求使二次表达式px²+4x+p对所有实数x恒大于0的p值

Alright, let's work through this problem logically to get the right range of p. Here's the breakdown:

• First, let's rule out non-quadratic cases first: If p = 0, the expression simplifies to 4x—a linear function that's negative whenever x is negative, so it can't be always positive for all real x. So we know p ≠ 0.
• For a quadratic expression f(x) = px² + 4x + p to be always positive for every real x, two critical conditions must hold:
  1. The parabola opens upwards: This requires p > 0. If p were negative, the parabola opens downward, and even if it has no real roots, it would trend to negative infinity as x approaches ±∞—so it can't stay positive everywhere.
  2. The parabola never touches or crosses the x-axis: Meaning the corresponding quadratic equation px² + 4x + p = 0 has no real roots.

The problem already tells us the equation has no real roots when p ∈ (-∞, -2) ∪ (2, +∞). Now we just need to combine this with our upward-opening requirement (p > 0).

Taking the overlap of these two conditions, we're left with p > 2.

Let's verify this to be thorough:

• When p > 2, the discriminant of the quadratic equation is Δ = 4² - 4*p*p = 16 - 4p² = 4(4 - p²). Since p > 2, p² > 4, so Δ < 0—no real roots. And with p positive, the parabola opens upward, so every value of f(x) is positive.
• If p were in (-∞, -2), even though there are no real roots, the downward-opening parabola would make f(x) negative for all x, which doesn't meet our requirement.

So the final valid range of p is (2, +∞).

内容的提问来源于stack exchange，提问作者Marva Jami