1. Interpret the given expression

The statement is intended as:

$p+q=2 \quad \text{and} \quad p^4+q^4=272.$

We need the quadratic equation whose roots are $p$ and $q$.

2. Use the identity for $p^4+q^4$

First find $pq$ using

$p^2+q^2=(p+q)^2-2pq=4-2pq.$

Now,

$p^4+q^4=(p^2+q^2)^2-2p^2q^2.$

Substitute $p^2+q^2=4-2pq$:

$p^4+q^4=(4-2pq)^2-2(pq)^2.$

Given $p^4+q^4=272$, so

$(4-2pq)^2-2(pq)^2=272.$

Expand:

$16-16pq+4p^2q^2-2p^2q^2=272$ $16-16pq+2p^2q^2=272$ $2p^2q^2-16pq-256=0$ $p^2q^2-8pq-128=0.$

Let

$t=pq.$

Then

$t^2-8t-128=0.$

Solve:

$t=\frac{8\pm\sqrt{64+512}}{2}=\frac{8\pm 24}{2}.$

So,

$t=16 \quad \text{or} \quad t=-8.$

Since $p,q>0$, we must have $pq>0$, hence

$pq=16.$

3. Form the quadratic equation with roots $p$ and $q$

For roots $p$ and $q$, the quadratic is

$x^2-(p+q)x+pq=0.$

Using

$p+q=2, \quad pq=16,$

we get

$x^2-2x+16=0.$

4. Match with the options

The correct option is:

$\boxed{x^2-2x+16=0}$

which is Option C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.