Positive and Negative Domains: Using roots

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

The function given is $y = -\left(x+4\right)^2 + 6$. This is in vertex form $y = a(x-h)^2 + k$ with vertex at $(-4, 6)$.

Step 1: To find the x-values for which the function is positive or negative, set $y = 0$:

$-\left(x+4\right)^2 + 6 = 0$

$\left(x+4\right)^2 = 6$

Step 2: Solve for $x$:

Take the square root of both sides:

$x + 4 = \pm \sqrt{6}$ i.e., $x = -4 \pm \sqrt{6}$

Step 3: Find where the function is positive or negative. The parabola opens downward, so the intervals are:

• Negative domain: $x < -4 - \sqrt{6}$ and $x > -4 + \sqrt{6}$; outside this interval.
• Positive domain: $-4 - \sqrt{6} < x < -4 + \sqrt{6}$; within this interval.

Conclusively:

$x > 0 : -4-\sqrt{26} < x < -4+\sqrt{26}$

$x > -4+\sqrt{26}$ or $x < 0 : x < -4-\sqrt{26}$

Therefore, the solution to this problem is as follows:

For $x > 0$: $-4-\sqrt{26} < x < -4+\sqrt{26}$

For $x < 0$: $x < -4-\sqrt{26}$ and $( x > -4 + \sqrt{26})$

To determine the positive and negative domains of the function, follow these steps:

• Step 1: Solve $(x+4)^2 = \frac{41}{4}$.
• Step 2: This implies $x+4 = \pm \frac{\sqrt{41}}{2}$.
• Step 3: Solving these gives the roots: $x = \frac{-8+\sqrt{41}}{2}$ and $x = \frac{-8-\sqrt{41}}{2}$.
• Step 4: Divide the real number line using these roots into intervals:
• Interval 1: $x < \frac{-8-\sqrt{41}}{2}$
• Interval 2: $\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}$
• Interval 3: $x > \frac{-8+\sqrt{41}}{2}$
• Step 5: Test each interval to see where the function is greater or less than zero, using the sign of $(x+4)^2 - \frac{41}{4}$.

Testing reveals that:

• For Interval 1, the function is positive.
• For Interval 2, the function is negative.
• For Interval 3, the function is positive.

Thus, the negative domain is $\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}$ and the positive domains are $x > \frac{-8+\sqrt{41}}{2}$ or $x < \frac{-8-\sqrt{41}}{2}$.

Therefore, the correct answer is:

$x < 0 :\frac{-8-\sqrt{41}}{2} < x < \frac{-8+\sqrt{41}}{2}$

$x > \frac{-8+\sqrt{41}}{2}$ or $x > 0 : x < \frac{-8-\sqrt{41}}{2}$

To determine where the function $y = \frac{1}{4}x^2 - \frac{4}{5}$ is positive or negative, we start by finding the roots of the function. These roots occur where

$\frac{1}{4}x^2 - \frac{4}{5} = 0$.

Multiply through by 4 to clear the fraction:

$x^2 = \frac{16}{5}$.

Take the square root of both sides:

$x = \pm \sqrt{\frac{16}{5}}$.

This simplifies to:

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

With roots at $x = \frac{4}{\sqrt{5}}$ and $x = -\frac{4}{\sqrt{5}}$, the x-axis is divided into three intervals:

• $x < -\frac{4}{\sqrt{5}}$,
• $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$,
• $x > \frac{4}{\sqrt{5}}$.

Analyze these intervals:

• Choose a test point $x = 0$ in $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$:
• $y(0) = \frac{1}{4}(0)^2 - \frac{4}{5} = -\frac{4}{5} < 0$. Thus, the function is negative in $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$.
• Choose a test point $x = -2$ in $x < -\frac{4}{\sqrt{5}}$:
• $y(-2) = \frac{1}{4}(-2)^2 - \frac{4}{5} = 1 - \frac{4}{5} = \frac{1}{5} > 0$. Thus, the function is positive in $x < -\frac{4}{\sqrt{5}}$.
• Choose a test point $x = 2$ in $x > \frac{4}{\sqrt{5}}$:
• $y(2) = \frac{1}{4}(2)^2 - \frac{4}{5} = 1 - \frac{4}{5} = \frac{1}{5} > 0$. Thus, the function is positive in $x > \frac{4}{\sqrt{5}}$.

Therefore, the negative domain is $-\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$, and the positive domains are $x < -\frac{4}{\sqrt{5}}$ or $x > \frac{4}{\sqrt{5}}$.

The correct choice based on this analysis is:

Choice 3: $x < 0 : -\frac{4}{\sqrt{5}} < x < \frac{4}{\sqrt{5}}$

and

$x > 0 : x < -\frac{4}{\sqrt{5}}$ or $x > \frac{4}{\sqrt{5}}$

To determine where the function $y = (x + 5)^2 - 6$ is positive and negative, we start by solving the equation:

$(x + 5)^2 - 6 = 0$

Adding 6 to both sides gives:

$(x + 5)^2 = 6$

Taking the square root of both sides, we obtain two solutions:

$x + 5 = \sqrt{6}$ or $x + 5 = -\sqrt{6}$

Solving these, we get:

$x = -5 + \sqrt{6}$ and $x = -5 - \sqrt{6}$

These roots divide the number line into three intervals: $(- \infty, -5 - \sqrt{6})$, $(-5 - \sqrt{6}, -5 + \sqrt{6})$, and $(-5 + \sqrt{6}, \infty)$.

Next, we determine the sign of the function in each interval:

• For $x < -5 - \sqrt{6}$, choose $x = -10$:

$(x + 5)^2 - 6 = ((-10) + 5)^2 - 6 = (-5)^2 - 6 = 25 - 6 = 19 > 0$. Therefore, the function is positive.

• For $-5 - \sqrt{6} < x < -5 + \sqrt{6}$, choose $x = -5$:

$(x + 5)^2 - 6 = ((-5) + 5)^2 - 6 = 0^2 - 6 = -6 < 0$. Therefore, the function is negative.

• For $x > -5 + \sqrt{6}$, choose $x = 0$:

$(x + 5)^2 - 6 = (0 + 5)^2 - 6 = 5^2 - 6 = 25 - 6 = 19 > 0$. Therefore, the function is positive.

Thus, the function is positive on the intervals $x < -5 - \sqrt{6}$ and $x > -5 + \sqrt{6}$, and negative on the interval $-5 - \sqrt{6} < x < -5 + \sqrt{6}$.

Therefore, the positive domain is $x > -5+\sqrt{6}$ or $x > 0 : x < -5-\sqrt{6}$, and the negative domain is $x < 0 : -5-\sqrt{6} < x < -5+\sqrt{6}$.

To find the positive and negative domains of the function $y = -\left(x-14\right)^2 + 8$, we'll start by identifying the roots of the quadratic equation.

Step 1: Find the roots of the equation:
To find when the function is zero, set $y = 0$:
$-\left(x-14\right)^2 + 8 = 0$.

Step 2: Solve for $x$:
Rearrange the equation:
$-\left(x-14\right)^2 = -8$
$(x-14)^2 = 8$.

Take the square root on both sides:
$x-14 = \pm\sqrt{8}$.
This simplifies to $x - 14 = \pm 2\sqrt{2}$.

Add 14 to both sides to solve for $x$:
$x = 14 \pm 2\sqrt{2}$.
So, the roots are $x = 14 + 2\sqrt{2}$ and $x = 14 - 2\sqrt{2}$.

Step 3: Analyze intervals between roots and outside:
The roots divide the $x$-axis into three intervals: $x < 14 - 2\sqrt{2}$, $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$, and $x > 14 + 2\sqrt{2}$.

- For $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$, $y > 0$ because points between roots are above the $x$-axis.
- For $x < 14 - 2\sqrt{2}$ or $x > 14 + 2\sqrt{2}$, $y < 0$ because points outside of roots are below the $x$-axis.

Conclusion:
The positive domain, where $y > 0$, is $14 - 2\sqrt{2} < x < 14 + 2\sqrt{2}$.
The negative domain, where $y < 0$, is $x < 14 - 2\sqrt{2}$ or $x > 14 + 2\sqrt{2}$.

Therefore, the solution is:
Positive domain: $x > 0 : 14-2\sqrt{2} < x < 14+2\sqrt{2}$.
Negative domain: $x > 14+2\sqrt{2}$ or $x < 0 : x < 14-2\sqrt{2}$.

To find the positive and negative domains of the function $y = -\left(x - 2.5\right)^2 + 0.5$, we analyze when $y$ is greater than and less than zero.

• Step 1: Solve for the positive domain ($y > 0$).

We need to solve the inequality:
$-\left(x - 2.5\right)^2 + 0.5 > 0$.

Rearrange this to:
$-\left(x - 2.5\right)^2 > -0.5$.

Remove the negative sign by multiplying by $-1$ (which flips the inequality sign):
$\left(x - 2.5\right)^2 < 0.5$.

Taking the square root of both sides gives:
$|x - 2.5| < \sqrt{0.5}$.
This implies:
$-\sqrt{0.5} < x - 2.5 < \sqrt{0.5}$.

Solve for $x$:
$2.5 - \sqrt{0.5} < x < 2.5 + \sqrt{0.5}$.

Step 2: Solve for the negative domain ($y < 0$).

From the inequality:
$-\left(x - 2.5\right)^2 + 0.5 < 0$.

Rearrange to:
$-\left(x - 2.5\right)^2 < -0.5$.

Again, multiply by $-1$:
$\left(x - 2.5\right)^2 > 0.5$.

Taking the square root gives:
$|x - 2.5| > \sqrt{0.5}$.
This implies:
$x - 2.5 < -\sqrt{0.5}$ or $x - 2.5 > \sqrt{0.5}$.

Solving gives:
$x < 2.5 - \sqrt{0.5}$ or $x > 2.5 + \sqrt{0.5}$.

Recall $\sqrt{0.5} = \frac{\sqrt{2}}{2}$, so:
The positive domain is: $2.5 - \frac{\sqrt{2}}{2} < x < 2.5 + \frac{\sqrt{2}}{2}$.
The negative domain is: $x < 2.5 - \frac{\sqrt{2}}{2}$ or $x > 2.5 + \frac{\sqrt{2}}{2}$.

Therefore, the correct answer based on the choices provided is:

$x<0:2\frac{1}{2}-\frac{\sqrt{2}}{2}$ or $x > 2\frac{1}{2} + \frac{\sqrt{2}}{2}$

$x > 0 : 2\frac{1}{2} - \frac{\sqrt{2}}{2} < x < 2\frac{1}{2} + \frac{\sqrt{2}}{2}$

To find the positive and negative domains of the function $y = (x-1)^2 - 2$, we need to determine the points where the function intersects the x-axis, as these will mark changes in sign.

Step 1: Set the function equal to zero to find the roots.
$(x-1)^2 - 2 = 0$

Step 2: Move -2 to the other side and solve:
$(x-1)^2 = 2$

Step 3: Solve for $x$ by taking the square root of both sides:
$x - 1 = \pm \sqrt{2}$

Step 4: Solve for $x$ by isolating it:
$x = 1 \pm \sqrt{2}$

The roots are $x = 1 + \sqrt{2}$ and $x = 1 - \sqrt{2}$. These roots divide the x-axis into three parts.

Step 5: Evaluate the function behavior in each interval defined by these roots.

• For $x < 1 - \sqrt{2}$, pick a point such as nearly approaching zero value and test the sign.
• For $1 - \sqrt{2} < x < 1 + \sqrt{2}$, pick a midpoint value and test.
• For $x > 1 + \sqrt{2}$, pick a value greater than root for testing function positivity.

Step 6: Determine where the function is positive and negative:

• Within the interval $[1-\sqrt{2}, 1+\sqrt{2}]$, the function lies below the x-axis and is negative.
• Outside this interval, specifically $x < 1 - \sqrt{2}$ or $x > 1 + \sqrt{2}$, the function lies above the x-axis and is positive.

The positive domain is $x < 1 - \sqrt{2}$ or $x > 1 + \sqrt{2}$ and the negative domain is $1 - \sqrt{2} < x < 1 + \sqrt{2}$.

Therefore, the solution is:

$x < 0 : 1-\sqrt{2} < x < 1+\sqrt{2}$

$x > 1+\sqrt{2}$ or $x > 0 : x < 1-\sqrt{2}$

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 find the positive and negative domains of the function $y = (x-6)^2 - 3$, follow these steps:

• Step 1: Set $y = 0$ to find the roots of the equation. This gives us $(x-6)^2 - 3 = 0$.
• Step 2: Add 3 to both sides to simplify, resulting in $(x-6)^2 = 3$.
• Step 3: Take the square root of both sides. We have two solutions: $x - 6 = \sqrt{3}$ and $x - 6 = -\sqrt{3}$.
• Step 4: Solve for $x$ from each equation:
• For $x - 6 = \sqrt{3}$, $x = 6 + \sqrt{3}$.
• For $x - 6 = -\sqrt{3}$, $x = 6 - \sqrt{3}$.
• Step 5: Determine the intervals:
• Interval 1: $x < 6 - \sqrt{3}$, insert a test point to determine sign.
• Interval 2: $6 - \sqrt{3} < x < 6 + \sqrt{3}$, insert a test point to determine sign.
• Interval 3: $x > 6 + \sqrt{3}$, insert a test point to determine sign.
• Step 6: Based on the intervals:
• For $x < 6 - \sqrt{3}$ and $x > 6 + \sqrt{3}$, $y > 0$.
• For $6 - \sqrt{3} < x < 6 + \sqrt{3}$, $y < 0$.

The positive domains are: $x < 6 - \sqrt{3}$ or $x > 6 + \sqrt{3}$.

The negative domain is: $6 - \sqrt{3} < x < 6 + \sqrt{3}$.

The correct answer to the problem is:

$x > 6+\sqrt{3}$ or $x < 0 : x < 6-\sqrt{3}$

$x < 0 : 6-\sqrt{3} < x < 6+\sqrt{3}$

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 find the roots of the quadratic equation $\frac{1}{2}x^2 - \frac{4}{9} = 0$, follow these steps:

• Set the equation to zero: $\frac{1}{2}x^2 - \frac{4}{9} = 0$.
• Multiply the entire equation by 9 to clear the fraction: $9 \times \frac{1}{2}x^2 - 4 = 0$, simplifying to $\frac{9}{2}x^2 = 4$.
• Multiply through by 2 to solve for $x^2$: $9x^2 = 8$.
• Divide both sides by 9: $x^2 = \frac{8}{9}$.
• Take the square root of both sides: $x = \pm \frac{\sqrt{8}}{3}$, simplifying $\sqrt{8}$ to $2\sqrt{2}$.
• Thus the roots are $x = \frac{2\sqrt{2}}{3}$ and $x = -\frac{2\sqrt{2}}{3}$.

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

Evaluate the sign of $y$ in each interval:

• When $x < -\frac{2\sqrt{2}}{3}$, choose a test point and check: $x = -1$, then $y = \frac{1}{2} \times (-1)^2 - \frac{4}{9} = \frac{1}{2} - \frac{4}{9} < 0$.
• When $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, choose a test point and check: $x = 0$, then $y = \frac{1}{2} \times 0^2 - \frac{4}{9} = -\frac{4}{9} < 0$.
• When $x > \frac{2\sqrt{2}}{3}$, choose a test point and check: $x = 1$, then $y = \frac{1}{2} \times 1^2 - \frac{4}{9} > 0$.

Therefore, $y$ is positive for $x > \frac{2\sqrt{2}}{3}$ and negative for $-\frac{2\sqrt{2}}{3} < x < \frac{2\sqrt{2}}{3}$, as well as $x < -\frac{2\sqrt{2}}{3}$.

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

Thus, the correct answer is:

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

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

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 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.

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) $x > \sqrt{2}$