Find the Domain of (x+10)²-3: Determining Valid Input Values

To solve this problem, we need to determine when $y = (x+10)^2 - 3$ is greater than and less than zero.

Start by finding the roots of the equation:

Set $y = 0$:

$(x+10)^2 - 3 = 0$

Rearrange the equation to find:

$(x+10)^2 = 3$

Take the square root of both sides:

$x + 10 = \pm\sqrt{3}$

Solving these gives:

• $x = -10 + \sqrt{3}$
• $x = -10 - \sqrt{3}$

These roots divide the number line into three intervals:

• $(-\infty, -10 - \sqrt{3})$
• $(-10 - \sqrt{3}, -10 + \sqrt{3})$
• $(-10 + \sqrt{3}, \infty)$

Test each interval to determine where the function is positive or negative:

For $x < -10 - \sqrt{3}$: Choose $x = -11$

Then: $y = ((-11 + 10)^2 - 3) = 1 - 3 = -2$

So, $y < 0$ in the interval $(-\infty, -10 - \sqrt{3})$.

For $-10 - \sqrt{3} < x < -10 + \sqrt{3}$: Choose $x = -10$

Then: $y = ((-10 + 10)^2 - 3) = 0 - 3 = -3$

So, $y < 0$ in the interval $(-10 - \sqrt{3}, -10 + \sqrt{3})$.

For $x > -10 + \sqrt{3}$: Choose $x = 0$

Then: $y = ((0 + 10)^2 - 3) = 100 - 3 = 97$

So, $y > 0$ in the interval $(-10 + \sqrt{3}, \infty)$.

Therefore, the positive domain is $x > -10 + \sqrt{3}$ while the negative domain is $x < -10 + \sqrt{3}$.

Using the analysis above and applying it to the choices, the correct response is:

$x < 0 : -10-\sqrt{3} < x < -10+\sqrt{3}$

$x > -10+\sqrt{3}$ or $x > 0 : x < -10-\sqrt{3}$