$$\begin{gathered} {\text{Let p and q be two positive numbers such that p+q=2, p}}^4+q^4=272. \\ Then p and q are the roots of the equation \end{gathered}$$

${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$

it implies 

$\begin{array}{rcl}{\left({\left(p+q\right)}^2-2pq\right)}^2-2p^2{q}^2& =& 272\\ {\left(4-2pq\right)}^2-2p^2{q}^2& =& 272\end{array}$

$16-16pq+2p^2{q}^2=272\Rightarrow {\left(pq\right)}^2-8pq-128=0$

${\left(pq\right)}^2-8pq-128=0$

$pq=\frac{8\pm 24}{2}=16,-8$

$pq=16, \sin ce both are positive real numbers$

$Therefore, the equation whose roots are p, q is x^2-2x+16=0$