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

roots are $\alpha \&\beta$

sum = $\alpha +\beta =-p$

product=$\alpha \beta$=q

${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta =p^2-2q{\alpha }^3+{\beta }^3=(\alpha +\beta )({\alpha }^2-\alpha \beta +{\beta }^2)=(-p)(p^2-2q-q)=3pq-p^3{\alpha }^4+{\alpha }^2{\beta }^2+{\beta }^4={\alpha }^4+2{\alpha }^2{\beta }^4-{\alpha }^2{\beta }^2={({\alpha }^2+{\beta }^2)}^2-{\left(\alpha \beta \right)}^2={(p^2-2q)}^2-q^2=p^4-4p^{2}q+{3q}^2=p^2(p^2-3q)-q(p^2-3q)=(p^2-q)(p^2-3q)$