Solve Quadratic Inequality: When is -x² + 2.5x - 1/4 Less Than Zero?

To solve the problem, we need to determine the values of $x$ for which the quadratic function $y = -x^2 + 2\frac{1}{2}x - \frac{1}{4}$ is less than zero.

Let's start by solving the equation $-x^2 + \frac{5}{2}x - \frac{1}{4} = 0$ to find the roots using the quadratic formula:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = \frac{5}{2}$, and $c = -\frac{1}{4}$.

Calculate the discriminant:

$b^2 - 4ac = \left(\frac{5}{2}\right)^2 - 4(-1)(-\frac{1}{4}) = \frac{25}{4} - 1 = \frac{21}{4}$.

Since the discriminant is positive, there are two distinct real roots.

Next, plug the discriminant back into the quadratic formula to find the roots:

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{21}{4}}}{2(-1)} = \frac{-\frac{5}{2} \pm \frac{\sqrt{21}}{2}}{-2} = \frac{5 \pm \sqrt{21}}{4}$.

Thus, the roots are $x = \frac{5 + \sqrt{21}}{4}$ and $x = \frac{5 - \sqrt{21}}{4}$.

The parabola opens downwards (since $a = -1$), so the function is positive between the roots and negative outside. Therefore, $f(x) < 0$ for $x > \frac{5+\sqrt{21}}{4}$ or $x < \frac{5-\sqrt{21}}{4}$.

The correct answer is $x > \frac{5+\sqrt{21}}{4}$ or $x < \frac{5-\sqrt{21}}{4}$.