Solve Quadratic Function: When is (1/5)x² + 1⅓x Less Than Zero?

Look at the following function:

$y=\frac{1}{5}x^2+1\frac{1}{3}x$

Determine for which values of $x$ the following is true:

$f(x) < 0$

To solve this problem, we'll proceed as follows:

• Step 1: Identify the quadratic formula and coefficients.
• Step 2: Find the roots using the quadratic formula.
• Step 3: Determine intervals and test signs.

Step 1: Identify the equation coefficients.

Given: $y = \frac{1}{5}x^2 + 1\frac{1}{3}x$. Let $a = \frac{1}{5}$, $b = \frac{4}{3}$, and $c = 0$.

Step 2: Find the roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Calculate the discriminant: $b^2 - 4ac = \left(\frac{4}{3}\right)^2 - 4\left(\frac{1}{5}\right)(0) = \frac{16}{9}$.

The roots of the equation are given by $x = \frac{-\frac{4}{3} \pm \sqrt{\frac{16}{9}}}{2 \times \frac{1}{5}}$.

Simplifying: $x = \frac{-\frac{4}{3} \pm \frac{4}{3}}{\frac{2}{5}} = \frac{-4 \pm 4}{\frac{2}{5}} \cdot \frac{1}{3}$.

The roots are $x = 0$ and $x = -6\frac{2}{3}$.

Step 3: Determine intervals created by the roots and test each interval.

The intervals are $(-\infty, -6\frac{2}{3})$, $(-6\frac{2}{3}, 0)$, and $(0, \infty)$.

- For $x = -7$ (or any point less than $-6\frac{2}{3}$), the expression is positive.

- For $x = -3$ (or any point between $-6\frac{2}{3}$ and $0$), test by substituting back into the function.
The function yields a negative result.

- For $x = 1$ (or any point greater than $0$), the expression is positive.

Conclusion: The function is negative between $-6\frac{2}{3}$ and $0$.

Thus, the solution to the problem is $-6\frac{2}{3} < x < 0$.