Expand $(x^2 - p)(x^2 - q)$ and write it in the form $x^4 + (\text{something})x^2 + (\text{constant})$. Then match its $x^2$ coefficient and its constant term with those in $x^4 - 14x^2 + 16$ to find $p+q$ and $pq$.

Once you know $p+q$ and $pq$, think about the identity for $(q-p)^2$ in terms of $p$ and $q$. How can you write $(q-p)^2$ using $p+q$ and $pq$ only?

After you find $(q-p)^2$, remember to take the positive square root (since $q>p$) and then simplify the radical by factoring out any perfect squares.

After you have found by hand that $p+q = 14$ and $pq = 16$, type sqrt(14^2 - 4*16) into a Desmos expression line. The value that Desmos shows is $q-p$ (take the positive root), and you can then rewrite that value in simplified radical form.

We are told

Multiply both sides by 2 to remove the fraction:

Now we can expand the right-hand side to compare coefficients.

Expand the product on the right:

This must equal the left-hand side:

So we match the coefficients of like powers of $x$:

• Coefficient of $x^2$: $-(p+q) = -14$, so $p+q = 14$.
• Constant term: $pq = 16$.

We now know $p+q$ and $pq$.

To find $q-p$, first find $(q-p)^2$ using the identity

Substitute $p+q = 14$ and $pq = 16$:

So $(q-p)^2 = 132$.

Since $q>p$, $q-p$ is positive, so

Simplify the square root:

Therefore, $q - p = 2\sqrt{33}$, which corresponds to choice C.