Solve the Quadratic: Where 1/4x² - 3.5x Is Positive

To solve the problem of finding for which values of $x$ the function $y = \frac{1}{4}x^2 - \frac{7}{2}x$ is greater than zero, follow these steps:

• Step 1: Identify the function. The quadratic function $y = \frac{1}{4}x^2 - \frac{7}{2}x$ can be rewritten as $y = \frac{1}{4}x^2 - \frac{7}{2}x$.
• Step 2: Use the quadratic formula to find the roots of the equation $y = 0$. The quadratic formula is:

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

Here, $a = \frac{1}{4}$, $b = -\frac{7}{2}$, and $c = 0$.

• Step 3: Calculate $b^2 - 4ac$:

$b^2 - 4ac = \left(-\frac{7}{2}\right)^2 - 4 \left(\frac{1}{4}\right)(0) = \frac{49}{4}$

• Step 4: Find the roots using the quadratic formula:

$x = \frac{-\left(-\frac{7}{2}\right) \pm \sqrt{\frac{49}{4}}}{2 \times \frac{1}{4}}$ $x = \frac{\frac{7}{2} \pm \frac{7}{2}}{\frac{1}{2}}$ $x = \frac{7 \pm 7}{1}$

This gives two roots: $x = 0$ and $x = 14$.

• Step 5: Determine the intervals defined by these roots. We have three intervals to check: $x < 0$, $0 < x < 14$, and $x > 14$.

• Step 6: Test the sign of $f(x)$ in each interval:
• For $x < 0$, choose $x = -1$:

$f(-1) = \frac{1}{4}(-1)^2 - \frac{7}{2}(-1) = \frac{1}{4} + \frac{7}{2} > 0$

• For $0 < x < 14$, choose $x = 1$:

$f(1) = \frac{1}{4}(1)^2 - \frac{7}{2}(1) = \frac{1}{4} - \frac{7}{2} < 0$

• For $x > 14$, choose $x = 15$:

$f(15) = \frac{1}{4}(15)^2 - \frac{7}{2}(15) = \frac{225}{4} - \frac{105}{2} > 0$

After confirming the signs, we find that $f(x) > 0$ for $x \lt 0$ and $x \gt 14$.

Therefore, the solution to the problem is $<strong>x > 14$ or $x < 0$.