Solving Quadratic Inequality: When is -x² + 2.5x - 1/4 > 0

Look at the function below:

$y=-x^2+2\frac{1}{2}x-\frac{1}{4}$

Then determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, follow these steps:

• Step 1: Use the quadratic formula to find the roots of $y = -x^2 + \frac{5}{2}x - \frac{1}{4}$.
• Step 2: Determine the intervals based on these roots and check where the function is positive.

Step 1: Find the roots using the quadratic formula:
The quadratic equation is $-x^2 + \frac{5}{2}x - \frac{1}{4} = 0$, with $a = -1$, $b = \frac{5}{2}$, and $c = -\frac{1}{4}$.
The roots are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values into the formula:

$x = \frac{-\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4(-1)\left(-\frac{1}{4}\right)}}{2(-1)}$

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{25}{4} - 1}}{-2}$

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{21}{4}}}{-2}$

$x = \frac{-\frac{5}{2} \pm \frac{\sqrt{21}}{2}}{-2}$

Simplify each root:

$x = \frac{5 \pm \sqrt{21}}{4}$

Step 2: Determine the intervals:
The roots are $x = \frac{5-\sqrt{21}}{4}$ and $x = \frac{5+\sqrt{21}}{4}$.

• The function is a downward-opening parabola (because $a = -1$), so it is positive between the roots and negative outside them.

Therefore, $f(x) > 0$ for $\frac{5-\sqrt{21}}{4} < x < \frac{5+\sqrt{21}}{4}$.

The solution is $\frac{5-\sqrt{21}}{4} < x < \frac{5+\sqrt{21}}{4}$.