If $x^2 + 2ax + 10 - 3a > 0$ for all $x ∈ R$, then
- A. -5 < a < 2
- B. a < -5
- C. a > 5
- D. 2 < a < 5
✅ Correct Answer: -5 < a < 2
Explanation
For quadratic always positive, discriminant < 0: $(2a)^2 - 4(10-3a) < 0$. Solving gives $-5 < a < 2$.
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