1. Stating the problem: We want to explain why the quadratic expression $3x^2 + kx - 1$ is never positive definite for any value of $k$. 2. Recall that a quadratic expression $ax^2 + bx + c$ is positive definite if $a > 0$ and the expression is strictly positive for all $x \neq 0$. 3. Here, $a = 3$, $b = k$, and $c = -1$. Since $a = 3 > 0$, the parabola opens upwards. 4. For the quadratic to be positive definite, its minimum value must be greater than zero. 5. The vertex of the parabola is at $x = -\frac{b}{2a} = -\frac{k}{2 \times 3} = -\frac{k}{6}$. 6. The minimum value is $f\left(-\frac{k}{6}\right) = 3\left(-\frac{k}{6}\right)^2 + k \left(-\frac{k}{6}\right) - 1 = 3 \frac{k^2}{36} - \frac{k^2}{6} - 1 = \frac{k^2}{12} - \frac{k^2}{6} - 1$. 7. Simplify: $\frac{k^2}{12} - \frac{k^2}{6} = \frac{k^2}{12} - \frac{2k^2}{12} = -\frac{k^2}{12}$. 8. Hence the minimum value is $-\frac{k^2}{12} - 1 = -\left(\frac{k^2}{12} + 1\right)$. 9. Since $\frac{k^2}{12} + 1 > 0$ for all $k$, the minimum value is always negative. 10. Therefore, the quadratic expression $3x^2 + kx - 1$ achieves a negative value for some $x$, and thus cannot be positive definite for any $k$. Final answer: $3x^2 + kx - 1$ is never positive definite because its minimum value is always negative regardless of $k$.