Question

Which equations have the solution set of {12, −3}
?

Select all that apply.

(3 points)
Responses

(4x - 2)(x + 3) = 0
(4x - 2)(x + 3) = 0

(x + 12)(x + 3) = 0
(x + 12)(x + 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
(x − 12)(x + 3) = 0

(x − 12)(x − 3) = 0
(x − 12)(x − 3) = 0

(4x + 2)(x + 3) = 0

Answer 1

To determine which equations have the solution set of \(\{12, -3\}\), we need to check the roots of each provided equation.

1. \((4x - 2)(x + 3) = 0\)
- Roots:
- \(4x - 2 = 0 \Rightarrow x = \frac{2}{4} = \frac{1}{2}\)
- \(x + 3 = 0 \Rightarrow x = -3\)
- **Roots: \(\left\{\frac{1}{2}, -3\right\}\)** (Not a match)

2. \((x + 12)(x + 3) = 0\)
- Roots:
- \(x + 12 = 0 \Rightarrow x = -12\)
- \(x + 3 = 0 \Rightarrow x = -3\)
- **Roots: \(\left\{-12, -3\right\}\)** (Not a match)

3. \((3x - 6)(3x - 6) = 0\)
- Roots:
- \(3x - 6 = 0 \Rightarrow x = 2\)
- **Roots: \(\{2\}\)** (Not a match)

4. \((6x - 3)(4x + 12) = 0\)
- Roots:
- \(6x - 3 = 0 \Rightarrow x = \frac{3}{6} = \frac{1}{2}\)
- \(4x + 12 = 0 \Rightarrow x = -3\)
- **Roots: \(\left\{\frac{1}{2}, -3\right\}\)** (Not a match)

5. \((x - 12)(x + 3) = 0\)
- Roots:
- \(x - 12 = 0 \Rightarrow x = 12\)
- \(x + 3 = 0 \Rightarrow x = -3\)
- **Roots: \(\{12, -3\}\)** (Match)

6. \((x - 12)(x - 3) = 0\)
- Roots:
- \(x - 12 = 0 \Rightarrow x = 12\)
- \(x - 3 = 0 \Rightarrow x = 3\)
- **Roots: \(\{12, 3\}\)** (Not a match)

7. \((4x + 2)(x + 3) = 0\)
- Roots:
- \(4x + 2 = 0 \Rightarrow x = -\frac{2}{4} = -\frac{1}{2}\)
- \(x + 3 = 0 \Rightarrow x = -3\)
- **Roots: \(\left\{-\frac{1}{2}, -3\right\}\)** (Not a match)

### Conclusion:
The equation that has the solution set of \(\{12, -3\}\) is:

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