CATSprint

CAT 2025Slot 2QAQues & Sol

AlgebraHard

Question

The set of all real values of x for which $(x^{2}-\mid x+9\mid+x)>0$, is

Options

Solution

1. Concept Used

  • Topic: Inequalities with Absolute Values — Case-Based Analysis
  • Formula: $$|x + 9| = \begin{cases} x+9, & \text{if } x \geq -9 \ -(x+9), & \text{if } x < -9 \end{cases}$$

2. Calculation

We need to solve $$x^2 - |x+9| + x > 0$$ Since the expression contains an absolute value, we split into two cases based on the critical point $x = -9$.

Case 1: $x \geq -9$

Here $|x+9| = x+9$, so the inequality becomes: $$x^2 - (x+9) + x > 0 \implies x^2 - x - 9 + x > 0 \implies x^2 - 9 > 0$$

This factors as $(x-3)(x+3) > 0$, which holds when $x < -3$ or $x > 3$.

Intersecting with the domain $x \geq -9$, we get: $$x \in [-9, -3) \cup (3, \infty)$$

Case 2: $x < -9$

Here $|x+9| = -(x+9) = -x-9$, so the inequality becomes: $$x^2 - (-x-9) + x > 0 \implies x^2 + x + 9 + x > 0 \implies x^2 + 2x + 9 > 0$$

The discriminant of $x^2 + 2x + 9$ is $\Delta = 4 - 36 = -32 < 0$. Since the discriminant is negative and the leading coefficient is positive, this quadratic is always positive for all real $x$.

So the inequality is satisfied for all $x < -9$, giving: $$x \in (-\infty, -9)$$

Combining Both Cases: $$x \in (-\infty, -9) \cup [-9, -3) \cup (3, \infty) = (-\infty, -3) \cup (3, \infty)$$

Notice that $x = -9$ satisfies $x^2 - 9 = 81 - 9 = 72 > 0$ ✅, so it is included, and the union simplifies cleanly to $(-\infty, -3) \cup (3, \infty)$.

3. Solution

Answer = Option 1 ✅

The final solution set is $(-\infty, -3) \cup (3, \infty)$.