CAT 2025 Slot2 Quant: The set of all real values of x for which

Case 1: $x \ge -9$
Expression: $x^2 - (x + 9) + x > 0$
$x^2 - x - 9 + x > 0 \implies x^2 - 9 > 0$
$(x-3)(x+3) > 0$. Roots are $3, -3$.
Solution for this case: $x \in [-9, -3) \cup (3, \infty)$.

Case 2: $x < -9$
Expression: $x^2 - (-(x + 9)) + x > 0$
$x^2 + x + 9 + x > 0 \implies x^2 + 2x + 9 > 0$
For $x^2 + 2x + 9$, $D = 2^2 - 4(9) = -32$. Since $D < 0$ and $a > 0$, the quadratic is always positive for all real $x$.
Solution for this case: $x \in (-\infty, -9)$.

Final Union:
Combining $(-\infty, -9)$ and $[-9, -3) \cup (3, \infty)$ gives us the complete set:
$x \in \mathbf{(-\infty, -3) \cup (3, \infty)}$.