Asked by Yes.
Determine whether the values -1 and 7/3 are solutions to the quadratic equation 3x2−4x−4=3
Responses
Both x=−1 and x=7/3 are solutions.
Only x=7/3 is a solution.
Neither x=−1 nor x=7/3 is a solution.
Only x=−1 is a solution.
Responses
Both x=−1 and x=7/3 are solutions.
Only x=7/3 is a solution.
Neither x=−1 nor x=7/3 is a solution.
Only x=−1 is a solution.
Answers
Answered by
GPT-4o mini
To determine whether the values \(-1\) and \(\frac{7}{3}\) are solutions to the equation \(3x^2 - 4x - 4 = 3\), we first simplify the equation. We can rewrite it as:
\[
3x^2 - 4x - 4 - 3 = 0
\]
This simplifies to:
\[
3x^2 - 4x - 7 = 0
\]
Now, we will substitute each candidate value into this simplified equation and check if it satisfies the equation.
### Testing \(x = -1\):
Substituting \(x = -1\):
\[
3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0
\]
Since the left-hand side equals 0, \(x = -1\) is a solution.
### Testing \(x = \frac{7}{3}\):
Substituting \(x = \frac{7}{3}\):
\[
3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7
\]
Calculating \(\left(\frac{7}{3}\right)^2\):
\[
\left(\frac{7}{3}\right)^2 = \frac{49}{9}
\]
Then,
\[
3\left(\frac{49}{9}\right) - 4\left(\frac{7}{3}\right) - 7 = \frac{147}{9} - \frac{28}{3} - 7
\]
Converting \(\frac{28}{3}\) and \(7\) to ninths:
\[
\frac{28}{3} = \frac{84}{9} \quad \text{and} \quad 7 = \frac{63}{9}
\]
Now substituting back:
\[
\frac{147}{9} - \frac{84}{9} - \frac{63}{9} = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0
\]
Since the left-hand side equals 0, \(x = \frac{7}{3}\) is also a solution.
### Conclusion:
Both \(x = -1\) and \(x = \frac{7}{3}\) are solutions.
Therefore, the response is:
**Both \(x = -1\) and \(x = \frac{7}{3}\) are solutions.**
\[
3x^2 - 4x - 4 - 3 = 0
\]
This simplifies to:
\[
3x^2 - 4x - 7 = 0
\]
Now, we will substitute each candidate value into this simplified equation and check if it satisfies the equation.
### Testing \(x = -1\):
Substituting \(x = -1\):
\[
3(-1)^2 - 4(-1) - 7 = 3(1) + 4 - 7 = 3 + 4 - 7 = 0
\]
Since the left-hand side equals 0, \(x = -1\) is a solution.
### Testing \(x = \frac{7}{3}\):
Substituting \(x = \frac{7}{3}\):
\[
3\left(\frac{7}{3}\right)^2 - 4\left(\frac{7}{3}\right) - 7
\]
Calculating \(\left(\frac{7}{3}\right)^2\):
\[
\left(\frac{7}{3}\right)^2 = \frac{49}{9}
\]
Then,
\[
3\left(\frac{49}{9}\right) - 4\left(\frac{7}{3}\right) - 7 = \frac{147}{9} - \frac{28}{3} - 7
\]
Converting \(\frac{28}{3}\) and \(7\) to ninths:
\[
\frac{28}{3} = \frac{84}{9} \quad \text{and} \quad 7 = \frac{63}{9}
\]
Now substituting back:
\[
\frac{147}{9} - \frac{84}{9} - \frac{63}{9} = \frac{147 - 84 - 63}{9} = \frac{0}{9} = 0
\]
Since the left-hand side equals 0, \(x = \frac{7}{3}\) is also a solution.
### Conclusion:
Both \(x = -1\) and \(x = \frac{7}{3}\) are solutions.
Therefore, the response is:
**Both \(x = -1\) and \(x = \frac{7}{3}\) are solutions.**
Answered by
GPT-4o mini
To solve the equation \(3x^2 = 21\), we can start by isolating \(x^2\). Here are the steps:
1. Divide both sides of the equation by 3:
\[
x^2 = \frac{21}{3} = 7
\]
2. Now we take the square root of both sides:
\[
x = \pm \sqrt{7}
\]
This gives us two solutions:
\[
x = \sqrt{7} \quad \text{and} \quad x = -\sqrt{7}
\]
### Conclusion:
The equation \(3x^2 = 21\) has **two** distinct solutions in the solution set.
Therefore, the correct response is:
**two**.
1. Divide both sides of the equation by 3:
\[
x^2 = \frac{21}{3} = 7
\]
2. Now we take the square root of both sides:
\[
x = \pm \sqrt{7}
\]
This gives us two solutions:
\[
x = \sqrt{7} \quad \text{and} \quad x = -\sqrt{7}
\]
### Conclusion:
The equation \(3x^2 = 21\) has **two** distinct solutions in the solution set.
Therefore, the correct response is:
**two**.
Answered by
GPT-4o mini
To find the number of solutions for the equation \(2(x-5)^2 + 2 = 20\), we will first simplify it step by step.
1. Start by isolating the squared term:
\[
2(x-5)^2 + 2 = 20
\]
Subtract 2 from both sides:
\[
2(x-5)^2 = 18
\]
2. Divide both sides by 2:
\[
(x-5)^2 = 9
\]
3. Now take the square root of both sides:
\[
x - 5 = \pm 3
\]
This gives us two equations to solve:
- \(x - 5 = 3\)
- \(x - 5 = -3\)
**Solving these equations:**
For the first equation:
\[
x - 5 = 3 \implies x = 8
\]
For the second equation:
\[
x - 5 = -3 \implies x = 2
\]
Thus, the solutions are \(x = 8\) and \(x = 2\).
### Conclusion:
The equation \(2(x - 5)^2 + 2 = 20\) has **two** distinct solutions in the solution set.
Therefore, the correct response is:
**two**.
1. Start by isolating the squared term:
\[
2(x-5)^2 + 2 = 20
\]
Subtract 2 from both sides:
\[
2(x-5)^2 = 18
\]
2. Divide both sides by 2:
\[
(x-5)^2 = 9
\]
3. Now take the square root of both sides:
\[
x - 5 = \pm 3
\]
This gives us two equations to solve:
- \(x - 5 = 3\)
- \(x - 5 = -3\)
**Solving these equations:**
For the first equation:
\[
x - 5 = 3 \implies x = 8
\]
For the second equation:
\[
x - 5 = -3 \implies x = 2
\]
Thus, the solutions are \(x = 8\) and \(x = 2\).
### Conclusion:
The equation \(2(x - 5)^2 + 2 = 20\) has **two** distinct solutions in the solution set.
Therefore, the correct response is:
**two**.
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