Let p, q (p \gt q) be the roots of the quadratic equation x^{2}+bx+c=0 where c \gt 0. If p^{2}+q^{2}-11pq=0, then what is p-q equal to?

Explanation

Let p and q be the roots. We have p+q = -b and pq = c. Given p^2+q^2-11pq = 0, we subtract 2pq from both sides to form a perfect square: p^2+q^2-2pq = 9pq. This implies (p-q)^2 = 9c. Since p \gt q, p-q must be positive, so p-q = 3\sqrt{c}.

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