{\left(q\right)}^2$

$={[p^2-2q]}^2-3q^2$

$=p^4-4p^{2}q+4q^2-q^2$

$=p^4-4p^{2}q+3q^2$

$=p^4-3p^{2}q-p^{2}q+3q^2$

$=p^2(p^2-3q)-q(p^2-3q)$

$=(p^2-q)(p^2-3q)$