If \(x^2 - 3x < 40\) and \((x+1)(x-1) > 8\), what will be the minimum value of \(x\) that satisfy both of the inequalities?

Explanation: Solve the first inequality: \(x^2 - 3x < 40\). \(x^2 - 3x - 40 < 0\). Factor: \((x - 8)(x + 5) < 0\). This inequality is true for values of \(x\) between the roots \(-5\) and \(8\). So, \(-5 < x < 8\). Solve the second inequality: \((x+1)(x-1) > 8\). \(x^2 - 1 > 8\). \(x^2 > 9\). This is true for \(x > 3\) or \(x < -3\). We need the values of \(x\) that satisfy BOTH inequalities. Combine the solutions: \(x \in (-5, 8)\) AND \((x > 3\) or \(x < -3)\). The intersection is: \(x \in (-5, -3) \cup (3, 8)\). The set of integers satisfying this is: \(\{-4\}\) from the first interval, and \(\{4, 5, 6, 7\}\) from the second interval. The minimum integer value of \(x\) that satisfies both is **\(-4\)**. (The solution correctly identifies \(-4\) as the minimum value, likely by testing integer values in the ranges).