Question

Which equations have the solution set of {14, −5} ? Select all that apply. (3 points) Responses (x + 14)(x + 5) = 0 (x + 14)(x + 5) = 0 (8x - 2)(x + 5) = 0 (8x - 2)(x + 5) = 0 (2x - 8)(x - 5) = 0 (2x - 8)(x - 5) = 0 (4x + 2)(x + 5) = 0 (4x + 2)(x + 5) = 0 (4x − 1)(x+5)=0 (4x − 1)(x+5)=0 (x − 14)(x − 5) = 0 (x − 14)(x − 5) = 0 (x − 14)(x + 5) = 0

Answer 1

To determine which equations have the solution set of \({14, -5}\), we need to analyze each equation and find its roots.

1. \((x + 14)(x + 5) = 0\)
   Roots: \(x = -14\), \(x = -5\) (Not valid)

2. \((8x - 2)(x + 5) = 0\)
   \(8x - 2 = 0 \implies x = \frac{2}{8} = \frac{1}{4}\) (Not valid)
   \(x + 5 = 0 \implies x = -5\)

   Roots: \(x = \frac{1}{4}\), \(x = -5\) (Not valid)

3. \((2x - 8)(x - 5) = 0\)
   \(2x - 8 = 0 \implies x = 4\) (Not valid)
   \(x - 5 = 0 \implies x = 5\)

   Roots: \(x = 4\), \(x = 5\) (Not valid)

4. \((4x + 2)(x + 5) = 0\)
   \(4x + 2 = 0 \implies x = -\frac{1}{2}\) (Not valid)
   \(x + 5 = 0 \implies x = -5\)

   Roots: \(x = -\frac{1}{2}\), \(x = -5\) (Not valid)

5. \((4x - 1)(x + 5) = 0\)
   \(4x - 1 = 0 \implies x = \frac{1}{4}\) (Not valid)
   \(x + 5 = 0 \implies x = -5\)

   Roots: \(x = \frac{1}{4}\), \(x = -5\) (Not valid)

6. \((x - 14)(x - 5) = 0\)
   \(x - 14 = 0 \implies x = 14\)
   \(x - 5 = 0 \implies x = 5\)

   Roots: \(x = 14\), \(x = 5\) (Not valid)

7. \((x - 14)(x + 5) = 0\)
   \(x - 14 = 0 \implies x = 14\)
   \(x + 5 = 0 \implies x = -5\)

   Roots: \(x = 14\), \(x = -5\) (Valid)

The only equation from the set given that has the solution set of \({14, -5}\) is:

• \((x - 14)(x + 5) = 0\)