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

Question

?

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 − 1/2)(x + 3) = 0
(x − 1/2)(x + 3) = 0

(x − 1/2)(x − 3) = 0
(x − 1/2)(x − 3) = 0

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

Answer 1

To find out which equations have the solution set of {12, −3}, we can evaluate each equation to see what roots they yield.

1. (4x - 2)(x + 3) = 0

   • This gives us \(4x - 2 = 0\) or \(x + 3 = 0\)
   • Solving:
     • \(4x - 2 = 0 \Rightarrow 4x = 2 \Rightarrow x = \frac{2}{4} = \frac{1}{2}\)
     • \(x + 3 = 0 \Rightarrow x = -3\)
   • So the roots are \({\frac{1}{2}, -3}\).
   • Not a solution set.

2. (x + 12)(x + 3) = 0

   • This gives us \(x + 12 = 0\) or \(x + 3 = 0\)
   • Solving:
     • \(x + 12 = 0 \Rightarrow x = -12\)
     • \(x + 3 = 0 \Rightarrow x = -3\)
   • So the roots are \({-12, -3}\).
   • Not a solution set.

3. (3x - 6)(3x - 6) = 0

   • This simplifies to \(3x - 6 = 0\)
   • Solving:
     • \(3x - 6 = 0 \Rightarrow 3x = 6 \Rightarrow x = \frac{6}{3} = 2\)
   • So the root is \({2}\).
   • Not a solution set.

4. (6x - 3)(4x + 12) = 0

   • This gives us \(6x - 3 = 0\) or \(4x + 12 = 0\)
   • Solving:
     • \(6x - 3 = 0 \Rightarrow 6x = 3 \Rightarrow x = \frac{3}{6} = \frac{1}{2}\)
     • \(4x + 12 = 0 \Rightarrow 4x = -12 \Rightarrow x = -3\)
   • So the roots are \({\frac{1}{2}, -3}\).
   • Not a solution set.

5. (x − 1/2)(x + 3) = 0

   • This gives us \(x - \frac{1}{2} = 0\) or \(x + 3 = 0\)
   • Solving:
     • \(x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{2}\)
     • \(x + 3 = 0 \Rightarrow x = -3\)
   • So the roots are \({\frac{1}{2}, -3}\).
   • Not a solution set.

6. (x − 1/2)(x − 3) = 0

   • This gives us \(x - \frac{1}{2} = 0\) or \(x - 3 = 0\)
   • Solving:
     • \(x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{2}\)
     • \(x - 3 = 0 \Rightarrow x = 3\)
   • So the roots are \({\frac{1}{2}, 3}\).
   • Not a solution set.

7. (4x + 2)(x + 3) = 0

   • This gives us \(4x + 2 = 0\) or \(x + 3 = 0\)
   • Solving:
     • \(4x + 2 = 0 \Rightarrow 4x = -2 \Rightarrow x = -\frac{1}{2}\)
     • \(x + 3 = 0 \Rightarrow x = -3\)
   • So the roots are \({-\frac{1}{2}, -3}\).
   • Not a solution set.

Based on this analysis, none of the equations provided yield the solution set of \({12, -3}\).