Solve the Quadratic Inequality: When is -1/7x² - 23/7x Greater Than Zero?

To solve this problem, we'll follow these steps:

• Step 1: Identify the coefficients from the quadratic function $y = -\frac{1}{7}x^2 - 2\frac{3}{7}x$. Set $a = -\frac{1}{7}$, $b = -\frac{17}{7}$, and $c = 0$.
• Step 2: Use the quadratic formula to find the roots of the equation $f(x) = 0$.
• Step 3: Apply the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Now, let's work through the calculations:

First, we calculate the discriminant: $b^2 - 4ac = \left(-\frac{17}{7}\right)^2 - 4\left(-\frac{1}{7}\right)(0) = \frac{289}{49}.$ The discriminant is positive, indicating two distinct real roots.

The roots are computed as: $x = \frac{-(-\frac{17}{7}) \pm \sqrt{\frac{289}{49}}}{2\left(-\frac{1}{7}\right)}.$ Simplify to: $x = \frac{\frac{17}{7} \pm \frac{17}{7}}{-\frac{2}{7}}.$ For $x = \frac{\frac{17}{7} + \frac{17}{7}}{-\frac{2}{7}}$, $x = \frac{\frac{34}{7}}{-\frac{2}{7}} = -17.$ And, for $x = \frac{\frac{17}{7} - \frac{17}{7}}{-\frac{2}{7}}$, $x = \frac{0}{-\frac{2}{7}} = 0.$

The roots are $x = -17$ and $x = 0$. Therefore, the intervals to consider are $(-\infty, -17)$, $(-17, 0)$, and $(0, \infty)$.

Step 4: Test values in these intervals:

• For $x < -17$: Choose $x = -18$ in $f(x) = -\frac{1}{7}x^2 - \frac{17}{7}x$. We get positive value, as substitute results in positive after sign evaluation.
• For $-17 < x < 0$: Choose $x = -1$ in $f(x)$, resulting in negative function value.
• For $x > 0$: Choose $x = 1$ in $f(x)$, yielding positive function value.

Thus, $f(x) > 0$ for $x > 0$ or $x < -17$.

Therefore, the solution to the problem is $x > 0$ or $x < -17$.