If the roots of the quadratic equation $x^2+6x+b=0$ are real and distinct and they differ by at most $4$, then the range of values of $b$ is:

Let the roots be p and q with p and q being real and distinct. By Vieta's formulas, p + q = -6 and pq = b. Since the roots differ by at most 4, we have |p - q| <= 4, so (p - q)^2 <= 16. Expanding gives (p + q)^2 - 4pq <= 16, which means (-6)^2 - 4b <= 16, yielding 36 - 4b <= 16 and b >= 5. Because the roots are distinct, the discriminant must be strictly positive, so 6^2 - 4b > 0, meaning b < 9. Thus, b must lie in the interval [5, 9).

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