Question

Which equations have the solution set of {12, −3} { 1 2 , − 3 } ? Select all that apply. (3 points) Responses (4x - 2)(x 3) = 0 (4x - 2)(x 3) = 0 (x 12)(x 3) = 0 ( 1 2 ) ( 3 ) = 0 (x 12)(x 3) = 0 ( 1 2 ) ( 3 ) = 0 (3x - 6)(3x - 6) = 0 (3x - 6)(3x - 6) = 0 (6x - 3)(4x 12) = 0 (6x - 3)(4x 12) = 0 (x − 12)(x 3) = 0 ( − 1 2 ) ( 3 ) = 0 (x − 12)(x 3) = 0 ( − 1 2 ) ( 3 ) = 0 (x − 12)(x − 3) = 0 ( − 1 2 ) ( − 3 ) = 0 (x − 12)(x − 3) = 0 ( − 1 2 ) ( − 3 ) = 0 (4x 2)(x 3) = 0

Answer 1

To find which equations have the solution set of {12, -3}, we can use the fact that if a polynomial equation has a certain set of roots, it can be factored accordingly. The roots can be expressed in the form of factors:

If the roots are \( x = 12 \) and \( x = -3 \), then the corresponding factors will be \( (x - 12) \) and \( (x + 3) \).

Thus, the polynomial can be represented as:

\[
(x - 12)(x + 3) = 0
\]

Let's analyze the given options:

1. **(4x - 2)(x + 3) = 0**:
- This does not yield a root of 12 or -3.

2. **(x - 12)(x + 3) = 0**:
- This matches perfectly with our expected factors, yielding the roots 12 and -3.

3. **(3x - 6)(3x - 6) = 0**:
- This simplifies to \( (3(x - 2))^2 = 0 \), giving x = 2 as the only solution.

4. **(6x - 3)(4x + 12) = 0**:
- Simplifying gives roots for \( x = \frac{1}{2} \) and \( x = -3 \), but not 12.

5. **(x - 12)(x + 3) = 0** (same as before):
- This also matches perfectly, yielding the roots 12 and -3.

6. **(x - 12)(x - 3) = 0**:
- This gives the roots 12 and 3, not -3.

7. **(4x + 2)(x + 3) = 0**:
- This gives no roots of 12 and -3.

Based on the analysis:

The equations that have the solution set of {12, -3} are:

- **(x - 12)(x + 3) = 0**
- **(x - 12)(x + 3) = 0** (duplicate; counted only once)

The selected equations are:

- **(x - 12)(x + 3) = 0** (from both occurrences).

So the final answer is:

1. \( (x - 12)(x + 3) = 0 \) (first occurrence)
2. \( (x - 12)(x + 3) = 0 \) (second occurrence)