Positive and Negative Domains: With fractions

To solve this problem, we need to analyze where the expression $(x - 4.4)(x - 2.3)$ is greater than zero. We have two roots, $x = 4.4$ and $x = 2.3$, which divide the number line into three intervals: $x < 2.3$, $2.3 < x < 4.4$, and $x > 4.4$.

Let's check these intervals:

• For $x < 2.3$, both $(x - 4.4)$ and $(x - 2.3)$ are negative, resulting in a positive product.
• For $2.3 < x < 4.4$, $(x - 4.4)$ is negative and $(x - 2.3)$ is positive, resulting in a negative product.
• For $x > 4.4$, both $(x - 4.4)$ and $(x - 2.3)$ are positive, resulting in a positive product.

Thus, the expression $(x - 4.4)(x - 2.3) > 0$ holds in the intervals $x < 2.3$ and $x > 4.4$.

Therefore, the solution is $x > 4.4$ or $x < 2.3$.

To solve this problem, we need to find the roots and determine the sign of the function on intervals between these roots:

• Step 1: Find the roots of the quadratic by solving each factor equal to zero.
• Step 2: $x - \frac{1}{3} = 0$ gives $x = \frac{1}{3}$.
  Solve $-x - 2\frac{1}{4} = 0$ gives $x = -2\frac{1}{4}$ or $x = -\frac{9}{4}$.
• Step 3: This defines the critical points, $x = \frac{1}{3}$ and $x = -\frac{9}{4}$.
• Step 4: Determine the sign of the function on intervals: $(-\infty, -\frac{9}{4})$, $(-\frac{9}{4}, \frac{1}{3})$, and $(\frac{1}{3}, \infty)$.
• Step 5: Test points in each interval:
  For $x < -\frac{9}{4}$, both factors are negative, the product is positive: Interval does not satisfy.
  For $-\frac{9}{4} < x < \frac{1}{3}$, signs will vary, and the product is negative: Interval satisfies $f(x) < 0$.
  For $x > \frac{1}{3}$, both factors are positive, the product is positive: Interval does not satisfy.

Thus, the solution is for values where the product is negative: $-2\frac{1}{4} < x < \frac{1}{3}$.

The correct answer choice is therefore Choice 1

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

• Step 1: Identify the roots of the quadratic expression.
• Step 2: Determine the sign of each factor in intervals defined by these roots.
• Step 3: Identify where the product of these factors is positive.

Now, let's work through each step:

Step 1: Identify the roots.
The given function is $y = \left(x - \frac{1}{3}\right)\left(-x - 2\frac{1}{4}\right)$. To find the roots, solve each factor for zero:

• $x - \frac{1}{3} = 0$ gives $x = \frac{1}{3}$.
• $-x - 2\frac{1}{4} = 0$ gives $x = -2\frac{1}{4}$.

Step 2: Determine the sign of each factor in the intervals defined by these roots.
The zeros divide the x-axis into three intervals: $(-∞, -2\frac{1}{4})$, $(-2\frac{1}{4}, \frac{1}{3})$, and $(\frac{1}{3}, ∞)$.

Step 3: Test the signs and find where the product is positive.

• For $x < -2\frac{1}{4}$, test with $x = -3$: - $x - \frac{1}{3} = -3 - \frac{1}{3} < 0$ - $-x - 2\frac{1}{4} = 3 - 2\frac{1}{4} > 0$ - Product: Negative
• For $-2\frac{1}{4} < x < \frac{1}{3}$, test with $x = 0$: - $x - \frac{1}{3} = 0 - \frac{1}{3} < 0$ - $-x - 2\frac{1}{4} = 0 - 2\frac{1}{4} > 0$ - Product: Positive
• For $x > \frac{1}{3}$, test with $x = 1$: - $x - \frac{1}{3} = 1 - \frac{1}{3} > 0$ - $-x - 2\frac{1}{4} = -1 - 2\frac{1}{4} < 0$ - Product: Negative

Therefore, the solution to $f(x) > 0$ is in the interval:

$-2\frac{1}{4} < x < \frac{1}{3}$.

The function $y = \left(\frac{1}{3}x + \frac{1}{6}\right)\left(-x - 4\frac{1}{5}\right)$ requires us to analyze the sign of the product for various $x$ values.

First, we must find the zeros of each factor:

• The zero of $\frac{1}{3}x + \frac{1}{6}$ is found by solving $\frac{1}{3}x + \frac{1}{6} = 0$:
  Subtract $\frac{1}{6}$ to get:
  $\frac{1}{3}x = -\frac{1}{6}$
  Multiply both sides by 3:
  $x = -\frac{1}{2}$.
• The zero of $-x - 4\frac{1}{5}$ is found by solving $-x - 4\frac{1}{5} = 0$:
  Add $4\frac{1}{5}$ to get:
  $-x = 4\frac{1}{5}$
  Multiply by $-1$ to find:
  $x = -4\frac{1}{5}$.

Next, we identify the intervals defined by these zeros: $x < -4\frac{1}{5}$, $-4\frac{1}{5} < x < -\frac{1}{2}$, and $x > -\frac{1}{2}$.

We will determine the sign of the function in each interval:

• In $x < -4\frac{1}{5}$:
  - $\frac{1}{3}x + \frac{1}{6}$ and $-x - 4\frac{1}{5}$ are both negative (since both points are below their respective roots), resulting in a positive product.
• In $-4\frac{1}{5} < x < -\frac{1}{2}$:
  - $\frac{1}{3}x + \frac{1}{6}$ is negative and $-x - 4\frac{1}{5}$ is positive, resulting in a negative product.
• In $x > -\frac{1}{2}$:
  - $\frac{1}{3}x + \frac{1}{6}$ and $-x - 4\frac{1}{5}$ are both positive, resulting in a positive product.

The function is negative in the interval $-4\frac{1}{5} < x < -\frac{1}{2}$. Thus, the correct answer corresponding to where the function is negative is the complementary intervals $x > -\frac{1}{2}$ or $x < -4\frac{1}{5}$, which matches choice 2.

Therefore, the solution is $x > -\frac{1}{2}$ or $x < -4\frac{1}{5}$.

To find when the function $y = \left(x - \frac{1}{2}\right)\left(x + 6\frac{1}{2}\right)$ is positive, we proceed as follows:

First, identify the roots of the expression by solving $x - \frac{1}{2} = 0$ and $x + 6\frac{1}{2} = 0$. These calculations give us the roots $x = \frac{1}{2}$ and $x = -6\frac{1}{2}$, or $x = -\frac{13}{2}$.

Next, determine the sign of the product $(x - \frac{1}{2})(x + \frac{13}{2})$ over the intervals defined by these roots:

• Interval 1: $x < -\frac{13}{2}$. In this region, both $x - \frac{1}{2}$ and $x + \frac{13}{2}$ are negative, so their product is positive.
• Interval 2: $-\frac{13}{2} < x < \frac{1}{2}$. In this region, $x + \frac{13}{2}$ is positive and $x - \frac{1}{2}$ is negative, so their product is negative.
• Interval 3: $x > \frac{1}{2}$. In this region, both $x - \frac{1}{2}$ and $x + \frac{13}{2}$ are positive, so their product is positive.

Therefore, the function is positive for $x < -6\frac{1}{2}$ and $x > \frac{1}{2}$.

Thus, the solution is:
$x > \frac{1}{2}$ or $x < -6\frac{1}{2}$

Let us solve the problem step by step to find: $x$ values for which $f(x) < 0$.

Firstly, identify the roots of the function $y = \left(x - \frac{1}{2}\right)\left(x + 6\frac{1}{2}\right)$:

• The root from $\left(x - \frac{1}{2}\right) = 0$ is $x = \frac{1}{2}$.
• The root from $\left(x + 6\frac{1}{2}\right) = 0$ is $x = -6\frac{1}{2}$.

These roots divide the real number line into three intervals:

• $x < -6\frac{1}{2}$
• $-6\frac{1}{2} < x < \frac{1}{2}$
• $x > \frac{1}{2}$

To determine where the function is negative, evaluate the sign in each interval:

• For $x < -6\frac{1}{2}$: Both factors $(x - \frac{1}{2})$ and $(x + 6\frac{1}{2})$ are negative, so their product is positive.
• For $-6\frac{1}{2} < x < \frac{1}{2}$: $(x - \frac{1}{2})$ is negative and $(x + 6\frac{1}{2})$ is positive, thus the product is negative.
• For $x > \frac{1}{2}$: Both factors are positive, so their product is positive.

Hence, the function is negative on the interval: $-6\frac{1}{2} < x < \frac{1}{2}$.

To solve this problem, we'll perform the following steps:

• Step 1: Identify the zeros of each factor.
• Step 2: Determine the sign of each factor across different intervals on the number line.
• Step 3: Identify where both factors give a positive product.

Now, let us work through each step:

Step 1: Find the values of $x$ where each factor equals zero:

• $5x - 1 = 0$ gives us $x = \frac{1}{5}$.
• $4x - \frac{1}{4} = 0$ gives us $x = \frac{1}{16}$.

These zeros divide the number line into intervals: $x < \frac{1}{16}$, $\frac{1}{16} < x < \frac{1}{5}$, and $x > \frac{1}{5}$.

Step 2: Analyze the sign of each factor in each interval:

• For $x < \frac{1}{16}$:
• $5x - 1$ is negative,
• $4x - \frac{1}{4}$ is negative,
• The product $(5x - 1)(4x - \frac{1}{4})$ is positive.
• For $\frac{1}{16} < x < \frac{1}{5}$:
• $5x - 1$ is negative,
• $4x - \frac{1}{4}$ is positive,
• The product $(5x - 1)(4x - \frac{1}{4})$ is negative.
• For $x > \frac{1}{5}$:
• $5x - 1$ is positive,
• $4x - \frac{1}{4}$ is positive,
• The product $(5x - 1)(4x - \frac{1}{4})$ is positive.

Step 3: Identify intervals where product is positive:

• $x < \frac{1}{16}$ and $x > \frac{1}{5}$.

Therefore, the solution to the inequality $y > 0$ is:
$x > \frac{1}{5}$ or $x < \frac{1}{16}$.

To solve the problem of determining for which values of $x$ the function $y = (5x-1)(4x-\frac{1}{4})$ is negative, we will follow these steps:

• Step 1: Determine the roots of the function by setting each factor equal to zero.
• Step 2: Analyze the sign of the function on intervals defined by these roots.
• Step 3: Identify where the function is negative based on this analysis.

Let's proceed with these steps:

Step 1: Find the roots of the function.

To find the roots, set each factor equal to zero:

$5x - 1 = 0 \Rightarrow x = \frac{1}{5}$

$4x - \frac{1}{4} = 0 \Rightarrow x = \frac{1}{16}$

Step 2: Determine the intervals on the number line.

The roots divide the number line into the following intervals: $(-\infty, \frac{1}{16})$, $(\frac{1}{16}, \frac{1}{5})$, and $(\frac{1}{5}, \infty)$.

Step 3: Analyze the sign of the function in each interval:

• For $x < \frac{1}{16}$: Both $5x - 1$ and $4x - \frac{1}{4}$ are negative. Hence, their product is positive.
• For $\frac{1}{16} < x < \frac{1}{5}$: Here, $5x - 1$ is negative, and $4x - \frac{1}{4}$ is positive, making the product negative.
• For $x > \frac{1}{5}$: Both $5x - 1$ and $4x - \frac{1}{4}$ are positive, resulting in a positive product.

Now, consolidate the findings:

The function $y = (5x-1)(4x-\frac{1}{4})$ is less than zero for values $\frac{1}{16} < x < \frac{1}{5}$.

Therefore, the solution to the given problem is $\frac{1}{16} < x < \frac{1}{5}$.

To solve this problem, we'll begin by finding the roots of the quadratic equation $y = \left(x - \frac{1}{2}\right)\left(-x + 3\frac{1}{2}\right)$.

First, set each factor equal to zero:

• $x - \frac{1}{2} = 0$ gives $x = \frac{1}{2}$
• $-x + 3\frac{1}{2} = 0$ gives $x = 3\frac{1}{2}$

This means the roots of the quadratic are $x = \frac{1}{2}$ and $x = 3\frac{1}{2}$.

Next, analyze the intervals determined by these roots:

• Interval 1: $x < \frac{1}{2}$
• Interval 2: $\frac{1}{2} < x < 3\frac{1}{2}$
• Interval 3: $x > 3\frac{1}{2}$

Perform a sign test within these intervals:

• For $x < \frac{1}{2}$: Both $x - \frac{1}{2}$ and $-x + 3\frac{1}{2}$ are negative, thus their product is positive (negative times negative is positive).
• For $\frac{1}{2} < x < 3\frac{1}{2}$: The factor $x - \frac{1}{2}$ is positive and $-x + 3\frac{1}{2}$ is positive as well, thus product is positive (positive times positive is positive).
• For $x > 3\frac{1}{2}$: $x - \frac{1}{2}$ is positive, but $-x + 3\frac{1}{2}$ is negative, so the product is negative (positive times negative is negative).

Therefore, the quadratic function is negative for: $x > 3\frac{1}{2}$ and $x < \frac{1}{2}$.

The solution to the problem is: $x > 3\frac{1}{2}$ or $x < \frac{1}{2}$.

To solve this problem, we'll determine when the product $(x - \frac{1}{2})(-x + 3\frac{1}{2})$ is positive. This involves finding the roots of the equation and testing the intervals between these roots:

Step 1: **Determine the roots of the factors.**
- The first factor $x - \frac{1}{2} = 0$ gives the root $x = \frac{1}{2}$.
- The second factor $-x + 3\frac{1}{2} = 0$ gives the root $x = 3\frac{1}{2}$.

Step 2: **Identify intervals based on these roots.**
- The roots divide the $x$-axis into three intervals: $x < \frac{1}{2}$, $\frac{1}{2} < x < 3\frac{1}{2}$, and $x > 3\frac{1}{2}$.

Step 3: **Analyze the sign of the function in each interval.**
- For $x < \frac{1}{2}$:
- $x - \frac{1}{2} < 0$ and $-x + 3\frac{1}{2} > 0$, so the product is negative.
- For $\frac{1}{2} < x < 3\frac{1}{2}$:
- Both $x - \frac{1}{2} > 0$ and $-x + 3\frac{1}{2} > 0$, so the product is positive.
- For $x > 3\frac{1}{2}$:
- $x - \frac{1}{2} > 0$ and $-x + 3\frac{1}{2} < 0$, so the product is negative.

Therefore, the intervals where $y > 0$ are $\frac{1}{2} < x < 3\frac{1}{2}$.

This matches the given correct answer choice: $\frac{1}{2} < x < 3\frac{1}{2}$.

To solve this problem, we will examine the intervals defined by the roots of the quadratic function.

Step 1: Find the roots of each factor:
For $2x - \frac{1}{2} = 0$, solve for $x$:
$2x = \frac{1}{2} \quad \Rightarrow \quad x = \frac{1}{4}$
For $x - 2\frac{1}{4} = 0$, solve for $x$:
$x = 2\frac{1}{4}$

Step 2: Determine the test intervals around these roots, which are $x < \frac{1}{4}$, $\frac{1}{4} < x < 2\frac{1}{4}$, and $x > 2\frac{1}{4}$.

Step 3: Test each interval to determine where the product is positive:

• For $x < \frac{1}{4}$, both factors $(2x - \frac{1}{2})$ and $(x - 2\frac{1}{4})$ are negative, so the product is positive.
• For $\frac{1}{4} < x < 2\frac{1}{4}$, one factor is positive $(2x - \frac{1}{2})$ and the other is negative $(x - 2\frac{1}{4})$, resulting in a negative product.
• For $x > 2\frac{1}{4}$, both factors $(2x - \frac{1}{2})$ and $(x - 2\frac{1}{4})$ are positive, so the product is positive.

Therefore, the solution for $f(x) > 0$ is when $x > 2\frac{1}{4}$ or $x < \frac{1}{4}$.

The correct choice that matches this analysis is:
$x > 2\frac{1}{4}$ or $x < \frac{1}{4}$.

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

• Step 1: Identify the roots of the quadratic function.
• Step 2: Analyze the intervals between these roots to determine where the function is negative.
• Step 3: Write the final solution as the interval where $f(x) < 0$.

Now, let's work through each step:
Step 1: Determine the roots by solving each factor for zero:
- $2x - \frac{1}{2} = 0 \Rightarrow 2x = \frac{1}{2} \Rightarrow x = \frac{1}{4}$.
- $x - 2\frac{1}{4} = 0 \Rightarrow x = 2\frac{1}{4}$.
Thus, the roots are $x = \frac{1}{4}$ and $x = 2\frac{1}{4}$.

Step 2: Analyze the intervals determined by the roots $x = \frac{1}{4}$ and $x = 2\frac{1}{4}$:

• Interval 1: $x < \frac{1}{4}$
• Interval 2: $\frac{1}{4} < x < 2\frac{1}{4}$
• Interval 3: $x > 2\frac{1}{4}$

Step 3: Test each interval:

• For interval 1 $x < \frac{1}{4}$: Choose $x = 0$. $f(0) = \left(2(0) - \frac{1}{2}\right)(0 - 2\frac{1}{4}) = \left(-\frac{1}{2}\right) \left(-2\frac{1}{4}\right) = \frac{9}{8} > 0$.
• For interval 2 $\frac{1}{4} < x < 2\frac{1}{4}$: Choose $x = 1$. $f(1) = \left(2(1) - \frac{1}{2}\right)(1 - 2\frac{1}{4}) = \left(\frac{3}{2}\right)(-\frac{5}{4}) = -\frac{15}{8} < 0$.
• For interval 3 $x > 2\frac{1}{4}$: Choose $x = 3$. $f(3) = \left(2(3) - \frac{1}{2}\right)(3 - 2\frac{1}{4}) = \left(\frac{11}{2}\right)\left(\frac{3}{4}\right) = \frac{33}{8} > 0$.

Therefore, the solution to $f(x) < 0$ is found in the interval $\frac{1}{4} < x < 2\frac{1}{4}$.

To find the positive and negative domains of the function $y = -x^2 + \frac{3}{2}x - \frac{21}{4}$, we first determine the roots of the equation:

Set $y = 0$, giving us:

$-x^2 + \frac{3}{2}x - \frac{21}{4} = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = \frac{3}{2}$, and $c = -\frac{21}{4}$, we calculate:

• First, compute the discriminant: $b^2 - 4ac = \left(\frac{3}{2}\right)^2 - 4(-1)\left(-\frac{21}{4}\right) = \frac{9}{4} - \frac{21}{1} = \frac{9}{4} - \frac{84}{4} = -\frac{75}{4}$.
• Since the discriminant is negative, the quadratic has no real roots.

This implies that the parabola does not intersect the $x$-axis and since the quadratic coefficient $a = -1$ is negative, the parabola opens downwards.

Thus, the function is always negative for all $x$. Therefore, the positive domain is empty, and the negative domain is the entire set of real numbers.

Conclusion: The solution to the problem is as follows:

$x < 0$: for all $x$

$x > 0$: none

To solve this problem, we'll determine where the quadratic function $y = -\frac{1}{2}x^2 + x - 1$ is positive and negative.

Step 1: Find the roots of the quadratic equation.
We'll use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -\frac{1}{2}$, $b = 1$, and $c = -1$.

Calculate the discriminant $b^2 - 4ac = 1^2 - 4(-\frac{1}{2})(-1) = 1 - 2 = -1$.
Notice that the discriminant is negative, meaning the quadratic equation has no real roots.

Step 2: Determine the sign of the quadratic function.
Since there are no real roots, the quadratic does not intersect the x-axis. Since $a = -\frac{1}{2}$, which is negative, the parabola opens downwards. Without real roots, it means it is always negative for all values of $x$.

Conclusion: The function is negative for all $x$.

Therefore, the positive and negative domains of the function are:

$x < 0 :$ for all $x$

$x > 0 :$ none

To solve this problem, we need to analyze the function $y = \frac{1}{3}x^2 + \frac{1}{2}x + \frac{2}{3}$ to determine where it is positive or negative. This function is quadratic, so it is a parabola. Let us find the domain of positive and negative values.

First, let's determine where the function is zero by finding its roots. This involves using the quadratic formula:

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

Here, we have $a = \frac{1}{3}$, $b = \frac{1}{2}$, and $c = \frac{2}{3}$. The discriminant is calculated as follows:

$b^2 - 4ac = \left(\frac{1}{2}\right)^2 - 4 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{1}{4} - \frac{8}{9}$

Calculating gives:

$\frac{1}{4} \approx 0.25$ and $\frac{8}{9} \approx 0.888$ which results in:

$0.25 - 0.888 = -0.638$

Since the discriminant is negative, there are no real roots. As the parabola opens upwards (since $a = \frac{1}{3} > 0$), the function never crosses the x-axis. Therefore, the function remains positive for all real values of $x$ and is never negative.

Thus, the positive domain consists of all real numbers, while there is no negative domain:

$x > 0$: \) for all $x$

$x < 0$: \) none

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

• Step 1: Understand the nature of the quadratic function
• Step 2: Determine when $y = x^2 + \frac{57}{20}$ is negative
• Step 3: Determine when it is positive

Now, let's work through each step:
Step 1: The function $y = x^2 + \frac{57}{20}$ is a parabola that opens upwards because $x^2$ is always non-negative.
Step 2: Since $x^2$ is always non-negative and $\frac{57}{20}$ is positive, $y = x^2 + \frac{57}{20}$ is always positive or zero.
Step 3: The function cannot be negative given that both terms $x^2$ and $\frac{57}{20}$ are both non-negative and positive, respectively.

However, for determining strictly positive values, it holds true for function value when $y > 0$. This occurs when $x \neq 0$, but since $y$ is indeed non negative for all reals, looking for positive domains here matters less for positive values.

Therefore, the solution to the problem is:

$x < 0 :$ none

$x > 0 :$ all x

Let's begin by finding the roots of the quadratic function $y = -x^2 + \frac{3}{4}x - 2$. This will help us determine the intervals where the function is positive or negative.

The coefficients of the quadratic are $a = -1$, $b = \frac{3}{4}$, and $c = -2$. Applying the quadratic formula:

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

Substitute in the values:

$x = \frac{-\frac{3}{4} \pm \sqrt{\left(\frac{3}{4}\right)^2 - 4(-1)(-2)}}{2(-1)}$ $x = \frac{-\frac{3}{4} \pm \sqrt{\frac{9}{16} - 8}}{-2}$

Calculate inside the square root:

$\frac{9}{16} - 8 = \frac{9}{16} - \frac{128}{16} = \frac{-119}{16}$

The discriminant $b^2 - 4ac$ is negative ($\frac{-119}{16}$), indicating there are no real roots. Therefore, the quadratic function doesn't intersect the x-axis.

Since the parabola is downward (the coefficient of $x^2$ is negative), it is negative for all $x \in \mathbb{R}$.

We conclude that:

• The function $y$ does not have a positive domain for any real $x$.
• The function $y$ is negative for all $x$.

Therefore, the positive and negative domains, based on choices given, are:

$x > 0$: none

$x < 0$: all $x$

To solve this problem, follow these steps:

• Step 1: Identify the coefficients from the quadratic function $y = -\frac{1}{6}x^2 - 3\frac{1}{9}x$. Here, $a = -\frac{1}{6}$, $b = -\frac{28}{9}$, and $c = 0$.
• Step 2: Find the roots using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Plugging in the values, we get:

$x = \frac{-\left(-\frac{28}{9}\right) \pm \sqrt{\left(-\frac{28}{9}\right)^2 - 4 \times \left(-\frac{1}{6}\right) \times 0}}{2 \times \left(-\frac{1}{6}\right)}$

Simplifying, we find:

$x = \frac{\frac{28}{9}}{-\frac{1}{3}}$

$x = \frac{28}{9} \times -3$

$x = -\frac{84}{9} = -9\frac{1}{3}$

• Step 3: Explore the intervals determined by these roots for where the function $y$ is negative:
• Consider intervals $(-\infty, -18\frac{2}{3})$, $(-18\frac{2}{3}, 0)$, and $(0, \infty)$.
• Evaluate the sign of $y$ over each interval:

For $(-\infty, -18\frac{2}{3})$: Choose a sample point like $x = -20$. Plug it in to see if $y < 0$. It turns out this interval is negative.

For $(-18\frac{2}{3}, 0)$: Choose a sample point like $x = -1$. Plug it in to see if $y > 0$. This interval turns out positive.

For $(0, \infty)$: Choose a sample point like $x = 1$. Plug it in to see if $y < 0$. This interval turns out negative.

Therefore, the solution is when $x > 0$ or $x < -18\frac{2}{3}$.

Thus, the quadratic function is negative outside the roots.

Therefore, the solution to the problem is $x > 0$ or $x < -18\frac{2}{3}$.