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(A) Given the inequalities $x^2  4$. First, solve $x^2 - 4x < 77$. Adding $4$ to both sides, we get $x^2 - 4x + 4 < 77 + 4$, whic

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$-6$

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(A) Given the inequalities $x^2  4$. First, solve $x^2 - 4x Adding $4$ to both sides, we get $x^2 - 4x + 4 Taking the square root, we have $-9 Next, solve $x^2 > 4$, which implies $x^2 - 4 > 0$, so $(x - 2)(x + 2) > 0$. This holds when $x > 2$ or $x Combining the two conditions, we need $x$ such that $(-7  2 	ext{ or } x The intersection is $(-7, -2) \cup (2, 11)$. The negative integers in this set are $\{-6, -5, -4, -3\}$. The smallest negative integer in this set is $-6$.

The smallest negative integer satisfying both the quadratic inequalities $x^2 < 4x + 77$ and $x^2 > 4$ is

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