Question

Use the square root property to solve.



Question 1 options:

v= -4, v= 4


v= -28, v = 28


v= -2, v=2


v = 2, v=0

Answer 1

To apply the square root property to solve an equation, we need a specific equation to work with. The square root property states that if \( x^2 = a \), then \( x = \sqrt{a} \) or \( x = -\sqrt{a} \).

Given the possible answer choices, it looks like they might be solutions to a quadratic equation.

Let's assume that the equation we are working with is in the form:

\( v^2 = a \)

For each of the answer options, let’s identify the value of \( a \):

1. **Option 1: \( v = -4, v = 4 \)**
- Setting \( v^2 = 4^2 = 16 \).
- \( a = 16 \).

2. **Option 2: \( v = -28, v = 28 \)**
- Setting \( v^2 = 28^2 = 784 \).
- \( a = 784 \).

3. **Option 3: \( v = -2, v = 2 \)**
- Setting \( v^2 = 2^2 = 4 \).
- \( a = 4 \).

4. **Option 4: \( v = 2, v = 0 \)**
- Setting \( v^2 = 2^2 = 4 \) and \( v \) could be \( 0 \).
- **However, zero is not generated by a positive square value.**

Assuming we want to solve for \( v \) using an equation like \( v^2 = a \):

- For options involving \( v = -2 \) and \( v = 2 \) or \( v = -4 \) and \( v = 4\), those are straightforward where \( a \) is \( 4 \) or \( 16 \).

So, if we want to conclude which solution corresponds to the previous calculations, we should note:

- \( v^2 = 16 \) leads to \( v = -4 \) or \( v = 4 \).
- \( v^2 = 4 \) leads to \( v = -2 \) or \( v = 2 \).

However, the last option \( v = 2, v = 0 \) does not fit the usual square property properly.

### Final Answer:
- **The correct answer corresponding to using the square root property would lead to:**
- \( v = -4, v = 4 \) or \( v = -2, v = 2 \), but if selecting the most straightforward pairs, choose:
- **Stating the result clearly: \( v = -4, v = 4 \) or \( v = -2, v = 2 \)**.