If $\alpha$ and $\beta$ are the roots of $x^2-px+q=0$ and ${\alpha }^4,{\beta }^4$ are the roots of $x^2-rx+s=0$, then the value of $r$ in terms of $p$ and $q$ is:

Solution

1. We have $\alpha +\beta =p$ and $\alpha \beta =q$.
2. The roots of the second equation are ${\alpha }^4$ and ${\beta }^4$, so $r={\alpha }^4+{\beta }^4$.
3. $$\begin{array}{rl}{\alpha }^2+{\beta }^2& =(\alpha +\beta {)}^2-2\alpha \beta \\ & =p^2-2q\end{array}$$.
4. ${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2(\alpha \beta {)}^2$.
5. Substituting the values, $$\begin{array}{rl}r& =(p^2-2q{)}^2-2q^2\\ & =p^4-4p^{2}q+4q^2-2q^2\\ & =p^4-4p^{2}q+2q^2\end{array}$$.
6. Hence the answer is (A).