Statement #1: If the numeric expression is zero, there are two solutions.

Statement #2: If the numeric expression is zero, there is one solution.

Statement #3: If the numeric expression is zero, there are no solutions

Statement #1: If the numeric expression is zero, there are two solutions.

This statement is not always true. It depends on the specific equation or problem being discussed. In some cases, a zero value may have two solutions, such as in the quadratic equation x^2 = 0, which has two solutions x = 0 and x = 0. However, not all equations or problems will have two solutions when the numeric expression is zero.

Statement #2: If the numeric expression is zero, there is one solution.
This statement is not always true either. Similar to statement #1, it depends on the specific equation or problem. There may be cases where a numeric expression being zero has only one solution, but it is not always the case.

Statement #3: If the numeric expression is zero, there are no solutions.
This statement is generally true. In most cases, if the numeric expression itself equals zero, then it indicates that the equation or problem has no solutions. However, exceptions may exist depending on the context of the equation or problem.

what is the correct answer?

The correct answer depends on the specific equation or context given. Without more information, it is not possible to determine which statement is correct.

Solve the following quadratic equation using square roots: −9 (x−3)2−4=−4 .(1 point)

To solve the quadratic equation −9(x−3)²−4=−4, we can follow these steps:

1. Move the constant term to the other side of the equation:
−9(x−3)² = 0

2. Divide both sides of the equation by -9 to isolate the squared term:
(x−3)² = 0

3. Take the square root of both sides of the equation to solve for x:
x−3 = ±√0

4. Simplify the square root of 0 to 0:
x−3 = ±0

5. Solve for x by adding 3 to both sides of the equation:
x = 3 ± 0

6. Simplify x = 3 ± 0 to get the solutions:
x = 3

So, the solution to the quadratic equation is x = 3.