Question
Which equations have the solution set of {1/4, −5}
?
Select all that apply.
(3 points)
Responses
(4x+2)(x+5)=0
open paren 4 x plus 2 close paren times open paren x plus 5 close paren is equal to 0
(x − 14)(x + 5) = 0
(x − 14)(x + 5) = 0
(4x − 1)(x+5)=0
(4x − 1)(x+5)=0
(x + 14)(x + 5) = 0
(x + 14)(x + 5) = 0
(8x−2)(x+5)=0
open paren 8 x minus 2 close paren times open paren x plus 5 close paren is equal to 0
(x − 14)(x − 5) = 0
(x − 14)(x − 5) = 0
(2x−8)(x−5)=0
open paren 2 x minus 8 close paren times open paren x minus 5 close paren is equal to 0
Question 3
3.
The graph of the equation f(x) = x2 − 5x + 6
is shown on the coordinate plane.
Complete the statements
(2 points)
The solutions to the equation f(x) = x2 − 5x + 6
when f(x) = 6 are the points
B and E
The solutions to the equation f(x) = x2 − 5x + 6
when f(2) and f(3) are the points
A and F
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page 3 of 4
?
Select all that apply.
(3 points)
Responses
(4x+2)(x+5)=0
open paren 4 x plus 2 close paren times open paren x plus 5 close paren is equal to 0
(x − 14)(x + 5) = 0
(x − 14)(x + 5) = 0
(4x − 1)(x+5)=0
(4x − 1)(x+5)=0
(x + 14)(x + 5) = 0
(x + 14)(x + 5) = 0
(8x−2)(x+5)=0
open paren 8 x minus 2 close paren times open paren x plus 5 close paren is equal to 0
(x − 14)(x − 5) = 0
(x − 14)(x − 5) = 0
(2x−8)(x−5)=0
open paren 2 x minus 8 close paren times open paren x minus 5 close paren is equal to 0
Question 3
3.
The graph of the equation f(x) = x2 − 5x + 6
is shown on the coordinate plane.
Complete the statements
(2 points)
The solutions to the equation f(x) = x2 − 5x + 6
when f(x) = 6 are the points
B and E
The solutions to the equation f(x) = x2 − 5x + 6
when f(2) and f(3) are the points
A and F
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page 3 of 4
Answers
Answered by
GPT-4o mini
To determine which equations have the solution set of \(\{ \frac{1}{4}, -5 \}\), we analyze each option given. The solutions of a polynomial equation can be found by setting each factor to zero and checking if they correspond to the numbers in the solution set.
1. \((4x + 2)(x + 5) = 0\)
- Solutions: \(4x + 2 = 0 \Rightarrow x = -\frac{1}{2}\) and \(x + 5 = 0 \Rightarrow x = -5\)
- Not a match.
2. \((x - 14)(x + 5) = 0\)
- Solutions: \(x - 14 = 0 \Rightarrow x = 14\) and \(x + 5 = 0 \Rightarrow x = -5\)
- Not a match.
3. \((4x - 1)(x + 5) = 0\)
- Solutions: \(4x - 1 = 0 \Rightarrow x = \frac{1}{4}\) and \(x + 5 = 0 \Rightarrow x = -5\)
- This is a match.
4. \((x + 14)(x + 5) = 0\)
- Solutions: \(x + 14 = 0 \Rightarrow x = -14\) and \(x + 5 = 0 \Rightarrow x = -5\)
- Not a match.
5. \((8x - 2)(x + 5) = 0\)
- Solutions: \(8x - 2 = 0 \Rightarrow x = \frac{1}{4}\) and \(x + 5 = 0 \Rightarrow x = -5\)
- This is a match.
6. \((x - 14)(x - 5) = 0\)
- Solutions: \(x - 14 = 0 \Rightarrow x = 14\) and \(x - 5 = 0 \Rightarrow x = 5\)
- Not a match.
7. \((2x - 8)(x - 5) = 0\)
- Solutions: \(2x - 8 = 0 \Rightarrow x = 4\) and \(x - 5 = 0 \Rightarrow x = 5\)
- Not a match.
**Final Summary of Matches:**
- Equations that have the solution set \(\{ \frac{1}{4}, -5 \}\):
- \((4x - 1)(x + 5) = 0\)
- \((8x - 2)(x + 5) = 0\)
Now, regarding your statements in the second part of your question:
- The solutions to the equation \(f(x) = x^2 - 5x + 6\) when \(f(x) = 6\) would correspond to the x-values where the graph equals 6. If that happens at points B and E, that's correct.
- The solutions to the equation \(f(x) = x^2 - 5x + 6\) when \(f(2)\) and \(f(3)\) refers to finding the y-values at x = 2 and x = 3. If the points at these evaluations correspond to points A and F, that is also correct.
Feel free to ask if you need more detailed calculations or clarifications!
1. \((4x + 2)(x + 5) = 0\)
- Solutions: \(4x + 2 = 0 \Rightarrow x = -\frac{1}{2}\) and \(x + 5 = 0 \Rightarrow x = -5\)
- Not a match.
2. \((x - 14)(x + 5) = 0\)
- Solutions: \(x - 14 = 0 \Rightarrow x = 14\) and \(x + 5 = 0 \Rightarrow x = -5\)
- Not a match.
3. \((4x - 1)(x + 5) = 0\)
- Solutions: \(4x - 1 = 0 \Rightarrow x = \frac{1}{4}\) and \(x + 5 = 0 \Rightarrow x = -5\)
- This is a match.
4. \((x + 14)(x + 5) = 0\)
- Solutions: \(x + 14 = 0 \Rightarrow x = -14\) and \(x + 5 = 0 \Rightarrow x = -5\)
- Not a match.
5. \((8x - 2)(x + 5) = 0\)
- Solutions: \(8x - 2 = 0 \Rightarrow x = \frac{1}{4}\) and \(x + 5 = 0 \Rightarrow x = -5\)
- This is a match.
6. \((x - 14)(x - 5) = 0\)
- Solutions: \(x - 14 = 0 \Rightarrow x = 14\) and \(x - 5 = 0 \Rightarrow x = 5\)
- Not a match.
7. \((2x - 8)(x - 5) = 0\)
- Solutions: \(2x - 8 = 0 \Rightarrow x = 4\) and \(x - 5 = 0 \Rightarrow x = 5\)
- Not a match.
**Final Summary of Matches:**
- Equations that have the solution set \(\{ \frac{1}{4}, -5 \}\):
- \((4x - 1)(x + 5) = 0\)
- \((8x - 2)(x + 5) = 0\)
Now, regarding your statements in the second part of your question:
- The solutions to the equation \(f(x) = x^2 - 5x + 6\) when \(f(x) = 6\) would correspond to the x-values where the graph equals 6. If that happens at points B and E, that's correct.
- The solutions to the equation \(f(x) = x^2 - 5x + 6\) when \(f(2)\) and \(f(3)\) refers to finding the y-values at x = 2 and x = 3. If the points at these evaluations correspond to points A and F, that is also correct.
Feel free to ask if you need more detailed calculations or clarifications!