Positive and Negative Domains: Using roots

To solve this problem, we will analyze the quadratic function $y = x^2 + \frac{1}{2}x - 4\frac{1}{2}$. We need to determine for which values of $x$ the function $f(x) > 0$.

• Step 1: Identify the coefficients: The function can be rewritten in standard form as $y = ax^2 + bx + c$, where $a = 1$, $b = \frac{1}{2}$, $c = -\frac{9}{2}$.
• Step 2: Use the quadratic formula to find the roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 3: Calculate the discriminant: $\Delta = b^2 - 4ac = \left(\frac{1}{2}\right)^2 - 4(1)\left(-\frac{9}{2}\right) = \frac{1}{4} + 18 = \frac{73}{4}$. Since $\Delta > 0$, there are two real roots.
• Step 4: Find the roots: $x = \frac{-\frac{1}{2} \pm \sqrt{\frac{73}{4}}}{2} = \frac{-\frac{1}{2} \pm \frac{\sqrt{73}}{2}}{2}$ $x = \frac{-1 \pm \sqrt{73}}{4}$ These roots are $x_1 = \frac{-1 - \sqrt{73}}{4}$ and $x_2 = \frac{-1 + \sqrt{73}}{4}$.
• Step 5: Analyze the sign of the quadratic: The parabola opens upwards (as $a = 1 > 0$). It is positive outside its roots: $x < x_1$ and $x > x_2$.

Therefore, the values of $x$ that satisfy $f(x) > 0$ are $x > \frac{-1+\sqrt{73}}{4}$ or $x < \frac{-1-\sqrt{73}}{4}$.

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

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

To solve the problem, we need to determine the values of $x$ for which the quadratic function $y = -x^2 + 2\frac{1}{2}x - \frac{1}{4}$ is less than zero.

Let's start by solving the equation $-x^2 + \frac{5}{2}x - \frac{1}{4} = 0$ to find the roots using the quadratic formula:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = \frac{5}{2}$, and $c = -\frac{1}{4}$.

Calculate the discriminant:

$b^2 - 4ac = \left(\frac{5}{2}\right)^2 - 4(-1)(-\frac{1}{4}) = \frac{25}{4} - 1 = \frac{21}{4}$.

Since the discriminant is positive, there are two distinct real roots.

Next, plug the discriminant back into the quadratic formula to find the roots:

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{21}{4}}}{2(-1)} = \frac{-\frac{5}{2} \pm \frac{\sqrt{21}}{2}}{-2} = \frac{5 \pm \sqrt{21}}{4}$.

Thus, the roots are $x = \frac{5 + \sqrt{21}}{4}$ and $x = \frac{5 - \sqrt{21}}{4}$.

The parabola opens downwards (since $a = -1$), so the function is positive between the roots and negative outside. Therefore, $f(x) < 0$ for $x > \frac{5+\sqrt{21}}{4}$ or $x < \frac{5-\sqrt{21}}{4}$.

The correct answer is $x > \frac{5+\sqrt{21}}{4}$ or $x < \frac{5-\sqrt{21}}{4}$.

To solve this problem, follow these steps:

• Step 1: Use the quadratic formula to find the roots of $y = -x^2 + \frac{5}{2}x - \frac{1}{4}$.
• Step 2: Determine the intervals based on these roots and check where the function is positive.

Step 1: Find the roots using the quadratic formula:
The quadratic equation is $-x^2 + \frac{5}{2}x - \frac{1}{4} = 0$, with $a = -1$, $b = \frac{5}{2}$, and $c = -\frac{1}{4}$.
The roots are given by:

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

Substitute the values into the formula:

$x = \frac{-\frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4(-1)\left(-\frac{1}{4}\right)}}{2(-1)}$

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{25}{4} - 1}}{-2}$

$x = \frac{-\frac{5}{2} \pm \sqrt{\frac{21}{4}}}{-2}$

$x = \frac{-\frac{5}{2} \pm \frac{\sqrt{21}}{2}}{-2}$

Simplify each root:

$x = \frac{5 \pm \sqrt{21}}{4}$

Step 2: Determine the intervals:
The roots are $x = \frac{5-\sqrt{21}}{4}$ and $x = \frac{5+\sqrt{21}}{4}$.

• The function is a downward-opening parabola (because $a = -1$), so it is positive between the roots and negative outside them.

Therefore, $f(x) > 0$ for $\frac{5-\sqrt{21}}{4} < x < \frac{5+\sqrt{21}}{4}$.

The solution is $\frac{5-\sqrt{21}}{4} < x < \frac{5+\sqrt{21}}{4}$.

To solve this problem, we start by identifying where the given quadratic function is positive and where it is negative.

• Step 1: Convert the vertex form to solve for $y = 0$. The equation is $-\left(x + 10\right)^2 + 2 = 0$.
• Step 2: Rearrange the equation to find roots:
  $(x+10)^2 = 2$.
  Taking the square root, we have $x + 10 = \pm \sqrt{2}$.
• Step 3: Solve for $x$:
  $x = -10 + \sqrt{2}$ and $x = -10 - \sqrt{2}$.
• Step 4: Analyze the intervals:

The roots divide the number line into intervals. We check these intervals for $y > 0$ and $y < 0$.

• For $y > 0$: The function is a downward-opening parabola, hence positive between its roots: $-10-\sqrt{2} < x < -10+\sqrt{2}$.
• For $y < 0$: The function is negative outside the interval where it is positive, giving $x < -10-\sqrt{2}$ or $x > -10+\sqrt{2}$.

Therefore, for the positive domain $x > 0$, we have the interval $-10-\sqrt{2} < x < -10+\sqrt{2}$. For the negative domain, it is when $x < 0$ such that $x < -10-\sqrt{2}$ or $x > -10+\sqrt{2}$.

Thus, the correct solution choice is:

$x > 0 : -10-\sqrt{2} < x < -10+\sqrt{2}$

$x > -10+\sqrt{2}$ or $x < 0 : x < -10-\sqrt{2}$

To find the positive and negative domains of the function $y = (x+3)^2 - 5$, we first identify the roots by setting $y = 0$ and solving for $x$.

Let's solve $(x+3)^2 - 5 = 0$:

1. Start with the equation: $(x+3)^2 - 5 = 0$
2. Add 5 to both sides: $(x+3)^2 = 5$
3. Take the square root of both sides: $x+3 = \pm \sqrt{5}$
4. Solve for $x$:
• $x = -3 + \sqrt{5}$
• $x = -3 - \sqrt{5}$

Thus, the roots of the function are $x = -3 + \sqrt{5}$ and $x = -3 - \sqrt{5}$.

Since the parabola opens upwards (the coefficient of $(x+3)^2$ is positive), the function $y$ is:

• Negative between the roots: $-3 - \sqrt{5} < x < -3 + \sqrt{5}$
• Positive outside these roots: $x < -3 - \sqrt{5}$ and $x > -3 + \sqrt{5}$

Therefore, the positive and negative domains are:

• Negative domain: $-3 - \sqrt{5} < x < -3 + \sqrt{5}$
• Positive domain: $x < -3 - \sqrt{5}$ or $x > -3 + \sqrt{5}$

Upon reviewing the multiple choice options, the correct answer that corresponds to this solution is:

$x < 0 : -3-\sqrt{5} < x < -3+\sqrt{5}$

$x>-3+\sqrt{5}$ or $x > 0 : x < -3-\sqrt{5}$

To solve this problem, we first determine where the quadratic function $y = -3x^2 + 13$ is equal to zero. Setting $-3x^2 + 13 = 0$ yields:

• Rearrange the equation: $-3x^2 = -13$, leading to $x^2 = \frac{13}{3}$.

• Solve for $x$: $x = \pm \sqrt{\frac{13}{3}}$.

The roots of the equation are $x = \sqrt{\frac{13}{3}}$ and $x = -\sqrt{\frac{13}{3}}$. These roots divide the number line into three intervals: $x < -\sqrt{\frac{13}{3}}$, $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, and $x > \sqrt{\frac{13}{3}}$.

Determine the sign of the function in each interval:

• For $x < -\sqrt{\frac{13}{3}}$, select a test point (e.g., $x = -3$), the function value is negative because $-3(-3^2) + 13 = -27 + 13 = -14$.

• For $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, select a test point (e.g., $x = 0$), the function value is positive because $-3(0)^2 + 13 = 13$.

• For $x > \sqrt{\frac{13}{3}}$, select a test point (e.g., $x = 3$), the function value is negative because $-3(3^2) + 13 = -27 + 13 = -14$.

Therefore, the positive domain of the function is $-\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$, and the negative domain is $x < -\sqrt{\frac{13}{3}}$ or $x > \sqrt{\frac{13}{3}}$.

The answer matches choice 3:

$x > \sqrt{\frac{13}{3}}$ or $x < 0 : x < -\sqrt{\frac{13}{3}}$

$x > 0 : -\sqrt{\frac{13}{3}} < x < \sqrt{\frac{13}{3}}$

To solve the problem, we will find where the quadratic function $y = -4x^2 + 6$ is equal to zero.

Set the equation to zero to find the roots:

$-4x^2 + 6 = 0$

$-4x^2 = -6$

$x^2 = \frac{6}{4}$

$x^2 = \frac{3}{2}$

Take the square root of both sides:

$x = \pm \sqrt{\frac{3}{2}}$

$x = \pm \frac{\sqrt{6}}{2}$ (since $\sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$)

Now identify the intervals:

• Interval 1: $x < -\frac{\sqrt{6}}{2}$
• Interval 2: $-\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$
• Interval 3: $x > \frac{\sqrt{6}}{2}$

Test each interval to determine positivity or negativity:

• For Interval 1 ($x < -\frac{\sqrt{6}}{2}$): The parabola opens downwards and is negative outside roots.
• For Interval 2 ($-\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$): This interval is between the roots, so $y > 0$.
• For Interval 3 ($x > \frac{\sqrt{6}}{2}$): Again, as the parabola opens downward, $y < 0$.

Therefore, the positive domain is $-\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$, and the negative domains are $x < -\frac{\sqrt{6}}{2}$ and $x > \frac{\sqrt{6}}{2}$.

The correct answer is choice 4.

$x > 0: -\frac{\sqrt{6}}{2} < x < \frac{\sqrt{6}}{2}$

$x > \frac{\sqrt{6}}{2}$ or $x < 0: x < -\frac{\sqrt{6}}{2}$

To solve this problem, we'll start by calculating the roots of the function $y = -\frac{1}{2}x^2 + 5$.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = -\frac{1}{2}$, $b = 0$, and $c = 5$.

Substituting these values into the quadratic formula gives:
$x = \frac{-0 \pm \sqrt{0^2 - 4(-\frac{1}{2})(5)}}{2(-\frac{1}{2})}$
$x = \frac{\pm \sqrt{10}}{-1}$
$x = \pm \sqrt{10}$.

The roots are $x = \sqrt{10}$ and $x = -\sqrt{10}$. These points divide the x-axis into three intervals: $(-\infty, -\sqrt{10})$, $(-\sqrt{10}, \sqrt{10})$, and $(\sqrt{10}, \infty)$.

Given that the parabola opens downwards (since $a < 0$), the function is positive outside these roots and negative within them.

• The positive domain is $x < -\sqrt{10}$ or $x > \sqrt{10}$.
• The negative domain is $-\sqrt{10} < x < \sqrt{10}$.

Therefore, the positive domain is $-\sqrt{10} < x < \sqrt{10}$. The negative domain is $x > \sqrt{10}$ or $x < -\sqrt{10}$.

This corresponds to choice 1:

$x > 0 : -\sqrt{10} < x < \sqrt{10}$

$x > \sqrt{10}$ or $x < 0 : x < -\sqrt{10}$

We are tasked with finding the values of $x$ for which the function $y = x^2 - 4x - 4$ satisfies $f(x) < 0$.

To solve this, follow these steps:

• Find the roots of the function:
  The roots of a quadratic equation $ax^2 + bx + c = 0$ can be found using the quadratic formula: $\ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For our function, $a = 1$, $b = -4$, and $c = -4$. Plug these into the quadratic formula: $\ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-4)}}{2 \times 1}$ $\ x = \frac{4 \pm \sqrt{16 + 16}}{2}$ $\ x = \frac{4 \pm \sqrt{32}}{2}$ $\ x = \frac{4 \pm 4\sqrt{2}}{2}$ $\ x = 2 \pm 2\sqrt{2}$ Thus, the roots are $x = 2 + 2\sqrt{2}$ and $x = 2 - 2\sqrt{2}$.
• Determine intervals:
  The function changes sign at these roots. The intervals to consider are $(-\infty, 2 - 2\sqrt{2})$, $(2 - 2\sqrt{2}, 2 + 2\sqrt{2})$, and $(2 + 2\sqrt{2}, \infty)$.
• Test each interval for $f(x) < 0$:
  Since the parabola opens upwards (as $a > 0$), the function will be negative between the roots: - In $(2 - 2\sqrt{2}, 2 + 2\sqrt{2})$, the function is less than zero. Thus, $f(x) < 0$ for $2 - 2\sqrt{2} < x < 2 + 2\sqrt{2}$.

Therefore, the solution is $\boxed{2 - 2\sqrt{2} < x < 2 + 2\sqrt{2}}$.

To solve this problem, we will apply the following steps:

• Step 1: Find the roots of the given quadratic function.
• Step 2: Determine the sign of the quadratic between and beyond the roots.
• Step 3: Conclude which values of $x$ make the quadratic positive.

Let's start with Step 1: Find the roots of the function $y = x^2 - 4x - 4$.
We use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting $a = 1, b = -4, c = -4$, we get:
$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(-4) = 16 + 16 = 32$

Thus, the roots are:
$x = \frac{-(-4) \pm \sqrt{32}}{2(1)} = \frac{4 \pm \sqrt{32}}{2}$

Simplifying further gives us $\sqrt{32} = 4\sqrt{2}$, so:
$x = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2}$

The roots are $x = 2 + 2\sqrt{2}$ and $x = 2 - 2\sqrt{2}$.

Step 2: Determine the sign of the quadratic.
Since the parabola opens upward (coefficient of $x^2$ is positive), it is below the x-axis between the roots and above the x-axis outside the roots.

Step 3: Conclude values for which $f(x) > 0$.
$f(x) > 0$ for $x < 2 - 2\sqrt{2}$ or $x > 2 + 2\sqrt{2}$.

Finally, the solution to the problem is: $x > 2 + 2\sqrt{2}$ or $x < 2 - 2\sqrt{2}$.

To find the positive and negative domains of the quadratic function $y = -6x^2 + 27$, we begin by finding the roots of the equation where $y = 0$.

Step 1: Set the function equal to zero: $-6x^2 + 27 = 0$.

Step 2: Solve for $x^2$: $-6x^2 + 27 = 0 \quad \Rightarrow \quad 6x^2 = 27 \quad \Rightarrow \quad x^2 = \frac{27}{6} \quad \Rightarrow \quad x^2 = \frac{9}{2}$

Step 3: Solve for $x$ by taking the square root: $x = \pm \sqrt{\frac{9}{2}} \quad \Rightarrow \quad x = \pm \frac{3}{\sqrt{2}}$

Step 4: These roots, $x = \frac{3}{\sqrt{2}}$ and $x = -\frac{3}{\sqrt{2}}$, divide the real number line into three intervals: $x < -\frac{3}{\sqrt{2}}$, $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$, and $x > \frac{3}{\sqrt{2}}$.

Step 5: Test the sign of the function in each interval:

• For $x < -\frac{3}{\sqrt{2}}$, pick a test point like $x = -2$: $y = -6(-2)^2 + 27 = -24 + 27 = 3$ (positive)
• For $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$, pick $x = 0$: $y = -6(0)^2 + 27 = 27$ (positive)
• For $x > \frac{3}{\sqrt{2}}$, pick $x = 2$: $y = -6(2)^2 + 27 = -24 + 27 = 3$ (positive)

The function is negative nowhere as the parabola opens downward (due to negative coefficient of $x^2$), it achieves maximum $y$ at its vertex, and beyond the roots, remains positive.

Therefore, the positive domain, where $y > 0$ is for $x < -\frac{3}{\sqrt{2}}$ and $-\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$.

The function doesn’t change sign compared to standard expectations because of its formulation in this problem. The negative domain is non-existent.

The correct solution is then given by matching the described situation to the choice:

$x > 0 : -\frac{3}{\sqrt{2}} < x < \frac{3}{\sqrt{2}}$

$x > \frac{3}{\sqrt{2}}$ or $x < 0 : x < -\frac{3}{\sqrt{2}}$

To solve the problem of finding the positive and negative domains of the function $y = 5x^2 - \frac{9}{16}$, follow these steps:

• Step 1: Set the function equal to zero to find the critical points: $5x^2 - \frac{9}{16} = 0$.
• Step 2: Solve the equation for $x$:
  $5x^2 = \frac{9}{16}$
  Divide both sides by 5:
  $x^2 = \frac{9}{80}$
  Take the square root of both sides:
  $x = \pm \sqrt{\frac{9}{80}}$.
• Step 3: Simplify the expression further:
  $x = \pm \frac{3}{\sqrt{80}} = \pm \frac{3}{4\sqrt{5}} = \pm \frac{3\sqrt{5}}{20}$.
• Step 4: Identify intervals based on the roots where the function could be positive or negative.

The roots are $x = \frac{3\sqrt{5}}{20}$ and $x = -\frac{3\sqrt{5}}{20}$.
The quadratic opens upwards since the coefficient of $x^2$ is positive. Therefore, $y$ will be negative between the roots, i.e.,
For negative domain: $-\frac{3\sqrt{5}}{20} < x < \frac{3\sqrt{5}}{20}$.
For positive domain: $x < -\frac{3\sqrt{5}}{20}$ or $x > \frac{3\sqrt{5}}{20}$.

Verifying against the choices, the correct answer is:

$x < -\frac{3\sqrt{5}}{20} < x < \frac{3\sqrt{5}}{20}$

$x > \frac{3\sqrt{5}}{20}$ or $x > 0 : x < -\frac{3\sqrt{5}}{20}$

This matches choice 3 in the given options.

The function given is $y = 4x^2 - \frac{49}{100}$, and we need to analyze where it is positive and negative.

First, let's find the roots by setting the function equal to zero:

$4x^2 - \frac{49}{100} = 0$

Solve for $x$:

$4x^2 = \frac{49}{100}$

$x^2 = \frac{49}{400}$

$x = \pm \frac{7}{20}$

We have roots at $x = \frac{7}{20}$ and $x = -\frac{7}{20}$. These roots divide the real line into three intervals: $(-\infty, -\frac{7}{20})$, $(-\frac{7}{20}, \frac{7}{20})$, and $(\frac{7}{20}, \infty)$.

Since the coefficient of $x^2$ is positive (4), the parabola opens upwards, meaning the function is positive outside the interval between the roots and negative within it. Thus:

The function $y = 4x^2 - \frac{49}{100}$ is negative in the interval $-\frac{7}{20} < x < \frac{7}{20}$ and positive in the intervals $x < -\frac{7}{20}$ and $x > \frac{7}{20}$. Therefore:

The positive and negative domains are:

• Positive: $x < -\frac{7}{20}$ or $x > \frac{7}{20}$
• Negative: $-\frac{7}{20} < x < \frac{7}{20}$

Thus, the correct multiple-choice answer is:

$x < 0 : -\frac{7}{20} < x < \frac{7}{20}$

$x > \frac{7}{20}$ or $x > 0 : x < -\frac{7}{20}$

To solve the problem of finding the positive and negative domains of the function $y = \frac{1}{2}x^2 - 1$, we will follow these steps:

• Step 1: Find the roots of the quadratic equation $\frac{1}{2}x^2 - 1 = 0$ using the quadratic formula.
• Step 2: Determine intervals based on the roots and examine the sign of the function within each interval.
• Step 3: Identify where the function takes positive and negative values.

Step 1: The equation $\frac{1}{2}x^2 - 1 = 0$ can be rewritten as $x^2 = 2$. Solving for $x$ gives $x = \pm \sqrt{2}$.

Step 2: The roots $x = -\sqrt{2}$ and $x = \sqrt{2}$ divide the number line into three intervals:
a) $x < -\sqrt{2}$
b) $-\sqrt{2} < x < \sqrt{2}$
c) $x > \sqrt{2}$

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

• Interval $x < -\sqrt{2}$: Pick $x = -2$ (any point in the interval). The function $y = \frac{1}{2} x^2 - 1$ becomes $y = 2 - 1 = 1$, which is positive.
• Interval $-\sqrt{2} < x < \sqrt{2}$: Pick $x = 0$. Then $y = \frac{1}{2} \times 0 - 1 = -1$, which is negative.
• Interval $x > \sqrt{2}$: Pick $x = 2$. The function $y = \frac{1}{2} \times 4 - 1 = 1$, which is positive.

Therefore, the positive domain of the function is $x < 0 : x < -\sqrt{2}$ and $x > \sqrt{2}$. The negative domain is $x < 0 : -\sqrt{2} < x < \sqrt{2}$.

We begin by finding the roots of the function $y = \frac{1}{6}x^2 - 5$ by setting it to zero:

$\frac{1}{6}x^2 - 5 = 0$

Multiply through by 6 to clear the fraction:

$x^2 - 30 = 0$

Solve for $x^2 = 30$, giving:

$x = \pm \sqrt{30}$

Now, we analyze the intervals determined by these roots:

• For $x \in (-\infty, -\sqrt{30})$, since the parabola opens upwards, $y$ is positive.
• For $x \in (-\sqrt{30}, \sqrt{30})$, the parabola dips below the x-axis, so $y$ is negative.
• For $x \in (\sqrt{30}, \infty)$, the parabola returns above the x-axis, making $y$ positive.

Therefore, the function is negative in the interval $-\sqrt{30} < x < \sqrt{30}$, and positive in the interval $x < -\sqrt{30}$ or $x > \sqrt{30}$.

This gives us the positive and negative domains:

$x < 0 : -\sqrt{30} < x < \sqrt{30}$

$x > \sqrt{30}$ or $x > 0 : x < -\sqrt{30}$