Solve x²+ ½x - 4½: Finding Values Where Function is Positive

To solve the problem, let's determine when $y = x^2 + \frac{1}{2}x - 4.5$ is greater than zero by following these steps:

• Step 1: Find the roots of the quadratic equation using the quadratic formula.
• Step 2: Determine the intervals defined by these roots.
• Step 3: Identify where the quadratic function is positive.

**Step 1**: Given the quadratic function $y = x^2 + \frac{1}{2}x - 4.5$, the coefficients are $a = 1$, $b = \frac{1}{2}$, and $c = -4.5$. Apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-\left(\frac{1}{2}\right) \pm \sqrt{\left(\frac{1}{2}\right)^2 - 4(1)(-4.5)}}{2(1)}$

Simplify further:

$x = \frac{-\frac{1}{2} \pm \sqrt{\frac{1}{4} + 18}}{2} = \frac{-\frac{1}{2} \pm \sqrt{\frac{73}{4}}}{2}$

This becomes:

$x = \frac{-1 \pm \sqrt{73}}{4}$

**Step 2**: The roots are $x_1 = \frac{-1 - \sqrt{73}}{4}$ and $x_2 = \frac{-1 + \sqrt{73}}{4}$. The function changes sign at the roots. Since the quadratic opens upwards (as $a = 1 > 0$), it will be positive between the roots:

**Step 3**: Identify the interval where the quadratic is positive:

$\frac{-1 - \sqrt{73}}{4} < x < \frac{-1 + \sqrt{73}}{4}$

Therefore, the values of $x$ for which the function is greater than zero are within this interval.

The correct solution is $\frac{-1 - \sqrt{73}}{4} < x < \frac{-1 + \sqrt{73}}{4}$.