Mathematics - Quadratic Equation Question with Solution | TestHub

${\alpha }^2=\alpha +1$
$\Rightarrow {\alpha }^4={({\alpha }^2)}^2$
$\Rightarrow {\alpha }^4={(\alpha +1)}^2$
$\Rightarrow {\alpha }^4={\alpha }^2+2\alpha +1$
$\Rightarrow {\alpha }^4=\alpha +1+2\alpha +1$
$\Rightarrow {\alpha }^4=3\alpha +2$

$Als\mathrm{o }\alpha +\beta =1$

$\therefore a_4=28 \Rightarrow p{\alpha }^4+q{\beta }^4=p\left(3\alpha +2\right)+q\left(3\beta +2\right)=28$

$\Rightarrow p\left(3\alpha +2\right)+q\left(3-3\alpha +2\right)=28$

$\Rightarrow \alpha \left(3p-3q\right)+2p+5q=28$     (as $\alpha \in Q^c$) (as it is root of quadratic equation)

$\Rightarrow p=q, 2p+5q=28 \Rightarrow p=q=4$

$\therefore p+2q=12$