$$\begin{gathered} Letpandqbetwopositivenumberssuchthatp+q=2and \\ {p}^4+q^4=272.Thenpandqarerootsoftheequation: \end{gathered}$$

$\begin{array}{l}Givenp+q=2and{p}^4+q^4=272\\ \Rightarrow {(p^2+q^2)}^2-2p^2{q}^2=272\\ \Rightarrow {({(p+q)}^2-2pq)}^2-2p^2{q}^2=272\\ \Rightarrow {(4-2pq)}^2-2p^2{q}^2=272\\ \Rightarrow 16+4p^2{q}^2-8pq-2p^2{q}^2=272\\ \Rightarrow {\left(pq\right)}^2-8pq-128=0\\ \Rightarrow (pq-16)(pq+8)=0\\ pq=16,-8\\ \mathrm{Re}quiredEquationis{x}^2-(p+q)x+pq=0\\ \Rightarrow x^2-2x+16=0or{x}^2-2x-8=0\end{array}$