Let$p,q$ be two positive number such that $p+q=2$and $p^4+q^4=272$. Then $p$ and $q$ are roots of the equation:

The correct option is D

$x^2–2x+16=0$

Explanation for the correct option:

We have, $p^4+q^4=272$ rearranging it, we get

$$\begin{gathered} {(p^2+q^2)}^2–2p^2{q}^2=272 \\ ({{(p+q)}^2–2pq)}^2–2p^2{q}^2=272[a^2+b^2=a^2+b^2-2ab] \\ \Rightarrow 16–16pq+2p^2{q}^2=272 \\ {\left(pq\right)}^2–8pq–128=0 \end{gathered}$$

Now,

$$\begin{gathered} \Rightarrow pq=\frac{(8\pm 24)}{2} \\ \Rightarrow pq=16,–8 \\ \Rightarrow pq=16 \end{gathered}$$

$$\begin{gathered} x^2–(p+q)x+pq=0 \\ {x}^2-2x+16=0 \end{gathered}$$

Hence, the correct option is (D)