To determine which equations have the solution set \({ \frac{1}{4}, -5 }\), we need to check each equation by solving for \(x\).
-
Equation: \((4x - 1)(x + 5) = 0\)
-
Set \(4x - 1 = 0\):
\(4x = 1\)
\(x = \frac{1}{4}\)
-
Set \(x + 5 = 0\):
\(x = -5\)
-
Solutions: \({ \frac{1}{4}, -5 }\)
-
-
Equation: \((8x - 2)(x + 5) = 0\)
-
Set \(8x - 2 = 0\):
\(8x = 2\)
\(x = \frac{1}{4}\)
-
Set \(x + 5 = 0\):
\(x = -5\)
-
Solutions: \({ \frac{1}{4}, -5 }\)
-
-
Equation: \((2x - 8)(x - 5) = 0\)
-
Set \(2x - 8 = 0\):
\(2x = 8\)
\(x = 4\)
-
Set \(x - 5 = 0\):
\(x = 5\)
-
Solutions: \({ 4, 5 }\) (not valid)
-
-
Equation: \((x - 14)(x - 5) = 0\)
-
Set \(x - 14 = 0\):
\(x = 14\)
-
Set \(x - 5 = 0\):
\(x = 5\)
-
Solutions: \({ 14, 5 }\) (not valid)
-
-
Equation: \((4x + 2)(x + 5) = 0\)
-
Set \(4x + 2 = 0\):
\(4x = -2\)
\(x = -\frac{1}{2}\)
-
Set \(x + 5 = 0\):
\(x = -5\)
-
Solutions: \({ -\frac{1}{2}, -5 }\) (not valid)
-
-
Equation: \((x - 14)(x + 5) = 0\)
-
Set \(x - 14 = 0\):
\(x = 14\)
-
Set \(x + 5 = 0\):
\(x = -5\)
-
Solutions: \({ 14, -5 }\) (not valid)
-
-
Equation: \((x + 14)(x + 5) = 0\)
-
Set \(x + 14 = 0\):
\(x = -14\)
-
Set \(x + 5 = 0\):
\(x = -5\)
-
Solutions: \({ -14, -5 }\) (not valid)
-
Based on the analysis, the equations that have the solution set \({ \frac{1}{4}, -5 }\) are:
- \((4x - 1)(x + 5) = 0\)
- \((8x - 2)(x + 5) = 0\)
So the responses with valid solution sets are:
- \((4x - 1)(x + 5) = 0\)
- \((8x - 2)(x + 5) = 0\)