Given the roots $p$ and $q$ of the quadratic equation $x^2 + bx + c = 0$ with the condition $p^2 + q^2 - 11pq = 0$, we start by expressing $p^2 + q^2$ in terms of $p$ and $q$:

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

Let $s = p + q$ and $r = pq$. The quadratic can also be expressed as $x^2 - sx + r = 0$, so we rewrite the given equation:

$(p + q)^2 - 2pq - 11pq = 0$

This simplifies to:

$s^2 - 13r = 0$

From this, we have:

$s^2 = 13r \quad \text{or} \quad r = \frac{s^2}{13}$

Next, we need to find $p - q$. We can express this as:

$p - q = \sqrt{(p + q)^2 - 4pq} = \sqrt{s^2 - 4r}$

Substituting $r = \frac{s^2}{13}$ into this expression:

$p - q = \sqrt{s^2 - 4\left(\frac{s^2}{13}\right)} = \sqrt{s^2 - \frac{4s^2}{13}} = \sqrt{s^2\left(1 - \frac{4}{13}\right)} = \sqrt{s^2\left(\frac{9}{13}\right)}$

This leads us to:

$p - q = \frac{3}{\sqrt{13}} s$

Now, since $r = pq = c > 0$, we can also find $s$:

Using $r = \frac{s^2}{13}$ gives:

$c = \frac{s^2}{13} \quad \Rightarrow \quad s^2 = 13c$

Thus:

$p - q = \frac{3}{\sqrt{13}} \sqrt{13c} = 3\sqrt{c}$

Now, looking at the answer choices provided:

1. $3\sqrt{c}$
2. $3c$
3. $9\sqrt{c}$
4. $9c$
5. $6\sqrt{c}$

The correct answer is $3\sqrt{c}$.

Summary

The value of $p - q$ is $3\sqrt{c}$.

Do you have any questions or would you like more details? Here are some related questions you might find interesting:

1. What is the relationship between the coefficients and roots of a quadratic equation?
2. How does the discriminant relate to the nature of the roots?
3. Can you derive the quadratic formula using the method of completing the square?
4. What is the significance of the roots being real and distinct in a quadratic equation?
5. How can you find the roots of a quadratic equation if you know the values of $b$ and $c$?

Tip: Always check if the quadratic equation has real roots by evaluating the discriminant $D = b^2 - 4ac$. If $D \geq 0$, the roots are real.