Explain why `3x^(2)+kx-1` is never always positive for any value of k.

Answer

To determine why the quadratic expression \(3x^2 + kx - 1\) is never always positive for any value of \(k\), we can analyze the properties of quadratic functions.

### Step-by-step Solution:

1. **Identify the Quadratic Function**:
The given expression is \(p(x) = 3x^2 + kx - 1\).

2. **Identify the Coefficients**:
In the standard form \(ax^2 + bx + c\), we have:
- \(a = 3\)
- \(b = k\)
- \(c = -1\)

3. **Check the Leading Coefficient**:
Since \(a = 3\) is positive, the parabola opens upwards.

4. **Calculate the Discriminant**:
The discriminant \(D\) of a quadratic equation is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = k^2 - 4(3)(-1) = k^2 + 12 \]

5. **Analyze the Discriminant**:
The discriminant \(D = k^2 + 12\) is always positive for all real values of \(k\) because \(k^2 \geq 0\) (the square of any real number is non-negative) and \(12\) is a positive constant. Therefore, \(D > 0\) for all \(k\).

6. **Conclusion about Roots**:
Since the discriminant is positive, the quadratic equation \(3x^2 + kx - 1 = 0\) has two distinct real roots. This means that the parabola intersects the x-axis at two points.

7. **Behavior of the Quadratic Function**:
Since the parabola opens upwards and has two distinct real roots, it will be negative between the two roots and positive outside of that interval. Therefore, there will always be some values of \(x\) for which \(p(x) < 0\).

### Final Conclusion:
Thus, \(3x^2 + kx - 1\) is not always positive for any value of \(k\) because it takes negative values for some \(x\).