We study conditions under which a quadratic inequality holds for all real numbers or has no solutions, as well as problems involving literal constants.
Example ①
Find the range of k such that x² + 2kx + k + 2 > 0 holds for all real numbers x.
Solution: Since a = 1 > 0, the condition is D < 0. D = 4k² - 4(k+2) < 0 → k² - k - 2 < 0 → (k+1)(k-2) < 0 → -1 < k < 2
Find the range of k such that x² + 2kx + k + 2 > 0 holds for all real numbers x.
Solution: Since a = 1 > 0, the condition is D < 0. D = 4k² - 4(k+2) < 0 → k² - k - 2 < 0 → (k+1)(k-2) < 0 → -1 < k < 2
Key Point
- “Positive for all x”: a > 0 and D < 0.
- “Negative for all x”: a < 0 and D < 0.
Practice Problems
- Find the range of k such that 2x² - 3x + k > 0 holds for all real numbers x.
Solutions and Explanations
- Solution: k > 9/8. D = 9 - 8k < 0 → k > 9/8.