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 \[ x^2 - |x+9| + x > 0, \] which can be rewritten as \[ x^2 + x - |x+9| > 0. \] Since the absolute value depends on the sign of \(x+9\), we consider two cases. Case 1: \(x+9 \ge 0\), that is \(x \ge -9\). In this case, \(|x+9| = x+9\). Substituting, \[ x^2 + x - (x+9) > 0 \Rightarrow x^2 - 9 > 0 \Rightarrow (x-3)(x+3) > 0. \] This inequality holds when \[ x < -3 \quad \text{or} \quad x > 3. \] Combining with the condition \(x \ge -9\), we obtain \[ [-9,-3) \cup (3,\infty). \] Case 2: \(x+9 < 0\), that is \(x < -9\). Here, \(|x+9| = -x-9\). Substituting, \[ x^2 + x - (-x-9) > 0 \Rightarrow x^2 + 2x + 9 > 0. \] The discriminant of this quadratic is \[ D = 2^2 - 4(1)(9) = -32 < 0, \] and since the coefficient of \(x^2\) is positive, the expression is always positive. Hence, the inequality is satisfied for all \[ x < -9. \] Combining the solutions from both cases, \[ (-\infty,-9) \cup [-9,-3) \cup (3,\infty) = (-\infty,-3) \cup (3,\infty). \] Therefore, the solution set is \[ \boxed{(-\infty,-3)\cup(3,\infty)}. \]

Answer 2

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: \[ \boxed{(-\infty,-3)\cup(3,\infty)}. \]