Question

If $\alpha ,\beta$ be the roots of the equation $x^2-px+q=0$ and $\alpha >0,\beta >0,$ then find the value of
${\alpha }^5+{\beta }^5$

Answer 1

Given equation is $x^2-px+q=0$ ...(1)

$\alpha$ and $beta$ are the roots, then

$\alpha +\beta =p$ ...(2)

and $\alpha \beta =q$ ...(3)

${(\alpha +\beta )}^2={\alpha }^5+5{\alpha }^4\beta +10{\alpha }^4\beta +10{\alpha }^3{\beta }^3+5\alpha {\beta }^4+{\beta }^5$

$=({\alpha }^5+{\beta }^5)+5\alpha \beta ({\alpha }^3+{\beta }^3)+10{\alpha }^2{\beta }^2(\alpha +\beta )$

$=({\alpha }^5+{\beta }^5)+5\alpha \beta ({(\alpha +\beta )}^3-3\alpha \beta (\alpha +\beta ))=10{\alpha }^2{\beta }^2(\alpha +\beta )$

$\Rightarrow {\alpha }^5+{\beta }^5={(\alpha +\beta )}^5-5\alpha \beta {(\alpha +\beta )}^3+15{\alpha }^2{\beta }^2(\alpha +\beta )-10{\alpha }^2{\beta }^2(\alpha +\beta )$

From (2) and (3)

$({\alpha }^5+{\beta }^5)=p^5-5q{p}^3+15q^{2}p-10q^{2}p=p^5-5q{p}^3+5q^{2}p$