Question

The set of all real values of $x$ for which $(x^2 - |x+9| + x)>0$ is:

• A. $(-\infty,-9)\cup(3,\infty)$
• B. $(-\infty,-3)\cup(9,\infty)$
• C. $(-\infty,-3)\cup(3,\infty)$
• D. $(-9,-3)\cup(3,9)$

Hint:

When solving inequalities involving absolute values, always split into cases based on the sign of the expression inside the absolute value. Analyze each case separately, then combine intervals carefully by intersection and union.

Answer 1

The Correct Option is C

We are given the inequality \[ x^2 - |x+9| + x>0 \quad\Longleftrightarrow\quad x^2 + x - |x+9|>0. \] Since the absolute value depends on the sign of \(x+9\), we split into two cases. Case 1: \(x+9 \ge 0 \;\Rightarrow\; x \ge -9\). Then \(|x+9| = x+9\). Substitute: \[ x^2 + x - (x+9)>0 \;\Rightarrow\; x^2 - 9>0 \;\Rightarrow\; (x-3)(x+3)>0. \] This product is positive when \[ x<-3 \quad \text{or} \quad x>3. \] But our case requires \(x \ge -9\). Intersection gives: \[ [-9,-3) \cup (3,\infty). \] Case 2: \(x+9<0 \;\Rightarrow\; x<-9\). Then \(|x+9| = -x-9\). Substitute: \[ x^2 + x - (-x-9)>0 \;\Rightarrow\; x^2 + 2x + 9>0. \] The discriminant is \[ D = 2^2 - 4\cdot 1 \cdot 9 = -32<0, \] so the quadratic is always positive (opens upward). Thus the inequality holds for all \(x<-9\). Case 2 gives: \((- \infty, -9)\).
Step 3: Combine both cases. \[ (-\infty,-9) \cup [-9,-3) \cup (3,\infty) = (-\infty,-3) \cup (3,\infty). \] Therefore, the solution set is: \[ \boxed{(-\infty,-3)\cup(3,\infty)}. \]