The set of all real values of \( x \) for which \( \left( x ^ { 2 } - | x + 9 | + x \right) > 0 \), is
Answer: B
Key idea: Split into cases at \(x = -9\), where \(x + 9\) changes sign.
Case 1: \(x \ge -9\). \(|x+9| = x + 9\), so the expression is \(x^2 + x - x - 9 = x^2 - 9 > 0\), giving \(|x| > 3\). Combined with \(x \ge -9\): \(x \in [-9, -3) \cup (3, \infty)\).
Case 2: \(x < -9\). \(|x+9| = -x - 9\), so the expression is \(x^2 + 2x + 9 = (x+1)^2 + 8\), which is always positive. So every \(x < -9\) works.
Combining: \((-\infty, -3) \cup (3, \infty)\).
Shortcut: Test \(x = -10\): \(100 - 1 - 10 = 89 > 0\), which rules out options (C) and (D). Test \(x = -5\): \(25 - 4 - 5 = 16 > 0\), which rules out (A).
Hence, the answer is (B) \((-\infty, -3) \cup (3, \infty)\).